Polarization Theory of Nuclear Reactions [1st ed. 2023] 3031118774, 9783031118777

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Polarization Theory of Nuclear Reactions

Qing-Biao Shen

Polarization Theory of Nuclear Reactions Revisers Chong-Hai Cai Ye Tian Bradley Kemp Wilson

Qing-Biao Shen China Institute of Atomic Energy Beijing, China

ISBN 978-3-031-11877-7 ISBN 978-3-031-11878-4 https://doi.org/10.1007/978-3-031-11878-4

(eBook)

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Summary

This book develops the polarization theory describing the nuclear reactions of the spin 1/2 and 1 incident particles with unpolarized target nuclei and residual nuclei with arbitrary spin, and studies the polarization theory of the nuclear reactions of two polarized light particles. The unitary transformation relations of the rank-2 and higher-rank tensors between the spherical basis coordinate system and the orthogonal Cartesian coordinate system are found, and the theoretical method for studying the polarization phenomena of the particles with spin equal to or greater than 3/2 is given. The continuum discretized coupled channels (CDCC) theory of the axissymmetric rotational nuclei describing the incident deuteron breakup reactions is developed. The relativistic nuclear reaction theories are introduced, and the relativistic nuclear reaction Dirac S matrix theory is proposed and developed. The polarization phenomena are not considered in the existing transport theory which describes the motion of the nucleons in nuclear medium, and it is considered that all particles are unpolarized, which is an approximate method from the view of microphysics. In order to describe the objective physical process in nature more vividly, this book presents for the first time the concrete scheme and method of establishing the polarization nuclear database. The Monte-Carlo method for describing the transport of the polarized particles and the transport equation of the polarized particles are proposed and developed for the first time. The book provides a technical way for the introduction of the polarization phenomenon in the micro-world into the research and design of the nuclear engineering projects closely related to human life.

v

Introduction

When a particle outside of the nucleus moves in the nuclear potential formed by the interactions between the nucleons located inside of the nucleus, not only will its coordinate position and momentum continue to change but the particle and the nucleus itself will also undergo self-rotation, called spin. The spin values of particles and nuclei are quantized and can only take integers or semi-odd numbers. Both 1 experiments and theories have proved that the nuclear potential acting on the spin 2 particle contains spin-orbit coupling potential, and the nuclear potential acting on the spin 1 particle contains not only spin-orbit coupling potential but also tensor potential. In the theory of nuclear structure, the spin-orbit coupling potential causes 1 the nuclear level to be split. Two nucleons with spin can be coupled to two 2 states of total spin S which is 0 or 1, and there is a tensor potential for the state which total spin equals 1. The spin of the deuteron is equal to 1. When studying the deuteron structure, it is found that not only the S state which orbit angular momentum L = 0 exists but the D state which orbit angular momentum L = 2 exists as well. The above results show that there is a tensor potential between the nucleon-nucleon. In nuclear reaction theory, because the spin-orbital coupling potential and tensor potential related to spin are included in the nuclear potential, the forces felt by particles with different spin magnetic quantum numbers are different, so their motion behavior will also be different. When all the magnetic quantum numbers of spins of the particles are equal and mutually incoherent, the particles are called unpolarized particles. Otherwise, they are polarized particles. In the case with spin-related nuclear potential, the polarized outgoing particles can be obtained through nuclear reaction induced by unpolarized incident particles. This is the polarization phenomenon of nuclear reactions. The microscopic system of the nucleus is understood through the study of various considerable measurements of nuclear reactions. Nuclear force, nuclear structure, and nuclear reaction mechanism can be studied by studying various cross sections, angular distributions, energy spectra, and double differential sections of nuclear reactions. However, if only the unpolarized states of particles and nuclei are studied, vii

viii

Introduction

the understanding for the nucleus has limitation. The polarization phenomena greatly enrich the data obtained from scattering and nuclear reactions, provide information on the interaction that can change spin orientation, and provide important verification data for the study of nuclear structures or reaction mechanisms. In addition, the effect of polarization phenomena on nuclear fusion rate is a promising research the directed neutron sources topic. Nowadays, some peoples are studying to produce * *  * *  by using the nuclear reactions d t , n α or d d , n 3 He introduced by two polarized light nuclei, and their civilian and military applications are being explored. The theory of calculating various polarization quantities from the nuclear reaction amplitude is called polarization theory of nuclear reactions. As early as the 1950s, Wolfinstein et al. conducted theoretical studies on the polarization of nuclear reactions. Later, many papers on polarization phenomena were published, some of them also contained some information on experiments of the polarization nuclear reactions. In 1972, Ohlsen published a paper in Rep. Prog. Phys. This paper is a classical article on the polarization theory of nuclear reactions. In 1974, Robson published a book titled The Theory of Polarization Phenomena with 119 pages. In their papers, the basic knowledge of polarization theory is introduced. More detailed 1 computational expressions are given for the polarization theory of spin particles 2 with spinless targets. A part of the computational expressions is given for the polarization theory of spin 1 particles with spinless targets. Some specific calculations were also made and compared with experimental data in spinless target case. In Ohlsen’s paper, a brief introduction of the polarization formulas of the two polarized light particles is given. In the relativistic optical model, the Dirac equation can be converted into a Schrodinger-like equation, and thus the polarization quantities of the elastic scattering between the nucleon and the target with spinless can also been calculated. For the nuclear reactions in which the target and the residual nucleus with non-zero spin, in 1965, Tamura gave a formula for calculating the polarization rate of the emitted nucleons using the method of averaging the Pauli matrix by the wave functions in his paper on coupling channel calculations. On this basis, some people made effort to study the corresponding analyzing power. In Robson’s book, the polarization problem of the nuclear reaction between an incident particle with arbitrary spin and a target nucleus with arbitrary spin is discussed by using the formal theory in the complex spherical basis coordinate, and it does not yet have the practical application value. However, before writing this book no one has ever given a systematic polarization theory and the clear theoretical formulas that can be used to calculate actually the polarization observables for elastic scattering and direct reac1 and 1 particles with targets with non-zero spin. For tion channels of spin 2 example, there are no clear theoretical formulas to calculate the polarization observ! ! → → 1 1 ables for 1 þ A → þ B and þ A → 1 þ B reactions which target or residual 2 2 ! 1 nucleus with non-zero spin certainly, where represents the polarized particle with 2 1 spin . Therefore, it is impossible to fully carry out the calculations of various 2

Introduction

ix

polarization quantities for targets or residual nuclei with non-zero spin. Since the spins of most nuclei (including all odd A nuclei) are not equal to 0, the development of polarization theory of nuclear reactions is constrained. There are only the theoretical formulas for calculating various polarization quantities of nuclear reactions using the nuclear reaction amplitude, but no specific research is doing on the nuclear reaction amplitude itself, one can only call this kind of theory as the formal theory. We know that the S matrix theory is a basic theory describing the nuclear reactions. The expression of the reaction amplitude given by S matrix theory satisfies the physical properties such as the angular momentum conservation and the parity conservation. When studying the two-body direct reac1 tion that occurs between spin or 1 particle with the unpolarized target with spin 2 I and the unpolarized residual nucleus with spin I', we found that there are some kind of internal relations between the matrix elements of the nuclear reaction amplitude for determined spin magnetic quantum numbers MI and M'I. Therefore, when making the summation of MI and M'I in the polarization formula, some items will cancel each other and some items will be merged due to equality. In this way, some clear expressions for calculating various polarization quantities are obtained. Taking ! ! 1 1 þ A → þ B reaction as an example, when the spins of A and B are both 0 and 2 2 the right hand screw rule is used to select the y-axis perpendicular to the reaction plane, the results given by the predecessors have clearly pointed out that only the y components of the polarization analyzing powers Ai (i = x, y, z) and the polarization rates Pi (i = x, y, z) of the outgoing particles for the unpolarized incident particles are not equal to 0, and only 5 of the 9 polarization transfer coefficients K ji ði, j = x, y, zÞ are not equal to 0. This book strictly proves that the above conclusion is still true when the spins of the unpolarized nuclei A and B are not equal to 0. This result has not been given before. We also use the above method to ! ! → → → → 1 1 study 1 þ A → 1 þ B reactions as well as 1 þ A → þ B and þ A → 1 þ B 2 2 reactions. The study for these reactions also found that some components of the polarization quantities are equal to 0 and have certain regularity. The above phenomenon is due to the fact that the nuclear reaction process satisfies the parity conservation. We know that optical model, coupled channel optical model, distorted wave Born approximation (DWBA), R matrix theory, phase-shift analysis method, and even some microscopic nuclear reaction theory can all be used to calculate S matrix elements of nuclear reactions. These theories belong to the usual nuclear reaction theory. Therefore, it can be said that based on the contents of this book the 1 polarization theory of spin or 1 particles and unpolarized targets and residual 2 nuclei with arbitrary spins has reached the stage that the actual calculations can be made.

x

Introduction

This book Polarization Theory of Nuclear Reactions comprises six chapters. The first chapter introduces the spin operators, irreducible tensors, polarization operators, density matrices etc., all of them are the basic knowledge of polarization theory of nuclear reactions. The second chapter introduces the polarization theory of nuclear reactions for 1 1 particle spin particles, such as the Pauli matrices, the polarization of spin 2 2 beams, the polarization theory of elastic scattering of unpolarized and polarized 1 particles with spinless target, polarization theory of triple elastic spin 2 1 scatterings between spin particle and spinless nucleus, two theoretical 2 methods to study the polarization phenomena of nuclear reactions. Then the polar! ! 1 1 ization theory of þ A → þ B reactions which can be used for various kinds of 2 2 1 two-body direct reactions is introduced, and the spin particle polarization 2 theory is also introduced when both target and residual nucleus are polarized. 1 polarized particles is studied Finally, the polarization problem of two spin 2 theoretically. The third chapter introduces the polarization theory of nuclear reactions for spin 1 particles, such as the spin operators and polarization operators of spin 1 particles, the vector polarization rate and tensor polarization rate of spin 1 incident particles, the elastic scattering amplitude of spin 1 particles with spinless target, the polarization theory of elastic scattering of spin 1 particle with both vector polarization and tensor polarization from a spinless target, the general form of the nuclear reaction polarization theory of spin 1 particles with non-zero spin target. Then the polariza! ! → → → → 1 1 tion theories of 1 þ A → 1 þ B, 1 þ A → þ B, þ A → 1 þ B reactions as 2 2 → → → → 1 well as þ 1 and 1 þ 1 reactions with unpolarized targets and residual nuclei 2 are introduced, respectively. In addition, the deuteron phenomenological optical potential with tensor term and its corresponding radial equation are introduced, the folding model of the deuteron-nucleus reactions is described, and the spherical nucleus continuum discretized coupled channels (CDCC) theory that only applies to the spinless targets describing the breakup reactions of the loosely bound light complex particles is introduced. This book also develops an axis-symmetric rotational nucleus CDCC theory describing the breakup reaction for loosely bound light complex particle. This theory can be used for targets with arbitrary spin and can consider the excited states of the targets. The above theories belong to the normal nuclear reaction theories of deuterons. 3 In Chap. 4, we first give the polarization theory of nuclear reactions for spin 2 particles in the spherical basis coordinate system. When we study the fivedimensional rank-2 tensor in spin 1 particle polarization theory, the unitary transformation relation of the rank-2 tensor between the spherical basis coordinate system

Introduction

xi

and the orthogonal Cartesian coordinate system is found. According to the regularity of the unitary transformation relations of the vector (rank-1 tensor) and the rank-2 tensor between the spherical basis coordinate system and the orthogonal Cartesian coordinate system, the unitary transformation relations of the rank-3 and higher-rank tensors between the spherical basis coordinate system and the orthogonal Cartesian coordinate system are found by recursion method. And the generalized Pauli 3 matrix is introduced, thus the theoretical method for studying spin ≥ particle 2 polarization phenomena is given. Finally, based on the Maxwell equation of electromagnetic field, the quantization method of electromagnetic field is introduced, and the polarization theory for photon beams is discussed. The fifth chapter introduces the polarization theory of relativistic nuclear reactions. It introduces the basic theory of relativistic quantum mechanics, transformation of relativistic coordinate systems, relativistic optical model and phenomenological optical potentials, Dirac S matrix theory of relativistic nuclear reactions, Dirac coupling channel theory including elastic scattering and collective inelastic scattering channels, relativistic collective deformation relativistic distorted wave Born approximation (RDWBA) method and calculation of nucleon polarization quantities, relativistic impulse approximation of elastic scattering, relativistic impulse approximation of inelastic scattering, relativistic impulse approximation of (p,n) reactions. Then the relativistic classical field theory and Lagrangian density in quantum hadron dynamics, relativistic Green function theory at zero temperature, real part of nucleon relativistic microscopic optical potential and relativistic nuclear matter properties, imaginary part of nucleon relativistic microscopic optical potential, and contribution to imaginary part of nucleon relativistic microscopic optical potential by fourth order exchange diagrams are introduced. Then it introduces the relativistic Bete-Salpeter equation, the Bonn one boson exchange potential, and the nucleon relativistic microscopic optical potential based on the Dirac-BruecknerHartree-Fock (DBHF) theory. Finally, the Proca relativistic dynamics equation of spin 1 particle and its application in elastic scattering calculations, Weinberg relativistic dynamics equation of spin 1 particles and discussion on its application, and the relativistic nuclear reaction theory considering the internal structure of the incident deuterons are introduced. The above theories are the basic relativistic nuclear reaction theories for the study of the high-energy nucleon and deuteron induced nuclear reactions. By solving the corresponding dynamic equations, the reaction amplitude can be obtained and then various microscopic physical quantities including polarization quantities can be calculated. The sixth chapter introduces the basis of the polarized particle transport theory. The polarized electron transport theory has been studied for a long time. This chapter proposes to carry out the polarized nucleus transport theory research. In order to simplify the theoretical formula, people usually select the y-axis along the direction which is perpendicular to the reaction plane, in this way there is the azimuthal angle φ = 0. If one wants to introduce the polarization process of the nuclear reactions into the particle transport theory, the polarization quantities of any (θ, φ) angles must be

xii

Introduction

! ! 1 1 known. This chapter introduces the polarization theories of þ A → þ B, 2 2 ! ! → → → → 1 1 1 þ A → 1 þ B, 1 þ A → þ B, and þ A → 1 þ B reactions related to the 2 2 polar angle θ and azimuthal angle φ at the same time, and puts forward the specific concept of establishing a polarization nuclear database. It is point out what polarization quantities that are only related to the polar angles θ should be stored in the polarization nuclear database for each considering nuclear reactions. The specific calculation formulas of the polarization quantities related to (θ, φ) angles at the same time by using the data stored in the polarization nuclear database are given. It ! ! → 1 1 is easier to deal with þ A → 0 þ B, 0 þ A → þ B, 1 þ A → 0 þ B, and 2 2 → 0 + A→ 1 þ B reactions related to spinless α particles, in which only the incident or outgoing channel is polarized. In addition, the Monte-Carlo methods for simulating the reactions including polarization channels are given and the method of the coordinate system rotation and the transformation of the polarization rates of transported particles are presented. The transport equation of polarized particles 1 with spin or 1 is also proposed and developed. Finally, the prospective problems 2 of the future development of the polarized particle transport theory are discussed.  * * In the normal particle transport theory, the particle flux function ψ r , p , t is studied, and in the polarized particle   transport theory proposed in this book the 1 1 * * particle flux function ψ r , p , s, t is studied, in which s = , - for nucleons 2 2 and s = 1, 0, -1 for deuterons. In the normal particle transport theory, only the usual nuclear data, which is not related to spin magnetic quantum numbers, are required. In the polarized nucleus transport theory proposed in this book, polarization nuclear data related to spin magnetic quantum numbers are needed for some reaction channels. Due to the existence objectively of spin-orbit forces and tensor forces related to spins of particles, the observable physical quantities of the emitted particles with different spin magnetic quantum numbers are different. Therefore, it can be said that the polarization particle transport theory can more realistically describe the physical processes that exist in nature. As to when it is necessary to consider the polarized particle transport process, this is a subject to be studied in the future. More than 50 % of the theoretical formulas that appear in this book are new derivation. The main innovations of this book are: 1. Developed a nuclear reaction polarization theory system that can be used for specific calculations for various types of elastic scattering and two-body direct 1 reaction between spin or 1 particle and unpolarized target nucleus with 2 arbitrary spin. It breaks through the bottleneck that previously the people could not fully calculate various polarization quantities when the spins of the target and residual nucleus are not equal to 0.

Introduction

xiii

2. The unitary transformation relations of the rank-2 and higher-rank tensors between the spherical basis coordinate system and the orthogonal Cartesian coordinate system are found, and the theoretical method for studying the spin 3 ≥ particle polarization phenomena is given. 2 3. The axis-symmetric rotational nucleus CDCC theory describing the breakup reaction of loosely bound light complex particles is developed. And the CDCC theory, which was originally applied only to spinless target, is developed to be used for the target with arbitrary spin and to include the inelastic scattering channels of the excited states. 4. The improper introduction of the coulomb potential into the Dirac equation was corrected. The Dirac S matrix theory of relativistic nuclear reactions is proposed 1 and developed, so that when the reaction amplitude of spin particles is solved 2 by the Dirac equation, various microscopic physical quantities including polarization quantities can be calculated using this theory. And when the incident particle energy is quite low, the relativistic Dirac S matrix theory will automat1 ically degenerate into the general non-relativistic S matrix theory of spin 2 particles. 5. For the first time, the idea of establishing a polarization nuclear database was proposed. Which polarization quantities, that are only related to the polar angle θ, ! ! ! → → → 1 1 1 þ A → þ B , 1 þ A → 1 þ B , 1 þ A → þ B , and are point out for 2 2 2 ! → 1 þ A → 1 þ B reactions in the polarization nuclear database, and the methods 2 for obtaining these data from experimental data and theoretical calculations are discussed. The theoretical formulas for calculating the various polarization quantities related to (θ, φ) angles are also given. 6. For the first time, the idea of carrying out research on polarized nucleus transport theory is put forward. The Monte-Carlo method describing the polarized particle transport process and the polarized particle transport equation are proposed and developed. The polarization phenomena of nuclear reactions are objectively existed in the microscopic world. In the existing neutron and other heavy particle transport theories the polarization problem is not included, the above reality can only be considered as an approximation. In order to describe the physical processes that exist in nature more realistically, the polarized nucleus transport theory should be studied, especially when the angular distribution of the emitted particles clearly presents anisotropy. Above circumstances show it is necessary to carry out the research on the polarized nucleus transport theory. Since this book 1 and gives a nuclear reaction polarization theory system in which the spin 2 1 particles react directly with unpolarized target and residual nucleus with 3 arbitrary spins, and the theoretical method to study the spin ≥ particle 2 polarization phenomena is also given, so these make it possible to carry out the

xiv

Introduction

research on polarized nucleus transport theory. That is the polarized nucleus transport theory research becoming feasible. For the first time, the establishment of a polarization nuclear database and the development of the polarized nucleus transport theory are proposed, that will inevitably encounter difficulties and challenges in the future. But after all, this idea is introducing the polarization phenomena of the microscopic world into the research and design of nuclear engineering projects that are closely related to human reality, and showing a way to explore the relationship of the nuclear engineering projects and the nuclear microscopic properties including the polarization phenomena. Professor Chong-Hai CAI of Nankai University in China made a very careful revision for the Chinese version of this book. During the revision process, most of the theoretical formulas in the manuscript were carefully deduced, many mistakes were discovered and corrected, and many valuable and even key modifications were proposed. When I handed Professor CAI the general formulas for calculating the ! ! 1 1 various polarization quantities of þ A → þ B reaction with the target and the 2 2 residual nucleus with no zero spins, he programmed it into the R matrix program and made the corresponding calculation. In the calculation results, it is found that for the case where the spins of the target and the residual nucleus are not equal to 0, there are also many polarization quantity components are equal to 0. It is also found that there is a certain relationship between certain matrix elements of reaction amplitude. Then, I carefully analyzed and deduced the relevant theoretical formulas, and theoretically proved that the relationship between these matrix elements of reaction amplitude seen in the calculation results does exist. And using these relations, it can be strictly proved that those components of polarization quantities with a calculated result of 0 should indeed be equal to 0. This is a critical step in the polarization theory of nuclear reactions given in this book. Some of the results obtained later in this book are closely related to this step. It can be seen from this issue that Professor CAI has made a very important contribution to this book. I would like to express my heartfelt thanks to Professor CAI, and I feel honored to have such an excellent partner in scientific research. The Nuclear Data Key Laboratory of China and the China Nuclear Data Center have provided good working conditions and necessary financial support for the writing of this book. Researcher Yin-Lu HAN provided sincere help and support to the writing and publication of this book from beginning to the end. I express my sincere thanks here. Researcher Hai-Rui GUO of the Beijing Institute of Applied Physics and Computational Mathematics and Professor Yong-Li XU of Shanxi Datong University of China read part of the manuscript, and I express my thanks to them for their advice and help. I am very grateful to Mr. Feng-Quan ZHAO for the large number of typing of Chinese manuscript and formulas and making charts for this book. My wife Ye TIAN, an associate researcher, has made an important contribution to the book. She and I both graduated from the Department of Modern Physics at the University of Science and Technology of China in July 1963 and were also assigned

Introduction

xv

to the China Institute of Atomic Energy, where she worked in reactor physics and then theoretical nuclear physics. Both of us were original members of the China Nuclear Data Center when it was founded in 1975. In the process of writing this book, she assisted the me to do some data collation and participated in the discussion of some related issues. When the Chinese version of the book was finished in July 2018, Ms. Ye TIAN used computer translation software and other means to produce the first English version of the book in 2018. Then, based on the first English draft, I carried out the revision sentence by paragraph, focusing on the use of special vocabulary and correct expression of the content of the book. At the same time, I also made some amendments and additions to the manuscript. After the draft of the English version of this book was completed, Dr. Bradley Kemp WILSON, a professor at the University of Saint Thomas in Saint Paul, Minnesota, USA, modified and refined most parts of the English version of this book. In September 2021, I formally submitted the English version of the book to Springer Nature press and applied for publication. I underwent lumbar spine surgery on November 9, 2021, and needed to spend more than 3 months in bed after discharge. Springer Nature press passed the reviews to the manuscript on November 20, 2021, and reviewers gave the book a lot of positive reviews, so there was an urgent need for me to do a lot of work with the publisher via e-mail, such as responding to reviewers’ comments, minor amendments, and discussing the content of the publishing agreement, among other things. Since it was difficult for me to get up to work on the computer, in this case, Ms. Ye TIAN discussed and agreed with me on some issues and then communicated with the publisher. This allowed the publication work of the book to proceed without delay. I take comfort in the fact that we 84-year-old couples are still working together professionally. I would like to express my sincere thanks to Associate Researcher Ye TIAN and Professor Bradley Kemp WILSON for their contributions to the English version of this book. Only this book commemorates my father Yu-Tian SHEN and my mother Qing-Wen KONG, who were ordinary farmers, they had worked for a lifetime in Shenzhuang Village, Gu’an County, Hebei Province in China. Beijing, China May 1, 2022

Qing-Biao Shen

Contents

1

Basic Knowledge of Polarization Theory of Nuclear Reactions . . . . . 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Spin Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Irreducible Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Polarization Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Density Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2

Polarization Theory of Nuclear Reactions for Spin 2.1 2.2 2.3 2.4 2.5 2.6

1 Particles . . . . . 2 Pauli Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Polarization of the Spin Particle Beams . . . . . . . . . . . . . . . . . . 2 Polarization Theory of Elastic Scattering of Unpolarized 1 Spin Particles with Spinless Targets . . . . . . . . . . . . . . . . . . . . 2 Polarization Theory of Elastic Scattering of Polarized 1 Spin Particles with Spinless Targets . . . . . . . . . . . . . . . . . . . . 2 Polarization Theory of Triple Elastic Scatterings Between 1 Spin Particles and Spinless Nuclei . . . . . . . . . . . . . . . . . . . . . . 2 Two Theoretical Methods to Study the Polarization Phenomena of Nuclear Reactions . . . . . . . . . . . . . . . . . . . . . . . . *

2.7 2.8

*

1 1 þ A → þ B Reactions . . . . . . . . . . . 2 2 1 Polarization Theory of Nuclear Reactions of Spin Particle 2 with Polarized Target and Residual Nucleus . . . . . . . . . . . . . . . . Polarization Theory for

→

→

1 1 4 9 12 16 21 23 23 26 33 39 47 60 62 73

1 1 þ Reactions . . . . . . . . . . . . . . . . . 77 2 2 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

2.9

Polarization Theory for

xvii

xviii

3

Contents

Polarization Theory of Nuclear Reactions for Spin 1 Particles . . . . . 3.1 Spin Operators and Polarization Operators of Spin 1 Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 General Expressions of S = 1 Spin Operators and Polarization Operators [1–3] . . . . . . . . . . . . . . . . . . 3.1.2 Specific Expressions in the Spherical Basis Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.3 Specific Expressions in the Cartesian Basis Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.4 Transformation Relations Between the Spherical Basis and Cartesian Coordinate Systems . . . . . . . . . . . . 3.1.5 Transformation Relations Between the Spherical Basis and Cartesian Basis Representations . . . . . . . . . . . . . . . 3.2 Some Mathematical Formulas Related to Spin Operators and Polarization Operators of Spin 1 Particles . . . . . . . . . . . . . . . 3.2.1 Coordinate Rotation of S = 1 Spin Wave Function and Spin Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 Action of the S = 1 Spin Operators and Polarization Operators on Basis Spin FUnctions . . . . . . . . . . . . . . . . 3.2.3 Products of S = 1 Spin Operators and Polarization Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.4 Traces of the S = 1 Spin Operators and Polarization Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.5 General Spin Wave Functions and Density Matrices for S = 1 Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Vector Polarization Rate and Tensor Polarization Rate of Spin 1 Incident Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Elastic Scattering Amplitude of Spin 1 Particles with Spinless Targets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Polarization Theory of Elastic Scattering of Unpolarized Spin 1 Particles with Spinless Targets . . . . . . . . . . . . . . . . . . . . 3.6 Polarization Theory of Elastic Scattering of Spin 1 Particle with both Vector Polarization and Tensor Polarization from a Spinless Target . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 General Form of Nuclear Reaction Polarization Theory of Spin 1 Particles with Non-zero Spin Target . . . . . . . . . . . . . . . →

→

101 101 101 107 110 112 114 116 116 117 117 120 123 130 134 139

143 159

3.8

Polarization Theory for 1 þ A → 1 þ B Reactions . . . . . . . . . . 172

3.9

Polarization Theory for 1 þ A →

→

*

3.10 3.11

*

1 þ B Reactions . . . . . . . . . . 178 2

→ 1 þ A → 1 þ B Reactions . . . . . . . . . . 192 2 → → 1 Polarization Theory for þ 1 Reactions . . . . . . . . . . . . . . . . . 204 2

Polarization Theory for

Contents

xix →

→

Polarization Theory for 1 þ 1 Reactions . . . . . . . . . . . . . . . . . Deuteron Phenomenological Optical Potentials with Tensor Terms and Corresponding Radial Equations . . . . . . . . . . . . . . . . 3.13.1 Deuteron Phenomenological Optical Potentials Containing Tensor Terms . . . . . . . . . . . . . . . . . . . . . . . 3.13.2 Deuteron Radial Equations for the Spherical Nuclei when Containing Tensor Potentials . . . . . . . . . . . 3.14 Folding Model Describing the Reaction Between Deuteron and Nucleus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.14.1 Deuteron Folding Model Without Breakup Channel [46, 52, 53] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.14.2 Deuteron Folding Optical Potentials Containing Break Channel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.15 The Spherical Nucleus CDCC Theory Describing the Breakup Reaction for Incident Loosely Bound Light Complex Particles . . . 3.15.1 Development of CDCC Theory . . . . . . . . . . . . . . . . . . . 3.15.2 Spherical Nucleus CDCC Equation . . . . . . . . . . . . . . . . 3.15.3 Angular Distribution of Deuteron Elastic Scattering . . . . 3.15.4 Double Differential Cross Section of the Ejected Nucleons After Deuteron Breaking . . . . . . . . . . . . . . . . 3.16 Axis Symmetric Rotational Nucleus CDCC Theory Describing the Breakup Reaction for Incident Loosely Bound Light Complex Particles . . . . . . . . . . . . . . . . . . . . . . . . . 3.16.1 Axis-Symmetric Rotational Nucleus CDCC Equation . . . 3.16.2 Angular Distribution of Deuteron Elastic and Inelastic Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.16.3 Double Differential Cross Section of the Ejected Nucleons After Deuteron Breaking . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.12 3.13

4

5

225 242 242 245 248 248 256 258 258 259 269 276

285 285 294 297 300

3 Particles 2 and Polarization Theory for Photon Beams . . . . . . . . . . . . . . . . . . . . 3 4.1 Polarization Theory of Nuclear Reactions for Spin Particles . . . 2 4.2 Polarization Theory for Photon Beams . . . . . . . . . . . . . . . . . . . . 4.2.1 Classic Electromagnetic Field Theory [3] . . . . . . . . . . . . 4.2.2 Hamilton Canonical Equation [3] . . . . . . . . . . . . . . . . . 4.2.3 Quantization of Electromagnetic Fields [3–5] . . . . . . . . . 4.2.4 Polarization Theory of Photon Beams . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

305 326 326 329 332 339 343

Polarization Theory of Relativistic Nuclear Reactions . . . . . . . . . . . 5.1 Basic Theory of Relativistic Quantum Mechanics . . . . . . . . . . . 5.1.1 Klein-Gordon Equation . . . . . . . . . . . . . . . . . . . . . . . 5.1.2 Dirac Equation [5] . . . . . . . . . . . . . . . . . . . . . . . . . . .

345 345 345 348

Polarization Theory of Nuclear Reactions for Spin

. . . .

305

xx

Contents

5.2 5.3 5.4 5.5 5.6 5.7 5.8

5.9 5.10

5.11

5.12

5.1.3 Pauli Metric and Bjorken-Drell Metric . . . . . . . . . . . . . . 5.1.4 Plane Wave Solution of the Dirac Equation . . . . . . . . . . 5.1.5 Lorentz Covariant of the Dirac Equation . . . . . . . . . . . . 5.1.6 The Trace Formulas for γ Matrix Products . . . . . . . . . . . Transformation of Relativistic Coordinate Systems . . . . . . . . . . . Relativistic Optical Model and Phenomenological Optical Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Dirac S Matrix Theory of the Relativistic Nuclear Reactions . . . . Dirac Coupling Channel Theory Including Elastic Scattering and Collective Inelastic Scattering Channels . . . . . . . . . . . . . . . . Relativistic Collective Deformation RDWBA Method and Calculation of Nucleon Polarization Quantities . . . . . . . . . . . Relativistic Impulse Approximation of Elastic Scattering and Calculation of Nucleon Polarization Quantities . . . . . . . . . . . Relativistic Impulse Approximation of Inelastic Scattering and Calculation of Nucleon Polarization Quantities . . . . . . . . . . . 5.8.1 Relativistic Distorted Wave Impulse Approximation of Single Particle State Inelastic Scattering [58, 60, 61] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8.2 Relativistic Distorted Wave Impulse Approximation of Collective State Inelastic Scattering [62–64] . . . . . . . Relativistic Impulse Approximation of (p, n) Reactions and Calculation of Nucleon Polarization Quantities . . . . . . . . . . . Relativistic Classical Field Theory and Lagrangian Density in Quantum Hadron Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . 5.10.1 Relativistic Classical Field Theory [2, 67, 68] . . . . . . . . 5.10.2 Lagrangian Density in Quantum Hadron Dynamics [2, 68–74] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.10.3 Relativistic Mean Field Equations . . . . . . . . . . . . . . . . . Relativistic Green Function Theory at Zero Temperature . . . . . . . 5.11.1 The Propagator of Neutral Scalar Bosons with Spinless [2, 67] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 5.11.2 The Propagator of Fermions with Spin [2, 6–8] . . . . . . 2 5.11.3 The Propagator of Neutral Vector Bosons with Spin 1 . . 5.11.4 Feynman Rules of Nucleon-Meson Interaction in Momentum Representation . . . . . . . . . . . . . . . . . . . . . . Real Part of Nucleon Relativistic Microscopic Optical Potential and Relativistic Nuclear Matter Properties . . . . . . . . . . . . . . . . . 5.12.1 Self-Energy Operator and Green Function in Nuclear Matter [26, 80] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.12.2 Real Part of Nucleon Relativistic Microscopic Optical Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.12.3 Relativistic Nuclear Matter Properties . . . . . . . . . . . . . .

352 355 361 370 371 382 395 411 419 425 441

441 442 443 448 448 451 458 461 462 466 477 484 486 486 488 506

Contents

5.13

5.14

5.15 5.16 5.17

5.18

xxi

Imaginary Part of Nucleon Relativistic Microscopic Optical Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.13.1 Contribution of σ - σ Meson Exchange Processes to the Imaginary Part of the Optical Potential . . . . . . . . . 5.13.2 Contribution of ω - ω Meson Exchange Processes to the Imaginary Part of the Optical Potential . . . . . . . . . 5.13.3 Contribution of σ - ω Meson Exchange Processes to the Imaginary Part of the Optical potential . . . . . . . . . 5.13.4 Contribution of πps - πps Meson Exchange Processes to the Imaginary Part of Optical Potential . . . . . . . . . . . 5.13.5 Contribution of πpv - πpv Meson Exchange Processes to the Imaginary Part of the Optical Potential . . . . . . . . . 5.13.6 Contribution of ρV - ρV Meson Exchange Processes to the Imaginary Part of the Optical Potential . . . . . . . . . 5.13.7 Contribution of ρT - ρT Meson Exchange Processes to the Imaginary Part of Optical Potential . . . . . . . . . . . 5.13.8 Contribution of ρV - ρT Meson Exchange Processes to the Imaginary Part of Optical Potential . . . . . . . . . . . Contribution to Imaginary Part of Nucleon Relativistic Microscopic Optical Potential by Fourth Order Exchange Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Relativistic Bethe-Salpeter Equation . . . . . . . . . . . . . . . . . . . . . . Bonn One Boson Exchange Potential . . . . . . . . . . . . . . . . . . . . . Nucleon Relativistic Microscopic Optical Potential Based on the Dirac-Brueckner-Hartree-Fock Theory . . . . . . . . . . . . . . . 5.17.1 Relativistic Brueckner Theory . . . . . . . . . . . . . . . . . . . . 5.17.2 Relativistic Pauli Incompatibility Operators . . . . . . . . . . 5.17.3 Calculation Formulas for T Matrix Elements and Nuclear Matter Properties in Symmetric Nuclear Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.17.4 Nucleon Self-Energy in Asymmetric Nuclear Matter [103, 104] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.17.5 Nucleon Relativistic Microscopic Optical Potential Based on DBHF Theory . . . . . . . . . . . . . . . . . . . . . . . . Proca Relativistic Dynamics Equation of Spin 1 Particle and Its Application in Elastic Scattering Calculations . . . . . . . . . 5.18.1 Proca Equation of Free Particles . . . . . . . . . . . . . . . . . . 5.18.2 Proca Equation with Interaction Potential . . . . . . . . . . . . 5.18.3 Some Expressions Related to S=1 Spin Operators [119] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.18.4 Proca Equation in the Form of Schrodinger-like Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

512 512 522 528 531 533 535 537 546

551 564 572 580 580 584

592 595 598 599 599 600 603 606

xxii

Contents

5.18.5

Discussion on the Application of the Proca Equation to Calculate the Elastic Scatterings Between Deuteron and Nucleus . . . . . . . . . . . . . . . . . . . . . . . . . 5.19 Weinberg Relativistic Dynamics Equation with Spin 1 Particles and Discussion on its Application . . . . . . . . . . . . . . . . . 5.19.1 Weinberg Equation of Free Particles . . . . . . . . . . . . . . . 5.19.2 Weinberg Equation with Interaction Potential . . . . . . . . . 5.19.3 Discussion on the Application of the Weinberg Equation to Calculate the Elastic Scatterings Between Deuteron and Nucleus . . . . . . . . . . . . . . . . . . . 5.20 The Relativistic Nuclear Reaction Theory Considering the Internal Structure of the Incident Deuterons . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

609 610 610 612

614 614 618

Basis of the Polarized Particle Transport Theory . . . . . . . . . . . . . . . 625 6.1 Introduction to the Polarized Particle Transport Theory . . . . . . . . 625 *

6.2

6.3

6.4

*

1 1 Polarized Particle Transport Theory for þ A → þ B 2 2 Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 6.2.1 Polarization Theory of Spin Particles Related 2 to Azimuthal Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 6.2.2 Polarization Nuclear Database of Spin Particles . . . . . . 2 6.2.3 Main Points of Polarized Particle Transport Theory 1 for Spin Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 → → Polarized Particle Transport Theory for 1 þ A → 1 þ B Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1 Polarization Quantities of Spin 1 Particles Related to Azimuthal Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.2 Polarization Nuclear Database of Spin 1 Particles . . . . . . 6.3.3 Main Points of Polarized Particle Transport Theory for Spin 1 Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . *

→

Polarized Particle Transport Theory for 1 þ A → *

630 630 635 636 640 640 658 659

1 þB 2

→ 1 þ A → 1 þ B Reactions . . . . . . . . . . . . . . . . . . . . . . . . . 670 2 * → 1 6.4.1 Polarization Quantities for 1 þ A → þ B 2 Reactions Related to Azimuthal Angle . . . . . . . . . . . . . . 670

and

6.4.2

→

*

1 þB 2 Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 680

Polarization Nuclear Database for 1 þ A →

Contents

xxiii *

6.4.3

→ 1 þ A → 1 þ B Reactions 2 Related to Azimuthal Angle . . . . . . . . . . . . . . . . . . . . . 680

Polarization Quantities for

*

6.4.4 6.4.5

→ 1 þ A→ 1 þ B 2 Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 688 Main Points of Polarized Particle Transport Theory

Polarization Nuclear Database for

→

*

*

→ 1 1 þ B and þ A → 1 þ B Reactions . . . 2 2 6.5 Polarized Particle Transport Equation . . . . . . . . . . . . . . . . . . . . . 6.6 Prospect of the Polarized Particle Transport Theory . . . . . . . . . . . 6.6.1 To Perfect the Polarization Theory of Nuclear Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6.2 To Establish a Polarization Nuclear Database . . . . . . . . . 6.6.3 To Research the Polarized Particle Transport Theory . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

for 1 þ A →

689 693 705 705 708 710 711

Appendix: Clebsch-Gordan Coefficients, Racah Coefficients, and 9j Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 713 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 717

About the Author

Qing-Biao Shen is a researcher at the China Institute of Atomic Energy (CIAE). He was born on September 1, 1938, in Gu’an County, Hebei Province. In July 1963, he graduated from the University of Science and Technology of China with a degree in physics under the tutelage of Mr. Hong-Yuan ZHU. In the same year, Mr. Shen joined the CIAE, and from September 1963 to January 1965, he participated in a group led by Min YU and Zu-Qia HUANG working on hydrogen bomb theory. Since then, Mr. Shen’s career has focused on low- and medium-energy nuclear reaction theory, nuclear many-body theory, and nuclear data theory, some of which he conducted while he was a visiting scholar at the University of Kentucky from 1985 to 1987. Mr. Shen is the recipient of 12 National or Ministerial Science and Technology Progress Awards, including 5 inaugural prizes, and he has published over 300 academic papers as either a lead author or co-author. He is the author of Nuclear Reaction Theory of Low and Medium Energies. In addition, Mr. Shen developed the Skyrme force microscopic optical potential theory, a theory that to this day best fits experimental data.

xxv

Chapter 1

Basic Knowledge of Polarization Theory of Nuclear Reactions

Abstract This chapter introduces the basic knowledge of polarization theory of nuclear reactions, including spin operators, irreducible tensors, polarization operators, density matrices, and so on. Keywords Density matrices · Irreducible tensors · Magnetic quantum numbers · Polarization operators · S matrix elements · Spin operators

1.1

Introduction

When a particle outside the nucleus moves under the action of the nuclear potential, its coordinate position and momentum change continuously. At the same time, the particle and the nucleus itself will also undergo self-rotation, called spin. The spin values of particles and nuclei are quantized. For example, the spin of α particles is 0; 1 the spin of neutrons, protons, tritium, and 3He is ; the spin of deuteron and 6Li is 1; 2 3 and the spin of 5He, 5Li, 7Li, 7Be, 9Be is . The unit of spin is ħ, and the spin of a 2 particle is typically represented by S. Its projection in a certain spatial direction is called the spin magnetic quantum number, and its value is S, S – 1, . . .,
S. The spin of the target is usually represented by I. In the nuclear structure theory, the spin-orbital coupling potential of the nucleons will cause the single particle level to be split. During the nuclear reaction process, the particles with different spin magnetic quantum numbers feel different nuclear forces. Its movement behavior will also be different, so the polarization phenomena will ! occur. The spin S of the particle is a vector in the coordinate space. Its average value ! hS i in the nuclear reaction system is called the particle polarization vector. If it is not equal to 0, then the particle is said to be polarized; otherwise it is unpolarized. 1 Two nucleons with spin can be coupled to two states of total spin, S ¼ 0 and 2 ! 1, and there is a tensor potential for the S ¼ 1 state. The deuteron spin S is a vector, ! and the vector S can also constitute a rank-2 tensor. Since the tensor potential is non-central, the total orbital angular momentum L is no longer a good quantum © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 Q.-B. Shen, Polarization Theory of Nuclear Reactions, https://doi.org/10.1007/978-3-031-11878-4_1

1

2

1 Basic Knowledge of Polarization Theory of Nuclear Reactions

number, and the coupling between different L waves will occur when solving the deuteron Schrodinger equation. Because of the tensor force existing between the nucleon-nucleon, it is found that when studying the deuterium structure not only the S state with L ¼ 0 exists, but the D state with L ¼ 2 exists as well. It is known that the spin of the deuteron is equal to 1 and its magnetic quantum numbers have three states: 1, 0, and
1. The action intensities of the spin-orbital potential on these three states are different: the action intensities of the tensor potential are also different. Therefore, there is not only vector polarization, but also tensor polarization for deuteron. They are all observables. For particles whose spin is greater than 1, there are more complex polarization observables. The polarization phenomenon greatly enriches the data obtained from scattering and nuclear reaction experiment; provides information on the interaction that can change spin orientation; and provides important verification data for the study of nuclear structures or reaction mechanisms. The theory of calculating various polarization quantities from the nuclear reaction amplitude is called polarization theory of nuclear reactions. In the 1950s, Wolfenstein et al. conducted theoretical studies on the polarization of nuclear reactions [1–3]. Subsequently, many papers [4–22] on polarization phenomena have been published, with some of them also containing information on experiments of polarization nuclear reactions. In 1972, a paper published in Rep. Prog. Phys. by Ohlsen is a classical article [6] on the polarization theory of nuclear reactions. In 1974, Robson published a short book titled The Theory of Polarization Phenomena [9]. In these studies, more detailed computational expressions are given for the 1 particles with spinless targets, and partial computapolarization theory of spin 2 tional expressions are given for the polarization theory of spin 1 particles with spinless targets. Some specific calculations were also made and compared with experimental data in the spinless target case. In Ohlsen’s paper, a brief introduction of the polarization formulas of the two polarized light particles was also given. In the relativistic optical model, the Dirac equation can be converted into a Schrodingerlike equation; thus, the polarization quantities of elastic scattering between the nucleon and the target with spinless can be calculated. For nuclear reactions with the non-zero spin target and the residual nucleus, in 1965, Tamura in his paper [23] on coupling channel calculations gave a formula for calculating the polarization rate of the emitted nucleons using the method of averaging the Pauli matrix by the wave functions. On this basis, effort was devoted by some people to studying the corresponding analyzing power [24, 25]. In Robson’s book [9], only in the complex spherical basis coordinate system, the polarization problem of the nuclear reaction between the incident particle with arbitrary spin and the target nucleus with arbitrary spin has been studied briefly by the formal theory, which has no yet the practical value. However, before this book, no one ever gave a systematic polarization theory with clear theoretical formulas that can be used to calculate the polarization observ1 ables for elastic scattering and direct reaction channels of spin and 1 particles 2 with targets with non-zero spins. For example, there is no clear theoretical formula to ! ! ! ! 1 1 þA! 1 þB calculate the polarization observables for 1 þ A ! þ B and 2 2

1.1

Introduction

3

! 1 represents the reactions with non-zero spin target or residual nucleus, where 2 1 polarized particle with spin . Therefore, it is impossible to carry out fully the 2 calculations of various polarization quantities for targets or residual nuclei with non-zero spins. Since the spins of most nuclei (including all odd A nuclei) are not equal to 0, the development of polarization theory of nuclear reactions is constrained. In order to perform a specific calculation of the polarization quantity, the nuclear reaction theory [26, 27] must be used to study how to get the corresponding reaction amplitude. We know that the S matrix theory with j–j coupling [26] or S-L coupling [27] can give expressions for scattering amplitude, or reaction amplitude of the two-body nuclear reactions, including the C-G coefficients, the spherical harmonic functions, the Legendre functions, and the S matrix elements. But the key issue is how to calculate the S matrix elements. In nuclear reaction theory, S matrix elements can be calculated by the phase shift analysis method for nucleon-nucleon or nucleondeuteron elastic scattering, and it is also possible to extend this method to other types of nuclear reactions. For the light nucleus resonance energy region, the R matrix theory can be used to calculate the S matrix elements. The S matrix elements in the case of elastic scattering of a particle with a spinless target can be calculated using a 1 spherical optical model. The polarization problem of elastic scattering of the spin 2 particles and spinless targets is introduced in Ref. [26], in which the S matrix elements are calculated by the spherical optical model. At present, two groups of authors have given the universal deuteron optical potentials in the energy region below 200 MeV [28, 29]. However, neither group considered the deuteron polarization data when determining the parameters of the two sets of deuteron optical potentials. Moreover, in another set of deuteron universal optical potential suitable for 12–90 MeV energy rangy [30], it only considered the vector polarization data. In order to calculate the deuteron observables that can fit vector and tensor polarization experimental data simultaneously, the tensor potentials should be included in the deuteron optical potentials. The S matrix elements of the inelastic scattering can be calculated by using the coupled channel optical model. The T matrix elements of the two-body direct reactions can be obtained by using the DWBA method and can be converted into the S matrix elements [26]. With the development of computers, coupled reaction channel theory considering elastic scattering, inelastic scattering, and other two-body direct reactions can simultaneously give the S matrix elements of these reaction channels. In addition, the three-body reaction theory starting from realistic nuclear forces can also give scattering amplitude or reaction amplitude, so it can also be considered to perform a specific calculation of the polarization observables. Since statistical average assumptions are made in the study of equilibrium and pre-equilibrium reaction theories, the polarization problem does not need to be considered in these theories. We know that the quantum theory of angular momentum is the main theoretical tool for studying how to obtain various polarization quantities from the amplitude of the nuclear reactions. In this respect, a very detailed and practical Ref. [31] can be found. Given this, it is possible now to develop a more widely applicable and

4

1 Basic Knowledge of Polarization Theory of Nuclear Reactions

practical polarization theory of nuclear reactions based on the existing polarization theory of nuclear reactions. 1 This book will study the polarization theory of nuclear reactions in which spin 2 or 1 particle reacts directly with the unpolarized or polarized targets and residual * * ! ! 1 1 nuclei with arbitrary spins, including 1 þ A ! þ B and þ A ! 1 þ B reac2 2 tions. This book will also give the polarization theory of nuclear reactions in which the two particles in the incident channel are polarized; and make study on the 3 polarization theory of nuclear reactions of the spin particles and the polarization 2 theory of the photon beams; and give the theory of relativistic nuclear reactions and the corresponding polarization theory. We know that in some energy regions, even if the target, the residual nucleus, and the incident particle are all unpolarized, the emitted particles are generally polarized. For example, for the neutron induced reaction, the nuclear forces felt by neutrons with spin up and spin down are different, so their motion behaviors are different. Although the polarization phenomena exist objectively in nuclear reactions, at present people only use unpolarized nuclear reaction data when studying the nuclear transport process, and do not consider the polarization effect. Obviously, this approach is only an approximation in some energy regions. Therefore, in the last chapter of this book, we will discuss how to establish polarization nuclear database and make studying and discussion on the development of polarized nucleus transport theory.

1.2

Spin Operators

In general, we adopt the natural unit system, that is, letting c ¼ ħ ¼ 1. In the actual calculation, the dimensionless spin and spin projection should all be multiplied by ħ. For coordinate space, the Cartesian coordinate system is generally used, that is, ! the rectangular coordinate system, and the coordinate position vector r can be represented as (x, y, z). Ohlsen, an authority on polarization theory, pointed out that formally constructing another coordinate system represented by complex num! bers, in which the coordinate position vector r can be represented as (r1, r0, r
1), has many advantages in theoretical derivation [7]. In the monograph Quantum Theory of Angular Momentum [31], the coordinate system represented by (r, θ, φ) is called the polar coordinate system, and the coordinate system represented by (r1, r0, r
1) is called the spherical coordinate system. In an article on polarization theory published in Phys. Rev. C [32], the coordinate system represented by (r1, r0, r
1) is also referred to as the spherical coordinate system. In Rose’s book Elementary Theory of Angular Momentum [33], the coordinate system represented by (r1, r0, r
1) is called the “spherical coordinate representation”, and the relations between the Cartesian basis functions and the spherical basis functions are given. In order to avoid confusion with the spherical coordinate system represented by (r, θ, φ), this book will refer to the coordinate system represented by (r1, r0, r
1) as the spherical basis

1.2

Spin Operators

5

coordinate system. Note that in the spin S space, there are 2S þ 1 spherical basis coordinate axes. If the spin of a particle is S, then its spin operator ^ S is a vector with three components and typically represented by a set of three square (2S þ 1)  (2S þ 1) matrices. The spin operator ^S is Hermitian ^Sþ ¼ ^S

ð1:1Þ

For the Cartesian coordinate components of ^S, Eq. (1.1) representing the Hermitian property has the form þ

ð^Si Þ ¼ ^Si ,

i ¼ x,y,z

ð1:2Þ

and for spherical basis coordinate components there are þ

μ

ð^Sμ Þ ¼ ^S ¼ ð
1Þμ ^S
μ ,

μ ¼ 1,0

ð1:3Þ

Spin operator ^S satisfies the following relation ^S  ^S ¼ i^S,

ð1:4Þ

and also satisfies the following commutation relation *

*

* * S ½ð a  ^SÞ,ðb  ^SÞ ¼ iða  b Þ  ^ *

*

*

ð1:5Þ *

*

*

where a and b are any constant vectors. When letting a ¼ e i , b ¼ e k (i, k ¼ x, y, * * z), where e i and e k are the basis vectors of the Cartesian coordinate system, the following commutation relations can be obtained from Eq. (1.5) 2 ½^Si , ^Sk  ¼ iεikl ^Sl , ½^S , ^Si  ¼ 0,

i,k,l ¼ x,y,z

ð1:6Þ

where

εikl

8 > < 1, ¼
1 > : 0

if the combinaton of ikl is obtained by an even permutation of xyz if the combinaton of ikl is obtained by an odd permutation of xyz if at least two indices among ikl are equal ð1:7Þ

The square of the spin operator ^S 2 may be expressed in terms of Cartesian coordinate components ^Si ði ¼ x, y, zÞ as

6

1

Basic Knowledge of Polarization Theory of Nuclear Reactions

X

^S 2 ¼

2 ^S 2 ¼ ^S 2 þ ^S 2 þ ^ Sz i x y

ð1:8Þ

i

In Ref. [26], Eq. (3.4.48) gives the definition of the spherical basis vectors, and Eqs. (3.4.51) and (3.4.52) give the spherical basis coordinate representations of any ! three-dimensional vector A as follows: !

A¼

1 X

!

ð
1Þμ Aμ e
μ

ð1:9Þ

μ¼
1

 1  A1 ¼
pffiffiffi Ax þ iAy , 2

 1  A0 ¼ Az , A
1 ¼ pffiffiffi Ax
iAy 2

ð1:10Þ

And we can find out ^S2 ¼

X μ

ð
1Þμ ^S
μ ^Sμ ¼

X μ ^ S ^ Sμ

ð1:11Þ

μ

The spin function of spin S particles can be expressed as χ(σ), where σ is the projection of the spin on the z-axis, σ ¼ S, S
1, ⋯,
S. The spin functions χ(σ) are commonly written as column matrices of 2S þ 1 elements 0

χ ðSÞ

1

B χ ð S
1Þ C B C ^χ ¼ B C @ ⋮ A

ð1:12Þ

χ ð
SÞ The corresponding Hermitian conjugate function ^ χ þ has the form of a row matrix ^χ þ ¼ ðχ  ðSÞ χ  ðS
1Þ ⋯ χ  ð
SÞÞ

ð1:13Þ

Its normalization condition is ^χ þ ^χ ¼

S X

j χ ðσ Þj 2 ¼ 1

ð1:14Þ

σ¼
S

The basis spin function is a spin state wave function that has definite spin and spin projection on the z-axis. The basis spin functions χ Sm are the common eigenfunctions 2 of spin operators ^ S and ^Sz , that is,

1.2

Spin Operators

7

^S2 χ Sm ¼ SðS þ 1Þχ Sm ,

^Sz χ Sm ¼ mχ Sm

ð1:15Þ

In the spherical basis representation, the dependence of the basis spin functions χ Sm(σ) on the spin variable σ is given by χ Sm ðσ Þ ¼ δmσ

ð1:16Þ

The basis spin functions of the spin S particles can be written as column matrices as follows: 0 χ SS

1

0

1

B C B 0 C B C C ¼B B ⋮ C, B C @ 0 A 0

χS

S
1

0

0

1

0

1

B C B C B 1 C B 0 C B C B C C, ⋯, χ S
S ¼ B ⋮ C ⋮ ¼B B C B C B C B C @ 0 A @ 0 A 0 1

ð1:17Þ

The collection of 2S þ 1 basis spin functions χ Sm (m ¼ S, S
1, ⋯,
S) constitutes a complete orthonormal set of functions, and its orthonormality condition is χþ Sm χ Sm0 ¼ δmm0

ð1:18Þ

Its completeness condition may be written in matrix form S X

^ χ Sm χ þ Sm ¼ I,

ð1:19Þ

m¼
S

where ^I is the unit (2S þ 1)  (2S þ 1) matrix. The spherical basis components of the spin operator ^ S may be expressed in terms of the basis spin functions χ Sm as [31] ^Sμ ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiX S m0 SðS þ 1Þ CSm 1μ χ Sm0 χ þ Sm ,

μ ¼ 1,0

ð1:20Þ

mm0

From Eq. (1.20) and Eq. (1.16), the matrix elements of the spherical basis component of the spin matrix in the spherical basis representation can be obtained as ð^Sμ Þσ0 σ ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Sσ0 SðS þ 1Þ C Sσ 1μ ,

σ,σ 0 ¼ S,S
1,⋯,
S

ð1:21Þ

The components of the spin operator ^S in the Cartesian coordinate system can be obtained from its components in the spherical basis coordinate system given by Eq. (1.20) by use of the following formulas:

8

1

Basic Knowledge of Polarization Theory of Nuclear Reactions

  ^Sx ¼ p1ffiffiffi ^S
^S ,
1 1 2

  ^Sy ¼ piffiffiffi ^S þ ^ Sz ¼ ^ S0 S1 , ^
1 2

ð1:22Þ

Its inverse transformation relations are   ^S1 ¼
p1ffiffiffi S^x þ i^Sy , 2

  ^S
1 ¼ p1ffiffiffi ^ ^ S x
i Sy 2

^S0 ¼ ^Sz ,

ð1:23Þ

In the Cartesian coordinate system, there are the following trace (tr) formulas for the products of spin matrices for an arbitrary spin S [31]:     SðS þ 1Þð2S þ 1Þ δik , tr ^Si ¼ 0, tr ^Si ^Sk ¼ 3   SðS þ 1Þð2S þ 1Þ tr ^Si ^Sk ^Sl ¼ i εikl , 6 (" # n o  SðS þ 1Þð2S þ 1Þ 1  tr ^Si ^Sk ^Sl ^Sj ¼ S ð S þ 1Þ þ δik δlj þ δij δkl 15 2 h

i

ð1:24Þ

)

þ SðS þ 1Þ
2 δil δkj ,

i, k, l, j ¼ x, y, z

The first symbol i on the right side of the third equation of the above expressions represents the imaginary number. In the spherical basis coordinate system, there are the following trace formulas for the products of spin matrices [31] n o tr ^Sμ ¼ 0,

n o SðS þ 1Þð2S þ 1Þ tr ^Sμ ^Sν ¼ ð
1Þμ δμ
ν , 3 ! n o SðS þ 1Þð2S þ 1Þ 1 1 1 SðS þ 1Þð2S þ 1Þ pffiffiffi pffiffiffi ¼ tr ^Sμ ^Sν ^Sλ ¼
ð
1Þ1þλ C11μ
λ1ν , 6 3 2 μ ν λ n o SðS þ 1Þð2S þ 1Þ nh i 1 SðS þ 1Þ þ ð
1Þμþλ tr ^Sμ ^Sν ^Sλ ^Sρ ¼ 15 2 o   δμ
ν δλ
ρ þ δμ
ρ δν
λ þ ½SðS þ 1Þ
2 ð
1Þμþν δμ
λ δν
ρ , μ, ν, λ, ρ ¼ 1, 0 ð1:25Þ where

1

1

1

μ

ν

λ

is the 3j symbol.

1.3

1.3

Irreducible Tensors

9

Irreducible Tensors

The Wigner D function DJMM 0 ðα, β, γ Þ is defined as the matrix elements of the ^ ðα, β, γ Þ with a definite angular momentum J and magnetic rotation operator D quantum number M ^ ðα, β, γ ÞjJ 0 M 0 i ¼ δJJ 0 DJMM 0 ðα, β, γ Þ hJMjD

ð1:26Þ

where α, β, γ are Euler angles. The wave function Ψ JM of a quantum system with angular momentum J and projection M rotates in coordinate space in the following way: Ψ JM 0 ðθ 0 , φ0 Þ ¼

J X

Ψ JM ðθ, φÞDJMM 0 ðα, β, γ Þ

ð1:27Þ

M¼
J

where θ, φ and θ 0 , φ0 are the polar and azimuthal angles of the initial and rotated coordinate systems. θ, φ and θ 0 , φ0 satisfy the following relations [31]: cos θ 0 ¼ cos θ cos β þ sin θ sin β cos ðφ
αÞ, cot θ sin β cot ðφ 0 þ γ Þ ¼ cot ðφ
αÞ cos β
sin ðφ
αÞ

ð1:28Þ

Its inverse relations are cos θ ¼ cos θ 0 cos β
sin θ 0 sin β cos ðφ 0 þ γ Þ, cot θ 0 sin β cot ðφ
αÞ ¼ cot ðφ 0 þ γ Þ cos β þ sin ðφ 0 þ γ Þ

ð1:29Þ

The spin projection on the linear momentum direction of a particle is called the helicity, and the helicity of the spin S particle is taken as λ ¼ S, S
1, ⋯,
S. Let * * n ¼ p =p denote the unit vector of the particle momentum direction, where θ and φ ! are the polar and azimuthal angles of n . The helicity basis functions χ Sλ(θ, φ) are the 2 * common eigenfunctions of the operators ^S and ^S  n , that is, ^S2 χ Sλ ðθ, φÞ ¼ SðS þ 1Þχ Sλ ðθ, φÞ,

^S  * n χ Sλ ðθ, φÞ ¼ λχ Sλ ðθ, φÞ

ð1:30Þ

The helicity basis functions can be obtained by rotating the general basis spin functions, which is defined to the z-axis [31]

10

1

Basic Knowledge of Polarization Theory of Nuclear Reactions

X

χ Sλ ðθ, φÞ ¼

DSmλ ðφ, θ, 0Þχ Sm ,

m

χ Sm

X ¼ DS
λ λ

X

χþ Sλ ðθ, φÞ ¼


m ð0,

θ, φÞχ Sλ ðθ, φÞ,

ð
1Þλ
m DS
m


λ ðφ,

θ, 0Þχ þ Sm ,

ð1:31Þ

m

X

χþ Sm ¼

λ

ð
1Þλ
m DSλ m ð0, θ, φÞχ þ Sλ ðθ, φÞ

The orthonormality condition for the helicity basis functions is   χþ Sλ0 θ, φ χ Sλ ðθ, φÞ ¼ δλ0 λ

ð1:32Þ

The completeness condition for the helicity basis functions has the following matrix form X λ

^ χ Sλ ðθ, φÞχ þ Sλ ðθ, φÞ ¼ I

ð1:33Þ

where ^I is a unit (2S þ 1)  (2S þ 1) matrix. Irreducible tensors mean that under rotations of coordinate systems these tensors transform in the same manner as eigenfunctions of the angular momentum operator. Assuming that MJ is an irreducible tensor of rank J, where J can be a non-negative integer or semi-odd number, MJ has 2J þ 1 components MJM, M ¼ –J, –J þ 1, ⋯, J. When the coordinate system rotates, a linear transformation of MJM is performed as follows: ^ ðα, β, γ ÞMJM 0 ðX Þ½D ^ ðα, β, γ Þ MJM 0 ðX 0 Þ ¼ D J X ¼ MJM ðX ÞDJMM 0 ðα, β, γ Þ


1

ð1:34Þ

M¼
J

where X and X0 denote sets of all arguments of the tensor in the initial and final coordinate systems, respectively. It can be seen that the linear transformation coefficient of this rotation is the Wigner D function. In fact, the angular momentum ^ ^S are the irreducible tensors of rank 1; the spherical harmonic function operators ^J,L, Ylm is an irreducible tensor of rank l; and the spin S wave function is an irreducible tensor of rank S. The spherical basis components of the angular momentum operator ^ J and the irreducible tensor MJM satisfy the following commutation relations:

1.3

Irreducible Tensors

11

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ^J 1 , MJM ¼  p1ffiffiffi eiδ J ðJ þ 1Þ
M ðM  1ÞMJ 2  ^J 0 , MJM ¼ MMJM

M1 ,

ð1:35Þ

The above formulas can be written in compact form as 

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ^J μ , MJM ¼ eiμδ J ðJ þ 1ÞC JJ Mþμ M 1μ MJ

Mþμ ,

μ ¼ 1,0

ð1:36Þ

From these relations it follows that h 2 i ^J , MJM ¼ J ðJ þ 1ÞMJM

ð1:37Þ

Let us adopt the phase δ ¼ 0, that is, eiδ ¼ 1. For the integer rank J, the overall phase of the MJM components is usually defined in such manner that ðMJM Þ ¼ ð
1Þ
M MJ

ð1:38Þ


M

This choice of phase coincides with that for spherical harmonics. You can also define another irreducible tensor as f JM ¼ iJ MJM M

ð1:39Þ

Then one has 

f JM M



fJ ¼ ð
1ÞJ
M M

ð1:40Þ


M

The irreducible tensor defined in this way can be used in the case where J is an integer or a semi-odd number; J
M is always an integer. According to Eq. (1.3), MJM and MM J respectively represent covariant and contravariant components of the irreducible tensor MJ, and we have the following relations: 
M MJ
M , MM J ¼ ðMJM Þ ¼ ð
1Þ    M f JM ¼ ð
1ÞJ
M M f ¼ M f J
M M

ð1:41Þ

J

The scalar product of two irreducible tensors MJ and N J of the same rank is defined as ðMJ  N J Þ ¼

X ð
1Þ
M MJM N J M


M

¼

X M

MJM N JM ¼

X MJM N M J M

ð1:42Þ

12



1

Basic Knowledge of Polarization Theory of Nuclear Reactions

 X f J  Ne J ¼ f JM Ne J ð
1ÞJ
M M M


M

¼

X

M

f JM Ne  ¼ M JM

M

X

f JM Ne M ð1:43Þ M J

M

The components of an irreducible tensor product LJ of two irreducible tensors MJ 1 and N J 2 of ranks J1 and J2 are defined as follows: LJM ¼

X

C JJ 1M M1

J 2 M 2 MJ 1 M 1 N J 2 M 2

ð1:44Þ

M1 M2

The irreducible tensor product is also denoted as   LJ M J 1 N J 2 J

ð1:45Þ

where MJ 1 M 1 N J 2 M 2 is the direct product of a set of (2J1 þ 1)(2J2 þ 1) components of two irreducible tensors MJ 1 and N J 2 . According to Eq. (1.44) one can obtain MJ 1 M 1 N J 2 M 2 ¼

JX 1 þJ 2 J¼jJ 1
J 2 j

C JJ 1M M1

J 2 M 2 LJM

ð1:46Þ

This formula shows that the direct product MJ 1 M 1 N J 2 M 2 is reducible and can be f J M and expanded with irreducible tensor LJM with different rank J. If one use M 1 1 e Ne J 2 M 2 defined by Eq. (1.39) to get LJM , one can prove 

e JM L



eJ ¼ ð
1ÞJ
M L


M

ð1:47Þ

e JM all satisfy the relation (1.40). f J M , Ne J M , and L This formula shows that M 1 1 2 2 Although MJ 1 M 1 and N J 2 M 2 satisfy relation (1.38), LJM does not except J1 þ J2
J ¼ even. From the above discussion, it is understandable why il Yl ml is used instead of using only Yl ml in the coupling wave function of the nuclear reactions.

1.4

Polarization Operators

The spin observables are different for particles with different spins. Table 1.1 gives the types of the spin observables corresponding to various spin particles. The polarization operator is used to describe the polarization state of particles. The polarization operator T^ LM ðSÞðM ¼
L,
L þ 1,⋯, L and L ¼ 0, 1, . . ., 2S; both L and M being integers) are square (2S þ 1)  (2S þ 1) matrices which act on spin functions. When the coordinate system is rotated, this operator is linearly transformed with the Wigner D function in the way given by Eq. (1.34), that is,

1.4

Polarization Operators

13

Table 1.1 Types of the spin observables corresponding to various spin particles 1 2

0 Spin observables √

Unpolarized, scalar, rank-0 tensor Vector, rank-1 tensor

Spin S 3 1 2

√

√

√

. . .. . . . . .. . .

√

√

√

. . .. . .

√

√

. . .. . .

√

. . .. . .

Rank-2 tensor Rank-3 tensor . . .. . .

. . .. . .

T^ LM ðSÞ are irreducible tensors of rank L. According to Eq. (1.36), the commutation relations between the spherical basis components of the spin operator ^ Sμ and the polarization operators T^ LM ðSÞ can be written as  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ^Sμ , T^ LM ðSÞ ¼ LðL þ 1ÞC L Mþμ T^ L LM 1μ

Mþμ ðSÞ

ð1:48Þ

Let us normalize the polarization operators by the condition n o þ tr T^ LM ðSÞT^ L 0 M 0 ðSÞ ¼ ð2S þ 1Þ δLL 0 δMM 0

ð1:49Þ

and choose the phase factors to satisfy the relations þ T^ LM ðSÞ ¼ ð
1ÞM T^ L


M ðSÞ

ð1:50Þ

The above three conditions completely determine the polarization operator T^ LM ðSÞ. The above definition of T^ LM ðSÞ in this book is consistent with that in the polarization theory articles [6, 34] in S¼1 case. But the defined polarization operator T^ LM ðSÞ in this book compared with that in the monograph pffiffiffiffiffiffiffiffiffiffiffiffiffiffititled Quantum Theory of Angular Momentum [31] has a more factor 2S þ 1 . Based on the trace formula n o  þ tr σ^μ σ^μ0 ¼ 2δμμ0 and Eq. (1.49), the defined T 1M 12 in this book is equal to the 2S P ð2L þ 1Þ ¼ ð2S þ 1Þ2 Pauli matrix σ^M . The polarization operator T^ LM ðSÞ has L¼0

linear independent components, and each component is a square (2S þ 1)  (2S þ 1) matrix. These (2S þ 1)2 matrices constitute a set of complete linear independent matrices. The polarization operator T^ LM ðSÞ can be represented by a basis spin function χ Sm T^ LM ðSÞ ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffi X S m 0 2L þ 1 C Sm LM χ Sm0 χ þ Sm m m0

Its inverse relation is

ð1:51Þ

14

1

χ Sm0 χ þ Sm

Basic Knowledge of Polarization Theory of Nuclear Reactions

¼

ffi X pffiffiffiffiffiffiffiffiffiffiffiffiffi 2L þ 1 2S þ 1

LM

0 C SSmm LM T^ LM ðSÞ

ð1:52Þ

Using Eq. (1.16) in the spherical basis representation, the matrix element of T^ LM ðSÞ in the spherical basis representation can be obtained as h

T^ LM ðSÞ

i σ0 σ

¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffi Sσ0 2L þ 1CSσ

LM ,

σ,σ 0 ¼
S,
S þ 1,⋯,S

ð1:53Þ

When L ¼ 0 from the above relation, one can obtain T^ 00 ðSÞ ¼ ^I

ð1:54Þ

where ^I is a unit (2S þ 1)  (2S þ 1) matrix. Comparing Eq. (1.20) with Eq. (1.51) in L ¼ 1 case, the following relation can be obtained: T^ 1M ðSÞ ¼

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 ^S , SðS þ 1Þ M

M ¼ 1,0

ð1:55Þ

1 According to Eq. (1.49), it can be seen that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi T^ LM ðSÞ constitutes a set of ð2S þ 1Þ complete linear independent matrices. Therefore, any square (2S þ 1)  (2S þ 1) 1 ^ can be expanded with a series of polarization operators pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi T^ LM ðSÞ matrix A ð2S þ 1Þ as 2S L X X 1 ^ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ALM T^ LM ðSÞ A 2S þ 1 L¼0 M¼
L

ð1:56Þ

By use of Eq. (1.49) and the above formula, we can get n o pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ^ T^ þ ðSÞ ¼ 2S þ 1 ALM tr A LM

ð1:57Þ

^ then by using Eqs. (1.50) and (1.56) one can ^ is Hermitian, that is, A ^ þ ¼ A, If A obtain X 1 ^¼A ^ þ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ALM ð
1ÞM T^ L
M ðSÞ A 2S þ 1 LM X 1 A ð
1ÞM T^ LM ðSÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2S þ 1 LM L
M

ð1:58Þ

Comparing Eq. (1.56) and Eq. (1.58), the following formulas can be obtained:

1.4

Polarization Operators

15

ALM ¼ ð
1ÞM AL


M ,

ALM ¼ ð
1ÞM AL

ð1:59Þ


M

Products of two polarization operators T^ L1 M 1 ðSÞ and T^ L2 M 2 ðSÞ may be written as [31] ^2 ^S ^1 L T^ L1 M 1 ðSÞT^ L2 M 2 ðSÞ ¼ L

X

ð
1ÞL1 þL2 þL W ðL1 L2 SS; LSÞC LM L1 M 1

^

L2 M 2 T LM ðSÞ

ð1:60Þ

LM

pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ^ 2L þ 1, ^S 2S þ 1 , and Racah coefficient W is used. Substituting where L Eq. (1.51) into the right end of Eq. (1.60), one can obtain ^2 ^S ^1 L IR L

X

^ ðL1 L2 SS; LSÞC LM ð
1ÞL1 þL2 þL LW L1 M 1

X L2 M 2

LM

mm0

CSm Sm

0

þ LM χ Sm0 χ Sm

ð1:61Þ

Substituting Eq. (1.51) into the left end of Eq. (1.60) and using Eq. (1.18), one can obtain

^2 ^1 L ¼ L

!

P

^2 ^1 L IL L

m1 m1 0

P

m1 m0 m

0 C SSmm11 L1 M 1 χ Sm1 0 χ þ Sm1

P

m2 m2 0 S m1 S m0 CSm1 L1 M 1 C Sm L2 M 2 χ Sm0 χ þ Sm

! 0 C SSmm22 L2 M 2 χ Sm2 0 χ þ Sm2

ð1:62Þ

Using the following formula including the C–G coefficient and the Racah coefficient X ε

Ceε aα

cγ bβ C eε dδ

¼

X ^e^f W ðabcd; ef ÞC fbβφ fφ

cγ dδ C aα f φ

ð1:63Þ

one can obtain P Sm0 C Sm1 m1

Sm1 L1 M 1 C Sm L2 M 2

¼

P LM

¼

P LM

^SLW ^ ðSL2 SL1 ; SLÞCLM L M 2

2

Sm0 L1 M 1 C Sm LM

^ ðL1 L2 SS; LSÞCLM ð
1ÞL1 þL2 þL ^SLW L1 M 1

Sm0 L2 M 2 C Sm LM

ð1:64Þ Substituting the above formula into Eq. (1.62), it is proved that the two ends of Eq. (1.60) are equal. The polarization operators satisfy the following commutation relations [31]: X   ^2 ^S ð
1ÞL1 þL2 þL3 1
ð
1ÞL1 þL2 þL3 ^1 L T^ L1 M 1 ðSÞ, T^ L2 M 2 ðSÞ ¼ L L3 3 W ðL1 L2 SS; L3 SÞC LL31 M M1

^ L2 M 2 T L3 M 3 ð SÞ

ð1:65Þ

Based on Eq. (1.55), we know that Eq. (1.48) is only a special case of Eq. (1.65) in the case of L1 ¼ 1. The anti-commutation relations of the polarization operators are

16

1

Basic Knowledge of Polarization Theory of Nuclear Reactions

X    ^2 ^S ð
1ÞL1 þL2 þL3 1 þ ð
1ÞL1 þL2 þL3 ^1 L T^ L1 M 1 ðSÞ, T^ L2 M 2 ðSÞ ¼ L L3 3 W ðL1 L2 SS; L3 SÞCLL31 M M1

^ L2 M 2 T L3 M 3 ð SÞ

ð1:66Þ

pffiffiffiffiffiffiffiffiffiffiffiffiffiffi Here note the introduced symbol S^ 2S þ 1. The following formulas can be obtained using Eqs. (1.49), (1.50), and (1.54) n o 2 tr T^ LM ðSÞ ¼ ^S δL0 δM0 n o 2 tr T^ L1 M 1 ðSÞT^ L2 M 2 ðSÞ ¼ ð
1ÞM 1 ^S δL1 L2 δM 1

ð1:67Þ
M 2

ð1:68Þ

Using Eq. (1.60) and Eq. (1.68), one can get   tr T^ L1 M 1 ðSÞT^ L2 M 2 ðSÞT^ L3 M 3 ðSÞ ^1 L ^2 ^S C LL3 M
ML3 M W ðL1 L2 SS; L3 SÞ ¼ ð
1ÞL1 þL2 þL3 þM 3 L 1 1 2 2 3

ð1:69Þ

Using Eq. (1.60) and Eq. (1.69), one can also get   tr T^ L1 M 1 ðSÞT^ L2 M 2 ðSÞT^ L3 M 3 ðSÞT^ L4 M 4 ðSÞ X 2 ^1 L ^2 ^SW ðL1 L2 SS; LSÞC LL MM 1 þM ¼ ð
1ÞL1 þL2 þL L 1 1 L2 M 2 L2S 4 ^L ^3 ^S C LL4 M
MþM ð
1ÞLþL3 þL4 þM 4 L 1 2

3

L3 M 3 W ðLL3 SS; L4 SÞ

ð1:70Þ

^1 L ^2 L ^3 ^S ¼ ð
1ÞL1 þL2 þL3 þL4 þM 4 L X L M þM L
M 4 ^ L M 1 L M2 CL4 M þM LC W ðL1 L2 SS; LSÞW ðLL3 SS; L4 SÞ 1 1 2 2 1 2 L3 M 3 4

L2S

1.5

Density Matrices

The density matrix is defined as [31] ^ρ ¼

X wα jψ α ihψ α j

ð1:71Þ

α

where jψ αi is the α sub-state of the system, wα is the weight and the normalization condition is given by X wα ¼ 1 α

ð1:72Þ

1.5

Density Matrices

17

Obviously the density matrix defined by Eq. (1.71) is Hermitian. Set {jφii} as a set of orthogonal complete basis, the matrix element of the density matrix ^ρ in the orthogonal basis is ρik ¼ hφi j^ρjφk i ¼

X α

ð1:73Þ

hφi jψ α iwα hψ α jφk i

^ is In general ^ρ is not a diagonal matrix. The definition of the trace of the observable O X ^ jφi i h φi j O

^g ¼ trfO

ð1:74Þ

i

The density matrix can be traced, using Eqs. (1.71), (1.72), and (1.74) as well as the orthonormal properties of the wave functions, as follows: trf^ρg ¼

X X X X X wα hψ α jφi ihφi jψ α i ¼ wα ¼ 1 hφi j wα jψ α ihψ α jφi i ¼ i

α

α

α

i

ð1:75Þ ^ is The average value of any physical quantity O P ^ jψ α i wα hψ α jO α ^i ¼ P hO wα hψ α jψ α i

ð1:76Þ

α

The complete condition of orthogonal basis |φii is X

jφi ihφi j ¼ ^I

ð1:77Þ

i

Here ^I is the unit matrix. Note that the matrix element is a value that can be exchanged with other physical quantities, so using Eqs. (1.77) and (1.73) one can rewrite Eq. (1.76) as X ^i ¼ hO

αik

X

^ jφi ihφi jψ α i wα hψ α jφk ihφk jO X ¼ wα hψ α jφi ihφi jψ α i αi

X X ik

α

 ^ j φi i hφi jψ α iwα hψ α jφk i hφk jO

X X i

α

hφi jψ α iwα hψ α jφi i



ρik Oki     ^ ^ρ tr ρ^O tr O^ X ¼ ¼ ¼ trfρ^g trfρ^g ρii ik

i

ð1:78Þ

18

1

Basic Knowledge of Polarization Theory of Nuclear Reactions

The polarization density matrix ρ^ of the spin S particles is a square (2S þ 1)  (2S þ 1) matrix defined by the spin wave functions χ(σ) in the spin space in the following way ρσσ0 ¼ hχ ðσ Þχ  ðσ 0 Þiξ

ð1:79Þ

ρ^ ¼ hχ^χ^ þ iξ

ð1:80Þ

or in matrix form by

where h iξ denotes the statistical average. In particular, for pure states, one gets ρ^ ¼ χ^χ^ þ ¼ jχ ihχ j

ð1:81Þ

From the above definition, it can be seen that the density matrix is Hermitian, that is, ρ^þ ¼ ρ^,

ρ^σσ 0 ¼ ρ^σ 0 σ

ð1:82Þ

and normalized, that is, trfρ^g ¼ 1,

X σ

ρσσ ¼ 1

ð1:83Þ

For pure states, from Eq. (1.81) and Eq. (1.14), one can prove ρ^2 ¼ χ^χ^þ χ^χ^þ ¼ χ^χ^ þ ¼ ρ^

ð1:84Þ

We use j χ ii and j χ fi to represent the initial and final states of the spin space, respectively. The reaction amplitude is defined by the following formula: ^ j χii j χf i ¼ F

ð1:85Þ

Using Eq. (1.81), the density matrix of the initial state and the final state can be expressed as ρ^in ¼ j χ i ih χ i j

ð1:86Þ

^þ ^ j χ i ih χ i jF ^þ ¼ F ^ ρ^in F ρ^out ¼ j χ f ih χ f j ¼ F

ð1:87Þ

^ in the final state can be obtained by The average value of any physical quantity O using the complete condition of the basis spin functions j χ Smi

1.5

Density Matrices

19

X ^i ¼ hO

^ j χ f i h χ i jF ^F ^ j χii ^ þO h χ f jO ¼ ¼ ^ þF ^ j χii hχ f jχ f i h χ i jF

mm0

    ^ j χ Sm0 i χ Sm0 ^ ^ þ j χ Sm ih χ Sm jO F  χi h χ i jF X ^ þ j χ Sm ih χ Sm jF ^ j χii h χ i jF m

ð1:88Þ Since the matrix element h. . .i is a numerical value, the matrix element can exchange positions with other physical quantities, so the above formula can be rewritten as X ^i ¼ hO

m m0

X

^ j χ Sm0 i ^ χ i ih χ i jF ^ þ jχ Sm ih χ Sm jO h χ Sm0 jFj X ^ j χ i ih χ i j F ^ þ j χ Sm i h χ Sm jF m

^ j χ Sm0 i ^ j χ i ih χ i j F ^ þO h χ Sm0 jF

0

m ¼ X

^þ

^ j χ i ih χ i jF j χ Sm i h χ Sm jF



¼







ð1:89Þ

^ ρ out ^ tr O^ tr ρ^out O ¼ trfρ^out g trfρ^out g

m

Equation (1.89) obtained in the spin space is the same as Eq. (1.78). If the polarization problem is studied by using the first equation of Eq. (1.88), it is called the expectation value method. If the polarization problem is studied by using the last right end of Eq. (1.89), it is called the tracing method. These two methods are equivalent. If the density matrix is normalized, an expectation value of any polarization operator T^ in a state described by the density matrix ρ^ may be evaluated by     hT^ i ¼ tr T^ ρ^ ¼ tr ρ^T^

ð1:90Þ

The normalized density matrix ρ^ of spin S particles can be expanded by a set of 1 ffi T^ ðSÞ that have been formed complete linear independent matrices pffiffiffiffiffiffiffiffiffiffiffiffiffi LM 2S þ 1 before, and the corresponding expansion coefficient can be represented by 1 ffi t ðSÞ: Given this, we have the following formula: pffiffiffiffiffiffiffiffiffiffiffiffiffi LM 2S þ 1 ρ^ ¼

2S 1 X t L ðSÞ  T^ L ðSÞ 2S þ 1 L¼0

ð1:91Þ

where T^ L is the polarization operator discussed earlier and tL is called the statistical tensor, which is a spatial coordinate function. This book does not use the polarization operator defined in Ref. [31], but uses the polarization operator normalized by Eq. (1.49). It can be seen from the above formula that there are some advantages by using our approach. When the density matrix given by Eq. (1.91) is used to find 1 the differential cross section, the factor , which should be multiplied when the 2S þ 1

20

1 Basic Knowledge of Polarization Theory of Nuclear Reactions

average of the initial states is carrying out, will appear automatically at the front of the expression. The point multiplication in Eq. (1.91) represents the scalar product of two irreducible tensors defined by Eq. (1.41), so Eq. (1.91) can be rewritten as 2S X L 1 X ð
1ÞM t L
M ðSÞT^ LM ðSÞ 2S þ 1 L¼0 M¼
L 2S X L 1 X t LM ðSÞð
1ÞM T^ L
M ðSÞ ¼ 2S þ 1 L¼0 M¼
L

ρ^ ¼

ð1:92Þ

We can obtain the following relation according to Eqs. (1.90), (1.92), and (1.68): D

E n o T^ LM ðSÞ ¼ tr ρ^ T^ LM ðSÞ ¼ t LM ðSÞ

ð1:93Þ

The above formula shows that tLM(S) is the expectation value of the polarization operator T^ LM ðSÞ in the state described by the normalized density matrix ρ^. In the spherical basis representation, using Eq. (1.53) we can get  ð
1ÞM T^ L




M ðSÞ σσ 0

pffiffiffiffiffiffiffiffiffiffiffiffiffiffi Sσ ¼ ð
1ÞM 2L þ 1C Sσ 0

L
M

¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffi Sσ0 2L þ 1CSσ

LM

ð1:94Þ

Substituting Eq. (1.94) into Eq. (1.92) we have ρσσ0 ¼

ffi X pffiffiffiffiffiffiffiffiffiffiffiffiffi 2L þ 1 LM

2S þ 1

CSσ Sσ

0

LM t LM ðSÞ

ð1:95Þ

Using the formula of the C-G coefficient 0 CSσ Sσ LM

S
σ

¼ ð
1Þ

pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2S þ 1 L
M pffiffiffiffiffiffiffiffiffiffiffiffiffiffi CSσ S 2L þ 1


σ 0

ð1:96Þ

and Eq. (1.95) we may obtain X σσ 0

CSσ Sσ

0

ffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi XX pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2S þ 1 2L0 þ 1 2S þ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffi p ffi ffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ρ ¼ 0 LM σσ 2L þ 1 σσ 0 L0 M 0 2S þ 1 2L0 þ 1 L
M L0
M 0 CSσ S
σ 0 C Sσ S
σ 0 t L0 M 0 ðSÞ

1 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi t LM ðSÞ 2L þ 1

ð1:97Þ

which yields t LM ðSÞ ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiX Sσ 0 2L þ 1 C Sσ σσ 0

LM

ρσσ 0

ð1:98Þ

References

21

By use of Eq. (1.83) the following relation is easily obtained from Eq. (1.98): t 00 ðSÞ ¼ 1

ð1:99Þ

It can easily be proven that the density matrix ρ^ of spin space defined by Eq. (1.91) or Eq. (1.92) is normalized using Eqs. (1.67), (1.99), and (1.54). According to Eq. (1.82) and Eq. (1.98), we can prove tLM ðSÞ ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiX 2L þ 1 ð
1ÞM C Sσ Sσ 0 σσ 0

L
M

ρσ 0 σ ¼ ð
1ÞM t L
M ðSÞ

ð1:100Þ

References 1. Wolfenstein L, Ashkin J. Invariance conditions on the scattering amplitudes for spin 1/2 particles. Phys Rev. 1952;85:947. 2. Wolfenstein L. Possible triple-scattering experimrnts. Phys Rev. 1954;96:1654. 3. Wolfenstein L. Polarization of fast nucleons. Ann Rev Nucl Sci. 1956;6:43. 4. Hoshizaki N. Appendix: Formalism of nucleon-nucleon scattering. Prog Theor Phys Suppl. 1968;42:107. 5. Binstock J, Bryan R. Nucleon-nucleon scattering near 50 MeV. II. Sensitivity of various n-p observables to the phase parameters. Phys Rev. 1974;D9:2528. 6. Ohlsen GG. Polarization transfer and spin correlation experiments in nuclear physics. Rep Prog Phys. 1972;35:717. 7. Ohlsen G G, Gammel J L, and Keaton P W. Description 4He(d,d)4He polarization-transfer experiments. Phys Rev., 1972, C5:1205 8. Salzman GC, Mitchell CK, Ohlsen GG. Techniques for polarization transfer coefficient determination. Nucl Inst Meth. 1973;109:61. 9. Robson BA. The theory of polarization phenomena. Oxford: Clarendon Press; 1974. 10. Sperisen F, Gruebler W, Konig V. A general formalism for polarization transfer measurements. Nucl Inst Meth. 1983;204:491. 11. Seiler F, Darden SE, Mcintyre LC, Weitkamp WG. Tensor polarization of deuterons scattered from He4 between 4 and 7.5 MeV. Nucl Phys. 1964;53:65. 12. Schwandt P, Haeberli W. Elastic scattering of polarized deuterons from 27Al, Si and 60Ni between 7 and 11 MeV. Nucl Phys. 1968;A110:585. 13. Cords H, Din GU, Ivanovich M, Robson BA. Tensor polarization of deuterons from 12C-d elastic scattering. Nucl Phys. 1968;A113:608. 14. Schwandt P, Haeberli W. Optical-model analysis of d-Ca polarization and cross-section measurements from 5 to 34 MeV. Nucl Phys. 1969;A123:401. 15. Gruebler W, Konig V, Schmelzbach PA, Mamier P. Elastic scatteriong of vector polarized deutrons on 4He. Nucl Phys. 1969;A134:686. 16. Djaloeis A, Nurzynski J. Tensor polarization of deuterons from the Mg(d,d)Mg elastic scattering at 7.0 MeV. Nucl Phys. 1971;A163:113. 17. Djaloeis A, Nurzynski J. Tensor polarization and differential cross sections for the elastic scattering of deuterons by Si at low energies. Nucl Phys. 1972;A181:280. 18. Irshad M, Robson BA. Elastic scattering of 15 MeV deuterons. Nucl Phys. 1974;A218:504. 19. Goddard RP, Haeberli W. The optical model for elastic scattering of 10 to 15 MeV polarized deuterons from medium-weight nuclei. Nucl Phys. 1979;A316:116.

22

1

Basic Knowledge of Polarization Theory of Nuclear Reactions

20. Matsuoka N, Sakai H, Saito T, et al. Optical model and folding model potential for elastic scattering of 56 MeV polarized deuterons. Nucl Phys, 1986, A455:413 21. Takei M, Aoki Y, Tagishi Y, Yagi K. Tensor interaction in elastic scattering of polarized deutrons from medium-weight nuclei near Ed¼22 MeV. Nucl Phys. 1987;A472:41. 22. Iseri Y, Kameyama H, Kamimura M, Yahiro M, Tanifuji M. Virtual breakup effects in elastic scattering of polarized deutrons. Nucl Phys. 1988;A490:383. 23. Tamura T. Analyses of the scattering of nuclear particles by collective nuclei in terms of the coupled-channel calculation. Rev Mod Phys. 1965;37:679. 24. Li R, Sun WL, Soukhovitskii ES, et al. Dispersive coupled-channels optical-model potential with soft-rotator couplings for Cr, Fe, and Ni isotopes. Phys Rev. 2013;C87:054611. 25. Sun WL, Li R, Soukhovitskii ES, et al. A fully Lane-consistent dispersive optical model potential for even Fe isotopes based on a soft-rotator model. Nucl Data Sheets. 2014;118:191. 26. Shen QB. Nuclear reaction theory of low and medium energy (upper volume). Beijing: Science Press; 2005. (in Chinese) 27. Shen QB. Nuclear reaction theory of low and medium energy (middle volume). Beijing: Science Press; 2012. (in Chinese) 28. An HX, Cai CH. Global deuteron optical model potential for the energy range up to 183 MeV. Phys Rev. 2006;C73:054605. 29. Han YL, Shi YY, Shen QB. Deuteron global optical model potential for energies up to 200 MeV. Phys Rev. 2006;C74:044615. 30. Daehnick WW, Childs JD, Vrcelj Z. Global optical model potential for elastic deuteron scattering from 12 to 90 MeV. Phys Rev. 1980;C21:2253. 31. Varshalovich DA, Moskalev AN, Khersonskii VK. Quantum theory of angular momentum. Singapore/New Jersey/Hong Kong: World Scientific; 1988. 32. Zhang JS, Liu KF, Shuy GW. Neutron suppression in polarized dd fusion reaction. Phys Rev. 1999;C60:054614. 33. Rose ME. Elementary theory of angular momentum. New York: Wiley; 1957. 34. Lakin W. Spin polarization of the deuteron. Phys Rev. 1955;98:139.

Chapter 2

Polarization Theory of Nuclear Reactions 1 for Spin Particles 2

Abstract This chapter will introduce the polarization theory of nuclear reactions for 1 1 spin particles, such as the polarization of spin particle beams, and the polari2 2 1 zation theory of elastic scattering of unpolarized and polarized spin particles 2 ! ! 1 1 with spinless target. The polarization theory of þ A → þ B reactions which 2 2 can be used for various kinds of two-body direct reactions will be introduced. 1 Finally, the polarization problem of two spin polarized particles will be studied 2 theoretically. Keywords Elastic scattering · Nuclear reactions · Pauli matrices · Polarization 1 phenomena · Residual nucleus · Spin particle · Spinless targets 2

2.1

Pauli Matrices

According to Eq. (1.17), the basis spin functions of a spin

1 particle may be 2

written as
 1 , χ 12 12 = 0


 0 χ 12 - 12 = 1

ð2:1Þ

Using Eq. (2.1) and Table 2.1, the spherical basis components of the spin operator ^ S 1 of a spin particle is obtained from Eq. (1.20) as 2


 1 0 1 1 1 1 0 0 0 , S^0 = ð2:2Þ S^1 = - pffiffiffi , S^ - 1 = pffiffiffi 2 0 -1 2 0 0 2 1 0

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 Q.-B. Shen, Polarization Theory of Nuclear Reactions, https://doi.org/10.1007/978-3-031-11878-4_2

23

24

2

Polarization Theory of Nuclear Reactions for Spin

Table 2.1 Clebsch-Gordan coefficient Cjl mm - μ μ=1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl þ mÞðl þ m þ 1Þ 2ðl þ 1Þð2l þ 1Þ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl þ mÞðl - m þ 1Þ 2lðl þ 1Þ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - mÞðl - m þ 1Þ 2lð2l þ 1Þ

j=l+1 j=l j=l-1

1μ

[1]

μ=0 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - m þ 1Þðl þ m þ 1Þ ðl þ 1Þð2l þ 1Þ m pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lðl þ 1Þ -

1 Particles 2

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - mÞðl þ mÞ lð2l þ 1Þ

μ = -1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - mÞðl - m þ 1Þ 2ðl þ 1Þð2l þ 1Þ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - mÞðl þ m þ 1Þ 2lðl þ 1Þ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl þ mÞðl þ m þ 1Þ 2lð2l þ 1Þ

Using Eq. (1.22) and the formulas above, the Cartesian basis components of the 1 spin operator S^ of a spin particle are 2


 1 0 1 1 0 -i 1 1 0 ^ ^ ^ Sx = , Sy = , Sz = ð2:3Þ 0 2 1 0 2 i 2 0 -1 For spin

1 particles, let 2 1 S^ = σ^ 2

ð2:4Þ

where σ^ is called the Pauli matrix. The spherical and Cartesian basis components of the Pauli matrix σ^ can be obtained by Eqs. (2.2) and (2.3), respectively

 pffiffiffi  1 0 0 0 - 2 , σ^ 0 = , σ^ - 1 = pffiffiffi 2 0 0 -1 0


 0 1 0 -i 1 0 σ^ x = , σ^ y = , σ^ z = 1 0 i 0 0 -1


σ^ 1 =

0 0

ð2:5Þ ð2:6Þ

The Pauli matrix σ^ has the following properties: ^i , σ^ þ i =σ μ σ^ þ ^ - μ, μ = ð - 1Þ σ

σ^ 2i = ^I ,

i = x, y, z μ =  1, 0 i = x, y, z

σ^ y σ^ z = - σ^ z σ^ y = i^ σ x,

ð2:8Þ ð2:9Þ

σ^ 2 = σ^ 2x þ σ^ 2y þ σ^ 2z = 3^I σ^ x σ^ y = - σ^ y σ^ x = i^ σ z,

ð2:7Þ

σ^ z σ^ x = - σ^ x σ^ z = i^ σy

ð2:10Þ ð2:11Þ

where ^I is the unit 2 × 2 matrix. The Eq. (2.9) can be combined with Eq. (2.11) as

2.1

Pauli Matrices

25

σ^ i σ^ k = δik ^I þ i

X

εikl σ^ l

ð2:12Þ

l

The definition of εikl is given by Eq. (1.7). The Pauli matrix satisfies the following commutation and anti-commutation relations:   ð2:13Þ σ^ × σ^ = 2i^ σ , σ^ i ; σ^ k = 2iεikl σ^ l , fσ^ i ; σ^ k g = 2δik ^I , i = x, y, z pffiffiffi 1λ     σ^ μ ; σ^ ν = - 2 2C 1μ 1ν σ^ λ , μ, ν, λ =  1, 0 σ^ μ ; σ^ ν = 2ð - 1Þμ δμ - ν^I , ð2:14Þ From Eq. (2.8) and Eq. (2.14), one can also obtain pffiffiffi  þ  σ^ μ ; σ^ ν = 2 2ð - 1Þ1þμ C 11

ν-μ ^ν - μ, - μ 1ν σ

μ, ν =  1, 0

ð2:15Þ

*

Letting a be a constant vector, the following formula can be proved using the anti-commutation relation (2.13): * *  2 a  σ^ a  σ^ = ax þ a2y þ a2z ^I = a2 ^I

ð2:16Þ

By means of Eq. (2.12), it can also be proved that [2] * * *

σ^ a  σ^ = a þ i a × σ^ ,

*

*

* a  σ^ σ^ = a - i a × σ^

ð2:17Þ

In addition, there is a vector formula * * * * * * * *

a × b ×c =b a  c -c a b

*

ð2:18Þ

Based on the above formulas, we have

* * * * σ^ × a × a = - a2 σ^ þ a  σ^ a

ð2:19Þ

Using Eq. (2.12) and the properties of εijk defined by Eq. (1.7), we can prove the following relation: * * * *

a  σ^ σ^ a  σ^ = 2 a  σ^ a - a2 σ^

ð2:20Þ

When the spherical basis components of the Pauli matrix σ^ act on the basis spin functions, we have the following relations:

26

2

pffiffiffi 1 mþμ 2 σ^μ χ 12m = - 3C 1μ 1 χ1 m 2 2

Polarization Theory of Nuclear Reactions for Spin

mþμ

pffiffiffi 1 = 2ð - 1Þ2 - m C 11 2

μ mþμ

1 2

-m

χ 12

1 Particles 2

ð2:21Þ

mþμ

and we can obtain hmjσ^jmi =

X μ

pffiffiffiX  * 1 ð - 1Þμ m σ^μ m e - μ = 2 ð - 1Þμþ2 - m C 11 μ

2

μ mþμ

* 1 2

-m

e -μ ð2:22Þ

The following tracing relations of products of the Pauli matrix Cartesian basis components are obtained, by use of Eqs. (2.6), (2.9), (2.11), and (2.12), as follows: trfσ^ i g = 0, trfσ^ i σ^ k g = 2δik , trfσ^ i σ^ k σ^ l g = 2iεikl , i, k, l, m = x, y, z trfσ^ i σ^ k σ^ l σ^ m g = 2ðδik δlm - δil δkm þ δim δkl Þ,

ð2:23Þ

For the Pauli matrix spherical basis components, the following tracing relations hold [1]   tr σ^ μ = 0,

  tr σ^ μ σ^ ν = 2ð - 1Þμ δμ - ν , ! pffiffiffi 1 1 1 pffiffiffi   λ tr σ^ μ σ^ ν σ^ λ = - 2 6 , = 2 2ð - 1Þ1þλ C11μ -1ν μ ν λ n   tr σ^ μ σ^ ν σ^ λ σ^ ρ = 2ð - 1Þμ ð - 1Þλ δμ - ν δλ - ρ - ð - 1Þν δμ - λ δν - ρ o þ ð - 1Þ v δμ - ρ δν - λ , μ, ν, λ, ρ =  1, 0

2.2

Polarization of the Spin

ð2:24Þ

1 Particle Beams 2

It is a commonly used experimental method to obtain the polarized particle beams through the previous scattering or reaction. Moreover, when the unpolarized particle beams pass through a uniform strong magnetic field, the particles with different magnetic quantum numbers are separated. It is also an experimental method to obtain the polarized particle beams. 1 Any spin wave function X 12 of a spin particle can be expanded using the basis 2 spin functions 2 X 1

X 12 =

m = - 12

1

a χ 12m = m

a2 1 a-2

! ð2:25Þ

2.2

Polarization of the Spin

1 Particle Beams 2

27

where am is the contravariant component of the spin wave function X 12 . The Hermitian conjugate wave function corresponding to Eq. (2.25) is 2 X 1

Xþ 1 = 2

 1 1 am χ þ a 2 a - 2 1 = m

ð2:26Þ

Xþ 1 X1 = 1 2

ð2:27Þ

2

m = - 12

From the normalization condition

2

the following expression can be obtained:

1 2 1 2

2

a þ a - 2 = 1

ð2:28Þ

Therefore, the coefficient am may be interpreted as the probability amplitude when the particle spin projection on the z-axis in a given state X 12 is equal to m. If the particle spin is directed along the z-axis, then a - 2 = 0,

1

1

a2 = 1,

X 12 = χ 12 12 =


 1 0

ð2:29Þ

In this case, the particles are completely polarized along the z-axis. If the particle spin is directed along the negative z-axis, then a - 2 = 1,

1

1

a2 = 0,

X 12 = χ 12

- 12

=


 0 1

ð2:30Þ

In this case, the particles are completely polarized along the negative z-axis. 1 The polarization density matrix for pure states of the spin particles can be 2 written as þ

ρ^ = X 12 X 1 = 2

X m

! a χ 12m m

X m0

! 0 am  χ þ 1 0 2m

ð2:31Þ

There are no requirements for the normalization of X 12 and ρ^. The matrix element of ρ^ is ρik = χ þ ρχ 12 k = ai ak 1 ^ i 2

When writing the above formula in matrix form, it becomes

ð2:32Þ

28

2

Polarization Theory of Nuclear Reactions for Spin

0 2 1 1 1

12 a 2 a - 2

a B C ^ρ = X 12 X þ 1 =@

2 A 2 1 1 1

a - 2 a 2 a - 2

1 Particles 2

ð2:33Þ

First we can obtain the relation

1 2 1 2



I  trf^ρg = a2 þ a - 2 Then we define the polarization rate of the spin Ipi = hσ i i,

ð2:34Þ

1 particles as 2

i = x, y, z

ð2:35Þ

The following polarization rates can be obtained by using the expectation value method given by Eq. (1.89) as well as Eqs. (2.35), (2.6), and (2.33):  1 1 1 1 1 1 Ipx = a2 a - 2 þ a - 2 a2 = 2 Re a2 a - 2 ,  1 1  1 1 1 1 Ipy = i a2 a - 2 - a - 2 a2 = - 2Im a2 a - 2 ,

1 2 1 2



Ipz = a2 - a - 2

ð2:36Þ

where Re and Im denote taking the real part and the imaginary part, respectively. The above results show that only the beam particle wave functions with different spin magnetic quantum numbers yield the same probabilities and there is no coherence between the different beam particle waves; the beam is unpolarized. Then we obtain the following relation from Eqs. (2.33) and (2.36): 1 ^ρ = I 2

1 þ pz

px - ipy

px þ ipy

1 - pz

! ð2:37Þ

1 particle coming from the polarized ion source has a 2 * * polarization vector p along the direction of the external magnetic field; p is just the particle polarization direction. Note that the particle polarization direction and the motion direction of the particle beam are two different things. Here let us choose the particle polarization direction as the z-axis. In the xyz coordinate system with the particle polarization direction as the z-axis, the components of the particle polarization vector must satisfy px = py = 0, so it can be seen from Eq. (2.37) that in this case the particle density matrix is a diagonal matrix. We use N 12 and N - 12 to represent the relative particle numbers corresponding to the spin-up and the spin-down, respectively, and require them to satisfy the normalization condition Assume that the spin

2.2

Polarization of the Spin

1 Particle Beams 2

29

N 12 þ N - 12 = 1 In this xyz coordinate system, the spin

^ ρ0 =

qffiffiffiffiffi !
 qffiffiffiffiffi N 12 N 12 0 þ 0

ð2:38Þ

1 particle density matrix is 2

0 qffiffiffiffiffiffiffiffiffi N - 12

!


qffiffiffiffiffiffiffiffiffi 0 N - 12 =

N 12 0

0 N - 12

! ð2:39Þ

Thus, we can find out the following results according to Eqs. (2.38), (2.39), and (2.6): I = trfρ0 g = N 12 þ N - 12 = 1,

Ipz = trfσ^z ρ0 g = N 12 - N - 12

Ipi = = trfσ^i ρ0 g = 0,

i = x,y

ð2:40Þ ð2:41Þ

This is a special polarized particle state, which different spin magnetic quantum number states are incoherent, but may have different shares. The above results are given in the xyz coordinate system where the z-axis is along the polarization direction of the incident particles. Now we choose a new x0 y0 z0 coordinate system where the z0 -axis is along the move direction of the incident particle beam, and assume that the z-axis of the original coordinate system in the new x0 y0 z0 coordinate system has the polar angle θ and the azimuthal angle φ. The I = N 12 þ N - 12 representing the number of particles does not change in the two coordinate systems, but the component Ipz = N 12 - N - 12 belonging to the original coordinate system becomes three components in the new coordinate system as    Ipx0 = N 12 - N - 12 sin θ cos φ, Ipy0 = N 12 - N - 12 sin θ sin φ, Ipz0 = N 12 - N - 12 cos θ

ð2:42Þ Using Eq. (2.42) and Eq. (2.37), the polarization density matrix in x0 y0 z0 coordinate system becomes 0

 N 12 þ N - 12 þ N 12 - N - 12 cos θ 1 B  ^ ρ0 = @ 2 N 12 - N - 12 sin θ eiφ

1 N 12 - N - 12 sin θ e - iφ C   A 1 1 1 1 N 2 þ N - 2 - N 2 - N - 2 cos θ 

ð2:43Þ Let χ 01 1 and χ 01 2 2

2

- 12

represent the components of the particle spin wave function in the

new x0 y0 z0 coordinate system; the corresponding weights of the particle number with spin up and spin down remain the N 12 and N - 12 , respectively. And assume

30

2

Polarization Theory of Nuclear Reactions for Spin


 u χ1 1 = , 2 2 v 0


 t χ1 - 1 = 2 2 s 0

1 Particles 2

ð2:44Þ

Then the following relation can be obtained: ^ρ0 =

X m

þ N m χ 1m χ 0 1m 2 2 0

=

N 12 juj2 þ N - 12 jt j2

N 12 uv þ N - 12 ts

N 12 u v þ N - 12 t  s

N 12 jvj2 þ N - 12 jsj2

! ð2:45Þ

Comparing Eq. (2.45) with Eq. (2.43), one can get 1 ð1 þ cos θÞ, 2 1 uv = sin θ e - iφ , 2 1  tt = ð1 - cos θÞ, 2 1  ts = - sin θ e - iφ , 2

uu =

1 ð1 - cos θÞ, 2 1 u v = sin θ eiφ , 2 1  ss = ð1 þ cos θÞ, 2 1  t s = - sin θ eiφ 2 vv =

ð2:46Þ

By using the triangular formulas, one can formalize them into θ uu = cos 2 , 2 θ θ  uv = sin cos e - iφ , 2 2  2θ tt = sin , 2 θ θ ts = - sin cos e - iφ , 2 2

θ vv = sin 2 , 2 θ θ  u v = sin cos eiφ , 2 2  2θ ss = cos , 2 θ θ t  s = - sin cos eiφ 2 2

ð2:47Þ

So the following results can be obtained: θ - iφ2 e , 2 φ θ t = - sin e - i2 , 2

u = cos

θ iφ2 e , 2 θ φ s = cos ei2 2 v = sin

ð2:48Þ

If u and v or t and s change the sign at the same time, Eq. (2.47) is still satisfied. This is only a phase extraction problem, whereas the square of the wave function has a physical meaning of probability. It follows that Eq. (2.44) can be rewritten as

2.2

Polarization of the Spin

0

1 Particle Beams 2

1 φ θ cos e - i 2 B C 2 χ 01 1 = @ A, 2 2 θ iφ2 sin e 2

31

0 χ 01 2

- 12

B =@

1 θ - iφ2 e C 2 A θ iφ2 cos e 2

- sin

ð2:49Þ

The spin projection on the linear momentum direction of a particle is called the

1 particle helicity. The helicity basis functions χ 12λ ðθ, φÞ λ =  12 of a spin 2 describe the spin states with spin projections λ along its linear momentum direction 2 → → S n ðθ, φÞ = p =p. These functions are the common eigenfunctions of operators ^ * and ^S  n , that is, ^S2 χ 1λ ðθ, φÞ = 3 χ 1λ ðθ, φÞ, 2 4 2

ð^S  * n Þχ 12λ ðθ, φÞ = λχ 12λ ðθ, φÞ

ð2:50Þ

According to Eq. (1.31) the corresponding helicity basis functions can be obtained from the usual basis spin functions as follows: χ 12λ ðθ, φÞ = χ 12m =

X

X 1 D2- λ λ

m

1

D2mλ ðφ, θ, 0Þχ 12m

- m ð0,

ð2:51Þ

θ, φÞχ 12λ ðθ, φÞ

ð2:52Þ

The corresponding Hermitian conjugate expressions are χþ 1 ðθ, φÞ = λ 2

χþ 1 = m 2

X 1 ð - 1Þλ - m D2- m

X m

m

- λ ðφ,

θ, 0Þχ þ 1 m

ð2:53Þ

2

1

ð - 1Þλ - m D2λ m ð0, θ, φÞχ þ 1 ðθ, φÞ λ

ð2:54Þ

2

Its orthogonal normalization condition is χþ 1 0 ðθ, φÞχ 1λ ðθ, φÞ = δλ0 λ λ 2

ð2:55Þ

2

And its completeness condition is X ^ χ 12λ ðθ, φÞχ þ 1 ðθ, φÞ = I λ λ

ð2:56Þ

2

The D function has the following expression: 0

DJMM 0 ðα, β, γ Þ = e - iMα dJMM 0 ðβÞ e - iM γ

ð2:57Þ

32

2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2

1

Table 2.2 Expressions of d2MM 0 ðβÞ [1] M0 M 1 2 1 2

1 2

-

β 2 β sin 2

1 2

β 2 β cos 2

- sin

cos

1

Table 2.2 gives the expressions of d 2MM 0 ðβÞ. The specific expressions of the helicity basis functions can be obtained based on Eqs. (2.51) and (2.53) [1]: 0

1 0 1 φ θ - iφ2 θ - sin e - i2 e B C B C 2 2 χ 1212 ðθ, φÞ = @ A, χ 12 - 12 ðθ, φÞ = @ A θ iφ2 θ iφ2 sin e cos e 2 2
 φ θ φ θ cos ei 2 sin e - i 2 , χþ 11 ðθ, φÞ = 2 2 22
 θ iφ2 θ - iφ2 χþ ð θ, φ Þ = sin cos e e 1 1 2 2 2 -2 cos

ð2:58Þ

ð2:59Þ

The results given by Eq. (2.58) and Eq. (2.49) are exactly the same. Based on Eq. (1.91) we can write the corresponding normalized polarization 1 density matrix for spin particles as 2 ^ρ =

h  *  i 1 1 1 t 00 T^ 00 þ t 1  T^ 1 2 2 2

ð2:60Þ

From Eq. (1.55) we obtain  1 T^ 1 = 2^S = σ^ 2 It can be seen that T^ 1

ð2:61Þ

 1 is equal to the Pauli matrix σ^. Letting 2 *

*

p= t1

ð2:62Þ

and using Eqs. (1.54) and (1.99), from Eq. (2.60), we can obtain ^ρ =

 1 ^ * I þ p  σ^ 2

The expression of the above formula in the Cartesian coordinate system is

ð2:63Þ

2.3

Polarization Theory of Elastic Scattering of Unpolarized. . .

^ρ =

33

1 ^ I þ px σ^x þ py σ^y þ pz σ^z 2

ð2:64Þ

Note that the density matrix ρ^ of the incident particles is normalized. Based on Eq. (1.89) and the tracing formula (2.23), the following relation can be obtained: * *  * σ = tr σ ^ρ = p

ð2:65Þ

*

It is clear that p is the polarization vector of the system, and its absolute value p is called polarization degree, and there is 0 ≤ p ≤ 1. Based on Eq. (2.62) and Eq. (1.98), one has pffiffiffiX 1σ0 pμ = 3 C 12σ σσ 0

2

1μ

ρσσ0

ð2:66Þ

1 * For p = 0 unpolarized state, the following result for spin particles can be obtained 2 from Eq. (2.63) 1 ^ρunpol = ^I 2

2.3

ð2:67Þ

Polarization Theory of Elastic Scattering 1 of Unpolarized Spin Particles with Spinless Targets 2

Equation (5.5.14) in Ref. [2] gives the differential cross section of the spin

1 2

particles with the spinless target in the unpolarization case as follows:

2 dσ 0 1 X

F 0 ðθ, φÞ = dΩ 2 μμ0 μ μ

ð2:68Þ

where the scattering amplitude is pffiffiffi

i π Xpffiffiffiffiffiffiffiffiffiffiffiffi 2iσl 2l þ 1e F μ0 μ ðθ, φÞ = f C ðθÞ δμ0 μ þ 1 - Sjl Cjl0μ 1μ C jlmμ l k lj 2

1 0 2μ

Ylml ðθ, φÞ ð2:69Þ

34

2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2

and k is the particle wave number. Slj is the S matrix element obtained by using the spherical optical model. Equation (3.3.41) in Ref. [2] presents the Coulomb scattering amplitude fC(θ) as follows: f C ðθ Þ = -

2θ η e - iη ln ð sin 2Þþ2iσ0 2θ 2k sin 2

ð2:70Þ

where η=

μzZe2 ħ2 k

ð2:71Þ

In above formula, z and Z are incident particle and target charges, respectively, μ is reduced mass. Phase shift factor σ l satisfies the relation e2iσ l =

Γðl þ 1 þ iηÞ , Γðl þ 1 - iηÞ

σ0 =

Γð1 þ iηÞ 1 ln 2i Γð1 - iηÞ

ð2:72Þ

where Γ denotes the gamma function. There are the following spherical harmonic function and associated Legendre function expressions: rffiffiffiffiffiffiffiffiffiffiffiffisffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2l þ 1 ðl - ml Þ! ml Ylml ðθ, φÞ = ð - 1Þml P ð cos θÞ eiml φ 4π ðl þ ml Þ! l Pl- m ðxÞ = ð - 1Þm

ðl - mÞ! m P ð xÞ ðl þ mÞ! l

ð2:73Þ ð2:74Þ

yielding rffiffiffiffiffiffiffiffiffiffiffiffirffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2l þ 1 1 Yl1 = P1 eiφ , 4π l ð l þ 1Þ l

rffiffiffiffiffiffiffiffiffiffiffiffirffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2l þ 1 1 Yl - 1 = P1 e - iφ 4π lðl þ 1Þ l

ð2:75Þ

The following C–G coefficients can be obtained from Table 2.3: rffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffi lþ1 l l - 12 12 , Cl0 11 = 11 = 2l þ 1 2l þ 1 22 22 rffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffi lþ1 l lþ12 - 12 l - 12 - 12 Cl0 1 - 1 = , C l0 1 - 1 = 2l þ 1 2l þ 1 2 2 2 2 rffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffi l lþ1 lþ12 12 l - 12 12 C l1 1 - 1 = , Cl1 1 - 1 = 2l þ 1 2l þ1 2 2 2 2 rffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffi l lþ1 lþ1 - 1 l-1 -1 , Cl -2 1 121 = C l 2- 1 211 = 2l þ 1 2l þ1 22 22 lþ1 C l0 2

1 2

ð2:76Þ

ð2:77Þ

2.3

Polarization Theory of Elastic Scattering of Unpolarized. . .

35

Table 2.3 Expressions of Clebsch-Gordan coefficient Cjl

j=l þ

1 2

j=l-

1 2

1 μ= 2 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u ul þ m þ 1 t 2 2l þ 1 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u ul - m þ 1 t 2 2l þ 1

m m-μ

1 2

μ

1 μ= 2 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u ul - m þ 1 t 2 2l þ 1 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u ul þ m þ 1 t 2 2l þ 1

The following results can be obtained substituting Eqs. (2.75) – (2.77) into Eq. (2.69): F 12

1 2

ðθ, φÞ = F - 12

F 12

1 2

φÞ = A ð θ Þ

ð2:78Þ

φÞ = - iBðθÞ e - iφ

- 12 ðθ,

F - 12

- 12 ðθ,

ð2:79Þ

ðθ, φÞ = iBðθÞ eiφ

ð2:80Þ

Writing the above three formulas in matrix form, one obtains ^ ðθ, φÞ = F

AðθÞ

- iBðθÞ e - iφ

iBðθÞ eiφ

AðθÞ

! þ

^ ðθ, φÞ = , F

A  ðθ Þ

- iB ðθÞ e - iφ

iB ðθÞ eiφ

A  ðθ Þ

!

ð2:81Þ where A ðθ Þ = f C ðθ Þ þ

h   i i X lþ1 l-1 ðl þ 1Þ 1 - Sl 2 þ l 1 - Sl 2 e2iσl Pl ð cos θÞ 2k l BðθÞ =

h i 1 X lþ12 l-1 Sl - Sl 2 e2iσl P1l ð cos θÞ 2k l

ð2:82Þ ð2:83Þ

Based on Eq. (2.63), the polarized incident particle density matrix can be written as ^ρin = →

 1 ^ → I þ p in  σ^ 2

ð2:84Þ

According to Eq. (2.65), p in is the polarization vector of incident particles. The physical significance of the scattering amplitude is that the final state can be obtained when letting the scattering amplitude act on the initial state. Therefore, when the

36

2 Polarization Theory of Nuclear Reactions for Spin

1 Particles 2

target spin is 0, the polarization density matrix of the elastic scattering outgoing particles can be written as ^þ ^ ρ^in F ρ^out = F

ð2:85Þ →

Assuming the incident particles are unpolarized, that is, p in = 0 , Eq. (2.84) is simplified to 1 ρ^0in = ^I 2

ð2:86Þ

1 ^ ^þ F ρ^0out = F 2

ð2:87Þ

and Eq. (2.85) is simplified to

From Eq. (2.81) we can find that the differential cross section of the outgoing particles is   dσ 0 = tr ρ^0out = jAðθÞj2 þ jBðθÞj2 dΩ

ð2:88Þ

In the center-of-mass (c.m.) coordinate system, the incident particle wave vector * * is k i; the outgoing particle wave vector is k f ; the reaction plane is formed by k i and * k f ; and the following unit vectors are introduced: *

*

*

k I = * i ,



k i *

*

*

k F = *f



k f

*

→

ki × kf

ð2:89Þ

*

I ×F n = * * =

* *



k i × k f I × F

*

*

*

I þF P =

* *

,

I þ F

*

*

ð2:90Þ *

F- I K =

* *

F - I *

ð2:91Þ

It is easy to see *

*

n  P = 0,

And we can prove

*

*

n K =0

ð2:92Þ

2.3

Polarization Theory of Elastic Scattering of Unpolarized. . .

* * * * * 1 P  K =

* *

* *

I  F - 1 þ 1 - I  F = 0

I þ F

F - I

*

37

ð2:93Þ

***

Then it can be seen that the unit vectors K ,n ,P form a Cartesian coordinate system. Since the Pauli matrix σ^ and the two-dimensional unit matrix ^I constitute a completeness independent two-dimensional matrix system, the two-dimensional ^ ðθ; φÞ given by Eq. (2.81) can be expanded as follows: scattering amplitude matrix F →

^ ðθ; φÞ = aðθ; φÞ^I þ bðθ; φÞ h  σ^ F

ð2:94Þ

We take the unit vectors of the x, y, and z axes of the Cartesian coordinate system as *** K , n ,P , respectively, so that we can write *

*

*

σ = hx K  σ^ þ hy n  σ^ þ hz P  σ^ h ^

→

ð2:95Þ

^ ðθ; φÞ must satisfy the parity In the case of spinless target, the scattering amplitude F conservation, that is, it must remain unchanged when space reflection is performed. It can be seen from Eqs. (2.89) – (2.91) that after substituting Eq. (2.95) into * * * * * * Eq. (2.94) and the space reflections K → - K , n → n , P → - P are performed, hx = hz = 0 should hold and hy = h = 1 can be taken since the parity conservation. → * So h = n is the unit vector in the polarization direction of the outgoing particles. Then Eq. (2.94) can be rewritten as ^ ðθ; φÞ = aðθ; φÞ^I þ bðθ; φÞ * F n  σ^

ð2:96Þ

Now let us reselect a new coordinate system. The direction of the incident particle * * (i.e., the k i direction) is the z-axis, the xy plane is vertical to k i, and the projection of * → k f on the xy plane is p f , which is located on the intersection of the reaction plane → and the xy plane. The unit vector n given by Eq. (2.90) is perpendicular to the → reaction plane. Since n is perpendicular to the z-axis and passes through the coordinate origin, it is also in the xy plane. Figure 2.1 shows the position relationship → → of p f and n in the xy plane. Fig. 2.1 Vector relationship in the xy plane

38

2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2 *

In this figure, φ is the azimuthal angle of the emitted particle wave vector k f . According to Fig. 2.1 we can write *

n  σ^ = - σ^ x sin φ þ σ^ y cos φ

ð2:97Þ

Using Eq. (2.6) the above formula can be written as a matrix *

n  σ^ =

0

- ie - iφ

ieiφ

0

! ð2:98Þ

Substituting the first equation of Eq. (2.81) and Eq. (2.98) into the Eq. (2.96), one can obtain aðθ, φÞ = AðθÞ,

bðθ, φÞ = BðθÞ

ð2:99Þ

Then Eq. (2.96) can be rewritten as ^ ðθ; φÞ = AðθÞ^I þ BðθÞ * F n  σ^

ð2:100Þ

*

When choosing the y-axis in the direction along n , Eqs. (2.100) and (2.81) become ^ = A^I þ B^ F σy ! A - iB ^þ = , F iB A

A

^= F

iB

- iB

ð2:101Þ

!

ð2:102Þ

A

^ is discussed below. By use of ^I and σ^ i ði = x; y; zÞ, F ^ Another approach to find F can be expanded as ^ = a^I þ u^ F σ x þ b^ σ y þ v^ σz

ð2:103Þ

By means of Eq. (2.102) and Eq. (2.6), the above formula becomes


A

- iB

iB

A




=a

1

0

0

1




þu

0

1

1

0




þb

0

-i

i

0




þv

1 0

0



-1 ð2:104Þ

Based on the four matrix elements of Eq. (2.104), one may obtain a þ v = A,

u - ib = - iB, u þ ib = iB,

a-v=A

ð2:105Þ

Thus, the following results can be obtained from the above formulas: v = 0,

u = 0,

a = A, b = B

ð2:106Þ

2.4

Polarization Theory of Elastic Scattering of Polarized. . .

39

That is, Eq. (2.101) can be obtained from Eq. (2.103) immediately. The above results show that the scattering amplitude obtained from the nuclear reaction theory has automatically satisfied various physical symmetry requirements and does not need to be additionally restricted. Substituting Eq. (2.100) into Eq. (2.87), for which the incident particle is also unpolarized, one obtains ρ^0out =



1 ^ * * AI þ Bn  σ^ A^I þ B n  σ^ 2

ð2:107Þ

Using Eq. (2.16) the above formula becomes ρ^0out =

* i 1 h 2 jAj þ jBj2 ^I þ ðAB þ A BÞ n  σ^ 2

ð2:108Þ

Since ρ^0out given by the above formula is not normalized, from Eqs. (1.78) and (2.17) as well as the tracing formula (2.23), the differential cross section and polarization vectors of the emitted particles can be obtained as follows:   I 0 = tr ρ^0out = jAj2 þ jBj2   → tr σ^ ρ^0out 2ReðAB Þ → AB þ A B → 0 0 P out = hσ^ iout = n = n = trfρ^0out g jAj2 þ jBj2 jAj2 þ jBj2

ð2:109Þ ð2:110Þ

where Re(A) denotes taking the real part of A. * It can be seen from Fig. 2.1 that if the y-axis is selected along the direction n , then there is an azimuthal angle φ = 0. In this case, the polarization vector of the emitted 0 ðθÞ of the polarization particle can be said to be in the y-axis direction. The value Pout vector of the emitted particles is also called the emitted particle polarization degree.

2.4

Polarization Theory of Elastic Scattering of Polarized 1 Spin Particles with Spinless Targets 2

If the incident particles are polarized, the polarization density matrix of the incident particles given by Eq. (2.84) should be used. Substituting Eq. (2.84) and Eq. (2.100) into Eq. (2.85), and noting that σ^ is a Hermitian matrix, one can obtain ρ^out =





1 ^ * * * AI þ Bn  σ^ ^I þ p in  σ^ A^I þ B n  σ^ = ρ^0out þ ρ^1out 2 *

ð2:111Þ

The item ρ^0out corresponding with p in = 0 has been given by Eq. (2.108). Using Eq. (2.17) and Eq. (2.20), the following formula can be obtained:

40



2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2



* * A^I þ Bn  σ^ σ^ A^I þ B n  σ^  *

*

* * = jAj2 σ^ þ AB σ^ n  σ^ þ A B n  σ^ σ^ þ jBj2 - σ^ þ 2 n  σ^ n h  *

* i* = jAj2 - jBj2 σ^ þ ðAB þ A BÞ^I þ 2jBj2 n  σ^ n þ iðAB - A BÞ n × σ^ ð2:112Þ

Subsequently yielding h * i * 1n  2 jAj - jBj2 σ^ þ ðAB þ A BÞ^I þ 2jBj2 n  σ^ n 2 *

o * þ iðAB - A BÞ n × σ^  p in

ρ^ 1out =

ð2:113Þ

Substituting Eq. (2.108) and Eq. (2.113) into Eq. (2.111), one obtains *

1n  2 jAj þ jBj2 ^I þ ðAB þ A BÞ n  σ^ 2 h  * * þ jAj2 - jBj2 σ^ þ ðAB þ A BÞ ^I þ 2jBj2 n  σ^ n *

i * o þiðAB - A BÞ n × σ^  p in

ρ^out =

ð2:114Þ

Using the tracing formula (2.23), the angular distribution of the emitted particles in the case of polarized incident particles can be obtained by Eq. (2.114). * *

dσ = trfρ^out g = jAj2 þ jBj2 þ 2ReðAB Þ n  p in dΩ

ð2:115Þ

*

From this formula it can be seen that only when the projection of p in in the direction * * n is not 0, p in has contributions to the angular distribution of the emitted particles. According to Eq. (2.88) we can rewrite Eq. (2.115) as [3] h i  dσ dσ 0 * * 1 þ n  p in Ay ðθÞ = dΩ dΩ

ð2:116Þ

where Ay(θ) is called the analyzing power and y refers to selecting the y-axis in the * direction n . Comparing Eq. (2.116) with Eq. (2.115), it can be seen that in the case 0 ðθ Þ of target with spinless Ay(θ) is exactly equal to the polarization vector value Pout of the emitted particles corresponding to the unpolarized incident particles given by Eq. (2.110). Now assume that the components of the incident particle polarization * * vector p in with spin up and down in the direction n are p+ and p-, respectively; that is,

2.4

Polarization Theory of Elastic Scattering of Polarized. . .

41



* p in = pþ - p - n

ð2:117Þ

 dσ  dσ 0  = 1  p Ay ðθÞ dΩ dΩ

ð2:118Þ

*

and define

From the above formulas, one can obtain

dσ 0

dσ dσ 0 = pþ 1 - p - Ay = p - pþ p - Ay , dΩ dΩ dΩ þ

dσ 0

dσ dσ 0 1 þ pþ A y = p þ p - pþ Ay p- þ = pdΩ dΩ dΩ -

pþ

ð2:119Þ

Using Eq. (2.118) and adding the two formulas in Eq. (2.119), we obtain Ay ðθÞ =

dσ þ dΩ

-

dσ dΩ

pþ þ p -

dσ 0 = dΩ

dσ þ dΩ pþ dσdΩ-

-

dσ dΩ

þ p-

dσ þ dΩ

ð2:120Þ

dσ  represent the complete polarization differential cross sections of the dΩ spin up ( p+ = 1, p- = 0) and spin down ( p+ = 0, p- = 1) of the incident particle, respectively, Eq. (2.118) can be rewritten as Letting

 dσ  dσ 0  = 1  A y ðθ Þ dΩ dΩ

ð2:121Þ

then we can obtain A y ðθ Þ =

dσ þ dΩ dσ þ dΩ

þ

dσ dΩ dσ dΩ

ð2:122Þ

This is the source of the analyzing power name. Using the vector expression *  * * * a  b ×* c = a×b  c

*

ð2:123Þ

we can derive the following formula:  * * * ð* n × σ^Þ  p in = - σ^  n × p in

ð2:124Þ

Using Eq. (2.22) and Eq. (2.124), the following is obtained from Eq. (2.114):

42

2

Polarization Theory of Nuclear Reactions for Spin

h  i  * * * * trfσ^^ρout g = AB þ A B þ 2jBj2 n  p in n þ jAj2 - jBj2 p in  * * - iðAB - A BÞ n × p in

1 Particles 2

ð2:125Þ

From Eq. (2.115) and Eq. (2.125), we can find that the polarization vector of the emitted particles in the case of the polarized incident particles is *

trfσ^ ρ^out g trhfρ^out g * *

* * i *  * 2 ReðAB Þ þ jBj2 n  p in n þ jAj2 - jBj2 p in þ 2ImðAB Þ n × p in = * *

jAj2 þ jBj2 þ 2ReðAB Þ n  p in

P out =

ð2:126Þ →

where Im(A) denotes taking the imaginary part of A. If p in is known, the value and → direction of P out can be obtained from the above expression, in which Pout is called the polarization degree of the emitted particles in the case of polarized incident → particles. When p in = 0, Eq. (2.126) degenerates to Eq. (2.110). * The following relation may be obtained by point multiplying Eq. (2.126) by n   2 2 * *  2 Re AB ð Þ þ A þ B  p n j j j j in * *  n  P out = 2 2  * * jAj þ jBj þ 2 Re ðAB Þ n  p in

ð2:127Þ

And above relation can be rewritten as 

*   P n out - 2 Re ðAB Þ * →  n  p in = * → jAj2 þ jBj2 - 2 Re ðAB Þ n  P out jAj2 þ jBj2



*

*

ð2:128Þ

*

This formula shows that if the projection of P out in the direction n is known, the * * projection of p in on the direction n can be derived out. When the emitted particle * polarization vector P out = 0, the formula above yields the following relation: *

→

n  p in = -

2 Re ðAB Þ jAj2 þ jBj2

ð2:129Þ

For the differential cross section and the polarization vector value of the emitted 1 particles corresponding to the spin unpolarized incident particles with the spinless 2 targets given by Eqs. (2.88) and (2.110), the following symbols are introduced, respectively:

2.4

Polarization Theory of Elastic Scattering of Polarized. . .

43

I 0 = jAj2 þ jBj2

ð2:130Þ

2 Re ðAB Þ jAj2 þ jBj2

ð2:131Þ

P=

Additionally, we introduce two symbols Q=

2ImðAB Þ jAj2 þ jBj2

ð2:132Þ

W=

jAj2 - jBj2 jAj2 þ jBj2

ð2:133Þ

And we can derive the following formula:  2 jAj2 - jBj2 þ ½2ImðAB Þ2 þ ½2 Re ðAB Þ2  2 = jAj4 þ jBj4 - 2jAj2 jBj2 þ 4jAj2 jBj2 = jAj2 þ jBj2

ð2:134Þ

Thus, we can prove the following relation: W 2 þ Q2 þ P2 = 1

ð2:135Þ

If P and Q are considered to be independent polarization functions, then W is not considered to be an independent polarization function. Eqs. (2.126) – (2.128) can also be rewritten as h  i  * * * * * * P þ ð1 - W Þ n  p in n þ W p in þ Q n × p in  P out = * * 1 þ P n  p in  * * P þ n  p in * *  n  P out = * * 1 þ P n  p in  * → n  P -P out * →  n  p in = * → 1 - P n  P out

→

ð2:136Þ

ð2:137Þ

ð2:138Þ *

We choose the direction of the incident particle as the z-axis; the direction of n as the y-axis; and the x-axis is chosen according to the right-handed screw rule. From Fig. 2.1 it can be seen that φ = 0 in the chosen coordinate system. This coordinate system is also a helicity coordinate system for incident particles. This section studies

44

2 Polarization Theory of Nuclear Reactions for Spin

1 Particles 2

Fig. 2.2 Polarization vector relationship diagram in scattering plane when * * p in = pz e z

the first reaction process in which incident and outgoing particles can be studied in * * * the same coordinate system. We use e x , e y , e z to represent the unit vectors of the x, y, * → * z axes, and there is n = e y. p in is the polarization vector of the incident particles and belongs to the nature of the incident particle beam before the reaction. The three → components px, py, pz of p in can be determined by Eq. (2.36). When taking * → p in = pz e z, based on Eqs. (2.115), (2.136), (2.130) – (2.133), the following relations can be obtained: I = I 0 = jAj2 þ jBj2 ,

*

*

*

*

P out = P e y þ pz W e z þ pz Q e x

ð2:139Þ

Figure 2.2 presents the polarization vector relationship diagram in scattering plane * * when P in = pz e z . It can be seen from Fig. 2.2 that tan β =

2ImðAB Þ 2ImðAB Þ Q = = 2 2 W I 0W jAj - jBj

ð2:140Þ

where β is the angle in the center-of-mass system. The above results show that when the incident particles have a polarization rate pz only in the z-axis direction, the outgoing particles have a polarization rate pzW in the z-axis direction and have a → * polarization rate pzQ in the x-axis direction. When taking p in = px e x , from Eqs. (2.115), (2.136), (2.130) – (2.133), one can obtain I = I 0 = jAj2 þ jBj2 ,

*

*

*

*

P out = P e y þ px W e x - px Q e z

ð2:141Þ

The above formulas show that when the incident particle has a polarization rate px only in the x-axis direction, the outgoing particles have a polarization rate pxW in the x-axis direction and have a polarization rate -pxQ in the z-axis direction. In view of the above results, we call Q a spin rotation function, and W (Wolfenstein) a spin continuation function [4, 5]. The above results also show that when the incident particles are polarized only in the x-axis direction and/or in the z-axis direction, the differential cross section of the outgoing particles is equal to that of the unpolarized * → incident particles. When taking p in = py e y , from Eqs. (2.115), (2.136), (2.130) – (2.133) one can obtain

2.4

Polarization Theory of Elastic Scattering of Polarized. . .



I = jAj2 þ jBj2 þ 2 Re ðAB Þpy = I 0 1 þ py P ,

45 →

P out =

P þ py * e 1 þ py P y

ð2:142Þ

*

→

It can be seen that when p in = py e y , the differential cross section is different from that of the unpolarized incident particle. And the polarization vector of the outgoing * particles only appears in the e y direction. * When the three components of P out are represented by Px, P y, Pz, from Eq. (2.126), one can obtain  Px = Py = Pz =

jAj2 - jBj2 px þ 2ImðAB Þpz

jAj2 þ jBj2 þ 2 Re ðAB Þpy  2 Re ðAB Þ þ jAj2 þ jBj2 py 2

2



2

2



jAj þ jBj þ 2 Re ðAB Þpy  jAj2 - jBj2 pz - 2ImðAB Þpx jAj þ jBj þ 2 Re ðAB Þpy

=

p x W þ pz Q , 1 þ py P

=

P þ py , 1 þ py P

=

pz W - p x Q 1 þ py P

ð2:143Þ

In the coordinate system selected above, from Eqs. (2.102), (2.23), and (2.131), one can get Pi ðθÞ 

  ^F ^þ tr σ^ i F = δiy P ^F ^ þg trfF

ð2:144Þ

where Pi(θ) represents the value of the ith component of the polarization vector obtained from the unpolarized incident particles. At the same time, one can get   ^ σ^ i F ^þ tr F = δiy P, Ai ðθÞ  ^F ^ þg trfF

Ay ðθÞ = P

ð2:145Þ

where Ai(θ) represents the ith component of the analyzing power polarization vector of the incident particles. Obviously, for spinless target there is P i ðθÞ = Ai ðθÞ = δiy P

ð2:146Þ

It means that both P i(θ) and Ai(θ) are not 0 only in the y-axis direction. We define the following symbol to represent the polarization transfer coefficient from the ith initial polarization component to the kth final polarization component

46

2 Polarization Theory of Nuclear Reactions for Spin Table 2.4 Polarization transfer coefficient K ki ðθÞ of spin

1 Particles 2

1 2

particles with spinless targets In i y

x W

Out k x y z

z Q

1 -Q

W

n o ^ σ^i F ^þ tr σ^k F  þ , K ki ðθÞ = ^F ^ tr F

i,k = x,y,z

ð2:147Þ

1 particles with spinless targets 2 derived from Eq. (2.147) by using Eqs. (2.102), (2.6), and (2.23). In this table, the empty cells are to be filled with 0. 1 The spin particle differential data including the differential cross section I and 2 → the polarization vector p can be represented by the following Stokes vector [6]:

Table 2.4 lists the specific results of K ki ðθÞ for spin

0

1

1

Bp C B xC Sc = I B C @ py A pz

ð2:148Þ

c

where c represents the selected coordinate system. Below we use ScðiÞ and Scðf Þ to represent the Stokes vectors for the incident particles and outgoing particles, respectively. In this section they are all given in the same coordinate system c. By use of the previous symbols, one can write the following formulas: 0

1 1 Bp C B xC SðciÞ = I 0 B C , @ py A pz

1 1 B Px C B C Sðcf Þ = I B y C @P A

c

0

Pz

ð2:149Þ

c

Using the simplified symbols introduced in this section, the physical quantities 0 * * dσ appeared in Eq. (2.116) are I = dΩ , I 0 = dσ dΩ , py = n  p in , P = Ay(θ), then according to Eqs. (2.142) and (2.143) the following relation can be obtained from Eq. (2.149): Sðcf Þ = Z^ cc SðciÞ where Z^ cc is a 4 × 4 matrix, called the Mueller matrix [6] , in the form of

ð2:150Þ

2.5

Polarization Theory of Triple Elastic Scatterings Between. . .

0

0 W

P 0

0

1

1 0 QC C C 0A

-Q

0

W

1 B0 B Z^ cc = B @P 0

2.5

47

ð2:151Þ

cc

Polarization Theory of Triple Elastic Scatterings 1 Between Spin Particles and Spinless Nuclei 2

This section studies the polarization theory of triple elastic scatterings between spin 1 particles and spinless nuclei. For the sake of simplicity, we constrain the incident 2 particles to be the same, but the targets can be the same or different. Considering that the calculations of the differential cross sections and the polarization physical quantities must be carried out in the center-of-mass (C) system, and the tracking and sampling of particles should be carried out in the laboratory (L ) system, the coordinate system transformation problem should be considered. Since the incident direction of the particles is generally selected as the z-axis, the transformation between C system and L system is also carried out along the z-axis direction. Therefore, only the polar angle θ changes under this transformation; the azimuthal angle φ is not affected. Let IC(θC, φ) and IL(θL, φ) denote the emitted particle angular distributions in C system and L system, respectively. By using Eq. (2.59) and Eq. (2.2.56) given in Ref. [2], the coordinate transformation relations between them can be obtained as I L ðθL , φÞ = X ðθC ÞI C ðθC , φÞ,

I C ðθC , φÞ = Y ðθL ÞI L ðθL , φÞ

ð2:152Þ

where 3=2

ð1 þ γ 2 þ 2γ cos θC Þ j1 þ γ cos θC j pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 - γ 2 sin 2 θL Y ðθL Þ = pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 - γ 2 sin 2 θL þ γ cos θL X ðθ C Þ =

ð2:153Þ ð2:154Þ

For elastic scattering γ=

m ≤1 M

ð2:155Þ

48

2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2

where m and M are masses of incident particle and target, respectively. For the polarization observables P, Q, and W introduced in Sect. 2.4, because the molecules and the denominators of their expressions require coordinate system transformation at the same time, so we have PC ðθC , φÞ = PL ðθL , φÞ, QC ðθC , φÞ = QL ðθL , φÞ, W C ðθC , φÞ = W L ðθL , φÞ ð2:156Þ The angular relations between the two coordinate systems are cos θC = cos θL cos θL =

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 - γ 2 sin 2 θL - γ sin 2 θL ,

γ þ cos θC

ð2:157Þ

ð1 þ γ 2 þ 2γ cos θC Þ1=2

For the elastic scattering, there are the following energy relations: EC =

M E mþM L

ð2:158Þ

εC =

M E mþM C

ð2:159Þ



εL = εC 1 þ γ 2 þ 2γ cos θC

ð2:160Þ

where EC and EL are the total kinetic energies of incident particle relative to the target in C system and L system, respectively, and εC and εL are the kinetic energies of the emitted particle in C system and L system, respectively. Substituting the first equation of Eq. (2.157) into Eq. (2.160), one can obtain  
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 εL = εC 1 þ γ þ 2γ cos θL 1 - γ sin θL - γ sin θL

ð2:161Þ

It is assumed that the incident particle beam of the first scattering is along the zaxis and is unpolarized. In order to make the polarization direction after the first scattering along the y-axis, we choose φ1 = 0. Thus, according to the discussion in Sect. 2.4, we can write that the Stokes vector in C1 system after the first scattering as 0

1

1

B 0 C B C ðf Þ SC 1 = I C 1 B C @ PC 1 A 0

ð2:162Þ

2.5

Polarization Theory of Triple Elastic Scatterings Between. . .

49

Fig. 2.3 Coordinates in the laboratory system for the particle first twice scatterings

Using Eq. (2.109) and Eq. (2.110), we can show I C1 = jA1 j2 þ jB1 j2

2 Re A1 B1 PC1 = jA1 j2 þ jB1 j2

ð2:163Þ ð2:164Þ

The expressions of Ai and Bi have been given by Eq. (2.82) and Eq. (2.83). By use of Eq. (2.152) the Stokes vector in L1 system after the first scattering can be obtained as 0

1

1

B 0 C B C ðf Þ S L1 = X ð θ C 1 Þ I C 1 B C @ PC1 A

ð2:165Þ

0 By use of Eq. (2.157) the angle θC1 at the right end of the above formula can be changed to angle θL1 in L1 system. Figure 2.3 shows coordinates in the laboratory system for the particle first twice scatterings. The axes y1 and y2 are parallel to each other and taken the vertical page upward direction. The direction of the second scattering incident particle is taken as z2-axis, and the angle between the z2-axis and the z1-axis is θL1 . The wave number of the first scattering incident particle beam in C1 system is pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2μ1 EC1 k1 = ħ

ð2:166Þ

where μ1 is the reduced mass μ1 =

m m þ M1

ð2:167Þ

Since the y-axes of L1 and L2 systems are parallel to each other, the Stokes vector of the incident particles in L2 system can be obtained directly from Eq. (2.165):

50

2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2

0

1 1 B 0 C B C ðiÞ S L2 = X ð θ C 1 Þ I C 1 B C @ PC1 A

ð2:168Þ

0 Transforming the above formula into C2 system yields 0

1

1

0

1

1

B 0 C B C B C ðiÞ ðiÞ B 0 C SC2 = Y ðθL1 ÞX ðθC1 Þ I C1 B C  I C2 B ðiÞ C @ PC1 A @ pC 2 A 0

ð2:169Þ

0

The above formula shows that the incident particle of the second scattering has no polarization component in x2 and z2 axis directions. The final state Stokes vector after the second scattering can be expressed as 0

1

1

B C ðf Þ ðf Þ B Px C SC 2 = I C 2 B 2 C @ P y2 A

ð2:170Þ

Pz2 The wave number of the incident particle steam of the second scattering in C2 system is pffiffiffiffiffiffiffiffiffiffiffiffiffi 2μ2 εC2 k2 = ħ

ð2:171Þ

And we can write out M2 ε m þ M 2 L1  
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi εL1 = εC1 1 þ γ 21 þ 2γ 1 cos θL1 1 - γ 2 sin 2 θL1 - γ 1 sin 2 θL1 εC 2 =

ð2:172Þ ð2:173Þ

εC 1 =

M1 E m þ M 1 C1

ð2:174Þ

E C1 =

M1 E m þ M 1 L1

ð2:175Þ

It can be seen that k2 is related to the angle θL1 .

2.5

Polarization Theory of Triple Elastic Scatterings Between. . .

51

Fig. 2.4 Projection of the wave vector of the emitted particle in C2 system on the x2y2 plane

*ðf Þ

The polar angle of the wave vector k 2 of the emitted particles of the second scattering in C2 system is θC2; its azimuthal angle is φ2; and its projection on the x2y2 ðf Þ plane is p2 (see Fig. 2.4). According to Eq. (2.81) the scattering amplitude of the second scattering can be written as ^ 2 ð θ C 2 , φ2 Þ = F ^þ F 2 ð θ C 2 , φ2 Þ =

A 2 ðθ C 2 Þ

- iB2 ðθC2 Þ e - iφ2

iB2 ðθC2 Þ eiφ2

A 2 ðθ C 2 Þ

A2 ðθC2 Þ

- iB2 ðθC2 Þ e - iφ2

iB2 ðθC2 Þ eiφ2

A2 ðθC2 Þ

! , !

ð2:176Þ

and the following expression can be obtained: ^ 2F ^þ F 2 =

jA2 j2 þ jB2 j2

i A2 B2 þ B2 A2 eiφ2

!

- i A2 B2 þ B2 A2 e - iφ2 jA2 j2 þ jB2 j2

ð2:177Þ

By using Eq. (2.177) and Eq. (2.6), the contributions of the unpolarized part of the second incident particle can be obtained as n o 1 ^þ ^2F = jA2 j2 þ jB2 j2 tr F 2 2 n o

1 ^ 2F ^þ I 0C2 P0x2 = tr σ^x F = - 2 Re A2 B2 sin φ2 = - I 0C2 PC2 sin φ2 2 2 n o

1 ^ 2F ^þ I 0C2 P0y2 = tr σ^y F = 2 Re A2 B2 cos φ2 = I 0C2 PC2 cos φ2 2 2 n o 1 ^2F ^þ I 0C2 P0z2 = tr σ^z F =0 2 2 I 0C2 =

where PC2 =



2 Re A2 B2 jA2 j2 þ jB2 j2

ð2:178Þ ð2:179Þ ð2:180Þ ð2:181Þ

ð2:182Þ

52

2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2

When considering the initial state of C2 system given by Eq. (2.169), one first finds the following expression based on Eqs. (2.176) and (2.6): ^þ ^ 2σy F F 2

= 0

A2

- iB2 e - iφ2

iB2 eiφ2

A2

!

A2 B2 eiφ2 þ B2 A2 e - iφ2 B =@  i jA2 j2 þ jB2 j2 e2iφ2

B2 eiφ2

- iA2

!

iA2 B2 e - iφ2  1 - i jA2 j2 þ jB2 j2 e - 2iφ2 C A  iφ2  - iφ2 B2 A2 e þ A2 B2 e

ð2:183Þ

It can also be found that the contributions of the polarized part of the second incident particles are n o 1 ^þ ^ 2 σ^y F = I 0C2 PC2 cos φ2 tr F 2 2 n o 1 1 ^ 2 σ^y F ^þ I 0C2 P1x2 = tr σ^x F = - I 0C2 ð1 - W C2 Þ sin ð2φ2 Þ 2 2 2 n o 1 ^ 2 σ^y F ^þ I 0C2 P1y2 = tr σ^y F = jA2 j2 þ jB2 j2 cos ð2φ2 Þ 2 2 i h 1 = I 0C2 1 - ð1 - W C2 Þ ð1 - cos ð2φ2 ÞÞ 2 n o

1 ^ 2 σ^y F ^þ I 0C2 P1z2 = tr σ^z F = - 2Im A2 B2 sin φ2 = - I 0C2 QC2 sin φ2 2 2 I 1C2 =

ð2:184Þ ð2:185Þ

ð2:186Þ

ð2:187Þ

where QC2 = W C2 =



2Im A2 B2 jA2 j2 þ jB2 j2 jA2 j2 - jB2 j2 jA2 j2 þ jB2 j2

ð2:188Þ ð2:189Þ

By using the above results, according to Eq. (2.169) one can obtain  ðf Þ ðiÞ ðiÞ I C2 = I C2 I 0C2 1 þ pC2 PC2 cos φ2

ð2:190Þ

ðiÞ ðiÞ 1 ð2:191Þ Px2 = P0x2 þ pC2 P1x2 = - PC2 sin φ2 - pC2 ð1 - W C2 Þ sin ð2φ2 Þ 2 h i 1 ðiÞ ðiÞ Py2 = P0y2 þ pC2 P1y2 = PC2 cos φ2 þ pC2 1 - ð1 - W C2 Þ ð1 - cos ð2φ2 ÞÞ 2 ð2:192Þ ðiÞ

ðiÞ

Pz2 = P0z2 þ pC2 P1z2 = - pC2 QC2 sin φ2

ð2:193Þ

In this way the final state Stokes vector after the second scattering given by Eq. (2.170) is obtained. Then transforming it to L2 system, we can get

2.5

Polarization Theory of Triple Elastic Scatterings Between. . .

53

0

1 1 B C ðf Þ ðf Þ ðf Þ B Px2 C SL2 = X ðθC2 ÞSC2 = X ðθC2 Þ I C2 B C @ P y2 A

ð2:194Þ

Pz2 Figure 2.5 shows laboratory system for the third scattering. The angle between the z3-axis and the z2-axis is θL2 ; the polar angle of the z3-axis in L2 system is θL2 ; and its azimuthal angle is φ2. From Eq. (2.194) we can see that the polarization vector in L2 system is *L 2

P

*

*

*

 Px2 e x þ Py2 e y þ Pz2 e z

ð2:195Þ

Based on Eq. (2.57) and Table 2.5, we can obtain 0 1 þ cos β e - iðαþγÞ 2 B B B sin β - iγ 1 ^ pffiffiffi e D ðα, β, γ Þ = B B 2 B @ 1 - cos β eiðα - γÞ 2

sin β - pffiffiffi e - iα 2 cos β sin β iα pffiffiffi e 2

1 1 - cos β iðγ - αÞ e 2 C C sin β iγ C C ð2:196Þ - pffiffiffi e C 2 C 1 þ cos β iðγþαÞ A e 2

According to Eq. (1.27) we can write A01M 0 =

X A1M D1MM 0 ðα, β, γ Þ

ð2:197Þ

M

Fig. 2.5 Laboratory system for the third scattering

Table 2.5 Expressions of d1MM 0 ðβÞ[1] M 1 0 -1

1 1 þ cos β 2 sin β pffiffiffi 2 1 - cos β 2

M0 0 sin β - pffiffiffi 2 cosβ sin β pffiffiffi 2

-1 1 - cos β 2 sin β - pffiffiffi 2 1 þ cos β 2

54

2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2

The following relations can be obtained by using Eqs. (2.197), (2.196), and (1.23):

sin β 1 þ cos β - iðαþγ Þ 1 pffiffiffi Ax þ iAy þ pffiffiffi e - iγ Az e 2 2 2

1 - cos β iðα - γÞ 1 pffiffiffi Ax - iAy þ e 2 2



sin β sin β 1 1 A010 = pffiffiffi e - iα pffiffiffi Ax þ iAy þ cos βAz þ pffiffiffi eiα pffiffiffi Ax - iAy 2 2 2 2

1 - cos β iðγ - αÞ 1 sin β pffiffiffi Ax þ iAy - pffiffiffi eiγ Az A01 - 1 = e 2 2 2

1 þ cos β iðγþαÞ 1 pffiffiffi Ax - iAy þ e 2 2 A011 = -

ð2:198Þ

Then by use of Eq. (1.22) the following relations can be obtained from Eq. (2.198):  



1 1- cos β iðγ - αÞ 1 þ cos β - iðαþγ Þ A0x = pffiffiffi A01 - 1 - A011 = þ e Ax þ iAy e 4 4 2    

1 þ cosβ iðγþαÞ 1- cos β iðα - γÞ sin β iγ sin β - iγ e e Ax - iAy Az þ e e þ 4 2 4 2   1 þ cos β 1- cos β = cos ðα þ γ Þ cos ðα- γ Þ Ax 2 2   1 þ cosβ 1 - cos β þ sin ðα þ γ Þsin ðα - γ Þ Ay - sin β cos γ Az 2 2 1 = ½ð1 þ cos βÞð cos αcos γ - sin αsin γ Þ - ð1- cos βÞð cos α cos γ þ sin αsin γ ÞAx 2 1 þ ½ð1 þ cos βÞð sin αcosγ þ cos αsin γ Þ - ð1- cos βÞð sin αcos γ - cosα sin γ ÞAy 2 - sin β cos γ Az = ð cos β cos αcosγ - sinα sinγ ÞAx þ ð cos β sin αcos γ þ cos αsin γ ÞAy - sin β cos γ Az

ð2:199Þ In the same way, the following relations can also be obtained: A0y = - ð cos β cos α sin γ þ sin α cos γ ÞAx A0z

- ð cos β sin α sin γ - cos α cos γ ÞAy þ sin β sin γAz = sin β cos αAx þ sin β sin αAy þ cos βAz

So that we can subsequently yield

ð2:200Þ

2.5

Polarization Theory of Triple Elastic Scatterings Between. . .

A0i =

X

Ak aki ,

55

i = x, y, z

ð2:201Þ

k = x,y,z

The rotation matrix composed by matrix elements aki is [1] 0

cos β cos α cos γ - sin α sin γ - cos β cos α sin γ - sin α cos γ sin β cos α

1

B C C ^aðα, β, γ Þ = B @ cos β sin α cos γ þ cos α sin γ - cos β sin α sin γ þ cos α cos γ sin β sin α A - sin β cos γ

sin β sin γ

cos β

ð2:202Þ If we take α = φ, β = θ, γ = 0, Eq. (2.202) degenerates into 0

cos θ cos φ

^ ðθ, φÞ = B B @ cos θ sin φ - sin θ

- sin φ cos φ 0

sin θ cos φ

1

C sin θ sin φ A cos θ

ð2:203Þ

h J i-1 ^ J ðα, β, γ Þ ^ ðα, β, γ Þ The inverse of a rotation matrix D is D = h J iþ ^ ðθ, φÞ given by Eq. (2.203) is a real number ^ ðα, β, γ Þ : Since the matrix B D matrix, its inverse is its transpose matrix, that is, 0

cos θ cos φ B ^bðθ, φÞ = B eðθ, φÞ = @ - sin φ sin θ cos φ

1 - sin θ C 0 A

cos θ sin φ cos φ sinθ sin φ

ð2:204Þ

cos θ

^ and ^b are all unit matrix. We can see that at θ = φ = 0, B When the object and the coordinate system rotate at the same time, the relative relationship between the object and the coordinate system will not change. Let ^ ðα, β, γ Þ be the rotation Euler angles operator; then we get D   ^ ðα, β, γ Þ  D =1 coord: system n o  J   J DMM 0 ðα, β, γ Þ object = D - 1 ðα, β, γ Þ MM 0 

^ ðα, β, γ Þ D



object

ð2:205Þ

coord: system

In the density matrix given by Eq. (1.91), t L ðSÞ  T^ L ðSÞ is a scalar with rotational invariance. We think of the particle vector spin operator or tensor polarization operator T^ L ðSÞ as a coordinate system, and the particle vector or tensor polarization rate tL(S) as a physical quantity of the object. Equation (2.201) gives the rotation relation of the coordinate system, while the particle polarization vector represents the property of the object, so its corresponding rotation relation is

56

2

p0i =

Polarization Theory of Nuclear Reactions for Spin

X

pk bki =

k = x,y,z

X

i = x, y, z

Bik pk ,

1 Particles 2

ð2:206Þ

k = x,y,z

For example, assuming that the components of the polarization vector before rotation are px = py = 0, pz = pZ and the particle polarization vector is rotated by angles (θ, φ), then we can obtain p0x = sin θ cos φ pZ , p0y = sin θ sin φ pZ , p0z = cos θ pZ from Eqs. (2.206) and (2.203). These formulas are the familiar results. It is also understood that in the Cartesian coordinate system XYZ taking the polarization vector as Z-axis, only the component pZ is not equal to 0; then the coordinate system is rotated reversely by angles (θ, φ), and the same results can also be obtained. → L2

P

given by Eq. (2.195) is the polarization vector in L2 system. L3 system is → L3

*L2

rotated by angles ðθL2 , φ2 Þ relative to L2 system; p is P described in L3 system. This situation is equivalent to only rotating the coordinate system and the object is not rotated, so there is → L3

p

*L 2

= ^bðθL2 , φ2 ÞP

ð2:207Þ

Using Eq. (2.204) we can obtain px3 = cos θL2 cos φ2 Px2 þ cos θL2 sin φ2 Py2 - sin θL2 Pz2 py3 = - sin φ2 Px2 þ cos φ2 Py2

ð2:208Þ

pz3 = sin θL2 cos φ2 Px2 þ sin θL2 sin φ2 Py2 þ cos θL2 Pz2 After the transformation of the polarization vector given by Eq. (2.208), from the final state of L2 system given by Eq. (2.194) we can obtain the initial state in L3 system as 0

1

1

B C ðiÞ ðf Þ B px C S L3 = X ð θ C 2 Þ I C 2 B 3 C @ py3 A

ð2:209Þ

pz3 ðiÞ

Then transforming SL3 to C3 system we have 1 0 1 1 1 B pðiÞ C B C x3 C ðiÞ ð f Þ B px 3 C ðiÞ B C SC3 = Y ðθL2 ÞX ðθC2 ÞI C2 B C  I C3 B ðiÞ C B @ py 3 A @ py 3 A 0

pz 3

ð2:210Þ

pzði3Þ

The final state Stokes vector after the third scattering can be expressed as

2.5

Polarization Theory of Triple Elastic Scatterings Between. . .

1 1 B C ðf Þ ðf Þ B Px3 C SC 3 = I C 3 B C @ P y3 A

57

0

ð2:211Þ

Pz3 The wave number of the incident particle beam of the third scattering in C3 system is pffiffiffiffiffiffiffiffiffiffiffiffiffi 2μ3 εC3 k3 = ħ

ð2:212Þ

We can also write out M3 ε m þ M 3 L2  
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 εL2 = εC2 1 þ γ 2 þ 2γ 2 cos θL2 1 - γ 2 sin θL2 - γ 2 sin θL2 εC 3 =

γ2 =

m M2

ð2:213Þ ð2:214Þ ð2:215Þ

where εC2 has been given by Eq. (2.172). It can be seen that k3 is the function of θL1 and θL2 . *ðf Þ

The polar angle of the emitted particle wave vector k 3 of the third scattering in ðf Þ C3 system is θC3 ; its azimuthal angle is φ3; and its projection on the x3y3 plane is p3 (see Fig. 2.6). The contributions of the unpolarized part of the third scattering are, according to Eqs. (2.178)–(2.182), as follows: n o 1 ^þ ^3F = jA3 j2 þ jB3 j2 tr F 3 2 n o 1 ^ 3F ^þ I 0C3 P0x3 = tr σ^x F = - I 0C3 PC3 sin φ3 3 2 n o 1 ^3F ^þ I 0C3 P0y3 = tr σ^y F = I 0C3 PC3 cos φ3 3 2 I 0C3 =

Fig. 2.6 Projection of the emitted particle wave vector on the x3y3 plane in C3 system

ð2:216Þ ð2:217Þ ð2:218Þ

58

2

Polarization Theory of Nuclear Reactions for Spin

I 0C3 P0z3 =

n o 1 ^3F ^þ =0 tr σ^z F 3 2

1 Particles 2

ð2:219Þ

where

2 Re A3 B3

PC3 =

ð2:220Þ

jA3 j2 þ jB3 j2

Comparing Eq. (2.210) with Eq. (2.169), we should note that the incident particle beams of the second scattering only in the y direction are polarized, i.e., ðiÞ pyði2Þ = pC2 , pðxi2Þ = pðzi2Þ = 0, but the incident particle beams of the third scattering in x, y, and z directions are all polarized, i.e., pðxi3Þ ,pðyi3Þ ,pzði3Þ are all not equal to 0. We can obtain the following results according to Eqs. (2.184)–(2.189): n o 1 ^ 3 σ^y F ^þ tr F = I 0C3 PC3 cos φ3 3 2 n o 1 1 ^ 3 σ^y F ^þ ^x F I 0C3 P1y = - I 0C3 ð1 - W C3 Þ sin ð2φ3 Þ 3 x3 = 2 tr σ 2 h i n o 1 1 ^ 3 σ^y F ^þ I 0C3 P1y = I 0C3 1 - ð1 - W C3 Þð1 - cos ð2φ3 ÞÞ ^y F y3 = 2 tr σ 3 2 n o þ 1 ^ ^ I 0C3 P1y = tr σ ^ σ ^ = - I 0C3 QC3 sin φ3 F F z 3 y 3 z3 2 I 1y C3 =

ð2:221Þ ð2:222Þ ð2:223Þ ð2:224Þ

where QC3 = W C3 =



2Im A3 B3

ð2:225Þ

jA3 j2 þ jB3 j2 jA3 j2 - jB3 j2 jA3 j2 þ jB3 j2

ð2:226Þ

According to Eq. (2.183) we can also obtain the following formulas from Eqs. (2.176) and (2.6): ^þ ^ 3 σ^x F F 3

=

A3

- iB3 e - iφ3

!

iB3 eiφ3 A3 0

i A3 B3 eiφ3 - B3 A3 e - iφ3 =@ jA3 j2 - jB3 j2 e2iφ3

iB3 eiφ3

A3

A3

- iB3 e - iφ3

!

jA3 j2 - jB3 j2 e - 2iφ3

1



A i B3 A3 eiφ3 - A3 B3 e - iφ3

ð2:227Þ

2.5

Polarization Theory of Triple Elastic Scatterings Between. . .

^ 3 σ^z F ^þ F 3

=

A3

- iB3 e - iφ3

iB3 eiφ3 A3 0 2 jA3 j - jB3 j2 @ =

i B3 A3 - A3 B3 eiφ3

!

A3

59

- iB3 e - iφ3

!

- iB3 eiφ3 - A3

1 - i A3 B3 - B3 A3 e - iφ3 A  - jA3 j2 - jB3 j2

ð2:228Þ

Given these, it can be found that the contributions of the polarization part of the incident particle of the third scattering in the x and z directions are n o 1 ^ 3 σ^x F ^þ tr F = - I 0C3 PC3 sin φ3 3 2 h i n o þ 1 1 0 ^ ^ F F = tr σ ^ σ ^ 1 ð 1 W Þ ð 1 þ cos ð 2φ Þ Þ I 0C3 P1x = I x 3 x 3 C3 3 x3 C3 2 2 n o 1 1 ^ 3 σ^x F ^þ I 0C3 P1x = - I 0C3 ð1 - W C3 Þ sin ð2φ3 Þ ^y F y3 = 2 tr σ 3 2 n o 1 ^ 3 σ^x F ^þ I 0C3 P1x = - I 0C3 QC3 cos φ3 ^z F 3 z3 = 2 tr σ n o 1 ^þ ^ ^z F I 1z =0 3 C3 = 2 tr F 3 σ n o þ 1 ^ ^ F F = σ ^ = I 0C3 QC3 cos φ3 I 0C3 P1z tr σ ^ x 3 z 3 x3 2 n o þ 1 ^ ^ I 0C3 P1z = tr σ ^ σ ^ = I 0C3 QC3 sin φ3 F F y 3 z 3 y3 2 n o 1 ^ 3 σ^z F ^þ I 0C3 P1z = I 0C3 W C3 ^z F 3 z3 = 2 tr σ I 1x C3 =

ð2:229Þ ð2:230Þ ð2:231Þ ð2:232Þ ð2:233Þ ð2:234Þ ð2:235Þ ð2:236Þ

The following physical quantities appearing in Eq. (2.211) can be obtained, by using Eqs. (2.216) – (2.219), (2.221) – (2.224), and (2.229) – (2.236): i h  ðf Þ ðiÞ I C3 = I C3 I 0C3 1 - PC3 pðxi3Þ sin φ3 - pyði3Þ cos φ3 h i 1 Px3 = - PC3 sin φ3 þ pðxi3Þ 1 - ð1 - W C3 Þð1 þ cos ð2φ3 ÞÞ 2 ðiÞ 1 - py3 ð1 - W C3 Þ sin ð2φ3 Þ þ pzði3Þ QC3 cos φ3 2 1 Py3 = PC3 cos φ3 - pðxi3Þ ð1 - W C3 Þ sin ð2φ3 Þ 2 i h 1 ðiÞ þpy3 1 - ð1 - W C3 Þ ð1 - cos ð2φ3 ÞÞ þ pzði3Þ QC3 sin φ3 2

ð2:237Þ

ð2:238Þ

ð2:239Þ

60

2

Polarization Theory of Nuclear Reactions for Spin

Pz3 = - pxði3Þ QC3 cos φ3 - pðyi3Þ QC3 sin φ3 þ pzði3Þ W C3

1 Particles 2

ð2:240Þ

In this way, the final state Stokes vector after the third scattering given by Eq. (2.211) can be obtained. And then transforming it to L3 system we have: 0

1 1 B C ðf Þ ðf Þ ðf Þ B Px3 C SL3 = X ðθC3 ÞSC3 = X ðθC3 Þ I C3 B C @ P y3 A

ð2:241Þ

Pz3 The method described in this section can be used to study the Monte-Carlo 1 transport process when spin particles and spinless targets are only undergoing 2 elastic scattering. Of course, the position of the emitted particle must be recorded at any time. For the first scattering, we select the x-axis on the scattering plane, which is equivalent to φ1 = 0, and for the second and third scatterings we can choose any azimuthal angles φ2 and φ3; so the process is more complicated to a certain extent. 1 When the spin particle elastic scattering with spinless target is considered only, if 2 the required physical quantities I0, P, Q, W can be obtained directly from the existing nuclear databases and the transport process calculation is only carried out in the laboratory system, the problem discussed above will be simplified. However, there are still many types of reactions in the actual particle transport process. The problems of the polarization nuclear databases and polarized nucleus transport theory will be studied in the final chapter of this book.

2.6

Two Theoretical Methods to Study the Polarization Phenomena of Nuclear Reactions

The polarization theory of the nuclear reactions described above is based on the density matrix and the tracing method. In fact, the study of polarization theory of nuclear reactions can also use the expectation value method based on the orthonormal properties of the spin wave function [2, 7, 8]. This section will introduce the second method of directly calculating the polarization observables. 1 Equations (2.81) – (2.83) give the scattering amplitude of the spin projectile 2 with the spinless target. Utilizing Eq. (2.100) and Eq. (2.16), one can obtain   ^ = jAj2 þ jBj2 ^I þ 2 Re ðAB Þ → ^þF n  σ^ F

ð2:242Þ *

The direction of the incident particle is usually taken as the z-axis; thus n is the unit vector located in the xy plane (see Fig. 2.1). If the incident particles are unpolarized

2.6

Two Theoretical Methods to Study the Polarization Phenomena of. . .

61

and there is no coherence between the spin magnetic quantum numbers of the incident particles, then the differential cross sections can be obtained by use of Eqs. (2.1) and (2.242) as follows: I0 

D þ E dσ 0 1 X

^ ^ F χ 12μ = jAj2 þ jBj2 χ 12μ F = dΩ 2 μ

ð2:243Þ

The following expression can be obtained utilizing Eqs. (2.100), (2.17), and (2.20):   * ^ = A^I þ B * ^ þ σ^ F n  σ^ σ^ A^I þ B n  σ^ F h i  ð2:244Þ →   → = jAj2 - jBj2 σ^ þ 2 Re ðAB Þ þ jBj2 ð → n  σ^Þ n - iðAB - BA Þ ð n × σ^Þ For the unpolarized incident particles, the polarization vector of the outgoing particles can be obtained by using Eqs. (2.1), (2.6), (2.98), and (2.244): *0 P out

= hσ^i0out =

D þ E 2 Re ðAB Þ 1 1X

^ → ^

χ 1μ = χ 12μ F σ^ F n 0 2 2 2 2 I jAj þ jBj μ

ð2:245Þ

Equation (2.243) and Eq. (2.245) are exactly the same as Eq. (2.88) and Eq. (2.110), respectively. 1 Equation (2.25) gives the arbitrary spin wave function X 12 of the spin particles, 2 which is a mixed state of the basis spin functions χ 1212 and χ 12 - 12. It indicates that in the spin wave function X 12 there are either couplings between the states with different spin magnetic quantum numbers, or they are coherent. In this case, it can be seen from Eq. (2.36) that the incident particle beam described by X 12 is polarized. The orthogonal normalization condition requires D E X 12 jX 12 = 1 1 incident particles is defined as 2 D E * p in = X 12 jσ^jX 12

ð2:246Þ

The polarization vector of the spin

ð2:247Þ

The following result can be obtained using Eqs. (2.242), (2.246), and (2.247): I

D þ E  dσ

^ ^ → * = X 12 F F X 12 = jAj2 þ jBj2 þ 2 Re ðAB Þ n  p in dΩ

ð2:248Þ

62

2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2

This expression is exactly the same as Eq. (2.115). It can be found that the polarization vector of the outgoing particles corresponding to the polarized incident particles is, by using Eqs. (2.244), (2.246), and (2.247), as follows: *

1D

^ þ ^

E X 1 F σ^ F X 12 Ih 2 → *

→ * i →  * 2 ReðAB Þ þ jBj2 n p in n þ jAj2 - jBj2 p in þ 2ImðAB Þ n × p in = → *

jAj2 þ jBj2 þ 2ReðAB Þ n p in

P out =

ð2:249Þ This expression is exactly the same as Eq. (2.126). The above results show that for spinless target, the polarization theoretical 1 formulas of the spin particles, obtained by use of the first equation of Eq. (1.88) 2 belonging to the expectation value method and the last two equations of Eq.(1.89) belonging to the tracing method, are exactly the same.

*

2.7

*

1 1 Polarization Theory for þ A → þ B Reactions 2 2

For the target and the residual nucleus with non-zero spin, Eq. (3.8.20) in Ref. [2] gives the differential cross section of the j–j coupling mode without considering the polarization as

2 dσ α0 n0 ,αn 1 X

= 2 2

f α0 n0 μ0 M 0I , αnμM I ðΩÞ dΩ ^i ^I μM I μ0 M 0 I

ð2:250Þ

It can be seen from the above equation that in the case of the unpolarized incident particle, target nucleus, and residual nucleus, the outgoing particles belong to a special polarized particle state, in which all the outgoing particle states with different spin quantum numbers of the incident particle, the outgoing particle, the target nucleus, and the residual nucleus are incoherent, but the shares of the outgoing particle states with different spin quantum numbers of the incident particle, the outgoing particle, the target nucleus, and the residual nucleus may be different. However, in Eq. (2.250) after summing the final state particle spin magnetic quantum numbers and averaging the initial state particle spin magnetic quantum numbers, it becomes the total differential cross section of the outgoing particles. The reaction amplitude appearing in Eq. (2.250) has been given by Eq. (3.8.12) in Ref. [2] as follows:

*

2.7 Polarization Theory for

*

63

1 1 þ A → þ B Reactions 2 2

f α0 n0 μ0 M 0I , αnμM I ðΩÞ = f C ðθÞ δα0 n0 , αn δμ0 M 0I , μM I pffiffiffi i π X^ iðσl þσl 0Þ  þ δα0 n0 l0 j0 , αnlj - SJα0 n0 l0 j0 , αnlj le k 00 ljl j J

Cjμ l0

j0 m0j J M J M 0 iμ C jμ IM I C l0 m0 l i0 μ0 C j0 m0 j I 0 M 0 I Yl m0 l ðθ,

ð2:251Þ

φÞ

pffiffiffiffiffiffiffiffiffiffiffiffi The simplification symbol ^i  2i þ 1 is introduced in the above expressions. α and α0 represent the types of incident particles and outgoing particles; n and n0 represent the level labels of the target and the residual nucleus, respectively. SJα0 n0 l0 j0 ,αnlj is a S matrix element of the nuclear reactions. For the S–L coupling mode, Eq. (2.250) can still be used. However its reaction amplitude needs to be changed to the form given by Eq. (13.8.13) in Ref. [9] as follows: f α0 n0 μ0 M 0I , αnμM I ðΩÞ = f C ðθÞ δα0 n0 ,αn δμ0 M 0I , μM I pffiffiffi i π X ^ i ðσl þσl 0 Þ  þ δα0 n0 l0 S0 ,αnlS - SJα0 n0 l0 S0 ,αnlS le k 0 0

ð2:252Þ

lS l S J

C Jl0 MSM CSiμ MIM I C Jl0 mM0 l

S0 M 0S 0 S0 M 0 S C i0 μ0 I 0 M 0 I Yl m0 l ðθ,

φÞ

Nucleon-nucleon elastic scattering is only a special case of the nuclear reaction described by Eq. (2.252), which is equivalent to taking α0 = α, n0 = n, i = i0 = I = I 0 = 12 in the above formula. Eq. (2.73) has given rffiffiffiffiffiffiffiffiffiffiffiffiffiffisffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0

l - m0l ! m0l 0 2l0 þ 1

P 0 ð cos θÞ eiml φ 0 Yl m0 l ðθ, φÞ = ð - 1Þ 4π l0 þ m0l ! l m0l

ð2:253Þ

From the C–G coefficients in Eq. (2.251) or Eq. (2.252), the following relation can be obtained: m0l = μ þ M I - μ0 - M 0I

ð2:254Þ

1 In the case of both incident particles and outgoing particles with spin , the following 2 symbols are introduced for the j–j coupling mode and S–L coupling mode, respectively:

64

2

Polarization Theory of Nuclear Reactions for Spin

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0

l - m0l ! X j μ 0

C 1 CJ M l þ m0l ! jj0 l0 2μ jμ IM I  J M J 0 0 C δ S 0 0 0 0 0 0 1 0 α0 n0 l j , αnlj α0 n0 l j , αnlj μ j mj I M I



0 f μll0 μJ m0l = ^l ^l eiðσ l þσ l0 Þ 0

j0 m0 C l0 m0 j l 0 f μll0 μJ

ð2:255Þ

2

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0

l - m0l ! X J M 0

C CS1 M l þ m0l ! SS0 l0 SM 2μ IM I  S0 M 0S J 0 S0 M 0 S C 1μ0 I 0 M 0 δα0 n0 l S0 , αnlS - Sα0 n0 l0 S0 , αnlS

0

0 ml = ^l ^l eiðσl þσl 0 Þ C Jl0 mM0 l

1 Particles 2

ð2:256Þ

I

2

Therefore, the reaction amplitude can be uniformly written in the following form under both angular momentum coupling modes: f α0 n0 μ0 M 0I , αnμM I ðΩÞ = f C ðθÞ δα0 n0 ,αn δμ0 M 0I , μM I þ

0X 0 i 0 ð - 1ÞμþM I - μ - M I f μll0 μJ μ þ M I - μ0 - M 0I ð2:257Þ 2k ll0 J μþM I - μ0 - M 0I

Pl0

ð cos θÞ eiðμþM I - μ

0

- M 0I Þφ

Note that δμ0 M 0I , μM I is contained in the first item on right end of the above formula, so 0 that it can be multiplied by a factor eiðM I - M I Þφ which is equal to 1. We define 0 i AM 0I M I ðθÞ = f C ðθÞ δα0 n0 , αn δM 0I , M I þ ð - 1ÞM I - M I 2k X 0

M I - M 0I ll J 0 f 1 1 M I - M I Pl0 ð cos θÞ

ll0 J

BM 0I M I ðθÞ = -

ð2:258Þ

2 2

X 0

M - M0 - 1 0 1 ð - 1ÞM I - M I - 1 f ll1 J- 1 M I - M 0I - 1 Pl0 I I ð cos θÞ 2 2 2k 0 ll J

ð2:259Þ C M 0I M I ðθÞ =

1 ð - 1Þ 2k

M I - M 0I þ1

X 0 f ll-J1 2

ll0 J

1 2



M I - M 0I þ 1 P

M I - M 0I þ1 l0

0 i DM 0I M I ðθÞ = f C ðθÞ δα0 n0 ,αn δM 0I , M I þ ð - 1ÞM I - M I 2k

M - M0 P ll0 J f - 1 - 1 M I - M 0I Pl0 I I ð cos θÞ

ll0 J

2

ð cos θÞ ð2:260Þ

ð2:261Þ

2

Now introducing the following simplification symbols F μ0 μ ðθ, φÞ  f α0 n0 μ0 M 0I , then we obtain

αnμM I ðΩÞ

ð2:262Þ

*

*

65

1 1 þ A → þ B Reactions 2 2

2.7 Polarization Theory for

F 12 12 ðθ, φÞ = AM 0I M I ðθÞ eiðM I - M I Þφ 0

- 12 ðθ,

F 12

ð2:263Þ

φÞ = - iBM 0I M I ðθÞ eiðM I - M I Þφ e - iφ 0

ð2:264Þ

ðθ, φÞ = iCM 0I M I ðθÞ eiðM I - M I Þφ eiφ 0

F - 12

1 2

F - 12

- 12 ðθ,

ð2:265Þ

φÞ = DM 0I M I ðθÞ eiðM I - M I Þφ 0

ð2:266Þ

When the foot note M 0I M I and argument (θ) of A, B, C, and D are omitted, the reaction amplitude with definite M 0I M I, in which arguments (θ, φ) are emitted, can be written as - iBe - iφ

A

^= F

iCe

iφ 

þ

^ = F

! eiðM I - M I Þφ , 0

D

A

- iC e - iφ

iB eiφ

D

!

ð2:267Þ e

- iðM I - M 0I Þφ

We first study the case where the target and the residual nucleus are all unpolarized. In this case the different MI components are incoherent and the different M 0I components are also incoherent. For the components with definite M 0I M I , from Eq. (2.267) one can obtain þ

^F ^ = F =

A

- iBe - iφ

iCeiφ

D

!

A

- iC  e - iφ

iB eiφ

D

!

jAj2 þ jBj2

- iðAC  þ BD Þ e - iφ

iðCA þ DB Þ eiφ

jC j2 þ jDj2

!

ð2:268Þ

→

For the sake of simplicity, we select the y-axis in the direction n perpendicular to the reaction plane. In this case there is φ = 0; thus Eqs. (2.267) and (2.268) are simplified, respectively, as ^= F




A

- iB

iC

D

^F ^þ = F

 ,

^þ = F

jAj2 þ jBj2 iðCA þ DB Þ

From Eq. (2.270) we can obtain




A

- iC

iB

D

- iðAC  þ BD Þ jC j2 þ jDj2

 ð2:269Þ ! ð2:270Þ

66

2

I0 =

Polarization Theory of Nuclear Reactions for Spin

 1  þ 1 ^F ^ = tr F jAj2 þ jBj2 þ jCj2 þ jDj2 2 2

1 Particles 2

ð2:271Þ

From Eq. (2.269) and Eq. (2.6) we can also obtain þ

^ = σ^x F

!

iB

D

A

- iC

þ

^ = , σ^y F

^þ = ^ σ^x F F þ

^ σ^y F ^ = F þ

^ σ^z F ^ = F

B

- iD

iA

C

! ,

þ

^ = σ^z F

iðAB - BA Þ

AD - BC 

- CB þ DA

iðCD - DC  Þ

A

- iC 

- iB

- D

!

ð2:272Þ

!

AB þ BA

- iðAD þ BC  Þ

iðCB þ DA Þ

CD þ DC 

jAj2 - jBj2

- iðAC  - BD Þ

iðCA - DB Þ

jC j2 - jDj2

! ð2:273Þ

!

Then from Eq. (2.273) the following results can be obtained: n o 1 ^ þ = - 1 ImðAB þ CD Þ, ^ σ^x F tr F 0 2I I0 n o 1 ^ þ = 1 Re ðAB þ CD Þ, ^ σ^y F Ay = 0 tr F I0 2I n o  1 ^ þ = 1 jAj2 - jBj2 þ jC j2 - jDj2 ^ σ^z F Az = 0 tr F 0 2I 2I Ax =

ð2:274Þ

From Eq. (2.270) and Eq. (2.6) we can also obtain n o 1 ^F ^þ = F tr σ ^ x 2I 0 n o 1 ^F ^þ = P0y = 0 tr σ^y F 2I n o 1 ^F ^þ = P0z = 0 tr σ^z F 2I

P0x =

i 1 ðCA þ DB - AC  - BD Þ = 0 ImðAC  þ BD Þ, 0 2I I 1 1     ðCA þ DB þ AC þ BD Þ = 0 Re ðAC  þ BD Þ, 0 2I I  1 ð2:275Þ jAj2 þ jBj2 - jC j2 - jDj2 2I 0

We define the polarization transfer coefficient as K ki =

n o þ 1 ^ ^ F σ ^ F tr σ ^ , k i 2I 0

i,k = x,y,z

ð2:276Þ

The following results can be obtained from Eqs. (2.276), (2.273), and (2.6):

*

2.7 Polarization Theory for

*

1 1 þ A → þ B Reactions 2 2

1 1 ð - CB þ DA þ AD - BC  Þ = 0 Re ðAD - BC  Þ, 2I 0 I i 1 K yx = 0 ðCB - DA þ AD - BC  Þ = - 0 ImðAD - BC  Þ, 2I I i 1 K zx = 0 ðAB - BA - CD þ DC  Þ = - 0 ImðAB - CD Þ, 2I I i 1 K xy = 0 ðCB þ DA - AD - BC  Þ = 0 ImðAD þ BC  Þ, 2I I 1 1 K yy = 0 ðCB þ DA þ AD þ BC  Þ = 0 Re ðAD þ BC  Þ, 2I I 1 1 K zy = 0 ðAB þ BA - CD - DC  Þ = 0 Re ðAB - CD Þ, 2I I i 1 K xz = 0 ðCA - DB - AC  þ BD Þ = 0 ImðAC  - BD Þ, 2I I 1 1 K yz = 0 ðCA - DB þ AC  - BD Þ = 0 Re ðAC  - BD Þ, 2I I  1 K zz = 0 jAj2 - jBj2 - jC j2 þ jDj2 2I

67

K xx =

ð2:277Þ

Since A, B, C, and D introduced by Eqs. (2.258) –(2.261) have foot note M 0I M I , it follows that I 0 , Ai , P0k , K ki ði, k = x, y, zÞ obtained previously also have foot note M 0I M I . In the case that both the target and residual nucleus are unpolarized, the simplification summation symbol for averaging to MI and summing to M 0I concurrently is introduced as e Σ

X 1 2I þ 1 0

ð2:278Þ

MI MI

The differential cross section of the unpolarized incident particle can be obtained by Eq. (2.271) as e I0 I =Σ 0

ð2:279Þ

And the analyzing power of the incident particles can be obtained by Eq. (2.274) as Ai =

1e 0 Σ I Ai , 0 I

i = x, y, z

ð2:280Þ

According to Eq. (2.63) and Eq. (2.34), the differential cross section of the polarized incident particles can be obtained as

68

2

Polarization Theory of Nuclear Reactions for Spin

I =I

0

1þ

X

!

1 Particles 2

ð2:281Þ

pi Ai

i=x,y,z

where pi (i = x, y, z) are the polarization rates of the incident particles. The polarization rates of the outgoing particles corresponding to the unpolarized incident particles can be obtained by Eq. (2.275) as 1 e 0 0k ΣI P , 0 I

0k

P =

k = x,y,z

ð2:282Þ

The polarization transfer coefficient can be obtained by Eq. (2.277): k

Ki =

1e 0 k Σ I Ki , 0 I

i,k = x,y,z

ð2:283Þ

From the above three formulas and Eq. (2.63), the polarization rates of the outgoing particles corresponding to the polarized incident particles can be obtained as 0

I P = I k

0k

P þ

X

!

k pi K i i = x,y,z

,

k = x,y,z

ð2:284Þ

According to Eq. (3.2.8) and Eq. (3.2.12) given in Ref. [2], when φ = 0 one can obtain sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2l þ 1Þðl - mÞ! m Pl ð cos θÞ Ylm ðθ, 0Þ = ð - 1Þm 4πðl þ mÞ! sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2l þ 1Þðl þ mÞ! - m m Pl ð cos θÞ Yl -m ðθ, 0Þ = ð - 1Þ 4πðl - mÞ!

ð2:285Þ

ð2:286Þ

Based on Eq. (3.2.10) given in Ref. [2], one can also obtain m Pm l ð cos θ Þ = ð - 1Þ

ðl þ mÞ! - m P ð cos θÞ ðl - mÞ! l

ð2:287Þ

Substituting Eq. (2.287) into Eq. (2.285) and using Eq. (2.286), the following relation can be obtained: Ylm ðθ, 0Þ = ð - 1Þm Yl

-m ðθ,

0Þ

ð2:288Þ

In the next analyses we will use Eq. (2.252) belonging to the S–L coupling mode. First we have the following relations based on the properties of the C–G coefficient:

*

2.7 Polarization Theory for

I C 1Sμ μþM = ð - 1ÞIþ2 - S C1S IM 1

I

2

C Jl0

M SM

*

69

1 1 þ A → þ B Reactions 2 2

2

= ð - 1ÞlþS - J C Jl0

S0

- μ - MI , - μ I - MI -M S - M,

C 1 μ0 2

μ0 þM 0I I0M0 I

0

S0 M 0 S

1

0

2

0

C Jl0 mM 0 l

S0

= ð - 1Þ I þ 2 - S C 1

= ð - 1Þl þS - J C Jl0 0

-M - m0l

- μ0 - M 0I , - μ0 I 0 - M 0I S0 - M 0S

ð2:289Þ We divide Eq. (2.252) into two items and study the second item first. Using Eq. (2.288) and Eq. (2.289), for the part which is not related with angle φ, the following phase factor comes out when all spin magnetic quantum numbers change their signs 0

ð - 1ÞlþS - Jþl þS

0

- JþIþ12 - SþI 0 þ12 - S0 þμþM I - μ0 - M 0I 0

0

= ð - 1Þ2J - 1þlþIþl þI þμþM I - μ

0

- M 0I

Let π i ,π 0i ,π I , and π 0I represent the parities of the incident particle, outgoing particle, target, and residual nucleus, respectively. If π i π I = π 0i π 0I , l and l0 should all be even or odd, and in this case let Π = 0. On the other hand, if π i π I = - π 0i π 0I , l and l0 should be that for which one is even and another is odd, and in this case let Π = 1. Moreover we can express the above results in a unified form as  Π=

0 1

when when

π i π I = π 0i π 0I π i π I = - π 0i π 0I

ð2:290Þ

1 In the case that the spins of both the incident particle and the outgoing particle are , 2 the target spin I and the residual nucleus spin I0 must be the integers or the semi-odd numbers at the same time. If I and I0 are all integers, then J is a semi-odd number, so (-1)2J - 1 = 1, and in this case let Λ = 0. If I and I0 are all semi-odd numbers, then J is an integer, so (-1)2J - 1 = - 1, and in this case let Λ = 1. Moreover we can express the above results in a unified form as  Λ=

0

when I and I 0 are all integers

1

when I and I 0 are all semi‐odd numbers

ð2:291Þ

Thus we can get the following relation. 0

ð2Þ

f α0 n0 μ0 M 0 ,αnμM I ðθÞ = ð - 1ÞΠþΛþIþI þμþM I - μ I

0

- M 0I ð2Þ f α0 n0 - μ0 - M 0 ,αn - μ - M I ðθÞ I

ð2:292Þ

Now we come back to study the first item on the right end of Eq. (2.252). For elastic scattering, we have Π = 0, I0 = I, (-1)Λ + 2I = 1, and the δ symbol requires μ = μ0 , M I = M 0I , so f (1) also satisfies the relation given by Eq. (2.292). Thus, for the independent part with angle φ of the total reaction amplitude f , there is the following relation:

70

2

Polarization Theory of Nuclear Reactions for Spin 0

f α0 n0 μ0 M 0I ,αnμM I ðθÞ = ð - 1ÞΠþΛþIþI þμþM I - μ

0

- M 0I

f α0 n0 - μ0 - M 0I ,αn - μ - M I ðθÞ

1 Particles 2

ð2:293Þ

The following relations can be obtained from Eqs. (2.263)–(2.266): AM 0I M I ðθÞ = f α0 n0 12M 0I ,αn12M I ðθÞ,

BM 0I M I ðθÞ = if α0 n0 12M 0I ,αn - 12M I ðθÞ,

C M 0I M I ðθÞ = - if α0 n0 - 12M 0I ,αn12M I ðθÞ, DM 0I M I ðθÞ = f α0 n0 - 12M 0I ,αn - 12M I ðθÞ

ð2:294Þ

According to Eq. (2.293) and Eq. (2.294) we have 0

0

0

0

DM 0I M I ðθÞ = ð - 1ÞΠþΛþIþI þM I - M I A - M 0I C M 0I M I ðθÞ = ð - 1ÞΠþΛþIþI þM I - M I B- M 0I

- M I ðθ Þ

ð2:295Þ

- M I ðθ Þ

ð2:296Þ

The phase factor L  Π þ Λ þ I þ I 0 þ M I - M 0I appearing in Eq. (2.295) and Eq. (2.296) must be an integer, so there must be (-1)2L = 1. In the summation e given by Eq. (2.278), the sum of MI and the sum of -MI are all from -I symbol Σ to I, and the sum of M'I and the sum of - M 0I are all from -I0 to I0 , so using Eqs. (2.295) and (2.296) the following results can be proved: e ðAB þ CD Þ = ΣIm e ðAB þ BA Þ = 0 ΣIm e Re ðAB þ BA Þ = 2Σ e Re ðAB Þ e Re ðAB þ CD Þ = Σ Σ e ðAC  þ BD Þ = ΣIm e ðAC  þ CA Þ = 0 ΣIm e Re ðAC  þ CA Þ = 2Σ e Re ðAC  Þ e Re ðAC  þ BD Þ = Σ Σ e ðAB - BA Þ = 2ΣIm e ðAB Þ e ðAB - CD Þ = ΣIm ΣIm

ð2:297Þ

e Re ðAB - BA Þ = 0 e Re ðAB - CD Þ = Σ Σ e ðAC  - CA Þ = 2ΣIm e ðAC  Þ e ðAC  - BD Þ = ΣIm ΣIm e Re ðAC  - CA Þ = 0 e Re ðAC  - BD Þ = Σ Σ e jAj2 , ejDj2 = Σ Σ

e jC j2 = Σ e jBj2 Σ

In addition, the following expression can also be proved: X M I M 0I

 1X AM 0I MI DM 0I M I þ A - M 0I -M D- M 0I 2 0 MI MI  1X = AM 0I MI DM 0I M I þ DM 0I M AM 0I M I 2 0

AM 0I MI DM 0I M I =

MI MI

-M I

ð2:298Þ

*

2.7 Polarization Theory for

*

1 1 þ A → þ B Reactions 2 2

71

From the above formula we can see that AD is a real number. Similarly, BC should be a real number. Given these, we can get e ðAD  BC  Þ = 0 ΣIm

ð2:299Þ

The following relation can be obtained utilizing Eqs. (2.279), (2.271), and (2.297):  0 e jAj2 þ jBj2 I =Σ

ð2:300Þ

The following relations can also be obtained by use of Eqs. (2.280), (2.274), and (2.297): Ax = Az = 0,

Ay =

2e Σ Re ðAB Þ 0 I

ð2:301Þ

It follows that Eq. (2.281) can be rewritten as

0 I = I 1 þ py Ay

ð2:302Þ

We can also obtain the following relations by using Eqs. (2.282), (2.275), and (2.297): 0x

0z

0y

P = P = 0,

P =

2e Σ Re ðAC  Þ 0 I

ð2:303Þ

And the following results can be obtained utilizing Eqs. (2.283), (2.277), (2.297), and (2.299): y

x

z

y

K x = K y = K y = K z = 0, 2e Σ ImðAB Þ, 0 I 2e x ImðAC  Þ, Kz = 0 Σ I z

Kx = -

1e Σ Re ðAD - BC  Þ, 0 I 1e y Ky = 0 Σ Re ðAD þ BC  Þ, I  1e z Kz = 0 Σ jAj2 - jBj2 I x

Kx =

ð2:304Þ

In addition Eq. (2.284) can be rewritten as 0 I x x

p x K x þ pz K z , I 0 I z z z

px K x þ p z K z P = I x

P =

y

P =

0

I I

 0y y P þ py K y ,

For a definite component M 0I M I , from Eq. (2.271) one can get

ð2:305Þ

72

2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2

 0 2 1h 4 I = jAj þ jBj4 þ jCj4 þ jDj4 þ 2 jAj2 jBj2 þ jAj2 jCj2 þ jAj2 jDj2 4 i ð2:306Þ þ jBj2 jCj2 þ jBj2 jDj2 þ jC j2 jDj2 From Eq. (2.277) the following relations can be obtained:

K zz

2

h  1 = 2 jAj4 þ jBj4 þ jC j4 þ jDj4 þ 2 jAj2 jDj2 þ jBj2 jC j2 4 I0 i - jAj2 jBj2 - jAj2 jCj2 - jBj2 jDj2 - jC j2 jDj2

x 2 1 K x = 2 ðAD þ A D - CB - C BÞ2 4 I0 h 1 = 2 ðAD þ A DÞ2 þ ðCB þ C  BÞ2 4 I0

ð2:307Þ

ð2:308Þ

i - 2AD CB - 2AD C  B - 2A DCB - 2A DC  B

 2 1 K yy = 2 ðAD þ A D þ CB þ C  BÞ2 4 I0 h 1 = 2 ðAD þ A DÞ2 þ ðCB þ C  BÞ2 4 I0

ð2:309Þ

i þ2AD CB þ 2AD C B þ 2A DCB þ 2A DC  B

Notice Re ðAB þ CD Þ = Re ðAB þ C  DÞ, ImðAB - CD Þ = ImðAB þ C DÞ

ð2:310Þ

From Eq. (2.274) and Eq. (2.277), one can obtain 2 z 2 1 Ay þ K x = 2 jAB þ DC  j2 I0 1  = 2 jAj2 jBj2 þ jDj2 jCj2 þ AB D C þ A BDC  I0

ð2:311Þ

And note Re ðAC  þ BD Þ = Re ðAC  þ B DÞ, ImðAC  - BD Þ = ImðAC  þ B DÞ Thus from Eq. (2.275) and Eq. (2.277), one can also obtain

ð2:312Þ

Polarization Theory of Nuclear Reactions of Spin. . .

2.8



P0y

2

2 1 þ K xz = 2 jAC  þ DB j2 I0  1 = 2 jAj2 jCj2 þ jDj2 jBj2 þ AC  D B þ A CDB I0

73

ð2:313Þ

Then the following relation is obtained when MI and M 0I are taking the definite values 2 z 2 z 2 0y 2 x 2 x 2  y 2 Ay þ K x þ K z þ P þ Kz þ Kx - Ky = 1

ð2:314Þ

Section 2.4 has already pointed out that when the spins of both the target and the residual nucleus are equal to 0 and the y-axis is selected to be perpendicular to the reaction plane, only their y components of the polarization analyzing powers Ai(i = x, y, z) and the polarization rate P0i(i=x, y, z) corresponding to the unpolarized incident particles are not equal to 0; only 5 of the 9 polarization transfer coefficient components K ji (i, j=x, y, z) are not equal to 0. The theoretical derivation of this section has proved that the conclusions mentioned above are still true when the spins of the unpolarized target and residual nucleus are not equal to 0. This is due to the fact that the nuclear reaction process satisfies the conservation of parity. If let e D = A, C = B in Eqs. (2.300) – (2.305) and remove the summation symbol Σ, they automatically degenerate into the result given in Sect. 2.4 in which the spins of the target and residual nucleus are equal to 0. Simultaneously, Eq. (2.314) will also automatically degenerate into Eq. (2.135) under the condition mentioned above.

2.8

1 2 Particle with Polarized Target and Residual Nucleus Polarization Theory of Nuclear Reactions of Spin

Setting ΦIM I as the basis spin function of the target along the z-axis, then according to Eq. (2.25) the arbitrary spin wave function ΦI of the target with spin I can be expressed as [7] ΦI =

X

aM I ΦIM I

ð2:315Þ

MI

where aM I is a contravariant component and can also be called the target polarization rate. The corresponding Hermitian conjugate function is Φþ I =

X aM I * Φþ IM I MI

The normalization condition requires

ð2:316Þ

74

2

Polarization Theory of Nuclear Reactions for Spin

X 2

aM I = 1

1 Particles 2

ð2:317Þ

MI

If the target spin wave function ΦI can be expanded with the basis spin functions ΦIN I along the z0 -axis (e.g., the direction of the target motion along the z0 axis), then we have ΦI =

X

aN I ΦIN I

ð2:318Þ

NI

If the expansion coefficient aN I is known, and the z0 -axis is not parallel to the z-axis, thus by use of the D function we can get the following relation: aM I =

X aN I DIN I M I ðφA , θA , 0Þ

ð2:319Þ

NI

where θA and φA are the polar and azimuthal angles of the z-axis in the z0 -axis coordinate system. The residual nucleus spin wave function ΦI' can be expressed as ΦI 0 =

X 0 bM I ΦI 0 M 0 I

ð2:320Þ

M 0I 0

where bM I is a contravariant component and can also be called the residual nucleus polarization rate. The normalization condition is X 0 2

bM I = 1

ð2:321Þ

M 0I

In the 2.7 section discussions, we assume that both the target and the residual nucleus are unpolarized, that is, the probabilities of the target and the residual nucleus at different spin magnetic quantum numbers are equal and they are incoherent to each other. However, the contributions of the different MI and M 0I states of the target and the residual nucleus to the reaction amplitude, after the interactions with the incident particles and the outgoing particles, may not be the same, but they still remain incoherent. In this section we assume that the target and the residual nucleus are in the states of the arbitrary spin wave functions described by Eqs. (2.315) and (2.320), that is, in the mixed states of different MI and M 0I . Thus the coherence effect between different MI and different M 0I should be considered, that is, it is assumed that both the target and the residual nucleus are in the polarized states. * If we select the y-axis along the direction of the unit vector n vertical to the reaction plane in Fig. 2.1, then the azimuthal angle φ = 0. According to Eq. (2.315) and Eq. (2.320) and refer to Eq. (2.269), we express the amplitude of the reaction as

Polarization Theory of Nuclear Reactions of Spin. . .

2.8

^= F

X

aM I b

AM 0I M I

- iBM 0I M I

iCM 0I M I

DM 0I M I

M 0I

M I M 0I

! ^þ = , F

X

75

0 

aM I b

M 0I 

@

M I M 0I

AM 0 M I

- iC M 0 M I

iBM 0 M I

DM 0 M I

I

I

I

1 A

I

ð2:322Þ Based on Eq. (2.271) one can write I 0M 0 M I M 0 M I = I

I

 1  AM 0 M I AM 0I M I þ BM 0 M I BM 0I M I þCM 0 M I CM 0I M I þ DM 0 M I DM 0I M I I I I I 2 ð2:323Þ

Based on Eq. (2.274) one can also write 1

Ax,M 0 M I M 0I M I = -

I 0M 0 M

I

I

Ay,M 0 M I M 0I M I = I

Az,M 0 M I M 0I M I = I

I

M 0I M I

I



1 I

I

I

I

 Re AM 0I M I BM 0 M þ C M 0I M I DM 0 M ,

1 I 0M 0 M M 0 M I I I I 2I 0M 0 M

 Im AM 0I M I BM 0 M I þ C M 0I M I DM 0 M I ,

M 0I M I

I

I

I

AM 0 M AM 0I M I - BM 0 M BM 0I M I þC M 0 M CM 0I M I - DM 0 M DM 0I M I I

I

I

I

I

I

I



I

ð2:324Þ

Based on Eq. (2.275) one can obtain P0x = M0 MI M0 MI I

I

= P0y M0 M M0 M I

I

I

I

1 I 0M 0 M M 0 M I I I I 1 I 0M 0 M I

= P0z M0 M M0 M I

I

I

I

 Im AM 0I M I CM 0 M I þ BM 0I M I DM 0 M I ,

I

M 0I M I

1 2I 0M 0 M M 0 M I

I

I

I

I



Re AM 0I M I CM 0 M I þ BM 0I M I DM 0 M I , I

I

 AM 0 M I AM 0I M I þ BM 0 M I BM 0I M I - C M 0 M I C M 0I M I - DM 0 M I DM 0I M I I

I

I

I

I

ð2:325Þ Based on Eq. (2.277) one can get

76

2

K xx,M 0 M I M 0 M I = I

I

I 0M 0 M M 0 M I

K yx,M 0 M I M 0 M I = I

K zx,M 0 M I M 0 M I = I

I

I

K xy ,M 0 M M 0 M = I

I

I

I

1 1 I 0M 0 M M 0 M I

I

I

I

I

I

I

=

I 0M 0 M M 0 M

I

I

I

I

I

I

I

I

I

Re AM 0I M I DM 0 M I þ BM 0I M I C M 0 M I , I

I

 Re AM 0I M I BM 0 M I - CM 0I M I DM 0 M I ,

I

Im AM 0I M I CM 0 M I - BM 0I M I DM 0 M I ,

I

ð2:326Þ

I



I

I

I

 Re AM 0I M I C M 0 M I - BM 0I M I DM 0 M I ,

1 I 0M 0 M M 0 M I I I I

I

I

I

I

I 0M 0 M M 0 M I

I



1

K zz ,M 0 M I M 0 M I = I

I

I 0M 0 M M 0 M

I

K yz ,M 0 M I M 0 M I =

I

I

 Im AM 0I M I BM 0 M I - CM 0I M I DM 0 M I ,

I

I

1 I

K xz ,M 0 M I M 0 M I =

I

1 I

K zy ,M 0 M M 0 M I I

I

I

 Im AM 0I M I DM 0 M I - BM 0I M I C M 0 M I ,

 Im AM 0I M I DM 0 M I þ BM 0I M I CM 0 M I ,

I

1 I 0M 0 M M 0 M I

K yy ,M 0 M M 0 M =

I

I

I 0M 0 M M 0 M I I I I

I

1 Particles 2

 Re AM 0I M I DM 0 M I - BM 0I M I C M 0 M I ,

1 I

I

Polarization Theory of Nuclear Reactions for Spin

I



1 2I 0M 0 M M 0 M I I I I

I

AM 0 M I AM 0I M I - BM 0 M I BM 0I M I I

I

- CM 0 M I CM 0I M I þDM 0 M I DM 0I M I I



I

Introduce the following simplification summation symbol Σ

1 2I þ 1

X



0

0

aM I aM I bM I bM I

ð2:327Þ

M I M I M 0I M 0I

Then based on Eq. (2.279) the following result can be obtained: 0

I = Σ I 0M 0 M I M 0 M I I

ð2:328Þ

I

Based on Eq. (2.280) we can obtain Ai =

1 I

0

Σ I 0M 0 M I M 0 M I Ai,M 0I M I M 0I M I ,

i = x, y, z

I

I

ð2:329Þ

Equation (2.281), which is still applicable here, is given again below: I =I

0

1þ

X i = x,y,z

Based on Eq. (2.282) we have

! pi Ai

ð2:330Þ

→

2.9 Polarization Theory for

1

0k

P =

→

77

1 1 þ Reactions 2 2 0

I

Σ I 0M 0 M I M 0 M I P0k , M0 MI M0 MI I

I

I

I

k = x,y,z

ð2:331Þ

i,k = x,y,z

ð2:332Þ

Based on Eq. (2.283) we can obtain k

Ki =

1 I

Σ I 0M 0 M I M 0 M I K ki,M 0 M I M 0 M I ,

0

I

I

I

I

Equation (2.284), which is still applicable here, is given again below: 0

I P = I k

0k

P þ

X i = x,y,z

! k pi K i

,

k = x,y,z

ð2:333Þ 0

In the expressions given in this section, only if the polarization rate aM I and/or bM I of the target and/or the residual nucleus are known, can the calculations of the 1 polarization physical quantities of the nuclear reactions with spin particles be done 2 for the polarized target and/or the polarized residual nucleus. The unpolarization conditions of the target and the residual nucleus are, respectively, 

a M I aM I = δM I M I 0

0

bM I bM I = δM 0I M 0I

ð2:334Þ ð2:335Þ

If substituting Eq. (2.334) and Eq. (2.335) into Eqs. (2.323) – (2.333), then these expressions mentioned above automatically degenerate into the expressions calculating the polarization physical quantities of the nuclear reactions for the unpolarized 1 target and residual nucleus with spin particles given by Eqs. (2.271), (2.274), 2 (2.275), (2.277) –(2.284), respectively.

→

2.9

→

1 1 Polarization Theory for þ Reactions 2 2

This section will study the polarization theory of the nuclear reactions between two 1 different polarized particles with spin , and take the n–p elastic scattering as an 2 example. The S–L angular momentum coupling mode is used for nucleon- nucleon interactions. The reaction amplitude of the n–p elastic scattering can be written from Eq. (2.252) as

78

2

Polarization Theory of Nuclear Reactions for Spin

pffiffiffi i π X ^ iðσl þσl0 Þ  le f μ0 ν0 ,μ ν ðΩÞ = δl0 S0 ,lS - SJl0 S0 ,lS k 0 0

ð2:336Þ

lS l S J

S0 M 0S CJl0 MSM C S1μ M1ν C Jl0 mM 0 S0 M 0 C 1 0 1 0 Yl0 m0 l ðθ, l S 2 2 2μ 2ν

1 Particles 2

φÞ

The S matrix element SJl0 S0 ,lS can be calculated by the phase shift analysis method of nucleon-nucleon elastic scattering. From the C–G coefficients in the above formula one can also see m0l = μ þ ν - μ 0 - ν 0

ð2:337Þ

where μ and μ0 , ν and ν 0 represent the spin magnetic quantum numbers of the incident neutron and the outgoing neutron, the target proton and the residual nucleus proton, respectively. Introducing the following symbols ^^ 0 iðσ l þσ l 0 Þ

0

f μll0 νJ0 ,μν = l l e

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0

l - m0l ! X J 0

C l þ m0l ! SS0 l0

S0 M 0S M S M J M SM C 12μ 12ν C l0 m0 l S0 M 0S C 1μ0 1ν0 2 2

 δl0 S0 ,lS - SJl0 S0 ,lS ð2:338Þ

and using Eq. (2.253), we can obtain the following expressions from Eq. (2.336): f μ0 ν0 ,μν ðΩÞ = F μ0 ν0 ,μν ðθÞeiðμþν - μ

0

- ν0 Þφ

X 0 i 0 0 - μ0 - ν0 f μll0 νJ0 ,μ ν Pμþν ð cos θÞ F μ0 ν0 ,μν ðθÞ = ð - 1Þμþν - μ - ν l0 2k 0

ð2:339Þ ð2:340Þ

ll J

Considering parity conservation, l and l 0 must be all even or odd numbers, so the following C-G coefficient relations can be obtained: C Jl0

μþν S μþν J μþν S μþν C 1μ 1ν C l0 μþν - μ0 - ν0 2 2

= C Jl0

C S1μ0 μ 1þν ν0 0

S0 μ0 þν0

2

0

0

2

-μ-ν S -μ-ν J -μ-ν S - μ - ν C 1 - μ 1 - ν C l0 - μ - νþμ0 þν0 2 2

S0 - μ 0 - ν0 S0 - μ0 - ν0 C 12 - μ0 12 - ν0

ð2:341Þ By using Eq. (2.287) the following relation can be obtained from Eqs. (2.338) and (2.340): F - μ0 - ν0 , - μ - ν ðθÞ = ð - 1Þμþν - μ

0

- ν0

F μ0 ν0 ,μ ν ðθÞ

*

ð2:342Þ

Taking the y-axis in the direction of the unit vector n perpendicular to the reaction plane, there is φ = 0. According to Eq. (2.342) we define the following matrix elements:

→

2.9 Polarization Theory for

AðθÞ = F 1212,1212 ðθÞ = F - 12

- 12, - 12 - 12 ðθÞ,

C ðθÞ = iF 1212, - 1212 ðθÞ = - iF - 12 E ðθÞ = iF 12 GðθÞ = F 12

- 12, - 12 - 12 ðθ Þ =

- 12,12

- 12

→

79

1 1 þ Reactions 2 2

BðθÞ = iF 1212,12

- 12,12 - 12 ðθ Þ,

- 12 ðθÞ =

DðθÞ = - F 1212, - 12

- iF - 1212,1212 ðθÞ,

ðθÞ = F - 1212, - 1212 ðθÞ,

F ðθÞ = F 12

- iF - 12

- 12 ðθ Þ =

- 12, - 1212 ðθ Þ,

- F - 12

- 12,1212 ðθÞ,

- 12, - 1212 ðθÞ = F - 1212,12 - 12 ðθ Þ,

H ðθÞ = iF - 1212, - 12

- 12 ðθ Þ =

- iF 12

- 12,1212 ðθÞ

ð2:343Þ Then the 4 × 4 reaction matrix given by Eq. (2.340) can be expressed as 0

A

B iH ^ =B F B @ iE -D

- iB

- iC

G F

F G

iC

iB

-D

0

1

- iE C C C, - iH A

A

B iB ^þ = B F B @ iC  - D

A

- D

1

- iH 

- iE 

G F

F G

- iC C C C - iB A

iE 

iH 

A ð2:344Þ

^ have double sets of subscripts: μ0 and The matrix elements F μ0 ν0 ,μν ðθÞ of the matrix F 0 ν correspond to the final states of neutron and proton, respectively, and μ and ν correspond to the initial states of neutron and proton, respectively. μ(μ0 ) and ν(ν0 ) 1 1 ^ is divided into 4 small 2 × 2 matrices, then we take on or - . If the 4 × 4 matrix F 2 2 1 will find that for the upper left 2 × 2 matrix μ0 = μ = ; for the lower right matrix 2 μ0 = μ = - 12 ; for the upper right matrix μ0 = 12 and μ = - 12 ; and for the lower left 1 1 matrix μ0 = and μ = . The second subscripts v0 and v of the four small 2 2 1 1 matrices are all arranged in the same way: in the upper left corner they are ; 2 2 1 1 1 1 in the lower right corner they are - - ; in the upper right corner they are - ; 2 2 2 2 1 1 and in the lower left corner they are . To facilitate the next operation, we 2 2 introduce the following 2 × 2 submatrices: ^t μ0 μ =

F μ0 12, μ12 F μ0 - 12, μ 12

F μ0 12, μ - 12 F μ0 - 12, μ - 12

! ð2:345Þ

1 For the sake of simplicity, in the following the foot note is replaced by + and 2 1 the foot note - is replaced by –; then according to Eq. (2.343) or Eq. (2.344) 2 we can write

80

2

^t þþ =

A - iB

, iH

^t - þ =

!

G

iE

F

- D iC

þ ^tþþ =

! þ , ^t - þ =

Polarization Theory of Nuclear Reactions for Spin

A - iH 

! , ^t þ - =

G

iB

F !

-iE - D - iC

F

-iC - D

, ^t - - =

!

- iE

G - iH iB

A

,

^tþþ-

=

! þ , ^t - - =

1 Particles 2

iC 

F

!

- D iE  G -iB iH 

!

A

ð2:346Þ ^ given by Eq. (2.344) can be expressed as and the matrix F


^t ^t þ ^ = þþ F ^t - þ ^t - -

 ,

þ

^ = F

þ þ ^t ^t þþ þ

^t þþ-

!

þ ^t -

ð2:347Þ

According to Eq. (2.64), the polarization density matrix of the normalized incident particle channel can be written as ^ρin =

  1 ^ I þ pnx σ^nx þ pny σ^ny þ pnz σ^nz ^I þ ppx σ^px þ ppy σ^py þ ppz σ^pz 4

ð2:348Þ

where pni and ppj (i, j = x, y, z) are the polarization rates of the neutrons and protons in the incident channel, respectively. σ^ni ði = x, y, zÞ act on the first set of subscripts μ0 , μ corresponding to the neutron states only, and σ^pj ðj = x, y, zÞ act on the second set of subscripts ν0 , ν corresponding to the proton states only. ^ and B ^ be the n- and m-dimensional square matrices such that their direct Let A product is defined as

^ ×B ^ A

ia,jb

 Aij Bab

ð2:349Þ

Let ^I be a 2 × 2 unit matrix and ^0 be a 2 × 2 zero matrix. Using the above direct product definition, we can obtain σ ni,þþ^I σ ni,þ - ^I σ i,n - þ^I σ i,n - - ^I ! 0 σ^jp ^ p p ^ j  ^I × σ^j = Σ ^0 σ^ p

^ ni  σ^ni × ^I = Σ

! ð2:350Þ

ð2:351Þ

j

^ ni only acts on the subscript μ0 , μ of It can be seen that the resulting 4 × 4 matrix Σ p ^ ni and Σ ^ pj are ^ j only acts on the subscript ν 0 , ν of protons. Σ neutrons, and Σ commutative of each other. Letting ^I 4 be the unit 4 × 4 matrix, then Eq. (2.348) can be rewritten as

→

2.9 Polarization Theory for

^ρin =

→

81

1 1 þ Reactions 2 2

  1 ^ ^ nx þ pny Σ ^ ny þ pnz Σ ^ nz ^I 4 þ ppx Σ ^ px þ ppy Σ ^ py þ ppz Σ ^ pz I 4 þ pnx Σ 4

ð2:352Þ

When both n and p in the incident channel are unpolarized, the differential cross sections of the emitted particles are I0 

n þo dσ 0 1 ^F ^ = tr F dΩ 4

ð2:353Þ

Substituting Eq. (2.344) into the above formula, the following result can be obtained immediately: I0 =

 1 jAj2 þ jBj2 þ jC j2 þ jDj2 þ jEj2 þ jF j2 þ jGj2 þ jH j2 2

ð2:354Þ

The polarization vector components of the outgoing particle n and p corresponding to the unpolarized incident channel are, respectively, n k þo ^F ^ ^ i0 F tr Σ ki0   , P0 = ^F ^þ tr F

i0 = x0 , y0 , z0

k = n,p;

ð2:355Þ

From Eq. (2.347) we can obtain ^F ^þ = F

þ þ ^t þþ^tþþ þ ^tþ - ^t þ -

þ ^t - þ^tþþ

þ þ ^t - - ^t þ -

!

þ ^ ^þ ^t þþ^t þ þ tþ - t - þ ^t - þ^t þ

þ þ ^t - - ^t - -




^ ^þ gþþ g ^-þ ^ g g- -



ð2:356Þ

Moreover using Eq. (2.356) and Eq. (2.346), we obtain ^gþþ = =

^ gþ - = =

A

- iB

iH

G

!

A

- iH 

iB

G

! þ

- iC

-D

F

- iE

!

iC

F

- D

iE

jAj2 þ jBj2 þ jC j2 þ jDj2

- iðAH  þ BG þ CF  þ DE  Þ

iðHA þ GB þ FC þ ED Þ

jH j2 þ jGj2 þ jF j2 þ jE j2

A

- iB

iH

G

!

- iE

- D

F

- iC 

! þ

- iC

-D

F

- iE

!

G iH 

! !

ð2:357Þ ! - iB A

- iðAE  þ BF  þ CG þ DH  Þ

- ðAD þ BC  þ CB þ DA Þ

HE  þ GF  þ FG þ EH 

- iðHD þ GC  þ FB þ EA Þ

!

ð2:358Þ

82

2

^g - þ = =

^g - - = =

iE

F

-D

iC

!

Polarization Theory of Nuclear Reactions for Spin

A

- iH 

iB

G

! þ

G

- iH

iB

A

!

iC

F

- D

iE 

1 Particles 2

!

!

iðEA þ FB þ GC  þ HD Þ

EH  þ FG þ GF  þ HE 

- ðDA þ CB þ BC  þ AD Þ

iðDH  þ CG þ BF  þ AE  Þ

iE

F

-D

iC

!

- iE 

- D

F

- iC 

! þ

G

- iH

iB

A

!

ð2:359Þ ! - iB

G

A

iH 

jEj2 þ jF j2 þ jGj2 þ jH j2

- iðED þ FC  þ GB þ HA Þ

iðDE  þ CF  þ BG þ AH  Þ

jDj2 þ jCj2 þ jBj2 þ jAj2

!

ð2:360Þ It can be seen that substituting Eqs. (2.356), (2.357), and (2.360) into Eq. (2.353) we get Eq. (2.354). First we obtain the following relations by using Eqs. (2.6), (2.350), (2.355), and (2.356) –(2.360): ^F ^þ ^ nx F Σ

=

^0 ^I ^I ^0

!


^gþþ g^ - þ

^gþ g^ - -




=

^ g-þ ^þþ g

^ g- ^þ g

 ð2:361Þ

0

^0 i^I

^F ^þ = ^ ny F Σ

- i^I ^0

!


^gþþ g^ - þ

Pnx 0 =0 
^-þ ^gþ g = -i ^þþ ^g - -g

ð2:362Þ ^ g- -^ gþ -

1 Re ðAE  þ BF  þ CG þ DH  Þ Io !

 ^0 ^ ^ ^gþþ ^gþ gþþ gþ = ^g - þ ^g - -^ g-þ - ^ g- - ^I 0

Pny 0 = ^F ^þ = ^ nz F Σ

^I ^0



0

Pnz 0 =0

ð2:363Þ ð2:364Þ ð2:365Þ ð2:366Þ

Because ^ pj F ^F ^þ Σ

=

σ^pj ^0

^0 σ^pj

!


^gþþ ^g - þ

^gþ ^g - -

 =

σ^pj ^ gþþ

gþ σ^pj ^

σ^pj ^ g-þ

σ^pj ^ g- -

! ð2:367Þ

^ j acting on the 4 × 4 matrix is equivalent to σ^pj acting on the 2 × 2 matrices therefore Σ 0 of its blocks. In order to find Ppi , first let us calculate the following terms: p

→

2.9 Polarization Theory for

σ^px ^gþþ =

0 1

!

→

83

1 1 þ Reactions 2 2

jAj2 þ jBj2 þ jC j2 þ jDj2

- iðAH  þ BG þ CF  þ DE  Þ

!

iðHA þ GB þ FC  þ ED Þ

jH j2 þ jGj2 þ jF j2 þ jE j2 ! iðHA þ GB þ FC  þ ED Þ jH j2 þ jGj2 þ jF j2 þ jE j2 = , jAj2 þ jBj2 þ jCj2 þ jDj2 - iðAH  þ BG þ CF  þ DE  Þ ! ! 0 1 jEj2 þ jF j2 þ jGj2 þ jH j2 - iðED þ FC  þ GB þ HA Þ p σ^x ^g - - = 1 0 iðDE  þ CF  þ BG þ AH  Þ jDj2 þ jC j2 þ jBj2 þ jAj2 ! iðDE  þ CF  þ BG þ AH  Þ jDj2 þ jC j2 þ jBj2 þ jAj2 = jEj2 þ jF j2 þ jGj2 þ jH j2 - iðED þ FC  þ GB þ HA Þ 1 0

ð2:368Þ Substituting Eq. (2.368) into Eq. (2.355), we get 0

Ppx 0 =0

ð2:369Þ

According to Eq. (2.368) we have  1 - i jH j2 þ jGj2 þ jF j2 þ jE j2 HA þ GB þ FC  þ ED C B σ^py ^gþþ = @  A, 2 2 2 2    i jAj þ jBj þ jC j þ jDj AH þ BG þ CF þ DE   1 0 - i jDj2 þ jC j2 þ jBj2 þ jAj2 DE  þ CF  þ BG þ AH  C B σ^py ^g - - = @  A 2 2 2 2     i jEj þ jF j þ jGj þ jH j ED þ FC þ GB þ HA 0

ð2:370Þ Substituting Eq. (2.370) into Eq. (2.355), the following result is obtained: 0

Ppy 0 =

1 Re ðAH  þ BG þ CF  þ DE  Þ I0

ð2:371Þ

According to Eq. (2.368) we have 0

1 - iðAH  þ BG þ CF  þ DE  Þ A  σ^pz ^gþþ = @ - iðHA þ GB þ FC  þ ED Þ - jH j2 þ jGj2 þ jF j2 þ jEj2 , 0 1 2 2 2 2     jE j þ jF j þ jGj þ jH j - iðED þ FC þ GB þ HA Þ A  σ^pz ^g - - = @ - iðDE  þ CF  þ BG þ AH  Þ - jDj2 þ jC j2 þ jBj2 þ jAj2 jAj2 þ jBj2 þ jCj2 þ jDj2

ð2:372Þ

84

2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2

Substituting Eq. (2.372) into Eq. (2.355), we get 0

Ppz 0 =0

ð2:373Þ

0

0

0

0

px pz py ny 0 nz0 The above results show that Pnx 0 = P0 = P0 = P0 = 0, P0 ≠ 0, P0 ≠ 0: The polarization analyzing power for n - p elastic scattering is defined as

n k þo ^ ^Σ ^i F tr F Aki = n þ o , ^F ^ tr F

k = n,p;

n n p þo ^j F ^ ^Σ ^i Σ tr F n o , = Anp ij ^F ^þ tr F

i = x, y, z

ð2:374Þ

i,j = x, y, z

ð2:375Þ

where Anp ij can also be called the correlation analyzing power [10]. We can obtain the following expression from Eqs. (2.6), (2.347), and (2.350): ^Σ ^ nx F ^þ = F = =

^t þþ

^t þ -

^t - þ ^t - ^t þþ

^t þ -

! !

^t - þ ^t - -

^0 ^I ^I ^0

!

þ ^t þþ

^t þ -þ

^t þþ-

þ ^t þþ- ^t ! ^t þ --

þ ^t þþ

^t þ -þ

!

þ ^ ^þ ^t þþ^t þ ^t þþ^t þþ- þ ^t þ - ^t þþ - - þ tþ - t - þ þ ^ ^þ ^t - þ^t þþ- þ ^t - - ^t þþ ^t - þ^t þ - - þ t- - t-þ

! 

^ nþþ X ^ nþ X

!

^ n- þ X ^ n- X

ð2:376Þ We can get the following results using Eqs. (2.376), (2.346), and (2.374): 

  
- iC - D A F - iH  A - iB iC þ - D iE iB G  F - iE
iH G       AF þ BE - CH - DG iðAC þ BD - CA - DB Þ =     - ðHC þ GD - FA - EB Þ iðHF  þ GE  - FH  - EG Þ

n ^ þþ X =

^ þn X




ð2:377Þ  

- iC - D - iE - D A - iB G - iB þ =    F - iC F - iE 
iH  G  iH  A      - iðAB þ BA - CD - DC Þ AG þ BH - CE - DF = HB þ GA - FD - EC iðHG þ GH  - FE  - EF  Þ





ð2:378Þ

→

2.9 Polarization Theory for

^ n- þ = X =

→

85

1 1 þ Reactions 2 2



  
G - iH A F - iH  iE F iC þ - D iE  iB G iB A - D iC
        - ðEC þ FD - GA - HB Þ iðEF þ FE - GH - HG Þ




X^ n- - = =

- iðDC  þ CD - BA - AB Þ











- iB iE F G þ  iH A - D iC
iðEG þ FH  - GE  - HF  Þ - ðDG þ CH  - BE  - AF  Þ

- ðDF  þ CE  - BH  - AG Þ


ð2:379Þ  G - iH - iE - D F - iC iB A     EB þ FA - GD - HC  




iðDB þ CA - BD - AC  Þ ð2:380Þ

^Σ ^ ny F ^þ = F =i

^t þ -

^t þþ

^t - þ ^t - ^t þþ

^t þ -

! !

^t - þ ^t - -

= -i

Axn

!

=0 ^t þ þþ

^0

- i^I

i^I

^0

^t þ þ-

- ^t þ -

- ^t - -

þ ^t þþ

^t þ -þ

þ

þ ^t þþ^t þþ- - ^t þ - ^t þþ

ð2:381Þ

þ

^t þ- þ !

!

^t þ- -

^t þþ^t þ- - - ^t þ - ^t þ- þ

^ ^þ ^t - þ^t þ- - - ^t - - ^t þ- þ ^t - þ^t þ þ - - t - - t þþ

! 

n n Y^ þþ Y^ þ -

!

n n Y^ - þ Y^ - -

ð2:382Þ We can also get the following results using Eqs. (2.382), (2.346), and (2.374):  



- iC - D A - iH  A - iB iC  F  n Y^ þþ þ i = -i - D iE iB G F - iE iH G
      - iðAF þ BE  þ CH þ DG Þ AC þ BD þ CA þ DB = iðHC  þ GD þ FA þ EB Þ HF  þ GE  þ FH  þ EG

Y^ þn -

ð2:383Þ 

 - iC - D - iE - iB - D A - iB G þ i = -i iH  A F - iC  F - iE iH G 
- iðAG þ BH  þ CE  þ DF  Þ - ðAB þ BA þ CD þ DC  Þ = HG þ GH  þ FE  þ EF  - iðHB þ GA þ FD þ EC  Þ











ð2:384Þ

86

2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2



 

G - iH A F - iH  iE F iC n ^ þi Y -þ = - i - D iE iB G iB A - D iC
        EF þ FE þ GH þ HG iðEC þ FD þ GA þ HB Þ = - ðDC  þ CD þ BA þ AB Þ iðDF  þ CE  þ BH  þ AG Þ

Y^ n- -

ð2:385Þ



   G - iH - iE - iB - D iE F G þi = -i iH  A F - iC  iB A - D iC
        - iðEB þ FA þ GD þ HC  Þ EG þ FH þ GE þ HF = iðDG þ CH  þ BE  þ AF  Þ DB þ CA þ BD þ AC  ð2:386Þ 1 ReðAC  þ BD þ EG þ FH  Þ I0 ! ! þ þ ! ^0 ^t ^t þþ ^I ^t þ þ þ ^0 - ^I ^t - ^tþ ^ t- þ! ! þ þ ^ ^ ^tþ t-þ t þþ

Ayn = ^t þþ

^Σ ^ nz F ^þ = F

^t - þ ^tþþ

=

þ

^t - þ ^t - -

=

þ

- ^t þ -

þ þ ^t þþ^t þþ - ^t þ - ^t þ -

^ ^þ ^t - þ^tþ þþ - t - - t þ -

ð2:387Þ

- ^t - -

!

^tþþ^tþ- þ - ^tþ - ^tþ- ^t - þ^t þ- þ - ^t - - ^t þ- þ



!

n n Z^ þþ Z^ þ n n Z^ - þ Z^ - -

ð2:388Þ The following results can be got using Eqs. (2.388), (2.346), and (2.374): n Z^ þþ =




=

Z^ þn - =





=

A

- iB

iH

G




A

- iH 

iB

G




-

jAj2 þ jBj2 - jC j2 - jDj2 iðHA þ GB - FC  - ED Þ A

- iB




- iE 



-D

 

-D

F

- iE




iC

F

- D

iE 

- iðAH  þ BG - CF  - DE  Þ jH j2 þ jGj2 - jF j2 - jE j2



iH G F - iC - iðAE  þ BF  - CG - DH  Þ

- iC


-

- iC

-D




G



 !

ð2:389Þ  - iB

F - iE iH  A     - ðAD þ BC - CB - DA Þ

- iðHD þ GC  - FB - EA Þ ð2:390Þ



    - iH F G - iH iE F iC A Z^ n- þ =    iB G - D iE  iB A - D iC
 iðEA þ FB - GC  - HD Þ EH  þ FG - GF  - HE  = - ðDA þ CB - BC  - AD Þ iðDH  þ CG - BF  - AE  Þ ð2:391Þ HE  þ GF  - FG - EH 

→

2.9 Polarization Theory for

Z^ n- - =




=

iE -D

F iC

→

87

1 1 þ Reactions 2 2




- iE  F

- D - iC




-

G iB

- iH A




G iH 

- iB A

- iðED þ FC  - GB - HA Þ

jE j2 þ jF j2 - jGj2 - jH j2 iðDE  þ CF  - BG - AH  Þ

 !

jDj2 þ jCj2 - jBj2 - jAj2 ð2:392Þ

=0

Azn

ð2:393Þ

We can write the following expressions according to Eqs. (2.356), (2.351), (2.346), and (2.374): ^Σ ^ px F ^þ = F

p = X^ þþ




!

^ ^px^t þ- þ þ ^t þ - σ^px^t þ- ^t þþ σ^px^t þ ^ ^px^t þ þþ þ t þ - σ þ - t þþ σ

^t - þ σ^px^t þ ^ ^px^t þ- þ þ ^t - - σ^px^t þ- ^ ^px^t þ þþ þ t - - σ þ - t - þσ 0 p 1 ^ pþ ^ þþ X X A @ p ^ -þ X ^ p- X

A

- iB




0 1




A

- iH 






- iC - D




ð2:394Þ

0 1




þ iB G F - iE 1 0 iH G 1 0 
 

 A - iB iB G - iC - D - D iE = þ iH G F - iE A - iH  iC  F 
       AG - BH þ CE  - DF  iðAB - BA þ CD - DC Þ = - ðHB - GA þ FD - EC  Þ iðHG - GH  þ FE  - EF  Þ




























p ^X þ=

iE

F

0 1















A 



F

- iH 

þ

G - iH

0 1



- D iE

ð2:395Þ 

- iC -D 0 1 G - iB A - iB 0 1 - iE - D þ   F - iE 1 0 iH G 1 0 F - iC iH  A


 
 - iC -D iH  A A - iB F - iC  þ =   F -iE iH G -iE -D G -iB 
      AF -BE þ CH - DG - iðAC - BD þ CA -DB Þ = iðHF  - GE  þ FH  - EG Þ HC - GD þ FA - EB

X^ þp - =



iC 



ð2:396Þ  iC  F  - D iE

iB G - D iC 1 0 iB A 1 0 

 
 G - iH - D iE  G iE F iB þ = A - iH  iC  F  iB A - D iC 
      - ðEB - FA þ GD - HC Þ iðEG - FH þ GE  - HF  Þ = - iðDB - CA þ BD - AC  Þ - ðDG - CH  þ BE  - AF  Þ


ð2:397Þ

88

2

p ^X -=




iE

F




0 1




Polarization Theory of Nuclear Reactions for Spin

- iE 

- D






G - iH




0 1




þ F - iC iB A 1 0  

  - iC A G - iH iH = þ - iE  - D G - iB - D iC iB A
 iðEF  - FE  þ GH  - HG Þ EC  - FD þ GA - HB = - ðDF  - CE  þ BH  - AG Þ iðDC - CD þ BA - AB Þ


- D iC 1 0 
 iE F F

1 Particles 2

G

- iB

iH 

A



ð2:398Þ Axp

=0

ð2:399Þ

We can also obtain the following expressions according to Eqs. (2.356), (2.351), (2.346), and (2.374): 0 ^þ ^Σ ^ py F F

=@



1

^t þþ σ^py^t þ ^ ^py^t þ ^ ^py^t þ- þ þ ^t þ - σ^py^t þ- þþ þ t þ - σ þ - t þþ σ þ ^t - þ σ^py^t þ ^ ^py^t þ þþ þ t - - σ ! p p Y^ þþ Y^ þ -

^t - þ σ^py^t þ- þ þ ^t - - σ^py^t þ- -

A ð2:400Þ

p p Y^ - þ Y^ - -




 


 A - iB 0 -i A - iH  - iC - D 0 -i iC  F  p Y^ þþ = þ iH G i 0 F - iE i 0 iB G - D iE 

 
 
  A - iB B - iG - iC - D iD E = þ iH G F - iE iA H  - C  iF  
AB þ BA þ CD þ DC  - iðAG þ BH  þ CE  þ DF  Þ = iðHB þ GA þ FD þ EC  Þ HG þ GH  þ FE þ EF 

ð2:401Þ





  A - iB 0 -i - iE - D - iC - D 0 -i G - iB þ Y^ þp - = iH G i 0 F - iE i 0 F  - iC iH  A


   
 A - iB - iC - D - iF - C H - iA = þ E - iD iG B iH G F - iE 
- iðAF  þ BE  þ CH  þ DG Þ - ðAC  þ BD þ CA þ DB Þ = HF  þ GE  þ FH  þ EG - iðHC  þ GD þ FA þ EB Þ










A












- iH G - iH 0 -i þ iB G - D iC iB A i 0 i 0



    iE F E G - iH B - iG iD = þ iA H  - C  iF  - D iC iB A 
iðEB þ FA þ GD þ HC  Þ EG þ FH  þ GE  þ HF  = - ðDB þ CA þ BD þ AC  Þ iðDG þ CH  þ BE  þ AF  Þ

p Y^ þ=

iE

F

0 -i

ð2:402Þ  iC F  - D iE 

ð2:403Þ

→

→

89

1 1 þ Reactions 2 2

2.9 Polarization Theory for




  

G - iH 0 -i G - iB iE F 0 -i - iE - D þ F  - iC  iH  A iB A i 0 - D iC i 0 



G - iH H  - iA iE F - iF  - C  þ =   iB A E - iD iG B - D iC 
      EF þ FE þ GH þ HG - iðEC þ FD þ GA þ HB Þ = iðDF  þ CE  þ BH  þ AG Þ DC  þ CD þ BA þ AB

p Y^ -=




ð2:404Þ Apy =

1 Re ðAB þ CD þ EF  þ GH  Þ I0

ð2:405Þ

And we can also obtain the following expressions according to Eqs. (2.356), (2.351), (2.346), and (2.374): ^þ ^Σ ^ pz F F

= 

p Z^ þþ =




A

!

^ ^pz ^t þ ^t þþ σ^pz ^t þ þþ þ t þ - σ þ-

^t þþ σ^pz^t þ- þ þ ^t þ - σ^pz ^t þ- -

þ ^ ^t - þ σ^pz ^t þ ^pz ^t þ þþ þ t - - σ ! p p Z^ þþ Z^ þ -

^t - þ σ^pz^t þ- þ þ ^t - - σ^pz ^t þ- -

- iB

p Z^ - þ




1

p Z^ - -

0




A

- iH 




þ

- iC

-D




1

0

ð2:406Þ




iC 

iB G iH G F - iE 0 -1 0 -1 

  
 - iH  F - iC - D iC A - iB A þ = F - iE iH G - iB - G D - iE ! - iðAH  - BG þ CF  - DE  Þ jAj2 - jBj2 þ jCj2 - jDj2 = jH j2 - jG j2 þ jF j2 - jE j2 iðHA - GB þ FC  - ED Þ

F



- D iE




ð2:407Þ Z^ þp - =




A - iB




1

0




- iE



-D






þ

- iC - D




1




0

F  - iC  iH G F - iE 0 -1 0 -1 


   - iB A - iB - iE - D - iC - D G = þ iH G F - iE - F  iC  - iH  - A
       - iðAE - BF þ CG - DH Þ - ðAD - BC þ CB - DA Þ = HE  - GF  þ FG - EH  - iðHD - GC  þ FB - EA Þ





















G - iH 1 0 iE F 1 0 A - iH þ   iB A 0 -1 - D iC 0 -1 iB G 


 
 G - iH iC iE F A - iH  F þ = iB A - D iC - iB - G D - iE 
       iðEA - FB þ GC - HD Þ EH - FG þ GF  - HE  = - ðDA - CB þ BC  - AD Þ iðDH  - CG þ BF  - AE Þ

p Z^ þ=




G



iH 

- iB



A

ð2:408Þ 

iC F - D iE 



ð2:409Þ

90

2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2






  iE F 1 0 - iE  - D G - iH 1 0 G - iB þ - D iC 0 -1 iB A 0 -1 F - iC iH  A



    - iB iE F G - iH - iE - D G = þ    -F iC - iH - A - D iC iB A ! 2 2 2 2 jE j - jF j þ jG j - jH j - iðED - FC  þ GB - HA Þ = iðDE  - CF  þ BG - AH  Þ jDj2 - jC j2 þ jBj2 - jAj2

p Z^ -=




ð2:410Þ Apz

=0

ð2:411Þ

Refer to Eq. (2.376) we can write ^ pi F ^þ ^Σ ^ nx Σ F

^t þþ σ^pi ^t þ ^ ^pi ^t þ þ - þ tþ - σ þþ

=

^t þþ σ^pi ^t þ- - þ ^t þ - σ^pi ^t þ- þ

!

^t - þ σ^p^t þ þ ^t - - σ^p^t þ ^t - þ σ^pi ^t þ- - þ ^t - - σ^pi ^t þ- þ 0 i i þ -i 1 i þþ ^ þþ X ^þ X A, i = x, y, z @ i i ^ ^ X -þ X - -

ð2:412Þ


 


 
 A - iB F - iC - D 0 1 0 1 iC A - iH  þ - D iE  iB G iH G F - iE 1 0 1 0



     A - iB G - iC - D - D iE iB = þ    iC F A - iH  iH G F - iE
       - ðAD - BC - CB þ DA Þ iðAE - BF - CG þ DH  Þ = - iðHD - GC  - FB þ EA Þ - ðHE  - GF  - FG þ EH  Þ

x ^ þþ X =




ð2:413Þ





 A - iB 0 1 G - iB - iC - D 0 1 - iE - D ^ þx - = þ X iH G 1 0 F - iE 1 0 iH  A F - iC



    A - iC - iC - D F A - iB iH þ =    F - iE iH G G - iB - iE - D 0 1 2 2     iðAH - BG - CF þ DE Þ jAj - jBj - jCj2 þ jDj2 A =@  2 - jH j - jGj2 - jF j2 þ jEj2 iðHA - GB - FC  þ ED Þ

^ x- þ = X







iE

F







0 1




iC



F 


þ

G

- iH




0 1




A



- D iE iB - D iC iB A 1 0 1 0 


     iE F G G - iH - D iE iB = þ    iC F A - iH  - D iC iB A 0 1  - iðED - FC - GB þ HA Þ - jE j2 - jF j2 - jGj2 þ jH j2 A =@ - iðDE  - CF  - BG þ AH  Þ j D j 2 - j C j 2 - j B j 2 þ jA j 2






ð2:414Þ  - iH  G

ð2:415Þ

→

2.9 Polarization Theory for x ^X -=




iE

F




→

1 1 þ Reactions 2 2

0 1




G

- iB

91






G - iH




0 1




- iE 

- D

þ iH  A F iB A 1 0 

   A - iC G - iH F = þ G - iB - iE  - D - D iC iB A
 - ðEH  - FG - GF  þ HE  Þ iðEA - FB - GC  þ HD Þ = - iðDH  - CG - BF  þ AE  Þ - ðDA - CB - BC  þ AD Þ




- iC 

- D iC 1 0 
 iE F iH

ð2:416Þ np =Axx

ð2:417Þ






  A - iB 0 -i iC  F  - iC - D 0 -i A - iH  þ iH G i 0 F - iE i 0 - D iE  iB G



    E A - iB - iC - D iD B - iG = þ - C  iF  iA H  iH G F - iE 
AE  þ BF  - CG - DH  iðAD þ BC  - CB - DA Þ = - ðHD þ GC  - FB - EA Þ iðHE  þ GF  - FG - EH  Þ

y ^ þþ = X




1 ReðAD - BC  þ EH  - FG Þ I0

ð2:418Þ





















A - iB 0 -i G - iB - iC - D 0 -i þ iH G i 0 F - iE i 0 iH  A

 

 A - iB H - iA - iC - D - iF  - C  = þ iH G F - iE iG B E - iD 0  1 AH  þ BG - CF  - DE  - i jAj2 þ jBj2 - jCj2 - jDj2 C B =@  A 2 2 2 2     HA þ GB - FC - ED i jH j þ jGj - jF j - jEj

X^ þy - =

- iE F



- D - iC 



ð2:419Þ













































- iH  G

F iE F 0 -i iC G - iH 0 -i A þ - D iC i 0 iB A i 0 - D iE  iB 

 

- iG G - iH B iE F iD E þ = iB A - D iC - C  iF  iA H 0  1 - ðED þ FC  - GB - HA Þ i jEj2 þ jF j2 - jGj2 - jH j2 B C =@  A 2 2 2 2     - ðDE þ CF - BG - AH Þ - i jDj þ jCj - jBj - jAj

^ y- þ = X



ð2:420Þ 

G - iH 0 -i iE F 0 -i G - iB þ iB A i 0 - D iC i 0 iH  A

 

 iE F H - iA G - iH - iF  - C = þ - D iC iB A iG B E - iD
        EA þ FB - GC - HD iðEH þ FG - GF - HE Þ = - ðDH  þ CG - BF  - AE  Þ iðDA þ CB - BC  - AD Þ

y ^X -=

- iE F



- D - iC 



ð2:421Þ

92

2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2

np Axy =0


















ð2:422Þ







A - iB 1 0 iC  F  - iC - D 1 0 þ iH G 0 -1 F - iE 0 -1 - D iE 

 

 F - iH  - iC - D A A - iB iC þ = F - iE iH G D - iE - iB G
         AF - BE - CH þ DG iðAC - BD - CA þ DB Þ = - ðHC  - GD - FA þ EB Þ iðHF  - GE  - FH  þ EG Þ

z ^ þþ = X




A - iH  iB G



ð2:423Þ 












- iC - D 1 0 A - iB 1 0 G - iB þ F - iE 0 -1 iH G 0 -1 iH  A



    A - iB G - iB - iC - D - iE - D = þ iH G F - iE - iH  - A - F  iC  
      - iðAB - BA - CD þ DC  Þ AG - BH - CE þ DF =  HB - GA - FD - EC  iðHG - GH  - FE  þ EF  Þ

X^ þz - =




- iE F



- D - iC



ð2:424Þ
































F iE F 1 0 iC G - iH 1 0 þ   - D iC 0 -1 iB A 0 -1 - D iE

 

  iE F iC F - iH  G - iH A = þ - D iC iB A D - iE  - iB - G
       - ðEC - FD - GA þ HB Þ iðEF - FE - GH  þ HG Þ = - iðDC  - CD - BA þ AB Þ - ðDF  - CE  - BH  þ AG Þ

z X^ þ=



A - iH  iB G



ð2:425Þ 


 iE F 1 0 G - iB G - iH 1 0 - iE - D z ^= þ X - D iC 0 -1 iB A 0 -1 iH  A F - iC



    - iB iE F G - iH G - iE - D = þ   - iH - A - F  iC - D iC iB A 
      EB - FA - GD þ HC  iðEG - FH - GE þ HF Þ = - ðDG - CH  - BE  þ AF  Þ iðDB - CA - BD þ AC  Þ

ð2:426Þ Axznp = -

1 ImðAC  - BD þ EG - FH  Þ I0

ð2:427Þ

We can write the following results according to Eqs. (2.382), (2.346), and (2.375):

→

2.9 Polarization Theory for

^Σ ^ ny Σ ^ pi F ^þ F

→

93

1 1 þ Reactions 2 2

þ ^t þþ σ^ip ^t þþ- - ^t þ - σ^ip ^t þþ

= -i

!

þ þ ^t þþ σ^ip ^t ^ ^ip ^t - - tþ - σ þ

þ þ þ ^ ^t - þ σ^ip ^t þþ- - ^t - - σ^ip ^t þþ ^t - þ σ^ip ^t ^ip ^t - þ - -t- -σ 1 i i Y^ þþ Y^ þ A, i = x, y, z @ i i Y^ - þ Y^ - -

0

ð2:428Þ








































F - iC - D 0 1 A - iB 0 1 iC þi   F - iE 1 0 iH G 1 0 - D iE



  A - iB - D iE  G - iC - D iB = -i þi iH G F - iE iC F  A - iH 
        AE - BF þ CG - DH  iðAD - BC þ CB - DA Þ = - ðHD - GC  þ FB - EA Þ iðHE  - GF  þ FG - EH  Þ

x = -i Y^ þþ



A - iH  iB G



ð2:429Þ 


G - iB - iC - D 0 1 - iE - D þi   F - iE 1 0 iH A F - iC 

    A - iC - iC - D F =-i þi    G - iB - iE - D iH G F - iE 0  1 AH  - BG þ CF  - DE  - i jAj2 - jBj2 þ jCj2 - jDj2 C B =@  A 2 2 2 2     i jH j - jG j þ jF j - jE j HA - GB þ FC - ED

Y^ þx - = - i

A - iB 0 1 iH G 1 0

 A - iB iH



ð2:430Þ Y^ x- þ = - i




iE

F




0 1




iC 

F




þi

G

- iH




0 1




A - iH 

- D iE  iB - D iC iB A 1 0 1 0 


     G iE F - D iE G - iH iB = -i þi - D iC iB A iC F  A - iH  0  1 - ðED - FC  þ GB - HA Þ i jE j2 - jF j2 þ jGj2 - jH j2 C B =@  A 2 2 2 2     - ðDE - CF þ BG - AH Þ - i jDj - jCj þ jBj - jAj




G

ð2:431Þ



















iE F 0 1 G - iB G - iH 0 1 - iE - D þi   - D iC 1 0 iB A 1 0 iH A F - iC 



    A - iC G - iH F iE F iH þi = -i iB A - D iC G - iB - iE  - D 
EA - FB þ GC  - HD iðEH  - FG þ GF  - HE  Þ = - ðDH  - CG þ BF  - AE  Þ iðDA - CB þ BC  - AD Þ

x Y^ - = -i



ð2:432Þ np Ayx =0

ð2:433Þ

94

2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2



 


  F A - iB 0 -i iC - iC - D 0 -i A - iH  þ i iH G i 0 F - iE i 0 - D iE iB G



    E A - iB - iC - D iD B - iG =-i þi   - C iF iA H  iH G F - iE 
      - iðAE þ BF þ CG þ DH  Þ AD þ BC þ CB þ DA = iðHD þ GC  þ FB þ EA Þ HE  þ GF  þ FG þ EH 

y =-i Y^ þþ




ð2:434Þ Y^ þy - =- i




A - iB




0 -i




G





- iB


þi

- iC - D




0 -i




iH  A iH G F - iE i 0 i 0 


   A - iB H - iA - iC - D - iF - C =- i þ i iH G F - iE iG B E  - iD 0 1  - iðAH  þ BG þ CF  þ DE  Þ - jAj2 þ jBj2 þ jCj2 þ jDj2 A =@ 2 2 2 2     - iðHA þ GB þ FC þ ED Þ jH j þ jGj þ jF j þ jE j

F




y Y^ þ = -i

iE

F

!

0 -i

!

iC 

F

!

G

- iH

!

0 -i

þi - D iE iB A i 0 ! ! ! !     E G - iH B - iG iE F iD þi = -i - C  iF  iA H  iB A - D iC 0 1 iðED þ FC þ GB þ HA Þ jEj2 þ jF j2 þ jGj2 þ jH j2 A =@  2 - jDj þ jCj2 þ jBj2 þ jAj2 iðDE þ CF  þ BG þ AH  Þ

y Y^ -

=- i =- i =

- D iC

iE

F

- D iC iE

F

- D iC

! !

i

0

0 -i i

!

0

H  - iA iG

G



- iB A

iH  ! 

B

EH  þ FG þ GF  þ HE  iðDH  þ CG þ BF  þ AE  Þ

! 

þi

þi

G - iH

G - iH iB !

!

A

0 -i i

- iF  - C

0 !

!

- iE

!



- D



- iC

ð2:435Þ A

- iH 

iB

G

!

ð2:436Þ - iE  - D F

!

- iC

- iD ! - iðEA þ FB þ GC  þ HD Þ iB

A

E

DA þ CB þ BC þ AD

ð2:437Þ np = Ayy

1 ReðAD þ BC  þ EH  þ FG Þ I0

ð2:438Þ

→

2.9 Polarization Theory for

A - iB

z =- i Y^ þþ

!

1

→

95

1 1 þ Reactions 2 2

!

0

iC 

F

!

- iC - D

!

1

!

0

þi F - iE 0 -1 - D iE  ! ! ! !    - iC - D A F - iH  A - iB iC þi =- i F - iE D - iE  -iB - G iH G ! AC  - BD þ CA - DB - iðAF  - BE  þ CH  - DG Þ = HF  - GE  þ FH  - EG iðHC  - GD þ FA - EB Þ iH

A -iB

z = -i Y^ þ-

!

1

0

!

G





-iB

!

-iC -D

!

1

0

!

þi F -iE 0 -1 iH  A ! ! ! !    -iC -D -iE -D A -iB G -iB þi = -i F -iE -iH  -A -F  iC iH G ! -iðAG -BH  þ CE  -DF  Þ - ðAB -BA þ CD -DC  Þ = -iðHB -GA þ FD -EC  Þ HG -GH  þ FE  -EF  iH

z Y^ þ =- i

iE

iB

0 -1

G

A - iH 

!

F

1

!

0

iC



F

!

G - iH

!

1

0

!

þi iB A 0 -1 - D iE  ! ! ! !    G - iH A iE F iC F - iH  þi =- i iB A - D iC D - iE  - iB - G ! iðEC  - FD þ GA - HB Þ EF  - FE  þ GH  - HG = - ðDC  - CD þ BA - AB Þ iðDF  - CE  þ BH  - AG Þ - D iC

0 -1

G

ð2:439Þ -iE -D F

0 -1

G

!

-iC

ð2:440Þ A



- iH 

iB

!

G

ð2:441Þ


















iE F 1 0 G - iB G - iH 1 0 þi - D iC 0 -1 iB A 0 -1 iH  A 


 
- iB iE F G - iH G - iE  - D =- i þ i - iH  - A - F  iC - D iC iB A
       - iðEB - FA þ GD - HC  Þ EG - FH þ GE - HF = iðDG - CH  þ BE  - AF  Þ DB - CA þ BD - AC 

z Y^ - =- i




- iE F



- D - iC



ð2:442Þ Ayznp = 0

ð2:443Þ

And we can obtain the following results according to Eqs. (2.388), (2.346), and (2.375):

!

96

2

^Σ ^ nz Σ ^ ip F ^þ = F

x Z^ þþ =




Polarization Theory of Nuclear Reactions for Spin

^ ^ip ^t þþ^t þþ σ^ip ^t þ þ þ - tþ - σ þ

!

^t þþ σ^ip ^t -þ þ - ^t þ - σ^ip ^t -þ -

þ

þ

þ

^t - þ σ^ p ^t - ^t - - σ^ip ^t þ - ^t - þ σ^ip ^t - þ - ^t - - σ^ip ^t - 0 i i þþ 1 i Z^ þþ Z^ þ A, i = x, y, z @ i i Z^ - þ Z^ - -

A

- iB




0 1




A - iH 






-

1 Particles 2

- iC - D




0 1




ð2:444Þ

iC  F 



iB G - D iE  iH G F - iE 1 0 1 0 
 
 
  G A - iB iB - iC - D - D iE = iH G F - iE A - iH  iC F  
      AG - BH - CE  þ DF  iðAB - BA - CD þ DC Þ = - ðHB - GA - FD þ EC  Þ iðHG - GH  - FE  þ EF  Þ


ð2:445Þ Z^ þx - =




A

- iB




0 1






- iE - D








- iC

-D




0 1




F  - iC iH G F - iE 1 0 1 0 

  
 - iC A - iC - D iH A - iB F = F - iE iH G - iE  - D G - iB 
      AF - BE - CH þ DG - iðAC - BD - CA þ DB Þ = iðHF  - GE  - FH  þ EG Þ HC  - GD - FA þ EB


G



iH 

- iB



A

ð2:446Þ























F iE F 0 1 A - iH G - iH 0 1 iC - D iC 1 0 iB A 1 0 iB G - D iE 


  
  G G - iH - D iE iE F iB = iB A - D iC A - iH  iC F 
       - ðEB - FA - GD þ HC Þ iðEG - FH - GE  þ HF  Þ = - iðDB - CA - BD þ AC  Þ - ðDG - CH  - BE  þ AF  Þ

x Z^ þ=



ð2:447Þ



















G - iH 0 1 iE F 0 1 - iE  - D iB A 1 0 - D iC 1 0 F - iC

 

  iE F F - iC A G - iH iH = - D iC iB A - iE - D G - iB
       EC - FD - GA þ HB iðEF - FE - GH þ HG Þ = - ðDF  - CE  - BH  þ AG Þ iðDC  - CD - BA þ AB Þ

x Z^ - =

G - iB iH  A



ð2:448Þ Azxnp = -

1 ImðAB - CD þ EF  - GH  Þ I0

ð2:449Þ

→

2.9 Polarization Theory for

97



 


 A - iB 0 -i A - iH  - iC - D 0 -i iC  F  iH G i 0 F - iE i 0 iB G - D iE

 
 
  E A - iB - iC - D B - iG iD =    iA H - C iF  iH G F - iE 
      - iðAG þ BH - CE  - DF  Þ AB þ BA - CD - DC = iðHB þ GA - FD - EC  Þ HG þ GH  - FE  - EF 

y = Z^ þþ




→

1 1 þ Reactions 2 2

ð2:450Þ Z^ þy - =




A

- iB




0 -i




- iE



-D






-

- iC - D




0 -i




F - iC iH G F - iE i 0 i 0 

   
 A - iB - iF - C - iC - D H - iA = iH G F - iE E - iD iG B
 - iðAF  þ BE  - CH  - DG Þ - ðAC  þ BD - CA - DB Þ = HF  þ GE  - FH  - EG - iðHC  þ GD - FA - EB Þ




G

- iB

iH 

A



ð2:451Þ y Z^ þ=




iE

F




0 -i




A

- iH




-

G






- iH




0 -i




iB G - D iC iB A i 0 i 0 
 
 
G - iH iD E iE F B - iG = iB A - D iC iA H  - C  iF 
 iðEB þ FA - GD - HC  Þ EG þ FH  - GE  - HF  = - ðDB þ CA - BD - AC  Þ iðDG þ CH  - BE  - AF  Þ


iC 

F



- D iE 

ð2:452Þ











iE F 0 -i - iE  - D G - iH 0 -i - D iC i 0 iB A i 0 F - iC  

 

G - iH H - iA iE F - iF  - C = iB A - D iC E - iD iG B
        - iðEC þ FD - GA - HB Þ EF þ FE - GH - HG = DC  þ CD - BA - AB iðDF  þ CE  - BH  - AG Þ

y Z^ -=




G - iB iH  A



ð2:453Þ Azynp = 0

ð2:454Þ



 


  F A - iB 1 0 A - iH  - iC - D 1 0 iC iH G 0 -1 F - iE 0 -1 iB G - D iE 

 
 
  - iH F A - iB A - iC - D iC = iH G F - iE - iB - G D - iE ! jAj2 - jBj2 - jC j2 þ jDj2 - iðAH  - BG - CF  þ DE  Þ = iðHA - GB - FC  þ ED Þ jH j2 - jG j2 - jF j2 þ jE j2

z Z^ þþ =




ð2:455Þ

98

2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2






  A - iB 1 0 - iE - D - iC - D 1 0 G - iB iH G 0 -1 F - iE 0 -1 F - iC  iH  A


  
 - iB A - iB - iC - D - iE - D G =    -F iC - iH - A iH G F - iE
       - iðAE - BF - CG þ DH Þ - ðAD - BC - CB þ DA Þ = HE  - GF  - FG þ EH  - iðHD - GC  - FB þ EA Þ

Z^ þz - =




ð2:456Þ z Z^ þ=




iE

F




1

0




A



- iH




-

G






- iH




1




0

iB G - D iC iB A 0 -1 0 -1 
 
 
  - iH F iE F A G - iH iC = - D iC iB A - iB - G D - iE  
EH  - FG - GF  þ HE  iðEA - FB - GC  þ HD Þ = - ðDA - CB - BC  þ AD Þ iðDH  - CG - BF  þ AE  Þ

iC



F



- D iE




ð2:457Þ z Z^ -=





= =

iE

F




1

0




- iE 

- D

-

G

- iH




1

0




F - iC - D iC iB A 0 -1 0 -1 
 
 
- iB G - iH G iE F - iE - D iB A - D iC - F  iC  - iH  - A ! - iðED - FC  - GB þ HA Þ jEj2 - jF j2 - jGj2 þ jH j2 iðDE  - CF  - BG þ AH  Þ

G

- iB

iH 

A

jDj2 - jCj2 - jBj2 þ jAj2

ð2:458Þ 1  2 Azznp = jAj - jBj2 - jC j2 þ jDj2 þ jEj2 - jF j2 - jGj2 þ jH j2 2I 0

ð2:459Þ

Based on the non-zero analyzing powers obtained above, the differential cross section of the emitted particles can be obtained by using Eq. (2.352) in the case of the polarized incident particle channel as X X   dσ np ^ þ = I0 1 þ ^ ρ^in F pky Ayk þ pny ppy Ayy þ pni pp j Aijnp I = tr F dΩ i, j = x, z k = n, p

!

ð2:460Þ np where Anp yy and Aij ði, j = x, zÞ are the correlation analyzing powers [10]. Define the following polarization transfer coefficients

 k  ^ ^ m ^þ ^ 0F tr Σ i Σi F   , k, m = n, p; i = x, y, z; i0 = x0 , y0 , z0 ^F ^þ tr F n o ^Σ ^ in Σ ^ jp F ^þ ^ k0 F tr Σ i 0  þ , k = n, p; i, j = x, y, z; i0 = x0 , y0 , z0 K knipi j  ^F ^ tr F 0

K kmii 

ð2:461Þ

ð2:462Þ



→

2.9 Polarization Theory for

→

99

1 1 þ Reactions 2 2

0 where K ki nipj can also be called the polarization transfer coefficient of the correlated incident particles. For the outgoing particles, the correlation problem may not be ^Σ ^m ^ þ (m = n, p; i = x, y, z) and considered. Using the matrix expressions of F i F n p þ ^j F ^ ði, j = x, y, zÞ obtained previously, the following polarization transfer ^Σ ^i Σ F coefficients can be obtained: 0     I 0 K nx nx = ReðAG þ BH - CE - DF Þ 0     I 0 K nx = - ImðAC  þ BD - EG - FH  Þ, I 0 K px nx = ReðAF þ BE - CH - DG Þ     pz0   ny0  I 0 K nx = - ImðAC þ BD þ EG þ FH Þ, I 0 K ny = ReðAG þ BH þ CE þ DF  Þ

nz0

0       nx0   I 0 K py ny = ReðAF þ BE þ CH þ DG Þ I 0 K nz = ImðAE þ BF - CG - DH Þ 1 2 0 I 0 K nz jAj þ jBj2 - jCj2 - jDj2 - jEj2 - jF j2 þ jGj2 þ jH j2 nz = 2 0     I 0 K px nz = ImðAH þ BG - CF - DE Þ  1 0 I 0 K pz jAj2 þ jBj2 - jCj2 - jDj2 þ jEj2 þ jF j2 - jGj2 - jH j2 nz = 2 0 0      = Re ð AF BE  þ CH  - DG Þ, I 0 K nz I 0 K nx px px = - ImðAB þ CD - EF - GH Þ     px0   pz0  I 0 K px = ReðAG - BH þ CE - DF Þ, I 0 K px = - ImðAB þ CD þ EF þ GH  Þ 0       py0   I 0 K ny py = ReðAF þ BE þ CH þ DG Þ, I 0 K py = ReðAG þ BH þ CE þ DF Þ   nx0   I 0 K pz = ImðAE - BF þ CG - DH Þ  1 0 2 2 2 2 2 2 2 2 I 0 K nz = j A j j B j þ j C j j D j j E j þ j F j j G j þ j H j pz 2 0     I 0 K px pz = ImðAH - BG þ CF - DE Þ  1 0 2 2 2 2 2 2 2 2 = A B þ C D þ E F þ G H I 0 K pz j j j j j j j j j j j j j j j j pz 2 0 0      I 0 K ny = Re ð AH - BG - CF  þ DE  Þ, I 0 K py nxpx nxpx = - ReðAE - BF - CG þ DH Þ       nx0  nz0  I 0 K nxpy = ReðAH þ BG - CF - DE Þ, I 0 K nxpy = - ImðAD þ BC - EH - FG Þ 0        pz0  I 0 K px nxpy = ReðAE þ BF - CG - DH Þ, I 0 K nxpy = - ImðAD þ BC þ EH þ FG Þ 0     I 0 K ny nxpz = - ImðAG - BH - CE þ DF Þ,    nx0  I 0 K nypx = ReðAH - BG þ CF - DE Þ, 0     I 0 K px nypx = ReðAE - BF þ CG - DH Þ, 0     I 0 K ny nypy = ReðAH þ BG þ CF þ DE Þ,

0     I 0 K py nxpz = - ImðAF - BE - CH þ DG Þ    nz0  I 0 K nypx = - ImðAD - BC - EH þ FG Þ 0     I 0 K pz nypx = - ImðAD - BC þ EH - FG Þ 0     I 0 K py nypy = ReðAE þ BF þ CG þ DH Þ

0     I 0 K nx nypz = ImðAG - BH þ CE - DF Þ,   px0  I 0 K nypz = ImðAF - BE þ CH - DG Þ, 0     I 0 K ny nzpx = - ImðAF - BE - CH þ DG Þ,   nx0  I 0 K nzpy = ImðAF þ BE - CH - DG Þ,

0     I 0 K nz nypz = ReðAC - BD - EG þ FH Þ   pz0  I 0 K nypz = ReðAC - BD þ EG - FH  Þ 0     I 0 K py nzpx = - ImðAG - BH - CE þ DF Þ    nz0  I 0 K nzpy = ReðAB - CD - EF þ GH Þ

0        pz0  I 0 K px nzpy = ReðAG þ BH - CE - DF Þ, I 0 K nzpy = ReðAB - CD þ EF - GH Þ      ny0   py0  I 0 K nzpz = ReðAE - BF - CG þ DH Þ, I 0 K nzpz = ReðAH - BG - CF þ DE Þ

ð2:463Þ The other polarization transfer coefficient components not given above are all equal to 0.

100

2

Polarization Theory of Nuclear Reactions for Spin

1 Particles 2

According to the non-zero polarization transfer coefficients obtained above, the components of the polarization vector of the emitted particles can be obtained by use of Eq. (2.352) in the polarized incident particle channel as P

ky0

! X X I0 ky0 ky0 ky0 ky0 P0 þ = pmy K my þ pny ppy K nypy þ pni pp j K nip j , k = n, p ð2:464Þ I m = n, p i, j = x, z

0

0

Pk i =

X

ki0

X

X

ki0

1

ki0

pmi K mi þ pny ppi K nypi þ ppy pni K nipy C I0 B B C, k = n,p; i0 = x0 ,z0 i = x, z i = x, z @ A I m = n, p i = x, z

ð2:465Þ →

If the y-axis is not selected in the direction n perpendicular to the reaction plane, then similar derivations can be made using Eq. (2.339) containing the azimuthal angle φ, and the formulas for the calculations of the various polarization physical quantities containing the θ and φ at the same time can also be obtained, but they are more complex.

References 1. Varshalovich DA, Moskalev AN, Khersonskii VK. Quantum theory of angular momentum. Singapore/New Jersey/Hong Kong: World Scientific; 1988. 2. Shen QB. Nuclear reaction theory of low and medium energy (upper volume). Beijing: Science Press; 2005. (in Chinese) → 3. Kocher DC, Bertrand FE, Gross EE, Newman E. 58Ni( p , p’) reaction at 60 MeV: study of the analyzing power for inelastic excitation of the giant resonance of the nuclear continuum and of low-lying bound states. Phys Rev. 1976; C14:1392 4. Wolfenstein L. Possible triple-scattering experimrnts. Phys Rev. 1954;96:1654. 5. Wolfenstein L. Polarization of fast nucleons. Ann Rev Nucl Sci. 1956;6:43. 6. Robson BA. The theory of polarization phenomena. Oxford: Clarendon Press; 1974. 7. Tamura T. Analyses of the scattering of nuclear particles by collective nuclei in terms of the coupled-channel calculation. Rev Mod Phys. 1965;37:679. 8. Joachain CJ. Quantum collision theory. North-Holland Publishing Company. 1975:488. 9. Shen QB. Nuclear reaction theory of low and medium energy (middle volume). Beijing: Science Press; 2012. (in Chinese) 10. Ohlsen GG. Polarization transfer and spin correlation experiments in nuclear physics. Rep Prog Phys. 1972;35:717.

Chapter 3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

Abstract This chapter will introduce the polarization theory of nuclear reactions for spin 1 particles, such as the spin operators and polarization operators of ! → → → 1 spin 1 particles. The polarization theories of 1 þ A → 1 þ B, 1 þ A → þ B, 2 ! → 1 þ A → 1 þ B reactions with unpolarized targets and residual nuclei as well as 2 → → → → 1 þ 1 and 1 þ 1 reactions will be introduced, respectively. This chapter will 2 develop an axis-symmetric rotational nucleus CDCC theory describing the breakup reaction for loosely bound light complex particle. Keywords Axisymmetric rotational nucleus · Breakup reaction · Cartesian coordinate systems · Deuteron folding model · Elastic scattering amplitude · Polarization operators · Polarization rate · Spherical basis coordinate systems · Spin 1 particle

3.1 3.1.1

Spin Operators and Polarization Operators of Spin 1 Particles General Expressions of S = 1 Spin Operators and Polarization Operators [1–3]

The basis spin functions χ m(m =  1, 0) are the common eigenfunctions of the 2 operators ^S and ^Sz , that is, ^S2 χ m = SðS þ 1Þχ m = 2χ m ,

^ Sz χ m = mχ m

ð3:1Þ

It can be seen that in S = 1 case there is

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 Q.-B. Shen, Polarization Theory of Nuclear Reactions, https://doi.org/10.1007/978-3-031-11878-4_3

101

102

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

^S 2 = 2^I

ð3:2Þ

where 0

1

^I = B @0 0

0

0

1

C 0A 1

1 0

ð3:3Þ

The transformation relation between the basis spin functions χ m(m = 1, 0) in spherical basis coordinate system and the basis spin functions χ i(i = x, y, z) in Cartesian coordinate system is as follows: 1 i χ x = pffiffiffi ðχ -1 - χ 1 Þ, χ y = pffiffiffi ðχ -1 þ χ 1 Þ, χ z = χ 0 2 2    1 1  χ 1 = - pffiffiffi χ x þ iχ y , χ 0 = χ z , χ -1 = pffiffiffi χ x - iχ y 2 2

ð3:4Þ ð3:5Þ

Their orthogonal normalization conditions are χþ m0 χ m = δm0 m ,

χþ i χ k = δik

ð3:6Þ

Their completeness conditions are X m = 1,0

^ χm χþ m = I,

X

^ χi χþ i =I

ð3:7Þ

i = x,y,z

When the particle spin S = 1, the spin operator ^ S and the polarization operator T^ LM ðL = 0, 1, 2; - L  M  LÞ are all square 3 × 3 matrices. In this case from Eq. (1.54) we can get T^ 00 = ^I

ð3:8Þ

and from Eq. (1.55) we can also get T^ 1M =

rffiffiffi 3^ S , 2 M

M = 1,0

ð3:9Þ

It follows that the following expressions can be obtained from Eqs. (3.9) and (1.23):

3.1

Spin Operators and Polarization Operators of Spin 1 Particles

pffiffiffi 3 ^ T^ 1 1 =  ðS  i^Sy Þ, 2 x

rffiffiffi 3^ T^ 10 = S 2 z

103

ð3:10Þ

Moreover from Eq. (1.51) we can obtain pffiffiffi X 1 m0 T^ 2M = 5 C 1m 2M χ m0 χ þ m

ð3:11Þ

mm0

and from Eq. (1.20) we can also obtain pffiffiffi X 1 m0 ^Sμ = 2 C 1m 1μ χ m0 χ þ m

ð3:12Þ

pffiffiffi X 2 M ^= 3 Sν C1μ 1ν ^Sμ ^ X

ð3:13Þ

mm0

Let

μν

Substituting Eq. (3.12) into Eq. (3.13), using Eq. (1.18) and the following relation X βδε

cγ fφ Ceε e^f C cγ aα bβ C eε dδ C bβ dδ = ^ aα f ϕ W ðabcd; ef Þ

ð3:14Þ

1 as well as noting W ð1111; 12Þ = , then comparing the obtained result with 6 ^ = T^ 2M , that is, Eq. (3.11), we can get X pffiffiffi X 2 M T^ 2M = 3 C 1μ 1ν ^Sμ ^Sν , M =  2,  1,0

ð3:15Þ

μν

The above formula is an expression of T^ 2M in the spherical basis coordinate system. It has a total of five linear independent components, each of which is a square 3 × 3 matrix. The expressions of the C-G coefficient C jl m m - μ 1μ with definite m value can be obtained from Table 2.1 and are listed in Table 3.1. The specific results of the C-G coefficient C11 m m - μ 1μ can be extracted from Table 3.1 and are listed in Table 3.2. And the specific results of the C-G coefficient C 21μ μþν 1ν can also be extracted from Table 3.1 and are listed in Table 3.3. The following results can be obtained utilizing Eqs. (3.15), (1.23), and (3.2) and Table 3.3:

104

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

Table 3.1 Expressions of C-G coefficient C jl mm- μ j

m-μ

l+1

0

l

0

l-1

0

l+1

1

l

1

l-1

1

l+1

-1

l

-1

l-1

1μ

μ=1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lþ2 2ð2l þ 1Þ

μ=0 rffiffiffiffiffiffiffiffiffiffiffiffi lþ1 2l þ 1

μ= -1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lþ2 2ð2l þ 1Þ

1 - pffiffiffi 2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi l-1 2ð2l þ 1Þ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl þ 2Þðl þ 3Þ 2ðl þ 1Þð2l þ 1Þ

0 rffiffiffiffiffiffiffiffiffiffiffiffi l 2l þ 1 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lðl þ 2Þ ðl þ 1Þð2l þ 1Þ

1 pffiffiffi 2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi l-1 2ð2l þ 1Þ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi l 2ð2l þ 1Þ

1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lðl þ 1Þ

1 pffiffiffi 2

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - 2Þðl - 1Þ 2lð2l þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - 1Þðl þ 1Þ lð2l þ 1Þ

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lþ1 2ð2l þ 1Þ

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi l 2ð2l þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lðl þ 2Þ ðl þ 1Þð2l þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl þ 2Þðl þ 3Þ 2ðl þ 1Þð2l þ 1Þ

1 - pffiffiffi 2

1 - pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lðl þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl þ 2Þðl - 1Þ 2lðl þ 1Þ

-1

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lþ1 2ð2l þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - 1Þðl þ 1Þ lð2l þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - 2Þðl - 1Þ 2lð2l þ 1Þ

l+1

2

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl þ 3Þðl þ 4Þ 2ðl þ 1Þð2l þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - 1Þðl þ 3Þ ðl þ 1Þð2l þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - 1Þl 2ðl þ 1Þð2l þ 1Þ

l

2

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - 2Þðl þ 3Þ 2lðl þ 1Þ

2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lðl þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - 1Þðl þ 2Þ 2lðl þ 1Þ

l-1

2

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - 3Þðl - 2Þ 2lð2l þ 1Þ

l+1

-2

l

-2

l-1

-2

-

-

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - 1Þðl þ 2Þ 2lðl þ 1Þ

-

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - 2Þðl þ 2Þ lð2l þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl þ 1Þðl þ 2Þ 2ð2l þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð l - 1Þ l 2ðl þ 1Þð2l þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - 1Þðl þ 3Þ ðl þ 1Þð2l þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl þ 3Þðl þ 4Þ 2ðl þ 1Þð2l þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - 1Þðl þ 2Þ 2lðl þ 1Þ

2 - pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lðl þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - 2Þðl þ 3Þ 2lðl þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl þ 1Þðl þ 2Þ 2lð2l þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - 2Þðl þ 2Þ lð2l þ 1Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl - 3Þðl - 2Þ 2lð2l þ 1Þ

3.1

Spin Operators and Polarization Operators of Spin 1 Particles Table 3.2 C-G coefficients C 11 μ 1

1 pffiffiffi 2

-1

0

1μ

m 0 1 - pffiffiffi 2

1 1 - pffiffiffi 2

0

m m-μ

105

0

-1 0 1 - pffiffiffi 2

1 pffiffiffi 2

1 pffiffiffi 2

Table 3.3 C-G coefficients C 21μ μþν 1ν ν 0 1 pffiffiffi 2

-1 1 pffiffiffi 6

1 pffiffiffi 2

2 pffiffiffi 6

1 pffiffiffi 2

1 pffiffiffi 6

1 pffiffiffi 2

1

μ

1

1

1

0 -1

pffiffiffi 2      1  T^ 2 0 = 2^Sz - pffiffiffi ^Sx þ i^Sy ^Sx - i^Sy þ ^ Sx - i ^ Sx þ i ^ Sy ^ Sy 2 2  2  h 2  2 i pffiffiffi 2 2 2 2 1 1 Sx þ ^ Sy þ ^ Sz = 2S^z - pffiffiffi 2S^x þ 2S^y = pffiffiffi 3^Sz - ^ 2 2 2  2  1 = pffiffiffi 3^Sz - 2 2 pffiffiffi    3 ^ ^ T 2 1 =  Sx  i ^ Sz ^ Sy Sx  i^Sy ^Sz þ ^ 2 pffiffiffi 2 3 ^ T^ 2 2 = Sx  i^Sy 2

ð3:16Þ

ð3:17Þ ð3:18Þ

Equations (3.10) and (3.16)–(3.18) have been given in References [4] and [5], so it can be seen that TLM determined by Eq. (1.51) is a common form in the polarization theory of nuclear reactions. ^ ik ði, k = x, y, zÞ in Cartesian coordinate We can use nine square 3 × 3 matrices Q ^ systems equivalent to T 2M ðM =  2, 1, 0Þ in spherical basis coordinate system, but we require the following conditions to be satisfied: X

^ ii = 0 Q

ð3:19Þ

i = x, y, z

^ ki , ^ ik = Q Q

i ≠ k;

i, k = x, y, z

ð3:20Þ

106

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

^ is not a trace Equations (3.19) and (3.20) require that in spatial coordinate Q tensor, but is a symmetric tensor, respectively. The two formulas give one and three ^ ik are constraints to tensor components, respectively, and therefore only five Q ^ linearly independent. If we select Qik according to the following relations [2] pffiffiffi  1 3 ^ T 2 2 þ T^ 2 -2 - pffiffiffi T^ 2 0 , 2 2 pffiffiffi   3 1 ^ yy = Q T^ 2 2 þ T^ 2 -2 - pffiffiffi T^ 2 0 , 2 2 pffiffiffi ^ zz = 2T^ 2 0 , Q pffiffiffi  ^ xx - Q ^ yy = 3 T^ 2 -2 þ T^ 2 2 ^ xx - yy  Q Q ^ xx = Q

pffiffiffi  i 3 ^ ^ ^ Qxy = Qyx = T 2 -2 - T^ 2 2 , 2 pffiffiffi  3 ^ ^ ^ Qxz = Qzx = T 2 -1 - T^ 2 1 , 2 pffiffiffi  i 3 ^ ^ ^ Qyz = Qzy = T 2 -1 þ T^ 2 1 : 2 ð3:21Þ

^ ik satisfies Eq. (3.19) and Eq. (3.20), and their inverse relations then the obtained Q can be obtained as follows: T^ 2

2

 1 ^ ^ ^ = pffiffiffi Q xx - yy  2iQxy , T 2 2 3

1

 1 ^ 1 ^ ^ ^ =  pffiffiffi Q xz  iQyz , T 2 0 = pffiffiffi Qzz 3 2 ð3:22Þ

^ xz ,Q ^ yz ,Q ^ xx - yy ,Q ^ zz form five linearly ^ xy ,Q From the above results it can be seen that Q independent tensor operators in the Cartesian coordinate system. We choose [2]   ^ ik = 3 ^Si ^Sk þ ^Sk ^Si - 4 δik^I , Q 2 3

i, k = x, y, z

ð3:23Þ

This formula was published as early as 1958 in a paper in Nucl. Phys., 7,622 by ^ ik given by the above formula satisfies Goldfarb. It can be seen directly that Q ^ ik also satisfies Eq. (3.19). Eq. (3.20), and using Eq. (3.2) we can prove that Q Substituting Eq. (3.15) into the first equation of Eq. (3.21) we can obtain " #  rffiffiffi X 3 3 2 2 2 -2 2 0 ^S ^ ^ xx = S C C 1μ 1ν þ C1μ 1ν Q 2 1μ 1ν μ ν μν 2  3 -2 ^ ^ = C 211211 ^S1 ^S1 þ C12 -1 1 -1 S -1 S -1 2 rffiffiffi  3 2 0 S -1 ^ S1 C11 1 -1 S^1 S^ -1 þ C 21 00 1 0 S^0 S^0 þ C21 0-1 11 ^ 2 Using Table 3.3, or Ref. [1], there are

ð3:24Þ

3.1

Spin Operators and Polarization Operators of Spin 1 Particles

1 0 C 21101 -1 = C21 -1 11 = pffiffiffi , 6

-2 C2112 11 = C 12 -1 1 -1 = 1,

107

2 C 210010 = pffiffiffi 6

ð3:25Þ

From Eq. (3.24) we can obtain the following expression utilizing Eqs. (3.25), (1.23), and (3.2): h 2  2 i ^ xx = 3 S^x þ i^ Sy þ ^Sx - i^Sy Q 4 h     i 2 1 ^ þ Sy Sx þ i ^ Sx þ i^Sy ^Sx - i^Sy - 4^Sz þ ^Sx - i^Sy ^ 4     2 2 2 2 1 2 3 ^2 ^ 2 = Sy - i^ Sx þ ^ Sz þ ^ Sx þ i ^ Sx ^ Sy ^ Sy Sx - Sy þ ^Sx - i^Sx ^Sy þ i^Sy ^Sx þ ^Sy - 4^ 4 2  2  2 2 2 2 3 ^ 2 3 ^ 2 1 ^ 2 1 ^2 ^ 2 = Sx - Sy þ Sx þ Sy - Sz = 3^Sx - ^Sx þ ^ Sy þ ^ Sz = 3 ^ Sx - 2^I 2 2 2 2 ð3:26Þ ^ xx obtained by Eq. (3.21) is exactly the same as Q ^ xx obtained by It can be seen that Q ^ can also be proved by a similar method, so Eq. (3.23). The other components of Q ^ ik given by Eq. (3.21) and Eq. (3.23) are exactly the same. that Q In S = 1 case the nine matrices T^ LM ðL = 0, 1, 2; - L  M  LÞ in the spherical basis coordinate system are equivalent to the nine linear independent matrices of ^ ik in the Cartesian coordinate system. ^I,^ S,Q Since in S = 1 case the basis spin functions may take χ 1m(m =  1, 0) or χ i(i = x, y, z), the spin operators and polarization operators can be presented in the spherical basis representation or the Cartesian basis representation. It is noted that the expressions given earlier in this section, including Eq. (3.23), are applicable to both representations.

3.1.2

Specific Expressions in the Spherical Basis Representation

The component of the basis spin functions of the spherical basis coordinate in the spherical basis representation is defined as χ Sm ðσ Þ = δmσ ,

m,σ = S, S - 1, . . . , - S þ 1, - S

ð3:27Þ

In S = 1 case, there are 0 1 1 B C χ 1 = @ 0 A, 0

0 1 0 B C χ 0 = @ 1 A, 0

Given these, the following expression is satisfied:

0 1 0 B C χ -1 = @ 0 A 1

ð3:28Þ

108

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

χ m = χ m ,

m =  1,0

ð3:29Þ

Using Eq. (1.22) or Eq. (3.4), from Eq. (3.28) we can find that the S = 1 particle basis spin functions of the Cartesian coordinate system in the spherical basis representation are as follows: 0

1 -1 1 B C χ x = pffiffiffi @ 0 A, 2 1

0 1 1 i B C χ y = pffiffiffi @ 0 A, 2 1

0 1 0 B C χz = @ 1 A

ð3:30Þ

0

Equation (1.21) gives the matrix element of the spin operator component in the spherical basis coordinate system and the spherical basis representation. In S = 1 case, this formula becomes pffiffiffi   ^Sμ 0 = 2C1σ0 , 1σ 1μ σσ

μ, σ, σ 0 =  1, 0

ð3:31Þ

In the spherical basis representation, using Table 3.2 and Eq. (3.31), the expressions of the S = 1 particle spin operator component in the spherical basis coordinate system can be obtained as follows: 0

0

^S1 = - B @0 0

1 0 0

0

1

0

C 1 A, 0

1

0

^S0 = B @0 0

0 0

1

0

0

C 0 A, -1

0

0 0

1

C 0 0 A ð3:32Þ 1 0

B ^ S-1 = @ 1 0

and using Eqs. (1.22) and (3.32) the expressions of the S = 1 particle spin operator components in the Cartesian spherical basis coordinate system can be obtained as follows: 0

0

B ^Sx = p1ffiffiffi B 1 2@ 0

1

0

0

1

0

C 1C A,

1

0

0

B ^Sy = piffiffiffi B 1 2@ 0

-1 0 1

0

1

C -1C A,

0

1

B ^ Sz = B @0

0

0

0 0 0

0

1

C 0 C A -1 ð3:33Þ

The above spin operator matrices satisfy the following relations: ^Sþ = ^Si , i ^Sþ = ð - 1Þμ ^S - μ , μ

i = x, y, z μ =  1, 0

ð3:34Þ ð3:35Þ

^ ik in the spherical basis representation Using Eq. (3.33) the specific expressions of Q can be obtained by Eq. (3.23) as [3]

3.1

Spin Operators and Polarization Operators of Spin 1 Particles

0

-1

0

3

1

0

-1

0

-3

1

109

0

1

B C C 1B B B ^ ^ 0 C 0 C 2 A, Qyy = 2 @ 0 A ,Qzz = @ 0 -3 0 -1 0 0 -1 0 1 0 0 0 -1 0 1 B C B 3i 3 B ^ yx = B 0 0 ^ ^ xy = Q ^ Q 0 C 0 A, Qxz = Qzx = 2pffiffi2ffi @ 1 2@ 1 0 0 0 -1 0 0 1 0 0 -1 0 B B C 3i B ^ zy = pffiffiffi B 1 ^ ^ ^ ^ yz = Q 0 1C Q A, Qxx - yy  Qxx - Qyy = 3@ 0 2 2@ 1 0 -1 0

B ^ xx = 1 B 0 Q 2@ 3

2

0

0

C 0C A,

-2 0 0

1

1 1

C -1C A, 0 0

1

1

0

C 0C A

0

0 ð3:36Þ

^ ik is a Hermitian matrix From the above formulas it can be seen that Q ^þ = Q ^ ik Q ik

ð3:37Þ

^ yy , Q ^ zz , Q ^ xz , Q ^ zx , Q ^ xx - yy are real numbers, but Q ^ xy , Q ^ yx , Q ^ yz , Q ^ zy are pure ^ xx , Q And Q imaginary numbers. The specific expressions of the polarization operators T^ LM in the spherical basis representation can be obtained by using Eqs. (3.8), (3.9), (3.32), (3.22), and (3.36): 0

1

0 1 0 1 rffiffiffi 0 1 0 rffiffiffi 1 0 0 B C C C 3B 3B C ^ B C ^ B C T^ 00 = B @ 0 1 0 A, T 11 = - 2@ 0 0 1 A, T 10 = 2@ 0 0 0 A, 0 0 1 0 0 0 0 0 -1 0 1 0 1 0 1 0 0 1 rffiffiffi 0 0 0 rffiffiffi 0 - 1 0 pffiffiffiB B C C C 3B B 1 0 0 C, T^ 22 = 3B 0 0 0 C, T^ 21 = 3B 0 0 1 C, T^ 1 -1 = @ A @ A @ A 2 2 0 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 rffiffiffi 0 0 0 pffiffiffiB C C C 1 B 3B B 1 0 0 C, T^ 2 -2 = 3B 0 0 0 C 0 -2 0C T^ 20 = pffiffiffi B , T^ 2 -1 = @ @ @ A A A 2 2 0 0 1 0 -1 0 1 0 0 1 0 0

ð3:38Þ From the above formulas it can be seen that all T^ LM ðL = 0, 1, 2; - L  M  LÞ are real numbers and satisfy

110

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles þ T^ L M = ð - 1ÞM T^ L

ð3:39Þ

-M

From Eq. (1.53) we can see that the matrix element of the S = 1 particle polarization operator in the spherical basis representation is 

T^ LM ð1Þ

 σ0 σ

pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 = 2L þ 1C1σ 1σ LM , L = 0, 1, 2;

- L  M  L; σ, σ 0 =  1, 0 ð3:40Þ

3.1.3

Specific Expressions in the Cartesian Basis Representation

In the Cartesian basis representation, the component of the S = 1 basis spin function in the Cartesian coordinate system is defined as χ i ðσ Þ = δiσ ,

i, σ = x, y, z

ð3:41Þ

The corresponding basis spin functions are 0 1 1 B C χ x = @ 0 A, 0

0 1 0 B C χ y = @ 1 A, 0

0 1 0 B C χz = @ 0 A 1

ð3:42Þ

From above formulas it can be seen that the following relation is satisfied: χ i = χ i ,

i = x, y, z

ð3:43Þ

Using Eq. (3.5) and Eq. (3.42) the expressions of the basis spin functions of the S = 1 particles in the spherical basis coordinate system and Cartesian basis representation can be obtained as follows: 0 1 1 1 B C χ 1 = - pffiffiffi @ i A, 2 0

0 1 0 B C χ 0 = @ 0 A, 1

0

1

1

1 B C χ - 1 = pffiffiffi @ - i A, 2 0

ð3:44Þ

And they satisfy the following relation: χ m = ð - 1Þm χ-m ,

m =  1,0

ð3:45Þ

Using Eq. (3.44) and the C-G coefficients given in Table 3.2, we can obtain the following expressions by Eq. (3.12):

3.1

Spin Operators and Polarization Operators of Spin 1 Particles

0

0

1

0 1

0

0 -i 0

1

111

0

0

0 1

1

B C B C B C ^S1 = p1ffiffiffi B 0 0 i C, ^S0 = B i 0 0 C, ^S -1 = p1ffiffiffi B 0 0 - i C ð3:46Þ @ A @ A @ A 2 2 0 0 0 -1 i 0 -1 -i 0 According to Eq. (1.22) we can also obtain the following expressions by Eq. (3.46): 0

0

B ^ Sx = @ 0 0

0

0

1

0

0

C - i A,

i

0

0

0

^Sy = B @ 0 -i

i

1

0

0

C 0 A,

0

0

0

B ^ Sz = @ i

-i 0

0

1

0

C 0 A ð3:47Þ

0

0

Equations (3.46) and (3.47) give the specific expressions of the spin operator components of the S = 1 particles in the spherical basis coordinate system and the Cartesian coordinate system in the Cartesian basis representation, respectively. The matrix element of ^Si can be written as   ^Si = - iεikl , kl

i, k, l = x, y, z

ð3:48Þ

where εikl is given by Eq. (1.7). In addition, there are the following relations:  þ ^Si = ^Si ,

i = x, y, z

 þ ^Sμ = ð - 1Þμ ^S - μ ,

ð3:49Þ

μ =  1,0

ð3:50Þ

^ ik Substituting Eq. (3.47) into Eq. (3.23), we can get the specific expressions of Q in the Cartesian basis representation as follows: 0

-2 0

B ^ xx = B 0 Q @

1

0

0 0

B ^ yz = 3 B ^ xy = Q Q 2@ 0

0 B 3 ^ zy = B 0 ^ yz = Q Q 2@ 0

0

1

0

1

0

0

1

0

1

0

0

1

B B C C C B B C ^ ^ 0C 0 C A, Qyy = @ 0 - 2 0 A, Qzz = @ 0 1 A, 1 0 0 1 0 0 -2 0 1 1 0 -1 0 0 0 -1 C C 3B B ^ ^ -1 0 0C 0 0 C A, Qxz = Qzx = 2 @ 0 A, 0 0 0 -1 0 0 0 1 1 -1 0 0 0 0 B C C B ^ ^ ^ 1 0C 0 -1C A A, Qxx - yy  Qxx - Qyy = 3@ 0 0 0 0 -1 0 ð3:51Þ

Its matrix elements can be summarized as

112

3

^ ik Þ = ðQ lm

Polarization Theory of Nuclear Reactions for Spin 1 Particles

3 2 ðδ δ þ δim δkl - δik δlm Þ, i, k, l, m = x, y, z 2 il km 3

ð3:52Þ

This formula is only applied in the Cartesian coordinate system and the Cartesian basis representation such that ^ ik , ^þ = Q Q ik

^ = Q ^ ik Q ik

ð3:53Þ

The specific expressions of the polarization operators T^ LM ð1Þ in the Cartesian basis representation can be obtained, by using Eqs. (3.8), (3.9), (3.46), (3.22), and (3.51), as follows: 0

0 1 0 1 0 0 1 rffiffiffi 0 - i 0 pffiffiffi B C C C 3B 3B C ^ B C B ^ 0 iC T^ 00 = B @ 0 1 0 A, T 11 = 2 @ 0 A, T 10 = 2@ i 0 0 A, 0 0 1 0 0 0 -1 -i 0 0 1 0 1 0 1 0 0 1 pffiffiffi pffiffiffi - 1 - i 0 pffiffiffi 0 0 1 C B C B C 3B B 0 0 - i C, T^ 22 = 3 B - i 1 0 C, T^ 21 = 3 B 0 0 i C, T^ 1 -1 = A A A 2 @ 2 @ 2 @ -1 i 0 0 0 0 1 i 0 0 1 0 0 1 1 1 0 0 0 0 -1 pffiffiffi pffiffiffi - 1 i 0 C B C C 1 B 3B B 0 0 i C, T^ 2 -2 = 3 B i 1 0 C T^ 20 = pffiffiffi B 0 1 0 C , T^ 2 -1 = @ A @ @ A A 2 2 2 0 0 -2 -1 i 0 0 0 0 1 0 0

1

ð3:54Þ where T^ LM satisfies the following relations: þ T^ LM = ð - 1ÞM T^ L

- M,

 T^ LM = ð - 1ÞLþM T^ L

- M,

L = 0,1,2;

-L  M  L ð3:55Þ

3.1.4

Transformation Relations Between the Spherical Basis and Cartesian Coordinate Systems

The coordinate system transformation here refers to the transformation relations of the components of the polarization quantities between the different coordinate systems. ^ can be used to transform the components Ax, Ay, The following unitary matrix U * Az of the vector A in the Cartesian coordinate system, and transform the contravariant

3.1

Spin Operators and Polarization Operators of Spin 1 Particles

113 *

components A1, A0, A-1 and the covariant components A1, A0, A-1 of the vector A in the spherical basis coordinate system 0 B B ^ U=B B @

1 - pffiffiffi 2 0 1 pffiffiffi 2

1 i pffiffiffi 0 C 2 C 0 1C C, A i pffiffiffi 0 2

0 B B þ -1 ^ ^ U =U =B B @

1 - pffiffiffi 2 i - pffiffiffi 2 0

1 1 pffiffiffi 2 C C i C - pffiffiffi C 2A 0

0 0 1

ð3:56Þ

Their transformation relations are as follows: 0

1 0 1 Ax A1 B 0 C ^B C @ A A = U @ Ay A, Az A -1 0 1 0 1 Ax A1 B C ^ B C @ A0 A = U @ A y A, A -1

Az

0

1 0 1 1 Ax A B C ^ -1 B 0 C @ Ay A = U @ A A Az A -1 0 1 0 1 Ax A1 B C ^ -1 B C @ Ay A = U @ A0 A Az

ð3:57Þ

ð3:58Þ

A -1

For example, using 0

1

0

^S1 B B ^ C B @ S0 A = B B @ ^S -1

1 - pffiffiffi 2 0 1 pffiffiffi 2

i - pffiffiffi 2 0 i - pffiffiffi 2

1

0 1 S C ^ CB x C ^ S @ 1C yA C A ^ Sz 0

0

ð3:59Þ

one can get Eq. (1.23), and using 0 1 1 - pffiffiffi ^Sx B 2 B^ C B i @ Sy A = B B pffiffiffi @ 2 ^Sz 0 0

1 1 0 pffiffiffi 0 ^ 1 S 2C CB 1 C C ^ i @ S0 A 0 pffiffiffi C 2A ^ S -1 1 0

ð3:60Þ

one can obtain Eq. (1.22). In addition, the transformation relations given by Eq. (3.4) and Eq. (3.5) can also be obtained by the above method. The transformation relations between the polarization tensor components T^ 2M ^ ij have been given by Eq. (3.21) and Eq. (3.22). If the spherical basis coordinate and Q system is used, then both the spin operator and the polarization operator can be

114

3 Polarization Theory of Nuclear Reactions for Spin 1 Particles

uniformly represented by irreducible tensors T^ LM ðL = 1, 2; M = - L, ⋯, LÞ . Each T^ LM is independent, easy to process, and suitable for theoretical calculations. Its disadvantage is that only T^ L0 is observable; T^ LM ðM ≠ 0Þ is not observable. In the 2 common eigenstates of ^S and ^Sz, the obtained expectation value is not a real number (it becomes a real number in the special case of azimuthal angle φ = 0). If the Cartesian coordinate system is used, the spin operator and the polarization operator cannot be represented in a unified mathematical form. They can only be expressed as ^ ij ði, j = x, y, zÞ, respecvectors ^Si ði = x, y, zÞ and rank-2 second reducible tensors Q ^ tively. Only five of the nine Qij are independent, but all expectation values of the above components are real numbers. All of them are observable physical quantities.

3.1.5

Transformation Relations Between the Spherical Basis and Cartesian Basis Representations

The representation transformation refers to the transformation relations between the 3 × 3 matrix expressions of the components of the same polarized physical quantity in the spherical basis coordinate system or Cartesian coordinate system in different representations. ^ C be the 3 × 3 matrix expressions of the same polarization physical ^ S and A Let A quantity components in the spherical basis representation and the Cartesian basis representation, respectively. The following relations should be satisfied between them: ^C, ^SU ^ =U ^A A

^S = U ^C U ^ - 1, ^A A

^SU ^C = U ^ ^ - 1A A

ð3:61Þ

Equations (3.32) and (3.46) yield, respectively 0

0 ^S1,S = - B @0 0

1 1 0 C 0 1 A, 0 0

0

0 ^S1,C = p1ffiffiffi B @ 0 2 -1

0 0

1 1 C iA

-i

0

^ given by Eq. (3.56): The following relations can be proved using matrix U

ð3:62Þ

3.1

Spin Operators and Polarization Operators of Spin 1 Particles

0

0 B ^S1,S U ^ = -B0 @ 0 0 B B B ^ ^ U S1,C = B B @

1 0 0

1 i - pffiffiffi pffiffiffi 2 2 0 0 1 i pffiffiffi pffiffiffi 2 2

115

0

1 0 1 1 i p ffiffi ffi p ffiffi ffi 0 0 0 1 0 B C 2 2 B 1 C C CB B pffiffiffi piffiffiffi 0 C B C 1C = B C, B C 0 0 1 AB 2 @ 2 A C @ 1 A i 0 pffiffiffi pffiffiffi 0 0 0 0 2 2 0 1 1 1 pffiffiffi 0 0 0 1 0 B 2C 0 0 1 CB C CB B 1 C i i C CB B C pffiffiffi C 0 C = - B pffiffiffi pffiffiffi 0 C 1 CB 0 2 CB @ 2 A 2C AB C A 0 @ 0 0 0 1 i - pffiffiffi - pffiffiffi 0 2 2 ð3:63Þ 1

S1 . For another example, Equation (3.63) indicates that Eq. (3.61) is valid for ^ Eqs. (3.36) and (3.51) yield, respectively 0

-1

^ xx,S = 1 B Q @ 0 2 3

0 2 0

3

1

C 0 A, -1

0

-2

^ xx,C = B Q @ 0 0

0 0

1

C 1 0A

ð3:64Þ

0 1

^ given by Eq. (3.56) one can also prove Using operator U 0

-1

B ^ xx,S U ^ = 1B 0 Q 2@ 3 0 B B ^ xx,C = B ^Q U B B @

0 2 0

1 - pffiffiffi 2 0 1 pffiffiffi 2

0

1 1 i 0 pffiffiffi p ffi p ffiffi ffi ffiffi 0 3 2 2 B C 2 2 B C CB C 1B C B 0 AB 0 0 1C= @ 0 B C 2 pffiffiffi @ 1 A i -1 -2 2 pffiffiffi 0 pffiffiffi 2 2 1 i 0 pffiffiffi pffiffiffi 0 0 -2 0 0 1 2 2 C 2 CB C 1B CB B 0 1 C@ 0 1 0 C A= 2@ 0 C pffiffiffi A i 0 0 1 -2 2 pffiffiffi 0 2 1

pffiffiffi 2i

0

0 pffiffiffi 2i

0

1

C 2C A,

1 pffiffiffi 2i 0 C 0 2C A pffiffiffi 2i 0 ð3:65Þ

^ xx . In fact Eq. (3.61) can Equation (3.65) indicates that Eq. (3.61) is also valid for Q be used for the representation transformation of any polarization physical quantity component.

116

3.2

3.2.1

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

Some Mathematical Formulas Related to Spin Operators and Polarization Operators of Spin 1 Particles Coordinate Rotation of S = 1 Spin Wave Function and Spin Operator

The S = 1 spin wave function χ m(m =  1, 0) in the spherical basis coordinate can be ^ 1 ðα, β, γ Þ, that is, rotated by Wigner D function D χ 0m0 =

X m = 1,0

^ 1m m0 ðα, β, γ Þ χmD

ð3:66Þ

Based on Eq. (2.57) and Table 2.5, Eq. (2.196) has given 0 1 þ cos β e - iðαþγÞ 2 B B B sin β - iγ ^ 1 ðα, β, γ Þ = B pffiffiffi e D B 2 B @ 1 - cos β eiðα - γ Þ 2

sin β - pffiffiffi e - iα 2 cos β sin β iα pffiffiffi e 2

1 - cos β iðγ - αÞ 1 e 2 C C C sin β - pffiffiffi eiγ C C ð3:67Þ 2 C 1 þ cos β iðγþαÞ A e 2

The S = 1 spin wave function χ i(i = x, y, z) in a Cartesian coordinate system can be rotated as follows: χ 0i =

X

χ k aki

ð3:68Þ

k = x, y, z

The rotation matrix composed of matrix elements aki has been given by Eq. (2.202) as [1] ^ aðα,0β, γ Þ = cos β cos α cos γ - sin α sin γ B @ cos β sin α cos γ þ cos α sin γ - sin β cos γ

1 - cos β cos α sin γ - sin α cos γ sin β cos α C - cos β sin α sin γ þ cos α cos γ sin β sin α A sin β sin γ

cos β ð3:69Þ

The corresponding inverse rotation operator of the coordinate system is [1]

3.2

h

Some Mathematical Formulas Related to Spin Operators. . .

^ 1 ðα, β, γ Þ D

i-1

117

h 1 iþ ^ ðα, β, γ Þ = D ^ 1 ð - γ, - β, - αÞ = D ^ 1 ðπ - γ, β, - π - αÞ = D ð3:70Þ

In addition, in S = 1 case the corresponding relations of the helicity basis functions can be obtained by Eqs. (1.30)–(1.33).

3.2.2

Action of the S = 1 Spin Operators and Polarization Operators on Basis Spin FUnctions

In the spherical basis representation, when the spin operator ^ Sμ ðμ =  1, 0Þ and polarization operator T^ 2M ðM =  2, 1, 0Þ in the spherical basis coordinate system act on the basis spin function χ m(m =  1, 0), we can obtain [1] pffiffiffi ^Sμ χ m = 2C 1 mþμ χ mþμ 1m 1μ pffiffiffi 1 mþM T^ 2M χ m = 5C 1m 2M χ mþM

ð3:71Þ ð3:72Þ

In the Cartesian basis representation, when the spin operator ^ Si ði = x, y, zÞ and the ^ ik ði, k = x, y, zÞ in the Cartesian coordinate system act on the polarization operator Q basis spin function χ i(i = x, y, z), we can obtain [1] ^Si χ k = iεikl χ l   ^ ik χ l = 3 2 δik χ l - δil χ k - δkl χ i Q 2 3

3.2.3

ð3:73Þ ð3:74Þ

Products of S = 1 Spin Operators and Polarization Operators

^ ik in Cartesian coordinate ^i and Q 1. Products of the operator component S system (i, k, l, et al. taking x, y, z) Equations (1.4) and (1.6) have given ^S × ^S = i^S,



 ^Si , ^Sk  ^Si ^Sk - ^ Si = iεikl ^ Sl Sk ^

From Eq. (3.23) and the second formula of Eq. (3.75), we can get

ð3:75Þ

118

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

^Si ^Sk = 2 δik^I þ i εikl ^Sl þ 1 Q ^ 3 3 ik 2 Using

ð3:76Þ

P ^ ii = 0 and the other equations, the following relations can be obtained Q i

from the above formula: ^S2 = ^S2 þ ^S2 þ ^S2 = 2^I x y z f^Si , ^Sk g  ^Si ^Sk þ ^Sk ^Si =

4 ^ 2^ δ I þ Qik 3 ik 3

ð3:77Þ ð3:78Þ

and X   ^Si ^Sk ^Sl = i εikl^I þ 1 δik ^Sl þ δkl ^Si þ i ^ ε Q 2 3 3 m ilm km  i X  1 i ^ lm þ εilm Q ^ km þ εklm Q ^ im = εikl^I þ δik ^Sl þ δkl ^Si þ εikm Q 6 m 2 3

ð3:79Þ

From the above formula we can also obtain ^Si ^Sk ^Sl þ ^Sl ^Sk ^Si = δik ^Sl þ δkl ^ Si X ^Si ^Sk ^Si = δik ^Si ðno sum to iÞ; ^ Si ^ Sk ^ Si = ^ Sk

ð3:80Þ ð3:81Þ

i

^Si ^Sk ^Sk þ ^Sk ^Sk ^Si = ^Si þ δik ^Sk ðno sum to k Þ ^Sx ^Sy ^Sz þ ^Sy ^ Sz ^Sx þ ^Sz ^Sx ^Sy = i^I, ^Sx ^Sz ^Sz = ^Sy ^Sy ^Sx = 1 ^Sx - i Q ^ , 2 3 yz ^Sy ^Sx ^Sx = ^Sz ^Sz ^Sy = 1 ^Sy - i Q ^ , 2 3 xz ^Sz ^Sy ^Sy = ^Sx ^Sx ^Sz = 1 ^Sz - i Q ^ , 2 3 xy

ð3:82Þ

^Sx ^Sz ^Sy þ ^Sz ^Sy ^ Sx þ ^ Sy ^ Sx ^ Sz = - i^I

ð3:83Þ

1 i^ ^Sx ^Sy ^Sy = ^Sz ^ Sz ^ Sx = ^ S þ Q 2 x 3 yz 1 i^ ^Sy ^Sz ^Sz = ^Sx ^ Sx ^ Sy = ^ S þ Q 2 y 3 xz 1 i^ ^Sz ^Sx ^Sx = ^Sy ^ S þ Q Sy ^ Sz = ^ 2 z 3 xy

ð3:84Þ

^S2nþ1 = ^ Si , n = 0, 1, 2⋯ i

ð3:87Þ

ð3:85Þ ð3:86Þ

Now, there are ^ ^S2n = 2 ^I þ Qii , n = 1, 2⋯ ; i 3 3 *

If n is a unit vector, then there are

Some Mathematical Formulas Related to Spin Operators. . .

3.2

2k 2 1X ^S  * ^ , k = 1, 2, 3⋯; n = ^I þ nnQ 3 3 il i l il     * 2kþ1 * ^ S n = ^S  n , k = 0, 1, 2⋯

119



ð3:88Þ

And there are   X  ^ ik ^Sl = 3 δil ^Sk þ δkl ^Si - 2 δik ^Sl þ i ^ km þ εklm Q ^ im εilm Q Q 2 m 4 3

ð3:89Þ

  X  ^Si Q ^ kl = 3 δik ^Sl þ δil ^Sk - 2 δkl ^Si þ i ^ lm þ εilm Q ^ km εikm Q 4 3 2 m

ð3:90Þ

  ^ ik Q ^ lm = 3 δil δkm þ δim δkl - 2 δik δlm ^I Q 2 3   3 ^ km þ δim Q ^ kl þ δkm Q ^ il þ δkl Q ^ im - 4 δik Q ^ lm - 4 δlm Q ^ ik δil Q 3 3 4  9i X Sp þ δ ε þ δim εklp þ δkl εimp þ δkm εilp ^ 8 p il kmp ð3:91Þ ^ and the polarization 2. The products of the components of the spin operator S ^ operator T LM in the spherical basis coordinate system (μ, ν =  1, 0) We have the following formula: [1] ^Sμ ^Sν = 2 ð - 1Þμ δμ 3

^ - p1ffiffiffi C1λ 1μ 2

- νI

^ þ p1ffiffiffi C 2M T^ 2M 1μ 1ν 3

1ν Sλ

ð3:92Þ

From the above formula we can obtain ^S2 = - ^S1 ^S - 1 þ S^0 ^S0 - ^S - 1 ^ S1 = 2^I pffiffiffi 1λ   ^Sμ , ^Sν  ^Sμ ^Sν - ^Sν ^Sμ = - 2C ^ 1μ 1ν Sλ

^Sμ , ^Sν  ^Sμ ^Sν þ ^Sν ^Sμ = 4 ð - 1Þμ δμ 3

^ þ p2ffiffiffi C2M ^ 1μ 1ν T 2M 3

- νI

ð3:93Þ ð3:94Þ ð3:95Þ

and ^S2 = p1ffiffiffi T^ 2 2 , 1 3

^S3 = ^S4 = ^S5 = . . . = 0 1 1 1

ð3:96Þ

120

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

pffiffiffi ^S2 = 2 ^I þ 2 T^ 20 0 3 3

ð3:97Þ

^S2n = ^S2 , n = 1, 2, . . . ; ^S2nþ1 = ^ S0 , n = 0, 1, 2, . . . 0 0 0 X pffiffiffi ^2 ð - 1ÞL1 þL2 þL W ðL1 L2 11; L1ÞC LM ^ ^1 L T^ L1 M 1 T^ L2 M 2 = 3L L M L M T LM 1

1

2

2

ð3:98Þ ð3:99Þ

L

From Eq. (3.99) we can also obtain T^ LM T^ 00 = T^ 00 T^ LM = T^ LM , L = 0, 1, 2; - L  M  L rffiffiffi pffiffiffi ^Sμ T^ 2M = - 5 C 1ν ^Sν - 3C 2N T^ 2N 2 1μ 2M 2 1μ 2M rffiffiffi pffiffiffi 5 1ν ^ 3 2N ^ S þ T C C T^ 2M ^Sμ = 2 1μ 2M ν 2 1μ 2M 2N rffiffiffi pffiffiffi 3 5 7 2Λ ^ M 1μ ^ T^ 2M T^ 2N = ð - 1Þ δM - N ^I þ C T S þ C 2 2 2M 2N μ 2 2M 2N 2Λ

3.2.4

ð3:100Þ ð3:101Þ ð3:102Þ ð3:103Þ

Traces of the S = 1 Spin Operators and Polarization Operators

When S = 1 one can obtain the following results from Eq. (1.24):





Sk ^ Sl = iεikl , tr S^i = 0, tr ^Si ^Sk = 2δik , tr ^ Si ^

tr S^i ^Sk ^Sl ^Sm = δik δlm þ δim δkl

ð3:104Þ

^ ik there are the following trace formulas: [1] For Q





^ ik = 0, ^ ik ^Sl = 0, ^ kl = 0, tr Q tr ^Si Q tr Q  

^ lm = 9 δil δkm þ δkl δim - 2 δik δlm , ^ ik Q tr Q 3 2





9 ^ ii Q ^ ik Q ^ ii = 6, ^ kk ^ ik ^ ii Q tr Q = - 3, tr Q = tr Q i≠k i≠k 2

ð3:105Þ

In the above two formulas, take i, k, l, m = x, y, z. Let aij and bij represent the matrix elements of the matrix A and B, respectively, so that we can prove

3.2

Some Mathematical Formulas Related to Spin Operators. . .

trfABg =

XX i

aij bji =

j

XX bji aij = trfBAg j

121

ð3:106Þ

i

Naturally we can prove the following relation: trfABC g = trfCABg = trfBCAg

ð3:107Þ

The following relation can be obtained utilizing Eqs. (3.107), (3.105), (3.75), and (3.76) and Qlm = Qml





^ ik ^Sl ^Sm = 1 tr Q ^ ik ^Sl = tr Q ^ lm ^ ik Q tr S^m Q 3  2 3 = δ δ þ δkl δim - δik δlm 3 2 il km

ð3:108Þ

And the following relation can also be obtained utilizing Eqs. (3.107), (3.105), (3.104), (3.89) (or (3.90), or (3.91)):



^ ik Q ^ ik Q ^ lm = tr Q ^ lm ^Sj tr ^Sj Q  9i  = δil εkmj þ δim εklj þ δkl εimj þ δkm εilj 4

ð3:109Þ

Equation (3.89) can be rewritten as   X  ^ ik ^Sj = 3 δij ^Sk þ δkj ^Si - 2 δik ^Sj þ i ^ kn þ εkjn Q ^ in εijn Q Q 2 n 4 3

ð3:110Þ

Using Eq. (3.105) and Eq. (3.109) we can obtain  X 

2 ^ ik ^Sj Q ^ lm = 9i εijn δkl δnm þ δnl δkm - δkn δlm tr Q 3 4 n   2 þεkjn δil δnm þ δnl δim - δin δlm 3  9i 2 2 = εijm δkl þ εijl δkm - εijk δlm þ εkjm δil þ εkjl δim - εkji δlm 4 3 3  9i  =δ ε þ δ ε þ δkl εimj þ δkm εilj 4 il kmj  im klj ^ ^ ik Q ^ lm = - tr Sj Q ð3:111Þ When S = 1 we can obtain the following results from Eq. (1.25):

122

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles



tr S^μ = 0,



tr S^μ ^Sν = 2ð - 1Þμ δμ -ν , ! pffiffiffi 1 1 1 pffiffiffi

1 -λ Sλ = - 6 = 2ð- 1Þ1þλ C 1μ tr ^Sμ ^Sν ^ 1ν , μ ν λ 

 Sλ ^Sρ = ð- 1Þμþλ δμ -ν δλ -ρ þ δμ -ρ δν -λ tr ^Sμ ^Sν ^

ð3:112Þ

where μ, ν, λ, ρ =  1, 0. We can also obtain the following trace formulas from Eqs. (1.67)–(1.70):



tr T LM ð1Þ = 3δL0 δM0 , tr T L1 M 1 ð1ÞT L2 M 2 ð1Þ = 3ð-1ÞM 1 δL1 L2 δM 1 -M 2 ,

tr T L1 M 1 ð1ÞT L2 M 2 ð1ÞT L3 M 3 ð1Þ pffiffiffi ^1 L ^2 CLL3 M-ML3 M W ðL1 L2 11; L3 1Þ, = 3 3ð-1ÞL1 þL2 þL3 þM 3 L 1 1 2 2

^1 L ^2 L ^3 tr T^ L1 M 1 ð1ÞT^ L2 M 2 ð1ÞT^ L3 M 3 ð1ÞT^ L4 M 4 ð1Þ = 9ð-1ÞL1 þL2 þL3 þL4 þM 4 L X 1 þM 2 4 ^ LL M LC CLL4M-M W ðL1 L2 11; L1ÞW ðLL3 11; L4 1Þ 1 M 1 L2 M 2 1 þM 2 L3 M 3 L2

ð3:113Þ Utilizing Eq. (3.9) we can further obtain the following relations from Eq. (3.113):



pffiffiffi tr ^SM 1 T^ L2 M 2 ð1Þ = tr T^ L2 M 2 ð1Þ^SM 1 = 6ð - 1ÞM 1 δ1L2 δM 1 -M 2 ð3:114Þ pffiffiffi

^2 CL3 -M 3 W ð1L2 11; L3 1Þ, tr ^SM 1 T^ L2 M 2 ð1ÞT^ L3 M 3 ð1Þ = - 3 6ð - 1ÞL2 þL3 þM 3 L 1M 1 L2 M 2 ð3:115Þ pffiffiffi 3 ^2 C LL3 M-M1M tr T^ L2 M 2 ð1Þ^SM 1 T^ L3 M 3 ð1Þ = - 3 6ð - 1ÞL2 þL3 þM 3 L W ðL2 111; L3 1Þ 2 2 1 pffiffiffi M3 ^ L3 -M 3 = 3 6ð - 1Þ L2 C 1M 1 L2 M 2 W ð1L2 11; L3 1Þ

= - ð - 1ÞL2 þL3 tr ^SM 1 T^ L2 M 2 ð1ÞT^ L3 M 3 ð1Þ



ð3:116Þ Based on Eq. (3.107) there is



tr T^ L2 M 2 ð1ÞT^ L3 M 3 ð1Þ^SM 1 = tr ^SM 1 T^ L2 M 2 ð1ÞT^ L3 M 3 ð1Þ

ð3:117Þ

We can also obtain pffiffiffi

3 -M 3 W ð1111; L3 1Þ tr S^M 1 ^SM 2 T^ L3 M 3 ð1Þ = 6 3ð-1ÞL3 þM 3 C L1M 1 1M 2

ð3:118Þ

3.2

Some Mathematical Formulas Related to Spin Operators. . .

123



2 ^3 C 11M-M tr S^M 1 T^ L3 M 3 ð1Þ^SM 2 = 6ð-1ÞL3 þM 3 L W ð1L3 11; 11Þ 1 L3 M 3 pffiffiffi

3 -M 3 SM 2 T^ L3 M 3 ð1Þ W ð1111; L3 1Þ = ð-1ÞL3 tr ^ SM 1 ^ = 6 3ð-1ÞM 3 C L1M 1 1M 2

ð3:119Þ

Based on Eq. (3.107) the following relations are satisfied:





tr T^ L3 M 3 ð1Þ^SM 1 ^SM 2 = tr ^SM 1 ^SM 2 T^ L3 M 3 ð1Þ = tr ^ SM 2 T^ L3 M 3 ð1Þ^ SM 1

3.2.5

ð3:120Þ

General Spin Wave Functions and Density Matrices for S = 1 Particles

Any spin wave function of the spin 1 particles can be expanded as follows: X

χ=

X

am χ m =

m = 1,0

X

ð-1Þm a - m χ m =

ai χ i

ð3:121Þ

i = x, y, z

m = 1,0

Its Hermitian conjugate wave function χ + is X

χþ =

m = 1,0

am χ þ m=

X m = 1,0

ð-1Þm a- m χ þ m=

X

ai χ þ i

ð3:122Þ

i = x, y, z

*

Usually one calls a the polarization vector of spin 1 particles. Its normalization condition χ +χ = 1 requires *2 * * a = a  a = 1 *

*

ð3:123Þ

*

Let a = a r þ ia i , so we can obtain     * * * * * * * a × a = - i a r × a i - a i × a r = - 2i a r × a i

*

*

ð3:124Þ

*

Then we can define a real vector p with a polarization vector a h i * * p i a×a

*

*

ð3:125Þ *

It can be seen that vectors p is perpendicular to vector a .

124

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

Using the spherical basis coordinate expansion formula *

A=

1 X

*

ð-1Þμ Aμ e - μ =

μ= -1

1 X

*

A-μ e -μ

ð3:126Þ

μ = -1

we can write the following expression from Eq. (3.125): *

p =i

X mn

  * * a - m a - n e - m × e - n

ð3:127Þ

Using expression pffiffiffi * * e m × e n = i 2C11mmþn 1n e mþn

*

ð3:128Þ

we can obtain pffiffiffi * * * mþn * e - m × e - n = ð-1Þ - n e - m × e n = i 2ð-1Þ - n C 11 - m 1n e -mþn pffiffiffi * m-n 1 -n = - i 2ð-1Þ C 1 -m 1 m-n e -mþn

*

ð3:129Þ

Substituting Eq. (3.129) into Eq. (3.127), we can get pffiffiffiP * p = 2 a -m a -n ð-1Þm-n C 11 -n -m 1 m-n e -mþn mn pffiffiffiP * = 2 am an ð-1Þ -mþn C11 nm 1 -mþn e m-n

*

ð3:130Þ

mn

Comparing this formula with Eq. (3.126), we have pffiffiffiX pμ = 2 C11 nm 1 μ am an ,

m, n, μ =  1, 0

ð3:131Þ

mn

Refer to Eqs. (1.98) and (1.79), we obtain rffiffiffi 2 t , pμ = 3 1μ

μ =  1, 0

ð3:132Þ

In fact, according to Eq. (1.93), the transformation from t1μ to pμ is the same as the transformation from T^ 1μ to ^Sμ, so Eq. (3.132) can be obtained directly from Eq. (3.9). When S = 1 the normalized density matrix ^ρ can be divided into three terms according to the L value based on Eqs. (1.81) and (1.92):

3.2

Some Mathematical Formulas Related to Spin Operators. . .

125

χχ þ = ^ρ = ^ρ0 þ ^ρ1 þ ^ ρ2

ð3:133Þ

We can obtain the following expression from Eqs. (1.92), (1.99), and (3.8): ^ρ0 =

1 ^ 1 t T = ^I 3 00 00 3

ð3:134Þ

The following expression can be obtained from Eqs. (1.92), (3.9), and (3.132): ^ρ1 =

1 1 X ð-1Þμ t 1 3 μ = -1

^

=

-μ T 1μ

1 1 X 1→ Sμ = p  ^ ð-1Þμ p -μ ^ S 2 μ = -1 2

ð3:135Þ

Next, order the rank-2 irreducible tensor p2 = t 2

ð3:136Þ

usually t2 or p2 is referred to as the statistical tensor or polarization tensor, and we can write ^ρ2 =

2 1 X ð-1ÞM t 2 3 μ= -2

^

-M T 2M

=

1 p  T^ 3 2 2

ð3:137Þ

where p2  T^ 2 in the above formula represents the scalar product of two rank-2 irreducible tensors defined by Eq. (1.41). We can get the following expression from Eqs. (1.98), (1.79), and (3.136): pffiffiffiX p2M = t 2M = 5 C 11 nm

2 Ma

m n

a ,

m, n =  1, 0; M =  2,  1, 0 ð3:138Þ

mn

Then in S = 1 case the density matrix can be expressed in compact form and in spherical basis coordinate system according to Eqs. (3.133)–(3.135) and Eq. (3.137), respectively: χχ þ = ^ρ =

  1 ^ 3* ^ I þ p  S þ p2  T^ 2 3 2

1 2 X 1 ^ 3 X Iþ ð-1Þμ p - μ ^Sμ þ ð-1ÞM p2 χχ = ^ρ = 3 2 μ = -1 M = -2 þ

ð3:139Þ ! ^

- M T 2M

ð3:140Þ

In the following context, Eq. (3.139) will be discussed in the Cartesian coordinate * system. Each term in Eq. (3.139) is a square 3 × 3 matrix in the spin space. p is a ^ vector in the spatial coordinate composed of three real components. S is a vector in the spatial coordinate composed of three square 3 × 3 matrices in spin space. p2 is a

126

3 Polarization Theory of Nuclear Reactions for Spin 1 Particles

rank-2 tensor in spatial coordinate composed of nine numerical components. T^ 2 is a rank-2 tensor in spatial coordinate composed of nine rank-2 tensors presented by → S in square 3 × 3 matrices in spin space. The three components of the vector p and ^ the Cartesian coordinate system are pi and ^Si ði = x, y, zÞ. The scalar product of two rank-2 tensors in a Cartesian coordinate system is defined as T  Q=

X ik

! * * T ik e k e i

:

X ** Qlm e l e m

! =

lm

X T ik Qik

ð3:141Þ

ik

where the symbol “:” represents taking the scalar product of two vectors, that are ^ ik in the close to each other, twice. The 3.1 section has pointed out that nine of Q ^ Cartesian coordinate system are equivalent to five of T 2M in the spherical basis coordinate systems, and the nine components of the rank-2 tensor p2 and T^ 2 in the ^ ik ði, k = x, y, zÞ, respectively. According to Cartesian coordinate system are pik and Q Eq. (3.21) it can be seen that the transformation relation of the five independent polarization tensor operators from a spherical basis coordinate system to a Cartesian coordinate system is 1 0 rffiffiffi 1^ 0 B 6Qxx - yy C 1 C B pffiffiffi C B ffiffi ffi r 2 0 1 B C B B ^ xx - yy 2^ H C B B Q 1 C B B C B 3 xy C B - i pffiffiffi B H ^ xy C B C B B 2 B C B rffiffiffi C B B C B C B 2 B H ^ xz C  B ^ C=B Q B C B 3 xz C B 0 B C B C B B H C rffiffiffi C B @ ^ yz A B C B B 2^ C B 0 B Q ^ zz yz C B B H 3 C @ B C B rffiffiffi A @ 0 1^ Qzz 2

0

0

0

0

1 - pffiffiffi 2 1 i pffiffiffi 2 0

0 0 1

1 1 pffiffiffi 2C 1 C0 ^ C T2 2 1 C C 0 i pffiffiffi CB B T^ 2 1 C 2C C CB C CB B T^ 2 0 C 1 C B C pffiffiffi 0 CB C CB ^ 2 C@ T 2 -1 C A C 1 i pffiffiffi 0 C C T^ 2 -2 A 2 0 0 0

ð3:142Þ ^ such that In the above formula the 5 × 5 matrix can be denoted by U

Some Mathematical Formulas Related to Spin Operators. . .

3.2

127

10 1 1 1 1 1 pffiffiffi pffiffiffi 0 0 0 p ffi p ffiffi ffi ffiffi i 0 0 0 B 2 2C C 2 2 CB B C CB B B C 1 1 CB B 1 1 ffiffi ffiffi p ffi p ffi i 0 i 0 0 CB 0 B pffiffiffi - i pffiffiffi 0 C 0 C B 2 2C B 2 2 C C B B C CB C ^U ^þ =B 1 1 U C B 0 0 0 0 1 0 B C 0 CB - pffiffiffi 0 pffiffiffi B C C B 2 2 1 1 B C CB 0 B pffiffiffi - i pffiffiffi 0 C 0 C B B C 1 1 2 2 B 0 C i pffiffiffi 0 i pffiffiffi 0 C CB B @ A A @ 2 2 1 1 pffiffiffi - i pffiffiffi 0 0 0 0 0 1 0 0 2 2 1 0 1 0 0 0 0 C B B0 1 0 0 0C C B C B C =B 0 0 1 0 0 C B C B B0 0 0 1 0C A @ 0 0 0 0 1 0

ð3:143Þ ^ i given by Eq. (3.142) is the unitary Apparently the transformation from T^ 2M to H  ^ xy , H ^ xz , H ^ yz , H ^ zz of the polari^ transformation, then the basis functions H xx - yy , H zation tensor form the unit vectors of the coordinate axes of the five-dimensional orthogonal Cartesian coordinate system, and the corresponding 



hxx - yy , hxy , hxz , hyz , hzz 

rffiffiffi ! rffiffiffi rffiffiffi rffiffiffi rffiffiffi 2 2 2 1 1 , p , p , p , p p 3 xy 3 xz 3 yz 2 zz 6 xx - yy

are the particle polarization tensor components in the five-dimensional orthogonal Cartesian coordinate system. From Eq. (1.93) we can see that pij should be the ^ ij , so that the transformation relation from p2M to pij is the expectation value of Q ^ ij . Therefore we can get same as the transformation relation from T^ 2M to Q P M

ð-1ÞM p2

^

- M T 2M

=

 2 ^ ^ xz þ pyz Q ^ yz p Q þ pxz Q 3 xy xy 1 ^ xx - yy þ 1 pzz Q ^ zz þ pxx - yy Q 6 2

ð3:144Þ

^ xx þ Q ^ yy þ Q ^ zz = 0 and pxx + pyy + pzz = 0 we can where pxx - yy  pxx - pyy. From Q obtain

128

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

    1 ^ yy þ 1 pzz Q ^ xx - Q ^ zz = 1 pxx Q ^ xx þ pyy Q ^ yy þ pzz Q ^ zz pxx - pyy Q 6 3 2

ð3:145Þ

So using Eq. (3.20) we can obtain the following result from Eq. (3.144): X M

ð-1ÞM p2

^

-M T 2M

=

 2 ^ ^ xz þ pyz Q ^ yz pxy Qxy þ pxz Q 3 þ

X  1 ^ ^ yy þ pzz Q ^ zz = 1 ^ p Q pxx Qxx þ pyy Q 3 3 ij ij ij

ð3:146Þ

If using the method given in the monograph Quantum Theory of Angular Momentum [1], p ffiffiffi the transformation coefficients in the right end of Eq. (3.21) are all divided by 3 , and the ptransformation coefficients in the right end of Eq. (3.22) are all ffiffiffi multiplied by 3, in this case the corresponding results obtained by the transforma^ 0 , and we can get tion relation Eq. (3.142) are Q ij X ð-1ÞM p2 M

^

- M T 2M

=

X

^0 p0ij Q ij

ð3:147Þ

ij

^0 . However, it should be noted that there are five items in T^ 2M and nine items in Q ij ^ 0 is 1.8 times smaller than the average value of Therefore the average value of p0ij Q ij pffiffiffiffiffiffiffi ^ 0 is 1:8 ≈ 1:34 times smaller than the ð-1ÞM p2 - M T^ 2M , and the average value of Q ij ^ ^ average r ffiffiffiffiffiffiffi value of T 2M . When using Eq. (3.146), the average value of Qij is 3 ≈ 1:29 times larger than the average value of T^ 2M . It can be seen that by 1:8 ^ ij is not much using any of the two methods, the range of the expectation value of Q different from the range of the expectation value of T^ 2M . Although the transformation method given by Eq. (3.147) based on Reference [1] is better in mathematical form, however the transformation method given by Eq. (3.146), Eq. (3.21), and Eq. (3.22) has been widely used by people engaged in polarization theory research [2, 3, 6]. Therefore, this book also uses the transformation method represented by Eq. (3.146), that is, using transformation formulas given by Eq. (3.21) and Eq. (3.22). According to the above choice in the Cartesian coordinate system, Eq. (3.139) can be rewritten as 1 ^ 3X ^ 1X ^ Iþ pS þ p Q χχ þ = ^ρ = 3 2 i i i 3 ik ik ik ^ ik are all square 3 × 3 matrices in spin space. where ^ρ, ^I, ^Si , Q In the Cartesian coordinate system there is

! ð3:148Þ

3.2

Some Mathematical Formulas Related to Spin Operators. . . *

*

*

e k × e l = εikl e i ,

i, k, l = x, y, z

129

ð3:149Þ

It follows that Eq. (3.125) can be rewritten as *

p =i

X * εikℓ ak aℓ e i

ð3:150Þ

ikl

that is, pi = i

X

εikℓ ak al ,

i, k, ℓ = x, y, z

ð3:151Þ

kℓ

In the following context, we will find the matrix element of the lth row and mth column in the spin space for each term of Eq. (3.148) in the Cartesian coordinate system and the Cartesian basis representation. First we can write the following expressions according to Eqs. (3.121) and (3.122): 

ρℓm = aℓ am 1^ 1 I = δ 3 ℓm 3 ℓm

ð3:152Þ ð3:153Þ

Using Eq. (3.151) and the expression for the matrix element of ^ Si in the Cartesian basis representation given by Eq. (3.48), we can obtain  1 XX 1 1 X ^  p S = ε a a ε = a a - am aℓ 2 i i i ℓm 2 i jn ijn j n iℓm 2 ℓ m

ð3:154Þ

^ ik in the Cartesian basis Then using the expression for the matrix element of Q representation given by Eq. (3.52), we can obtain  1 X ^  1X 2 pik Qik ℓm = pik δiℓ δkm þ δim δkℓ - δik δℓm 9 ik 6 ik 3 " ! # 2 X 1 =pii δℓm pℓm þ pmℓ 3 6 i

ð3:155Þ

From Eq. (3.148) the following formula can be obtained using Eqs. (3.152)–(3.155): !  1 1 1  1 X  pii δℓm 0 = δℓm - aℓ am þ am aℓ - ðpℓm þ pmℓ Þ þ 6 3 2 9 i From the above formula one can immediately see

ð3:156Þ

130

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

pℓm = pmℓ ,

ℓ ≠ m;

ℓ, m = x, y, z

ð3:157Þ

which means that p2 is a symmetric tensor in spatial coordinate. Let X

pii = 0

ð3:158Þ

i = x, y, z

that is, require that p2 is not a trace tensor in spatial coordinate. So from Eq. (3.156) we can obtain  3 2   pℓm = a a þ am aℓ - δℓm 2 ℓ m 3

ð3:159Þ

From this formula we can see that pℓm is a real number. And only five components pℓm are independent. Here Eq. (3.159) describing pℓm was proved in the Cartesian basis representation. We can obtain the following results according to Eq. (1.89) and using Eqs. (3.140), (3.112)–(3.114)

  ^S = χ þ ^Sχ = tr ^ρ^S = * p  

T^ 2M = χ þ T^ 2M χ = tr ^ρT^ 2M = p2M

ð3:160Þ ð3:161Þ

We can also obtain Eq. (3.160) using Eq. (3.148), Eq. (3.104), and Eq. (3.105). The following relation can also be obtained:  X 9  

2 ^ ℓm = 1 ^ ℓm χ = tr ^ρ Q ^ ℓm = χ þ Q pik δiℓ δkm þ δkℓ δim - δik δℓm = pℓm Q 3 9 ik 2 ð3:162Þ Note that Eqs. (3.157) and (3.158) are needed when deriving the above formula.

3.3

Vector Polarization Rate and Tensor Polarization Rate of Spin 1 Incident Particles →

Select the direction k in of the incident particles as the z-axis, where the angle → between the directionk out of the  outgoing particles →  and the z axis is θ. Next, define → → →   → n is perpendicular to the the unit vector n = k in × k out = k in × k out  , where ^ →

→

→

reaction plane formed by k in and k out . If the y-axis is selected in the direction n ,

3.3

Vector Polarization Rate and Tensor Polarization Rate of Spin. . .

131

→

and k out is on the xz plane, then the azimuthal angle of the outgoing particle φ = 0. The results of the second chapter of this book have pointed out that in the xyz 1 coordinate system mentioned above, for the problem of spin particle scattering, if 2 the incident particles are unpolarized, then only the polarization rate of the outgoing particles in the y-axis direction is not equal to 0, that is, the outgoing particles are → → polarized along the direction n . Therefore, n is sometimes referred to as the polarization direction of the outgoing particles. In fact, if the incident particles are polarized, then the polarization rates of the outgoing particles may not be equal to 0 in x, y, and z directions simultaneously. The previous discussion was carried out in the xyz coordinate system with the incident particle direction along the z-axis, that is, it is the helicity coordinate system of the incident particles. Let us now discuss the polarization rate of spin S = 1 incident particles. Polarization particle beam is provided by the polarization ion sources, and the quantum axis of incident particles is consistent with the magnetic field direction of the experimental device. The direction of the quantum axis of the particles is just the polarization direction of the particles. We select the Z axis along the quantum axis of the particles, and the X axis and the Y axis are arbitrary in XYZ Cartesian coordinate system. The xyz and XYZ systems are both different spatial coordinate system. First, we will carry out studying in the XYZ coordinate system. We use Nμ to represent the share of spin 1 particles in the spin magnetic quantum number μ (=  1, 0) subspace, and obviously it should be satisfied that N 1 þ N 0 þ N -1 = 1:

ð3:163Þ

1 particles in Sect. 2.2, in XYZ coordinate system 2 taking spin S = 1 particle polarization direction as the Z-axis, the three components of the polarization vector and the five independent components of the polarization tensor of the incident particles should all be equal to 0 except pZ and pZZ, and in this case the particle density matrix should be a diagonal matrix. Refer to Eq. (2.39) in this case the density matrix should be Refer to the discussion on spin

0 pffiffiffiffiffiffi 1 0 1 0 N1 B C pffiffiffiffiffiffi  B pffiffiffiffiffiffi C  pffiffiffiffiffiffi  C C N1 0 0 þ B ρ^0 = B N0 0 @ 0 A @ N0 A 0 0 0 0 1 0 1 0 N1 0 0 B C C pffiffiffiffiffiffiffiffiffi B C N -1 = B 0 C þB @ 0 A 0 0 @ 0 N0 A pffiffiffiffiffiffiffiffiffi 0 0 N -1 N -1

ð3:164Þ

The following results can be obtained according to Eqs. (3.163), (3.164), and (3.33):

132

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles



^Z ^ I = trf^ρ0 g = N 1 þ N 0 þ N -1 = 1, pZ = tr S ρ0 = N 1 - N -1 ,

pi = tr ^Si ^ρ0 = 0, i = X, Y

ð3:165Þ

We can also obtain the following results according to Eqs. (3.163), (3.164), and (3.36):

^ ZZ ^ ρ0 = N 1 - 2N 0 þ N -1 , pZZ = tr Q

^ i ^ρ0 = - 1 ðN 1 - 2N 0 þ N -1 Þ = - 1 pZZ , i = XX, YY pi = tr Q 2 2

^ j ^ρ0 = 0, j = XY, XZ, YZ, XX - YY pj = tr Q

ð3:166Þ

This case also belongs to the special polarized particle states, for which the states with different spin magnetic quantum numbers are incoherent, but they may have different shares. It can be seen that the result given by Eq. (3.166) satisfies pXX + pYY + pZZ = 0 given by Eq. (3.158). We can also see that 1 ≥ pZ ≥ - 1 1 and 1 ≥ pZZ ≥ - 2. When N 1 = N 0 = N - 1 = , there are pZ = 0 and pZZ = 0; in this 3 case the particles are unpolarized. And notice that as long as there is N1 = N-1, there 1 is pZ = 0; as long as there is N 0 = , there is pZZ = 0 3 It is also noted that if the coordinate transformation method represented by ^ 0 = p1ffiffi Q ^ and based on Eq. (3.166) there is p0 = Eq. (3.147) is used, since Q zz ZZ 3 zz p1ffiffi p , then it can be ensured that the results given by the right end of the second ZZ 3 equal sign of Eq. (3.146) and the right end of Eq. (3.147) are equal to each other. Assuming that the angle between the Z axis and the z axis is β, the angle between the projection of the Z axis on the xy plane of the xyz coordinate system and the x axis is ψ. This is the usual spherical coordinate selection method. It is easy to write px = pZ sin β cos ψ,

py = pZ sin β sin ψ,

pz = pZ cos β

ð3:167Þ

The above equations can be written in a matrix form 0

1

0

U xx B C B p @ y A = @ U yx

U xy U yy

U zx

U zy

px pz

10 1 sin β cos ψ 0 CB C sin β sin ψ A@ 0 A cos β pZ

ð3:168Þ

And it can be rewritten by the vector and matrix form as *

^* p xyz = U p XYZ

ð3:169Þ

^ is a real 3 × 3 matrix. It can be seen from Eq. (3.168) that only Uxz, Uyz, Uzz where U e as the transpose matrix of U ^ , and require that U ^ is a unitary are known. Let us set U ^U e = I. matrix, that is, U

3.3

Vector Polarization Rate and Tensor Polarization Rate of Spin. . . →

133

→

For vector A and B , we define the outer product 1 0 Ax Bx Ax By Ax Bz   C B C B ðABÞ  A B = @ Ay A Bx By Bz = @ Ay Bx Ay By Ay Bz A Az Bx Az By Az Bz Az 0

** e

Ax

1

ð3:170Þ

* * * e *e e so we can obtain ^ A, we have A 0 = A U, Then, letting A 0 = U

^ ðABÞU e ðABÞ0 = U

ð3:171Þ

From Eq. (3.166) we can see that the tensor polarization rate in the coordinate system XYZ is 0 B B p p XYZ = B @

-

**

1

1 p 2 ZZ

0

0

1 - pZZ 2 0

0 0

C C 0 C A

ð3:172Þ

pZZ

And it can be rewritten as [2] 0

0 3 ** B p p XYZ = pZZ @ 0 2 0

1 0 0 0 1 C 1 B 0 0 A - pZZ @ 0 2 0 1 0

0 1

1 0 C 0A

0

1

ð3:173Þ

Transforming the above formula into the xyz coordinate system with Eq. (3.171), we can obtain 0

0 0

3 ^B p p xyz = pZZ U @0 0 2 0 0

**

0

1

0

1 0

B Ce 1 0 AU - pZZ @ 0 1 2 0 0 1

0

1

C 0A 1

ð3:174Þ

^U e = I is used in the above formula. The following relation can be obtained where U by Eqs. (3.174) and (3.168): 0 **

p p xyz =

B 3 ^B pZZ U @ 2

0

0

0

0

sin β cos ψ

sin β sin ψ

0

1

0

1 0

C 1 B C - pZZ B 0 1 A 2 @ cos β 0 0 0

0

1

C 0C A 1 ð3:175Þ

Further we can get the following results from the above formula and Eq. (3.168):

134

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

 1 3 sin 2 β cos 2 ψ - 1 pZZ , 2  1 pzz = 3 cos 2 β - 1 pZZ , 2 3 pxz = sin β cos β cos ψpZZ, 2 pxx =

 1 3 sin 2 β sin 2 ψ - 1 pZZ , 2 3 pxy = sin 2 β cos ψ sin ψpZZ , 2 3 pyz = sin β cos β sin ψpZZ 2

pyy =

ð3:176Þ

From the above formula it can be seen that pxx + pyy + pzz = 0 and we have pxx - yy  pxx - pyy =

3 sin 2 β cos ð2ψ ÞpZZ 2

ð3:177Þ

We can find that the tensor polarization rates of the incident particles in the spherical basis coordinate system are, using Eqs. (3.22), (3.176), and (3.177), as follows: p2 2 =

3.4

pffiffiffi pffiffiffi 3 3 sin 2 β e2iψ pZZ , p2 1 =  sin ð2βÞ eiψ pZZ , 4 4  1  p20 = pffiffiffi 3 cos 2 β - 1 pZZ 2 2

ð3:178Þ

Elastic Scattering Amplitude of Spin 1 Particles with Spinless Targets

Based on Eq. (5.5.51) in Ref. [7], the scattering amplitude of an incident particle with spin 1 and a spinless target can be written as pffiffiffi  i πX j μ j μ j  2iσ ℓ C C Yℓ μ - μ0 ðθ, φÞ F μ0 μ ðθ, φÞ = f C ðθÞδμ0 μ þ 0 0 1 - Sℓ e k ℓj ℓ0 1μ ℓ μ -μ 1μ ð3:179Þ where Sjℓ is the S matrix element. Let   2X j μ Cℓ0 1μ C jℓ μμ -μ0 1μ0 1 - Sjℓ , j

f ℓμ0 μ = ^ℓ here ^ℓ 

ð3:180Þ

pffiffiffiffiffiffiffiffiffiffiffiffiffi 2ℓ þ 1. Next we introduce the following symbols: ℓþ1 0 ℓ ℓ-1 αþ ℓ = 1 - Sℓ , αℓ = 1 - Sℓ , αℓ = 1 - Sℓ

ð3:181Þ

Using the C-G coefficients given in Table 3.1, we can obtain the following expressions by Eqs. (3.180) and (3.181):

3.4

Elastic Scattering Amplitude of Spin 1 Particles with Spinless Targets

f ℓ00 = ðℓ þ 1Þαþ ℓ þ ℓαℓ

135



ð3:182Þ

1 0 ð3:183Þ ðℓ þ 2Þαþ ℓ þ ð2ℓ þ 1Þαℓ þ ðℓ - 1Þαℓ 2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ℓ ðℓ þ 1Þ  þ ℓ ℓ f 10 = f -1 0 = ð3:184Þ αℓ - αℓ2 h i 1 0 f ℓ01 = f ℓ0 -1 = pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ℓðℓ þ 2Þαþ ð 2ℓ þ 1 Þα ð ℓ 1 Þ ð ℓ þ 1 Þα ℓ ℓ ℓ 2ℓðℓ þ 1Þ f ℓ11 = f ℓ-1

-1

=

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi i h 1 ð ℓ - 1Þ ðℓ þ 2Þ þ ℓ ℓ f 1 -1 = f -1 1 = ℓαℓ - ð2ℓ þ 1Þα0ℓ þ ðℓ þ 1Þαℓ2 ℓ ðℓ þ 1Þ

ð3:185Þ ð3:186Þ

We can get the following relations according to Eqs. (3.73), (3.75), and (3.77) in Ref. [7]: rffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffirffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ℓ þ 1 2ℓ þ 1 1 P1 eiφ , P , Yℓ 1 =  Yℓ0 = 4π ℓ 4π ℓðℓ þ 1Þ ℓ rffiffiffiffiffiffiffiffiffiffiffiffiffirffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ℓ þ 1 1 P2 e2iφ Yℓ 2 = 4π ðℓ - 1Þℓðℓ þ 1Þðℓ þ 2Þ ℓ

ð3:187Þ

Next we introduce the following symbols according to Eqs. (3.179)–(3.187): A ðθ Þ = f C ðθ Þ þ BðθÞ = f C ðθÞ þ e

2iσ ℓ

h i i X ðℓ þ 1Þαþ þ ℓα e2iσℓ Pℓ ð cos θÞ ℓ ℓ 2k ℓ

h i i X1 þ ð2ℓ þ 1Þα0ℓ þ ðℓ - 1Þαℓðℓ þ 2Þαþ ℓ 2k ℓ 2

ð3:189Þ

Pℓ ð cos θÞ C ðθ Þ = -

D ðθ Þ = -

ð3:188Þ

 1 X 1  þ pffiffiffi αℓ - αℓ- e2iσℓ P1ℓ ð cos θÞ 2k ℓ 2

ð3:190Þ

h i 1 X 1 1 pffiffiffi ℓðℓ þ 2Þαþ - ð2ℓ þ 1Þα0ℓ - ðℓ - 1Þðℓ þ 1Þαℓℓ 2k ℓ 2 ℓðℓ þ 1Þ e2iσ ℓ P1ℓ ð cos θÞ

ð3:191Þ h i X 1 i 0 E ðθ Þ = e2iσ ℓ P2ℓ ð cos θÞ ℓαþ ℓ - ð2ℓ þ 1Þαℓ þ ðℓ þ 1Þαℓ 2k ℓ 2ℓðℓ þ 1Þ ð3:192Þ Given these, the following relations can be obtained by Eq. (3.179):

136

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

F 00 ðθ, φÞ = AðθÞ, F 11 ðθ, φÞ = F - 1 - 1 ðθ, φÞ = BðθÞ, F 10 ðθ, φÞ = - iCðθÞe - iφ , F - 1 0 ðθ, φÞ = iC ðθÞeiφ , F 0 1 ðθ, φÞ = iDðθÞeiφ , F 0 -1 ðθ, φÞ = - iDðθÞe - iφ , F1

-1 ðθ,

φÞ = - E ðθÞe - 2iφ , F - 1 1 ðθ, φÞ = - EðθÞe2iφ ð3:193Þ

In this way, the matrix elements given by Eq. (3.179) can form the following scattering matrix in the spherical basis representation: 0

B ðθ Þ

B ^ ðθ, φÞ = B iDðθÞeiφ F @ 0

- EðθÞe2iφ B  ðθ Þ

B ^ þ ðθ, φÞ = B iC  ðθÞeiφ F @ - E  ðθÞe2iφ

- iC ðθÞe - iφ AðθÞ iC ðθÞeiφ - iD ðθÞe - iφ A ðθÞ iD ðθÞeiφ

- EðθÞe - 2iφ

1

C - iDðθÞe - iφ C A, B ðθ Þ - E ðθÞe - 2iφ

1

ð3:194Þ

C - iC ðθÞe - iφ C A B  ðθ Þ

Skipping the argument (θ) of A, B, C, D, and E, the following expression can be obtained by Eq. (3.194): 0

10 1 B B - iCe - iφ - Ee - 2iφ - iD e - iφ - E  e - 2iφ B C B C  iφ B ^F ^ þ = B iDeiφ F A - iDe - iφ C A - iC  e - iφ C @ A@ iC e A - Ee2iφ iCeiφ B - E  e2iφ iD eiφ B   1 0 - iðBD þ CA þ ED Þe - iφ - BE  þ jC j2 þ EB e - 2iφ jBj2 þ jC j2 þ jE j2 C B C B 2 2     iφ   - iφ C =B Þ i ð DB þ AC þ DE Þe 2 j D j þ j A j i ð DB þ AC þ DE e C B A @   2 2 2 2      iφ 2iφ - EB þ jC j þ BE e iðBD þ CA þ ED Þe jBj þ jC j þ jE j

ð3:195Þ And it can be found that the differential cross section of the unpolarized incident particles is I0 =

n þo h  i 1 ^F ^ = 1 jAj2 þ 2 jBj2 þ jCj2 þ jDj2 þ jE j2 tr F 3 3

ð3:196Þ

According to Eq. (3.140) the scattering matrix given by Eq. (3.194) can be expanded in the spherical basis coordinate system X 1X ^ = 1 u ^I þ 1 F ð-1Þμ v - μ ^Sμ þ ð-1ÞM w - M T^ 2M 3 M 3 2 μ

ð3:197Þ

From Eq. (3.194) and Eq. (3.197), the following relations can be obtained by using ^ Sμ given by Eq. (3.32) and T^ 2M given by Eq. (3.38) in the spherical basis representation

3.4

Elastic Scattering Amplitude of Spin 1 Particles with Spinless Targets

1 1 1 u þ v0 þ pffiffiffi w0 , 3 2 3 2 1 - Ee - 2iφ = pffiffiffi w - 2 , 3 pffiffiffi 2 1 w , A= u3 0 3 1 - Ee2iφ = pffiffiffi w2 , 3 1 1 1 B = u - v0 þ pffiffiffi w0 3 2 3 2

B=

1 1 v þ pffiffiffi w - 1 , 2 -1 6 1 1 iDeiφ = - v1 - pffiffiffi w1 , 2 6 1 1 - iDe - iφ = v - 1 - pffiffiffi w - 1 , 2 6 1 1 iφ iCe = - v1 þ pffiffiffi w1 , 2 6

137

- iCe - iφ =

ð3:198Þ

From the above formulas we can obtain u = A þ 2B, v -1 = - iðD þ CÞe - iφ , v0 = 0, v1 = - iðD þ C Þeiφ , rffiffiffi pffiffiffi pffiffiffi -2iφ 3 w -2 = - 3Ee , w -1 = i ðD - CÞe - iφ , w0 = - 2ðA - BÞ, 2 rffiffiffi pffiffiffi 3 ðD - C Þeiφ , w2 = - 3Ee2iφ w1 = - i 2

ð3:199Þ

^ contains the vector polarization terms ^ The above results show that F S1 and ^ S -1 ^ as well as tensor polarization terms T 2M ðM =  2, 1, 0Þ. If φ = 0 we have that S1 for the vector v-1 = v1, w-2 = w2, w-1 = -w1, then there is only one term ^ polarization and there are only three terms T^ 20 , T^ 21 , T^ 22 for the tensor polarization. We can get the following expressions utilizing Eqs. (1.23) and (3.199): 1X ð-1Þμ v - μ ^Sμ 2 μ   1   1 1 = pffiffiffi v -1 ^Sx þ i^Sy þ v0 ^Sz - pffiffiffi v1 ^Sx - i^ Sy 2 2 2 2 2 1 1 i = pffiffiffi ðv -1 - v1 Þ^Sx þ v0 ^Sz þ pffiffiffi ðv -1 þ v1 Þ^ Sy 2 2 2 2 2     i 1 Sy = - pffiffiffi ðD þ C Þ e - iφ - eiφ S^x þ pffiffiffi ðD þ CÞ e - iφ þ eiφ ^ 2 2 2 2 1 1 = - pffiffiffi ðD þ C Þ sin φ ^Sx þ pffiffiffi ðD þ C Þ cos φ ^ Sy 2 2   1 = pffiffiffi ðD þ C Þ cos φ S^y - sin φ S^x 2

ð3:200Þ

We can also get the following expressions utilizing Eqs. (3.22) and (3.199):

138

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

1X ð-1ÞM w - M T^ 2M 3 M

    1 1 1 ^ xy - p ^ yz þ p ^ xx - yy -2iQ ^ xz - iQ ^ zz ffiffiffi w1 Q ffiffiffi w0 Q = pffiffiffi w2 Q 3 3 3 2 6 3     1 1 ^ yz þ p ^ xy ^ xz þ iQ ^ xx - yy þ 2iQ ffiffiffi w -2 Q þ pffiffiffi w -1 Q 3 3 6 3   1 i ^ xz ^ zz þ pffiffiffi ðD - CÞ eiφ þ e - iφ Q = - ðA - BÞQ 3 3 2     1 ^ yz - 1 E e2iφ þ e -2iφ Q ^ xx - yy þ pffiffiffi ðD - C Þ eiφ - e - iφ Q 6 3 2  i  ^ xy þ E e2iφ - e -2iφ Q 3 pffiffiffi   1 ^ xz þ sin φ Q ^ yz ^ zz þ i 2 ðD - C Þ cos φ Q = - ðA - BÞQ 3 3  1  ^ xy ^ xx - yy þ 2 sin ð2φÞQ - E cos ð2φÞQ 3

ð3:201Þ

Substituting the first equation of Eq. (3.199), Eq. (3.200), and Eq. (3.201) into Eq. (3.197), we can obtain   ^ = 1 ðA þ 2BÞ^I þ p1ffiffiffi ðD þ CÞ cos φ^Sy - sin φ^ Sx F 3 2 pffiffiffi   2 ^ yz ^ xz þ sin φQ þi ðD - C Þ cos φQ 3  1  ^ xy - 1 ðA - BÞQ ^ xx - yy þ 2 sin ð2φÞQ ^ zz - E cos ð2φÞQ 3 3

ð3:202Þ

*

If one takes the y-axis in the direction n , in this case there is φ = 0, then Eq. (3.202) is simplified to pffiffiffi ^ xz ^ = 1 ðA þ 2BÞ^I þ p1ffiffiffi ðD þ C Þ^Sy þ i 2 ðD - C ÞQ F 3 3 2 1 ^ 1 ^ - EQ xx - yy - ðA - BÞQzz 3 3

ð3:203Þ

This formula is consistent with the included terms in the result given in Reference [3]. From Eq. (3.203) we can obtain the following result by using the tracing formulas Eqs. (3.104) and (3.105):

3.5

Polarization Theory of Elastic Scattering of Unpolarized Spin. . .

n þo 1 1 1 ^ ^ I 0 = tr F F = ðA þ 2BÞðA þ 2B Þ þ ðD þ C ÞðD þ C  Þ 3 3 3  2 2     þðD - CÞðD - C Þ þ ðA - BÞðA - B Þ þ 2jE j 3 h  i 1 = jAj2 þ 2 jBj2 þ jC j2 þ jDj2 þ jEj2 3

139

ð3:204Þ

This formula is exactly the same as Eq. (3.196). ^ is given by Eq. (3.194) in the spherical basis represenThe scattering matrix F tation, and it can be expanded in the spherical basis coordinate system using Eq. (3.197). Their expansion coefficients have been given by Eq. (3.199). And ^ that has been transformed into the Cartesian Eq. (3.202) gives the scattering matrix F coordinate system. We use F μ0 μ ðμ, μ0 =  1, 0Þ to represent the matrix elements in the spherical basis coordinate system and the spherical basis representation. We can obtain the following results according to Eq. (3.202) as well as Eqs. (3.33) and (3.36) in the spherical basis representation: 1 1 F 1 1 = ðA þ 2BÞ - ðA - BÞ = B, 3 3 1 1 F 1 0 = - ðD þ C Þði cos φ þ sin φÞ þ ðD - CÞði cos φ þ sin φÞ = - iCe - iφ , 2 2 F 1 -1 = - E ½ cos ð2φÞ - i sin ð2φÞ = - Ee -2iφ , 1 1 F 0 1 = ðD þ CÞði cos φ - sin φÞ þ ðD - CÞði cos φ - sin φÞ = iDeiφ , 2 2 1 2 F 0 0 = ðA þ 2BÞ þ ðA - BÞ = A, 3 3 1 1 F 0 -1 = ðD þ CÞð - i cos φ - sin φÞ þ ðD - C Þð - i cos φ - sin φÞ = - iDe - iφ , 2 2 F -1 1 = - E½ cos ð2φÞ þ i sin ð2φÞ = - Ee2iφ , 1 1 F -1 0 = ðD þ C Þði cos φ - sin φÞ - ðD - C Þði cos φ - sin φÞ = iCeiφ , 2 2 1 1 F -1 -1 = ðA þ 2BÞ - ðA - BÞ = B 3 3 ð3:205Þ The above results are consistent with Eq. (3.194). And the above discussions are carried out in the spherical basis representation.

3.5

Polarization Theory of Elastic Scattering of Unpolarized Spin 1 Particles with Spinless Targets

According to Eq. (3.148) we will write out the density matrix of the incident particles ^ ik = Q ^ ki in i ≠ k case, in the density in the Cartesian coordinate system. Because Q matrix expression we only keep one of them but multiply it by 2. It follows that

140

3

^ρin =

Polarization Theory of Nuclear Reactions for Spin 1 Particles

h  2  1 ^ 3 ^ ^ xy þ pxz Q ^ xz þ pyz Q ^ yz I þ px Sx þ py ^Sy þ pz ^Sz þ pxy Q 3 2 3 ð3:206Þ i 1 ^ ^ ^ þ p þ p Q Q þ pxx Q xx yy yy zz zz 3

where pi is the component of the incident particle polarization vector and pik is the component of the incident particle polarization tensor. Eq. (3.19) requires 0

0 0 0

1

C ^ yy þ Q ^ zz = B ^ xx þ Q Q @0 0 0A 0 0 0

ð3:207Þ

pxx þ pyy þ pzz = 0

ð3:208Þ

and Eq. (3.158) requires

By using Eq. (3.104) and Eq. (3.105), the following results are easily demonstrated from Eqs. (3.206) and (3.208):  

pi = ^Si = tr ^Si ^ρin ,  

^ ik = tr Q ^ ik ^ρin , pik = Q

i = x, y, z i,k = x, y, z

ð3:209Þ ð3:210Þ

We can obtain the following relations according to Eqs. (3.207) and (3.208):   1 ^ yy þ 3 pzz Q ^ xx - Q ^ zz = pxx Q ^ xx þ pyy Q ^ yy þ pzz Q ^ zz pxx - pyy Q 2 2   1 ^ xx - Q ^ zz þ 3 pyy Q ^ yy = pxx Q ^ xx þ pyy Q ^ yy þ pzz Q ^ zz p - pzz Q 2 xx 2   1 ^ zz þ 3 pxx Q ^ yy - Q ^ xx = pxx Q ^ xx þ pyy Q ^ yy þ pzz Q ^ zz pyy - pzz Q 2 2

ð3:211Þ ð3:212Þ ð3:213Þ

^ xx - yy appearing in the matrix F ^ given by Eq. (3.203) and the transformation Noting Q formula Eq. (3.22), using Eq. (3.211) we can rewrite Eq. (3.206) as ^ρin =

 2  1 ^ 3 ^ ^ xy þ pxz Q ^ xz þ pyz Q ^ yz I þ px Sx þ py ^Sy þ pz ^Sz þ pxy Q 3 2 3  ð3:214Þ   1 1 ^ ^ ^ þ pxx - pyy Qxx - Qyy þ pzz Qzz 6 2

From Eq. (3.214) we can still obtain Eq. (3.209) and Eq. (3.210) by using Eq. (3.104) and Eq. (3.105). We can also obtain

3.5

Polarization Theory of Elastic Scattering of Unpolarized Spin. . .

141

   

^ xx - Q ^ xx - Q ^ yy = tr Q ^ yy ^ ρin pxx - yy  pxx - pyy = Q

ð3:215Þ

  ^ xx - yy  Q ^ xx - Q ^ yy , Q ^ xz , Q ^ yz , Q ^ zz form 5 independent tensor ^ xy , Q In this case, Q operators. In the practical applications, Eq. (3.206) or Eq. (3.214) can all be used. This book generally uses Eq. (3.214) which contains only 5 independent tensor operators. If the incident particle beam is polarized only in the y-axis direction, that is, 1 ^ xx = Q ^ zz = - 1 Q ^ , then px = pz = 0, pxy = pxz = pyz = 0, pxx = pzz = - pyy , Q 2 2 yy according to Eqs. (3.206) and (3.212), we can write ^ρin =

  1 ^ 3 ^ 1 ^ I þ py Sy þ pyy Q yy 3 2 2

ð3:216Þ

Notice Eq. (3.216) is only a special case. If the incident particles are unpolarized, then we have 1 ^ρ0in = ^I 3 1 ^F ^þ ^ρ0out = F 3

ð3:217Þ ð3:218Þ

In the 3.4 section, Eq. (3.196) has given  i   1 n þo 1 h 2 dσ 0 ^F ^ = = I 0 = tr ^ρ0out = tr F jAj þ 2 jBj2 þ jC j2 þ jDj2 þ jEj2 dΩ 3 3 ð3:219Þ *

When the y-axis is taken along the direction n , there is φ = 0, then Eqs. (3.194) and (3.195) can be simplified, respectively, as 0

B

- iC

-E

1

0

B

- iD

- E

1

C C ^þ B ^ =B ð3:220Þ F = @ iC  A - iC  A A - iD A, F @ iD    -E iD B -E iC B  1 0 -iðBD þCA þED Þ - BE  þ jCj2 þEB jBj2 þ jC j2 þ jEj2 C B C þ B 2 2       B ^ ^ F F = B iðDB þAC þDE Þ 2jDj þ jAj -iðDB þAC þDE Þ C C A @   2 2 2 2      - EB þ jC j þBE iðBD þCA þED Þ jBj þ jCj þ jE j ð3:221Þ

142

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

Next we define the following as the component of the polarization vector of the outgoing particles corresponding to the unpolarized incident particles: n o ^Si0 F ^F ^þ tr 0 Pi0 = n þ o , ^F ^ tr F

i 0 = x0 , y0 , z 0

ð3:222Þ

We also define the following as the component of the polarization tensor of the outgoing particles corresponding to the unpolarized incident particles: n o ^ i0 F ^F ^þ tr Q 0 n þo , Pi0 = ^F ^ tr F

i0 = x0 y0 , x0 z0 , y0 z0 , x0 x0 - y0 y0 , z0 z0

ð3:223Þ

From Eq. (3.222) and Eq. (3.223), we can obtain the following results by using Eqs. (3.33), (3.36), (3.219), and (3.221): h i 0 1 Px0 = pffiffiffi iðDB þ AC  þ DE  Þ - iðDB þ AC  þ DE  Þ = 0 3 2I 0 h 0 i Py0 = pffiffiffi - iðDB þ AC  þ DE  Þ -2iðBD þ CA þ ED Þ 3 2I 0 pffiffiffi i 2 2    - iðDB þ AC þ DE Þ = Re ðDB þ AC  þ DE  Þ 3I 0 h   i 0 1 Pz0 = jBj2 þ jCj2 þ jEj2 - jBj2 þ jCj2 þ jEj2 = 0 3I 0 h   i 0 0 i EB þ jC j2 þ BE  - BE  þ jC j2 þ EB = 0 Px0 y = 2I 0 h 0 0 1 Px0 z = pffiffiffi iðDB þ AC  þ DE  Þ -2iðBD þ CA þ ED Þ 2 2I 0 pffiffiffi i 2 þiðDB þ AC  þ DE  Þ = ImðDB þ AC  þ DE  Þ I0 h i 0 0 i - iðDB þ AC  þ DE  Þ þ iðDB þ AC  þ DE  Þ = 0 Py0 z = pffiffiffi 2 2I 0 h    i 0 0 0 0 1 Px0 x - y y = - EB þ jCj2 þ BE  - BE  þ jC j2 þ EB I0 h i 2 =jCj2 þ 2 Re ðEB Þ I0

ð3:224Þ

ð3:225Þ

ð3:226Þ ð3:227Þ

ð3:228Þ

ð3:229Þ

ð3:230Þ

3.6

Polarization Theory of Elastic Scattering of Spin 1 Particle. . . 0 0

143

h     i 1 jBj2 þ jC j2 þ jE j2 -2 2jDj2 þ jAj2 þ jBj2 þ jCj2 þ jE j2 3I 0   2 = jBj2 þ jC j2 þ jEj2 - jAj2 -2jDj2 3I 0 ð3:231Þ

Pz0 z =

0

From the above results we can see that only Pi0 ði0 = y0 , x0 z0 , x0 x0 - y0 y0 , z0 z0 Þ are not equal to 0. The above results also show that even if the spin 1 incident particle is unpolarized, when the spins of the target and the residual nucleus are equal to 0, the vector polarization rate and tensor polarization rate of the spin 1 outgoing particle are generally not equal to 0. That is, the outgoing particles will also be polarized.

3.6

Polarization Theory of Elastic Scattering of Spin 1 Particle with both Vector Polarization and Tensor Polarization from a Spinless Target

If the spin 1 incident particle contains both vector and tensor polarization components, then the items following the first item in the incident particle density matrix given by Eq. (3.214) should be considered. We define the following as the vector polarization analyzing power n o ^þ ^ ^Si F tr F Ai = n þ o , ^F ^ tr F

i = x, y, z

ð3:232Þ

and define the tensor polarization analyzing power as n o ^ iF ^þ ^Q tr F Ai = n þ o , ^F ^ tr F

i = xy, xz, yz, xx - yy, zz

ð3:233Þ

From Eq. (3.232) and Eq. (3.233), we can obtain the following expressions using Eqs. (3.33), (3.36), (3.219), and (3.220): 0 iC  1 þ  ^ = pffiffiffi @ B - E  S^x F 2 iC 

A 0 A

1 - iC B  - E  A, - iC

144

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

^ ^Sx F ^ þ = p1ffiffiffi F 2 1 0  BA - EA - iðBC  þ CB - CE  - EC Þ iðBC - CB þ CE  - EC  Þ C B C B AB - AE  0 AB - AE  A @           - iðEC - CB þ CE - BC Þ - EA þ BA iðEC þ CB - CE - BC Þ ð3:234Þ 1 Ax = pffiffiffi ½iðBC  - CB þ CE  - EC  Þ þ iðEC  þ CB - CE  - BC  Þ = 0 3 2I 0 ð3:235Þ 0 1    -A iC -iC B  C þ i ^Sy F ^ = pffiffiffi B B þE  -2iD - ðE  þB Þ C, @ A 2 iC A -iC  ^y F ^ þ = piffiffiffi ^S F 2 1 0  -iðBC þCB þCE  þEC  Þ - ðBA þ2CD þEA Þ iðBC  þCE  þCB þEC Þ C B B 2DC  þAB þAE  -2iðDA þAD Þ - ð2DC  þAE  þAB Þ C A @            EA þ2CD þBA -iðEC þCE þCB þBC Þ iðEC þCB þCE þBC Þ ð3:236Þ i Ay = pffiffiffi ½ - iðBC  þ CB þ CE  þ EC  Þ -2iðDA þ AD Þ 3 2I 0 pffiffiffi 2 2 - iðEC  þ CE  þ CB þ BC  Þ = Re ðBC  þ CE  þ DA Þ 3I 0 0  1 - iD - E  B B C ^Sz F ^þ = B 0 0 0 C @ A,    E - iD -B 0 1 2 2 - iðBD - ED Þ - ðBE  - EB Þ jBj - jE j B C ^ þ = B iðDB - DE  Þ ^ ^Sz F F 0 - iðDE  - DB Þ C @ A - ðEB - BE  Þ iðED - BD Þ jE j2 - jBj2 h   i 1 Az = jBj2 - jEj2 þ jEj2 - jBj2 = 0 3I 0

ð3:237Þ

ð3:238Þ

ð3:239Þ

3.6

Polarization Theory of Elastic Scattering of Spin 1 Particle. . .

0

E

B ^ xy F ^ þ = 3i B 0 Q 2@ B 0

- iD 0 - iD

- B

1

C 0 C A,  -E

BE  - EB

B ^ xy F ^Q ^ þ = 3i B iðDE  - DB Þ F 2@ - jE j2 þ jBj2

145

- iðBD - ED Þ 0 iðED - BD Þ

- jBj2 þ jE j2

1 ð3:240Þ

C - iðDB - DE  Þ C A EB - BE 

i ½ðBE  - EB Þ þ ðEB - BE  Þ = 0 ð3:241Þ 2I 0 0 1 A -iC  iC C 3 B ^ xz F ^þ = p ffiffiffi B B þE  -2iD - ðE  þB Þ C Q @ A, 2 2 iC  -iC  -A 3 ^ xz F ^þ = p ^Q ffiffiffi F 2 2 0 1 iðBC  -CB -CE þEC Þ BA -2CD þEA -iðBC  -CE -CB þEC  Þ B C B C -2DC  þAB þAE  2iðDA -AD Þ 2DC  -AE  -AB @ A Axy =

-iðEC  -CB -CE þBC  Þ -EA þ2CD -BA iðEC  -CE  -CB þBC  Þ ð3:242Þ 1 Axz = pffiffiffi ½iðBC  - CB - CE  þ EC  Þ þ 2iðDA - AD Þ 2 2I 0 pffiffiffi ð3:243Þ 2 þiðEC  - CE  - CB þ BC  Þ = ImðBC  þ DA þ EC  Þ I0 0 1    -A iC - iC C 3i B ^ yz F ^þ = p ffiffiffi B B - E  0 - E  þ B C Q @ A, 2 2 - iC  - A iC 3i ^ yz F ^þ = p ^Q ffiffiffi F 2 2 1 0 iðBC  þ CE  - CB - EC  Þ - iðBC  þ CB - CE  - EC  Þ - BA þ EA C B C B AB - AE  0 - AE  þ AB A @ iðEC  þ CB - CE  - BC  Þ EA - BA - iðEC  þ CE - CB - BC  Þ ð3:244Þ

146

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

i Ayz = pffiffiffi ½ - iðBC  þ CB - CE  - EC  Þ - iðEC  þ CE  - CB - BC  Þ = 0 2 2I 0 ð3:245Þ 0 1    iD B -E B C þ ^ xx - yy F ^ = 3B 0 Q 0 0 C @ A,    B - iD -E 0 1 iðBD þ ED Þ - BE  - EB jBj2 þ jE j2 B C ^ xx - yy F ^ þ = 3 B - iðDE  þ DB Þ ^Q F -2jDj2 iðDB þ DE  Þ C @ A - EB - BE  ð3:246Þ h i h i 1 2 Axx - yy = -2ðBE  þ EB Þ -2jDj2 = 2 Re ðBE  Þ þ jDj2 ð3:247Þ I0 I0 0 1 B - iD - E  B C ^ zz F ^ þ = B -2iC  -2A 2iC  C, Q @ A    -E iD B jE j2 þ jBj2

^ zz F ^þ = ^Q F 0 jBj2 -2jC j2 þ jE j2 B B iðDB -2AC  þ DE  Þ @ - EB þ 2jCj2 - BE 

Azz

- iðED þ BD Þ

- iðBD -2CA þ ED Þ 2jDj2 -2jAj2

- BE  þ 2jCj2 - EB

1

C - iðDE  -2AC  þ DB Þ C A

iðED -2CA þ BD Þ

jEj2 -2jC j2 þ jBj2 ð3:248Þ h   i 1 = 2 jBj2 -2jC j2 þ jE j2 þ 2jDj2 -2jAj2 3I 0   ð3:249Þ 2 jBj2 þ jE j2 þ jDj2 - jAj2 -2jC j2 = 3I 0

From the above results we can see that only Ai (i = y, xz, xx - yy, zz) are not equal to 0. According to Eq. (3.214) it can be written that the differential cross section of the outgoing particle corresponding to the polarized incident particle is   3 2 1 1 dσ = I = I 0 1 þ py Ay þ pxz Axz þ pxx - yy Axx - yy þ pzz Azz 2 3 6 2 dΩ

ð3:250Þ

The polarization transfer coefficient from the ith initial polarization vector component of the incident particles to the i’th final polarization vector component of the outgoing particles is defined as

3.6

Polarization Theory of Elastic Scattering of Spin 1 Particle. . .

n o ^Si0 F ^ ^Si F ^þ tr 0 n þo , K ii = ^F ^ tr F

i = x, y, z;

i 0 = x0 , y0 , z 0

147

ð3:251Þ

The polarization transfer coefficient from the ith initial polarization vector component to the i’th final polarization tensor component is defined as n o ^ i0 F ^^ ^þ S F tr Q i 0 n þ o , i = x, y, z; i0 = x0 y0 , x0 z0 , y0 z0 , x0 x0 - y0 y0 , z0 z0 K ii = ^F ^ tr F

ð3:252Þ

The polarization transfer coefficient from the ith initial polarization tensor component to the i’th final polarization vector component is defined as n o ^Si0 F ^ iF ^Q ^þ tr 0 n þ o , i = xy, xz, yz, xx - yy, zz; i0 = x0, y0, z0 K ii = ^F ^ tr F

ð3:253Þ

Finally, the polarization transfer coefficient from the ith initial polarization tensor component to the i’th final polarization tensor component is defined as n o ^ i0 F ^ iF ^Q ^þ tr Q 0 n þo , K ii = ^F ^ tr F

i = xy, xz, yz, xx - yy, zz;

ð3:254Þ

i 0 = x0 y0 , x0 z 0 , y0 z 0 , x0 x0 - y0 y0 , z 0 z 0 The following results can be obtained from Eqs. (3.251)–(3.254) and utilizing ^ iF ^ þ and F ^ þ obtained ^^ ^Q Eqs. (3.33), (3.36), and (3.219) as well as the matrices F Si F early in this section 0

1 ½ðAB - AE  Þ þ 2ðBA - EA Þ þ ðAB - AE  Þ 6I 0 2 = Re ðAB - AE  Þ 3I 0

K xx =

0

K yx =

i ½ð - AB þ AE  Þ þ ðAB - AE  Þ = 0 6I 0

ð3:255Þ

ð3:256Þ

148

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

0 1 K zx = pffiffiffi ½iðBC  - CB þ CE  - EC  Þ þ iðBC  - CB þ CE  - EC  Þ 3 2p I 0ffiffiffi 2 2 ImðBC  þ CE  Þ =3I 0 ð3:257Þ 0 0 i K xx y = pffiffiffi ½ - iðBC  þ CB - CE  - EC  Þ - iðBC  þ CB - CE  - EC  Þ 2 ffiffiffi2I 0 p 2 = Re ðCB - EC  Þ I0 ð3:258Þ 0 0

K xx z = 0 0

1 ½ðAB - AE  Þ - ðAB - AE  Þ = 0 4I 0

i ½ - ðAB - AE  Þ þ 2ðBA - EA Þ - ðAB - AE  Þ 4I 0 1 = ImðAB - AE  Þ I0

K yx z =

ð3:259Þ

ð3:260Þ

0 0 0 0 1 K xx x - y y = pffiffiffi ½ - iðEC  - CB þ CE  - BC  Þ - iðBC  þ CB - CE  - EC  Þ = 0 2I 0 ð3:261Þ 0 0 1 K zx z = pffiffiffi ½iðBC  - CB þ CE  - EC  Þ - iðBC  - CB þ CE  - EC  Þ = 0 3 2I 0 ð3:262Þ

i ½ð2DC  þ AB þ AE  Þ - ð2DC  þ AB þ AE  Þ = 0 6I 0

ð3:263Þ

1 ½ð -2DC  - AB - AE  Þ þ ð -2BA - 4CD -2EA Þ 6I 0 2 Re ðAB þ AE  þ 2DC  Þ þð -2DC  - AB - AE  Þ = 3I 0

ð3:264Þ

0

K xy = 0

K yy = -

0 i K zy = pffiffiffi ½ - iðBC  þ CB þ CE  þ EC  ÞþiðBC  þ CB þ CE  þ EC  Þ = 0 3 2I 0 ð3:265Þ 0 0 1 K xy y = - pffiffiffi ½ - iðBC  þ CB þ CE  þ EC  ÞþiðBC  þ CB þ CE  þ EC  Þ = 0 2 2I 0 ð3:266Þ

3.6

Polarization Theory of Elastic Scattering of Spin 1 Particle. . . 0 0

K xy z =

0 0

K yy z = 0 0

K xy x

149

i ½ðAB þ AE  þ 2DC  Þ -2ðBA þ EA þ 2CD Þ 4I 0 1 þðAB þ AE  þ 2DC  Þ = - ImðAB þ AE  þ 2DC  Þ I0

ð3:267Þ

1 ½ - ðAB þ AE  þ 2DC  Þ þ ðAB þ AE  þ 2DC  Þ = 0 4I 0

ð3:268Þ

- y0 y0

i = pffiffiffi ½iðEC  þ CB þ CE  þ BC  Þ 2I 0 þiðBC  þ CE  þ CB þ EC  Þ = -

pffiffiffi 2 2 Re ðCB þ EC  Þ I0 ð3:269Þ

0 0 i K zy z = pffiffiffi ½ - iðBC  þ CB þ CE  þ EC  Þ þ 4iðAD þ DA Þ 3 2I 0 pffiffiffi ð3:270Þ 2 2 - iðBC  þ CB þ CE  þ EC  Þ = Re ðBC  þ CE  -2DA Þ 3I 0 0 1 K xz = pffiffiffi ½iðDB - DE  Þ -2iðBD - ED Þ þ iðDB - DE  Þ 3 2p I 0ffiffiffi 2 2 =ImðDB - DE  Þ 3I 0 0 i K yz = pffiffiffi ½ - iðDB - DE  Þ þ iðDB - DE  Þ = 0 3 2I 0 h    i   0 1 2 K zz = jBj2 - jE j2 þ jBj2 - jEj2 = jBj2 - jE j2 3I 0 3I 0 0 0

K xz y =

i 2 ½ðEB - BE  Þ - ðBE  - EB Þ = - ImðEB Þ 2I 0 I0 0 0 1 K xz z = pffiffiffi ½iðDB - DE  Þ - iðDB - DE  Þ = 0 2 2I 0

0 0 i K yz z = pffiffiffi ½ - iðDB - DE  Þ -2iðBD - ED Þ - iðDB - DE  Þ 2 2I 0 pffiffiffi 2 = Re ðDB - DE  Þ I0

1 ½ - ðEB - BE  Þ - ðBE  - EB Þ = 0 I0 h   i 0 0 1 K zz z = jBj2 - jEj2 þ jE j2 - jBj2 = 0 3I 0 0 0

K xz x

- y0 y0

=

ð3:271Þ

ð3:272Þ ð3:273Þ ð3:274Þ ð3:275Þ

ð3:276Þ

ð3:277Þ ð3:278Þ

150

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

0 i K xxy = pffiffiffi ½ - iðDB - DE  Þ -2iðBD - ED Þ - iðDB - DE  Þ 2pffiffiffi2I 0 2 = Re ðBD - DE  Þ I0 0 1 K yxy = - pffiffiffi ½iðDB - DE  Þ - iðDB - DE  Þ = 0 2 2I 0

i 2 ½ðBE  - EB Þ - ðEB - BE  Þ = - ImðBE  Þ 2I 0 I0 h   i   0 0 3 3 K xxyy = jE j2 - jBj2 þ - jBj2 þ jE j2 = jBj2 - jEj2 2I 0 4I 0 0

K zxy =

0 0 3i K xxyz = pffiffiffi ½iðDE  - DB Þ þ iðDB - DE  Þ = 0 4 2I 0 0 0 3 K yxyz = - pffiffiffi ½ - iðDE  - DB Þ -2iðBD - ED Þ þ iðDB - DE  Þ 4 2I 0 3 = - pffiffiffi ImðBD þ DE  Þ 2I 0   0 0 0 00 3i - jE j2 þ jBj2 - jBj2 þ jEj2 = 0 K xxyx - y y = 2I 0 0 0

K zxyz = 0

K xxz =

i ½ðBE  - EB Þ þ ðEB - BE  Þ = 0 2I 0

ð3:279Þ

ð3:280Þ ð3:281Þ ð3:282Þ ð3:283Þ

ð3:284Þ

ð3:285Þ ð3:286Þ

1 ½ð -2DC  þ AB þ AE  Þ þ ð2DC  - AE  - AB Þ = 0 4I 0

ð3:287Þ

i ½ð2DC  - AB - AE  Þ þ 2ðBA -2CD þ EA Þ 4I 0 1 þð2DC  - AE  - AB Þ = ImðAB þ AE  -2DC  Þ I0

ð3:288Þ

0

K yxz =

0 1 K zxz = pffiffiffi ½iðBC  - CB - CE  þ EC  Þ - iðBC  - CB - CE  þ EC  Þ = 0 2 2I 0 ð3:289Þ 0 0 3i K xxzy = pffiffiffi ½iðEC  - CB - CE  þ BC  Þ - iðBC  - CE  - CB þ EC  Þ = 0 4 2I 0 ð3:290Þ 0 0

K xxzz =

0 0

K yxzz =

3 ½ðAB þ AE  -2DC  Þ þ 2ðBA þ EA -2CD Þ 8I 0 3 Re ðAB þ AE  -2DC  Þ þðAB þ AE  -2DC  Þ = 2I 0

3i ½ð2DC  - AB - AE  Þ - ð2DC  - AE  - AB Þ = 0 8I 0

ð3:291Þ

ð3:292Þ

Polarization Theory of Elastic Scattering of Spin 1 Particle. . .

3.6

0 0

K xxzx

151

3 = pffiffiffi ½ - iðEC  - CB - CE  þ BC  Þ 2 2I 0 pffiffiffi ð3:293Þ 3 2       ImðCB - EC Þ - iðBC - CE - CB þ EC Þ = I0

- y0 y0

0 0 1 K zxzz = pffiffiffi ½ - iðCB - BC  þ CE  - EC  Þ þ 4iðAD - DA Þ 2 2I 0 pffiffiffi 2 ImðBC  - CE  -2DA Þ - iðCB - BC  þ CE  - EC Þ = I0 ð3:294Þ 0

i ½ðAB - AE  Þ -2ðBA - EA Þ þ ðAB - AE  Þ 4I 0 1 = - ImðAB - AE  Þ I0

K xyz =

0

K yyz = -

1 ½ - ðAB - AE  Þ þ ðAB - AE  Þ = 0 4I 0

ð3:295Þ

ð3:296Þ

0 i K zyz = pffiffiffi ½ - iðBC  þ CB - CE  - EC  Þ þ iðEC  þ CE  - CB - BC  Þ 2pffiffiffi2I 0 2 = Re ðBC  - CE  Þ I0 ð3:297Þ 0 0 3 K xyzy = - pffiffiffi ½ - iðEC  þ CB - CE  - BC  Þ þ iðBC  þ CE  - CB - EC  Þ 4 2I 0 3 = - pffiffiffi ImðCB þ EC  Þ 2I 0

ð3:298Þ 0 0

K xyzz = 0 0

K yyzz = = 0 0

K xyzx

- y0 y0

3i ½ðAB - AE  Þ - ðAB - AE  Þ = 0 8I 0

3 ½ - ðAB - AE  Þ -2ðBA - EA Þ - ðAB - AE  Þ 8I 0

3 Re ðAB - AE  Þ 2I 0

ð3:299Þ

ð3:300Þ

3i = pffiffiffi ½iðEC  þ CB - CE  - BC  ÞþiðBC  þ CE  - CB - EC  Þ = 0 2 2I 0 ð3:301Þ

0 0 i K zyzz = pffiffiffi ½ - iðBC  þ CB - CE  - EC  Þ - iðEC  þ CE  - CB - BC  Þ = 0 2 2I 0 ð3:302Þ

152

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

0 1 K xxx -yy = pffiffiffi ½ - iðDE þ DB Þ þ iðBD þ ED Þ - iðED þ BD Þ þ iðDB þ DE  Þ = 0 2I 0 ð3:303Þ 0 i K yxx - yy = pffiffiffi ½iðDE  þ DB Þ þ iðBD þ ED Þ þ iðED þ BD Þ þ iðDB þ DE  Þ 2Ip 0 ffiffiffi 2 2 Re ðBD þ DE  Þ =I0 ð3:304Þ

1 ½ - BE  - EB þ EB þ BE   = 0 I0 h i 0 0 3i - jE j2 - jBj2 þ jBj2 þ jEj2 = 0 K xxxy- yy = 2I 0 0

K zxx - yy =

0 0 3 K xxxz- yy = pffiffiffi ½ - iðDE  þ DB Þ þ iðBD þ ED Þ þ iðED þ BD Þ 2 2I 0 pffiffiffi 3 2   ImðBD - DE  Þ - iðDB þ DE Þ = I0

ð3:305Þ ð3:306Þ

ð3:307Þ

0 0 3i K yxxz- yy = pffiffiffi ½iðDE  þ DB Þ þ iðBD þ ED Þ - iðED þ BD Þ- iðDB þ DE  Þ = 0 2 2I 0 ð3:308Þ     0 0 0 0 6 3 K xxxx--yyy y = ð3:309Þ jE j2 þ jBj2 þ jBj2 þ jE j2 = jBj2 þ jEj2 I0 I0   h i 0 0 4 1 - BE  - EB þ 4jDj2 - EB - BE  = Re ðBE  Þ - jDj2 K zxxz- yy = I0 I0 ð3:310Þ 0 1 K xzz = pffiffiffi ½iðDB -2AC  þ DE  Þ - iðDE  -2AC  þ DB Þ = 0 3 2I 0 0 i K yzz = pffiffiffi ½ - iðDB -2AC  þ DE  Þ -2iðBD -2CA þ ED Þ 3 2I 0 pffiffiffi 2 2    Re ðDB þ DE  -2AC  Þ - iðDB -2AC þ DE Þ = 3I 0 h   i 0 1 K zzz = jBj2 -2jC j2 þ jEj2 þ - jE j2 þ 2jC j2 - jBj2 = 0 3I 0 h   i 0 0 i EB -2jC j2 þ BE  - BE  -2jC j2 þ EB = 0 K xzzy = 2I 0

ð3:311Þ

ð3:312Þ

ð3:313Þ ð3:314Þ

3.6

Polarization Theory of Elastic Scattering of Spin 1 Particle. . . 0 0 1 K xzzz = pffiffiffi ½iðDB -2AC  þ DE  Þ -2iðBD -2CA þ ED Þ 2 2I 0 pffiffiffi 2    ImðDB þ DE  -2AC  Þ þiðDB -2AC þ DE Þ = I0

153

ð3:315Þ

0 0 i K yzzz = pffiffiffi ½ - iðDB -2AC  þ DE  Þ þ iðDB -2AC  þ DE  Þ = 0 ð3:316Þ 2 2I 0   0 0 0 0 1 - EB þ 2jC j2 - BE  - BE  þ 2jCj2 - EB K xzzx - y y = I0 h i ð3:317Þ 4 =Re ðEB Þ - jC j2 I0 h      i 0 0 1 K zzzz = jBj2 -2jC j2 þ jEj2 - 4 jDj2 - jAj2 þ jEj2 -2jC j2 þ jBj2 3I 0   2 = jBj2 þ jEj2 þ 2jAj2 -2jC j2 -2jDj2 3I 0 ð3:318Þ

From the above results we can see 0 0 9 z0 9 0 K z , K yxyz = - K xz , 4 4 9 z0 9 x0 x0 y0 z 0 z0 x0 y0 x0 y0 y0 z 0 K yz = - K x , K yz = K x , K yz = - K x , K yz = K x 4 4 0

0 0

0

0 0

0 0

K xxy = K yz z , K zxy = - K xz y , K xxyy =

ð3:319Þ

We divide the Cartesian coordinate indexes of spin 1 particles into two groups ε1 = y, xz, xx - yy, zz;

ε2 = x, z, xy, yz

0

ð3:320Þ 0

For the polarization transfer coefficient K ii , due to the conservation of parity, K ii is 0 not equal to 0 only when i and i0 are in the same group; otherwise K ii must be equal to 0 0 0. There are a total of 16 K ii of ε1 which are not equal to 0, and 16 K ii of ε2 that are 0 not equal to 0; however, taking into account Eq. (3.319), only 8 non-zero K ii of ε2 are 0 independent, that is, only a total of 24 non-zero K ii are independent. Also note that 0 Pi0 and Ai given above are not equal to 0 only in the first group ε1 of the coordinate index. According to Eq. (3.214) the polarization vector and polarization tensor components of the outgoing particles corresponding to the polarized incident particles can be written as follows:  0  0 0 0 0 0 3 2 1 1 Pi = ðI 0 =I Þ Pi0 þ py K iy þ pxz K ixz þ pxx - yy K ixx - yy þ pzz K izz , 2 3 6 2 i0 = ε01 = y0 , x0 z0 , x0 x0 - y0 y0 , z0 z0

ð3:321Þ

154

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

  0 0 0 0 0 3 3 2 2 Pi = ðI 0 =I Þ px K ix þ pz K iz þ pxy K ixy þ pyz K iyz , 2 2 3 3 i0 = ε02 = x0 , z0 , x0 y0 , y0 z0

ð3:322Þ

Using Eq. (1.99) and Eq. (3.8), in S = 1 case the following normalized incident particle density matrix can be written as from Eq. (1.92): "    X 1 ^ X ^ρin = Iþ ð-1Þμ - it 1 - μ iT^ 1μ þ ð-1ÞM t 2 3 μ M

# ^

ð3:323Þ



T^ 2μ = tr T^ 2M ^ ρin = t 2M

ð3:324Þ

- M T 2M

Utilizing Eq. (3.113) we can prove 



iT^ 1μ = tr iT^ 1μ ^ρin = it 1μ ,



We define

^ iT^ 1μ F þ tr F , μ =  1, 0; iT 1μ = ^ þg trfFF



^ T^ 2M F þ tr F T 2M = , M =  2,  1, 0 ^ þg trfFF ð3:325Þ

Here iT1μ and T2M are no longer the operators; referring to Eqs. (3.232) and (3.233), iT1μ and T2M can be called the polarization analyzing powers in the spherical basis coordinate system. * When the y-axis is not parallel to the unit vector n , which is perpendicular to ^ has been given by the reaction plan, the corresponding scattering amplitude F þ ^ and F ^ can be simplified as Eq. (3.194), and F 0

B

^ =B F @ iDeiφ - Ee2iφ

- iCe - iφ - Ee -2iφ A iCeiφ

1

0

B

C ^þ B = @ iC  eiφ - iDe - iφ A, F B - E e2iφ

- iD e - iφ - E e -2iφ A iD eiφ

1

C - iC  e - iφ A B ð3:326Þ

We can find the following expressions from Eqs. (3.38), (3.326), (3.196), and (3.325):

3.6

Polarization Theory of Elastic Scattering of Spin 1 Particle. . .

155

0 1 rffiffiffi iC  eiφ A - iC  e - iφ C ^þ = - 3 i B iT^ 11 F B @ - E  e2iφ iD eiφ A, 2 0 0 0 pffiffiffi  3i    iφ iT 11 = - pffiffiffi iðBC þ CE Þe þ iðDA þ AD Þeiφ þ iðEC þ CB Þeiφ p3ffiffiffi 2I 0 2 = pffiffiffi Re ðCB þ CE þ AD Þeiφ 3I 0 ð3:327Þ 0 1 rffiffiffi B - iD e - iφ - E  e -2iφ þ 3 B C ^ = iT^ 10 F i@ 0 0 0 A, 2  2iφ  iφ  ð3:328Þ - iD e -B pffiffiffi hE e   i 3i iT 10 = pffiffiffi jBj2 - jE j2 þ jE j2 - jBj2 = 0 3 2I 0 0 1 rffiffiffi 0 0 0 C ^þ = 3 i B iT^ 1 -1 F @ B - iD e - iφ - E  e -2iφ A, 2  iφ A - iC e - iφ pffiffiffi iC e  3i  iT 1 -1 = pffiffiffi - iðCB þ EC  Þ e -iφ - iðAD þ DA Þ e - iφ - iðCE þ BC  Þ e -iφ 3pffiffi2ffi I 0 2 = pffiffiffi Re ðCB þ CE þ AD Þ e - iφ 3I 0 ð3:329Þ 1 0 - E  e2iφ iD eiφ B pffiffiffiB þ C ^ = 3@ T^ 22 F 0 0 0 A, 0 0 0 i   h 1 1 2  T 22 = - pffiffiffi BE þ jDj þ EB e2iφ = - pffiffiffi 2 Re ðBE  Þ þ jDj2 e2iφ 3I 0 3I 0 ð3:330Þ 0 1    rffiffiffi - iC eiφ - A iC e - iφ þ 3 B C   2iφ iφ ^ = T^ 21 F iD e B A, @ -E e 2 0 0 0  1    iφ T 21 = pffiffiffi - iðBC - CE Þe - iðDA - AD Þeiφ - iðEC  - CB Þeiφ 6Ip 0 ffiffiffi 2 = - pffiffiffi ImðCB þ CE  þ AD Þ eiφ 3I 0 ð3:331Þ

156

3

0

B

Polarization Theory of Nuclear Reactions for Spin 1 Particles

- iD e - iφ - E  e -2iφ

1

C ^ þ= p1ffiffiffi B T^ 20 F -2A 2iC  e - iφ A, @ -2iC  eiφ 2  2iφ iD eiφ B

E e        1 T 20 = pffiffiffi jBj2 -2jC j2 þ jE j2 þ jDj2 -2jAj2 þ jDj2 þ jE j2 -2jC j2 þ jBj2 3 ffiffiffi2I 0 p  2 2 = jBj þ jE j2 - jAj2 þ jDj2 -2jCj2 3I 0 ð3:332Þ

0 1 rffiffiffi 0 0 0 C ^ þ = 3B T^ 2 -1 F - iD e - iφ - E  e -2iφ A, @ B 2  iφ - A iC e - iφ pffiffiffi - iC e  3  T 2 -1 = pffiffiffi - iðCB - EC Þe - iφ - iðAD - DA Þe - iφ - iðCE  - BC  Þe - iφ 3pffiffi2ffi I 0 2 = pffiffiffi ImðCB þ CE þ AD Þ e - iφ 3I 0 ð3:333Þ 1 0 0 0 0 pffiffiffi C ^ þ = 3B T^ 2 -2 F 0 0 A, @0   - iφ  -2iφ -E e B - iD e   h i 1 1 2  T 2 -2 = - pffiffiffi EB þ jDj þ BE  e -2iφ = - pffiffiffi 2Re ðBE  Þ þ jDj2 e -2iφ 3I 0 3I 0 ð3:334Þ Therefore, according to Eq. (3.323), the differential cross section of the elastic scattering of the polarized spin 1 incident particle beam with the spinless targets can be written as n o dσ ^ þ = I 0 ðθÞ ½1 þ it 1 ^ ρin F = I ðθ, φÞ = tr F^ dΩ þt 2 -2 T 22 - t 2 -1 T 21 þ t 20 T 20 - t 21 T 2

-1 ðiT 11 Þ -1

þ it 11 ðiT 1

þ t 22 T 2

-1 Þ

-2 

ð3:335Þ

The analyzing powers in the spherical basis coordinate system appearing in the above formula have been given by Eqs. (3.327) and (3.329)–(3.334). We can obtain the following results from Eqs. (3.327)–(3.329) and using Eqs. (3.232), (1.22), and (3.9): -i Ax = pffiffiffi ðiT 1 3

-1 - iT 11 Þ = -

pffiffiffi 2 2 Re ðCB þ CE  þ AD Þ sin φ 3I 0

ð3:336Þ

3.6

Polarization Theory of Elastic Scattering of Spin 1 Particle. . .

1 Ay = pffiffiffi ðiT 1 3

pffiffiffi 2 2 Re ðCB þ CE  þ AD Þ cos φ -1 þ iT 11 Þ = 3I 0 rffiffiffi 2 ð- iÞ iT 10 = 0 Az = 3

157

ð3:337Þ ð3:338Þ

Noting Eq. (3.324) we can obtain the following results utilizing Eqs. (1.22) and (3.132): 1 px = pffiffiffi ðt 1 3

1 -1 - t 11 Þ, py = pffiffiffi ðit 1 3

rffiffiffi 2 t -1 þ it 11 Þ, pz = 3 10

ð3:339Þ

and their inverse relations are pffiffiffi  3 it 11 = py - ipx , 2

rffiffiffi pffiffiffi  3 3 it 1 -1 = py þ ipx , t 10 = p 2 2 z

ð3:340Þ

In φ = 0 case Eqs. (3.336), (3.337), and (3.338) are reduced to Eqs. (3.235), (3.237), and (3.239), respectively. In this case, because we have Ax = 0, then iT1 -1 = iT11, so from Eq. (3.337) we can obtain 2 Ay = pffiffiffi iT 11 , 3

pffiffiffi 3 iT 11 = A 2 y

ð3:341Þ

Using Eq. (3.233) and Eq. (3.21) we can obtain the following results from Eqs. (3.330)–(3.334): pffiffiffi h i i 3 1 ðT 2 -2 - T 22 Þ = 2 Re ðBE  Þ þ jDj2 sin ð2φÞ 2 I0 pffiffiffi pffiffiffi 2 3 ðT 2 -1 - T 21 Þ = ImðCB þ CE  þ AD Þ cos φ Axz = I0 2 pffiffiffi pffiffiffi 2 i 3 ðT 2 -1 þ T 21 Þ = ImðCB þ CE  þ AD Þ sin φ Ayz = I0 2 h i pffiffiffi 2 2 Re ðBE  Þ þ jDj2 cos ð2φÞ Axx - yy = 3ðT 2 -2 þ T 22 Þ = I0   pffiffiffi 2 jBj2 þ jE j2 - jAj2 þ jDj2 -2jC j2 Azz = 2T 20 = 3I 0 Axy =

ð3:342Þ ð3:343Þ ð3:344Þ ð3:345Þ ð3:346Þ

And we can find the following expressions based on Eqs. (3.21), (3.136), and (3.324):

158

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

pffiffiffi pffiffiffi pffiffiffi i 3 i 3 3 pxy = ðt - t Þ, pxz = ðt - t Þ, pyz = ðt 2 pffiffiffi2 -2 22 2 2 2pffiffiffi 2 -1 21 pxx - yy = 3ðt 2 -2 þ t 22 Þ, pzz = 2t 20

-1

þ t 21 Þ, ð3:347Þ

Using Eq. (3.22) we can also obtain   1  1  1 t 2 2 = pffiffiffi pxx - yy  2ipxy , t 2 1 =  pffiffiffi pxz  ipyz , t 20 = pffiffiffi pzz ð3:348Þ 3 2 3 2 If φ = 0, Eqs. (3.342)–(3.346) will degenerate to Eqs. (3.241), (3.243), (3.245), (3.247), and (3.249), respectively, and in this case because Axy = 0, Ayz = 0, we can see T2 -2 = T22, T2 -1 = - T21 from Eq. (3.21), and we can obtain pffiffiffi Axz = - 3T 21 ,

pffiffiffi Axx - yy = 2 3T 22 ,

pffiffiffi Azz = 2T 20

ð3:349Þ

From the above formulas we can also get 1 T 21 = - pffiffiffi Axz , 3

1 T 20 = pffiffiffi Azz , 2

1 T 22 = pffiffiffi Axx - yy 2 3

ð3:350Þ

In References [5, 8–18] the measurement and theoretical calculation results of deuteron elastic scattering vector polarization and tensor polarization analyzing powers of some spinless targets are given, and most of them are given in the form of iT11, T20, T21, T22. If the incident particle beam is polarized only in the y-axis direction, then the initial density matrix should be in the form given by Eq. (3.216), and the corresponding differential cross section of the outgoing particles is

 n o dσ ^ ρ^in F ^ þ = I 0 ðθÞ 1 þ 3 Py Ay ðθÞ þ 1 Pyy Ayy ðθÞ = I ðθÞ = tr F dΩ 2 2

ð3:351Þ

References [19–21] give the measurement and calculation results of deuteron elastic scattering polarization analyzing power Ay(θ) and Ayy(θ) of some targets in this type of polarization case. If we use the spherical optical model or the R matrix theory to calculate the S matrix elements of elastic scattering between the deuteron and the spinless target, the polarization data can be calculated using the theory described in this section.

3.7

3.7

General Form of Nuclear Reaction Polarization Theory of Spin 1. . .

159

General Form of Nuclear Reaction Polarization Theory of Spin 1 Particles with Non-zero Spin Target

Equations (2.251) and (2.252) give the reaction amplitude expressions of the j-j coupling [7] and S-L coupling [22], respectively. In the case where the spins of the incident particles and the outgoing particles are both 1, the following symbols are introduced for the j-j coupling and the S-L coupling, respectively: sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0  l - m0l ! X j μ J M 0  C C l þ m0l ! jj0 l0 1μ j μ IM I   j0 m0 C l0 m0 j 1μ0 C Jj0 mM0j I 0 M 0I δα0 n0 l0 j0 , αnlj - SJα0 n0 l0 j0 , αnlj

0 f μll0 μJ

 0 0 ml = ^l^l eiðσl þσl0 Þ

ð3:352Þ

l

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0  l - m0l ! X J M S M 0  C C l þ m0l ! SS0 l0 SM 1μ IM I   S0 M 0 C Jl0 mM0 S0 M 0 C 1μ0 IS0 M 0 δα0 n0 l0 S0 , αnlS - SJα0 n0 l0 S0 , αnlS

  0 f μll0 μJ m0l = ^l^l eiðσl þσl0 Þ 0

l

S

ð3:353Þ

I

Given these, the reaction amplitudes can be uniformly written in the following form under the two angular momentum coupling modes according to Eqs. (2.251)– (2.254): 0 0 i f α0 n0 μ0 M 0I ,αnμM I ðΩÞ = f C ðθÞ δα0 n0 ,αn δμ0 M 0I ,μM I þ ð -1ÞμþM I - μ - M I 2k X 0   μþM - μ0 - M 0I 0 0 f μll0 μJ μ þ M I - μ0 - M 0I Pl0 I ð cos θÞ eiðμþM I - μ - M I Þφ

ð3:354Þ

ll0 J

Note that δM 0I ,M I is contained in the first item of the above formula so that this item 0 can be multiplied by eiðM I - M I Þφ , which is equal to 1 since δ 0 . Let M I ,M I

AM 0I M I ðθÞ = f C ðθÞ δα0 n0 ,αn δM 0I ,M I  M - M0 0X 0  i þ ð-1ÞM I - M I f 0ll J0 M I - M 0I Pl0 I I ðcos θÞ 2k 0

ð3:355Þ

BM 0I M I ðθÞ = f C ðθÞ δα0 n0 ,αn δM 0I ,M I  M - M0 0X 0  i þ ð-1ÞM I - M I f 1ll J1 M I - M 0I Pl0 I I ðcos θÞ 2k 0

ð3:356Þ

ll J

ll J

160

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

H M 0I M I ðθÞ = f C ðθÞ δα0 n0 ,αn δM 0I ,M I   M - M0 0X 0 i þ ð-1ÞM I - M I f ll-1J -1 M I - M 0I Pl0 I I ð cos θÞ 2k 0

ð3:357Þ

ll J

X 0   M - M 0 -1 0 1 f 1ll J0 M I - M 0I -1 Pl0 I I ð cos θÞ ð3:358Þ ð-1ÞM I - M I -1 2k 0

C M 0I M I ðθÞ = -

ll J

GM 0I M I ðθÞ =

1 ð-1ÞM I 2k

DM 0I M I ðθÞ = -

X - M 0 þ1 I

ll0 J

  M - M 0 þ1 0 f ll-1J 0 M I - M 0I þ 1 Pl0 I I ð cos θÞ ð3:359Þ

X 0   M - M0 - 1 0 1 f 0ll J-1 M I - M 0I -1 Pl0 I I ð cos θÞ ð3:360Þ ð-1ÞM I - M I -1 2k 0 ll J

OM 0I M I ðθÞ =

1 ð-1ÞM I 2k

E M 0I M I ðθÞ = -

X - M 0 þ1 I

ll0 J

 M - M 0 þ1 0  f 0ll J1 M I - M 0I þ 1 Pl0 I I ð cos θÞ

ð3:361Þ

X 0   M - M0 - 2 0 i ð-1ÞM I - M I -2 f 1ll J-1 M I - M 0I -2 Pl0 I I ð cos θÞ ð3:362Þ 2k 0 ll J

F M 0I M I ðθÞ = -

X 0   M - M 0 þ2 0 i ð-1ÞM I - M I þ2 f ll-1J 1 M I - M 0I þ 2 Pl0 I I ð cos θÞ 2k 0 ll J

ð3:363Þ Introducing the following symbol F μ0 μ ðθ, φÞ = f α0 n0 μ0 M 0I ,αnμM I ðΩÞ

ð3:364Þ

F 0 0 ðθ, φÞ = AM 0I M I ðθÞeiðM I - M I Þφ

ð3:365Þ

F 1 1 ðθ, φÞ = BM 0I M I ðθÞeiðM I - M I Þφ

ð3:366Þ

then one can get 0

0

F -1

-1 ðθ,

φÞ = H M 0I M I ðθÞeiðM I - M I Þφ 0

ð3:367Þ

F 1 0 ðθ, φÞ = - iCM 0I M I ðθÞeiðM I - M I Þφ e - iφ

ð3:368Þ

F -1 0 ðθ, φÞ = iGM 0I M I ðθÞeiðM I - M I Þφ eiφ

ð3:369Þ

0

0

F0

-1 ðθ,

φÞ = - iDM 0I M I ðθÞeiðM I - M I Þφ e - iφ 0

F 0 1 ðθ, φÞ = iOM 0I M I ðθÞeiðM I - M I Þ eiφ 0

F1

-1 ðθ,

φÞ = - E M 0I M I ðθÞeiðM I - M I Þ e -2iφ 0

ð3:370Þ ð3:371Þ ð3:372Þ

3.7

General Form of Nuclear Reaction Polarization Theory of Spin 1. . .

161

F -1 1 ðθ, φÞ = - F M 0I M I ðθÞeiðM I - M I Þ e2iφ 0

ð3:373Þ

If the footnote M 0I M I and argument (θ) of A, B, H, C, G, D, O, E, F are omitted, ^ with definite footnote M 0I M I , but which footnote M 0I M I then the reaction amplitude F and argument (θ, φ) are omitted, can be written as 0

B

^ =B F @ iOeiφ - Fe2iφ

- iCe - iφ A iGeiφ

- Ee -2iφ

1

0 C - iDe - iφ AeiðM I - M I Þφ H

ð3:374Þ

Comparing this formula with Eq. (3.194), it shows that when the target spin I = 0 and the residual nucleus spin I0 = 0 there are AM 0I M I = A, H M 0I M I = BM 0I M I = B, GM 0I M I = C M 0I M I = C, OM 0I M I = DM 0I M I = D,

ð3:375Þ

F M 0I M I = E M 0I M I = E

When the target and the residual nucleus are both unpolarized, the magnetic ^ and F ^ þ appearing in the quantum number footnotes of the matrix elements of F polarization theory formulas are the same, i.e., they are incoherent. So from Eq. (3.374) we can obtain 0

B þ B ^ = @ iC eiφ F - E e2iφ ^F ^þ = F 0

jBj2 þ jCj2 þ jE j2

B B iðOB þ AC þ DE  Þ eiφ @ - ðFB þ GC  þ HE  Þ e2iφ

- iO e - iφ A  iφ

iD e

1 - F  e -2iφ 0 C - iG e - iφ Ae - iðM I - M I Þφ H

ð3:376Þ



- iðBO þ CA þ ED Þ e - iφ - ðBF  þ CG þ EH  Þ e -2iφ jOj2 þ jAj2 þ jDj2 iðFO þ GA þ HD Þ eiφ

1

C - iðOF  þ AG þ DH  Þ e - iφ C A jF j2 þ jGj2 þ jH j2

ð3:377Þ For the definite M'IMI from Eq. (3.377) we can obtain   1 þ 1 ^F ^ = I 0 = tr F jAj2 þ jBj2 þ jH j2 þ jCj2 þ jGj2 þ jDj2 þ jOj2 þ jE j2 þ jF j2 3 3 ð3:378Þ *

For the sake of simplicity, choose the y-axis along the direction n . In this case there is φ = 0. Then the following formulas can be obtained from Eqs. (3.374), (3.376), and (3.377):

162

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

0

1 0  1 B - iO - F  B - iC - E C C ^ =B ^þ = B A - iG A ð3:379Þ F A - iD A, F @ iC @ iO    -E iD H -F iG H 0 1 - iðBO þ CA þ ED Þ - ðBF  þ CG þ EH  Þ jBj2 þ jCj2 þ jEj2 C ^F ^þ = B F @ iðOB þ AC  þ DE  Þ - iðOF  þ AG þ DH  Þ A jOj2 þ jAj2 þ jDj2 - ðFB þ GC  þ HE  Þ iðFO þ GA þ HD Þ

jF j2 þ jGj2 þ jH j2 ð3:380Þ

For definite M 0I M I the component of the outgoing particle polarization vector corresponding to the unpolarized incident particles is defined as 0 Pi0



^F ^þ tr ^Si0 F = þ , ^F ^ tr F

i 0 = x0 , y0 , z 0

ð3:381Þ

And the component of the outgoing particle polarization tensor corresponding to the unpolarized incident particles is defined as n o ^ i0 F ^F ^þ tr Q 0 þ , Pi0 = ^F ^ tr F

i0 = x0 y0 , x0 z0 , y0 z0 , x0 x0 - y0 y0 , z0 z0

ð3:382Þ

From Eq. (3.381) and Eq. (3.382), the following results can be obtained utilizing Eqs. (3.33), (3.36), (3.378), and (3.380): 0 1 Px0 = pffiffiffi ½iðOB þ AC  þ DE  Þ - iðBO þ CA þ ED Þ 3 2I 0 þiðFO þ GA þ HD Þ - iðOF  þ AG þ DH  Þ pffiffiffi 2 Im½ðOB þ AC  þ DE  Þ þ ðFO þ GA þ HD Þ =3I 0 0 i Py0 = pffiffiffi ½ - iðOB þ AC  þ DE  Þ - iðBO þ CA þ ED Þ 3 2I 0 - iðFO þ GA þ HD Þ - iðOF  þ AG þ DH  Þ pffiffiffi 2 = Re ½ðOB þ AC  þ DE  Þ þ ðFO þ GA þ HD Þ 3I 0 h   i 0 1 Pz0 = jBj2 þ jC j2 þ jEj2 - jH j2 þ jGj2 þ jF j2 3I 0

ð3:383Þ

ð3:384Þ

ð3:385Þ

3.7

General Form of Nuclear Reaction Polarization Theory of Spin 1. . . 0 0

i ½ðFB þ GC  þ HE  Þ - ðBF  þ CG þ EH  Þ 2I 0 1 = - ImðFB þ GC  þ HE  Þ I0

Px0 y =

163

ð3:386Þ

0 0 1 Px0 z = pffiffiffi ½iðOB þ AC  þ DE  Þ - iðBO þ CA þ ED Þ 2 2I 0

- iðFO þ GA þ HD Þ þ iðOF  þ AG þ DH  Þ

ð3:387Þ

1 = - pffiffiffi Im½ðOB þ AC  þ DE  Þ - ðFO þ GA þ HD Þ 2I 0 0 0 i Py0 z = pffiffiffi ½ - iðOB þ AC  þ DE  Þ - iðBO þ CA þ ED Þ 2 2I 0

þiðFO þ GA þ HD Þ þ iðOF  þ AG þ DH  Þ

ð3:388Þ

1 = pffiffiffi Re ½ðOB þ AC  þ DE  Þ - ðFO þ GA þ HD Þ 2I 0 1 ½ - ðFB þ GC  þ HE  Þ - ðBF  þ CG þ EH  Þ I0 ð3:389Þ 2 = - Re ðFB þ GC  þ HE  Þ I0 h  i 0 0 1 Pz0 z = jBj2 þ jCj2 þ jE j2 þ jF j2 þ jGj2 þ jH j2 -2 jOj2 þ jAj2 þ jDj2 3I 0 ð3:390Þ 0 0

Px0 x

- y0 y0

=

And in I = I0 = 0 case, using Eq. (3.375) we can see that Eqs. (3.383)–(3.390) will automatically degenerate to Eqs. (3.224)–(3.231). For definite M 0I M I the polarization vector analyzing power and polarization tensor analyzing power are defined as, respectively

^þ ^ S^i F tr F Ai = þ , i = x, y, z ^F ^ tr F

^ iF ^þ ^Q tr F Ai = þ , i = xy, xz, yz, xx - yy, zz ^F ^ tr F

ð3:391Þ ð3:392Þ

From Eq. (3.391) and Eq. (3.392), we can obtain the following expressions utilizing Eqs. (3.33), (3.36), (3.378), and (3.379):

164

3

0

iC

A

Polarization Theory of Nuclear Reactions for Spin 1 Particles

-iG

1

C B ^ ^ þ = p1ffiffiffi B B -E  iðD -O Þ H  -F  C, Sx F A @ 2 iC A -iG ^x F ^S ^ þ = p1ffiffiffi F 2 1 0 iðBC  -CB þCE -EC Þ BA -EA þCD -CO -iðBG -EG þCH  -CF  Þ C B B AB -AE  -OC  þDC  iðOA -DA þAD -AO Þ OG -DG þAH  -AF  C A @ -iðFC -GB þGE  -HC Þ -FA þHA -GD þGO iðFG -HG þGH  -GF  Þ

pffiffiffi 2 Ax = ImðBC  þ CE  þ OA þ AD þ FG þ GH  Þ 3I 0 0

-iC 

-A

iG

ð3:393Þ ð3:394Þ

1

C B ^ ^ þ = piffiffiffi B B þE  -iðD þO Þ - ðF  þH  Þ C, Sy F A @ 2 iC A -iG ^^ ^ þ = piffiffiffi F Sy F 2 1 0 -iðBC  þCB þCE  þEC Þ - ðBA þEA þCD þCO Þ iðBG þEG þCF  þCH  Þ C B B DC  þOC  þAB þAE  -iðOA þDA þAD þAO Þ - ðOG þDG þAF  þAH  Þ C A @ iðFC þHC þGB þGE  Þ

FA þHA þGD þGO

-iðFG þHG þGF  þGH  Þ

pffiffiffi 2 Ay = Re ðBC  þ CE  þ OA þ AD þ FG þ GH  Þ 3I 0 0  1 B - iO - F  B C ^Sz F ^þ = B 0 0 0 C @ A, E  - iD - H  0 1 - iðBO - ED Þ - ðBF  - EH  Þ jBj2 - jE j2 B C ^ ^Sz F ^ þ = B iðOB - DE  Þ F - iðOF  - DH  Þ C jOj2 - jDj2 @ A jF j2 - jH j2 - ðFB - HE  Þ iðFO - HD Þ   1 Az = jBj2 - jH j2 þ jF j2 - jE j2 þ jOj2 - jDj2 3I 0

ð3:395Þ ð3:396Þ

ð3:397Þ

ð3:398Þ

3.7

General Form of Nuclear Reaction Polarization Theory of Spin 1. . .

0

E

B ^ xy F ^ þ = 3i B 0 Q 2@ B 0

- iD

- H

0

0

- iO

- F

BE  - EB

B ^ xy F ^Q ^ þ = 3i B iðOE  - DB Þ F 2@ - FE  þ HB

165

1 C C, A

- iðBD - EO Þ OD - DO

- BH  þ EF 

1

C - iðOH  - DF  Þ C A

iðFD - HO Þ

FH  - HF  ð3:399Þ

Axy = 0

iC

A

1 ImðBE  þ OD þ FH  Þ I0 -iG

ð3:400Þ

1

C 3 B ^ xz F ^þ = p ffiffiffi B Q B þE -iðO þD Þ - ðF  þH  Þ C A, @ 2 2    -iC -A iG þ 3 ^ xz F ^Q ^ = pffiffiffi F 2 2 1 0 iðBC  -CB -CE þEC Þ BA -CO -CD þEA -iðBG -CF  -CH  þEG Þ C B B -OC  -DC  þAB þAE  iðOA þDA -AO -AD Þ OG þDG -AF  -AH  C A @             -iðFC -GB -GE þHC Þ -FA þGO þGD -HA iðFG -GF -GH þHG Þ

ð3:401Þ 1 Axz = - pffiffiffi ImðBC  - CE  þ OA - AD þ FG - GH  Þ 2I 0

ð3:402Þ

1 -iC -A iG C B 3i B  ^ yz F ^þ = p ffiffiffi @ B -E iðD -O Þ - ðF  -H  Þ C Q A, 2 2  -iC -A iG þ 3i ^ yz F ^Q ^ = pffiffiffi F 2 2 0 -iðBC  þCB -CE -EC Þ -BA þEA þCD -CO iðBG -EG þCF  -CH  Þ B B AB -AE  þOC  -DC  -iðOA -DA -AD þAO Þ -AF  þAH  -OG þDG @ 0

iðFC -HC þGB -GE  Þ

FA -HA -GD þGO

1 C C A

-iðFG -HG þGF  -GH  Þ

ð3:403Þ 1 Ayz = pffiffiffi Re ðBC  - CE  þ OA - AD þ FG - GH  Þ 2I 0

ð3:404Þ

166

3

0

- E

B ^ xx - yy F ^þ = 3 B 0 Q @ 0

Polarization Theory of Nuclear Reactions for Spin 1 Particles

iD 0

B

- iO

- BE  - EB

H

C 0 C A,  -F

B ^ xx - yy F ^Q ^ þ = 3 B - iðOE  þ DB Þ F @ FE þ HB

1

iðBD þ EO Þ

BH  þ EF 

- ðOD þ DO Þ - iðFD þ HO Þ

1

C iðOH  þ DF  Þ C A   - ðFH þ HF Þ ð3:405Þ

2 Re ðBE  þ OD þ FH  Þ I0 1 - F C 2iG C A,  H

Axx - yy = 0

B

B ^ zz F ^ þ = B -2iC  Q @ - E

- iO -2A iD

^ zz F ^þ = ^Q F 0 jBj2 -2jC j2 þ jE j2 B B iðOB -2AC  þ DE  Þ @ - FB þ 2GC  - HE 

- iðBO -2CA þ ED Þ jOj2 -2jAj2 þ jDj2

ð3:406Þ

- BF  þ 2CG - EH 

1

C - iðOF  -2AG þ DH  Þ C A

iðFO -2GA þ HD Þ

jF j2 -2jGj2 þ jH j2 ð3:407Þ h  i 1 Azz = jBj2 þ jH j2 þ jE j2 þ jF j2 þ jOj2 þ jDj2 -2 jC j2 þ jAj2 þ jGj2 3I 0 ð3:408Þ In I = I0 = 0 case, using Eq. (3.375) we can see that Eqs. (3.393)–(3.408) will automatically degenerate to Eqs. (3.234)–(3.249). For the definite M 0I M I , we still define the polarization transfer coefficients in the form of the expressions given by Eqs. (3.251)–(3.254). The following results can be obtained from Eqs. (3.251)–(3.254) and utilizing Eqs. (3.33), (3.36), and (3.378) as ^ iF ^ þ and F ^ þ obtained earlier in this section ^ ^Si F ^Q well as the matrices F 0

K xx = 0

K yx =

1 Re ðAB þ DC  þ GO þ HA - OC  - AE  - FA - GD Þ 3I 0

ð3:409Þ

1 ImðAB þ DC  þ GO þ HA - OC  - AE  - FA - GD Þ 3I 0 pffiffiffi 0 2 ImðBC  þ CE  - FG - GH  Þ K zx = 3I 0

ð3:410Þ ð3:411Þ

3.7

General Form of Nuclear Reaction Polarization Theory of Spin 1. . . 0 0 1 K xx y = pffiffiffi Re ðGB þ HC  - FC  - GE  Þ 2I 0 0 0

K xx z = 0 0

167

ð3:412Þ

1 Re ðAB þ DC  - GO - HA - OC  - AE  þ FA þ GD Þ ð3:413Þ 2I 0

1 ImðAB þ DC  - GO - HA - OC  - AE  þ FA þ GD Þ ð3:414Þ 2I 0 pffiffiffi 2 x0 x0 - y0 y0 =ImðGB þ HC  - FC  - GE  Þ ð3:415Þ Kx I0 pffiffiffi 0 0 2 Im½BC  þ CE  þ FG þ GH  -2ðOA þ AD Þ ð3:416Þ K zx z = 3I 0

K yx z =

0

K xy = -

0

K yy =

1 ImðAB þ DC  þ GO þ HA þ OC  þ AE  þ FA þ GD Þ 3I 0 ð3:417Þ

1 Re ðAB þ DC  þ GO þ HA þ OC  þ AE  þ FA þ GD Þ ð3:418Þ 3I 0 pffiffiffi 2 z0 Re ðBC  þ CE  - FG - GH  Þ ð3:419Þ Ky = 3I 0 0 0 1 K xy y = - pffiffiffi ImðGB þ HC  þ FC  þ GE  Þ 2I 0

0 0

K xy z = -

0 0

ð3:420Þ

1 ImðAB þ DC  - GO - HA þ OC  þ AE  - FA - GD Þ 2I 0 ð3:421Þ

1 Re ðAB þ DC  - GO - HA þ OC  þ AE  - FA - GD Þ 2I 0 pffiffiffi 2 x0 x0 - y0 y0 =Re ðGB þ HC  þ FC  þ GE  Þ Ky I0 pffiffiffi 0 0 2 Re ½BC  þ CE  þ FG þ GH  -2ðOA þ AD Þ K zy z = 3I 0 pffiffiffi 2 x0 ImðOB - DE  þ FO - HD Þ Kz = 3I 0 pffiffiffi 2 y0 Re ðOB - DE  þ FO - HD Þ Kz = 3I 0   0 1 K zz = jBj2 - jE j2 þ jH j2 - jF j2 3I 0

K yy z =

0 0

K xz y = -

1 ImðFB - HE  Þ I0

ð3:422Þ ð3:423Þ ð3:424Þ ð3:425Þ ð3:426Þ ð3:427Þ ð3:428Þ

168

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

0 0 1 K xz z = - pffiffiffi ImðOB - DE  - FO þ HD Þ 2I 0 0 0 1 K yz z = pffiffiffi Re ðOB - DE  - FO þ HD Þ 2I 0

2 Re ðFB - HE  Þ I0 h  i 0 0 1 K zz z = jBj2 - jEj2 - jH j2 þ jF j2 -2 jOj2 - jDj2 3I 0 0 0

K xz x

- y0 y0

=-

0 1 K xxy = pffiffiffi Re ðBD þ OH  - OE  - FD Þ 2I 0 0 1 K yxy = - pffiffiffi ImðBD þ OH  þ OE  þ FD Þ 2I 0 0

0 0

0 0

- y0 y0

K zxyz = 0

K xxz = 0

K yxz =

=-

3 ImðHB - FE  Þ I0

1 ImðBE  þ FH  -2OD Þ I0

ð3:434Þ

ð3:437Þ ð3:438Þ ð3:439Þ ð3:440Þ

1 Re ðAB - DC  þ GO - HA - OC  þ AE  - FA þ GD Þ ð3:441Þ 2I 0 1 ImðAB - DC  þ GO - HA - OC  þ AE  - FA þ GD Þ 2I 0

0 0 3 K xxzy = pffiffiffi Re ðGB - HC  - FC  þ GE  Þ 2 2I 0 0 0

ð3:433Þ

ð3:436Þ

0 1 K zxz = - pffiffiffi ImðBC  - CE  - FG þ GH  Þ 2I 0

K xxzz =

ð3:432Þ

3 Re ðHB - FE  Þ 2I 0

0 0 3 K yxyz = - pffiffiffi ImðBD - OH  þ OE  - FD Þ 2 2I 0 0 0

ð3:431Þ

ð3:435Þ

0 0 3 K xxyz = pffiffiffi Re ðBD - OH  - OE  þ FD Þ 2 2I 0

K xxyx

ð3:430Þ

1 ImðBE  - FH  Þ I0

K zxy = K xxyy =

ð3:429Þ

ð3:442Þ ð3:443Þ ð3:444Þ

3 Re ðAB - DC  - GO þ HA - OC  þ AE  þ FA - GD Þ ð3:445Þ 4I 0

General Form of Nuclear Reaction Polarization Theory of Spin 1. . .

3.7

0 0

K yxzz =

3 ImðAB - DC  - GO þ HA - OC  þ AE  þ FA - GD Þ ð3:446Þ 4I 0 0 0

K xxzx

- y0 y0

3 = - pffiffiffi ImðGB - HC  - FC  þ GE  Þ 2I 0

0 0 1 K zxzz = - pffiffiffi Im½BC  - CE  þ FG - GH  -2ðOA - AD Þ 2I 0 0

K xyz = 0

K yyz =

K xyzz = -

0 0

ð3:448Þ

1 Re ðAB - DC  þ GO - HA þ OC  - AE  þ FA - GD Þ ð3:450Þ 2I 0

0 0 3 K xyzy = - pffiffiffi ImðGB - HC  þ FC  - GE  Þ 2 2I 0

K yyzz =

ð3:447Þ

1 ImðAB - DC  þ GO - HA þ OC  - AE  þ FA - GD Þ ð3:449Þ 2I 0

0 1 K zyz = pffiffiffi Re ðBC  - CE  - FG þ GH  Þ 2I 0

0 0

169

ð3:451Þ ð3:452Þ

3 ImðAB - DC  - GO þ HA þ OC  - AE  - FA þ GD Þ 4I 0 ð3:453Þ

3 Re ðAB - DC  - GO þ HA þ OC  - AE  - FA þ GD Þ ð3:454Þ 4I 0 0 0

K xyzx

- y0 y0

3 = - pffiffiffi Re ðGB - HC  þ FC  - GE  Þ 2I 0

0 0 1 K zyzz = pffiffiffi Re ½BC  - CE  þ FG - GH  -2ðOA - AD Þ 2I 0 pffiffiffi 0 2 K xxx - yy = ImðBD þ OH  - OE  - FD Þ I0 pffiffiffi 2 y0 Re ðBD þ OH  þ OE  þ FD Þ K xx - yy = I0 0

K zxx - yy = 0 0

K xxxy- yy =

2 Re ðBE  - FH  Þ I0

3 ImðHB þ FE  Þ I0

0 0 3 K xxxz- yy = - pffiffiffi ImðBD - OH  - OE  þ FD Þ 2I 0 0 0 3 K yxxz- yy = - pffiffiffi Re ðBD - OH  þ OE  - FD Þ 2I 0

ð3:455Þ ð3:456Þ ð3:457Þ ð3:458Þ ð3:459Þ ð3:460Þ ð3:461Þ ð3:462Þ

170

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

0 0

6 Re ðHB þ FE  Þ I0

ð3:463Þ

2 Re ðBE  þ FH  -2OD Þ I0

ð3:464Þ

0 0

K xxxx--yyy y = 0 0

K zxxz- yy = -

pffiffiffi 2 =ImðOB þ DE  -2AC  þ FO þ HD -2GA Þ 3I 0 pffiffiffi 2 y0 Re ðOB þ DE  -2AC  þ FO þ HD -2GA Þ K zz = 3I 0 h  i 0 1 K zzz = jBj2 - jH j2 þ jE j2 - jF j2 -2 jCj2 - jGj2 3I 0

0 K xzz

0 0

K xzzy = -

1 ImðFB þ HE  -2GC  Þ I0

0 0 1 K xzzz = - pffiffiffi ImðOB þ DE  -2AC  - FO - HD þ 2GA Þ 2I 0 0 0 1 K yzzz = pffiffiffi Re ðOB þ DE  -2AC  - FO - HD þ 2GA Þ 2I 0 0 0

K xzzx 0 0

K zzzz =

- y0 y0

=-

2 Re ðFB þ HE  -2GC  Þ I0

ð3:465Þ ð3:466Þ ð3:467Þ ð3:468Þ ð3:469Þ ð3:470Þ ð3:471Þ

  1 jBj2 þ jH j2 þ jE j2 þ jF j2 -2jCj2 -2jGj2 þ 4jAj2 -2jOj2 -2jDj2 3I 0 ð3:472Þ

In I = I0 = 0 case, using Eq. (3.375) we can see that Eqs. (3.409)–(3.472) will automatically degenerate to Eqs. (3.255)–(3.318). In the case where both the target and the residual nucleus are polarized and the y* axis along to direction n is chosen, so there is φ = 0, the footnote of the matrix ^ element of F in Eq. (3.379) is marked by M 0I M I, and the footnote of the matrix element ^ þ in Eq. (3.379) is marked by M 0I M I : And the of complex conjugate matrix F following changes for the previous results are made: jX j2 → X M 0 M X M 0I M I , I

I

X = A,B,C,D,E,F,G,H,O

ð3:473Þ

So according to Eq. (3.378) we can write  1  A 0 A 0 þ BM 0 M BM 0I M I þ H M 0 M H M 0I M I 3 M MI MI MI I I I I    þ C M 0 M C M 0I M I þ GM 0 M GM 0I M I þ DM 0 M DM 0I M I I I I I I I     þ OM 0 M OM 0I M I þ EM 0 M E M 0I M I þ F M 0 M F M 0I M I

I 0,M 0 M 0 M I M I = I

I

I

I

I

I

I

I

ð3:474Þ

3.7

General Form of Nuclear Reaction Polarization Theory of Spin 1. . .

171

Using the form of the target and the residual nucleus polarization wave functions introduced by Eq. (2.315) and Eq. (2.320), while still using the simplification summation symbol Σ for the polarized target and residual nucleus introduced by Eq. (2.327), the angular distribution of the outgoing particles corresponding to the unpolarized incident particles can be expressed as dσ 0α0 n0 ,αn = Σ I 0,M 0 M 0 M I M I I I dΩ

0

I  I 0α0 n0 ,αn =

ð3:475Þ

Next introduce the following Cartesian coordinate index set ε = x, y, z, xy, xz, yz, xx - yy, zz

ð3:476Þ

The components of the polarization vector and polarization tensor corresponding to the unpolarized incident particles are 0i0

P

0

 P0i α0 n0 ,αn =

1 I

0

0

Σ I 0,M 0 M 0 M I M I Pi0,M 0 M 0 M M , I

I

I

I

I

I

i 0 = ε0

ð3:477Þ

When the normal variables are added with the footnote M 0I M I and the complex 0 conjugate variables are added with the footnote M I M I in Eqs. (3.383)–(3.390), and 0 using Eq. (3.473), the various Pi0,M 0 M 0 M M in Eq. (3.477) can be obtained. I

I

I

I

When the incident particles, targets, and residual nuclei are all polarized, the components of the vector polarization analyzing power and tensor polarization analyzing power of the nuclear reaction are Ai  Ai,α0 n0 ,αn =

1 I

Σ I 0,M 0 M 0 M I M I Ai,M 0 M 0 M I M I ,

0

I

I

I

I

i=ε

ð3:478Þ

where Ai,M 0 M 0 M I M I has been given by Eqs. (3.394)–(3.408). Referring to Eq. (3.214), I I we can write the corresponding outgoing particle differential cross section as h  dσ 0 0 3 0 I  I α0 n0 ,αn = α n ,αn = I 1 þ px Ax þ py Ay þ pz Az 2 dΩ i  1 2 1 þ pxy Axy þ pxz Axz þ pyz Ayz þ pxx - yy Axx - yy þ pzz Azz 3 6 2

ð3:479Þ

The corresponding polarization transfer coefficient is i0

0

K i  K ii,α0 n0 ,αn =

1 I

0

0

ΣI 0,M 0 M 0 M I M I K ii, M 0 M 0 M M , I

I

I

I

I

I

i = ε;

i 0 = ε0

ð3:480Þ

172

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

0

where K ii, M 0 M 0 M M has been given by Eqs. (3.409)–(3.472). Referring to Eq. (3.214), I

I

I

I

the corresponding components of the polarization vector and polarization tensor of the outgoing particles can be written as 0h   0 I 3 i0 i0 i0 i0 1 þ p x K x þ py K y þ pz K z P  Piα0 n0 ,αn = 2 I   i 0 1 2 1 i i0 i0 i0 i0 þ pxy K xy þ pxz K xz þ pyz K yz þ pxx - yy K xx - yy þ pzz K zz , i0 = ε0 6 3 2 ð3:481Þ

3.8

→

→

Polarization Theory for 1 þ A → 1 þ B Reactions

Equations (2.334) and (2.335) give the conditions for the target and the residual nucleus to be unpolarized, respectively. Using the abovementioned two conditions e introduced in Eq. (2.278) for the and the simplification summation symbol Σ unpolarized target and residual nucleus, Eqs. (3.474), (3.475), (3.477), (3.478), and (3.480) can be simplified as follows, respectively:  2  2  2  2  2 1  AM 0I M I  þ BM 0I M I  þ H M 0I M I  þ C M 0I M I  þ GM 0I M I  3  2  2  2  2         0 0 0 0 þ DM I M I þ OM I M I þ EM I M I þ F M I M I ð3:482Þ

I 0,M 0I M I =

e 0,M 0 M I = ΣI I I 0

0 i0

ð3:483Þ

0 1e i0 = ε0 ΣI 0,M 0I M I Pi0,M 0I M I , 0 I 1e i=ε Ai = 0 ΣI 0,M 0I M I Ai,M 0I M I , I 1e i0 i0 K i = 0 ΣI i = ε; i 0 = ε0 0,M 0I M I K i,M 0I M I , I

P

=

ð3:484Þ ð3:485Þ ð3:486Þ

Next we analyze Eq. (3.353) which belongs to the S-L coupling mode. The following relations can be obtained according to the C-G coefficient properties: S0 μ0 þM 0

0

0

S0 -μ0 -M 0I I 0 -M 0I

Iþ1 - S S -μ-M I I C S1μμþM C1 -μ I -M I , C 1μ0 I 0 M 0 II = ð-1ÞI þ1 - S C 1 -μ0 IM I = ð-1Þ 0

0

l þS - J J -M J M C Jl0 MSM = ð-1ÞlþS - J C Jl0 -M Cl0 -m0 S0 S -M , C l0 m0 S0 M 0 = ð-1Þ l

S

l

ð3:487Þ

-M 0S

Dividing Eq. (2.252) with i = i0 = 1 into two items, we study first the second item. For the part which is not related to angle φ, the following phase factor comes out

3.8

→

→

173

Polarization Theory for 1 þ A → 1 þ B Reactions

using Eqs. (2.288) and (3.487) when all spin magnetic quantum numbers change their signs 0

ð-1ÞlþS - Jþl þS

0

- JþIþ1 - SþI 0 þ1 - S0 þμþM I - μ0 - M 0I

0

0

= ð-1Þ2JþlþIþl þI þμþM I - μ →

0

- M 0I

→

Using the same procedure to get Eq. (2.290), for 1 þ A → 1 þ B reaction we can obtain the following similar expression:  Π=

when π i π I = π 0i π 0I when π i π I = - π 0i π 0I

0 1

ð3:488Þ

In the case that the spins of both the incident particle and the outgoing particle are 1, the target spin I and the residual nucleus spin I0 must be either the integers or the semi-odd numbers at the same time. If I and I0 are all integers, J is an integer, so (-1)2J = 1, and in this case let Λ = 0; if I and I0 are all semi-odd numbers, J is a semiodd number, so (-1)2J = - 1, and in this case let Λ = 1. Moreover, we can express the above results in a unified form as  Λ=

0 1

when I and I 0 are all integers when I and I 0 are all semi‐odd numbers

ð3:489Þ

This last expression (3.489) is similar to Eq. (2.290). Given this, we can get the following relation: 0

ð2Þ

f α0 n0 μ0 M 0 ,αnμM I ðθÞ = ð-1ÞΠþΛþIþI þμþM I - μ

0

I

- M 0I ð2Þ f α0 n0 - μ0 - M 0 ,αn - μ - M I ðθÞ I

ð3:490Þ

We can now study the first item of Eq. (2.252) with i = i0 = 1. For elastic scattering, we have Π = 0, I0 = I, (-1)Λ+2I = 1, and the δ symbol requires μ = μ0 , M I = M 0I , so f (1) also satisfies the relation given by Eq. (3.490). Thus, for the independent part with angle φ of the total reaction amplitude f, we have 0

f α0 n0 μ0 M 0I ,αnμM I ðθÞ = ð-1ÞΠþΛþIþI þμþM I - μ

0

- M 0I

f α0 n0 - μ0 - M 0I ,αn - μ - M I ðθÞ

ð3:491Þ

The following relations can be obtained from Eqs. (3.354)–(3.363): AM 0I M I ðθÞ = f α0 n0 0M 0I ,αn0M I ðθÞ, BM 0I M I ðθÞ = f α0 n0 1M 0I ,αn1M I ðθÞ, C M 0I M I ðθÞ = if α0 n0 1M 0I ,αn0M I ðθÞ,

H M 0I M I ðθÞ = f α0 n0 -1M 0I ,αn -1M I ðθÞ, GM 0I M I ðθÞ = - if α0 n0 -1M 0I ,αn0M I ðθÞ,

DM 0I M I ðθÞ = if α0 n0 0M 0I ,αn -1M I ðθÞ,

OM 0I M I ðθÞ = - if α0 n0 0M 0I ,αn1M I ðθÞ,

ð3:492Þ

E M 0I M I ðθÞ = - f α0 n0 1M 0I ,αn -1M I ðθÞ, F M 0I M I ðθÞ = - f α0 n0 -1M 0I ,αn1M I ðθÞ Given this, we have the following relations according to Eqs. (3.491) and (3.492):

174

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles 0

0

AM 0I M I ðθÞ = ð-1ÞΠþΛþIþI þM I - M I A - M 0I

- M I ðθ Þ

ð3:493Þ

B - M 0I

- M I ðθ Þ

ð3:494Þ

GM 0I M I ðθÞ = ð-1ÞΠþΛþIþI þM I - M I C - M 0I

- M I ðθ Þ

ð3:495Þ

D - M 0I

- M I ðθ Þ

ð3:496Þ

F M 0I M I ðθÞ = ð-1ÞΠþΛþIþI þM I - M I E - M 0I

- M I ðθ Þ

ð3:497Þ

0

þM I - M 0I

0

0

0

þM I - M 0I

0

0

H M 0I M I ðθÞ = ð-1ÞΠþΛþIþI

OM 0I M I ðθÞ = ð-1ÞΠþΛþIþI

By using the method to prove Eq. (2.297), and using once again the simplification e introduced by Eq. (2.278), the following relations can be summation symbol Σ proved based on Eqs. (3.493)–(3.497): e ðBC  þ GH  Þ = 0, ΣIm e ðCE  þ FG Þ = 0, ΣIm e ðOA þ AD Þ = 0, ΣIm e Re ðBC  þ GH  Þ = 2Σ e Re ðBC  Þ, Σ e Re ðCE  þ FG Þ = 2Σ e Re ðCE  Þ, Σ

e Re ðOA þ AD Þ = 2Σ e Re ðOA Þ, ΣIm e ðBC  - GH  Þ = 2ΣIm e ðBC  Þ, Σ e ðCE - FG Þ = 2ΣIm e ðCE  Þ, ΣIm e ðOA - AD Þ = 2ΣIm e ðOA Þ, ΣIm

e Re ðBC  - GH  Þ = 0, Σ e Re ðCE  - FG Þ = 0, Σ e Re ðOA - AD Þ = 0, Σ e jH j2 = Σ e jBj2 , Σ e jGj2 = Σ e jC j2 , Σ e jOj2 = Σ e jDj2 , Σ e jF j2 = Σ e jE j2 Σ ð3:498Þ

In addition, other expressions similar to the above formulas can also be obtained. By using the method used to prove Eq. (2.298), we can prove e  , ΣOD e  , ΣFE e , Σ e ðBC  þ GH  Þ, Σ e ðBD þ OH  Þ, e  , ΣGC ΣHB e ðCD þ OG Þ, Σ eðCE  þ FG Þ, Σ e ðDE  þ FO Þ e ðBE  þ FH  Þ, Σ Σ

ð3:499Þ

such that their corresponding complex conjugate terms are all real numbers. Now, defining h          i eI 0, M 0 M = 1 AM 0 M 2 þ 2 BM 0 M 2 þ C M 0 M 2 þ DM 0 M 2 þ EM 0 M 2 I I I I I I I I I I I I 3 ð3:500Þ we can prove the following expression from Eqs. (3.482), (3.483), and (3.498): e eI 0,M 0 M I =Σ I I 0

ð3:501Þ

From Eq. (3.484) we can obtain the following results according to Eqs. (3.383)– (3.390) and by using Eqs. (3.493)–(3.499):

3.8

→

→

175

Polarization Theory for 1 þ A → 1 þ B Reactions 0 x0

0 z0

0 x0 y0

0 y0 z0

=P =P =P =0 pffiffiffi 2 2e 0 y0 P = 0 Σ Re ðOB þ AC  þ DE  Þ 3I pffiffiffi 0 0 2e 0 xz = - 0 ΣIm ðOB þ AC  þ DE  Þ P I 2e 0 x0 x0 - y0 y0 P = - 0Σ Re ð2FB þ GC  Þ I   2 e 0 z0 z0 = 0Σ P jBj2 þ jC j2 þ jE j2 - jAj2 -2jDj2 3I P

ð3:502Þ ð3:503Þ ð3:504Þ ð3:505Þ ð3:506Þ

And from Eq. (3.485) the following results can be obtained according to the Eqs. (3.394)–(3.408): Ax = Az = Axy = Ayz = 0 pffiffiffi 2 2e Ay = 0 Σ Re ðBC  þ CE  þ OA Þ 3I pffiffiffi 2e ðBC  þ OA þ FG Þ Axz = - 0 ΣIm I 2e Axx - yy = - 0 Σ Re ð2BE  þ OD Þ I   2 e Azz = 0 Σ jBj2 þ jEj2 þ jDj2 - jAj2 -2jCj2 3I

ð3:507Þ ð3:508Þ ð3:509Þ ð3:510Þ ð3:511Þ

i0

From Eq. (3.486), K i can be found according to Eqs. (3.409)–(3.472). First, we can see y0

x0 z0

x0 x0 - y0 y0

Ki = Ki = Ki x

0

0

z

0 0

xy

Ki = Ki = Ki

z0 z0

= K i = 0,

0 0

yz

= K i = 0,

i = x, z, xy, yz

ð3:512Þ

i = y, xz, xx - yy, zz

ð3:513Þ

Then we can still get x0

Kx =

2 e Σ Re ðAB - AE  þ DC  - OC  Þ 0 3I pffiffiffi 2 2e z0 K x = - 0 ΣImðBC  þ CE  Þ 3I

ð3:514Þ ð3:515Þ

176

3 x0 y0 Kx

Polarization Theory of Nuclear Reactions for Spin 1 Particles

pffiffiffi 2e = 0 Σ Re ðGB - FC  Þ I

y0 z0

1e ΣImðAB - AE  þ DC  - OC  Þ 0 I 2 e y0 K y = 0 Σ Re ðAB þ AE  þ DC  þ OC  Þ 3I 0 0 1e xz K y = - 0 ΣIm ðAB þ AE  þ DC  þ OC  Þ I pffiffiffi 0 0 2 2e x x - y0 y0 Ky =- 0 Σ Re ðGB þ FC  Þ I pffiffiffi 0 0 2 2e zz Ky = 0 Σ Re ðBC  þ CE  -2OA Þ 3I pffiffiffi 2 2e x0 K z = - 0 ΣImðOB - DE  Þ 3I   0 2 e z Kz = 0 Σ jBj2 - jE j2 3I 2e x0 y0 K z = - 0 ΣIm ðFB Þ I pffiffiffi 2e y0 z 0 Kz = 0 Σ Re ðOB - DE  Þ I pffiffiffi 2e x0 K xy = 0 Σ Re ðBD - OE  Þ I 2e z0 K xy = - 0 ΣIm ðBE  Þ I 3 e x0 y0 K xy = 0 Σ Re ðHB - FE  Þ 2I 0 0 3 e yz K xy = - pffiffiffi 0 ΣIm ðBD þ OE  Þ 2I Kx =

y0

1e ΣImðAB þ AE  - DC  - OC  Þ 0 I 3 e x0 z 0 K xz = 0 Σ Re ðAB þ AE  - DC  - OC  Þ 2I pffiffiffi 0 0 3 2e x x - y0 y0 K xz = - 0 ΣImðGB - FC  Þ I K xz =

ð3:516Þ ð3:517Þ ð3:518Þ ð3:519Þ ð3:520Þ ð3:521Þ ð3:522Þ ð3:523Þ ð3:524Þ ð3:525Þ ð3:526Þ ð3:527Þ ð3:528Þ ð3:529Þ ð3:530Þ ð3:531Þ ð3:532Þ

3.8

→

→

Polarization Theory for 1 þ A → 1 þ B Reactions

pffiffiffi 2e = - 0 ΣIm ðBC  - CE  -2OA Þ I 1e x0 K yz = - 0 ΣIm ðAB - AE  - DC  þ OC  Þ I pffiffiffi 2e z0 K yz = 0 Σ Re ðBC  - CE  Þ I 3 e x0 y0 K yz = - pffiffiffi 0 ΣIm ðGB þ FC  Þ 2I z0 z0 K xz

y0 z 0

3 e Σ Re ðAB - AE  - DC  þ OC  Þ 0 2I pffiffiffi 2 2e y0 K xx - yy = - 0 Σ Re ðBD þ OE  Þ I pffiffiffi 3 2e x0 z0 K xx - yy = - 0 ΣImðBD - OE  Þ I 0 0 0 0 6e x x -y y K xx - yy = 0 Σ Re ðHB þ FE  Þ I 4e z0 z0 K xx - yy = - 0 Σ Re ðBE  - OD Þ I pffiffiffi 2 2e y0 K zz = 0 Σ Re ðOB þ DE  -2AC  Þ 3I pffiffiffi 2e x0 z 0 ðOB þ DE  -2AC  Þ K zz = - 0 ΣIm I 4e x0 x0 - y0 y0 K zz = - 0Σ Re ðFB - GC  Þ I   2 e z0 z0 K zz = 0 Σ jBj2 þ jEj2 þ 2jAj2 -2jC j2 -2jDj2 3I K yz =

177

ð3:533Þ ð3:534Þ ð3:535Þ ð3:536Þ ð3:537Þ ð3:538Þ ð3:539Þ ð3:540Þ ð3:541Þ ð3:542Þ ð3:543Þ ð3:544Þ ð3:545Þ

Here, and still according to Eq. (3.320), the component labels of the Cartesian coordinate of the spin-1 particle are divided into two groups ε1 = y, xz, xx - yy, zz and ε2 = x, z, xy, yz. From the results obtained previously in this section, we can see 0 i0



P

 Ai

≠0 =0

≠0 =0

i0 = ε01 i0 = ε02 i = ε1 i = ε2

ð3:546Þ ð3:547Þ

178

3

( i0 Ki

≠0 =0

Polarization Theory of Nuclear Reactions for Spin 1 Particles



   i = ε01 and i0 = ε01 or i = ε2 and i0 = ε02     i = ε1 and i0 = ε02 or i = ε2 and i0 = ε01

ð3:548Þ

Now when the spins of the unpolarized target and the unpolarized residual nucleus → → are not equal to 0, it can be seen that for 1 þ A → 1 þ B reaction the presence circumstance of the non-zero polarization physical quantities is consistent with the case where the spins of the target and the residual nucleus are equal to 0. It shows that the influence of parity conservation exists in both cases, and we can obtain from Eqs. (3.479) and (3.481) I =I i0



3 2 1 1 1 þ py Ay þ pxz Axz þ pxx - yy Axx - yy þ pzz Azz 2 3 6 2



 3 2 1 1 i0 i0 i0 i0 þ py K y þ pxz K xz þ pxx - yy K xx - yy þ pzz K zz , 2 3 6 2

  0    2 I 3 i0 i0 i0 i0 i0 P = px K x þ pz K z þ pxy K xy þ pyz K yz , i0 = ε2 0 3 I 2 0

P =

I I



0

0 i0

P

ð3:549Þ i 0 = ε1 0 ; ð3:550Þ

If we let H = B, G = C, O = D, F = E and we remove the simplification e the expressions obtained in this section will automatically summation symbol Σ, degenerate to the corresponding expressions given in Sects. 3.5 and 3.6 where the spins of the target and residual nucleus are all equal to 0.

3.9

→

Polarization Theory for 1 þ A →

*

1 2

þ B Reactions

! 1 þ B reactions such 2 that the spins of the target and the residual nucleus are not equal to 0, but both are unpolarized. The expressions for the reaction amplitude of the j-j coupling and the S-L coupling are given in Eq. (2.251) and Eq. (2.252), respectively. In the case where 1 the incident particle spin is 1 and the outgoing particle spin is , the following 2 symbols are introduced for j-j coupling and S-L coupling, respectively: →

In this section, we study the polarization theory for 1 þ A →

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl0 - ml 0 Þ! ðl0 þ ml 0 Þ!   X jμ j0 m 0 C l0 1μ C Jjμ MIM I Cl0 m 0j 1μ0 CJj0 mMj 0 I 0 M 0I δα0 n0 l0 j0 ,αnlj - SJα0 n0 l0 j0 ,αnlj

0 0 f llμ0 μJ ðml 0 Þ = ^l^l eiðσl þσl0 Þ

jj0

l 2

ð3:551Þ

3.9

→

Polarization Theory for 1 þ A →

^^0 iðσl þσl0 Þ

0

f μll0 μJ ðml 0 Þ = l l e

*

1 2

179

þ B Reactions

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl0 - ml 0 Þ! ðl0 þ ml 0 Þ!

  X S0 M 0 C Jl0 MSM CS1μMIM I C Jl0 mMl 0 S0 M 0 C1μ0 IS0 M 0 δα0 n0 l0 S0 ,αnlS - SJα0 n0 l0 S0 ,αnlS S

SS0

2

ð3:552Þ

I

pffiffiffiffiffiffiffiffiffiffiffiffi where ^l  2l þ 1. According to Eqs. (2.251)–(2.254), it follows that the reaction amplitude can be uniformly written in the following form under the two angular momentum coupling schemes: 0 0 i f α0 n0 μ0 M I 0 ,αnμM I ðΩÞ = f C ðθÞδα0 n0 ,αn δμ0 M I 0 ,μM I þ ð-1ÞðμþM I - μ - M I Þ 2k X 0 0 0 ð3:553Þ 0 0 I - μ - MI f μll0 μJ ðμ þ M I - μ0 - M I 0 ÞPμþM ð cos θÞ eiðμþM I - μ - M I Þφ l0

ll0 J

! 1 þ B reactions, 2 so the first term representing Coulomb scattering in the above formula does not appear. Let →

Because there can be no elastic scattering channel for 1 þ A →

A M I 0 M I ðθ Þ =

  0 1X 0 i 1 M - M 0 þ1 ð-1ÞM I - M I þ2 f ll1 J1 M I - M I 0 þ Pl0 I I 2 ð cos θÞ ð3:554Þ 2 2k 2 0 ll J

BM I 0 M I ðθÞ =

0 1X 0 i ð-1ÞM I - M I - 2 f ll-J1 2 2k 0

 -1

MI - MI 0 -

ll J

 1 M I - M I 0 - 21 ð cos θÞ P0 2 l

ð3:555Þ   0 1X 0 1 1 M -M 0 -1 C M I 0 M I ðθÞ = - ð-1ÞM I - M I - 2 f ll1 J0 M I - M I 0 - Pl0 I I 2 ð cos θÞ 2 2k 2 0 ll J

ð3:556Þ   X 0 1 0 1 0 1 1 M -M þ DM I 0 M I ðθÞ = ð-1ÞM I - M I þ2 f ll-J1 0 M I - M I 0 þ Pl0 I I 2 ð cos θÞ 2 2k 2 0 ll J

ð3:557Þ   X 0 3 0 i 3 M -M 0 -3 EM I 0 M I ðθÞ = - ð-1ÞM I - M I - 2 f ll1 J-1 M I - M I 0 - Pl0 I I 2 ð cos θÞ 2 2k 2 0 ll J

F M I 0 M I ðθ Þ = -

0 3X 0 i ð-1ÞM I - M I þ2 f ll-J1 2 2k 0

ll J

 1

ð3:558Þ  3 M - M 0 þ3 M I - M I 0 þ Pl0 I I 2 ð cos θÞ 2 ð3:559Þ

And introduce the following simplification symbols

180

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

F μ0 μ ðθ, φÞ = f α0 n0 μ0 M I 0 ,αnμM I ðΩÞ

ð3:560Þ

then we can get F 12 1 ðθ, φÞ = AM I 0 M I ðθÞeiðM I - M I þ2Þφ 0

1

F 12 0 ðθ, φÞ = - iCM I 0 M I ðθÞeiðM I - M I

ð3:561Þ

- 12Þφ

ð3:562Þ

- 32Þφ

ð3:563Þ

F - 12 1 ðθ, φÞ = - F M I 0 M I ðθÞeiðM I - M I þ2Þφ

ð3:564Þ

F - 12 0 ðθ, φÞ = iDM I 0 M I ðθÞeiðM I - M I þ2Þφ

ð3:565Þ

F 12

-1 ðθ,

0

φÞ = - E M I 0 M I ðθÞeiðM I - M I

0

0

0

F - 12

-1 ðθ,

φÞ = BM I 0 M I ðθÞeiðM I - M I

0

3

1

- 21Þφ

ð3:566Þ

When the foot note M 0I M I and argument (θ) of A, B, C, D, E, and F are omitted, the ^ in which arguments (θ, φ) are emitted, with definite M 0I M I can reaction amplitude F, be written as Ae2φ

- iCe -2φ

i

^= F

- Fe

i

3i 2φ

iDe

i 2φ

- Ee -2 φ 3i

Be

-2i φ

! 0

eiðM I - M I Þφ

ð3:567Þ

^ is no longer a square Matrix, but rather a rectangular From here we can see that F matrix with an unequal numbers of rows and columns. According to Eqs. (2.334) and (2.335), it can be seen that the magnetic quantum numbers as the footnotes of the ^ and F ^ þ are the same for each term in the case where both the matrix elements of F target and the residual nucleus are unpolarized, and so from Eq. (3.567) we can obtain 0

i

A e - 2 φ

B ^ þ = B iC e2i φ F @

3i - E e 2 φ

þ

^F ^ = F

3i

- Fe - 2 φ

1

i C - iðM I - M I 0 Þ φ - iD e - 2 φ C Ae

jAj2 þ jC j2 þ jEj2 - ðFA þ DC  þ BE  Þ eiφ

i B e2 φ

- ðAF  þ CD þ EB Þ e - iφ jBj2 þ jDj2 þ jF j2

ð3:568Þ ! ð3:569Þ

For the definite MI0MI, the following result can be obtained from Eq. (3.569):

3.9

→

Polarization Theory for 1 þ A →

*

1 2

181

þ B Reactions

n þo   1 ^F ^ = 1 jAj2 þ jBj2 þ jC j2 þ jDj2 þ jEj2 þ jF j2 I 0 = tr F 3 3

ð3:570Þ

*

For the sake of simplicity, we select the y-axis along the direction n ; in this case φ = 0. Then we can obtain the following expressions from Eqs. (3.567)–(3.569): ^= F



^F ^þ = F

A -F

- iC iD

-E , B

0

A ^þ = B F @ iC  - E

jAj2 þ jC j2 þ jE j2 - ðFA þ DC  þ BE  Þ

1 - F C - iD A

ð3:571Þ



B

- ðAF  þ CD þ EB Þ jBj2 þ jDj2 þ jF j2

! ð3:572Þ

For the definite MI0MI we define the following as the components of the polarization vector of the outgoing particles corresponding to the unpolarized incident particles: n o ^F ^þ tr σ^i0 F Pi0 = n þ o , ^F ^ tr F 0

i0 = x0 , y0 , z0

ð3:573Þ

From Eq. (3.573) we can obtain the following results using Eqs. (2.6), and (3.570)– (3.572): 1 ½ - ðFA þ DC  þ BE  Þ - ðAF  þ CD þ EB Þ 3I 0 2 =Re ðFA þ DC  þ BE  Þ 3I 0

0

Px0 =

0

1 ½iðFA þ DC  þ BE  Þ - iðAF  þ CD þ EB Þ 3I 0 2 ImðFA þ DC  þ BE  Þ =3I 0   0 1 Pz0 = jAj2 þ jCj2 þ jE j2 - jBj2 - jDj2 - jF j2 3I 0

Py0 =

ð3:574Þ

ð3:575Þ

ð3:576Þ

For the definite MI0MI we also define n o ^þ ^ ^Si F tr F Ai = n þ o , ^F ^ tr F and

i = x, y, z

ð3:577Þ

182

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

n o ^ iF ^Q ^þ tr F Ai = n þ o , ^F ^ tr F

i = xy, xz, yz, xx - yy, zz

ð3:578Þ

as the polarization vector analyzing powers and the polarization tensor analyzing powers, respectively. By means of Eqs. (3.33), (3.36), (3.570), and (3.571), we can obtain the following expressions from Eqs. (3.577) and (3.578): 0

iC 

B ^ ^ þ = p1ffiffiffi B A - E  Sx F 2@ iC  þ

^^ ^ = p1ffiffiffi F Sx F 2

- iD

1

C - F  þ B C A - iD

iðAC  - CA þ CE  - EC  Þ

- iðAD - CF  þ CB - ED Þ

- iðFC - DA þ DE  - BC  Þ

iðFD - DF  þ DB - BD Þ

pffiffiffi 2 Ax = ImðAC  þ CE  þ FD þ DB Þ 3I 0 0 1 iD - iC  B C ^Sy F ^ þ = pi ffiffiffi B A þ E  - ðF  þ B Þ C @ A 2   - iD iC ^^ ^ þ = p1ffiffiffi F Sy F 2

ð3:579Þ ð3:580Þ

AC  þ CA þ CE  þ EC

- ðAD þ CF  þ CB þ ED Þ

- ðFC  þ DA þ DE  þ BC  Þ

FD þ DF  þ DB þ BD

pffiffiffi 2 Ay = Re ðAC  þ CE  þ FD þ DB Þ 3I 0 0  1 A - F B C ^þ = B 0 Sz F 0 C @ A E  - B ! - AF  þ EB jAj2 - jE j2 þ ^ zF ^ = FS - FA þ BE  jF j2 - jBj2   1 jAj2 - jBj2 - jEj2 þ jF j2 Az = 3I 0

!

!

ð3:581Þ ð3:582Þ

ð3:583Þ

ð3:584Þ

3.9

→

Polarization Theory for 1 þ A →

*

1 2

0

E

- B

B ^ xy F ^ þ = 3i B 0 Q 2@ A ^ xy F ^Q ^ þ = 3i F 2

183

þ B Reactions



AE - EA 

C 0 C A,  -F 

- FE þ BA

Axy = -

1



- AB þ EF 



!

ð3:585Þ

FB - BF 

1 ImðAE  þ FB Þ I0

0 1 iC  - iD 3 ^ xz F ^ þ = pffiffiffi@ A þ E  - ðF  þ B Þ A, Q 2 2  iD  - iC   3 i AC ð CA - CE þ EC  Þ ^ xz F ^Q ^ þ = pffiffiffi F     2 2 - iðFC - DA - DE þ BC Þ

ð3:586Þ

- iðAD - CF  - CB þ ED Þ iðFD - DF  - DB þ BD Þ



ð3:587Þ 1 Axz = - pffiffiffi ImðAC  - CE  þ FD - DB Þ 2I 0 0 1  iD - iC C 3i B ^ yz F ^þ = p ffiffiffi B A  - E  - F  þ B C Q @ A, 2 2   - iC iD þ

3 ^ yz F ^ = p ^Q ffiffiffi F 2 2

ð3:588Þ

AC  þ CA - CE - EC

- ðAD þ CF  - CB - ED Þ

- ðFC  þ DA - DE  - BC  Þ

FD þ DF  - DB - BD

!

ð3:589Þ 1 Ayz = pffiffiffi Re ðAC  - CE  þ FD - DB Þ 2I 0 0 1 - E B B C ^ xx - yy F ^ þ = 3B 0 0 C Q @ A, - F A ^ xx - yy F ^þ = 3 ^Q F



- AE - EA FE  þ BA

Axx - yy = -





AB þ EF



ð3:590Þ

!

ð3:591Þ

- FB - BF 

2 Re ðAE  þ FB Þ I0

ð3:592Þ

184

3

0

A

B ^ zz F ^ þ = B -2iC  Q @ - E ^ zz F ^Q ^þ = F

2

Polarization Theory of Nuclear Reactions for Spin 1 Particles

- F

1

C 2iD C A  B 2

jAj -2jC j þ jE j

2







- AF þ 2CD - EB

!

- FA þ 2DC  - BE  jF j2 -2jDj2 þ jBj2 h i 1 Azz = jAj2 þ jBj2 þ jEj2 þ jF j2 -2jC j2 -2jDj2 3I 0

ð3:593Þ

ð3:594Þ

Now define n o ^ ^Si F ^þ 0F tr σ ^ i 0 n þ o , i = x, y, z; K ii = ^F ^ tr F

i 0 = x0 , y0 , z 0

ð3:595Þ

as the components of the polarization transfer coefficients from the incident particle polarization vector to the outgoing particle polarization vector, and define n o ^ iF ^Q ^þ 0F tr σ ^ i 0 n þo , K ii = ^F ^ tr F

i = xy, xz, yz, xx - yy, zz;

i 0 = x0 , y0 , z 0

ð3:596Þ

as the components of the polarization transfer coefficients from the incident particle polarization tensor to the outgoing particle polarization vector. Using Eqs. (2.6) and ^ iF ^ þ and F ^ þ obtained in this section previously, ^ ^Si F ^Q (3.570) as well as the matrices F we can obtain from Eqs. (3.595) and (3.596) pffiffiffi 2 = ImðAD þ CB þ FC  þ DE  Þ 3I 0 pffiffiffi 2 y0 Re ðAD þ CB - FC  - DE  Þ Kx = 3I 0 pffiffiffi 0 2 ImðAC  þ CE  - FD - DB Þ K zx = 3I 0 pffiffiffi 2 x0 Re ðAD þ CB þ FC  þ DE  Þ Ky = 3I 0 pffiffiffi 2 y0 ImðAD þ CB - FC  - DE  Þ Ky = 3I 0 0 K xx

ð3:597Þ ð3:598Þ ð3:599Þ ð3:600Þ ð3:601Þ

3.9

→

Polarization Theory for 1 þ A → 0 K zy

*

1 2

185

þ B Reactions

pffiffiffi 2 = Re ðAC  þ CE  - FD - DB Þ 3I 0 0

K xz = -

2 Re ðFA - BE  Þ 3I 0

ð3:603Þ

2 ImðFA - BE  Þ 3I 0

ð3:604Þ

0

K yz = 1 3I 0

0

K zz =



jAj2 þ jBj2 - jE j2 - jF j2



1 ImðAB þ FE  Þ I0

ð3:606Þ

0

1 Re ðAB - FE  Þ I0

ð3:607Þ

K yxy = 0

K zxy = -

1 ImðAE  - FB Þ I0

0 1 K xxz = pffiffiffi ImðAD - CB þ FC  - DE  Þ 2I 0 0 1 K yxz = pffiffiffi Re ðAD - CB - FC  þ DE  Þ 2I 0 0 1 K zxz = - pffiffiffi ImðAC  - CE  - FD þ DB Þ 2I 0 0 1 K xyz = - pffiffiffi Re ðAD - CB þ FC  - DE  Þ 2I 0 0 1 K yyz = pffiffiffi ImðAD - CB - FC  þ DE  Þ 2I 0 0 1 K zyz = pffiffiffi Re ðAC  - CE  - FD þ DB Þ 2I 0

K xxx - yy =

2 Re ðAB þ FE  Þ I0

0

K yxx - yy = 0

K zxx - yy = 0

K xzz = -

ð3:605Þ

0

K xxy =

0

ð3:602Þ

ð3:608Þ ð3:609Þ ð3:610Þ ð3:611Þ ð3:612Þ ð3:613Þ ð3:614Þ ð3:615Þ

2 ImðAB - FE  Þ I0

ð3:616Þ

2 Re ðAE  - FB Þ I0

ð3:617Þ

2 Re ðFA -2DC  þ BE  Þ 3I 0

ð3:618Þ

186

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

0

K yzz = 0

K zzz =

1 3I 0



2 ImðFA -2DC  þ BE  Þ 3I 0

jAj2 - jBj2 þ jE j2 - jF j2 -2jC j2 þ 2jDj2

ð3:619Þ 

ð3:620Þ

In the case where both the target and the residual nucleus are polarized, and * * continuing to choose the y-axis along the direction n corresponding to n , where, in ^ is M I 0 M I and the foot note of the Eq. (3.571), the footnote of the matrix element of F þ 0 ^ matrix element of F is taken as M I M I , and making the following changes to the previous results jX j2 → X M I 0 M I X M I 0 M I ,

X = A, B, C, D, E, F

ð3:621Þ

Then according to Eq. (3.570) we can write I 0,M I 0 M I 0 M I M I =

1  ðA 0 A 0 þ BM I 0 M I BM I 0 M I þ CM I 0 M I C M I 0 M I 3 MI MI MI MI þDM I 0 M I DM I 0 M I þ E M I 0 M I E M I 0 M I þ F M I 0 M I F M I 0 M I Þ

ð3:622Þ

By means of the form of the polarization wave functions of the target and the residual nucleus introduced by Eqs. (2.315) and (2.320) and continuing to use the simplification summation symbol Σ for the polarized target and residual nucleus introduced by Eq. (2.327), the angular distribution of the emitted particles corresponding to the unpolarized incident particles can be expressed as 0

I  I 0α0 n0 ,

αn =

dσ 0α0 n0 ,αn = Σ I 0, dΩ

MI 0 MI 0 MI MI

ð3:623Þ

The component of the emitted particle polarization vector corresponding to the unpolarized incident particle is 0 i0

P

0

 P0α0 ni 0 ,

αn

=

1 I

0

Σ I 0,

i0 M I 0 M I 0 M I M I P0, M I 0 M I 0 M I M I ,

i 0 = x0 , y0 , z 0

ð3:624Þ

When the normal variable in the Eqs. (3.574)–(3.576) is added with the footnote 0 M 0I M I , and the complex conjugate variable is added with the footnote M I M I , and i0 using Eq. (3.621), then every P0, M 0 M 0 M M in Eq. (3.624) can be obtained. I I I I When the incident particles, targets, and residual nuclei are all polarized, the components of the vector polarization analyzing power and the tensor polarization analyzing power of the nuclear reactions are

→

Polarization Theory for 1 þ A →

3.9

Ai  Ai,

α0 n0 , αn

=

1 I

0

ΣI 0,

*

1 2

187

þ B Reactions

M I 0 M I 0 M I M I Ai, M I 0 M I 0 M I M I ,

i=ε

ð3:625Þ

where ε is the symbol of the label set of the components of the Cartesian coordinate introduced by Eq. (3.476). Here Ai,M I 0 M I 0 M I M I have been given by Eqs. (3.580), ρin (3.582), (3.584), (3.586), (3.588), (3.590), (3.592), and (3.594). Referring to ^ given by Eq. (3.214) and ^ρout given by Eq. (2.85), the corresponding differential cross section of the emitted particles can be written as h  dσ 0 0 3 0 I  I α0 n0 , αn = α n , αn = trf^ρout g = I 1 þ px Ax þ py Ay þ pz Az 2 dΩ i  1 1 2 þ pxy Axy þ pxz Axz þ pyz Ayz þ pxx - yy Axx - yy þ pzz Azz 2 6 3

ð3:626Þ

The corresponding polarization transfer coefficients are i0

0

K i  K ii, α0 n0 , αn =

1 I

0

ΣI 0,

i0 M I 0 M I 0 M I M I K i, M I 0 M I 0 M I M I ,

i = ε,

i0 = x0 , y0 , z0 ð3:627Þ

0

where K ii, M 0 M 0 M M have been given by Eqs. (3.597)–(3.620). Referring to I I I I Eqs. (3.214) and (2.85), the corresponding components of the polarization vector of the emitted particles can be written as follows: i0

P 

0   trfσ^i0 ^ρout g I 3 0 i0 i0 i0 i0 P þ p x K x þ py K y þ pz K z = = 2 trf^ρout g I    0 0 0 1 1 2 i i i i0 i0 þ pxy K xy þ pxz K xz þ pyz K yz þ pxx - yy K xx - yy þ pzz K zz , i0 = x0 , y0 , z0 2 6 3

0 Piα0 n0 , αn

ð3:628Þ In the case where both the target and the residual nucleus are unpolarized, and e using Eqs. (2.334) and (2.335) as well as the simplification summation symbols Σ given by Eq. (2.278), the Eqs. (3.622)–(3.625), and (3.627) can be simplified, respectively, as follows: I 0,

MI 0 MI

=

 2  2  2  2  2  2  1  AM I 0 M I  þ BM I 0 M I  þ CM I 0 M I  þDM I 0 M I  þ E M I 0 M I  þ F M I 0 M I  3 ð3:629Þ e 0, I = ΣI 0

0i0

P =

1e ΣI 0, 0 I

ð3:630Þ

MI 0 MI

i0 M I 0 M I P0, M I 0 M I ,

i0 = x0 , y0 , z0

ð3:631Þ

188

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

Ai = i0

Ki =

1e ΣI 0, 0 I

1e ΣI 0, 0 I

M I 0 M I Ai, M I 0 M I ,

i0 M I 0 M I K i, M I 0 M I ,

i=ε

i = ε;

ð3:632Þ

i 0 = x 0 , y0 , z 0

ð3:633Þ

Next, we will analyze Eq. (3.552) which belongs to S-L coupling scheme. Based on the properties of the C-G coefficient, there are the following relations: Iþ1 - S S I CS1μμþM C1 IM I = ð-1Þ

CJℓ0MSM

ℓþS -J

= ð-1Þ

-μ -M I - μ I -M I ,

J -M Cℓ0 S -M ,

0

0

0

0

0

0

0

0

-μ -M I I C 1Sμ0 μI 0þM = ð-1ÞI þ2 - S C 1S -μ 0 I 0 -M 0 , M0 1

I

2

I

2

CJℓ0 mMℓ 0 S0 M S 0

= ð-1Þ

ℓ 0 þS0 - J

CℓJ0 -M -mℓ 0 S0 -M S 0 ð3:634Þ

Now we only need to study the second item of Eq. (3.553) with i = 1 and i0 = 12 . Using Eqs. (2.288) and (3.634), for the part which is not associated with angle φ, the following phase factor comes out when all spin magnetic quantum numbers change signs: 0

0

0

0

ð-1ÞℓþS - Jþℓ þS - JþIþ1 - SþI þ2 - S þμþM I - μ 1

0

- M 0I

0

0

= ð-1Þ2JþℓþIþℓ þI þ2þμþM I - μ 3

0

- M 0I

0

The symbol Π introduced by Eq. (3.488) can also be used here, so ð-1Þℓþℓ = ð-1ÞΠ . ! → 1 For 1 þ A → þ B reaction, I and I0 must be that one is an integer and another is a 2 semi-odd number. When I is an integer and I0 is a semi-odd number, J is an integer; when I is a semi-odd number and I0 is an integer, J is a semi-odd number. Let  Λ=

0

when I is an integer and I 0 is a semi‐odd number

1

when I is a semi‐odd number and I 0 is an integer

ð3:635Þ

thus (-1)2J can be replaced by (-1)Λ, and we can get 0

f α0 n0 μ0 M I 0 ,αnμM I ðθÞ = ð-1ÞΠþΛþIþI þ2þμþM I - μ 3

0

- MI 0

f α0 n0 - μ0 - M I 0 ,αn - μ - M I ðθÞ ð3:636Þ

The following relations can be seen from Eqs. (3.553)–(3.559): AM I 0 M I ðθÞ = f α0 n0 12M I 0 ,αn1M I ðθÞ,

BM I 0 M I ðθÞ = f α0 n0 - 12M I 0 ,αn -1M I ðθÞ,

C M I 0 M I ðθÞ = if α0 n0 12M I 0 ,αn0M I ðθÞ,

DM I 0 M I ðθÞ = - if α0 n0 - 12M I 0 ,αn0M I ðθÞ, ð3:637Þ

E M I 0 M I ðθÞ = - f α0 n0 12M I 0 ,αn -1M I ðθÞ, F M I 0 M I ðθÞ = - f α0 n0 - 12M I 0 ,αn1M I ðθÞ And the following relations can be obtained according to Eqs. (3.636) and (3.637):

3.9

→

*

Polarization Theory for 1 þ A →

1 2

189

þ B Reactions 0

0

0

0

0

0

BM I 0 M I ðθÞ = ð-1ÞΠþΛþIþI þM I - M I þ1 A - M I 0 DM I 0 M I ðθÞ = ð-1ÞΠþΛþIþI þM I - M I þ1 C - M I 0 F M I 0 M I ðθÞ = ð-1ÞΠþΛþIþI þM I - M I þ1 E - M I 0

- M I ðθ Þ

ð3:638Þ

- M I ðθ Þ

ð3:639Þ

- M I ðθ Þ

ð3:640Þ

e given by Here we continue to use the simplification summation symbol Σ Eq. (2.278). Note that it must be (-1)2X = 1 when X is an integer, and also note that the sum of MI and the sum of -MI are all from -I to I, and the sum of M 0I and the sum of - M 0I are all from -I0 to I0, so using the method to prove Eq. (2.297), from Eqs. (2.423)–(2.425) the following results can be proved: ~ ImðAE  þ FB Þ = 0, Σ ~ ImðCE  þ FD Þ = 0, Σ~ ImðAC  þ DB Þ = 0, Σ ~ ReðAC  Þ, Σ~ ReðAE  þ FB Þ = 2Σ ~ ReðAE  Þ, Σ~ ReðAC  þ DB Þ = 2Σ ~ ReðCE  Þ, Σ ~ ImðAC  - DB Þ = 2Σ ~ ImðAC  Þ, Σ~ ReðCE  þ FD Þ = 2Σ ~ ImðCE  Þ, Σ~ ImðAE  - FB Þ = 2Σ~ ImðAE  Þ, Σ~ ImðCE - FD Þ = 2Σ

ð3:641Þ

~ ReðAE  - FB Þ = 0, Σ ~ ReðCE  - FD Þ = 0, Σ~ ReðAC  - DB Þ = 0, Σ 2 2 2 2 2 ~ jA j , Σ ~ jDj = Σ ~ jC j , Σ ~ jF j = Σ ~ jE j2 Σ~ jBj = Σ And by using the method to prove Eq. (2.298), we can prove e  , ΣDC e  , ΣFE e , Σ e ðAC  þ DB Þ, Σ e ðAE  þ FB Þ, Σ e ðCE  þ FD Þ ð3:642Þ ΣBA such that their corresponding complex conjugate terms are all real numbers. There are other items that also satisfy the above two expressions. Defining        eI 0,M 0 M = 2 AM 0 M 2 þ C M 0 M 2 þ EM 0 M 2 I I I I I I I I 3

ð3:643Þ

then from Eqs. (3.629), (3.630), and (3.641), we can prove e eI 0,M 0 M I =Σ I I 0

ð3:644Þ

From Eq. (3.631), and according to Eqs. (3.574)–(3.576), we can obtain the following results using Eqs. (3.638)–(3.642): 0 x0

P

=-

2 e Σ Re ð2FA þ DC  Þ 3I 0 0 y0

P

=P

0 z0

=0

ð3:645Þ ð3:646Þ

190

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

And from Eq. (3.632) we can obtain the following results according to Eqs. (3.580), (3.582), (3.584), (3.586), (3.588), (3.590), (3.592), and (3.594): Ax = Az = Axy = Ayz = 0 pffiffiffi 2 2e Ay = Σ Re ðAC  þ CE  Þ 3I 0 pffiffiffi 2e Axz = ΣImðAC  - CE  Þ I0 Axx - yy = Azz =

4e Σ Re ðAE  Þ I0

  2 e Σ jAj2 þ jE j2 -2jC j2 3I 0

ð3:647Þ ð3:648Þ ð3:649Þ ð3:650Þ ð3:651Þ

Further from Eq. (3.633) we can obtain the following results according to Eqs. (3.597)–(3.620): i0

K i = 0,

ði = ε1 and i0 = y0 , z0 Þ or ði = ε2 and i0 = x0 Þ

ð3:652Þ

such that the definition of the subscripts ε1 and ε2 are given by Eq. (3.320). Given this, we can obtain pffiffiffi 2 2e = Σ Re ðAD - FC  Þ 3I 0 pffiffiffi 2 2e z0 Kx = ΣImðAC  þ CE  Þ 3I 0 pffiffiffi 2 2e x0 Ky = Σ Re ðAD þ FC  Þ 3I 0 y0 Kx

y0

4 e ΣImðFA Þ 3I 0   2 e z0 Σ jAj2 - jE j2 Kz = 3I 0 Kz = -

y0

K xy = z0

1e Σ Re ðAB - FE  Þ I0

K xy = -

2e ΣImðAE  Þ I0

ð3:653Þ ð3:654Þ ð3:655Þ ð3:656Þ ð3:657Þ ð3:658Þ ð3:659Þ

*

3.10

Polarization Theory for

1 2

→

191

þ A → 1 þ B Reactions

pffiffiffi 2e = ΣImðAD þ FC  Þ I0 pffiffiffi 2e y0 K yz = ΣImðAD - FC  Þ I0 pffiffiffi 2e z0 K yz = Σ Re ðAC  - CE  Þ I0 x0 K xz

ð3:660Þ ð3:661Þ ð3:662Þ

x0

2e Σ Re ðAB þ FE  Þ I0

ð3:663Þ

x0

4 e Σ Re ðFA - DC  Þ 3I 0

ð3:664Þ

K xx - yy = K zz = -

From the results obtained above, it can be seen that P

0 i0



Ai i0 Ki



i 0 = x0

≠0

ð3:665Þ

i0 = y0 , z0

=0  ≠0

i = ε1

=0

i = ε2

ð3:666Þ

≠0

ði = ε1 and i0 = x0 Þ or ði = ε2 and i0 = y0 , z0 Þ

=0

ði = ε1 and i0 = y0 , z0 Þ or ði = ε2 and i0 = x0 Þ

ð3:667Þ

And we can obtain the following results from Eqs. (3.626) and (3.628): h i 3 2 1 1 0 I = I 1 þ py Ay þ pxz Axz þ pxx - yy Axx - yy þ pzz Azz 2 3 6 2 0 h i I 3 2 1 1 i0 0i0 i0 i0 i0 i0 P þ py K y þ pxz K xz þ pxx - yy K xx - yy þ pzz K zz , P = 2 3 6 2 I i0 = x 0 i0

P =

0h    i 2 I 3 i0 i0 i0 i0 px K x þ pz K z þ pxy K xy þ pyz K yz , 3 I 2

*

3.10

Polarization Theory for

1 2

i0 = y0 ,z0

ð3:668Þ ð3:669Þ

ð3:670Þ

→

þ A → 1 þ B Reactions

! * 1 This section studies the polarization theory for þ A → 1 þ B reactions, where 2 target and residual nucleus spins are not equal to 0, but are both unpolarized.

192

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

1 incident particle and a spin 1 outgoing particle, and 2 referring to Eqs. (2.251) and (2.252), we introduce the following symbols for j-j coupling and S-L coupling, respectively: In the case of a spin

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl0 - ml 0 Þ! X j μ J M C 1 C ðl0 þ ml 0 Þ! jj0 l0 2μ jμ IM I   J M J 0 1μ0 C j0 mj 0 I 0 M I 0 δα0 n0 l j0 ,αnlj - Sα0 n0 l0 j0 , αnlj

0 0 f μll0 μJ ðml 0 Þ = ^l^l eiðσ l þσ l0 Þ

j0 m 0 C l0 mjl 0

^^ 0 iðσ l þσ l0 Þ

0

f μll0 μJ ðml 0 Þ = l l e CJl0 mMl 0

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl0 - ml 0 Þ! X ðl0 þ ml 0 Þ! SS0

CJl0 MSM C S1μMIM I 2

  S0 M 0 S J 0 0 C δ S 0 0 0 0 0 0 0 0 0 0 0 α n l S , αnlS S M S 1μ I M I α n l S , αnlS

ð3:671Þ

ð3:672Þ

Given this, the reaction amplitude can be uniformly written in the form of Eq. (3.553) under two angular momentum coupling schemes according to Eqs. (2.251)–(2.254). ! * 1 Because there is no elastic scattering channel for þ A → 1 þ B reactions, the 2 first term representing Coulomb scattering in Eq. (3.553) does not appear. Let AM I 0 M I ðθÞ =

  1X 0 0 i 1 M -M 0 -1 ð-1ÞM I - M I - 2 f 1ll J1 M I - M I 0 - Pl0 I I 2 ð cos θÞ ð3:673Þ 2 2k 2 0 ll J

BM I 0 M I ðθÞ =

0 1X 0 i ð-1ÞM I - M I þ2 f ll-1J 2k 0

ll J

- 12

  1 M - M 0 þ1 M I - M I 0 þ Pl0 I I 2 ð cos θÞ 2

ð3:674Þ   0 1X 0 1 1 M - M 0 þ1 C M I 0 M I ðθÞ = - ð-1ÞM I - M I þ2 f ll0 J1 M I - M I 0 þ Pl0 I I 2 ð cos θÞ 2 2k 2 0 ll J

DM I 0 M I ðθÞ =

1 ð-1Þ 2k

0

M I - M I - 12

X ll0 J

0

f ll0 J- 1 2



ð3:675Þ  0 1 1 M -M M I - M I 0 - Pl0 I I 2 ð cos θÞ 2

ð3:676Þ   X 0 3 0 i 3 M - M 0 þ3 EM I 0 M I ðθÞ = - ð-1ÞM I - M I þ2 f ll-1J 1 M I - M I 0 þ Pl0 I I 2 ð cos θÞ 2 2k 2 0 ll J

ð3:677Þ

*

3.10

Polarization Theory for

F M I 0 M I ðθ Þ = -

1 2

→

193

þ A → 1 þ B Reactions

  3X 0 0 i 3 M -M 0 -3 ð-1ÞM I - M I - 2 f 1ll J- 1 M I - M I 0 - Pl0 I I 2 ð cos θÞ 2 2k 2 0 ll J

ð3:678Þ Using the simplified symbols introduced in Eq. (3.560), we can get F 1 12 ðθ, φÞ = AM I 0 M I ðθÞeiðM I - M I - 12 ðθ,

F1

0

- 12Þφ

φÞ = - F M I 0 M I ðθÞeiðM I - M I

0

ð3:679Þ

- 23Þφ

ð3:680Þ

F 0 12 ðθ, φÞ = - iC M I 0 M I ðθÞeiðM I - M I þ2Þφ 0

F0

- 12 ðθ,

φÞ = iDM I 0 M I ðθÞeiðM I - M I

0

1

ð3:681Þ

- 12Þφ

ð3:682Þ

F -1 12 ðθ, φÞ = - E M I 0 M I ðθÞeiðM I - M I þ2Þφ

ð3:683Þ

φÞ = BM I 0 M I ðθÞeiðM I - M I þ2Þφ

ð3:684Þ

0

F -1

- 12 ðθ,

0

3

1

When the footnote M 0I M I and argument (θ) of A, B, C, D, E, and F are omitted, the ^ in which foot note M 0I M I and arguments (θ, φ) are emitted, too, reaction amplitude F, can be written as 0

Ae - 2φ i

- Fe - 2 φ 3i

i ^ =B F @ - iCe2φ 3i - Ee 2 φ

1

C i M - M0 φ i iDe - 2φ A e ð I I Þ i Be2φ

ð3:685Þ

In the case of that both the target and the residual nucleus are unpolarized, from Eqs. (2.334) and (2.335), we can see that the magnetic quantum number footnotes of ^ and F ^ þ are always the same in the polarization theory the matrix element of F formulas, so from Eq. (3.685) we can obtain ^þ = F 0

A  e 2φ i



-F e

3i 2φ

iC  e - 2φ i



- iD e

jAj2 þ jF j2

B ^F ^ þ = B - iðCA þ DF  Þ eiφ F @ - ðEA þ BF  Þ e2iφ

i 2φ

- E e - 2 φ 3i

Be

- 2i φ

! 0

e - iðM I - M I Þφ

iðAC  þ FD Þe - iφ jCj2 þ jDj2 - iðEC þ BD Þ eiφ

ð3:686Þ

- ðAE  þ FB Þe -2iφ

1

C iðCE þ DB Þ e - iφ C A jEj2 þ jBj2 ð3:687Þ

For the definite MI0MI, from Eq. (3.687) we can obtain

194

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

n þo   1 ^F ^ = 1 jAj2 þ jBj2 þ jC j2 þ jDj2 þ jEj2 þ jF j2 I 0 = tr F 2 2

ð3:688Þ

*

For the sake of simplicity, select the y-axis along the direction n , such that φ = 0. Given this, the following expressions can be obtained from Eqs. (3.685)–(3.687): 0

1 A -F   A iC - E C ^þ = ^ =B F F @ - iC iD A, - F  - iD B -E B 0 1 iðAC  þ FD Þ - ðAE  þ FB Þ jAj2 þ jF j2 C ^F ^þ = B F @ - iðCA þ DF  Þ iðCE  þ DB Þ A jC j2 þ jDj2 - ðEA þ BF  Þ - iðEC  þ BD Þ jE j2 þ jBj2

ð3:689Þ

ð3:690Þ

For the definite MI0MI, we define the following as the component of the outgoing particle polarization vector corresponding to the unpolarized incident particle, n o ^F ^þ tr S^i0 F Pi0 = n þ o , ^F ^ tr F 0

i 0 = x0 , y0 , z 0

ð3:691Þ

and we define n o ^ i0 F ^F ^þ tr Q n þo , Pi0 = ^F ^ tr F 0

i0 = x0 y0 , x0 z0 , y0 z0 , x0 x0 - y0 y0 , z0 z0

ð3:692Þ

as the component of the outgoing particle polarization tensor corresponding to the unpolarized incident particle. Utilizing Eqs. (3.33), (3.36), (3.688), and (3.690), the following results can be obtained from Eqs. (3.691) and (3.692): 0 1 Px0 = pffiffiffi ½ - iðCA þ DF  Þ þ iðAC  þ FD Þ 2 2I 0 - iðEC  þ BD Þ þ iðCE  þ DB Þ pffiffiffi 2 ImðCA þ DF  þ EC  þ BD Þ = 2I 0

ð3:693Þ

*

3.10

Polarization Theory for

1 2

→

0 i Py0 = pffiffiffi ½iðCA þ DF  Þ þ iðAC  þ FD Þ 2 2I 0 þiðEC  þ BD Þ þ iðCE  þ DB Þ pffiffiffi 2 =Re ðCA þ DF  þ EC  þ BD Þ 2I 0 h   i 0 1 Pz0 = jAj2 þ jF j2 - jE j2 þ jBj2 2I 0 0 0

195

þ A → 1 þ B Reactions

3i ½ðEA þ BF  Þ - ðAE  þ FB Þ 4I 0 3 =ImðEA þ BF  Þ 2I 0

Px0 y =

ð3:694Þ

ð3:695Þ

ð3:696Þ

0 0 3 Px0 z = pffiffiffi ½ - iðCA þ DF  Þ þ iðAC  þ FD Þ 4 2I 0 þiðEC  þ BD Þ - iðCE  þ DB Þ 3 = pffiffiffi ImðCA þ DF  - EC  - BD Þ 2 2I 0

ð3:697Þ

0 0 3i Py0 z = pffiffiffi ½iðCA þ DF  Þ þ iðAC  þ FD Þ 4 2I 0 - iðEC  þ BD Þ - iðCE  þ DB Þ 3 = - pffiffiffi Re ðCA þ DF  - EC  - BD Þ 2 2I 0

ð3:698Þ

3 ½ - ðEA þ BF  Þ - ðAE  þ FB Þ 2I 0 3 = - Re ðEA þ BF  Þ I0   0 0 1 Pz0 z = jAj2 þ jBj2 þ jE j2 þ jF j2 -2jC j2 -2jDj2 2I 0 0 0

Px0 x

- y0 y0

=

ð3:699Þ

ð3:700Þ

For the definite MI0MI, we define the polarization vector analyzing power as n o ^þ ^ σi F tr F^ Ai = n þ o , ^F ^ tr F

i = x, y, z

ð3:701Þ

The following expressions can be obtained from Eq. (3.701) and utilizing Eqs. (2.6), (3.688), and Eq. (3.689)

196

3

þ

^ = σ^x F

Polarization Theory of Nuclear Reactions for Spin 1 Particles

!

- F

- iD

B

A

iC 

- E

0

- ðAF  þ FA Þ

- iðAD þ FC  Þ

B ^ σx F ^ þ = B iðCF  þ DA Þ F^ @

- ðCD þ DC Þ iðED þ BC Þ

Ax = ^ = σ^y F

iF  - D - iB





EF  þ BA

þ

,

!

1

AB þ FE 

C - iðCB þ DE Þ C A   - ðEB þ BE Þ 

1 Re ðAF  þ CD þ EB Þ I0

, iA - C  - iE  0 iðAF  - FA Þ B ^ σyF ^ þ = B CF  - DA F^ @ - iðEF  - BA Þ

- AD þ FC 



iðCD - DC Þ ED - BC 

ð3:703Þ

- iðAB - FE  Þ

1

C - CB þ DE C A   iðEB - BE Þ 

A

iC





F 0

iD

-E

1 ImðAF  þ CD þ EB Þ I0 ! 

- B

jAj2 - jF j2

B ^ σz F ^ = B - iðCA - DF  Þ F^ @

ð3:704Þ



Ay = ^þ = σ^z F

ð3:702Þ



ð3:705Þ

, iðAC  - FD Þ

þ

jC j2 - jDj2

- AE  þ FB

1

C iðCE  - DB Þ C A

- EA þ BF  - iðEC - BD Þ jE j2 - jBj2 h i 1 Az = jAj2 þ jC j2 þ jEj2 - jF j2 - jDj2 - jBj2 2I 0

ð3:706Þ

ð3:707Þ

Next we define the polarization transfer coefficient from the incident particle polarization vector to the outgoing particle polarization vector as n o ^i0 F^ ^ σiF ^þ tr S 0 n þo , K ii = ^F ^ tr F

i = x, y, z;

i0 = x0 , y0 , z0

ð3:708Þ

and define. n o ^ i0 F^ ^ σiF ^þ tr Q 0 n þo , K ii = ^F ^ tr F

i = x, y, z;

i0 = x0 y0 , x0 z0 , y0 z0 , x0 x0 - y0 y0 , z0 z0

ð3:709Þ

*

3.10

Polarization Theory for

1 2

→

197

þ A → 1 þ B Reactions

as the polarization transfer coefficient from the incident particle polarization vector to the outgoing particle polarization tensor. Using Eqs. (3.33), (3.36), and (3.688), as ^ þ ði = x, y, zÞ obtained previously in this section, the ^ σi F well as the matrices F^ following results can be obtained by Eqs. (3.708) and (3.709): 0 1 K xx = - pffiffiffi ImðDA þ BC  þ CF  þ ED Þ 2I 0 0 1 K xy = - pffiffiffi Re ðDA þ BC  - CF  - ED Þ 2I 0 0 1 K xz = pffiffiffi ImðCA - DF  þ EC  - BD Þ 2I 0 0 1 K yx = pffiffiffi Re ðDA þ BC  þ CF  þ ED Þ 2I 0 0 1 K yy = - pffiffiffi ImðDA þ BC  - CF  - ED Þ 2I 0 0 1 K yz = - pffiffiffi Re ðCA - DF  þ EC  - BD Þ 2I 0 0

K zx = 0

K zy = 0

K zz =

1 2I 0



ð3:710Þ ð3:711Þ ð3:712Þ ð3:713Þ ð3:714Þ ð3:715Þ

1 Re ðAF  - EB Þ I0

ð3:716Þ

1 ImðAF  - EB Þ I0

ð3:717Þ

jAj2 - jF j2 - jE j2 þ jBj2



ð3:718Þ

0 0

3 ImðBA þ EF  Þ 2I 0

ð3:719Þ

0 0

3 Re ðBA - EF  Þ 2I 0

ð3:720Þ

K xx y = K xy y = 0 0

K xz y = -

3 ImðEA - BF  Þ 2I 0

0 0 3 K xx z = - pffiffiffi ImðDA - BC  þ CF  - ED Þ 2 2I 0 0 0 3 K xy z = - pffiffiffi Re ðDA - BC  - CF  þ ED Þ 2 2I 0 0 0 3 K xz z = pffiffiffi ImðCA - DF  - EC  þ BD Þ 2 2I 0

ð3:721Þ ð3:722Þ ð3:723Þ ð3:724Þ

198

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

0 0 3 K yx z = pffiffiffi Re ðDA - BC  þ CF  - ED Þ 2 2I 0 0 0 3 K yy z = - pffiffiffi ImðDA - BC  - CF  þ ED Þ 2 2I 0 0 0 3 K yz z = - pffiffiffi Re ðCA - DF  - EC  þ BD Þ 2 2I 0 0 0

K xx x

0 0

K xz x

ð3:729Þ

- y0 y0

=-

3 Re ðEA - BF  Þ I0

ð3:730Þ

K zy z = 0 0

ð3:728Þ

3 ImðBA - EF  Þ I0

0 0

K zz z =

3 Re ðBA þ EF  Þ I0

ð3:727Þ

=-

K zx z = 0 0

=

ð3:726Þ

- y0 y0

0 0

K xy x

- y0 y0

ð3:725Þ

1 Re ðAF  þ EB -2CD Þ I0

ð3:731Þ

1 ImðAF  þ EB -2CD Þ I0

ð3:732Þ

h i 1 jAj2 - jF j2 þ jE j2 - jBj2 þ 2jDj2 -2jC j2 2I 0

ð3:733Þ

In the case where both the target and the residual nucleus are polarized, and * continuing to choose the y-axis along to direction n , that is, φ = 0, the footnote of ^ in Eq. (3.689) is marked by M 0I M I , and the footnote of the the matrix element of F ^ þ in Eq. (3.689) is marked by matrix element of the complex conjugate matrix F 0 M I M I . Also, we make the following changes to the previous results: jX j2 → X M I 0 M I X M I 0 M I ,

X = A, B, C, D, E, F

ð3:734Þ

Given this, and according to Eq. (3.688), we can write I 0,

0

0

MI MI MI MI

=

 1  A 0 A 0 þ BM I 0 M I BM I 0 M I þ CM I 0 M I C M I 0 M I 2 MI MI MI MI 

þDM I 0 M I DM I 0 M I þ E M I 0 M I E M I 0 M I þ F M I 0 M I F M I 0 M I

ð3:735Þ

By means of the form of the polarization wave functions of the target and the residual nucleus introduced by Eqs. (2.315) and (2.320), and continuing to use the simplification summation symbol Σ for the polarized target and residual nucleus introduced by Eq. (2.327), the angular distribution of the emitted particles corresponding to the unpolarized incident particles can be expressed as

*

3.10

1 2

Polarization Theory for

→

199

þ A → 1 þ B Reactions

dσ 0α0 n0 , αn = ΣI 0, dΩ

0

I  I 0α0 n0 , αn =

ð3:736Þ

MI 0 MI 0 MI MI

The components of the polarization vector and polarization tensor corresponding to the unpolarized incident particles are 0 i0

1

0

 P0α0 ni 0 , αn =

P

I

0

ΣI 0,

i0 M I 0 M I 0 M I M I P0, M I 0 M I 0 M I M I ,

i 0 = ε0

ð3:737Þ

where ε0 is the Cartesian coordinate index set introduced by Eq. (3.476). When the normal variables are added with the footnote M 0I M I and the complex conjugate 0 variables are added with the footnote M I M I in Eqs. (3.693)–(3.700) and using i0 Eq. (3.734), the various P0, M 0 M 0 M M in Eq. (3.737) can be obtained. I

I

I

I

When the incident particles, targets, and residual nuclei are all polarized, the component of the nuclear reaction polarization vector analyzing power is Ai  Ai,α0 n0 ,αn =

1 I

0

ΣI 0,M I 0 M I 0 M I M I Ai,M I 0 M I 0 M I M I ,

i = x, y, z

ð3:738Þ

where Ai,M I 0 M I 0 M I M I have been given by Eqs. (3.703), (3.705), and (3.707). Referring to ^ρin given by Eq. (3.214) and ^ρout given by Eq. (2.85), then the differential cross section of the emitted particles corresponding to polarized incident particles can be written as I  I α0 n0 ,αn =

h i dσ α0 n0 ,αn 3 = trf^ρout g = I 0 1 þ px Ax þ py Ay þ pz Az 2 dΩ

ð3:739Þ

The corresponding polarization transfer coefficient is i0

1

0

K i  K ii,α0 n0 ,αn =

I

0

0

ΣI 0,M I 0 M I 0 M I M I K ii,M I 0 M I 0 M I M I ,

i = x, y, z;

i 0 = ε0

ð3:740Þ

0

where K ii,M I 0 M I 0 M I M I have been given by Eqs. (3.710)–(3.733). Referring to Eqs. (3.214) and (2.85) we can write the components of the polarization vector and polarization tensor of the outgoing particles as i0

0

P  Piα0 n0 ,αn =

0h  i I 3 0 i0 i0 i0 i0 P þ p x K x þ py K y þ pz K z , 2 I

i 0 = ε0

ð3:741Þ

Now, using Eqs. (2.334) and (2.335), as well as the simplification summation e given by Eq. (2.278), Eqs. (3.735)–(3.738), and (3.740) can be simplified, symbols Σ respectively, in the case where both the target and the residual nucleus are unpolarized, as follows:

200

3

I 0,M I 0 M I =

Polarization Theory of Nuclear Reactions for Spin 1 Particles

 2  2  2  2  2  2  1  AM I 0 M I  þ BM I 0 M I  þ C M I 0 M I  þDM I 0 M I  þ EM I 0 M I  þ F M I 0 M I  2 ð3:742Þ e 0,M 0 M I = ΣI I I 0

P

0 i0

Ai = 0

K ii =

=

ð3:743Þ

1e ΣI 0,M I 0 M I Pi00,M I 0 M I , 0

I

1e ΣI 0,M I 0 M I Ai,M I 0 M I , 0 I

0 1e ΣI 0,M I 0 M I K ii,M I 0 M I , 0 I

i 0 = ε0

ð3:744Þ

i = x, y, z

ð3:745Þ

i0 = ε0

i = x, y, z;

ð3:746Þ

Next, we will analyze Eq. (3.672) which belongs to S-L coupling scheme. Based on the properties of the C-G coefficient, there are the following relations: 0

0

0

0

0

0

I I CS1μμþM = ð-1ÞIþ2 - S C1S --μμ -I M-I M , C S1μ0μIþM = ð-1ÞI þ1 - S C S1 0 MI 0 IM 1

2

I

2

I

CJl0 MSM = ð-1ÞlþS - J C Jl0 -SM- M , C Jl0 mMl 0

0

S0 M S 0

= ð-1Þl þS

0

-J

CJl0

- μ0 - M I 0 - μ0 I 0 - M I 0

-M - ml 0 S0 - M S 0

ð3:747Þ Now we only need to study the second item of Eq. (3.553) with i = 12 and i0 = 1. Using Eqs. (2.288) and (3.747), for the part which is not associated with angle φ, the following phase factor comes out when all spin magnetic quantum numbers change signs: 0

ð-1ÞlþS - Jþl þS

0

- JþIþ12 - SþI 0 þ1 - S0 þμþM I - μ0 - M I 0

0

0

= ð-1Þ2JþlþIþl þI þ2þμþM I - μ 3

0

- MI 0

The symbol Π introduced in Eq. (3.488) can also be used here, so we have ! * 1 l + l' Π = (-1) . In þ A → 1 þ B reactions, I and I0 must be that one is an (-1) 2 integer and another is a semi-odd number. When I is a semi-odd number and I0 is an integer, J is an integer. When I is an integer and I0 is a semi-odd number, J is a semiodd number. Letting  Λ=

0 1

when I is a semi‐odd number and I 0 is an integer when I is an integer and I 0 is a semi‐odd number

ð3:748Þ

then one can use (-1)Λ to replace (-1)2J in this case. Note that the definition of Λ given by Eq. (3.748) and the definition of Λ given by Eq. (3.635) are not the same; thus, we can get

*

3.10

Polarization Theory for

f α 0 n 0 μ0 M I 0 ,

1 2

→

201

þ A → 1 þ B Reactions

αnμM I ðθ Þ = ð-1Þ

ΠþΛþIþI 0 þ32þμþM I - μ0 - M I 0

f α0 n0 - μ0 - M I 0 , αn - μ - M I ðθÞ ð3:749Þ

Now, we can obtain the following relations from Eqs. (3.553) and (3.673)– (3.678): AM I 0 M I ðθÞ = f α0 n0 1M I 0 ,

αn12M I ðθ Þ,

C M I 0 M I ðθÞ = if α0 n0 0M I 0 ,

BM I 0 M I ðθÞ = f α0 n0 -1M I 0 , αn - 12M I ðθÞ,

αn12M I ðθ Þ,

E M I 0 M I ðθÞ = - f α0 n0 -1M I 0 ,

DM I 0 M I ðθÞ = - if α0 n0 0M I 0 ,

αn12M I ðθ Þ,

αn - 12M I ðθ Þ,

F M I 0 M I ðθÞ = - f α0 n0 1M I 0 , αn - 12M I ðθÞ ð3:750Þ

According to Eqs. (3.749) and (3.750), we have 0

0

0

0

0

0

BM I 0 M I ðθÞ = ð-1ÞΠþΛþIþI þM I - M I A - M I 0 DM I 0 M I ðθÞ = ð-1ÞΠþΛþIþI þM I - M I C - M I 0 F M I 0 M I ðθÞ = ð-1ÞΠþΛþIþI þM I - M I E - M I 0

- M I ðθ Þ

ð3:751Þ

- M I ðθ Þ

ð3:752Þ

- M I ðθ Þ

ð3:753Þ

One can see that Eqs. (3.641) and (3.642) are still applicable in this section by comparing Eqs. (3.751)–(3.753) with Eqs. (3.638)–(3.640). Letting       eI 0,M 0 M ðθÞ = AM 0 M 2 þ CM 0 M 2 þ E M 0 M 2 I I I I I I I I

ð3:754Þ

then according to Eq. (3.641) the following relation can be obtained from Eqs. (3.742) and (3.743): e eI 0,M 0 M I =Σ I I 0

ð3:755Þ

Given this, the following results can be obtained from Eq. (3.744), according to Eqs. (3.693)–(3.700), and using Eqs. (3.751)–(3.753), (3.641), and (3.642): 0 x0

=P

0 z0

0 x0 y0

0 y0 z0

=P =P =0 pffiffiffi 2e 0 y0 P =- 0 Σ Re ðCA þ DF  Þ I 3 e 0 x0 z 0 P = pffiffiffi 0 ΣImðCA þ DF  Þ 2I P

ð3:756Þ ð3:757Þ ð3:758Þ

202

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

0 x0 x0 - y0 y0

P

0 z0 z0

P

=

=-

6e Σ Re ðEA Þ 0 I

 1 e 2 Σ jAj þ jEj2 - 2jC j2 0 I

ð3:759Þ ð3:760Þ

Further, according to Eqs. (3.703), (3.705), and (3.707), we can obtain the following results by Eq. (3.745): Ax = -

1e Σ Re ð2AF  þ CD Þ 0 I Ay = Az = 0

ð3:761Þ ð3:762Þ

It follows that from Eq. (3.746) we can obtain the following results according to Eqs. (3.710)–(3.733): i0

K i = 0,



   i = x and i0 = ε02 or i = y, z and i0 = ε01

ð3:763Þ

where the subscript ε1 and ε2 are given by Eq. (3.320). Further we can obtain pffiffiffi 2e Σ Re ðDA - CF  Þ 0 I pffiffiffi 2e x0 K z = 0 ΣIm ðCA - DF  Þ I pffiffiffi 0 2e y Kx = 0 Σ Re ðDA þ CF  Þ I 2e z0 K y = - 0 ΣIm ðAF  Þ I  1 e 2 z0 Kz = 0 Σ jAj - jF j2 I 3 e x0 y0 K y = 0 Σ Re ðBA - EF  Þ 2I 0 0 3e xy K z = - 0 ΣIm ðEA Þ I 3 e x0 z 0 K x = - pffiffiffi 0 ΣImðDA þ CF  Þ 2I x0

Ky = -

ð3:764Þ ð3:765Þ ð3:766Þ ð3:767Þ ð3:768Þ ð3:769Þ ð3:770Þ ð3:771Þ

*

3.10

Polarization Theory for

1 2

→

203

þ A → 1 þ B Reactions

3 e y0 z0 K y = - pffiffiffi 0 ΣIm ðDA - CF  Þ 2I

ð3:772Þ

3 e y0 z0 K z = - pffiffiffi 0 Σ Re ðCA - DF  Þ 2I

ð3:773Þ

x0 x0 - y0 y0

3e Σ Re ðBA þ EF  Þ 0 I 2e z0 z0 Kx = - 0 Σ Re ðAF  - CD Þ I

Kx

=

ð3:774Þ ð3:775Þ

Now from the results obtained above, it can be seen that 0i0



P

 Ai i0

Ki



≠0 =0

≠0

i 0 = ε1 0

=0

i 0 = ε2 0

≠0

i=x

=0

i = y, z

ð3:776Þ ð3:777Þ

ði = x and i0 = ε1 0 Þ or ði = y, z and i0 = ε2 0 Þ ði = x and i0 = ε2 0 Þ or ði = y, z and i0 = ε1 0 Þ

ð3:778Þ

And we can obtain the following results from Eqs. (3.739) and (3.741): 

3 1 þ px A x 2 0   I 3 i0 0 i0 i0 P þ px K x , P = 2 I 0   0 I 3 i0 i0 P = py K y þ pz K iz , I 2 I =I

0



ð3:779Þ i 0 = ε1 0

ð3:780Þ

i 0 = ε2 0

ð3:781Þ

References [23–29] gave the experiment data of the polarization analyzing powers for the (p, d) reaction with the targets 12C,13C,28Si, 24Mg, 208Pb, separately. The transition matrix element T can be calculated by using the DWBA method, and it can be converted into the S matrix element. It follows that the theoretical analyses of the corresponding polarization physical quantities can be carried out using the methods described in this section.

204

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles →

3.11

Polarization Theory for

1 2

→

þ 1 Reactions

In this section we will study the polarization theory of the reactions between * * polarized spin 12 particles and polarized spin 1 particles, and take N þ d elastic scattering as an example, where N may be neutron or proton. In order to simplify this * problem, people generally only study either the N þ d reactions of polarized nucleon → and unpolarized deuteron [30–35], or the d þ N reactions of polarized deuteron and unpolarized nucleon [32–35] . With the theory described in this book, these two cases can be treated together. We know that for the few-body reactions, the angular distribution anisotropy is obvious even in the keV energy region, and its polarization effect should not be ignored. In particular, some people are currently studying whether fusion reactions using polarized charged particles as fuel are likely to increase the fusion reaction rate ! * * * [6, 36–40], and there are also people studying d þ t → n þ α and d þ 3 He → p þ α * * reactions [2, 39, 40], both of which belong to 1 þ 12 reaction, where the spin of the emitted particles is 0 or 12, and their reaction amplitude constitutes a 2 × 6 matrix. The elastic scattering of the polarized nucleons (deuterons) and deuterons (nucleons) that are the secondary particles of the radioactive nuclear beam, or both * → as the secondary particles in the device, belongs to 12 þ 1 reaction. Of course, the polarization experiments can also be done directly using already polarized nucleons and polarized deuterons. The N + d elastic scattering can be studied by using the phenomenological phase shift analysis method to obtain the corresponding S matrix element, or using the microscopic three-body theory [41–45]; in this way the corresponding T matrix elements are first obtained, and then the corresponding S matrix elements are obtained from the T matrix elements. For the nucleon-deuteron interactions, the S-L angular momentum coupling mode is used. The reaction amplitude of the N - d elastic scattering can be written by Eq. (2.252) as f μ0 ν0 ,μν ðΩÞ = f C ðθÞδμ0 ν0 ,μν þ

pffiffiffi  i π X ^ iðσl þσl0 Þ  δl0 S0 ,lS - SJl0 S0 ,lS le k 00 lSl S J

S0 M 0 C Jl0 MSM CS1μM1ν CJl0 mM0 S0 M 0 C1μ0 1νS 0 Yl0 m0l ðθ, S l 2 2

One can see that

φÞ

ð3:782Þ

→

3.11

Polarization Theory for

1 2

→

205

þ 1 Reactions

ml 0 = μ þ ν - μ0 - ν0

ð3:783Þ

where μ and μ0, ν and ν0 represent the spin magnetic quantum numbers of the incident nucleon N and the target deuteron d in incident channel and the outgoing channel, respectively. If N is a proton, the first item on the right side of Eq. (3.782) representing the Coulomb interaction cannot be ignored. Now, introduce the following symbols 0

^^ 0 iðσl þσl0 Þ

f μll0 νJ0 ,μν = l l e

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðl0 - ml 0 Þ! X J M S M J M C C1 C 0 0 ðl0 þ ml 0 Þ! SS0 l0 SM 2μ 1ν l ml

  S0 M S 0 J 0 0 C δ S 0 0 1 0 0 l S ,lS l S ,lS S MS μ 1ν 0

0

2

ð3:784Þ Using Eq. (2.253), one can obtain the following from Eq. (3.782): f μ0 ν0 ,μν ðΩÞ = F μ0 ν0 ,μν ðθÞ eiðμþν - μ F μ0 ν0 ,μν ðθÞ = f C ðθÞ δμ0 ν0 ,μν þ

0

- ν0 Þφ

ð3:785Þ

0 0X 0 i - μ0 - ν0 f μll0 νJ0 ,μν Pμþν ð cos θÞ ð-1Þμþν - μ - ν l0 2k 0

ll J

ð3:786Þ Considering parity conservation, l and l0 must simultaneously be even or odd, and because 2J is odd, the following C-G coefficient relations can be obtained: S μþν J μþν C Jl0 μþν S μþν C 1μ 1ν C l0 μþν - μ0 - ν0 2

0

S0 μ0 þν0

= C Jl0 -Sμ -- νμ - ν CS1 --μμ -1ν - ν C Jl0 2

0

C1Sμ0 μ1νþν0

0

2

-μ-ν - μ - νþμ0 þν0

0

S0 - μ0 - ν0

C 1S --μμ0

0

2

- ν0 1 - ν0

ð3:787Þ

Noting the relation expression (2.288), we can obtain the following from Eqs. (3.782)–(3.787): F μ0 ν0 ,μν ðθÞ = ð-1Þμþν - μ

0

- ν0

F - μ0 *

- ν0 , - μ - ν ðθ Þ

ð3:788Þ

Next, take the y-axis along the direction n , so that φ = 0, and introduce the following matrix element symbols according to Eq. (3.788):

206

3

AðθÞ = F 121, 121 ðθÞ = F - 12

-1, - 12 -1 ðθ Þ,

C ðθÞ = - F 121, 12 -1 ðθÞ = - F - 12 E ðθÞ = F 121, - 12 0 ðθÞ = F - 12 GðθÞ = F 12 0, - 12

-1, - 12 1 ðθ Þ,

-1, 12 1 ðθ Þ =

F ðθÞ = iF 121, - 12

0, 12 -1 ðθÞ,

-1 ðθ Þ =

0, 12 0 ðθ Þ,

K ðθÞ = iF 12 0, 12 -1 ðθÞ = - iF - 12

0, - 12 1 ðθÞ,

- F - 12

OðθÞ = - iF 12

-1, - 12 0 ðθÞ =

- F - 12

0, - 12 -1 ðθÞ,

-1, 12 0 ðθÞ = iF - 12 1, - 12 0 ðθ Þ,

RðθÞ = iF 12 -1, - 12 1 ðθÞ = - iF - 12

1, 12 0 ðθ Þ,

-1, 12 1 ðθ Þ,

H ðθÞ = - iF 12 0, - 12 0 ðθÞ = iF - 12

M ðθÞ = - iF 12 0, 12 1 ðθÞ = iF - 12

1, - 12 -1 ðθ Þ,

-1, 12 -1 ðθ Þ,

- iF - 12

0, - 12 0 ðθ Þ,

-1, 12 -1 ðθ Þ = F - 12 1, - 12 1 ðθÞ,

SðθÞ = - F 12

-1, - 12 0 ðθ Þ,

DðθÞ = - iF 121, - 121 ðθÞ = iF - 12

-1 ðθ Þ = F - 12 0, 12 1 ðθ Þ,

LðθÞ = F 12 0, 12 0 ðθÞ = F - 12

QðθÞ = F 12

BðθÞ = iF 121, 120 ðθÞ = - iF - 12

-1, 12 0 ðθ Þ,

J ðθÞ = - F 12 0, - 12 1 ðθÞ = - F - 12

N ðθÞ = - F 12

Polarization Theory of Nuclear Reactions for Spin 1 Particles

T ðθÞ = - iF 12

1, 12 -1 ðθ Þ,

-1, - 12 -1 ðθ Þ = iF - 12 1, 12 1 ðθ Þ

ð3:789Þ Then the 6 × 6 reaction matrix given by Eq. (3.786) can be expressed as 0 B B B B B B ^ F=B B B B B @ 0 B B B B B þ B ^ F =B B B B B @

A iM -N - iT G iF A iB - C - iD E iF 

- iB

-C

iD

E

L

- iK

-J

iH

- iF

1

C G C C C iO Q - iR - S iT C C C, -S iR Q - iO - N C C C - iH - J iK L - iM C A E - iD - C iB A 1 - iM  - N  iT  G - iF  C L - iO - S iH  E C C C      C iK Q - iR -J iD C C - J iR Q - iK  - C C C C     C - iH -S iO L - iB A G

- iT 

- N

ð3:790Þ

A

iM 

Now introduce the following 3 × 3 submatrices: 0

F μ0 1, μ1

^t μ0 μ = B @ F μ0 0, μ1 F μ0 -1, μ1

F μ0 1, μ0

F μ0 1, μ

-1

F μ0 0, μ0

F μ0 0, μ

-1

F μ0

-1, μ0

F μ0

-1, μ -1

1 C A

ð3:791Þ

→

3.11

Polarization Theory for

1 2

→

207

þ 1 Reactions

1 For the sake of simplicity, the footnote is represented by the footnote +, and the 2 1 footnote - is represented by the footnote -, so that the following expressions can 2 be written according to Eq. (3.790) as follows: 0

A

B ^t þþ = B @ iM 0

- iB

-C

0

A

- iM 

C þ B  B - iK C A, ^t þþ = @ iB

L

-N

iO

Q

iD

E

- iF

B ^t þ - = B @ -J - iR 0 - iT B ^t - þ = B @ G iF 0 Q B ^t - - = B @ iK -C

1

iH -S -S - iH E - iO L iB

0

1

L

- N

C - iO C A

- C

iK 

Q

- iD

- J

iR

B C þ  B G C - iH  A, ^t þ - = @ E iT iF  G 1 0 G iR iT  C þ B  B -J C iH  A, ^t - þ = @ - S - iR - J  - iD 1 0  -N Q - iK  C þ B  B - iM C L A, ^t - - = @ iO A - N iM 

1

1

C - S C A - iT  1 ð3:792Þ - iF  C E C A  iD 1 - C C - iB C A  A

þ

^ and F ^ given by Eq. (3.790) can be expressed as and the matrix F 

^t þþ ^t þ ^= F ^t - þ ^t - -

,

þ

^ = F

^t þ ^t þ- þ þþ

^t þ þ-

!

^t þ- -

ð3:793Þ

The polarization density matrix of the normalized incident channel can be written according to Eqs. (2.64) and (3.214) as follows: ^ρin =

 h   1 ^ 3 d^ N N d^ d^ ^ I S S S I 2 þ pN σ ^ þ p σ ^ þ p σ ^ þ þ p þ p p 3 x x y y z z y y z z 6 2 x x   i 2 ^ xy þ pd Q ^ xz þ pd Q ^ yz þ 1 pd ^ xx - yy þ 1 pd Q ^ zz þ pdxy Q Q xz yz 6 xx - yy 3 2 zz

ð3:794Þ

d where ^I 2 and ^I 3 are the unit 2 × 2 and 3 × 3 matrices. pN i and pi ði = x, y, zÞ are the vector polarization rates of the incident nucleons and deuterons, respectively, and pdi ði = xy, xz, yz, xx - yy, zzÞ are the tensor polarization rates of the incident deuterons. Now, let ^02 and ^03 be the zero 2 × 2 and 3 × 3 matrices, respectively. According to the definition of the matrix direct product given by Eq. (2.349), we have

208

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

σ i,þþ^I 3 σ i, - þ^I 3

^ i  σ^i × ^I 3 = Σ

^ i  ^I 2 × ^Si = Γ ^i = ^ i  ^I 2 × Q Ω

! ^ i ^03 Q , ^03 Q ^i

^Si ^03

! σ i,þ - ^I 3 , i = x, y, z σ i, - - ^I 3 ! ^03 , i = x, y, z ^Si

i = xy, xz, yz, xx - yy, zz

ð3:795Þ

ð3:796Þ

ð3:797Þ

Also, setting ^I 6 to be the unit 6 × 6 matrix, Eq. (3.794) can be rewritten as ^ρin =

    1 ^ N^ N^ ^I 6 þ 3 pdx Γ ^ ^ x þ pdy Γ ^ y þ pdz Γ ^z Σ Σ Σ I 6 þ pN þ p þ p x y z x y z 6 2  ð3:798Þ   2 ^ xy þ pd Ω ^ xz þ pd Ω ^ yz þ 1 pd ^ xx - yy þ 1 pd Ω ^ zz þ pdxy Ω Ω xz yz 2 zz 6 xx - yy 3

From Eqs. (3.790) and (3.798), when N and d in the incident channel are both unpolarized, we can obtain I0 

n þo  dσ 0 1 ^F ^ = 1 jAj2 þ jBj2 þ jCj2 þ jDj2 þ jE j2 þ jF j2 þ jGj2 þ jH j2 = tr F 3 dΩ 6  þjJ j2 þ jK j2 þ jLj2 þ jM j2 þ jN j2 þ jOj2 þjQj2 þ jRj2 þ jSj2 þ jT j2 ð3:799Þ

The polarization vector component of the outgoing nucleon N corresponding to the unpolarized incident channel is 0

i PN 0 =

^F ^ þg ^ i0 F trfΣ , ^F ^ þg trfF

i0 = x0 , y0 , z0

ð3:800Þ

The corresponding polarization vector component of the outgoing d is n o ^F ^þ ^ i0 F tr Γ 0 Pd0 i = n þ o , ^F ^ tr F

i 0 = x0 , y0 , z 0

ð3:801Þ

The corresponding polarization tensor component of the outgoing d is 0

Pd0 i =

þ

^ i0 F ^F ^ g trf Ω , þ ^F ^ g trfF

From Eq. (3.793) we can obtain

i0 = x0 y0 ,x0 z0 ,y0 z0 ,x0 x0 - y0 y0 ,z0 z0

ð3:802Þ

→

3.11

Polarization Theory for þ

^F ^ = F

1 2

→

209

þ 1 Reactions

^ ^þ ^t þþ^t þ þþ þ t þ - t þ -

^t þþ^t þ- þ þ ^t þ - ^t þ- -

^ ^t - þ^t þ ^þ þþ þ t - - t þ -

^t - þ^t þ- þ þ ^t - - ^t þ- -

! 

^ gþþ ^ gþ -

!

^ g-þ ^ g- ð3:803Þ

Using Eq. (3.792), we can further obtain 0

A -iB -C

10

A -iM  -N 

1 0

iD

E -iF

10

-iD -J 

iR

1

CB C B C CB B CB C B C CB B ^ gþþ = B iM L -iK CB iB L -iO C þ B -J iH G CB E  -iH  -S C A@ A @ A A@ @       -iR -S iT -C iK Q iF G -iT -N iO Q 0 2 2 2 2 2 2 -iðAM  þBL þCK  þDJ  þEH  þFG Þ jAj þ jBj þ jC j þ jDj þ jE j þ jF j B B ¼B iðMA þLB þKC  þJD þHE  þGF  Þ jM j2 þ jLj2 þ jK j2 þ jJ j2 þ jH j2 þ jGj2 @

- ðNA þOB þQC  þRD þSE þTF  Þ iðNM  þOL þQK  þRJ  þSH  þTG Þ 1 - ðAN  þBO þCQ þDR þES þFT  Þ C -iðMN  þLO þKQ þJR þHS þGT  Þ C A 2 2 2 2 2 2 jN j þ jOj þ jQj þ jRj þ jSj þ jT j

ð3:804Þ 1 iT  G -iF Q -iK  -C  A -iB -C iD E -iF B CB C C B CB     C B  C CB CB ^ gþ- =B @ iM L -iK A@ -S iH E Aþ@ -J iH G A@ iO L -iB A -N iO Q -iR -S iT -iR -J  iD -N  iM  A 0

10

1 0 

10

0

iðAT  þBS þCR þDQ þEO þFN  Þ AG þBH  þCJ  þDK  þEL þFM  B B = B - ðMT  þLS þKR þJQ þHO þGN  Þ iðMG þLH  þKJ  þJK  þHL þGM  Þ @ -iðNT  þOS þQR þRQ þSO þTN  Þ - ðNG þOH  þQJ  þRK  þSL þTM  Þ 1 -iðAF  þBE  þCD þDC  þEB þFA Þ C MF  þLE  þKD þJC  þHB þGA C A       i ðNF þOE þQD þRC þSB þTA Þ

ð3:805Þ

210

3

0

-iT -S

iR

10

Polarization Theory of Nuclear Reactions for Spin 1 Particles

A -iM  -N 

1 0

Q -iO -N

10

-iD -J 

iR

1

CB  C C B CB  B  C  CB B CB ^ g -þ = B L -iO C A þ @ iK L -iM A@ E -iH -S A @ G -iH -J A@ iB -C iB A -C iK  Q iF  G -iT  iF E -iD 0 -iðTA þSB þRC  þQD þOE  þNF  Þ - ðTM  þSL þRK  þQJ  þOH  þNG Þ B B = B GA þHB þJC  þKD þLE  þMF  -iðGM  þHL þJK  þKJ  þLH  þMG Þ @

iðFA þEB þDC  þCD þBE  þAF  Þ 1 iðTN  þSO þRQ þQR þOS þNT  Þ C - ðGN  þHO þJQ þKR þLS þMT  Þ C A -iðFN  þEO þDQ þCR þBS þAT  Þ 0

-iT -S

iR

10

iT 

G -iF

FM  þEL þDK  þCJ  þBH  þAG

1 0 

Q -iO -N

10

ð3:806Þ Q -iK  -C 

1

C B CB C B CB C B CB C B CB ^g -- = B G -iH -J CB -S iH  E  C þ B iK L -iM CB iO L -iB C A @ A@ A @ A@ iF E -iD -C iB A -iR -J  iD -N  iM  A 0 jT j2 þ jSj2 þ jRj2 þ jQj2 þ jOj2 þ jN j2 -iðTG þSH  þRJ  þQK  þOL þNM  Þ B 2 2 2 2 2 2       =B @ iðGT þHS þJR þKQ þLO þMN Þ jGj þ jH j þ jJ j þ jK j þ jLj þ jM j - ðFT  þES þDR þCQ þBO þAN  Þ iðFG þEH  þDJ  þCK  þBL þAM  Þ 1 - ðTF  þSE  þRD þQC  þOB þNA Þ C -iðGF  þHE  þJD þKC  þLB þMA Þ C A jF j2 þ jE j2 þ jDj2 þ jC j2 þ jBj2 þ jAj2

ð3:807Þ Next, we can obtain first the following expression from Eqs. (2.6), (3.795), and (3.803): þ

^ xF ^F ^ = Σ

^03 ^I 3

^I 3 ^03

!

^gþþ ^gþ ^g - þ ^g - -



 =

^ g-þ ^ g- ^ gþþ ^ gþ -

ð3:808Þ

Then we can obtain the following result from Eqs. (3.800), (3.805), and (3.806): 0

x PN =0 0

Using the same method we can obtain

ð3:809Þ

→

3.11

Polarization Theory for þ

^ yF ^F ^ = Σ

^ 03 - i^I 3 i^I 3 ^03

1 2

→

211

þ 1 Reactions

!

^gþþ g^þ g^ - þ ^g - -



 = -i

^ ^- g-þ g ^þþ - g ^þ -g

ð3:810Þ

2 Re ðAT  þ BS þ CR þ DQ þ EO þ FN  þ GM  þ HL þ JK  Þ 3I 0 ð3:811Þ !  ^ ^ ^ ^gþþ ^gþ gþþ ^ gþ ^F ^ þ = I 3 03 ^ zF = ð3:812Þ Σ ^03 - ^I 3 ^g - þ ^g - -^ g-þ - ^ g- -

0

y PN =0

0

z PN =0 0

ð3:813Þ

0

Now in order to calculate Pd i ði0 = x0 , y0 , z0 Þ, first we find the following expressions from Eqs. (3.33), (3.796), (3.804), and (3.807):

1 tr S^x ^gþþ = pffiffiffi ½iðMA þ LB þ KC  þ JD þ HE  þ GF  Þ 2 - iðAM  þ BL þ CK  þ DJ  þ EH  þ FG Þ þiðNM  þ OL þ QK  þ RJ  þ SH  þ TG Þ       pffiffiffi iðMN þ LO þ KQ þ JR þ HS þ GT Þ = 2ImðAM þ BL þ CK þ DJ þ EH þ FG þGT  þ HS þ JR þ KQ þ LO þ MN  Þ

ð3:814Þ



1 tr ^Sx g^ - - = pffiffiffi ½iðGT  þ HS þ JR þ KQ þ LO þ MN  Þ 2 - iðTG þ SH  þ RJ  þ QK  þ OL þ NM  Þ þiðFG þ EH  þ DJ  þ CK  þ BL þ AM  Þ -piðffiffiffiGF  þ HE þ JD þ KC  þ LB þ MA Þ = - 2ImðAM  þ BL þ CK  þ DJ  þ EH  þ FG þGT  þ HS þ JR þ KQ þ LO þ MN  Þ

ð3:815Þ

Then from Eq. (3.801) we find 0

Pd0 x = 0

ð3:816Þ

Using the same method we can obtain 0

Pd0 y =

pffiffiffi

2 1 ^ tr Sy ^ Re ðAM  þ BL þ CK  þ DJ  gþþ þ ^Sy ^g - - = 3I 0 6I 0 þEH  þ FG þ GT  þ HS þ JR þ KQ þ LO þ MN  Þ 0

Pd0 z =

1 ^ tr Sz ^gþþ þ ^Sz ^g - - = 0 6I 0

ð3:817Þ ð3:818Þ

212

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

And using the same method we can obtain the following results according to Eqs. (3.36) and (3.797):

1 ^ ^ x0 y0 ^ tr Qx0 y0 ^gþþ þ Q g- - = 0 6I 0

0 0 1 ^ ^ x0 z0 ^g - tr Qx0 z0 ^gþþ þ Q Pd0 x z = 6I 0 1 = pffiffiffi Im½ðAM  þ BL þ CK  þ DJ  þ EH  þ FG Þ 2I 0 - ðGT  þ HS þ JR þ KQ þ LO þ MN  Þ 0 0

Pd0 x y =

ð3:819Þ

ð3:820Þ

1 ^ ^ y0 z0 ^ g- - = 0 tr Qy0 z0 ^gþþ þ Q ð3:821Þ 6I 0

0 0 0 0 1 ^ ^ x0 x0 - y0 y0 ^g - tr Qx0 x0 - y0 y0 ^gþþ þ Q Pd0 x x - y y = 6I 0 ð3:822Þ 2 Re ðAN  þ BO þ CQ þ DR þ ES þ FT  Þ =3I 0 h 0 0 1 Pd0 z z = jAj2 þ jBj2 þ jC j2 þ jDj2 þ jEj2 þ jF j2 þ jN j2 þ jOj2 þ jQj2 3I 0  i þjRj2 þ jSj2 þ jT j2 -2 jGj2 þ jH j2 þ jJ j2 þ jK j2 þ jLj2 þ jM j2 0 0

Pd0 y z =

ð3:823Þ Next, introduce a simplification symbol of the coordinate index set: γ  xy, xz, yz, xx - yy, zz

ð3:824Þ

Given this, the following polarization analyzing powers can be defined for → * * → N d , d N reaction: n o ^þ ^Σ ^ iF tr F n þo , AN i = ^F ^ tr F

i = x, y, z

n o ^þ ^Γ ^ iF tr F Adi = n þ o , i = x, y, z ^F ^ tr F n o ^ iF ^Ω ^þ tr F = n þo , i=γ ^F ^ tr F

ð3:825Þ

ð3:826Þ

→

3.11

Polarization Theory for

1 2

→

213

þ 1 Reactions

n o ^Σ ^ iΓ ^ jF ^þ tr F d n þo , AN i,j = ^F ^ tr F n o ^ jF ^þ ^Σ ^ iΩ tr F n þo , = ^F ^ tr F

i,j = x, y, z ð3:827Þ i = x, y, z;

j=γ

where ANi,j d can also be called a correlation analyzing power. The following polarization transfer coefficients can also be defined: n o ^Σ ^ iF ^þ ^ i0 F tr Σ i0 n þo , KN Ni = ^F ^ tr F n o ^Σ ^ iF ^þ ^ i0 F tr Γ n þo , K dN ii = ^F ^ tr F n o ^ i0 F ^Σ ^ iF ^þ tr Ω n þo , = ^F ^ tr F 0

n o ^Γ ^ iF ^þ ^ i0 F tr Σ i0 n þo , KN di = ^F ^ tr F n o ^ iF ^Ω ^þ ^ i0 F tr Σ n þo , = ^F ^ tr F 0

K dd ii =

=

=

=

i = x, y, z;

i = x, y, z;

i 0 = x0 , y0 , z 0

ð3:828Þ

i0 = x0 , y0 , z0 ð3:829Þ

i = x, y, z;

i = x, y, z;

i0 = γ 0

i 0 = x0 , y0 , z 0 ð3:830Þ

i = γ;

i0 = x0 , y0 , z0

n o ^Γ ^ iF ^þ ^ i0 F tr Γ n þ o , i = x, y, z; i0 = x0 , y0 , z0 ^ ^ tr F F n o ^ i0 F ^Γ ^ iF ^þ tr Ω n þ o , i = x, y, z; i0 = γ 0 ^ ^ tr F F n o ^ iF ^ i0 F ^Ω ^þ tr Γ n þ o , i = γ; i 0 = x 0 , y0 , z 0 ^F ^ tr F n o ^ iF ^ i0 F ^Ω ^þ tr Ω n þ o , i = γ; i0 = γ 0 ^F ^ tr F

ð3:831Þ

214

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

Now the polarization transfer coefficients defined by the above four formulas are all introduced assuming that only one particle in the incident channel is polarized. If both particles in the incident channel are polarized, then a correlation polarization transfer coefficient needs to be introduced. For the outgoing particles, the correlation problem need not be considered. If it is not necessary to study the polarization rate of the outgoing particles corresponding to the polarized incident particles, then there is no need to find the polarization transfer coefficient; therefore, this section will not deduce the polarization transfer coefficient. It follows that when deriving the polarization analyzing power, it will no longer be necessary to prepare the matrices for deriving the polarization transfer coefficients. Instead, we first obtain the following relation from Eqs. (2.6), (3.795), and (3.793): þ

^ = ^ xF Σ

^03 ^I 3 ^I 3 ^03

!

^t þ ^þ þþ t - þ

^t þ ^þ þ- t- -

! =

^t þ ^þ þ- t - -

! ð3:832Þ

^t þ ^þ þþ t - þ

Now we can obtain the following result from Eqs. (3.792), (3.793), and (3.825): n o   1 ^ þ = 1 tr ^t þþ^t þ ^Σ ^ xF ^ ^þ ^ ^þ ^ ^þ tr F þ - þ t þ - t þþ þ t - þ t - - þ t - - t - þ 6I 0 6I 0 1 ½ - iðAD þ BE  þ CF  Þ - iðMJ  þ LH  þ KG Þ - iðNR þ OS þ QT  Þ = 6I 0 þiðDA þ EB þ FC  Þ þ iðJM  þ HL þ GK  Þ þ iðRN  þ SO þ TQ Þ

AN x =

- iðTQ þ SO þ RN  Þ - iðGK  þ HL þ JM  Þ - iðFC  þ EB þ DA Þ þiðQT  þ OS þ NR Þ þ iðKG þ LH  þ MJ  Þ þ iðCF  þ BE  þ AD Þ = 0 ð3:833Þ Using the same method we can obtain þ

^ yF ^ = Σ

^03 - i^I 3 i^I 3 ^03

!

^t þ ^þ þþ t - þ

^t þ ^þ þ- t- -

! = -i

^t þ þ-

þ

^t þ- þ

- ^t þþ - ^t - þ

! ð3:834Þ

Next, referring to the results given by Eq. (3.833), we can obtain n o   1 ^ þ = - i tr ^t þþ^t þ ^Σ ^ yF ^ ^þ ^ ^þ ^ ^þ tr F þ - - t þ - t þþ þ t - þ t - - - t - - t - þ 6I 0 6I 0 2 Re ðAD þ BE  þ CF  þ MJ  þ LH  þ KG þ NR þ OS þ QT  Þ =3I 0 ð3:835Þ

AN y =

→

3.11

Polarization Theory for

1 2

→

215

þ 1 Reactions

We can also obtain þ

^ zF ^ = Σ

^I 3 ^03 ^03 - ^I 3

!

^t þ ^þ þþ t - þ

!

^t þ þþ

=

^t þ ^þ þ- t- -

þ

!

^t þ- þ

ð3:836Þ

þ

- ^t þ - - ^t - -

Now from Eq. (3.792) we can directly see AN z =

n o

1 þ þ þ ^ þ = 1 tr ^t þþ^t þ ^Σ ^ zF tr F - ^t þ - ^t þ - þ ^t - þ^t - þ - ^t - - ^t - - = 0 þþ 6I 0 6I 0 ð3:837Þ

and we can write the following expression according to Eqs. (3.803) and (3.796): n o

^ ^þ ^ þ = tr ^t þþ ^Si^t þ ^Γ ^iF ^ ^ ^þ ^ ^ ^þ ^ tr F þþ þ t þ - Si t þ - þ t - þ Si t - þ þ t - - Si t - - , i = x, y, z ð3:838Þ Also, we can obtain the following expressions from Eqs. (3.33) and (3.792): 0

iB

B ^Sx^t þ = p1ffiffiffi B A - C  þþ 2@ iB 0

- i ðM  - K  Þ

E

- S

B ^Sx^t þ = p1ffiffiffi B iðT  - R Þ -þ 2@ - S 0

iO

B ^Sx^t þ = p1ffiffiffi B Q - N  -2@ iO

1

C - ðN  - Q  Þ C A, - iO

L

B ^Sx^t þ = p1ffiffiffi B - iðD - F  Þ þ2@ E 0

- iO

L

- iH 

- S

- iH 

- S

1

C - ðJ  - G Þ iðR - T  Þ C A, iH 

G - J 

E

C - iðF  - D Þ C A,

iH  L - iðK  - M  Þ L

ð3:839Þ

1

E - iB

1

C - ðC  - A  Þ C A - iB

216

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

n o 1 ^þ ^Γ ^ xF tr F 6I 0 1 = pffiffiffi ½iðAB - BA þ BC - CB Þ þ iðML - LM  þ LK  - KL Þ 6 2I 0 þiðNO - ON  þ OQ - QO Þ þ iðDE  - ED þ EF  - FE Þ ð3:840Þ þiðJH  - HJ  þ HG - GH  Þ þ iðRS - SR þ ST  - TS Þ

Adx =

þiðTS - ST  þ SR - RS Þ þ iðGH  - HG þ HJ  - JH  Þ þiðFE  - EF  þ ED - DE  Þ þ iðQO - OQ þ ON  - NO Þ þiðKL - LK  þ LM  - ML Þ þ iðCB - BC  þ BA - AB Þ = 0 From Eq. (3.839), a comparison of ^Sx and ^Sy in Eq. (3.33) permits us to directly write 0

- iB

- L

iO

1

C ^Sy^t þ = piffiffiffi B @ A þ C  - iðM  þ K  Þ - ðN  þ Q Þ A, þþ 2  L - iO 0 iB  1  -E iH S C ^Sy^t þ = piffiffiffi B @- iðD þ F  Þ - ðJ  þ G Þ iðR þ T  Þ A, þ2 E - iH  - S1 0    S - iH -E C     ^Sy^t þ = piffiffiffi B - iðF  þ D Þ A, @iðT þ R Þ G þ J -þ 2  - iH  E 0 -S  1  -L iB - iO C ^Sy^t þ = piffiffiffi B @ Q þ N  - iðK  þ M  Þ - ðC  þ A Þ A -2 L - iB iO

ð3:841Þ

and referring to Eq. (3.840), we can immediately write n o 1 ^Γ ^ yF ^þ tr F 6I 0 i = pffiffiffi ½ - iðAB þ BA þ BC þ CB Þ - iðML þ LM  þ LK  þ KL Þ 6 2I 0 - iðNO þ ON  þ OQ þ QO Þ - iðDE  þ ED þ EF  þ FE  Þ

Ady =

- iðJH  þ HJ  þ HG þ GH  Þ - iðRS þ SR þ ST  þ TS Þ - iðTS þ ST  þ SR þ RS Þ - iðGH  þ HG þ HJ  þ JH  Þ - iðFE þ EF  þ ED þ DE  Þ - iðQO þ OQ þ ON  þ NO Þ - iðKL þ LK  þ LM  þ ML Þ - iðCB þ BC  þ BA þ AB Þ pffiffiffi 2 = Re ðAB þ BC  þ DE  þ EF  þ GH  þ HJ  3I 0 þKL þ LM  þ NO þ OQ þ RS þ ST  Þ ð3:842Þ

→

3.11

Polarization Theory for

1 2

→

217

þ 1 Reactions

Further we can also obtain the following expressions from Eqs. (3.33) and (3.792): 0

A

B ^Sz^t þ = B 0 þþ @

- iM 

- N

0

0

1

0

C C, A

B ^Sz^t þ = B þ@

C  - iK  - Q 0  1 iT G - iF  B C ^Sz^t þ = B 0 0 0 C -þ @ A, iR J  - iD

^Sz^t þ --

- iD

- J

0

0

iR

1

C 0 C A, - iF  - G iT  0  1 Q - iK  - C  B C =B 0 0 C @ 0 A    N - iM -A ð3:843Þ

Now the following results can be obtained based on Eqs. (3.792) and (3.838): Adz =

n o 1 ^ þ = 1 ð jAj2 - jC j2 þ jM j2 - jK j2 þ jN j2 - jQj2 ^Γ ^ zF tr F 6I 0 6I 0

þjDj2 - jF j2 þ jJ j2 - jGj2 þ jRj2 - jT j2 þ jT j2 - jRj2 þ jGj2 - jJ j2 þjF j2 - jDj2 þ jQj2 - jN j2 þ jK j2 - jM j2 þ jC j2 - jAj2 Þ = 0

ð3:844Þ

And we can also write the following expression according to Eqs. (3.797) and (3.803): n o

^ i^t þ þ ^t þ - Q ^ i^t þ þ ^t - þ Q ^ i^t þ þ ^t - - Q ^ i^t þ ^ iF ^ þ = tr ^t þþ Q ^Ω tr F þþ þ-þ -- ,

i=γ ð3:845Þ

where the definition of the index set γ has been given by Eq. (3.824). Moreover the following expressions can be obtained from Eqs. (3.36) and (3.792): 0

1 0 1 C - iK  - Q - iF  - G iT  ^ xy^t þ = 3i @ 0 ^ xy^t þ = 3i @ 0 Q 0 0 A, Q 0 0 A, þþþ 2 2       iM N J iR A iD 0  1 0 1  J  - iD iR N  - iM  - A A, Q ^ xy^t þ = 3i @ 0 ^ xy^t þ = 3i @ 0 0 0 0 0 A Q --þ 2 2 iT  G - iF  Q - iK  - C  ð3:846Þ n o 1 ^ xy F ^ þ = i ðAC  - CA þ MK  - KM  þ NQ - QN  ^Ω Adxy = tr F 4I 0 6I 0 þDF  - FD þ JG - GJ  þ RT  - TR þ TR - RT  þ GJ  - JG ð3:847Þ þFD - DF  þ QN  - NQ þ KM  - MK  þ CA - AC  Þ = 0

218

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

We can also obtain 0

iB

L

- iO

1

C 3 B ^ xz^t þ ¼ p ffiffiffi B A  þ C  - i ðM  þ K  Þ - ðN  þ Q  Þ C Q þþ @ A, 2 2 - L iO - iB 0 1 - iH  - S E C 3 B   B ^ xz^t þ - ðJ  þ G Þ iðR þ T  Þ C Q þ - ¼ pffiffiffi @ - iðD þ F Þ A, 2 2 - E iH  S 0 1    iH E -S C 3 B ^ xz^t þ = p ffiffiffi B iðT  þ R Þ G þ J  - iðF  þ D Þ C Q -þ @ A, 2 2    S - iH -E 0 1   iO L - iB C 3 B ^ xz^t þ = p ffiffiffi B Q þ N  - iðK  þ M  Þ - ðC þ A Þ C Q -@ A 2 2    - iO -L iB

ð3:848Þ

and we can obtain the following result utilizing Eqs. (3.845), (3.848), and (3.792): n o 1 ^ xz F ^Ω ^þ tr F 6I 0 1 = pffiffiffi ½iðAB - BA - BC  þ CB Þ þ iðML - LM  - LK  þ KL Þ 4 2I 0 þiðNO - ON  - OQ þ QO Þ þ iðDE  - ED - EF  þ FE  Þ

Adxz =

þiðJH  - HJ  - HG þ GH  Þ þ iðRS - SR - ST  þ TS Þ þiðTS - ST  - SR þ RS Þ þ iðGH  - HG - HJ  þ JH  Þ þiðFE  - EF  - ED þ DE  Þ þ iðQO - OQ - ON  þ NO Þ þiðKL - LK  - LM  þ ML Þ þ iðCB - BC  - BA þ AB Þ 1 = - pffiffiffi ImðAB - BC  þ DE  - EF  þ GH  - HJ  2I 0 þKL - LM  þ NO - OQ þ RS - ST  Þ ð3:849Þ ^ yz in Eq. (3.36), from Eq. (3.848), we can directly write ^ xz and Q Comparing Q

→

3.11

Polarization Theory for

0

1 2

→

219

þ 1 Reactions

- iB

- L

1

iO

C 3i B ^ yz^t þ = p ffiffiffi B A - C - iðM  - K  Þ - ðN  - Q Þ C Q þþ @ A, 2 2    -L iO - iB 0 1   iH S -E C 3i B ^ yz^t þ = p ffiffiffi B - iðD - F  Þ - ðJ  - G Þ iðR - T  Þ C Q þ@ A, 2 2 - E iH  S 0 1 - iH  - E S C 3i B ^ yz^t þ = p ffiffiffi B iðT  - R Þ G - J  - iðF  - D Þ C Q -þ @ A, 2 2 S - iH  - E 0 1   -L iB - iO C 3i B ^ yz^t þ- - = p ffiffiffi B Q - N  - iðK  - M  Þ - ðC - A Þ C Q @ A 2 2 - L iB - iO

ð3:850Þ

Next, comparing Eq. (3.850) with Eq. (3.848), from Eq. (3.849), we can directly write n o 1 ^ yz F ^þ ^Ω tr F 6I 0 i = pffiffiffi ½ - iðAB þ BA - BC - CB Þ - iðML þ LM  - LK  - KL Þ 4 2I 0 - iðNO þ ON  - OQ - QO Þ - iðDE  þ ED - EF  - FE  Þ

Adyz =

- iðJH  þ HJ  - HG - GH  Þ - iðRS þ SR - ST  - TS Þ - iðTS þ ST  - SR - RS Þ - iðGH  þ HG - HJ  - JH  Þ - iðFE  þ EF  - ED - DE  Þ - iðQO þ OQ - ON  - NQ Þ - iðKL þ LK  - LM  - ML Þ - iðCB þ BC  - BA - AB Þ = 0 ð3:851Þ We can also obtain the following expressions from Eqs. (3.36) and (3.792): 0

- C

B ^ xx - yy^t þ = 3B 0 Q þþ @ 0 B ^ xx - yy^t þ- þ = 3B Q @

A

iK  0 - iM 

- iR - J  0

0

iT 

G

Q

1

0

iF 

G

- iT 

1

B C C þ B ^ 0 C 0 C 0 A, Qxx - yy^t þ - = 3@ 0 A, - iD - J  iR - N 1 0 1 iD A - N  iM  C B C þ B ^ 0 C 0 0 C A, Qxx - yy^t - - = 3@ 0 A     - iF - iK - C Q ð3:852Þ

220

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

n o 1 ^ xx - yy F ^ þ = 1 ð - AC  - CA - MK  - KM  - NQ ^Ω tr F 2I 0 6I 0 - QN  - DF  - FD - JG - GJ  - RT  - TR - TR - RT  - GJ  - JG - FD - DF  - QN  - NQ - KM  - MK  - CA - AC  Þ 2 = - Re ðAC  þ MK  þ NQ þ DF  þ JG þ RT  Þ I0 ð3:853Þ

Adxx - yy =

The following expressions can also be obtained from Eqs. (3.36) and (3.792): 0

A

B ^ zz^t þ = B -2iB Q þþ @ 0

- iM  -2L

- C

iK 



G

iT

B ^ zz^t þ- þ = B 2S Q @ - iR

-2iH  - J

- N

1

0

- iD

- J

iR

1

B C C ^ zz^t þ = B -2E 2iH  2iO C 2S C Q þ@ A, A, Q iF  G - iT  0 1 1    - iF - iK - C Q B C C  B ^ þ -2E C -2L 2iB C A, Qzz^t - - = @ -2iO A iD - N iM  A

ð3:854Þ n o 1 ^ zz F ^ þ = 1 ð jAj2 -2jBj2 þ jCj2 þ jM j2 -2jLj2 þ jK j2 þ jN j2 ^Ω tr F Adzz = 6I 0 6I 0 2 2 -2jOj þ jQj þ jDj2 -2jEj2 þ jF j2 þ jJ j2 -2jH j2 þ jGj2 þ jRj2 -2jSj2 þ jT j2 þjT j2 -2jSj2 þ jRj2 þ jGj2 -2jH j2 þ jJ j2 þ jF j2 -2jE j2 þ jDj2 þ jQj2 -2jOj2 þ jN j2 þ jK j2 -2jLj2 þ jM j2 þ jC j2 -2jBj2 þ jAj2 Þ 1 = ½ jAj2 þ jC j2 þ jDj2 þ jF j2 þ jGj2 þ jJ j2 þ jK j2 þ jM j2 þ jN j2 þ jQj2 3I 0   þjRj2 þ jT j2 -2 jBj2 þ jEj2 þ jH j2 þ jLj2 þ jOj2 þ jSj2  ð3:855Þ There are also the following expressions according to Eqs. (3.833), (3.835), (3.837), (3.796), and (3.797): n o   ^ ^þ ^Σ ^ xΓ ^ iF ^ þ = tr ^t þþ ^Si^t þ ^ ^ ^þ ^ ^ ^þ ^ tr F þ - þ t þ - Si t þþ þ t - þ Si t - - þ t - - Si t - þ , n o   ^ ^þ ^ iF ^ þ = - i tr ^t þþ ^Si^t þ ^Σ ^ yΓ ^ ^ ^þ ^ ^ ^þ ^ tr F þ - - t þ - Si t þþ þ t - þ Si t - - - t - - Si t - þ , n o   ^Σ ^ zΓ ^ iF ^ þ = tr ^t þþ ^Si^t þ ^ ^ ^þ ^ ^ ^þ ^ ^ ^þ tr F þþ - t þ - Si t þ - þ t - þ Si t - þ - t - - Si t - - , i = x, y, z ð3:856Þ

→

3.11

Polarization Theory for

1 2

→

221

þ 1 Reactions

n o   ^ iF ^ i^t þ þ ^t þ - Q ^ i^t þ þ ^t - þ Q ^ i^t þ þ ^t - - Q ^ i^t þ , ^Σ ^ xΩ ^ þ = tr ^t þþ Q tr F þþþ --þ n o   ^ iF ^ i^t þ - ^t þ - Q ^ i^t þ þ ^t - þ Q ^ i^t þ - ^t - - Q ^ i^t þ , ^ þ = - i tr ^t þþ Q ^Σ ^ yΩ tr F þþþ --þ n o   ^ iF ^ i^t þ - ^t þ - Q ^ i^t þ þ ^t - þ Q ^ i^t þ - ^t - - Q ^ i^t þ ^ þ = tr ^t þþ Q ^Σ ^ zΩ tr F þþ þ-þ -- , i=γ ð3:857Þ Utilizing Eqs. (3.839) and (3.792), we can find the following result from Eqs. (3.827), (3.795), (3.796), and (3.856): ANd x,x ¼

n o 1 ^Σ ^ xΓ ^ xF ^ þ ¼ p1ffiffiffi ðAE  - BD þ BF  - CE  þ MH  - LJ  tr F 6I 0 6 2I 0

þLG - KH  þ NS - OR þ OT  - QS - DB þ EA - EC  þ FB - JL þHM  - HK  þ GL - RO þ SN  - SQ þ TO þ TO - SQ þ SN  - RO þGL - HK  þ HM  - JL þ FB - EC  þ EA - DB - QS þ OT  - OR þNS - KH  þ LG - LJ  þ MH  - CE þ BF  - BD þ AE  Þ pffiffiffi 2 = Re ðAE  - BD þ BF  - CE  þ MH  - LJ  þ LG - KH  3I 0 þNS - OR þ OT  - QS Þ

ð3:858Þ Now, according to Eqs. (3.856) and (3.858), we get the following result: ANd y,x = 0

ð3:859Þ

Comparing the third equation of Eq. (3.856) with Eq. (3.838), from Eq. (3.840), we can obtain ANd z,x =

pffiffiffi n o 1 ^ xF ^ þ = - 2 ImðAB þ BC  - DE  - EF  ^Σ ^ zΓ tr F 3I 0 6I 0 þGH  þ HJ  - KL - LM  þ NO þ OQ - RS - ST  Þ

ð3:860Þ

Using the same method, we can obtain the following result via Eqs. (3.792), (3.795), (3.796), (3.827), (3.841), and (3.856):

222

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

i        ANd x,y = pffiffiffi ð - AE - BD - BF - CE - MH - LJ - LG 6 2I 0 - KH  - NS - OR - OT  - QS þ DB þ EA þ EC þ FB þJL þ HM  þ HK  þ GL þ RO þ SN  þ SQ þ TO - TO - SQ - SN  - RO - GL - HK  - HM  - JL - FB - EC - EA - DB þ QS þ OT  þ OR þ NS þ KH  þ LG þ LJ 

ð3:861Þ

þMH  þ CE þ BF  þ BD þ AE  Þ = 0 We can also obtain the following result according to Eqs. (3.856) and (3.861): pffiffiffi 2 Re ðAE  þ BD þ BF  þ CE  þ MH  þ LJ  3I 0 þLG þ KH  þ NS þ OR þ OT  þQS Þ

ANd y,y = -

ð3:862Þ

Comparing the third equation of Eq. (3.856) with Eq. (3.838), from Eq. (3.842), we can obtain ANd z,y = 0

ð3:863Þ

Next, the following result can be obtained utilizing Eqs. (3.792), (3.795), (3.796), (3.827), (3.843), and (3.856): 1 ð - iAD þ iCF  - iMJ  þ iKG - iNR þ iQT  þ iDA - iFC 6I 0 þiJM  - iGK  þ iRN  - iTQ - iTQ þ iRN  - iGK  þ iJM  - iFC

ANd x,z =

þiDA þ iQT  - iNR þ iKG - iMJ  þ iCF  - iAD Þ 2 ImðAD - CF  þ MJ  - KG þ NR - QT  Þ = 3I 0 ð3:864Þ and we can get the following according to Eqs. (3.856) and (3.864): ANd y,z = 0

ð3:865Þ

Comparing the third equation of Eq. (3.856) with Eq. (3.838), from Eq. (3.844) we can obtain ANd z,z =

1 3I 0



jAj2 - jCj2 þ jF j2 - jDj2 þ jGj2 - jJ j2 þ jM j2 - jK j2  þjN j2 - jQj2 þ jT j2 - jRj2

ð3:866Þ

→

3.11

Polarization Theory for

1 2

→

223

þ 1 Reactions

and the following results can be obtained by using Eqs. (3.846), (3.857), (3.792), and (3.797): i ð - iAF  þ iCD - iMG þ iKJ  - iNT  þ iQR þ iDC  4I 0 - iFA þ iJK  - iGM  þ iRQ - iTN  - iTN  þ iRQ - iGM  þ iJK 

ANd x,xy =

- iFA þ iDC  þ iQR - iNT  þ iKJ  - iMG þ iCD - iAF  Þ 1 = Re ðAF  - CD þ MG - KJ  þ NT  - QR Þ I0 ð3:867Þ ANd y,xy = 0

ð3:868Þ

Using the third equation of Eq. (3.857) and Eq. (3.847), we can obtain ANd z,xy = -

1 ImðAC  - DF  þ MK  - JG þ NQ - RT  Þ I0

ð3:869Þ

The following results can also be obtained by using Eqs. (3.848), (3.857), (3.792), and (3.797): 1         ANd x,xz = pffiffiffi ðAE - BD - BF þ CE þ MH - LJ - LG þ KH 4 2I 0 þNS - OR - OT  þ QS - DB þ EA þ EC  - FB - JL þ HM  þ HK  - GL - RO þ SN  þ SQ - TO þTO - SQ - SN  þ RO þ GL - HK  - HM  þ JL

ð3:870Þ

þFB - EC  - EA þ DB - QS þ OT  þ OR - NS - KH  þ LG þ LJ  - MH  - CE þ BF  þ BD - AE  Þ = 0 1       ANd y,xz = pffiffiffi ImðAE - BD - BF þ CE þ MH - LJ 2I 0 - LG þ KH  þ NS - OR - OT  þ QS Þ

ð3:871Þ

Next, using the third equation of Eq. (3.857) and Eq. (3.849), we can obtain ANd z,xz = 0

ð3:872Þ

We can also obtain the following results by using Eqs. (3.850), (3.857), (3.792), and (3.797):

224

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

i         ANd x,yz = pffiffiffi ð - AE - BD þ BF þ CE - MH - LJ þ LG þ KH 4 2I 0 - NS - OR þ OT  þ QS þ DB þ EA - EC  - FB þJL þ HM  - HK  - GL þ RO þ SN  - SQ - TO - TO - SQ þ SN  þ RO - GL - HK  þ HM  þ JL - FB - EC  þ EA þ DB þ QS þ OT  - OR - NS

ð3:873Þ

þKH  þ LG - LJ  - MH  þ CE þ BF  - BD - AE  Þ 1 = pffiffiffi ImðAE  þ BD - BF  - CE  þ MH  þ LJ  2I 0 - LG - KH  þ NS þ OR - OT  - QS Þ ANd y,yz = 0

ð3:874Þ

Now, using the third equation of Eq. (3.857) together with Eq. (3.851), we can obtain 1 ANd Re ðAB - BC  - DE  þ EF  þ ML - LK  z,yz = pffiffiffi 2I 0 - JH  þ HG þ NO - OQ - RS þ ST  Þ

ð3:875Þ

Next, we can obtain the following results by using Eqs. (3.852), (3.857), (3.792), and (3.797): ANd x,xx - yy =

1 ð - iDC  - iFA - iJK  - iGM  - iRQ - iTN  2I 0 þiAF  þ iCD þ iMG þ iKJ  þ iNT  þ iQR - iQR - iNT  - iKJ  - iMG - iCD - iAF 

ð3:876Þ

þiTN  þ iRQ þ iGM  þ iJK  þ iFA þ iDC  Þ = 0 ANd y,xx - yy = ANd z,xx - yy =

2 Re ðAF  þ CD þ MG þ KJ  þ NT  þ QR Þ I0

1 ½ - AC  - CA - MK  - KM  - NQ - QN  2I 0 - ð - DF  - FD - JG - GJ  - RT  - TR Þ - TR - RT  - GJ  - JG - FD - DF  - ð - QN  - NQ - KM  - MK  - CA - AC  Þ = 0

ð3:877Þ

ð3:878Þ

The following results can also be obtained by using Eqs. (3.854), (3.857), (3.792), and (3.797):

3.12

→

→

225

Polarization Theory for 1 þ 1 Reactions

ANd x,zz =

1 ðiDA -2iEB þ iFC þ iJM  -2iHL þ iGK  6I 0 þiRN  -2iSO þ iTQ - iAD þ 2iBE  - iCF  - iMJ  þ 2iLH  - iKG - iNR þ 2iOS - iQT  þiQT  -2iOS þ iNR þ iKG -2iLH  þ iMJ 

ð3:879Þ

þiCF  -2iBE  þ iAD - iTQ þ 2iSO - iRN  - iGK  þ 2iHL - iJM  - iFC  þ 2iEB - iDA Þ = 0 ANd y,zz = -

2 Re ½AD þ CF  þ MJ  þ KG þ NR þ QT  3I 0 -2ðBE  þ LH  þ OS Þ

ð3:880Þ

Now referring to Eq. (3.855), we have ANd z,zz = 0

ð3:881Þ

^in given by According to the non-zero analyzing powers obtained above and ρ Eq. (3.798), it can be seen that when N and d of the incident channel are polarized the differential cross section is n o dσ ^þ ^ ρin F = tr F^ dΩ h 3 d d 2 d d 1 d 1 d d N d = I 0 1 þ pN y Ay þ 2 py Ay þ 3 pxz Axz þ 6 pxx - yy Axx - yy þ 2 pzz Azz   1 d Nd 3 d Nd 2 d Nd 1 d Nd þpN y 2 py Ay,y þ 3 pxz Ay,xz þ 6 pxx - yy Ay,xx - yy þ 2 pzz Ay,zz ! X X X 3 2 þ pN pd ANd þ pd ANd i 2 j = x, z j i, j 3 j = xy, yz j i, j i = x, z

I

ð3:882Þ

where ANd i, j is called a correlation analyzing power.

→

→

3.12 Polarization Theory for 1 þ 1 Reactions The deuteron fusion reaction d(d, p)t and d(d, n)3He belong to the typical rearrangement collision process, and are the representative reactions to be used to discuss whether using the polarized charged particles as fuel can increase the fusion reaction rate. Some individuals have conducted theoretical research [2, 6, 39, 40] on

226

3 Polarization Theory of Nuclear Reactions for Spin 1 Particles

**  **  * * the reactions d d , p t and d d , n 3 He that belong to the 1 þ 1 reaction. This   * * section will take the d d , p t reaction as an example for theoretical derivation.

Using μ, ν, μ0, ν0 to represent the spin magnetic quantum numbers for the first d, second d, p, and t in d + d → p + t reaction, respectively, and noting that the types of particles before and after the reaction are different, then according to Eq. (2.252) the reaction amplitude of d + d → p + t reaction can be written as pffiffiffi  i π X ^ iðσl þσl0 Þ  le δl0 S0 ,lS - SJl0 S0 ,lS f μ0 ν0 ,μν ðΩÞ = f C ðθÞδμ0 ν0 ,μν þ k 00 lSl S J

S0 M 0 C Jl0 MSM C S1μM1ν C Jl0 mM0 S0 M S 0 C1μ0 1νS0 Yl0 ml 0 ðθ, l 2

2

ð3:883Þ

φÞ

From this formula we can see m0l = μ þ ν - μ0 - ν0

ð3:884Þ

Now, introduce the following symbol 0

^^ 0 iðσl þσl0 Þ

f μll0 νJ0 ,μν = l l e

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0    l - m0l ! X J M S M J M S0 M 0S J 0  C C C 0 0 0 0 C 1 0 1 0 δl0 S0 ,lS - S 0 0 l0 SM 1μ 1ν l ml S M S 2μ 2ν l S ,lS l þ m0l ! SS0 ð3:885Þ

Using Eq. (2.253), we can obtain the following expressions from Eq. (3.883) f μ0 ν0 ,μν ðΩÞ = F μ0 ν0 ,μν ðθÞ eiðμþν - μ

0

- ν0 Þφ

0 0X 0 i - μ 0 - ν0 f μll0 νJ0 ,μν Pμþν ð cos θÞ F μ0 ν0 ,μν = f C ðθÞδμ0 ν0 ,μν þ ð-1Þμþν - μ - ν l0 2k 0

ð3:886Þ ð3:887Þ

ll J

Now for parity conservation, l and l0 must be even or odd simultaneously, and because S and S0 are all integers, 2J must be even. Given this, the following C-G coefficient relation can be obtained: S μþν J μþν C Jl0 μþν S μþν C 1μ 1ν C l0 μþν - μ0 - ν0 S0

= - CJl0 -S μ--μν- ν CS1

0

μ0 þν0

0

0

C1Sμ0 μ þν 1 0 ν 2

2

-μ-ν J -μ-ν S0 - μ0 -ν0 - μ 1 - ν C l0 - μ - νþμ0 þν0 S0 - μ0 - ν0 C 1 - μ0 1 -ν0 2 2

ð3:888Þ

3.12

→

→

227

Polarization Theory for 1 þ 1 Reactions

Noting Eq. (2.287), we can obtain the following relation from Eqs. (3.885) and (3.887): 0

F μ0 ν0 ,μν ðθÞ = ð-1Þμþν - μ

- ν0 þ1

F - μ0 - ν 0 , - μ - ν ð θ Þ

ð3:889Þ

*

Once again, if we take the y-axis along the direction n , then φ = 0. Next, introduce the following matrix element symbols according to Eq. (3.889): AðθÞ = F 1212,11 ðθÞ = F - 12 - 12, -1 -1 ðθÞ,

BðθÞ = iF 1212,10 ðθÞ = - iF - 12 - 12, -10 ðθÞ,

C ðθÞ = - F 1212,1 -1 ðθÞ = - F - 12 - 12, -11 ðθÞ, DðθÞ = iF 1212,01 ðθÞ = - iF - 12 - 12,0 -1 ðθÞ, E ðθÞ = - F 1212,00 ðθÞ = - F - 12 - 12,00 ðθÞ,

F ðθÞ = - iF 1212,0 -1 ðθÞ = iF - 12 - 12,0 1 ðθÞ,

GðθÞ = - F 1212, -11 ðθÞ = - F - 12 - 12,1 -1 ðθÞ, H ðθÞ = - iF 1212, -10 ðθÞ = iF - 12 - 12,10 ðθÞ, J ðθÞ = F 1212, -1 -1 ðθÞ = F - 12 - 12,11 ðθÞ,

K ðθÞ = iF 12 - 12,11 ðθÞ = - iF - 1212, -1 -1 ðθÞ,

LðθÞ = - F 12 - 12,10 ðθÞ = - F - 1212, -10 ðθÞ,

M ðθÞ = - iF 12 - 12,1 -1 ðθÞ = iF - 1212, -11 ðθÞ,

N ðθÞ = - F 12 - 12,01 ðθÞ = - F - 1212,0 -1 ðθÞ,

OðθÞ = - iF 12 - 12,00 ðθÞ = iF - 1212,00 ðθÞ,

QðθÞ = F 12 - 12,0 -1 ðθÞ = F - 1212,01 ðθÞ,

RðθÞ = - iF 12 - 12, -11 ðθÞ = iF - 1212,1 -1 ðθÞ,

SðθÞ = F 12 - 12, -10 ðθÞ = F - 1212,10 ðθÞ,

T ðθÞ = iF 12 - 12, -1 -1 ðθÞ = - iF - 1212,11 ðθÞ ð3:890Þ

1 In the following, the footnote is represented by the footnote +, and the footnote 2 1 - is represented by the footnote -, so the 4 × 9 reaction amplitude matrix 2 given by Eq. (3.887) can be expressed as 0

F þþ,11 F þþ,10 F þþ,1-1 F þþ,01 F þþ,00 F þþ,0-1 F þþ,-11 F þþ,-10 F þþ,-1-1

1

B C B F þ-,11 F þ-,10 F þ-,1-1 F þ-,01 F þ-,00 F þ-,0-1 F þ-,-11 F þ-,-10 F þ-,-1-1 C B C ^ =B F C B F -þ,11 F -þ,10 F -þ,1-1 F -þ,01 F -þ,00 F -þ,0-1 F -þ,-11 F -þ,-10 F -þ,-1-1 C @ A F - -,11 F - -,10 F - -,1-1 F - -,01 F - -,00 F - -,0-1 F - -,-11 F - -,-10 F - -,-1-1 0 1 A -iB -C -iD -E iF -G iH J B C B -iK -L iM -N iO Q iR S -iT C B C =B C B iT C S -iR Q -iO -N -iM -L iK @ A J -iH -G -iF -E iD -C iB A ð3:891Þ

228

3

0

F þþ,11 B F B þþ,10 B  B F þþ,1 -1 B B F B þþ,01 B þ ^ = B F þþ,00 F B B F B þþ,0 -1 B  B F þþ, -11 B B  @ F þþ, -10 0

F þ - ,11 F þ - ,10

J

F - þ,11 F - þ,10

F þ - ,1 -1

F - þ,1 -1

F þ - ,0 -1 F þ - , -11

F - þ,0 -1 F - þ, -11

F þ - , -1 -1

F - þ, -1 -1 1 J C iH  C C  C -G C C  iF C C - E C C C - iD C C - C C C C - iB A A

F þ - ,01 F þ - ,00

F þ - , -10

F þþ, -1 -1

A B iB B B B - C B B B iD B  =B B -E B B - iF  B B - G B B @ - iH 

Polarization Theory of Nuclear Reactions for Spin 1 Particles

iK  - L

- iT  S

- iM 

iR



-N - iO 

Q iO 

Q - iR

-N iM 

S

- L

iT 

- iK 

F - þ,01 F - þ,00

F - þ, -10

1 F - - ,11 F - - ,10 C C C F - - ,1 -1 C C F - - ,01 C C C F - - ,00 C C F - - ,0 -1 C C C F - - , -11 C C C F - - , -10 A F - - , -1 -1

ð3:892Þ

Now, from Eqs. (3.891) and (3.892) we can find that the differential cross section is as follows when the two deuterons of the incident channel are unpolarized: dσ 0 1 = trfFF þ g dΩ 9  2 2 = jAj þ jBj2 þ jCj2 þ jDj2 þ jE j2 þ jF j2 þ jGj2 þ jH j2 þ jJ j2 9 

I0 

ð3:893Þ

þjK j2 þ jLj2 þ jM j2 þ jN j2 þ jOj2 þ jQj2 þ jRj2 þ jSj2 þ jT j2

According to Eq. (3.214) the polarization density matrix of the normalized incident channel can be written as ^ρin =

h  ðk Þ  ðk Þ ðk Þ 1 Y ^ 3 I 3 þ pkx ^Sx þ pky ^Sy þ pkz ^Sz 9 k = 1,2 2   i ð3:894Þ 2 1 k ðk Þ ðk Þ ^ ð k Þ þ pk Q ^ ð k Þ þ pk Q ^ ð k Þ þ 1 pk þ pkxy Q Q þ Q p xy xz xz yz yz 6 xx - yy xx - yy 2 zz zz 3

where k = 1, 2 such that the two multiplied items represent the first deuteron and the second deuteron of the incident channel, respectively.

3.12

→

→

229

Polarization Theory for 1 þ 1 Reactions

Next, define the following polarization analyzing powers: 8 n o ðk Þ þ > ^ ^ ^ > F tr F S > i > > n þo , > > > ^F ^ < tr F ðk Þ Ai = n ðk Þ þ o > ^ F ^ ^Q > tr F > i > > n o , > > > ^F ^þ : tr F

k = 1, 2;

i = x, y, z ð3:895Þ

k = 1, 2;

8 n ð1Þ ð2Þ þ o ^ ^ ^Si ^Sj F > tr F > > > n o > , > > ^F ^þ > tr F > > > > n ð1Þ ð2Þ þ o > > > ^ F ^ ^ ^Si Q > tr > F j > > n o , > > > ^F ^þ < tr F ð12Þ Ai,j = n ð1Þ ð2Þ þ o > ^^ ^ ^ > tr > > F Q i Sj F > n þo > , > > > ^F ^ tr F > > > > > > n ^ ^ ð1Þ ^ ð2Þ ^ þ o > > tr F Qi Qj F > > > n þo , > > : ^F ^ tr F

i=γ

i, j = x, y, z

i = x, y, z;

j=γ ð3:896Þ

i = γ;

j = x, y, z

i, j = γ

where the definition of the footnote set γ has been given by Eq. (3.824), that is, ð12Þ γ = xy, xz, yz, xx - yy, zz. Ai,j is also known as correlation polarization analyzing power. It follows that according to Eq. (3.894) the general form of the differential **  cross section for d d , p t reaction can be written as X dσ I = trfF^ρin F þ g = I 0 1 þ dΩ k = 1, 2

X i = x, y, z, γ

ðk Þ pki Ai

þ

X i, j = x, y, z, γ

! ð12Þ p1i p2j Ai,j

ð3:897Þ where

230

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

8 3 k > > p, > > 2 i > > > > 2 > < pki , k pi = 3 1 > > > pki , > > 6 > >1 > > : pk , 2 i

i = x, y, z i = xy, xz, yz

ð3:898Þ

i = xx - yy i = zz

If the expectation value method given by Eq. (1.88) is used to find the mean value * * ð1Þ ^ ð1Þ in of the physical quantity for d þ d → p þ t reaction, the operators ^ Si and Q i ð2Þ this section can only act on the spin wave function of the first deuteron, and ^ S and i

^ ð2Þ act on the spin wave function of the second deuteron. When we use the tracing Q i ð1Þ ^ ð1Þ only works on the first deuteron magnetic quantum number μ method, S^ and Q i

i

0 0 ^ð2Þ ^ ð2Þ of F þ μ0 ν0 ,μν in the same μ ν ν case, but Si and Qi only works on the second deuteron 0 0 magnetic quantum number ν of F þ μ0 ν0 ,μν in the same μ ν μ case. Next, define the following vectors:

0

F lm,n1

1

0

F lm,1n

1

C * ð1Þ B C * ð2Þ ð2Þ B  C, U C ^ð1Þ B  F U i,lmn = ^Si B lm,n0 i,lmn = Si @ F lm,0n A, @ A F lm,n -1 F lm, -1 n 0  0  1 1 F lm,n1 F lm,1n B B C *ð1Þ C *ð2Þ C ^ ð2Þ B F lm,n0 C, V ^ ð1Þ B  V i,lmn = Q i,lmn = Qi @ F lm,0n A, i @ A F lm,n -1 F lm, -1 n → ðk Þ

i = x, y, z ð3:899Þ i=γ

→ ðk Þ

The b (=1, 0, -1) components of the vectors U i,lma and V i,lma are represented by ðk Þ ðk Þ U i,lma,b and V i,lma,b separately below. Now redefine 0 1 8 ð2Þ U j,lm1,n > > > B C > > C ð1Þ B > ð2Þ > ^ B C, i, j = x, y, z > S U > i j,lm0,n B C > > @ A > > ð2Þ > < U j,lm -1,n * ð12Þ 0 1 W ij,lmn = ð2Þ > > V > > B j,lm1,n C > > C > ð1Þ B > ^S B V ð2Þ C, i = x, y, z; j = γ > > i B j,lm0,n C > > @ A > > : ð2Þ V j,lm -1,n

ð3:900Þ

3.12

→

→

Polarization Theory for 1 þ 1 Reactions

0 1 8 ð2Þ U > > > B j,lm1,n C > > C ð1Þ B > > C ^ B U ð2Þ > j = x, y, z >Q i B j,lm0,n C, i = γ; > > @ A > > ð2Þ > < U j,lm -1,n *ð12Þ 0 1 Z ij,lmn = ð2Þ > > V > j,lm1,n > B C > > C > ð1Þ B >Q ð2Þ B C ^ > > i B V j,lm0,n C, i, j = γ > > @ A > > : ð2Þ V j,lm -1,n

231

ð3:901Þ

where in the above three formulas l, m = + , - ; n = 1, 0, - 1. For the second deuteron, one can obtain the following results from Eqs. (3.33), (3.892), and (3.899): 0

10  1 0 1 0 1 0 A iB ð2Þ B 1 B 1 B C CB C C U x,þþ1 = ^Sx @ F þþ,10 A = pffiffiffi @ 1 0 1 A@ iB A = pffiffiffi @ A - C A, 2 2 F þþ,1 -1 - C iB 0 1 0 1 1 0 0 - E - iH  * ð2Þ * ð2Þ 1 B 1 B C C U x,þþ -1 = pffiffiffi @ J  - G A, U x,þþ0 = pffiffiffi @ iðD - F  Þ A, 2 2 - iH  - E 0 1 0 1 - L - iO ð 2 Þ * ð2Þ * 1 B 1 B C C U x,þ -1 = pffiffiffi @ iðK  - M  Þ A, U x,þ - 0 = pffiffiffi @ Q - N  A, 2 2 - iO - L 0 1 0 1 S S ð 2 Þ ð 2 Þ * * 1 B 1 B C C U x,þ - -1 = pffiffiffi @ iðT  - R Þ A, U x, - þ1 = pffiffiffi @ iðR - T  Þ A, 2 2 S S 0 1 0 1 iO - L * ð2Þ * ð2Þ 1 B 1 B C C U x, - þ0 = pffiffiffi @ Q - N  A, U x, - þ -1 = pffiffiffi @ iðM  - K  Þ A, 2 2 - L iO 1 1 0 0 iH  - E * ð2Þ * ð2Þ 1 B 1 B C C U x, - - 0 = pffiffiffi @ iðF  - D Þ A, U x, - -1 = pffiffiffi @ J  - G A, 2 2 iH  - E 0 1 - iB * ð2Þ 1 B C U x, - - -1 = pffiffiffi @ A - C  A ð3:902Þ 2  - iB * ð2Þ

F þþ,11

1

0

232

3 Polarization Theory of Nuclear Reactions for Spin 1 Particles

such that one can now write 0

iB - L S    B A - C  i ðK - M Þ i ðR - T  Þ B  B iB - L S B  B - E - iO iO 1 B ð2Þ ^ þ      ^ B S x F = pffiffiffi B iðD - F Þ Q -N Q - N 2 B - E  - iO iO B  B - iH  S - L B    @ J - G i ðT - R Þ i ðM  - K  Þ  - iH S - L

1 iH  J  - G C C C iH  C  C -E C   C iðF - D Þ C - E C C - iB C C A - C A - iB

ð3:903Þ

Substituting Eqs. (3.891) and (3.903) into Eq. (3.895), one can obtain the following result: Aðx2Þ =

n ð2Þ o 1 i tr F ^Sx F þ = pffiffiffi ðAB - BA þ BC  - CB þ DE  þ EF  9I 0 9 2I 0 - ED - FE þ GH  - HG þ HJ  - JH  þ KL - LK  þ LM  - ML

þNO þ OQ - ON  - QO þ RS - SR þ ST  - TS þ TS - ST  þSR - RS þ QO - OQ þ ON  - NO þ ML - LM  þ LK  - KL þJH  þ HG - HJ  - GH  þ FE þ ED - EF  - DE  þ CB þ BA - BC  - AB Þ = 0 ð3:904Þ One can also obtain 0

F þþ,11

1

0

0 -1

0

10

A

1

0

B

1

C C CB  C * ð2Þ ð2Þ B i B 1 B   C C C B  B CB U y,þþ1 = ^Sy B @ F þþ,10 A = pffiffi2ffi @ 1 0 -1 A@ iB A = pffiffi2ffi @ iðA þ C Þ A, F þþ,1 -1 - B - C 0 1 0 1 1 0 0 iE - H ð 2 Þ ð 2 Þ * * 1 B 1 B C C U y,þþ -1 = pffiffiffi @ - iðG þ J  Þ A, U y,þþ0 = pffiffiffi @ - ðD þ F  Þ A, 2 2 - iE  H 0 1 0 1 iL - O * ð2Þ 1 B 1 B C * ð2Þ C U y,þ -1 = pffiffiffi @ - ðK  þ M  Þ A, U y,þ - 0 = pffiffiffi @ - iðN  þ Q Þ A, 2 2 - iL O 0 1 0 1 - iS - iS * ð2Þ * ð2Þ 1 B 1 B C C U y,þ - -1 = pffiffiffi @ R þ T  A, U y, - þ1 = pffiffiffi @ R þ T  A, 2 2 iS iS

3.12

→

→

233

Polarization Theory for 1 þ 1 Reactions

0

1 0 1 O iL * ð2Þ * ð2Þ 1 B 1 B C C U y, - þ0 = pffiffiffi @ iðQ þ N  Þ A, U y, - þ -1 = pffiffiffi @ - ðK  þ M  Þ A, 2 2 - O - iL 1 1 0 0 H iE  * ð2Þ * ð2Þ 1 B 1 B C C U y, - - 0 = pffiffiffi @ - ðD þ F  Þ A, U y, - -1 = pffiffiffi @ iðG þ J  Þ A, 2 2 - H - iE  1 0 - B * ð2Þ 1 B C U y, - - -1 = pffiffiffi @ - iðA þ C Þ A ð3:905Þ 2  B such that one can now write 0

B

iL

-iS

H

1

B C B iðA þC Þ - ðK  þM  Þ R þT  iðG þJ  Þ C B C B C     B C -iL iS -H -B B C B C     B C iE -O O iE B C B C þ 1 B         C ^Sð2Þ F ^ p ffiffi ffi = ð D þF Þ -i ð N þQ Þ i ð Q þN Þ ð D þF Þ y B C 2B C    B -iE  C O -O -iE B C B C     B C -H -iS iL -B B C B C B -iðG þJ  Þ R þT  - ðK  þM  Þ -iðA þC  Þ C @ A H

iS

-iL

ð3:906Þ

B

Substituting Eqs. (3.891) and (3.906) into Eq. (3.895), one can obtain the following result: n ð2Þ o 1 1 tr F ^Sy F þ = pffiffiffi ðAB þ BA þ BC  þ CB þ DE  þ ED 9I 0 9 2I 0 þEF  þ FE þ GH  þ HG þ HJ  þ JH  þ KL þ LK  þ LM  þ ML

Aðy2Þ =

þNO þ ON  þ OQ þ QO þ RS þ SR þ ST  þ TS þ TS þ ST  þSR þ RS þ QO þ OQ þ ON  þ NO þ ML þ LM  þ LK  þ KL þJH  þ HJ  þ HG þ GH  þ FE þ EF  þ ED þ DE  þ CB þ BC  þBA þ AB Þ pffiffiffi 2 2 ðAB þ BC  þ DE  þ EF  þ GH  þ HJ  þ KL þ LM  þ NO = 9I 0 þOQ þ RS þ ST  Þ ð3:907Þ

234

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

and one can still get 0

1 0 10  1 0  1 F þþ,11 1 0 0 A A F þþ,10 A = @ 0 0 0 A@ iB A = @ 0 A,  F - C0 0  1þþ,1 -1 00 0  -1 1 1C  iD -G iK * ð2Þ * ð2Þ * ð2Þ U z,þþ0 = @ 0 A, U z,þþ -1 = @ 0 A, U z,þ -1 = @ 0 A,   - J iM 0iF  1 0 1 0 1  -N - iR - iT  * ð2Þ * ð2Þ * ð2Þ U z,þ - 0 = @ 0 A, U z,þ - -1 = @ 0 A, U z, - þ1 = @ 0 A,    0 -Q1 0 -1iT 0  1- iR Q iM J * ð2Þ * ð2Þ * ð2Þ U z, - þ0 = @ 0 A, U z, - þ -1 = @ 0 A, U z, - -1 = @ 0 A,   G 0N  1 0iK  1 iF -C * ð2Þ * ð2Þ U z, - - 0 = @ 0 A, U z, - - -1 = @ 0 A iD - A * ð2Þ U z,þþ1 = S^ðz2Þ @

ð3:908Þ

It follows that 0 B B B B B B B B þ ^Sð2Þ F B ^ = z B B B B B B B @

A

iK 

- iT 

0 C

0 iM 

0 - iR

iD 0

- N 0

Q 0







iF

-Q 

N

-G 0

- iR 0



iM  0

- J

- iT 

iK 

J

1

C C C C C  C iF C C 0 C C C iD C C - C C C C 0 A - A 0 G

ð3:909Þ

Substituting Eqs. (3.891) and (3.909) into Eq. (3.895), one can obtain the following result: Aðz2Þ =

n ð2Þ o 1 1 tr F ^Sz F þ = ð jAj2 - jC j2 þ jDj2 - jF j2 þ jGj2 9I 0 9I 0 - jJ j2 þ jK j2 - jM j2 þ jN j2 - jQj2 þ jRj2 - jT j2 þ jT j2 - jRj2 þ jQj2 - jN j2 þ jM j2 - jK j2 þ jJ j2 - jGj2 þ jF j2 - jDj2 þ jC j2 - jAj2 Þ = 0 ð3:910Þ

Now for the first deuteron, one can obtain the following results from Eqs. (3.33), (3.892), and (3.899):

3.12

→

→

235

Polarization Theory for 1 þ 1 Reactions

0 1 1 0 10  1 F þþ,11 010 iD A * ð1Þ ð1Þ B C 1 B CB C 1 B C U x,þþ1 = ^Sx @ F þþ,01 A = pffiffiffi @ 1 0 1 A@ iD A = pffiffiffi @ A -G A, 2 2  F 010 -G 0 iD1 0 þþ,-11 1   -E -iF * ð1Þ * ð1Þ 1 B 1 B C C U x,þþ0 = pffiffiffi @ iðB -H  Þ A, U x,þþ -1 = pffiffiffi @ - ðC  -J  Þ A, 2 2  -iF  1 0 -E  1 0 -N -iO ð1Þ * ð1Þ * 1 B  1 C C B U x,þ-0 = pffiffiffi @ - ðL -S Þ A, U x,þ-1 = pffiffiffi @ iðK -R Þ A, 2 2   0 -N  1 0 -iO 1 Q Q ð 1 Þ ð 1 Þ * 1 B 1 B C * C U x,þ- -1 = pffiffiffi @ -iðM  -T  Þ A, U x,-þ1 = pffiffiffi @ iðM  -T  Þ A, 2 2  Q 1 0 0 Q  1  iO -N * ð1Þ * ð1Þ 1 B 1 B C C U x,-þ0 = pffiffiffi @ - ðL -S Þ A, U x,-þ -1 = pffiffiffi @ -iðK  -R Þ A, 2 2 iO -N  1 1 0 0  iF -E  * ð1Þ * ð1Þ 1 B 1 C C B  U x,--0 = pffiffiffi @ -iðB -H  Þ A, U x,--1 = pffiffiffi @ - ðC -J  Þ A, 2 2  -E  0 iF  1 -iD * ð1Þ 1 B  C U x,-- -1 = pffiffiffi @ A -G A 2 -iD 0

ð3:911Þ

and the following can be written: 0

iD

- N

Q

iF 

1

B C B - E - iO iO - E C B C B C    B - iF  Q -N - iD C B C B  C  C      B A - G i ð K R Þ i ð M T Þ J C B C C þ 1 B         ^Sð1Þ F B ^ = pffiffiffi BiðB - H Þ S -L S -L iðH - B Þ C x C 2B C   C     B J  - C i ð M T Þ i ð R K Þ A G B C B C    B iD C -N Q iF B C B C    B - E C iO iO E @ A     - iF Q -N - iD

ð3:912Þ

Substituting Eqs. (3.891) and (3.912) into Eq. (3.895), one can obtain the following result:

236

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

n ð1Þ o 1 i tr F ^Sx F þ = pffiffiffi ðAD þ BE  þ CF  - DA þ DG - EB 9I 0 9 2I 0 þEH  - FC þ FJ  - GD - HE - JF  þ KN  þ LO þ MQ - NK  þ NR - OL þ OS - QM  þ QT  - RN  - SO - TQ þ TQ

Aðx1Þ =

þSO þ RN  - QT  þ QM  - OS þ OL - NR þ NK  - MQ - LO - KN  þ JF  þ HE  þ GD - FJ  þ FC  - EH  þ EB - DG þ DA - CF  - BE  - AD Þ = 0 ð3:913Þ One can also obtain 0

F þþ,11

1

0

0

-1

0

10

A

1

0

D

1

C C C CB * ð1Þ ð1Þ B i B 1 B   C  C B B  B C U y,þþ1 = ^Sy B -1 C @ F þþ,01 A= pffiffi2ffi @ 1 0 A@ iD A = pffiffi2ffi @ iðA þ G Þ A, F þþ, -11 - G - D 0 1 0 0 0 1 1 iE  - F ð 1 Þ ð 1 Þ B B C C * * 1 1 - ðB þ H  Þ C - iðC  þ J  Þ C U y,þþ0 = pffiffiffi B , U y,þþ -1 = pffiffiffi B @ @ A A, 2 2   - iE F 0 0 1 1  iN - O C * ð1Þ C * ð1Þ 1 B 1 B - ðK  þ R Þ C - iðL þ S Þ C , U y,þ - 0 = pffiffiffi B U y,þ -1 = pffiffiffi B @ @ A A, 2 2 - iN  O 0 1 0 1 - iQ - iQ * ð1Þ * ð1Þ 1 B 1 B C C U y,þ - -1 = pffiffiffi@ M  þ T  A, U y, - þ1 = pffiffiffi@ T  þ M  A, 2 2 iQ iQ 1 1 0 0 O iN  * ð1Þ * ð1Þ 1 B 1 B C C U y, - þ0 = pffiffiffi@ iðS þ L Þ A, U y, - þ -1 = pffiffiffi@ - ðR þ K  Þ A, 2 2 - O - iN  0 1 0 1 F iE  * ð1Þ * ð1Þ 1 B 1 B C C U y, - -1 = pffiffiffi@ iðJ  þ C  Þ A, U y, - - 0 = pffiffiffi@ - ðH  þ B Þ A, 2 2 - F - iE  0 1 - D * ð1Þ 1 B C U y, - - -1 = pffiffiffi@ - iðG þ A Þ A ð3:914Þ 2  D

3.12

→

→

237

Polarization Theory for 1 þ 1 Reactions

therefore, the following can be written: 0

D

B B iE B B B B - F B B B iðA þ G Þ B B ð1Þ þ 1 B ^ ^ Sy F = pffiffiffi B - ðB þ H  Þ 2B B B - iðC  þ J  Þ B B B - D B B B B - iE @ F

iN 

- iQ

- O

O

- iQ

iN 

- ðK  þ R  Þ

T  þ M

- i ð L þ S  Þ

iðS þ L Þ

M þ T 

- ðR þ K  Þ

- iN 

iQ

O

- O

iQ

- iN 

1

F

C C C C C C - D C C   i ðJ þ C Þ C C C   C - ðH þ B Þ C C C - iðG þ A Þ C C C  C -F C C C C - iE  A  D iE 

ð3:915Þ Substituting Eqs. (3.891) and (3.915) into Eq. (3.895), one can obtain the following result: Aðy1Þ =

n ð1Þ o 1 1 tr F ^Sy F þ = pffiffiffi ðAD þ BE  þ CF  þ DA þ DG þ EB 9I 0 9 2I 0 þEH  þ FC þ FJ  þ GD þ HE þ JF  þ KN  þ LO þ MQ þ NK  þNR þ OL þ OS þ QM  þ QT  þ RN  þ SO þ TQ þ TQ þ SO þRN  þ QT  þ QM  þ OS þ OL þ NR þ NK  þ MQ þ LO þ KN  þJF  þ HE  þ GD þ FJ  þ FC  þ EH  þ EB þ DG þ DA þ CF 

þBE  þ AD Þ pffiffiffi 2 2 = Re ðAD þ BE  þ CF  þ DG þ EH  þ FJ  þ KN  þ LO þ MQ 9I 0 þNR þ OS þ QT  Þ ð3:916Þ and we can still get

238

3

0

1 0 F þþ,11 1 F þþ,01 A = @ 0 F þþ, -11 0 1

* ð1Þ ð1Þ U z,þþ1 = S^z @

0

iB 0 A,  0iH  1 -L * ð1Þ U z,þ - 0 = @ 0 A,  0 -S 1 S * ð1Þ U z, - þ0 = @ 0 A,  0L  1 iH * ð1Þ U z, - - 0 = @ 0 A, iB * ð1Þ U z,þþ0 = @

Polarization Theory of Nuclear Reactions for Spin 1 Particles

10  1 0  1 0 0 A A 0 0 A@ iD A = @ 0 A, G 0 -1 -G 0 1 0 1  -C iK * ð1Þ * ð1Þ U z,þþ -1 = @ 0 A, U z,þ -1 = @ 0 A,  -J  iR 0 1 0 1  -iM - iT  * ð1Þ * ð1Þ U z,þ - -1 = @ 0 A, U z, -þ1 = @ 0 A,   0 -1iT 0  1-iM iR J * ð1Þ * ð1Þ U z, - þ -1 = @ 0 A, U z, - -1 = @ 0 A, iK  C 0 1 -G * ð1Þ =@ 0 A U z, - - -1

ð3:917Þ

- A

So one can write 0 B B B B B B B B ð1Þ þ ^S F B ^ = z B B B B B B B @

A

iK 

- iT 

iB - C

- L - iM 

S iR

0 0

0 0

0 0

0

0

0





G iH 

iR - S

- iM  L

- J

- iT 

iK 

J

1

iH  C C C C -G C C 0 C C 0 C C C 0 C C C C C C iB A

ð3:918Þ

- A

Substituting Eqs. (3.891) and (3.918) into Eq. (3.895), one can obtain the following result: n ð1Þ o 1 1 tr F ^Sz F þ = ð jAj2 þ jBj2 þ jCj2 - jGj2 - jH j2 - jJ j2 9I 0 9I 0 þjK j2 þ jLj2 þ jM j2 - jRj2 - jSj2 - jT j2 þ jT j2 þ jSj2 þ jRj2 - jM j2

Aðz1Þ =

- jLj2 - jK j2 þ jJ j2 þ jH j2 þ jGj2 - jCj2 - jBj2 - jAj2 Þ = 0

ð3:919Þ

3.12

→

→

239

Polarization Theory for 1 þ 1 Reactions

Also, one can obtain the following expressions from Eqs. (3.900) and (3.902): 0

0 1 0

10

iB

1

0

- E

1

B C C CB 1B B 1 0 1 CB - E  C = 1 B iðB - H  Þ C, @ @ @ A A A 2 2   - iH -E 0 1 0 0 1 0 1   iðD - F Þ - E C C * ð12Þ * ð12Þ 1B  1B  i ðB - H  Þ C A - C þ J  - G C , W xx,þþ -1 = B W xx,þþ0 = B @ A, @ A 2 2 - E iðD - F  Þ 0 1 0 1   - iO Q - N C C * ð12Þ * ð12Þ 1B 1B     C B - L þ S C W xx,þ -1 = B A, W xx,þ - 0 = 2 @ iðK - M þ T - R Þ A, 2@ Q - N  - iO 0 0 1 1 - iO iO C C * ð12Þ * ð12Þ 1B 1B  - L  þ S C S - L C W xx,þ - -1 = B , W xx, - þ1 = B @ A A, 2 2@   - iO iO 0 1 0 1   Q -N iO C C * ð12Þ * ð12Þ 1B  1B  i ðR - T  þ M  - K  Þ C S - L C , W xx, - þ -1 = B W xx, - þ0 = B @ A @ A, 2 2 iO Q - N  0 1 0 1 - E iðF  - D Þ C C * ð12Þ * ð12Þ 1B 1B  W xx, - -1 = B - iðB - H  Þ C J - G þ A - C C , W xx, - - 0 = B @ A @ A, 2 2 - E iðF  - D Þ 0 1  -E C * ð12Þ 1B - iðB - H  Þ C W xx, - - -1 = B A 2@ - E * ð12Þ

W xx,þþ1 =

ð3:920Þ Therefore, one can write

240

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

^Sð1Þ ^Sð2Þ F ^þ = 1 x x 2 0 1 -E  -iO iO -E  B C B iðD -F  Þ Q -N Q -N  -iðD -F  Þ C B C B C   B -E  -iO C iO -E B C B C B iðB -H  Þ      S -L S -L -iðB -H Þ C B C B C B      C           B A -C þJ -G iðK -M þT -R Þ -iðK -M þT -R Þ A -C þJ -G C B C B C B iðB -H  Þ S -L S -L -iðB -H  Þ C B C B C     B C -E -iO iO -E B C B C         B iðD -F Þ Q -N Q -N -iðD -F Þ C @ A -iO

-E 

iO

-E  ð3:921Þ

Substituting Eqs. (3.891) and (3.921) into Eq. (3.896), one can obtain the following results: 12Þ = Aðx,x

n ð1Þ ð2Þ o 1 1 tr F ^ Sx ^Sx F þ = ½ - AE  þ BðD - F  Þ þ CE  þ DðB - H  Þ 9I 0 18I 0 - EðA - C þ J  - G Þ - F ðB - H  Þ þ GE  - H ðD - F  Þ- JE  - KO - LðQ - N  Þ þ MO - N ðS - L Þ- OðK  - M  þ T  - R Þ þQðS - L Þ þ RO þ SðQ - N  Þ - TO - TO þ SðQ - N  Þ þ RO þQðS - L Þ- OðK  - M  þ T  - R Þ- N ðS - L Þ þ MO - LðQ - N  Þ

- KO - JE  - H ðD - F  Þ þ GE  - F ðB - H  Þ- EðA - C þ J  - G Þ þDðB - H  Þ þ CE  þ BðD - F  Þ- AE   1 ½ - E ðA - C  þ J  - G Þ- ðA - C þ J - GÞE = 9I 0 - OðK  - M  þ T  - R Þ- ðK - M þ T - RÞO þ ðB - H ÞðD - F  Þ þðD - F ÞðB - H  Þ þ ðS- LÞðQ - N  Þ þ ðQ - N ÞðS - L Þ 2 Re ½ ðA - C þ J - GÞE  þ ðK - M þ T - RÞO - ðB - H ÞðD - F  Þ =9I 0 - ðL - SÞðN  - Q Þ ð3:922Þ ð12Þ

Other items of Ai,j ði, j = x, y, zÞ can be derived by the above method. By using ðk Þ

Eq. (3.36) further, Ai ðk = 1, 2, i = γ Þ and all remaining items of ð12Þ Ai,j ði, j = x, y, z, γ Þ can be derived. Due to parity conservation, there must be some analyzing powers with a value of 0.

3.12

→

→

241

Polarization Theory for 1 þ 1 Reactions

Referring to Eqs. (3.795)–(3.797), the following 9 × 9 matrices can be introduced: 0

0

Si,1 1^I 3 ^S1i  B @ Si,0 1^I 3 Si, -11^I 3

Si,1 0^I 3 Si,0 0^I 3 Si, -1 0^I 3

Gi,1 1^I 3 B ^ G1i  @ Gi,0 1^I 3 Gi, -11^I 3

1 Si,1 -1^I 3 C Si,0 -1^I 3 A, i = x, y, z Si, -1 -1^I 3 1

ð3:923Þ

Gi,1 0^I 3 Gi,1 -1^I 3 C Gi,0 0^I 3 Gi,0 -1^I 3 A, i = xy,xz,yz,xx - yy,zz Gi, -1 0^I 3 Gi, -1 -1^I 3 1 0 ^Si ^03 ^03 C ^S2i  B @ ^03 ^Si ^03 A, i = x, y, z 0^3 0^3 S^i 1 0 ^ i ^03 ^03 G ^ 2i  B ^ i ^03 C G A, i = xy,xz,yz,xx - yy,zz @ ^03 G ^03 ^03 G ^i

ð3:924Þ

ð3:925Þ

ð3:926Þ

^ 1i act on the first deuteron functions, and the matrices ^ The matrices ^S1i and G S2i and * * ^ 2i act on the second deuteron functions. It follows that the 1 þ 1 reaction can be G processed according to the method described in Sect. 3.11. The previous method used for the arrangement of reaction amplitude footnotes belongs to the usual method. In order to facilitate the operation, the reaction amplitude matrix element F μ0 ν0 ,μν given by Eq. (3.887) can also be rewritten as *

*

F μ0 μ,ν0 ν [6], such that for the d þ d → p þ t reaction the footnote μ0 is the line number of p, the footnote μ is the column number of the first d, the footnote ν0 is the line number of t, and the footnote ν is the column number of the second d. For the definite ^ ðν10 νÞ ; for the definite μ0μ, the ν0ν, the footnotes μ0μ constitute a 2 × 3 submatrix F ð2Þ ^ ð1Þ act on F ^ ð20 Þ ; ^Sð1Þ and Q ^ ð10 Þþ ; ^ S and footnotes ν0ν constitute a 2 × 3 submatrix F μμ

i

*

*

i

νν

i

^ ð2Þ act on F ^ ðμ20 μÞþ . In order to find the correlation analyzing powers, one must first Q i ^ð2Þ or Q ^ ð2Þ acting on all F ^ ðμ20 μÞ , and then calculate ^þ form a new 9 × 4 matrix X j j after Sj ð1Þþ ^Sð1Þ or Q ^ ð1Þ acting on the X ^ j,ν 0ν . i i In fact, because the matrix operations of 1 þ 1 reaction are too large, the above complex formula derivation cannot be performed; only the numerical calculation program including the tracing needs to be compiled. As for how to calculate the S matrix elements of a few-body reaction, there are several options which can be employed, including, but not limited to, direct reaction theory, R matrix theory, the phase-shift analysis method, resonance group theory, Faddeev equation, and CDCC theory. The theoretical approach described in this section can also be applied to polari* *  zation nuclear reaction d d , n 3 He.

242

3

3.13

Polarization Theory of Nuclear Reactions for Spin 1 Particles

Deuteron Phenomenological Optical Potentials with Tensor Terms and Corresponding Radial Equations

3.13.1

Deuteron Phenomenological Optical Potentials Containing Tensor Terms

If the optical potential only contains the spin-independent central potential, the incident particle will not induce any spin polarization phenomenon. In order to consider the particle polarization phenomenon, only the spin-orbital coupling potential needs to be introduced for the reactions of S = 12 particles with spinless targets. For S = 1 particles, not only does the spin-orbital coupling potential of two vectors need to be considered, it is also necessary to consider the tensor potential coupled by two rank 2 tensors of spin space and coordinate space. The tensor potential to be included in the deuteron optical potential must be a scalar, that is, it satisfies the invariance of rotation, and it should also satisfy the parity conservation condition. *

*

In the coordinate space, we have three vectors: coordinate r , momentum p = *

*

*

*

*

*

- i∇, and angular momentum L = - i r × ∇. Here p is the wave vector k . Now, we have the following expression: [22, 46] *

*

*

*

ða  c Þð b  d Þ =

1 * * * * 1 * * * * * * * * ða  b Þ ð c  d Þ þ ða × b Þ  ð c × d Þ þ T 2 ð a , b Þ  T 2 ð c , d Þ 3 2 ð3:927Þ →

→

→

→

*

*

*

*

If in the above formula we let c = d = ^S and a = b = r , or a = b = p , or * → * a = b = L , and note that *

*

*

*

r × r = p × p = 0,

*

*

*

L × L = iL ,

^S × ^S = i^ S,

2 ^ S = 2^I

ð3:928Þ

then the following tensor potentials of the optical potential can be obtained: [46] 1 2 * * * T ð r , r Þ  T 2 ð^S, ^SÞ = ð^ S  r Þ2 =r 2 3 r2 2 2 * * * T p  T 2 ð p , p Þ  T 2 ð^S, ^SÞ = ð^S  p Þ2 - p2 3 * * * * ^ SÞ ^ = ðL  SÞ ^ 2þ1L S ^ - 2 L2 T L  T 2 ðL , L Þ  T 2 ðS, 2 3 Tr 

The component of T 2 ð^S, ^SÞ appearing in the above three formulas is

ð3:929Þ ð3:930Þ ð3:931Þ

3.13

Deuteron Phenomenological Optical Potentials with Tensor Terms. . .

T 2M ð^S, ^SÞ =

X μν

Sμ ^ Sν C21μM 1ν ^

243

ð3:932Þ

Comparing the above formula with Eq. (3.15) we can see 1 T 2M ð^S, ^SÞ = pffiffiffi T^ 2M ð1Þ 3

ð3:933Þ

Using Eqs. (3.16)–(3.18) and (1.23), we can obtain the following expressions: 2 1 T 20 ð^S, ^SÞ = pffiffiffi ð3^S0 -2Þ 6

^ ^SÞ = p1ffiffiffi ð^S1 ^S0 þ ^ S1 Þ S0 ^ 2

ð3:935Þ

^ ^SÞ = ^S2 1

ð3:936Þ

1 ðS,

T2

ð3:934Þ

T2

2 ðS,

* *

The component of T 2 ð r , r Þ appearing in Eq. (3.929) is * *

T 2M ð r , r Þ =

X μν

C21μM 1ν r μ r ν

ð3:937Þ

Using Table 3.3 the following results can be obtained from Eq. (3.937): rffiffiffi   2 2 1  T 20 ð r , r Þ = r þ r þ1 r -1 = pffiffiffi 3r 20 - r 2 3 0 6 pffiffiffi * * T 2 1 ð r , r Þ = 2r 0 r 1 * *

T2

* * 2 2 ð r , r Þ = r 1

ð3:938Þ ð3:939Þ ð3:940Þ

Now the three-dimensional spherical harmonic function is defined as   * Ylm r = r l Ylm ð^r Þ

ð3:941Þ

Also, there is the following relation rffiffiffiffiffi   3 * r Y1μ r = rY1μ ð^r Þ = 4π μ For the spherical harmonic function, there is the following relation

ð3:942Þ

244

3

Yl1 m1 Yl2 m2 = where ^l 

Polarization Theory of Nuclear Reactions for Spin 1 Particles

X ^l1^l2 l m þm pffiffiffiffiffi C l1 m11 l2 m22 C ll100 4π ^l l

l2 0 Yl m1 þm2

ð3:943Þ

pffiffiffiffiffiffiffiffiffiffiffiffi 2l þ 1. Using the C-G coefficient properties and Eq. (3.943) we have X m1 m2

Cll1mm1

Ylm =

l2 m2 Yl1 m1 Yl2 m2

^l ^l = p1ffiffiffiffiffi2 Cll100 l2 0 Ylm , 4π ^l

pffiffiffiffiffi 4π ^l X l m C ^l1^l2 Cl 0 m1 m2 l1 m1 l1 0 l2 0

ð3:944Þ

l2 m2 Yl1 m1 Yl2 m2

Checking Table 3.3 we can get rffiffiffiffiffiffiffiffi 10πX 2M Y2M ð^r Þ = C Y ð^r ÞY1ν ð^r Þ, 3 μν 1μ 1ν 1μ rffiffiffiffiffiffiffiffi 10πX 2M → → → Y2M ð r Þ = C Y ð r ÞY1ν ð r Þ 3 μν 1μ 1ν 1μ

ð3:945Þ

Using Eqs. (3.937), (3.942), and (3.945), we can then obtain the following relation: rffiffiffiffiffi 8π → Y ðrÞ T 2M ð r , r Þ = 15 2M →

→

ð3:946Þ

Further, we can obtain the following relation according to Eqs. (3.929), (3.933), and (3.946): rffiffiffiffiffi 8π Y ð^r Þ  T^ 2 ð1Þ Tr = 45 2

ð3:947Þ

Now the general form of the deuteron spherical phenomenological optical potential can be written as →

*

Vð r Þ = V C ðr Þ þ V c ðr Þ þ iW c ðr Þ þ ½V LS ðr Þ þ iW LS ðr ÞL  ^ S þ½V TR ðr Þ þ iW TR ðr ÞT r þ ½V TL ðr Þ þ iW TL ðr ÞT L

1 þ T p ½V TP ðr Þ þ iW TP ðr Þ þ ½V TP ðr Þ þ iW TP ðr ÞT p 2

ð3:948Þ

where VC(r) is the Coulomb potential, Vc(r) is the real part of the center potential, and Wc(r) is the imaginary part of the center potential which includes both the volume absorption and the surface absorption parts. Nuclear forces include two-body force and three-body force, whereas the optical potential is equivalent to

3.13

Deuteron Phenomenological Optical Potentials with Tensor Terms. . .

245

the average field, and is a one-body potential. Let Ur(r) be the coefficient of the Tr term in the optical potential, and we have *

*

½U r ðr ÞT 2M ð r , r Þþ = ð-1ÞM T 2

* *  - M ð r , r ÞU r ðr Þ

ð3:949Þ

*

Because r and Ur(r) is commutative, U r ðr Þ can be moved to the front of * * T 2 - M ð r , r Þ in the above formula. It can be seen that the real part VTR(r)Tr is Hermitian, and iWTR(r)Tr is anti-Hermitian. Because TL and the corresponding UL(r) is also commutative, it follows that VTL(r)TL is Hermitian, and iWTL(r)TL is anti* Hermitian. However, because p and corresponding Up(r) is not commutative, it is impossible to move U p ðr Þ to the front of T þ p for the momentum term. Nevertheless, using the form of the Tp term in Eq. (3.948) can ensure which real part is Hermitian and which imaginary part is anti-Hermitian. In fact, the central potential and spinorbit coupling potential also have the above properties. In References [47–49], the deuteron universal phenomenological optical potentials are given separately, but none of them contain tensor potential. Reference [49] gives the calculations of the vector polarization quantities. The deuteron optical potentials given in References [13–17, 50] all contain the Tr term, and in References [8–10, 12] the deuteron optical potentials contain the Tr term and the TL term simultaneously. In Reference [51], the necessity of the inclusion of the Tp term in the deuteron optical potential is discussed, and in Reference [17] there is also some discussion on the Tp term, but at present there is a question about whether or not the Tp term needs to be included; besides, the handling of the Tp term in the deuteron optical potential has yet to be further studied, so we will no longer include the Tp term in subsequent discussions. This noted, it is generally believed that the Tr term in the three tensor potentials is the most important, and it plays an important role in fitting experimental data of S = 1 particle polarization analyzing powers.

3.13.2

Deuteron Radial Equations for the Spherical Nuclei when Containing Tensor Potentials *

The stationary Schrodinger equation satisfied by the deuteron wave function ΦJM ð r Þ with total angular momentum J is

-

  ħ2 * 2 * * * ∇ þ V r ΦJM ð r Þ = EΦJM ð r Þ 2μ

ð3:950Þ *

where μ is the deuteron reduced mass. The optical potential Vð r Þ takes the form given by Eq. (3.948), but it does not include the last Tp term. When the target spin is * equal to 0, ΦJM ð r Þ can be written as

246

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles *

ΦJM ð r Þ =

X uJ ð r Þ L

L

LJM =

X C JLMML MLν

r

LJM

L rÞχ 1ν 1ν i YLM L ð^

ð3:951Þ ð3:952Þ

The coordinate tensor operator between the two nucleons is defined as S12 =

3 * * ð^ σ 1  r Þ ð^ σ 2  r Þ - ðσ^1  σ^2 Þ r2

ð3:953Þ

and let ^S = 1 ðσ^1 þ σ^2 Þ 2

ð3:954Þ

Now we can obtain the following expression from Eqs. (3.953) and (3.929): [22]

 

    1 ^2 2 → 2 2 → 2 2 ^ ^ = 6T r S12 = 6 S  r =r - S = 6 S  r =r 3 3

ð3:955Þ

  D E →  In Reference [22], the matrix elements L0 JM  L  ^ SLJM and hL0 JM jS12 jLJM i have been obtained, and we list them in Tables 3.4 and 3.5, respectively: Utilizing Eq. (3.931) we can obtain T L LJM =

n

o 1 1 2 ½J ðJ þ 1Þ -LðL þ 1Þ-22 þ ½J ðJ þ 1Þ- LðL þ 1Þ-2 - LðL þ 1Þ LJM 4 4 3 ð3:956Þ

Note that LJM is an orthogonal normalized wave function, and it follows that the matrix elements hL0 JM jT L jLJM i can be obtained and are listed in Table 3.6 [46]. * The above results show that L  ^S and TL only have the non-zero diagonal elements for the orbital angular momentum L, but Tr has non-zero off-diagonal matrix elements with a difference of 2 for L, so in this case L is no longer a good quantum number. For a definite J, L = J  1, and two states will be coupled, but due to the parity conservation, the L = J state and the L = J  1 states are not coupled, and Table 3.5 clearly shows the above results. Since the deuteron spin S = 1, then it must be that for a definite total angular momentum J, L = J  1 and L = J, and the three states of J must be solved at the same time. When J = 0, there are only two states of L, namely, L = 0, 1. Substituting Eq. (3.951) into Eq. (3.950), multiplying þ L0 JM on both sides of the equation from the left concurrently, and integrating the direction angles of the coordinate space to the obtained equation, the following three equations can be obtained.

3.13

Deuteron Phenomenological Optical Potentials with Tensor Terms. . .

247

→  D E   Table 3.4 Matrix elements L0 JM  L  b SLJM L0 J+1 -(J + 2) 0 0

L J+1 J J-1

J-1 0 0 J-1

J 0 -1 0

Table 3.5 Matrix elements hL0 JM jS12 jLJM i L0 J 0

J+1 2ðJ þ 2Þ 2J þ 1 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6 J ðJ þ 1Þ 2J þ 1

L J+1 J J-1

2 0

J-1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi J ðJ þ 1Þ 2J þ 1 0 2ðJ - 1Þ 2J þ 1 6

Table 3.6 Matrix elements hL0 JM jT L jLJM i

J

J+1 1 ðJ þ 2Þð2J þ 5Þ 6 0

J-1

0

L J+1

L0 J 0 -

1 ð2J - 1Þð2J þ 3Þ 6 0

J-1 0 0 1 ðJ - 1Þð2J - 3Þ 6

 d2 J ðJ -1Þ 2 J J J þ k - U 0 ðr Þ- U LS ðr ÞCJ -1 - U TL ðr ÞBJ -1 - U TR ðr ÞAJ -1,J -1 uJJ -1 dr 2 r2

= U TR ðr ÞAJJ -1,Jþ1 uJJþ1



ð3:957Þ

J ð J þ 1Þ d2 þ k2 - U 0 ðr Þ - U LS ðr ÞC JJ - U TL ðr ÞBJJ - U TR ðr ÞAJJ,J uJJ = 0 r2 dr 2

2

ð3:958Þ

3 2 ð J þ 1 Þ ð J þ 2 Þ d 4 þ k 2 -U 0 ðr Þ -U LS ðr ÞC JJþ1 - U TL ðrÞBJJþ1 - U TR ðr ÞAJJþ1,Jþ1 5uJJþ1 r2 dr 2 ¼ U TR ðr ÞAJJþ1,J -1 uJJ -1 ð3:959Þ

where

248

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

pffiffiffiffiffiffiffiffi 2μE k= ħ U 0 ðr Þ =

2μ ½V C ðr Þ þ V c ðr Þ þ iW c ðr Þ ħ2

2μ ½V x ðr Þ þ iW x ðr Þ, ħ2 8 > < J -1 CJL = -1 > : - ð J þ 2Þ 8 1 > > ðJ -1Þ ð2J - 3Þ > > >

6 > > > 1 > : ðJ þ 2Þ ð2J þ 5Þ 6 U x ðr Þ =

ð3:960Þ

x = LS,TL,TR

ð3:961Þ ð3:962Þ

L = J -1 L=J

ð3:963Þ

L=J þ 1 L = J -1 L=J

ð3:964Þ

L=J þ 1

J -1 1 Jþ2 , AJJ,J = , AJJþ1,Jþ1 = , 3 3ð2J þ 1Þ 3ð2J þ 1Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi J ð J þ 1Þ AJJ -1,Jþ1 = AJJþ1,J -1 = 2J þ 1

AJJ -1, J -1 = -

ð3:965Þ

Equations (3.957) and (3.959) are simultaneous, and Eq. (3.958) is independent. For each total angular momentum J, the above three equations are solved concurrently, and the boundary conditions are linked to the incident channel, so the required S matrix elements can be obtained. It is worth noting that it is not uncommon for the off-diagonal term of Tr to be neglected, as it is only an approximate method [8].

3.14 Folding Model Describing the Reaction Between Deuteron and Nucleus 3.14.1 Deuteron Folding Model Without Breakup Channel [46, 52, 53] The basic assumption of the deuteron single folding model is that the interaction between the deuteron and the target nucleus can be approximately described by a * * potential field U 0 ðR Þ , where U 0 ðR Þ is obtained based on the interactions of the neutron and proton in the deuteron with the target after being averaged by the * deuteron internal wave function φð r Þ . The effective Hamiltonian of the system composed by the neutron and proton in the deuteron and target nucleus is

3.14

Folding Model Describing the Reaction Between Deuteron and Nucleus

       ħ2 * 2 * 2 * * * * H= ∇n þ ∇p þ V n r n þ V p r p þ v r n - r p 2m *

249

ð3:966Þ

*

where V n ð r n Þ and V p ð r p Þ are the potential fields felt by the neutron and proton in the deuteron and are usually regarded as theoptical potentials of the neutron and  * * proton in the deuteron, respectively. Also, v r n - r p are the n - p interactions within the deuteron that determine the deuteron binding energy; this potential belongs to nuclear force. Next we define *

R=

  1 * * rn þ rp , 2

*

*

*

r = rn- rp

ð3:967Þ

where the two expressions represent the center of mass coordinate of the deuteron and the relative coordinate between the neutron and proton, respectively (see Fig. 3.1). Now we can obtain *2

*2

∇n þ ∇ p =

*2 1 *2 ∇R þ 2∇r 2

ð3:968Þ

* *

Letting Ψ ðR , r Þ be the system wave function, we can write the stationary Schrodinger equation of the whole system as

-

 *  *  * * * * 1* 1* ħ2 * 2 ħ 2 * 2 * ∇R - ∇r þ V n R þ r þ V p R - r þ vð r Þ Ψ ðR , r Þ= EΨ ðR , r Þ 4m m 2 2 ð3:969Þ

where E represents the total energy of the system. Next, let *  *  * * 1* 1* VðR , r Þ = V n R þ r þ V p R - r 2 2

Fig. 3.1 Single folding model coordinate system of deuteron

ð3:970Þ

250

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

*

* *

Now, when R is relatively large, it is possible that the variables of Ψ ðR, r Þ can be separated and becomes * *

*

*

Ψ ðR , r Þ = φð r ÞΦðR Þ

ð3:971Þ

*

where ΦðR Þ is a movement wave function of the center of mass of the deuteron with * respect to the target nucleus, and φð r Þ is a relative motion wave function of the neutron and proton inside the deuteron. Now it may be that the deuteron is first absorbed by the target, followed by breakage in the target, with the separated neutron and proton again synthesizing a deuteron, subsequently emitted from the target. This process corresponds to the influence of the non-local potential. Here we ignore this process, for this process can be described by the imaginary part of the optical potential. Therefore, in Eq. (3.971), the components, for which the variables cannot be separated, can be approximately ignored. At the same time, we also assume that the target spin is equal to 0. It follows that Eq. (3.969) can be rewritten as

 * * * * ħ2 * 2 ħ2 * 2 * * * ∇ þ VðR , r Þ - ∇r þ vð r Þ φð r ÞΦðR Þ = Eφð r ÞΦðR Þ 4m R m

ð3:972Þ *

Now the equation satisfied by the internal motion wave function φð r Þ of the deuteron is h

-

i ħ2 * 2 * * * ∇r þ vð r Þ φð r Þ = εφð r Þ m

ð3:973Þ

*

The deuteron has a ground bound state φ0 ð r Þ, ε0 = -2:225 MeV. The other * solutions of Eq. (3.973) are the deuteron breakup states φk ð r Þ with continuous * positive energy εk, where k is a n-p relative motion wave vector such that εk =

ħ2 k 2 , 2μ

μ=

m 2

ð3:974Þ

In this case Eq. (3.971) can be rewritten as * *

*

*

Ψ ðR , r Þ = φ0 ð r ÞΦ0 ðR Þ þ *

*

Z

*

*

*

d k φk ð r ÞΦk ðR Þ

ð3:975Þ *

*

where φ0 ð r Þ and Φ0 ðR Þ all have dimensions fm-3/2, though φk ð r Þ and Φk ðR Þ are all dimensionless. It is only meaningful to calculate the optical potential when there is no deuteron breakage, but there is indeed the possibility of such a breakup when studying deuteron scattering. Therefore, it is necessary to study the influence of the deuteron breakup on the deuteron optical potential. Note that when the energy of the incident

3.14

Folding Model Describing the Reaction Between Deuteron and Nucleus

251

deuteron is relatively low, the probability of deuteron breakup is very small. Therefore, we first study the case that Eq. (3.975) does not include a breakup channel. * * * * * Substituting Ψ ðR , r Þ = φ0 ð r ÞΦ0 ðR Þ into Eq. (3.972), multiplying φþ 0 ð r Þ from * the left on both sides of the equation, and integrating to r , we obtain

-

 * * * ħ2 * 2 ∇R þ V 0 ðR Þ Φ0 ðR Þ = ðE - ε0 ÞΦ0 ðR Þ 4m

ð3:976Þ

where Z

*

V 0 ðR Þ =

*

* *

*

*

φþ 0 ð r Þ VðR , r Þφ0 ð r Þd r

ð3:977Þ

is the deuteron folding optical potential and E0  E - ε0 represents the deuteron incident energy in the c.m. system. Next, we can prove [22] 1 4π

Z    r2 * * σ^1  r σ^2  r dΩ = ðσ^1  σ^2 Þ 3

ð3:978Þ

Now, there is the following relation for S12 given by Eq. (3.953) Z S12 dΩ = 0

ð3:979Þ

The n-p interaction potential containing the tensor term in the deuteron can be written as v = v0 ðr Þ þ vσ ðr Þ^ σ 1  σ^2 þ vT ðr ÞS12

ð3:980Þ

Since v is a two-body nuclear force, tensor terms can be included. The Schrodinger equation satisfied by the deuteron ground state wave function is

 ħ2 * 2 * * σ 1  σ^2 þ vT ðr ÞS12 φ0 ð r Þ = ε0 φ0 ð r Þ ∇ þ v0 ðr Þ þ vσ ðr Þ^ 2μ

ð3:981Þ

For the deuteron, we have that spin S = 1, total angular momentum J = 1, orbital angular momentum l = 0 (S state), and l = 2 (D state); the deuteron ground state wave function can be written as *

φ0 ð r Þ =

uð r Þ w ðr Þ 01M þ 21M r r

ð3:982Þ

252

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

 l1M =

X C 1lmlM1ν il Ylml ðΩÞχ 1ν

ð3:983Þ

ml ν

and note ðσ^1  σ^2 Þχ 1ν = χ 1ν ,

ðσ^1  σ^2 Þχ 00 = - 3χ 00

ð3:984Þ

þ Now, substitute Eq. (3.982) into Eq. (3.981), multiply þ 01M and 21M from the left of the two sides of the equation, respectively, and integrate to the direction angles of the spherical coordinate system. Next, use the results given by Eq. (3.984) and *2

Þ Table 3.5, and note that ∇ = 1r drd 2 r - lðlþ1 r2 . Now when l = 0 there is l(l+1) = 0, and when l = 2 there is l(l+1) = 6; in this way we can obtain 2

pffiffiffi ħ2 d 2 u þ ½v0 ðr Þ þ vσ ðr Þu - ε0 u = 8vT ðr Þw ð3:985Þ 2 2μ dr  pffiffiffi 6 ħ 2 d2 w w þ ½ v ð r Þ þ v ð r Þ w -2v ð r Þw ε w = 8vT ðr Þ u ð3:986Þ 0 σ T 0 2μ dr 2 r2 -

From Eq. (3.983) we can see 1 01M = pffiffiffiffiffi χ 1M 4π

ð3:987Þ

Assuming that Eq. (3.982) can be rewritten as [16, 54]

 1 uðr Þ wðrÞ þ pffiffiffi S12 χ 1M φ0 ð r Þ = pffiffiffiffiffi 4π r 8r *

ð3:988Þ

and noting that the square of any component of the Pauli matrix in Cartesian coordinate system is 1 and satisfies the anti-commutation relation, it can be proved that 

 * 2

σ^  r r2

=1

ð3:989Þ

For χ 1M we can get

S212 = 9 - 6

   * * σ^1  r σ^2  r r2

þ 1 = 8 -2S12

ð3:990Þ

Also, we can obtain the following relations utilizing Eqs. (3.953), (3.979), and (3.990):

3.14

Folding Model Describing the Reaction Between Deuteron and Nucleus

1 4π

1 4π

Z

253

Z

ðS12 Þ3 dΩ =

ðS12 Þ2 dΩ = 8 1 4π

ð3:991Þ

Z ð8 -2S12 ÞS12 dΩ = -16

ð3:992Þ

ffi χ þ or In the process of substituting Eq. (3.988) into Eq. (3.981) and multiplying p1ffiffiffi 4π 1M S12ffiffi þ p1ffiffiffiffi p χ 1M from the left of the above mentioned equations, then integrating the 4π

8

direction angles of the spherical coordinate system and utilizing Eqs. (3.984), (3.979), (3.991), and (3.992), we can obtain Eqs. (3.985) and (3.986), respectively. It can be seen that Eq. (3.988) is the deuteron ground state wave function and satisfies the following normalization condition: Z

1



 u2 ðr Þ þ w2 ðr Þ dr = 1

ð3:993Þ

0 *

*

The neutron optical potential V n ð r n Þ and proton optical potential V p ð r p Þ in 1 Eq. (3.966) are the one-body potentials of spin particles. They can contain spin2 orbital coupling potentials and cannot contain tensor potentials. For spherical optical potentials, they are taken as the following form: *

V n ðr n Þ = V cn ðr n Þ þ V lsn ðr n Þ l n  ^sn        → V p r p = V Cp r p þ V cp r p þ V lsp r p l p  ^sp

ð3:994Þ ð3:995Þ

    where V Cp r p is the proton Coulomb potential and V cp r p is the proton central potential. Now we have the following relations: 1=2 *   1 * r2  2 , rn =  R þ r  = R þ þ Rr cos θ 2 4 1=2 *   1 * r2  r p =  R - r  = R2 þ - Rr cos θ 2 4

ð3:996Þ

and perform the following expansion: V xi ðr i Þ =

X

ð2l þ 1Þ V xi,l ðR, r Þ Pl ð cos θÞ, x = c,ls, i = n;

x = C,c,ls, i = p

l

ð3:997Þ where

254

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

V xi,l ðR, r Þ =

1 2

Z V xi ðr i ÞPl ð cos θÞd cos θ

ð3:998Þ

such that Eq. (3.997) can be rewritten as V xi ðr i Þ = 4π

X ^ lm ð^r Þ V xi,l ðR, r ÞYlm ðRÞY

ð3:999Þ

lm

According to Eqs. (3.955), (3.929), and (3.946) and the definition of the scalar product of the two irreducible tensors given by Eq. (1.41), we can get the following relation: rffiffiffiffiffi 6πX  ^ ^ T ðS, SÞ Y2M ð^r Þ S12 = 4 5 M 2M

ð3:1000Þ

where ^S represents an operator, ^r represents the direction angle in spherical coordinate system, and T 2M ð^S, ^SÞ has been given by Eq. (3.932). Firstly, we can obtain the following expression from Eqs. (3.988) and (3.990):

  1 w w 2 2 p ffiffi ffi p ffiffi ffi ð3:1001Þ u þw þ uS12 4πr2 8 2     For V cn ðr n Þ, V Cp r p , and V cp r p we can obtain the following relations using Eqs. (3.999)–(3.1001) and (3.955): * * φþ 0 ð r Þφ0 ð r Þ =

*

V xi ðR Þ =

Z

Z 1  2  * * * x φþ ð r Þ V ð r Þφ ð r Þd r = u þ w2 V xi,0 ðR, r Þdr 0 0 i i 0   pffiffiffi Z 1 x w V i,2 ðR, r Þw u - pffiffiffi dr T R þ3 2 8 0

ð3:1002Þ

where

TR =

 *2 ^S  R R

2

-

2 3

ð3:1003Þ *

For spin-orbit coupling potential, Eq. (3.1002) cannot be used because l i  ^si and S12 * are not commutable. Set L as the orbital angular momentum of the deuteron center of mass

3.14

Folding Model Describing the Reaction Between Deuteron and Nucleus *

*

*

L = - iR × ∇R

255

ð3:1004Þ

* * *

where l n , l p ,L are all orbital angular momentum relative to the target center of *

mass. Let l be the orbital angular momentum of the relative motion between n-p *

*

*

l = - i r × ∇r

ð3:1005Þ

Now we know that the nucleon spin-orbital potential Vls is a small amount relative to the central potential Vc, and the deuteron D state is a small amount relative to the deuteron S state. Therefore, when we use Eq. (3.977) to find the deuteron folding optical potential, we ignore the influence on Vls by the w part representing the D state * in φ0 ð r Þ. That is, when finding the deuteron spin-orbital coupling potential VLS only * the effect on Vls by the u part representing the S state in φ0 ð r Þ is considered. When only the S-state in the deuteron is considered, l = 0. At this time, for the most * * * extreme situation where r = 0, r n = r p = R , it follows that when l = 0 we should have *

*

*

ln= lp=L

ð3:1006Þ

Now, with some loss of generality, though not unreasonable, let us assume the following special provision:   V lsn ðr n Þ = V lsp r p

ð3:1007Þ

At this time we can obtain *

*

*

S l n  ^sn þ l p  ^sp = L  ^

ð3:1008Þ

where ^S has been given by Eq. (3.954). From here we can get [46] Z V LS ðRÞ =

*

*

*

ls φþ 0 ð r Þ V n ðr n Þφ0 ð r Þd r ffi

Z 0

1

u2 V lsn,0 ðR, r Þdr

ð3:1009Þ

When taking into account only the contribution of the deuteron ground state, according to Eq. (3.977) the folding optical potential of the deuteron is obtained in the following form: *

*

V 0 ðR Þ = V C ðRÞ þ V c ðRÞ þ V LS ðRÞL  ^S þ V T ðRÞT R Substituting Eq. (3.998) into the results obtained above, we can obtain

ð3:1010Þ

256

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

Z   1 1 u2 þ w2 C * 1 * * V ðRÞ = V R r dr   p 4π 0 2 r2 Z   i h  * 1 1 u2 þ w2 c * 1 * 1 *  * V n  R þ r  þ V cp  R - r  d r V c ðRÞ = 2 4π 0 2 2 r Z   1 1 u2 ls * 1 * * dr V R þ V LS ðRÞ = r   4π 0 r 2 n 2 h  Z 1     6 w 1 * 1 * w C * c * p ffiffi ffi V V T ðRÞ = pffiffiffi u R r R þ r þ V     p n 2 2 8 4π 2 0 r 2 i * 1   * * þV cp  R - r  P2 ð cos θÞd r 2 C

ð3:1011Þ ð3:1012Þ ð3:1013Þ

ð3:1014Þ

References [16, 53] present the contribution of the nucleon spin-orbit coupling potential Vls to deuteron optical potential in the presence of deuteron D state, but there are still some problems to be examined. When the l = 2 D-state exists in the deuteron ground state structure, only the tensor force is contained in the nucleon-nucleon interaction, and in this case the experimental data of the deuteron magnetic moment can be explained by theory. Because the sum of the spins of the two nucleons can be 1, there can be tensor force in the two-body forces between the nucleon-nucleon. The optical potentials for the neutrons and protons in the deuteron with target are one-body potential; the tensor potential cannot exist in them. But since the deuteron wave function Eq. (3.988) contains the contribution of tensor force, the tensor potential VT(R)TR appears in the deuteron optical potential given by Eq. (3.1010) obtained through folding the deuteron ground state wave functions by using Eq. (3.977). The discussion above gave out the theoretical basis for the introduction of the tensor potential TR in the deuteron phenomenological optical potential.

3.14.2

Deuteron Folding Optical Potentials Containing Break Channel

When the break channel is included, then substituting Eq. (3.975) into Eq. (3.972) * and multiplying φþ 0 ð r Þ from the left on both sides of the equation, or multiplying * þ * φk ð r Þ from the left on both sides of the equation, and then integrating on r , one can yield the following equations, respectively:

 Z * * * ħ2 * 2 ∇R þ h0jV j0i - ðE - ε0 Þ Φ0 ðR Þ = - d k h0jV jk iΦk ðR Þ 4m

ð3:1015Þ

3.14

Folding Model Describing the Reaction Between Deuteron and Nucleus

257

 Z *0 * * * ħ2 * 2 ∇R þ hk jV jki - ðE - εk Þ Φk ðR Þ = - hk jV j0iΦ0 ðR Þ - d k hkjV jk0 iΦk0 ðR Þ 4m ð3:1016Þ

where Z h0jV j0i =

* *

*

*

*

*

φþ 0 ð r Þ VðR , r Þφ0 ð r Þd r = V 0 ðR Þ

hk jV jk 0 i =

Z

*

* *

*

ð3:1017Þ

*

φþ k ð r Þ VðR , r Þφk 0 ð r Þd r

ð3:1018Þ

When the contribution of the break channel is ignored, Eq. (3.1015) will be reduced to Eq. (3.976). It should be clear that the greatest contribution of the break channel likely comes from the area where k ≈ 0 and k0 ≈ 0. Now, if we omit the hk|V|k0i (k0 ≠ k) term in Eq. (3.1016), then we get

-

 * * ħ2 * 2 ∇R þ hk jV jki - ðE - εk Þ Φk ðR Þ = - hk jV j0iΦ0 ðR Þ 4m

ð3:1019Þ

Referring to Eq. (3.976) again and assuming that the following formula is also approximately true

 * ħ2 * 2 ∇ þ h0jV j0i - ðE - ε0 Þ Φk ðR Þ = 0 4m R

ð3:1020Þ

then from the above two equations we can obtain *

*

Φk ðR Þ =

hkjV j0iΦ0 ðR Þ h0jV j0i - hk jV jk i - εk þ ε0

ð3:1021Þ

Substituting the above equation into Eq. (3.1015), we can get

 * * ħ2 * 2 ∇ þ VðR Þ - ðE - ε0 Þ Φ0 ðR Þ = 0 4m R *

*

ð3:1022Þ

*

VðR Þ = V 0 ðR Þ þ V 1 ðR Þ *

ð3:1023Þ *

where V 0 ðR Þ has been given by Eq. (3.977) or Eq. (3.1017), V 0 ðR Þ represents the contribution of the deuteron bound state, and

258

3 *

V 1 ðR Þ =

Polarization Theory of Nuclear Reactions for Spin 1 Particles

Z

*

dk

h0jV jkihk jV j0i h0jV j0i - hk jV jki - εk þ ε0

ð3:1024Þ

*

It can be seen that V 1 ðR Þ represents the contribution of the deuteron break channel to the deuteron folding optical potential. Referring to Eqs. (3.994), (3.995), and (3.1010), h0|V|ki or hk|V|0i all have potential → S terms; in Eq. (3.1024) they appear by multiplication. Therefore, to contain the l  ^  → 2 in a certain sense, it provides a theoretical basis for including the L  ^ S terms, or angular momentum tensor TL terms, in the deuteron optical potential.

3.15

3.15.1

The Spherical Nucleus CDCC Theory Describing the Breakup Reaction for Incident Loosely Bound Light Complex Particles Development of CDCC Theory

The microscopic three-body theory usually describes the elastic scattering or threebody breakup reaction between the nucleon and the deuteron. The nucleon-nucleon interaction uses the realistic nuclear force. Since the nucleons are fermions, the total wave function of the three-nucleon system must be completely asymmetric. The continuous discretization coupling channel (CDCC) theory [18, 55–72] is usually used to describe the reaction process of the deuteron (or other loosely bound light complex particles) elastic scattering or breakup reaction with a heavier nucleus A. After the deuteron breaking, it becomes a n + p + A three-body problem. n A and p - A interactions are usually described with an optical potential, and the n p interaction is usually described by nuclear force. In CDCC theory, the asymmetry problem is not considered generally. This kind of method can be one considered as an approximation, and the excitation of the target is not considered for spherical nuclei. Therefore, it can be said that CDCC theory only describes the direct reaction process. Since the S-matrix elements of deuteron elastic scattering can be calculated in CDCC theory, the polarization problem of the deuteron can also be studied using CDCC theory [18, 21, 60]. In References [63–65], the elastic scattering and three-body breakup reaction caused by the deuteron are studied. If a loosely bound light complex particle that is heavier than a deuteron is used as an incident particle, the complex particle may split into two groups during the reaction process, and the complex particle in the elastic scattering outgoing channel can be in the ground state or in the excited state. For example, the incident particle 6He can be regarded as the (2n + α) two-body weakly bound state of the double neutron nucleus and α particle; thus they can be described by CDCC theory [66]. It is also possible to regard the 6He as the (n + n +α) threebody loosely bound state of two neutrons and an α particle; then the CDCC theory of

3.15

The Spherical Nucleus CDCC Theory Describing the Breakup Reaction. . .

259

four-body breakup [67–69] is developed in which 6He can be a ground state 0+ or an excited state 2+. Some studies have also introduced the internal freedom of the target in the CDCC theory, using it to study inelastic scattering and two-step reaction processes [70]. In References [71] and [72], the target 6Li(7Li) is considered as the α + d(t) loosely bound state, and the ground state and excited state of the target 6Li (7Li) can be solved with the internal interaction potential between the α and d(t). One then uses the three-body CDCC theory to solve the 6Li(7Li)+n(p) reaction, in which outgoing channel 6Li(7Li) may be in the ground state (elastic scattering), or excited state (inelastic scattering), or breakup. However, this book only introduces the CDCC theory of the deuteron induced reaction, as the deuteron does not have any excited bound state. This section will introduce the CDCC theory of the spherical nuclei, and the next section will give the CDCC theory of the axis-symmetric rotating nuclei.

3.15.2

Spherical Nucleus CDCC Equation

In the center of mass system, the spatial coordinate positions of the neutron and * * proton of the deuteron relative to the target center of mass are r n and r p , respectively. Let →

R=

  1 * * rn þ rp , 2

*

*

*

r = rn- rp

ð3:1025Þ *

Since the masses of neutrons and protons are basically equal, R represents the * coordinate of the deuteron center of mass, and r is the relative coordinate between n and p. The total Hamiltonian of the system is H= -

    ħ2 * 2 ∇R þ U n ðr n Þ þ U p r p þ U C r p þ H np 2μR H np = -

ħ2 * 2 → ∇ þ V np ð r Þ 2μr r

ð3:1026Þ ð3:1027Þ

where the reduced masses are μR = →

md mA , md þ mA

μr =

mn mp mn þ mp

ð3:1028Þ

and V np ð r Þ is the n–p interaction potential, and Un(rn) and Up(rp) are the nucleon potentials of n and p relative to the target center of mass; for the deuteron, they can be replaced by the nucleon optical potential of half the kinetic energy of the deuteron. UC(rp) is the Coulomb potential of the proton relative to the target center of mass; there is no Coulomb interaction between n–p.

260

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

Since the CDCC theory describing the reaction between the deuteron and the spherical nucleus does not include the physical mechanism of the target excited by the internal degree of freedom of the target, that is, there is no operator that can change the target state in the interaction potential, the target spin is assumed to be 0 [18, 60] in the spherical nucleus CDCC theory, as in the spherical optical model. If * the relative orbital angular momentum l inside the deuteron is coupled with the spin → * S of the deuteron, then it is coupled with the deuteron orbital angular momentum L * relative to the target center of mass, and finally it is coupled with the target spin I A, it will be found that taking different value for IA has no effect to the last results. That is, * * * L þ l þ S is the total angular momentum of the system. This shows that target spin IA = 0 should be selected in the spherical nucleus CDCC theory. Therefore, using the spherical nucleus CDCC theory for odd A nucleus is only an approximation. * * We stipulate that l and L are the deuteron orbital angular momentum relative to * * * * r and R , respectively, the spins of n and p are expressed by s n and s p , and we define *

*

*

*

S = s n þ s p,

*

*

j=l þS

*

ð3:1029Þ

*

where S represents the deuteron spin, and j is the total angular momentum of the deuteron. Now, redefine *

*

*

J =Lþ j

ð3:1030Þ

*

where J is the total angular momentum of the system, and νn, νp, ν, m, μ, ML, M are * * ***** used to represent the z components of the angular momentum s n , s p ,S , l , j ,L , J . Let ε0 be the deuteron bound energy such that ε0 = -2.225 MeV. Using ε0 to represent the n-p continuous state relative motion energy after the deuteron breaking, the total energy conservation of the system in the system center of mass frame requires E = E0 þ ε0 = 2 þε,

2 =

ħ2 K 2 2μR

ð3:1031Þ

where the incident deuteron kinetic energy is E0 =

ħ2 K 20 2μR

ð3:1032Þ

and K0 and K are the deuteron momentum wave number of the bound state and the continuous state in the deuteron center of mass system, respectively. Next, let

3.15

The Spherical Nucleus CDCC Theory Describing the Breakup Reaction. . .

ε=

ħ2 k 2 2μr

261

ð3:1033Þ

where k represents the n–p relative motion wave number of the deuteron continuous state. We can see from Eq. (3.1031) that the maximum value of ε is ε max = E0 þ ε0 =

ħ2 k2max 2μr

ð3:1034Þ

We know that deuteron spin S = 1 and total angular momentum j = 1. Let → ψ 0l11μ ð r Þ be the deuteron binding state wave function, its subscript l1 represent the internal orbital angular momentum l and spin S = 1, the subscript 1μ are the deuteron → total angular momentum j = 1 and the corresponding z component μ. ψ 0l11μ ð r Þ satisfies the following Schrodinger equation: 

 → H np - ε0 ψ 0l11μ ð r Þ = 0

ð3:1035Þ

The method to obtain the detailed solution of the above equation is given in * Section 13.5 of Reference [22]. Its solution ψ 0l11μ ð r Þ can be written formally as ϕ0 ð r Þ * ψ 0l11μ ð r Þ = l l11μ ðΩr Þ r X jμ lSjμ ðΩr Þ = C lm Sν il Ylm ðΩr Þχ Sν

ð3:1036Þ ð3:1037Þ

mν

and ϕ0l ðr Þ is the deuteron bound state normalized radial wave function, and χ Sν is the deuteron spin wave function. The deuteron continuous state equation with n–p relative motion wave number k can be written as 

 * H np - ε ψ klSjμ ð r Þ = 0

ð3:1038Þ

where jμ are the total angular momentum and its z component of the deuteron continuous state, and its solution can be formally written as *

ψ klSjμ ð r Þ =

ϕklSj ðr Þ lSjμ ðΩr Þ r

ð3:1039Þ

where ϕklSj ðr Þ is the normalization radial wave function of the deuteron continuous * state. For the deuteron bound state, if only the central potential is included in V np ð r Þ, then only l = 0 S wave is obtained in its solution. If the tensor force belonging to the * non-central potential is also contained in V np ð r Þ, then the l = 2 D wave will be included in its solution in addition to the S wave. Now the key question is how to

262

3 Polarization Theory of Nuclear Reactions for Spin 1 Particles *

select V np ð r Þ. Some studies have chosen the Reid soft core potential for the deuteron bound state [18] , with the following Gaussian nuclear force Vnp(r) for the deuteron continuum state adopted V np ðr Þ = vlSj e

- ða r Þ2

ð3:1040Þ

lSj

In Reference [18], the specific values for the used vlSj and alSj are given. The orbital angular momentum wave function of the deuteron relative to the target center of mass, coupled further with lSjμ given by Eq.(3.1037), becomes φlSjLJM ðΩR , Ωr Þ =

X ML μ

CJLMML

jμ i

L

YLM L ðΩR ÞlSjμ ðΩr Þ

ð3:1041Þ

The total wave function of a system with total angular momentum JM satisfies the following Schrodinger equation: * *

ðH - E Þ Ψ JM ðR , r Þ = 0

ð3:1042Þ

For the bound state, the JM component of the total wave function of the system is * *

Ψ 0JM ðR , r Þ =

X uJ

0l11L ðRÞ

lL

R

ϕ0l ðr Þ φl11LJM ðΩR , Ωr Þ r

ð3:1043Þ

where uJ0l11L ðRÞ is the radial wave function of the bound state deuteron center of mass moving relative to the target center of mass. For the continuous states, the JM component of the total wave function of the system is * *

Ψ kJM ðR , r Þ =

X uJklSjL ðRÞ ϕklSj ðr Þ φlSjLJM ðΩR , Ωr Þ R r lSjL

ð3:1044Þ

where uJklSjL ðRÞ is the radial wave function of the continuous state deuteron center of mass moving relative to the target center of mass. It follows that the JM component of the total wave function of the system can be written as * *

* *

Ψ JM ðR , r Þ = Ψ 0JM ðR , r Þ þ

Z 0

k max

* *

Ψ kJM ðR , r Þ dk

ð3:1045Þ

For the deuteron bound state, there are S = 1 and j = 1, and according to parity conservation, l = 0, 2 can only be taken. For the deuteron continuous state, there are S = 0, 1, l = 0, 2, 4, ⋯, lmax, and j = jl - Sj, . . ., l + S. The larger the incident deuteron energy E0, the larger the lmax; lmax is also related with the weight of the target. When E0 is only few tens of MeV, lmax= 4 is generally taken for the light

3.15

The Spherical Nucleus CDCC Theory Describing the Breakup Reaction. . .

263

target. We then dissociate the integral of k appearing in Eq. (3.1045) and divide 0– kmax into N equal parts. Let Δ=

k max , N

k i = iΔ,

i = 0,1,2,⋯,N

ð3:1046Þ

thus the continuously changing k is divided into [0, k1], [k1, k2], ⋯, [ki-1, ki] , ⋯, [kN-1, kN] N equal parts. Next, introduce the following symbols for the deuteron continuous state wave functions: ϕilSj ðr Þ =

1 pffiffiffi Δ

1 uJilSjL ðRÞ = pffiffiffi Δ

Z Z

ki ki -1 ki

k i -1

ϕklSj ðr Þ dk

ð3:1047Þ

uJklSjL ðRÞ dk

ð3:1048Þ

where Eqs. (3.1039) and (3.1044) can be now rewritten as *

ψ ilSjμ ð r Þ = * *

Ψ iJM ðR , r Þ =

ϕilSj ðr Þ lSjμ ðΩr Þ r

X uJilSjL ðRÞ ϕilSj ðr Þ φlSjLJM ðΩR , Ωr Þ R r lSjL

ð3:1049Þ ð3:1050Þ

Now, concurrently introduce the following symbols for the deuteron bound state wave functions: ( ϕ0lSj ðr Þ =  uJ0lSjL ðRÞ =

ϕ0l ðr Þ, 0,

when S = j = 1, l = 0, 2 otherwise

uJ0l11L ðRÞ,

when S = j = 1, l = 0, 2

0,

otherwise

ð3:1051Þ ð3:1052Þ

Thus Eqs. (3.1036) and (3.1043) can be written in the form of Eqs. (3.1049) and (3.1050) only needing to let i = 0. In this way, Eq. (3.1045) can be rewritten as * *

Ψ JM ðR , r Þ =

N X i=0

* *

Ψ iJM ðR , r Þ

ð3:1053Þ

where i = 0 corresponds to the deuteron bound state and i = 1, 2,. . ., N correspond to the deuteron continuous states. In addition, we define

264

3

εi =

Polarization Theory of Nuclear Reactions for Spin 1 Particles

 ħ2  2 k þ k 2i , 4μr i -1

i = 1,2,⋯,N

ð3:1054Þ

and assume that the radial wave functions after average processing with Eqs. (3.1047) and (3.1048) still satisfy the orthogonal normalization condition, that is, Z

0

i ϕi lSj ðr Þϕl0 S0 j0 ðr Þdr = δilSj,i0 l0 S0 j0

ð3:1055Þ

J uJ ilSjL ðRÞui0 l0 S0 j0 L0 ðRÞdR = δilSjL,i0 l0 S0 j0 L0

ð3:1056Þ

Z

Let VC(R) be the spherical symmetric Coulomb potential felt by the deuteron in the center of mass when assuming that the deuteron charge is at its center of mass. Then substituting Eq. (3.1026) into Eq. (3.1042), we can obtain the following equation for the spherical target:

  * * Lð L þ 1 Þ ħ 2 1 d2 V R ð R Þ H þ E Ψ JM ðR , r Þ C np 2μR R dR2 R2       * * = U n ðr n Þ þ U p r p þ U C r p - V C ðRÞ Ψ JM ðR , r Þ

ð3:1057Þ

In the above equation, we do not ignore the Coulomb distortion effect caused by * * (UC(rp) - VC(R)). Substituting Ψ JM ðR , r Þ given by Eq. (3.1053) into Eq. (3.1057); ϕi ðrÞ

referring to Eq. (3.1050) and multiplying R lSjr φþ lSjLJM ðΩR , Ωr Þ on both sides of the * above equation from the left; integrating over d r dΩR for the obtained equation; and using the orthogonal normalization relation given by Eq. (3.1055) and other relations, we can obtain  2  Lð L þ 1 Þ ħ2 d - V C ðRÞ þ ðE - εi Þ uJilSjL ðRÞ 2μR dR2 R2 X = F JilSjL, i0 l0 S0 j0 L0 ðRÞ uJi0 l0 S0 j0 L0 ðRÞ

ð3:1058Þ

i0 l0 S0 j0 L0

where Z F JilSjL,i0 l0 S0 j0 L0 ðRÞ =

0

i ϕi * * ϕl0 S0 j0 ðr Þ * lSj ðr Þ þ φl0 S0 j0 L0 JM ðΩR , Ωr Þd r dΩR φilSjLJM ðΩR , Ωr ÞUðR , r Þ r r

ð3:1059Þ

3.15

The Spherical Nucleus CDCC Theory Describing the Breakup Reaction. . .

    * * UðR , r Þ = U n ðr n Þ þ U p r p þ U C r p - V C ðRÞ

265

ð3:1060Þ

From here, we can see that Eq. (3.1058) is a coupled channel equation that separates the continuous momentum wave number k of the deuteron breakup channel. An appropriate method can be chosen for finding its numerical solution [73]. Nucleon optical potential can usually be expressed as *

→

SO U N ðr N Þ = U CE NA ðr N Þ þ U NA ðr N Þ l N  s N ,

N = n,p

ð3:1061Þ

* *

Therefore, UðR , r Þ given by Eq. (3.1060) can be divided into two parts. * *

* *

* *

UðR , r Þ = U 1 ðR , r Þ þ U 2 ðR , r Þ     * * CE U 1 ðR , r Þ = U CE nA ðr n Þ þ U pA r p þ U C r p - V C ðRÞ

ð3:1062Þ ð3:1063Þ

 * → * * * * SO U 2 ðR , r Þ = U SO nA ðr n Þ l n  s n þ U pA r p l p  s p

ð3:1064Þ

*

*

Let θ be an angle between r and R . Then we have 1=2 *   1 * r2  2 r n =  R þ r  = R þ þ Rr cos θ , 4 2 1=2 *   1 * r2  r p =  R - r  = R2 þ - Rr cos θ 4 2

ð3:1065Þ

For U CE nA ðr n Þ we can do the following expansion: U CE nA ðr n Þ =

X λ

U CE nA,λ ðR, r Þ =

ð2λ þ 1Þ U CE nA,λ ðR, r Þ Pλ ð cos θÞ 1 2

ð3:1066Þ

Z U CE nA ðr n Þ Pλ ð cos θ Þ d cos θ

ð3:1067Þ

such that Eq. (3.1066) can be rewritten as U CE nA ðr n Þ = 4π

X λ mλ

 U CE nA,λ ðR, r ÞYλ mλ ðΩR ÞYλ mλ ðΩr Þ

ð3:1068Þ

  Also,U CE pA r p and UC(rp) can be expanded in the same way. Now Eq. (3.1059) can be divided into two parts according to Eq. (3.1062):

266

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles J ð1Þ

J ð2Þ

F JilSjL,i0 l0 S0 j0 L0 = F ilSjL,i0 l0 S0 j0 L0 þ F ilSjL,i0 l0 S0 j0 L0

ð3:1069Þ

* *

Since there is no spin operator in U 1 ðR , r Þ , it follows that according to Eqs. (3.1041) and (3.1037) we can write the following expression: ! X jμ X ϕi * * lSj ðr Þ J M -L  -l  C LM L jμ i YLM L ðΩR Þ Clm Sν i Ylm ðΩr Þ U 1 ðR , r Þ r ν m ML μ 1 i0 ϕl0 S j0 ðrÞ X j0μ0 0 * L0 l 0 0 Cl0 m0 Sν i Yl0 m0 ðΩr ÞAd r dΩR j0 μ0 i YL M L ðΩR Þ r 0 m

J ð1Þ F ilSjL,i0 l0 S0 j0 L0 ðRÞ = δSS0

0 @

X

M 0L μ0

M CJL0 M 0 L

XZ

ð3:1070Þ * *

Next, we first find the contribution to the above formula by the first item of U 1 ðR , r Þ given by Eq. (3.1063). Using Eq. (3.1068) and paying attention to the following relations Z

^l 0^λ Ylm ðΩr ÞYλmλ ðΩr ÞYl0 m0 ðΩr ÞdΩr = ð-1Þmλ pffiffiffiffiffi Cll0 mm0 4π ^l Z

^0^λ L ML YLM L ðΩR ÞYλmλ ðΩR ÞYL0 M 0 L ðΩR ÞdΩR = pffiffiffiffiffi C LL0 M 0 L ^ 4π L

l 0 λ - mλ C l0 0 λ0

ð3:1071Þ

L 0 λmλ C L0 0 λ0

ð3:1072Þ

and ordering Z

Sλð1, 1Þ

Bilj,i0 l0 j0 ðRÞ = 4π

0

1

0

CE i ϕi lSj ðr ÞU nA,λ ðR, r Þϕl0 Sj0 ðr Þdr

ð3:1073Þ

we can obtain the following expression according to Eq. (3.1070): J ð1, 1Þ

F ilSjL,i0 l0 S0 j0 L0 ðRÞ = δSS0 iL X

0

- Lþl0 - l

X Sλð1, 1Þ ^l 0 L ^0^λ2 l0 Bilj,i0 l0 j0 ðRÞ Cl0 0 λ0 C L0 L0 0 λ0 ^lL ^ λ

C JLMML jμ C jlmμ Sν ð -1Þmλ C ll0 mm0

j0 μ0 L ML J M λ - mλ C L0 M 0L λmλ C L0 M 0L j0 μ0 C l0 m0 Sν

ð3:1074Þ

ν MLμ m mλ M 0L μ0 m0

Because Eq. (3.1074) contains the factor Cl0 l0 0 λ0 , λ is an integer and there is l + l0 + λ = even; thus, using Eq. (3.14) we can obtain

3.15

The Spherical Nucleus CDCC Theory Describing the Breakup Reaction. . .

X mm0 ν

Cjlmμ Sν C ll0 mm0

j 0 μ0 λ - mλ C l0 m0 Sν

0

= ^l^j Cjλ μ- mλ

j0 μ0 W ðλl

0

jS; lj0 Þ

267

ð3:1075Þ

And there is the following C-G coefficient formula X

C eε aα

cγ fφ cγ bβ C eε dδ C bβ dδ C aα f φ

= ^e^f W ðabcd; ef Þ

ð3:1076Þ

αβ δεφ

using this formula we can get X

CJLMML

jμ ð-1Þ

mλ

ML C LL0 M 0 L

λ mλ

M L μM 0L μ0 mλ

M C JL0 M 0 L

jμ j0 μ0 C λ - mλ j0 μ0

0

^ ^jW ðL0 λJj; lj0 Þ = ð-1Þj - j L ð3:1077Þ

Next, the following expression can be obtained substituting Eqs. (3.1075) and (3.1077) into Eq. (3.1074): J ð1, 1Þ

F ilSjL,i0 l0 S0 j0 L0 ðRÞ = δSS0 iL

0

- Lþl0 - l

X0 0 20 ^l L ^ ^λ ^j ^j C l00 C L00 l 0 λ0 L 0 λ0

0

ð-1Þj - j

λ

0

0

0

0

Sλð1, 1Þ

W ðλl jS; lj ÞW ðL λJj; Lj ÞBilj,i0 l0 j0 ðRÞ = δSS0 iL

0

- Lþl0 - l

0

ð-1Þj - j

X - S - J ^ ^ 0^^ 0^^0 λ0 Cλ0 LL l l jj l0 l0 0 C L0 L0 0 λ Sλð1, 1Þ

W ðll0 jj0 ; λSÞW ðLL0 jj0 ; λJ ÞBilj,i0 l0 j0 ðRÞ ð3:1078Þ   Similar derivations can be done for the second and third items U CE pA r p and UC(rp) of Eq. (3.1063). Based on Eq. (3.1070) the fourth item -VC(R) can be easily obtained as J ð1, 4Þ

F ilSjL,i0 l0 S0 j0 L0 ðRÞ = - V C ðRÞδilSjL,i0 l0 S0 j0 L0

ð3:1079Þ

Now, we can obtain the following results using Eqs. (3.1078) and (3.1079): J ð1Þ

F ilSjL,i0 l0 S0 j0 L0 ðRÞ = δSS0 iL

0

- Lþl0 - l

ð-1Þj - j

0

X - S - J ^ ^ 0^^ 0^^0 λ0 Cλ0 LL l l jj l0 l0 0 C L0 L0 0 λ

Sλð1Þ

W ðll0 jj0 ; λSÞW ðLL0 jj0 ; λJ ÞBilj,i0 l0 j0 ðRÞ - V C ðRÞδilSjL,i0 l0 S0 j0 L0 ð3:1080Þ where

268

3

Sλð1Þ Bilj,i0 l0 j0 ðRÞ = 4π

Z

1

0

Polarization Theory of Nuclear Reactions for Spin 1 Particles

h i 0 CE CE i ϕi lSj ðr Þ U nA,λ ðR, r Þ þ U pA,λ ðR, rÞ þ U C,λ ðR, rÞ ϕl0 Sj0 ðr Þdr ð3:1081Þ

where U CE pA,λ ðR, r Þ and UC, λ(R, r) can be found using the same method used to find Eq. (3.1067). For the spin-orbit coupling potential given by Eq. (3.1064), Eq. (3.1029) yields *

*

*

S = sn þ sp =

1 → →  σnþ σp 2

ð3:1082Þ

In the deuteron bound state, the S wave is the main component. The deuteron continuous state evolves from the deuteron bound state, so the S wave is also the main component. Therefore, for the deuteron spin-orbit coupling potential, the S-wave approximation method given in the previous section can be used. By limiting   SO U SO nA ðr n Þ = U pA r p

ð3:1083Þ

and assuming that the deuteron has a spin-orbit coupling potential only when l = 0, then the following relation can be obtained from Eq. (3.1064) using Eqs. (3.1029) and (3.1006): * * * * ð r Þ L  S δl0 U 2 ðR , r Þ = U SO n nA →

*

ð3:1084Þ *

where S has been given by Eq. (3.1082). Because L  S and U SO nA ðr n Þ are not commutated, the above equation can be rewritten as * *

U 2 ðR , r Þ =

1 2

h* * * *i SO L  S U SO ð r Þ þ U ð r Þ L  S δl0 n n nA nA

ð3:1085Þ

Now we have the following relations according to Eqs. (3.1029) and (3.1030): *

*

*

*

J =Lþ l þ S

  * * 1 * 2 * 2 *2 *2 * * * LS= J - L - l - S -2 l  L þ S 2

ð3:1086Þ ð3:1087Þ

Referring to Eq. (3.1080), we can write the following expression using Eqs. (3.1085) and (3.1087):

3.15

The Spherical Nucleus CDCC Theory Describing the Breakup Reaction. . .

J ð2Þ

F ilSjL,i0 l0 S0 j0 L0 ðRÞ = δSS0 iL

0

- Lþl0 - l

ð-1Þj - j

0

269

X - S - J ^ ^ 0^^ 0^^0 C λl00l0 0 C λL00 L0 0 W ðll0 jj0 ; λSÞ LL l l jj λ



 1 1 Sλð2Þ 0 0 W ðLL jj ; λJ Þ J ð J þ 1Þ LðL þ 1Þ þ L ðL þ 1Þ þ 2SðS þ 1Þ Bilj,i0 l0 j0 δl0 δl0 0 2 2 0 0

ð3:1088Þ where Sλð2Þ Bilj,i0 l0 j0 ðRÞ = 4π

Z 0

1

0

SO i ϕi lSj ðr ÞU nA,λ ðR, r Þϕl0 Sj0 ðr Þdr

ð3:1089Þ

where U SO nA,λ ðR, r Þ can be obtained using the same method to find Eq. (3.1067). Equations (3.1080) and (3.1088) show that S is a conservation quantity during the entire reaction process. Because S=1 for the deuteron bound state, S =1 is maintained throughout the reaction process.

3.15.3

Angular Distribution of Deuteron Elastic Scattering

The spherical nucleus CDCC equation (3.1058) describes the process of elastic scattering and breakup of the deuteron. When R is large enough, the nuclear potential contained in the term on the right side of this equation will tend to 0, and UC(rp) VC(R) will also tend to 0. In this case, the solutions of this equation become its asymptotic solutions FL(KiR) and GL(KiR), where Ki =

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2μR ðE - εi Þ , ħ

i = 0,1,⋯,N

ð3:1090Þ

and F1 and GL are the regular and irregular Coulomb wave functions, respectively. The speed of the deuteron center of mass is Vi =

ħK i μR

ð3:1091Þ

In the center of mass system, the incident wave function of the deuteron relative to the target A is *    → * 1 * * Ψ μðinÞ R , r = pffiffiffiffiffiffi ei K 0 R ψ μ r V0

ð3:1092Þ

270

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

  * where ψ μ r is a normalization bound state wave function of the deuteron with the total angular momentum magnetic quantum number μ. Referring to Eq. (3.1036), we can write ψ

  X ϕ0 ð r Þ * l r = l11μ ðΩr Þ r μ l

ð3:1093Þ

thus Eq. (3.1092) can be rewritten as [7] 



* * Ψ ðμinÞ R , r

rffiffiffiffiffiffi X ϕ0 ð r Þ 4π X^ iσðL0Þ F L ðK 0 RÞ L l Le = i YL0 ðθR , 0Þ l11μ ðΩr Þ ð3:1094Þ K0R V0 L r l

ð0Þ

where σ L is a Coulomb phase shift corresponding to the deuteron bound state. Using Eq. (3.1041) we can write iL Yl0 l11μ =

X

CJL0μ 1μ φl11LJμ

ð3:1095Þ

J

and Eq. (3.1094) can be rewritten as Ψ ðμinÞ

rffiffiffiffiffiffi ϕ0 ð r Þ 4π X^ J μ iσðL0Þ 1 LC L0 1μ e F L ðK 0 RÞ l = φl11LJμ r K 0 R V 0 lLJ pffiffiffi ϕ0 ð r Þ i π X^ J μ iσðL0Þ pffiffiffiffiffiffi = LC L0 1μ e ½ðGL - iF L Þ - ðGL þ iF L Þ l φl11LJμ r K 0 R V 0 lLJ ð3:1096Þ

In an isolated system where n–p elastic scattering occurs, the total angular momentum jμ is the conserved quantity. If the n–p interaction potential given by Eq. (3.1040) is taken, for which the tensor force is not included, and in this case, the n–p relative motion orbital angular momentum l is also a good quantum number. For discrete energy εi, from Eq. (3.1038) we have 

   * H np - εi ψ iljμ r = 0

ð3:1097Þ

  * The form of ψ iljμ r has been given by Eq. (3.1049), but S = 1 is no longer marked here. Assuming that the n–p collision takes place along the z-axis direction, the initial total wave function for n–p scattering is [7]

3.15

The Spherical Nucleus CDCC Theory Describing the Breakup Reaction. . .

Ψ ðνinn νÞp

!   rffiffiffiffiffi X 4π * l ^ l ðki r Þ i Yl0 ðθ, 0Þ χ 1ν χ 1ν Lj r = 2 n 2 p νi l

271

ð3:1098Þ

The n–p relative motion speed is vi =

ħki μr

ð3:1099Þ

where jl is the spherical Bessel function. Taking the spin coupled wave functions as χ 1ν =

X νn νp

C 11νnν 1νp χ 12νn χ 12νp

ð3:1100Þ

X C 11νnν 1νp χ 1ν

ð3:1101Þ

χ 12νn χ 12νp =

2

ν

2

2

2

and utilizing Eq. (3.1037), we can get il Yl0 χ 1ν =

X

C jν l0

1ν l1jν

ð3:1102Þ

j

Note that when there is no Coulomb interaction Eq. (3.1098) can be rewritten as  

* Ψ ðνinn νÞp r

F l ðkr Þ kr

→ jl ,

Gl ðkr Þ kr

→ - nl ðkr Þ; thus,

rffiffiffiffiffi pffiffiffi i π X^ j ν 1 ν 4πX^ j ν 1 ν p lC lC C 1 1 j ðk r Þl1jν = ffiffiffiffi C1 1 = νi ljν l0 1ν 2νn 2νp l r νi ljν l0 1ν 2νn 2νp f½ - nl ðki rÞ - ijl ðki r Þ - ½ - nl ðki rÞ þ ijl ðk i r Þgl1jν ð3:1103Þ

If n–p elastic scattering occurs, and its energy εi, angular momentum l, and S = 1 have not changed, then, when r is large enough to ignore the nuclear potential, the corresponding total wave function becomes   ipffiffiπffi X * ^lC j ν C 11 ν 1 Ψ ðνtnÞνp r = pffiffiffiffi νi ljν l0 1ν 2νn 2νp

½ - n ðk r Þ - ijl ðk i r Þ - silj ½ - nl ðki r Þ þ ijl ðki r Þ l1jν rffiffiffiffiffi l i    4πP^ j ν 1 ν i lC = C j ðk r Þ þ 1 - silj ½ - nl ðk i r Þ þ ijl ðki r Þ l1jν νi ljν l0 1ν 12νn 12νp l i 2 ð3:1104Þ where silj is a n–p elastic scattering S-matrix element. Letting the outer boundary of r be rm, then when r ≥ rm, the nuclear potential between n–p can be ignored.

272

3 Polarization Theory of Nuclear Reactions for Spin 1 Particles

We stipulate that the symbol “ 0 ” represents the derivative for kir. According to Eq. (3.1049) and the first equation of Eq. (3.1104), the boundary condition can be written as 

0   0 0 0 0  ϕilj ðr Þ   = - ðki r nl ðk i r ÞÞ -iðki r jl ðk i rÞÞ - silj - ðk i r nl ðk i rÞÞ þ iðk i r jl ðk i r ÞÞ  ½ - ðk i r nl ðk i r ÞÞ -iðki r jl ðk i rÞÞ - silj ½ - ðk i r nl ðk i rÞÞ þ iðki r jl ðk i r ÞÞ rm ϕilj ðr Þ rm ð3:1105Þ

where ϕilj ðr Þ is obtained by numerically solving the Schrodinger equation in the n–p system inner region of the existence of the nuclear potential. The S matrix element silj of the n–p elastic scattering can be obtained by using the boundary conditions given by the above equation. Note that r→1

- nl ðki rÞ þ i jl ðk i r Þ

→ ð - iÞl

eiki r ki r

ð3:1106Þ

thus we can find that the final state total wave function is       * * * Ψ ðνfnÞνp r = Ψ ðνtnÞνp r - Ψ ðνinn νÞp r pffiffiffi  i π X^ j ν 1 ν  = pffiffiffiffi lC l0 1ν C 1νn 1νp 1 - silj ½ - nl ðki r Þ þ ijl ðk i r Þl1jν 2 2 νi ljν pffiffiffi X  eiki r r→1 i π ^ jν 1ν 



→ pffiffiffiffi lC l0 1ν C 1νn 1νp 1 - silj 2 2 ki r νi ljν X jν 0 0 1 ν þν C l ν - ν0n - ν0p 1 ν0n þν0p C 1ν0 n1ν0 p Yl v - ν0n - ν0p ðΩr Þχ 12ν0n χ 12ν0p

ð3:1107Þ

2 n 2 p

ν0n ν0p

From the above expression, the n–p elastic scattering amplitude can be written as ðiÞ

f νn 0 νp 0 ,νn νp ðΩr Þ =

pffiffiffi  i π X^ j ν 1 v  lC l0 1ν C 1νn 1νp 1 - silj 2 2 k i ljν

1 ν0 þν0 Cjl νν - ν0n - ν0p 1 ν0n þν0p C 1ν0 n1ν0 p Yl ν - ν0n - ν0p ðΩr Þ 2 n 2 p

ð3:1108Þ

If the deuteron breaks under the interaction with the target, the outgoing channels may be i = 1, 2, . . ., N. We assume that n–p is an isolated system during the derivation process of Eqs. (3.1097)–(3.1108) of n–p elastic scattering. However, n–p is not an isolated system in the d + A → n + p + A reaction process. In fact, the interactions of n and p with A cause the deuteron to be changed from a bound state to a continuous state after breaking, and jμ are no longer the conservation quantities; but here we introduce the elastic scattering channel approximation [61, 74, 75]. That is, the radial wave function ϕilj ðr Þ of the n–p relative motion of the deuteron breakup

The Spherical Nucleus CDCC Theory Describing the Breakup Reaction. . .

3.15

273

channel is obtained by the occurrence of n–p elastic scattering. The corresponding S matrix element silj is only related to the outgoing channel subscripts, and the solution method has been given by Eq. (3.1105). If there is only a central potential between the deuteron and the spherical target nucleus, the orbital angular momentum L is a conservation quantity when the deuteron undergoes an elastic scattering reaction. However, in the CDCC theory, the deuteron feels the folding potential. It can be seen from Eq. (3.1058) that this potential is non-central, and the elastic scattering channel is coupled with the breakup channel, so L is no longer a conservation quantity. It follows that, according to Eq. (3.1096), the total wave function of the system after the reaction can be written as Ψ ðμtÞ

pffiffiffi ϕ0 ð r Þ i π X^ J μ iσðL0Þ 1 pffiffiffiffiffiffi ðGL ðK 0 RÞ - iF L ðK 0 RÞÞ l LC L0 1μ e = φl11LJM K 0 R lLJ r V0 -

X 1 ϕ00 ðr Þ pffiffiffiffiffiffi SJ0l0 11L0 ,0l11L ðGL0 ðK 0 RÞ þ iF L0 ðK 0 RÞÞ l φl0 11L0 JM r V0 l 0 L0

N X X 1 pffiffiffiffiffiffi SJi0 l0 1j0 L0 ,0l11L si0 l0 j0 ðGL0 ðK i0 RÞ þ iF L0 ðK i0 RÞÞ V i0 i0 = 1 l0 j0 L0  0 0 0 0 0 0 0 ð - nl ðki r Þ þ ijl ðki r ÞÞ φl 1j L JM

ð3:1109Þ

The first item in the square brackets of the above expression is the spherical incident wave; the second item is the spherical outgoing wave of the elastic scattering channel; and the third item is the spherical outgoing wave of the breakup channel in which si0 l0 j0 is used, indicating that the elastic scattering channel approximation is used. The n–p continuous state is actually created because of the breakup of the deuteron, but here it is approximated that when R is large enough the n-p continuous state is obtained by the occurrence of n–p elastic scattering. Letting the outer boundary of R be Rm, then when R ≥ Rm the nuclear potential and Coulomb distortion potential between the deuteron and the target can be ignored. We stipulate that the symbol “ 0 ” represents the derivative for K0R, so the boundary conditions for the elastic scattering process of the deuteron center of mass can be written according to Eqs. (3.1043) and (3.1109) as follows:  ðu0l0 11L0 ðRÞÞ0  u0l0 11L0 ðRÞ Rm   0   GL ðK 0 RÞ - iF 0L ðK 0 RÞ δl0 L0 , lL - S0lJ 0 11L0 ,0l11L G0L0 ðK 0 RÞ þ iF 0L0 ðK 0 RÞ  ¼  ðGL ðK 0 RÞ - iF L ðK 0 RÞÞδl0 L0 , lL - S0lJ 0 11L0 ,0l11L ðGL0 ðK 0 RÞ þ iF L0 ðK 0 RÞÞ 

Rm

ð3:1110Þ

274

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

where uJ0l0 11L0 ðRÞ is obtained by numerically solving the coupling channel equation (3.1058) in the internal region of the existence of the nuclear potential, which includes the influence of the breakup channel on the elastic scattering channel. In the actual calculation, the elastic scattering channel and breakup channel must be solved simultaneously, and Eq. (3.1110) is just the equation to determine the S matrix element SJ0l0 11L0 ,0l11L . Now, we can rewrite Eq. (3.1109) as Ψ ðμtÞ =

pffiffiffiffiffi X

0 ð0Þ 4π ^ JL0μ1μ eiσ L p1ffiffiffiffiffiffi F L ðK 0 RÞ ϕl ðr Þ φl11LJM LC r K 0 R lLJ V0 þ

X l0 L0

  ϕ00 ðr Þ i pffiffiffiffiffiffi δl0 L0 ,lL - SJ0l0 11L0 ,0l11L ðGL0 ðK 0 RÞ þ iF L0 ðK 0 RÞÞ l φl0 11L0 JM r 2 V0

N X X

i pffiffiffiffiffiffi SJi0 l0 1j0 L0 ,0l11L si0 l0 j0 ðGL0 ðK i0 RÞ þ iF L0 ðK i0 RÞÞ 2 V i0 i =1 l j L  ð - nl0 ðki0 r Þ þ ijl0 ðki0 r ÞÞ φl0 1j0 L0 JM

-

0

0 0 0

ð3:1111Þ and note R→1

0

ð i0 Þ

GL0 ðK i0 RÞ þ iF L0 ðK i0 RÞ

→ eiK i0 R ð - iÞL eiσL0

ð3:1112Þ

Recalling Eq. (3.1106), we can get the final state total wave function from Eqs. (3.1111) and (3.1096) as follows: = Ψ ðμtÞ

X

  ϕ00 ðr Þ 1 pffiffiffiffiffiffi δl0 L0 ,lL - SJ0l0 11L0 ,0l11L ðGL0 ðK 0 RÞ þ iF L0 ðK 0 RÞÞ l φl0 11L0 JM r V0

l0 L0

N X X

- Ψ ðμinÞ

pffiffiffi i π X ^ J μ iσ ðL0Þ LC L0 1μ e = K 0 R lLJ

Ψ ðμoutÞ

1 pffiffiffiffiffiffi SJi0 l0 1j0 L0 ,0l11L si0 l0 j0 ðGL0 ðK i0 RÞ þ iF L0 ðK i0 RÞÞ V i0 i0 = 1 l0 j0 L0  ð - nl0 ðki0 r Þ þ ijl0 ðki0 r ÞÞφl0 1j0 L0 JM

-

3.15

The Spherical Nucleus CDCC Theory Describing the Breakup Reaction. . .

→

275

ð 0Þ pffiffiffi  i π X ^ J μ iσðL0Þ X eiσL0  pffiffiffiffiffiffi δl0 L0 ,lL - SJ0l0 11L0 ,0l11L LC l0 1μ e K 0 ljJ V0 0 0 lL

e

iK 0 R

R

ϕ0l0 ðr Þ X J μ CL0 μ - μ0 1μ0 YL0 μ - μ0 ðΩR Þl0 11μ0 ðΩr Þ r μ0 i0

ð Þ N X X eiσL0 eiK i0 R eiki0 r pffiffiffiffiffiffi SJi0 l0 1j0 L0 ,0l11L si0 l0 j0 R r V i0 k i0 i0 = 1 l0 j0 L0

X Jμ 0 0 C L0 μ - μ0 j0 μ0 Cjl0 μμ0 - ν0 μ0 ν 0

1ν0

YL0 μ - μ0 ðΩR ÞYl0



μ0 - ν0 ðΩr Þχ 1ν0

ð3:1113Þ

where Eqs. (3.1041) and. (3.1037) were also used in the final step of the derivation of Eq. (3.1113). Now it is known that the outgoing wave of the Coulomb scattering is [7]   1 * Ψ ðμCÞ → pffiffiffiffiffiffi f C ðθR Þ eiK 0 R ψ μ r V 0R

ð3:1114Þ

It follows that from the above two expressions, the final state wave function of the deuteron elastic scattering can be written as

pffiffiffi X ϕ0 ð r Þ i π X^ J μ iσðL0Þ X iσð00Þ 1 l LCL0 1μ e e L Ψ ðμrÞ = pffiffiffiffiffiffi f C ðθR Þ l11μ ðΩr Þ þ K 0 lLJ r V0 l l0 L0  iK R  X ϕ00 ðr Þ e 0 l0 11μ0 ðΩr Þ δl0 L0 ,lL - SJ0l0 11L0 ,0l11L CJL0μμ - μ0 1μ0 YL0 μ - μ0 ðΩR Þ l r R μ0 ð3:1115Þ and from this expression, the deuteron elastic scattering amplitude can be written as pffiffiffi i π X^ J μ iσðL0Þ LC L0 1μ e K 0 LJ  X ð 0Þ  eiσ L0 δl0 L0 ,lL - SJ0l0 11L0 ,0l11L CJL0μμ - μ0 1μ0 YL0 μ - μ0 ðΩR Þ

f l0 μ0 ,lμ ðΩr Þ = f C ðθR Þ δl0 μ0 ,lμ þ

ð3:1116Þ

L0

such that both incident and outgoing waves can be S waves or D waves. With this elastic scattering amplitude, various corresponding polarization quantities of the deuteron can be studied. In the center of mass system, the deuteron elastic scattering angular distribution is

276

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

2 1 X  dσ = f l0 μ0 ,lμ ðΩR Þ  dΩR 3 0 0

ð3:1117Þ

lμ l μ

where the deuteron elastic scattering angular distribution in the laboratory system can be obtained through the coordinate system transformation for the above expression [7]. The deuteron elastic scattering angular distribution obtained here includes the influence of the breakup channel; that calculated by the ordinary deuteron optical model does not take into account the influence of the breakup channel. The problem of calculating the absorption cross section of the deuteron induced nuclear reaction, using the obtained S-matrix element SJ0l0 11L,0l11L , can be further studied here.

3.15.4 Double Differential Cross Section of the Ejected Nucleons After Deuteron Breaking Stipulating that the symbol “ 0 ” represents the derivative for Ki'R, the boundary conditions for the deuteron breakup channel (1  i  N ) in the case of the introduction of elastic scattering channel approximation can be written according to Eqs. (3.1050) and (3.1109) as follows: 0  uiJ0 l0 1j0 L0 ðRÞ   uiJ0 l0 1j0 L0 ðRÞ   Rm   0   GL ðK i0 RÞ - iF 0L ðK i0 RÞ δi0 l0 j0 L0 ,0l1L - SiJ0 l0 1j0 L0 ,0l11L G0L0 ðK i0 RÞ þ iF 0L0 ðK i0 RÞ  =  ðGL ðK i0 RÞ - iF L ðK i0 RÞÞδi0 l0 j0 L0 ,0l1L - SiJ0 l0 1j0 L0 ,0l11L ðGL0 ðK i0 RÞ þ iF L0 ðK i0 RÞÞ 



Rm

ð3:1118Þ It is important to note that the boundary condition Eq. (3.1110) of elastic scattering is only a special case of the above equation, and the boundary condition Eq. (3.1118) can be used uniformly. The above equation is the equation to determine the S matrix element SJi0 l0 1j0 L0 ,0l11L . The reaction amplitude of the breakup channel can be written by Eq. (3.1113) as ð i0 Þ pffiffiffi N i π X^ J μ iσðL0Þ X X eiσL0 J LC L0 1μ e f ν0 ,μ ðΩR , Ωr Þ = S00 0 0 s000 K 0 lLJ ki0 i l 1j L ,0l11L i l j i0 = 1 l0 j0 L0 X Jμ 0 0 CL0 μ - μ0 j0 μ0 C jl0 μμ0 - ν0 1ν0 YL0 μ - μ0 ðΩR Þ Yl0 μ0 - ν0 ðΩr Þ μ0

ð3:1119Þ

3.15

The Spherical Nucleus CDCC Theory Describing the Breakup Reaction. . .

277

Recall that εi is defined by Eq. (3.1054). If ki-1  k  ki, then the amount related to k is obtained through interpolation in the interval [ki-1, ki]. Using c = 0 to represent the bound state, and c ≠ 0 to represent the continuous state, Eq. (3.1119) can be rewritten as f ν0 ,μ ðΩR , Ωr Þ = -

pffiffiffi 0 0 i π X^ J μ iσðL0Þ X J μ LC L0 1μ e C L0 μ - μ0 j0 μ0 Cjl0 μμ0 - ν0 1ν0 K 0 lLJ 0 0 0 0 ljLμ

Z k max eiσ L0 ðk Þ J YL0 μ - μ0 ðΩR ÞYl0 μ0 - ν0 ðΩr Þ ðk Þscl0 j0 ðkÞdk S 0 0 0 kΔ cl 1j L ,0l11L 0

ð3:1120Þ

and stipulate that SJcl0 1j0 L0 ,0l11L ð0Þ = SJcl0 1j0 L0 ,0l11L ðk max Þ = scl0 j0 ð0Þ = scl0 j0 ðk max Þ = 0

ð3:1121Þ

Note that k = 0 still belongs to the c ≠ 0 continuous state, and the difference of the c = 0 bound state and the k = 0 continuous state is the binding energy ε0. So according to Eq. (3.1121), the five-dimensional reaction amplitude of the breakup channel can be written as pffiffiffi i π X^ J μ iσ ðL0Þ X J μ LC L0 1μ e f ν0 ,μ ðk, ΩR , Ωr Þ = C L 0 μ - μ0 K 0 lLJ 0 0 0 0 ljLμ

0

j0 μ0

C jl0

μ0 μ0 - ν0 1ν0

eiσ L0 ðkÞ J ðk Þscl0 j0 ðkÞYL0 μ - μ0 ðΩR ÞYl0 μ0 - ν0 ðΩr Þ S 0 0 0 kΔ cl 1j L ,0l11L ð3:1122Þ The corresponding five-differential cross sections in the center of mass system are 2 d5 σ 1 X = f ν0 ,μ ðk, ΩR , Ωr Þ  dk dΩR dΩr 3 μ ν0

ð3:1123Þ

Also, note the relation Z

Yl00 μ00 - ν0 ðΩr ÞYl0

μ0 - ν 0 ð Ω r Þ

dΩr = δl00 μ00 ,l0 μ0

ð3:1124Þ

From Eq. (3.1123) we can find that the double differential cross section of the deuteron center of mass in the center of mass system is

278

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

2 d2 σ 1 X  = f 0 0 0 ðk, ΩR Þ  dk dΩR 3 0 0 0 l μ ν ,μ

ð3:1125Þ

μν l μ

pffiffiffi i π X^ J μ iσðL0Þ X J μ LCL0 1μ e C L 0 μ - μ0 f l0 μ0 ν0 ,μ ðk, ΩR Þ = K 0 lLJ 0 0

0

j0 μ 0

0

Cjl0 μμ0 - ν0

jL

eiσ L0 ðkÞ J ðk Þscl0 j0 ðkÞYL0 μ - μ0 ðΩR Þ S 0 0 0 kΔ cl 1j L ,0l11L

1ν0

ð3:1126Þ

Now because the entire reaction system is axisymmetric, the above two equations will not change with azimuth angle φR. The double differential cross section of the deuteron center of mass in the laboratory system can be obtained from Eq. (3.1125) through the coordinate system transformation. The momentum wave vectors of neutrons, protons, and target nuclei in the * * * breakup channel in laboratory system are k n , k p , and k A , respectively, and the total momentum wave vector of the system is *

*

*

*

kt = kn þ kp þ kA

ð3:1127Þ

*

where k t is along the z-axis direction, i.e., the incident deuteron direction. The total kinetic energy of the system in the laboratory system after the deuteron breaking is k2p k2 ħ2 k 2n þ þ A Et = 2 mn mp mA

! ð3:1128Þ

In the laboratory system, the total energy of the incident channel is E in =

md þ mA E 0 þ ε0 = E L þ ε0 mA

ð3:1129Þ

where EL is the deuteron incident energy in the laboratory system, and we have pffiffiffiffiffiffiffiffiffiffiffiffiffi 2md E L kin = ħ

ħ2 k2in EL = , 2md

ð3:1130Þ

In the laboratory system, the momentum conservation requires *

*

k t = k in

ð3:1131Þ

and it can be seen that kt is a known amount. It follows that Eq. (3.1129) can be rewritten as

3.15

The Spherical Nucleus CDCC Theory Describing the Breakup Reaction. . .

E in =

ħ2 k2t þ ε0 2md

279

ð3:1132Þ

Using the energy conservation condition given by Eq. (3.1031), Eq. (3.1123) can be rewritten as !Z X  * * 1 2 f ðk, ΩR , Ωr Þ  δðE 0 þ ε0 - 2 - εÞ d k dK d5 σ = 3 μv0 v0,μ δ

ð3:1133Þ

where 2 =

ħ2 K 2 2μR

ð3:1134Þ

The δ function R indicates that the energy conservation is satisfied in the center of mass system, and δ indicates that the integral will be done only to the argument of the δ function. Now introduce the integral amount that appeared in Eq.(3.1133) as Z dW 

* *

δ

δðE0 þ ε0 - 2 - εÞd k dK

ð3:1135Þ

Since dW is an integral amount that is independent with the coordinate system, so the above equation can be rewritten in the laboratory system as Z dW =

δ

* * * *  * * * δ k t - k n - k p - k A δðEin - Et Þd k n d k p d k A

ð3:1136Þ

The two δ functions in the above equation represent the momentum conservation and the energy conservation in the laboratory system. From the momentum conservation we have *

*

*

*

kA = kt - kn - kp

ð3:1137Þ

and substituting this equation into Eq. (3.1128) we can get ħ2 Et = 2

(

h * * k 2p * * * * i k2n 1 2 þ þ kt þ k 2n þ k2p þ 2 k n  k p - k n  k t - k p  k t mn mp mA

)

ð3:1138Þ It follows that Eq. (3.1136) can be rewritten as

280

3

Polarization Theory of Nuclear Reactions for Spin 1 Particles

  1 ħ2 1 1 1 2 k þ k2 dW = δðε0 - ½ þ þ 2 mn mA n mp mA p δ    * 1 1 2 * * 2 * * * * k2t kt - kn  kp k  k Þd k n d k p md mA mA mA n t Z 2mp mA 2mp * *  * mp ðmn þ mA Þ 2  kn  2 δðk2p = kt - kn  kp þ  m þ m m p A mp þ mA ħ n mp þ mA * * 2mp * * mp ðmA - md Þ 2 2mp mA k -   2 ε0 Þd k n d k p k k -  mp þ mA n t md mp þ mA t mp þ mA ħ Z

ð3:1139Þ When the δ function is integrated in the above equation, kp is taken as the integral * variable. It is known that k t is along the z-axis direction, and let *

*

k n  k p = k n kp cos θnp



cos θnp = cos θn cos θp þ sin θn sin θp cos φn - φp



ð3:1140Þ ð3:1141Þ

The following equation can be obtained from the δ function in Eq. (3.1139): k2p -

 2mp  kt cos θp - k n cos θnp k p mp þ mA  mp mA - md 2 2mA mn þ mA 2 þ kn -2kn k t cos θn k t - 2 ε0 = 0 mp þ mA mn md ħ ð3:1142Þ

and its solution is k p =

   2 mp f kt cos θp - kn cos θnp  ½ k t cos θp - kn cos θnp mp þ mA  mp þ mA mn þ mA 2 m - md 2 2mA kn -2k n k t cos θn - A kt - 2 ε0 1=2 g mp mn md ħ ð3:1143Þ

The condition that the above equation has a solution is   2 mp þ mA mn þ mA 2 kn k t cos θp - kn cos θnp mp mn -2kn k t cos θn -

mA - md 2 2mA k t - 2 ε0 ≥ 0 md ħ

ð3:1144Þ

3.15

The Spherical Nucleus CDCC Theory Describing the Breakup Reaction. . .

281

and areas that do not meet this condition are forbidden. Only the kp ≥ 0 solutions are taken in the above equation. If kp- particle polarization theory can also be carried out the 2 specific studying work. However, the higher the spin value of the particle, the more complex the polarization theory is, and the practical value will become smaller. Therefore, under normal circumstances, there is no need to study the polarization 3 problem that the particle spin is greater than . 2 the spin

326

4

Polarization Theory of Nuclear Reactions. . .

Table 4.4 Expressions of d3MM 0 ðβÞ [1] M0 3 2 1 0 1 2 3 M 2

M0 0

M=3

0

1

4.2

3 1 8 ð1 þ cos βÞ pffiffi  86 sin βð1 þ pffiffiffiffi 2 15 8 sin β ð1 þ pffiffi  45 sin 3 β pffiffiffiffi 2 15 8 sin β ð1  pffiffi 6  8 sin βð1  3 1 8 ð1  cos βÞ

cos βÞ2

1

cos βÞ

2

M=2

pffiffiffiffi 2 30 4 sin β cos β pffiffiffiffi 10  8 sin βð1 þ 2 cos β  3 cos 2 2 1 4 ð1  cos βÞ ð2 þ 3 cos β Þ

βÞ

M0

M=1

cos βÞ

1

cos βÞ2

0

 18 ð1 þ cos βÞð1 þ 10 cos β  15 cos 2 βÞ pffiffi 3 2 4 sin β ð1  5 cos βÞ

1

M=2

M 0

 14 ð1 þ cos βÞ2 ð2  3 cos βÞ pffiffiffiffi 10 2 8 sin β ð1  2 cos β  3 cos β Þ

 18 ð1  cos βÞð1  10 cos β  15 cos 2 βÞ

0

M=0  12 cos βð3  5 cos 2 βÞ

Polarization Theory for Photon Beams

This section will consider electromagnetic fields in a vacuum and use Gaussian units.

4.2.1

Classic Electromagnetic Field Theory [3]

The force of the electromagnetic field felt by a particle with charge q (Lorentz force) is given by the following formula: h*   1 * * *  i F r, t ¼ q E * r, t þ v  B r, t c

**

ð4:99Þ

 ! * *  * where v is particle velocity, c is the speed of light, and E * r , t and B r , t are the electric field strength and magnetic induction strength, respectively, which satisfy the following Maxwell equations: *

*

∇ E ¼ 4πρ *

*

*

∇E ¼ *

ð4:100Þ

*

1 ∂B c ∂t

∇ B ¼0

ð4:101Þ ð4:102Þ

4.2

Polarization Theory for Photon Beams *

327 *

*

∇B ¼

4π * 1 ∂E j þ c c ∂t

ð4:103Þ

  **  where ρ * r , t and j r , t are the charge density and electric current density, respectively. Now, because of the charge conservation, we have the following continuous equation: ∂ρ * * þ∇ j ¼0 ∂t

ð4:104Þ

*

*

The divergence of B given in Eq. (4.102) is 0, that is, B is the transverse field, so  * it can be expressed as a kind of curl of a vector potential A * r, t *

*

*

B ¼∇A

ð4:105Þ

Substituting this formula into Eq. (4.101), we can get *

∇

*!

1 ∂A Eþ c ∂t

*

¼0

ð4:106Þ

Therefore, the amount  in the brackets above must be expressed as a gradient of a scalar potential φ * r , t , that is, *

*

* 1 ∂A ¼ ∇φ c ∂t

*

1 ∂A *  ∇φ c ∂t

Eþ

ð4:107Þ

which can be rewritten as *

E¼

ð4:108Þ

Substituting this formula into Eq. (4.100), we can get *2

∇ φþ


 1 ∂ * * ¼ 4πρ ∇ A c ∂t

ð4:109Þ

We can also obtain the following relation substituting Eqs. (4.105) and (4.108) into Eq. (4.103)

328

4

Polarization Theory of Nuclear Reactions. . .

*
*  4π * 1 ∂ * 1 ∂A * ∇ ∇A ¼ j  ∇φ þ c c ∂t c ∂t *

! ð4:110Þ

Now using the vector formula
*  * * 2* *
* * * ∇  ∇  A ¼ ∇ A þ ∇ ∇ A

ð4:111Þ

Equation (4.110) can be translated into * 2*

∇ A

 2* 1 ∂ A 4π * * * * 1 ∂φ ¼  j þ ∇ ∇ A þ c c ∂t c2 ∂t 2

ð4:112Þ *

Next we will prove that the above equation cannot guarantee that φ and A have * * definite solutions, that is, by using different φ and A, we can obtain the same E and
 * * B : In order to prove this statement, we introduce a scalar potential ψ r , t and define the following: *0

*

*

A ¼ A þ ∇ψ

ð4:113Þ

1 ∂ψ c ∂t

ð4:114Þ

φ0 ¼ φ  *

*

When we use φ0 and A 0 given by the above two formulas instead of φ and A in Eqs. (4.105), (4.108), (4.109), and (4.112), it can be seen that all ψ terms appearing in these four formulas disappear; that is, they all degenerate into the original forms, and Eqs. (4.105), (4.108), (4.109), and (4.112) are equivalent with Eqs. (4.100) – * * (4.103). The transformation from φ and A to φ0 and A 0 given by Eqs. (4.113) and (4.114) is called the gauge transformation. The invariance of the four Maxwell * * * equations is called the gauge invariance, that is, E and B produced after φ and A are gauge transformed are still unchanged. Given this, it is permissible to apply * certain constraints on the relationship between φ and A. The Lorentz gauge, which is typically applied, is *

*

∇ Aþ

1 ∂φ ¼0 c ∂t

ð4:115Þ

With this constraint, we can obtain the following expressions substituting Eq. (4.115) into Eqs. (4.109) and (4.112):

4.2

Polarization Theory for Photon Beams

329 2

*2

1 ∂ φ ¼ 4πρ c2 ∂t 2

* 2*

1 ∂ A 4π * ¼ j 2 2 c c ∂t

∇ φ

ð4:116Þ

2*

∇ A

ð4:117Þ

Another constraint that has been employed is the Coulomb gauge (or transverse gauge), that is, *

*

∇ A ¼0

ð4:118Þ

*

where this formula requires that A is a transverse potential, and using this formula, Eqs. (4.109) and (4.112) can be converted to *2

∇ φ ¼ 4πρ * 2*

∇ A


 1 ∂ A 4π * 1 ∂ * ¼  þ φ j ∇ c c ∂t c2 ∂t 2

ð4:119Þ

2*

ð4:120Þ

*

In the vacuum region ρ ¼ 0 and j ¼ 0, applying the Coulomb gauge condition given by Eq. (4.118), and selecting φ ¼ 0 according to Eq. (4.119), which is also a gauge condition, then Eq. (4.120) becomes 2*

1 ∂ A ¼0 ∇ A 2 c ∂t 2 * 2*

4.2.2

ð4:121Þ

Hamilton Canonical Equation [3]

The Lagrange function of the system is denoted as Lðq1 , ⋯, qn , q_ 1 , ⋯, q_ n , t Þ , or _ t Þ, where q_ i is a generalized speed. For conservative systems abbreviated as Lðq, q, that do not change over time, time t in function L does not appear. Now set the system to begin at time t0 , reaching the end point at time t00 through a certain track q(t), and define the action quantity Z S½qðt Þ ¼

t 00 t0

Lðq, q_ Þdt

ð4:122Þ

Note that the principle of minimum action quantity requires that the actual orbit of the particle should let S be minimized. Next, set q(t)! q(t) + δq(t) in the condition

330

4

Polarization Theory of Nuclear Reactions. . .

δqðt 0 Þ ¼ δqðt 00 Þ ¼ 0

ð4:123Þ

δS¼0

ð4:124Þ

such that we require

Now, using the partial integral method and noting δq_ i ¼ dtd δqi , we can obtain the following expression according to Eqs. (4.122) and (4.123) P ∂L ∂L dt δq þ δq_ δS ¼ ∂q i ∂q_ i i t0

 t00 Z t00 i i P P ∂L ∂L d ∂L ¼ δqi þ dt  δq _ i i t0 ∂q dt ∂q_ i i t0 i ∂q

 Z t00 i P ∂L d ∂L δqi ¼ 0 ¼ dt  ∂qi dt ∂q_ i i t0 Z

t 00

ð4:125Þ

where, since δqi is arbitrary, one requires

 ∂L d ∂L ¼ 0,  ∂qi dt ∂q_ i

i ¼ 1,2,⋯,n

ð4:126Þ

and this formula is the Lagrange equation. Introducing the generalized momentum pi ¼

∂L ∂q_ i

ð4:127Þ

the definition of the Hamilton quantity of the system is H ðq, pÞ ¼

X

pi q_ i  Lðq, q_ Þ

ð4:128Þ

i

Using Eqs. (4.126) – (4.128), we can prove the following equations: q_ i ¼

∂H , ∂pi

p_ i ¼ 

∂H , ∂qi

i ¼ 1,2, . . . ,n

ð4:129Þ

where these equations are the Hamilton canonical equations. Now, the Poisson bracket of any two mechanical quantities is defined as X ∂A ∂B ∂A ∂B   fA, Bg ∂qi ∂pi ∂pi ∂qi i and from this formula it is easy to prove that

ð4:130Þ

4.2

Polarization Theory for Photon Beams

331

 qi , pj ¼ δij ,

  qi , q j ¼ p i , p j ¼ 0

ð4:131Þ

Using Poisson bracket, the canonical equation (4.129) can be expressed as p_ i ¼ fpi , H g

q_ i ¼ fqi , H g,

ð4:132Þ

Given the following set of transformations q ! Qðq, pÞ,

p ! Pðq, pÞ

ð4:133Þ

we can first obtain the following expression: Q_ j ¼

X ∂Qj i

∂qi

q_ i þ

∂Qj p_ ∂pi i

 ¼

X ∂Qj ∂H ∂Qj ∂H   ∂qi ∂pi ∂pi ∂qi i

ð4:134Þ

Next, transform H(q, p) into H(Q, P) ¼ H(q, p); thus, we can obtain

 ∂H ðq, pÞ ∂H ðQ, PÞ X ∂H ∂Qk ∂H ∂Pk , ¼ ¼ þ ∂pi ∂Qk ∂pi ∂Pk ∂pi ∂pi k

 ∂H ðq, pÞ ∂H ðQ, PÞ X ∂H ∂Qk ∂H ∂Pk ¼ ¼ þ ∂Qk ∂qi ∂Pk ∂qi ∂qi ∂qi k

ð4:135Þ

Substituting Eq. (4.135) into Eq. (4.134), and after sorting, we can get Q_ j ¼

 X ∂H  ∂H  Qj , Qk þ Qj , Pk ∂Qk ∂Pk k

ð4:136Þ

In a similar way we can also get P_ j ¼

 X ∂H  ∂H  Pj , Qk þ Pj , Pk ∂Pk ∂Qk k

ð4:137Þ

Now if the following conditions are met

 Qj , Pk ¼ δjk ,

  Qj , Qk ¼ Pj , Pk ¼ 0

ð4:138Þ

then we can obtain the following equations from Eqs. (4.136) and (4.137) ∂H Q_ i ¼ , ∂Pi

∂H P_ i ¼  ∂Qi

ð4:139Þ

At this time, (q, p) and (Q, P) are all canonical, and the transformation given by Eq. (4.133) is a canonical transformation.

332

4.2.3

4

Polarization Theory of Nuclear Reactions. . .

Quantization of Electromagnetic Fields [3–5]

In order to avoid the difficulty of non-normalization in the calculation process, it is assumed that the radiation field is limited to a square box with a volume V, and it is also assumed that the electromagnetic field satisfies the periodic boundary conditions; thus V ! 1 is finally ordered. The special solutions of Eq. (4.121) can be obtained using the following separation variable method:  *  A r , t ¼ qðt ÞA * r

**

ð4:140Þ

The general solution can be expressed as a linear superposition of these special solutions. Substituting Eq. (4.140) into Eq. (4.121), we can obtain * 2 *  ** 2 ∇ A * r þk A r ¼0

ð4:141Þ

€qðt Þ þ ω2 qðt Þ ¼ 0

ð4:142Þ

where €qðt Þ represents a quadratic derivative of q(t) to t; ω and ω ¼ kc in the above * formulas are constants that do not depend on r and t; and c is the speed of light. * We study the plane wave light beams that propagate along the k direction. Here, * k is the wave vector of the light beams. For a specific k, we use the following unit vectors to compose a right-handed rectangular coordinate system corresponding to x, y, z axes *

ε k1

0 1 1 B C ¼ @ 0 A, 0

*

ε k2

0 1 0 B C ¼ @ 1 A, 0

*

ε k3

0 1 0 k B C ¼ ¼ @0A k 1 *

ð4:143Þ

such that the following relation is satisfied: *

*

ε kμ ε kμ0 ¼ δμμ0

ð4:144Þ

Note that Eq. (4.141) can take the following plane wave solutions:
 1 * ** * A λ r ¼ pffiffiffiffi ε λ ei k r V

*

*

ð4:145Þ

where λ ¼ kμ (μ ¼ 1, 2, 3), and ε λ is a unit vector that describes the direction of * photon radiation vibration. We require that A λ is periodic with changes in space, and * since A λ is required to have an entire number of cycles in the square box, the ! desirable values of k are

4.2

Polarization Theory for Photon Beams *

k ¼

2π ðl, m, nÞ, V 1=3

333

l, m, n ¼ 0,  1,  2,⋯ ðexcept l ¼ m ¼ n ¼ 0 Þ ð4:146Þ *

*

where l, m, n represent the integer value taken in the x, y, z direction; thus k r is an integer multiple of 2π at the boundary of the square box of the volume V. Each group l, m, n corresponds to a wave vector, so it follows that k is a separate variable, whereas k becomes a continuous variable only when V ! 1. The following relation is readily seen using the periodic boundary condition Eq. (4.146): 1 V

Z

* *  * 0 * d r ei k  k r ¼ δ!k !k 0

V

ð4:147Þ

In addition, we have the following δ function formula: 1 ð2πÞ3

Z

*

*



* i k  k 0 * r

dre


* *  ¼ δ k  k0

ð4:148Þ

Comparing the above two formulas, we can see that there is the following correspondence:
* *  V 0 ! ! , δ k  k δ , 0 ð2πÞ3 k k

when V ! 1

ð4:149Þ

Now, we can obtain the following relations using Eqs. (4.144), (4.145), and (4.147): Z V

Z

* V

*

*

*

*

A λ A λ0 d r ¼ δλλ0 *

Z

A λ A λ0 d r ¼

* V

*

ð4:150Þ *

A λ A λ0 d r ¼ 0

ð4:151Þ

Substituting Eq. (4.145) into Eq. (4.140), and then using the vector formula *  * * ∇ * a r ¼ a , from Eq. (4.118) we can obtain *

*

ελ k ¼ 0

ð4:152Þ *

*

This is the transverse wave condition, which requires that ε λ is vertical with k . If we * * * take k along the z-axis direction, then the photons have only ε k1 and ε k2 two polarization waves, and there is no longitudinal polarization wave along the z-axis direction. The solution for Eq. (4.142) is

334

4

Polarization Theory of Nuclear Reactions. . .

qðt Þ / eiωt

ð4:153Þ

where we take q(t) ~ eiωt. Since the solution of Eq. (4.121) should be a real number, its general solution can be represented by the linear superposition of its special solutions as follows: *
*

A r, t



*
* *
*i Ph qλ ðt ÞA λ r þ qλ ðt ÞA λ r λ   * *  1 X* i *k *r ωt i k r ωt  pffiffiffiffi þe ελ e V λ

¼

ð4:154Þ

Using Eq. (4.154) and taking the scalar potential φ ¼ 0, then from Eqs. (4.108) and (4.105) we can obtain *
* *  1 ∂A i X ω qλ A λ  qλ A λ ¼ c λ c ∂t

*

E¼

*

*

*

B ¼∇A ¼

X
λ

*

*

*

* 

qλ ∇  A λ þ qλ ∇  A λ

ð4:155Þ ð4:156Þ

Utilizing Eq. (4.151) we can obtain the following result from Eq. (4.155): 1 8π

Z  2 Z
* * 
* *  * 1 X 0 * * ωω qλ A λ  qλ A λ qλ0 A λ0  qλ0 A λ0 d r E  d r ¼ 2 8πc V V λλ0 Z
* * * *  * 1 X 0 ¼ kk qλ qλ0 A λ A λ0 þ qλ qλ0 A λ A λ0 d r 8π 0 V λλ

ð4:157Þ We can also obtain the following result from Eq. (4.156): 1 8π

Z h Z  2
* * 
* * 
* * 
* *  1 X ! * qλ qλ0 ∇  A λ ∇  A λ0 þ qλ qλ0 ∇  A λ ∇  A λ0 B d r ¼ 8π 0 V V λλ
* * 
* * 
* * 
* * i * þqλ qλ0 ∇  A λ ∇  A λ0 þ qλ qλ0 ∇  A λ ∇  A λ0 d r

ð4:158Þ where we have the following vector formula:
*  *
* * *
* * * * ∇ a  b ¼ ∇ a b  a ∇ b

ð4:159Þ

4.2

Polarization Theory for Photon Beams

335

*

It is stipulated that the operator ∇ in the parentheses only acts on the quantities in the *

*

*

parentheses. If we take b ¼ ∇  c , then we have
* i
* 
*  h*
* i * h* * * * * * ∇ a  ∇ c ¼ ∇ a ∇ c  a ∇ ∇ c ð4:160Þ Utilizing the above formula we can get
* * 
* *  * h*
* * i * h *
* * i ∇  A λ ∇  A λ 0 ¼ ∇ A λ  ∇  A λ0 þ A λ ∇  ∇  A λ 0 ð4:161Þ Also, from the Gaussian theorem, that is, ZZ ZZZ
* * * * *

f dσ ¼ ∇ f dr

ð4:162Þ

and noting that there is no outflow on the surface of infinite space, the first item on the right-hand side of Eq. (4.161) has no effect. Now, we can obtain the following relation using Eqs. (4.111), (4.118), and (4.141):
* *  * 2* ∇  ∇  A λ 0 ¼ k 0 A λ0

ð4:163Þ

Using Eqs. (4.161) – (4.163), as well as noting Eq. (4.151), we can obtain the following relation from Eq. (4.158): 1 8π

Z
Z  2 * * * *  * 1 X 02 !  * k qλ qλ0 A λ A λ0 þ qλ qλ0 A λ A λ0 d r B d r ¼ 8π 0 V V

ð4:164Þ

λλ

Note that the items marked λ0 as footnote are all behind the items marked λ as footnote in Eq. (4.158). Using the orthogonal relation Eq. (4.150), the total energy of the radiation field can be obtained by Eqs. (4.157) and (4.164) as 1 8π

Z  2  2   1 X 2  * * * ω qλ qλ þ qλ qλ E  þ  B  d r ¼ 2 4πc λ V

ð4:165Þ

Since qλ and qλ are not real variables, the following real variables are defined to facilitate the quantification of the radiation field    1  1  iω  Qλ ¼ pffiffiffiffiffiffiffiffiffi qλ þ qλ , Pλ ¼ pffiffiffiffiffiffiffiffiffi q_ λ þ q_ λ ¼  pffiffiffiffiffiffiffiffiffi qλ  qλ 4πc2 4πc2 4πc2 where their inverse transformations are

ð4:166Þ

336

4

pffiffiffiffiffiffiffiffiffi

 4πc2 i qλ ¼ Qλ þ Pλ , ω 2

qλ

Polarization Theory of Nuclear Reactions. . .

pffiffiffiffiffiffiffiffiffi

 4πc2 i ¼ Qλ  Pλ ω 2

ð4:167Þ

Substituting Eq. (4.167) into Eq. (4.165) we can obtain H¼

 1 X 2 Pλ þ ω2 Q2λ 2 λ

ð4:168Þ

Now it is known that the one-dimensional harmonic oscillator Hamiltonian is H¼

p2 1 þ mω2 q2 2m 2

ð4:169Þ

and if we let p P ¼ pffiffiffiffi , m

Q¼

pffiffiffiffi mq

ð4:170Þ

then Eq. (4.169) becomes H¼

 1 2 P þ ω2 Q2 2

ð4:171Þ

It can be seen from Eq. (4.168) that the radiation field can be seen as a system consisting of an infinite number of harmonic oscillators; the oscillator frequency ω ¼ kc. In addition, it can be seen from Eq. (4.146) that the system tends to change continuously when V ! 1. Using the canonical equation (4.139) and Eq. (4.168), we can obtain ∂H ¼ Pλ , Q_ λ ¼ ∂Pλ € λ ¼ P_ λ ¼ ω2 Qλ , Q € λ ¼ ω2 Q_ λ ¼ ω2 Pλ , P

∂H P_ λ ¼  ¼ ω2 Qλ ∂Qλ € λ þ ω2 Q λ ¼ 0 Q

ð4:172Þ

€ λ þ ω2 P_ λ ¼ 0 P

The above equations are in the same form as Eq. (4.142), where Qλ and Pλ can be regarded as the coordinates and momentum that are canonically conjugated to each other. Only after changing qi and pj to Qλ and Pλ0 can obtain the following relations using Eqs. (4.130) and (4.131): fQλ , Pλ0 g ¼ δλλ0 ,

fQλ , Qλ0 g ¼ fPλ , Pλ0 g ¼ 0

ð4:173Þ

The basic idea of field quantification is to find a complete set of canonical “coordinates” and “momentum” describing the classical field and then treat them as corresponding operators to satisfy the commutative relations of canonical

4.2

Polarization Theory for Photon Beams

337

coordinates and momentum; thus, they are quantized. Next we establish the following equivalence relation between the Poisson bracket {} in classical electromagnetic field and the commutative relation [ ] in quantum theory. ½F, G ¼ iħfF, Gg

ð4:174Þ

From here, and according to Eq. (4.173), the following commutative relations can be obtained in quantum theory. ½Qλ , Pλ0  ¼ iħδλλ0 ,

½Qλ , Qλ0  ¼ ½Pλ , Pλ0  ¼ 0

ð4:175Þ

Next, introduce the dimensionless operators in the following way: rffiffiffiffiffiffi  ħ Qλ ¼ a þ aþ λ , 2ω λ

rffiffiffiffiffiffi  ħω Pλ ¼ i a  aþ λ 2 λ

ð4:176Þ

and then find their inverse solutions. It follows that using Eq. (4.166) together with these solutions, we can obtain rffiffiffiffiffiffi rffiffiffiffiffi

 ω i 2ω 1 Q þ P ¼ pffiffiffiffiffiffiffiffiffi aλ ¼ q, 2 2ħ λ ω λ ħ λ 4πc rffiffiffiffiffiffi rffiffiffiffiffi

 ω i 2ω  1 þ Q  P ¼ pffiffiffiffiffiffiffiffiffi q aλ ¼ 2 2ħ λ ω λ ħ λ 4πc

ð4:177Þ

Given this, the following relations are straightforward to prove by using Eqs. (4.177) and (4.175):   aλ , aþ ¼ δλλ0 , λ0

  þ ½ aλ , aλ 0  ¼ a þ λ , a λ0 ¼ 0

ð4:178Þ

which are precisely the commutative relations satisfied by the generation and annihilation operators of bosons. Substituting qλ and qλ obtained by Eq. (4.177) into the first formula of Eq. (4.154) and writing out the harmonic changing factors in qλ(t) and qλ ðt Þ with time by use of Eq. (4.153), thus we can obtain  pffiffiffiffiffiffiffiffiffiXrffiffiffiffiffiffih*
 i *
* ħ * A λ r eiωt aλ þ A λ r eiωt aþ A r , t ¼ 4πc2 λ 2ω λ

*
*

ð4:179Þ

In addition, substituting Eq. (4.145) into the above formula, we can also obtain

338

4

Polarization Theory of Nuclear Reactions. . .

rffiffiffiffiffiffiffiffiffiffiffiffi  * *  * * * 4πħc2 X ε λ pffiffiffiffiffiffi ei k r ωt aλ þ ei k r ωt aþ A r, t ¼ λ V λ 2ω

*
*



ð4:180Þ

where aλ and aþ λ are the Boson annihilation and generation operators that are time invariant. Now, we can obtain the following expressions substituting Eq. (4.180) into Eqs. (4.108) and (4.105) with φ ¼ 0 rffiffiffiffiffiffiffiffiffiffiffiffi *  * *  * 2πħck X* ε λ ei k r ωt aλ  ei k r ωt aþ E r, t ¼ i λ V λ


* *



rffiffiffiffiffiffiffiffiffiffi
* *   * *  2πħcX * * i k r ωt i k r ωt þ aλ  e aλ B r, t ¼ i k  ελ e Vk λ


* *



ð4:181Þ

ð4:182Þ

where λ ¼ kμ such that μ ¼ 1, 2 represents the transverse polarized light in the x, y axis directions, respectively. According to the theory of quadratic quantification, in the occupy number representation we have [3, 6] aþ λ jnλ i ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffi nλ þ 1 jnλ þ 1i,

aλ j n λ i ¼

pffiffiffiffiffi n λ j nλ  1 i

ð4:183Þ

such that it is straightforward to verify aþ λ aλ jnλ i ¼ nλ jnλ i

ð4:184Þ

where nλ is the number of photons in λ state. Substituting Eq. (4.176) into Eq. (4.168) and using the commutative relation Eq. (4.178), we can obtain H¼

 X
1 nλ þ ħω, 2 λ

nλ ¼ 0, 1, 2,⋯

ð4:185Þ

Now, the energy and momentum of each photon on λ state are as follows: E ¼ ħω,

*

!

p ¼ ħk

ð4:186Þ

and because ω ¼ kc, we can see E 2  p2 c2 ¼ 0 This is the reflection of a photon with a stationary mass of 0.

ð4:187Þ

4.2

Polarization Theory for Photon Beams

4.2.4

339

Polarization Theory of Photon Beams

In the particle occupy number representation, according to Eq. (4.183), we have h1jaþ λ j 0i ¼ 1

h0jaλ j1i ¼ 1,

ð4:188Þ

where the measurement of photon beams belongs to the photon annihilation process. According to Eq. (4.181), we order *ð  Þ Eμ




rffiffiffiffiffiffiffiffiffiffiffiffi   2πħck* i *k *r ωt akμ , k, r , t ¼ i ε kμ e V *



μ ¼ 1, 2

ð4:189Þ

Also, according to the Golden rule obtained by perturbation theory, the system transition rate is [2, 3, 5] W¼

2π 2 jT j ρf ħ

ð4:190Þ

The square of absolute value of the transition matrix element T in above formula is  2   ffiffiffiffiffiffiffiffiffiffiffiffi    X D  *   E2 1 r2πħck * *   1 * *       * ei k r ωt ε k1 þ ε k2  0E ðμÞ k; r ; t 1  ¼  i jT j 2 ¼      2 μ 2 V  rffiffiffiffiffiffiffiffiffiffiffiffi   2  1 X  2πħck i *k r*ωt *  2πħck ε kμ  ¼ ¼ e i  2 μ  V V ð4:191Þ 1 represents the averaging of the two transverse vibration 2 directions. According to the principle of phase space, the number of states in the spherical shell, which volume is V and momentum wave number is k ! k þ dk, is where the coefficient

dn ¼

V4π k 2 dk ð2πÞ3

ð4:192Þ

For photon dE ¼ ħdω ¼ ħcdk, the photon energy state density is ρf ¼

dn Vk2 ¼ 2 dE 2π ħc

ð4:193Þ

Next, we can obtain the following result substituting Eqs. (4.191) and (4.193) into Eq. (4.190):

340

4

W¼

Polarization Theory of Nuclear Reactions. . .

2k3 ħ

ð4:194Þ

Now, we have the following explanation for the above transition rate W

I 0 ðk Þ ¼


 * dM 0 k, r , t *

dEd r dt

¼W¼

2k3 ħ

ð4:195Þ


 * where M 0 k, r , t represents the number of the measured photons and I0(k) represents the number of photons with energy ħck measured in unit time, unit volume, and in the unit energy interval; because we are studying the stable plane wave light beam, which has no relationship with position and time, but its intensity is proportional to the third power of photon energy. Now suppose that we put an ideal plane wave photon detector in the vertical direction with the photon beam. The wavelength of the photon is known as LðkÞ ¼ 1k. When the photon is injected into the detector from the surface of the detector and its depth reaches L(k), then, as an ideal detector, it should be able to obtain all the information of the photons and record the photons; thus, the sensitive detection thickness of the ideal detector to photons is L(k). If the detector has a very high energy resolution and only detects photons with an average energy ħck in a very small energy range ΔE ¼ ħcΔk, then, using an ideal detector with an effective detection area of S through the detection time T, it can be seen that the number of photons recorded should be, according to Eq. (4.195), N 0 ðk Þ ¼ I 0 ðkÞSLðkÞTΔE ¼ 2cSTk 2 Δk

ð4:196Þ

It can be seen from
Eqs.  (4.189) and (4.191) that the amplitudes of the electric * field strength E ðμÞ k, r , t in the two directions μ ¼ 1 and μ ¼ 2 are equal, and *

*

since ε k1 ε k2 ¼ 0, there is no coherent term, so that satisfies the unpolarization condition. It means that the original light beam is unpolarized. The normalization density matrix of the unpolarized light beam is 0 1 3 20 1 0 1 0 1 1 0 0 1 6B C B C 7 1B C ρ^0 ¼ 4@ 0 A ð1 0 0Þ þ @ 1 A ð0 1 0Þ5 ¼ @ 0 1 0 A 2 2 0 0 0 0 0

ð4:197Þ

such that, apparently,   1 X * 2  ε kμ  ¼ trf^ρ0 g 2 μ It follows that Eq. (4.191) can be rewritten as

ð4:198Þ

4.2

Polarization Theory for Photon Beams

jT j2 ¼

341

2πħck trf^ρ0 g V

ð4:199Þ

and Eq. (4.195) can be rewritten as

I 0 ðk Þ ¼


 * dM 0 k, r , t *

dEd r dt

¼

2k 3 trf^ ρ0 g ħ

ð4:200Þ

*
*  Since photons can be described by three-dimensional vector potential A r , t ,

the photon spin is equal to 1. Eq. (4.143) is the spin S ¼ 1 particle basis spin function in the rectangular basis representation and rectangular coordinate system given by Eq. (3.42). The spin S ¼ 1 particle spin operators in the rectangular basis representation and rectangular coordinate system are given by Eq. (3.47) as follows: 0

0 0

^Sx ¼ B @0 0 0 i

0

1

C i A, 0

0

0

0

^Sy ¼ B @ 0 0 i 0

i

1

C 0 A, 0

0

0

B ^ Sz ¼ @ i 0

i 0 0

0

1

C 0A 0

ð4:201Þ

Now assume that the photon beam propagated along the z-axis passes through both a compensation prism and a polarizer, then we measure the number of photons with a photon detector [7, 8]. Note that at this time the intensity of the light beam may have changed. Although the light beam is still a transverse wave, its amplitude ratio in the x and y directions may have changed, that is, the light beam is polarized, and the phase of the light beam x and y components may also have changed, respectively. According to Eq. (4.189) the total electric field strength of the wave number k in this case can be written as

 rffiffiffiffiffiffiffiffiffiffiffiffi * *  2πħck i k r ωt * iβx * iβy * k, r , t ¼ ax e ε k1 ak1 þ ay e ε k2 ak2 i e E V

 rffiffiffiffiffiffiffiffiffiffiffiffi * *  !ð  Þ 2πħck i k r ωt * iβx * iβy * k, r , t j1i ¼ ax e ε k1 þ ay e ε k2 i e h0jE V

!ðÞ

ð4:202Þ ð4:203Þ

Now introduce the following symbol of the polarized photon wave function 0

ax eiβx

1

* * B C jk i ¼ ax eiβx ε k1 þ ay eiβy ε k2 ¼ @ ay eiβy A

ð4:204Þ

0 where it is now possible to define the density matrix of the polarized light beam as

342

4

0

ax eiβx

Polarization Theory of Nuclear Reactions. . .

1

B C
 iβ  iβ iβy C x a e y ^ρ ¼ jki hkj ¼ B y @ ay e A a x e 0

 0

0 jax j2

B B ¼ B a a eiðβx βy Þ @ y x

ax ay eiðβx βy Þ

0

jay j2 0

1 0

ð4:205Þ

C C 0C A 0

such that, apparently, we have trf^ρg ¼ jax j2 þ jay j2

ð4:206Þ

where the density matrix ^ρ is not necessarily normalized. Referring to Eq. (4.200), the transition rate of the polarized light beam can be written as

I ðk Þ ¼


 * dM k, r , t *

dEd r dt

¼


 3 2k 2k3 trf^ρg ¼ jax j2 þ jay j2 ħ ħ

ð4:207Þ

Using the photon spin operators given by Eq. (4.201) and the density matrix given by Eq. (4.205), the polarization rates of the light beam in x, y, and z directions can be obtained, respectively, as follows:



 tr ^Sx ρ^ tr ^Sy ^ ρ Px ¼ ¼ 0, Py ¼ ¼0 trf^ρg trf^ρg


  iðβ β Þ iay ax eiðβx βy Þ þ iax ay eiðβx βy Þ 2Im ay ax e x y tr ^Sz ^ρ ¼ Pz ¼ ¼ trf^ρg jax j2 þ jay j2 jax j2 þ jay j2 ð4:208Þ such that 1  Pz  1

ð4:209Þ

The above results show that the light beam polarization direction is perpendicular to the vibration direction of the light wave. From the above discussion we know that the polarization of the particles with mass not equal to 0 is caused by the particle spin, and the photon polarization with mass equal to 0 is caused by the transverse vibration of the photons.In this section, the polarization problem of plane wave light beam is preliminarily discussed, and this subject still needs to be studied from experimental and theoretical aspects. In Refs. [7] and [8] the polarization problem of the light beams is studied theoretically, but photon spin operators are not used in that work. In Reference [9] the polarization problem of the light beams is also discussed.

References

343

References 1. Varshalovich DA, Moskalev AN, Khersonskii VK. Quantum theory of angular momentum. Singapore/New Jersey/Hong Kong: World Scientific; 1988. 2. Shen QB. Nuclear reaction theory of low and medium energy (upper volume). Beijing: Science Press; 2005. (in Chinese) 3. Zeng JY. Quantum mechanics, volume II (third board). Beijing: Science Press; 2003. (in Chinese) 4. Eisenberg JM, Greiner W. Nuclear theory-excitation mechanisms of the nucleus. Amsterdam/ New York/Oxford: North-Holland Publishing Company; 1976. p. 58. 5. Gottfried K, Yan TM. Quantum mechanics: fundamentals, vol. Springer. 2nd ed; 2004. p. 437, 463. 6. Shen QB. Nuclear reaction theory of low and medium energy (middle volume). Beijing: Science Press; 2012. (in Chinese) 7. Glauber RJ. The quantum theory of optical coherence. Phy Rev. 1963;130:2529. 8. Lahiri M, Wolf E. Quantum analysis of polarization properties of optical beams. Phys Rev. 2010; A82:043805. 9. Robson BA. The theory of polarization phenomena. Oxford: Clarendon Press; 1974.

Chapter 5

Polarization Theory of Relativistic Nuclear Reactions

Abstract This chapter will introduce the polarization theory of relativistic nuclear reactions. A Dirac S matrix theory of relativistic nuclear reactions will be developed. Then the relativistic classical field theory and Lagrangian density in quantum hadron dynamics as well as relativistic Green function theory at zero temperature will be introduced. The nucleon relativistic microscopic optical potentials by relativistic mean field theory and based on DBHF theory will be introduced. Finally, the relativistic dynamics equation of spin 1 particle and its application in elastic scattering calculations will be discussed. Keywords Bjorken-dreel metric · Dirac-Brueckner-Hartree-Fock · Dirac equation · Dirac S matrix theory · Feynman rules · Klein-Gordon equation · Lagrangian density · Lorentz covariance · Pauli metric · Relativistic mean field · Self-energy operator

5.1 5.1.1

Basic Theory of Relativistic Quantum Mechanics Klein-Gordon Equation

The restricted theory of relativity requires that all physical equations remain unchanged under the Lorentz transformation. That is, when we transform the space-time coordinate system S(x, y, z, t) to another space-time coordinate system S0(x0, y0, z0, t0), under the condition that the following equation is maintained x2 þ y2 þ z2  c2 t 2 ¼ x0 2 þ y0 2 þ z02  c2 t 0 2

ð5:1Þ

the physical equation form should remain the same, where c is the speed of light. This is Lorentz covariant [1, 2].

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 Q.-B. Shen, Polarization Theory of Nuclear Reactions, https://doi.org/10.1007/978-3-031-11878-4_5

345

346

5

Polarization Theory of Relativistic Nuclear Reactions

The Einstein’s free particle mass-energy relation is m0 c2 m 0 c2 m0 c2 ffi ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi E ¼ mc2 ¼ rffiffiffiffiffiffiffiffiffiffiffiffi ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p2 c 2 v2 p2 1 2 1 2 1 2 2 c m c E

ð5:2Þ

where m0 is the particle stationary mass. From the above relation we can obtain E 2 ¼ p2 c2 þ m20 c4

ð5:3Þ

When v  c, from Eq. (5.2) we can get 1 E  m0 c2 þ m0 v2 2

ð5:4Þ

The second item at the right end of the above equation is the non-relativistic kinetic energy. Next Eq. (5.3) is interpreted using operators as follows: E ! iħ

∂ , ∂t

*

*

p ! iħ∇

ð5:5Þ

Substituting them into Eq. (5.3) and then acting on the wave function ψ, we can obtain   m 2 c2 ψ ¼0 □ þ 02 ħ

ð5:6Þ

where 2

□ ¼ *

1 ∂ Δ c2 ∂t 2

ð5:7Þ

Δ ¼ ∇2 and □ are called Laplace operator and D’Alembert operator, respectively. Equation (5.6) is the Klein-Gordon equation of free particles and apparently satisfies the Lorentz covariant. The Klein-Gordon equation is not a relativistic motion equation to describe a single particle, but a field equation. Since the equation has only one component, it describes a field with spinless. In 1935, Yukawa [3] proposed that the nuclear force between the nucleons is propagated by a field. It can be assumed that this field satisfies the Klein-Gordon equation. The stationary mass m0 in Eq. (5.6) corresponds to the mass of a meson described by this field. The nucleon is the source of this field. Since this source is independent of time, the field produced by it is also independent of time, and the wave function ψ should also be independent of time. Refer to the electromagnetic field scalar potential equation with a stationary mass of 0 given by Eq. (4.116), the static Klein-Gordon equation with source can be written by Eq. (5.6) as

5.1

Basic Theory of Relativistic Quantum Mechanics

347

  * m20 c2 2 ∇  2 ψ ¼ ρ ħ

ð5:8Þ

Assuming the nucleon is heavy enough to make a classical approximation, and the nucleon is at the origin of the coordinate, then this point source can be expressed as   * ρ¼gδ r

ð5:9Þ

where g is the action constant of the nucleon and this meson field, so that Eq. (5.8) can be rewritten as 

*

∇2 

   m20 c2 * * ψ r ¼ g δ r 2 ħ

ð5:10Þ

The following equation has been given in Section 3.11 of the Ref. [4]: *      * * * * ∇2 þ k 2 G r , r 0 ¼ δ r  r 0

ð5:11Þ

Its solution is * *   0 1 eikj r  r j * * * *  G r, r0 ¼  4π  r  r 0 

ð5:12Þ

So we can write the solution of Eq. (5.10) as m0 c   g e ħ r * ψ r ¼ 4π r

This formula is called Yukawa potential. If the distance to decay to

ð5:13Þ 1 is defined as e

the force distance r0, there is r0 ¼

ħ ¼ λC m0 c

ð5:14Þ

here λC is called the Compton wavelength. If taking nuclear force distance r0 ¼ 1.4 fm, m0  140 MeV can be obtained by Eq. (5.14). The particles predicted by Yukawa were discovered in cosmic rays in 1947 and are called π mesons.

348

5

5.1.2

Polarization Theory of Relativistic Nuclear Reactions

Dirac Equation [5]

In order to establish a relativistic motion equation that describes a single particle, the equation should contain only a one order derivative term of the time. In order to satisfy the covariant of the relativistic Lorentz transformation, the equation should also contain only a one order derivative term of the space coordinate. Dirac linearized the relative mass-energy relation given by Eq. (5.3) using operators in the following way: sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi X X 2 2 2 2 2 2 pi þ m0 c ¼ c αi pi þ β m0 c2 E ¼ c p þ m0 c ¼ c i

ð5:15Þ

i

The operators αi(i ¼ 1, 2, 3) and β that are independent of coordinates and momentums are introduced. In order to satisfy the mass-energy relation, taking the square of the above equation and divide it by c2, then we can get X

þ

p2i

i

¼

m20 c2

X

X ¼ αi pi þ βm0 c i

!

X

! αj pj þ βm0 c

j

α2i p2i þ β2 m20 c2 þ

i

X ðαi β þ βαi Þpi m0 c þ ðα1 α2 þ α2 α1 Þp1 p2

ð5:16Þ

i

þðα1 α3 þ α3 α1 Þp1 p3 þ ðα2 α3 þ α3 α2 Þp2 p3 In order to make the two sides of the above formula to be equal, the following equations must be established: α21 ¼ α22 ¼ α23 ¼ β2 ¼ 1

 αi αj þ αj αi  αi , αj ¼ 2δij , αi β þ βαi ¼ fαi , βg ¼ 0,

ð5:17Þ i, j ¼ 1, 2, 3 ð5:18Þ !

The above two formulas indicate that the three components of α and β satisfy the anti-commutative relations to each other. According to Eq. (5.15), the Hamiltonian operator of the free particle can be written as *

*

*

*

H ¼ cα  p þ β m0 c2 ¼ icħα  ∇ þ β m0 c2 

*

ð5:19Þ



Let ψ r , t be a free particle wave function, and the equation satisfied by it is Hψ ¼ Eψ

ð5:20Þ

We can obtain the following equations according to Eqs. (5.5), (5.19), and (5.20):

5.1

Basic Theory of Relativistic Quantum Mechanics

iħ 

349

∂ψ ¼ Hψ ∂t

ð5:21Þ

   ∂ * * * 2 iħ  icħα  ∇ þ β m0 c ψ r , t ¼ 0 ∂t

ð5:22Þ

The above two formulas are the Dirac equation. In order to ensure that H is * Hermitian, α and β must also be Hermitian, that is, *þ

*

βþ ¼ β

α ¼ α,

ð5:23Þ

The conjugate equation of Eq. (5.21) is iħ

∂ψ þ ¼ ψ þH ∂t

ð5:24Þ

The following relations can be proved utilizing Eqs. (5.21) and (5.24): ∂ ∂t

Z

þ

*

Z 

ψ ψ dr ¼

∂ψ þ ∂ψ ψ þ ψþ ∂t ∂t



i dr ¼ ħ *

Z

*

ðψ þ Hψ  ψ þ Hψ Þ d r ¼ 0 ð5:25Þ

The above formula shows that the Dirac equation satisfies the total spatial probability conservation relationship. In order to facilitate the representation of the equation, the γ matrices are introduced as follows: γ k ¼ iβαk or αk ¼ iβγ k ,

k ¼ 1, 2, 3;

γ 4 ¼ β, γ 5 ¼ γ 1 γ 2 γ 3 γ 4

ð5:26Þ

The following relations can be proved using Eqs. (5.17), (5.18), (5.23), and (5.26): γþ k ¼ iαk β ¼ iβαk ¼ γ k ,

þ k ¼ 1, 2, 3; γ þ 4 ¼ γ4, γ5 ¼ γ5  γ i , γ j ¼ 2δij , i, j ¼ 1, 2, 3

ð5:28Þ

μ, ν ¼ 1, 2, 3, 4

ð5:29Þ

f γ i , γ 4 g ¼ 0,  γ μ , γ v ¼ 2δμν ,

ð5:27Þ

And the following equation can be got by multiplying β/ħc ¼ γ 4/ħc from the left of Eq. (5.22): "  #  X ∂ i ∂ þ γi þκ ψ ¼0 γ4  c ∂t ∂xi i where

ð5:30Þ

350

5

Polarization Theory of Relativistic Nuclear Reactions

m0 c ħ

κ¼

ð5:31Þ

The definitions of the four-dimensional Minkowski coordinate and its derivative are xμ ¼





 r , ict ,

*

∂μ ¼

∂ ∂ ∂ i ∂ , , , c ∂t ∂x ∂y ∂z

 ð5:32Þ

Then the Dirac equation (5.30) can be expressed as

 γ μ ∂μ þ κ ψ ¼ 0

ð5:33Þ

It should be noted that according to Einstein’s agreement, two identical Greek letter subscripts represent the summation of four-dimensional coordinates, and two identical English letter subscripts represent the summation of three-dimensional space coordinates. It is known that the two-rank Pauli matrix b σ satisfies the following anticommutative relation: 

σ^i , σ^j ¼ 2δij ,

i, j ¼ 1, 2, 3

ð5:34Þ

According to the definition of the square matrix direct product given by Eq. (2.349), we define the two 4  4 matrices in the following way. τ ¼ ^I  σ^,

ρ ¼ σ^  ^I

ð5:35Þ

where ^I is the unit 2  2 matrix. We can further obtain τi ¼ ρ1 ¼

! ^0 ^I , ^I ^0

σ^i ^0

ρ2 ¼

^0 σ^i

! i ¼ 1, 2, 3

,

! ^0 i^I , ^0 i^I

ρ3 ¼

ð5:36Þ ^I ^ 0

^ 0 ^I

! ð5:37Þ

where ^0 is a 2  2 zero matrix. We define α i  ρ1 τ i ¼

^0 σ^i σ^i ^0

So from Eq. (5.26) we can obtain

! ,

β ¼ γ 4  ρ3 ¼

^I ^ 0

^ 0 ^I

! ð5:38Þ

5.1

Basic Theory of Relativistic Quantum Mechanics

! i^ σi , ^0

^0

γi ¼

i^ σi

i ¼ 1, 2, 3;

351

γ5 ¼

^ 0 ^I ^I ^ 0

! ð5:39Þ

One can verify that Eqs. (5.28) and (5.29) are valid by utilizing Eqs. (5.34), (5.38), and (5.39). And from Eq. (5.39) one can obtain 

γ μ , γ 5 ¼ 0,

μ ¼ 1, 2, 3,4

ð5:40Þ

And it can be seen that γ μ and γ 5 are Hermitian. In Eq. (5.33), γ μ is four-rank matrix, so ψ in the equation must be four-component wave function, which is called Dirac spinor and can be expressed as  1 * ψ1 r ,t B C B  C Bψ * C   B 2 r ,t C * B C ψ r ,t ¼ B   C, B C * B ψ3 r ,t C B C @  A * ψ4 r ,t            * * * * * ψ þ r ,t ¼ ψ 1 r , t ψ 2 r , t ψ 3 r , t ψ 4 r , t 0

ð5:41Þ

Define a new symbol          * * * * ψ  ψ þ γ 4 ¼ ψ 1 r , t ψ 2 r , t  ψ 3 r , t  ψ 4 r , t

ð5:42Þ

Multiplying ψ + on the left side of the Dirac equation (5.22), there is iħψ þ

∂ * * ψ  icħψ þ α  ∇ψ þ ψ þ βm0 c2 ψ ¼ 0 ∂t

ð5:43Þ

The conjugate equation of Eq. (5.22) is iħ

* ∂ þ * ψ þ icħ∇ψ þ  α þ ψ þ βm0 c2 ¼ 0 ∂t

ð5:44Þ

Multiplying ψ on the right side of Eq. (5.44), there is  iħ

 *  ∂ þ * ψ ψ þ icħ ∇ψ þ  α ψ þ ψ þ βm0 c2 ψ ¼ 0 ∂t

ð5:45Þ

352

5

Polarization Theory of Relativistic Nuclear Reactions

Then we subtract Eq. (5.45) from Eq. (5.43), and multiply i/ħ on the obtained result; thus the following equation can be obtained:  *  ∂ þ * ðψ ψ Þ þ ∇  ψ þ cα ψ ¼ 0 ∂t

ð5:46Þ

In the Dirac equation, the probability density and probability flow density are as follows, respectively:   * ρ r , t ¼ ψ þ ψ,

  * j r , t ¼ cψ þ α ψ

* *

ð5:47Þ

So Eq. (5.46) can be rewritten as   * ∂ρ r , t ∂t

* **  þ ∇ j r, t ¼ 0

ð5:48Þ

This is the continuity equation. Therefore the Dirac equation is a relativistic motion equation describing a single particle that satisfies Lorentz covariant and guarantees probability conservation. From Eq. (5.47) we can write ji ¼ cψ þ αi ψ ¼ icψ þ γ 4 γ i ψ ¼ icψγ i ψ

ð5:49Þ

and according to Eq. (5.32) take j4 ¼ icρ ¼ icψ þ ψ ¼ icψγ 4 ψ

ð5:50Þ

Then the continuity equation (5.48) can be rewritten as ∂μ jμ ¼ 0

ð5:51Þ

The definition of the symbol ∂μ has been given by Eq. (5.32).

5.1.3

Pauli Metric and Bjorken-Drell Metric

The four-dimensional space-time vectors and γ matrices in the Minkowski space discussed above are represented under the Pauli metric; however there is another widely used representation method called Bjorken-Drell metric [6, 7]. The ħ ¼ c ¼ 1 unit system will be adopted below. In the Bjoken-Drell metric, the time component of the four-dimensional spacetime vector in Minkowski space is defined as component of 0, and the space-time vector is expressed by the contravariant components as follows:

5.1

Basic Theory of Relativistic Quantum Mechanics

353

0

1 0 1 t x0 B x1 C B x C B C B C x¼B 2C¼B C @x A @yA x3

ð5:52Þ

z

The Greek letters μ, ν, . . . represent the four components of the Minkowski space in the order of 0, 1, 2, and 3. Introduce the Bjoken-Drell metric tensor 0

1

B0 B g¼B @0

1

0

0

0

1 0

0 1

0 0

0

0

1

0

C C C A

ð5:53Þ

and let its covariant component gμν equal to its contravariant component gμν, that is, gμν ¼ gμν

ð5:54Þ

The relations between covariant components xμ and contravariant components xμ of the space-time vector are as follows: xμ ¼ gμν xν ,

xμ ¼ gμν xν

ð5:55Þ

Thus the space-time vector can be expressed as covariant components based on Eqs. (5.52), (5.53), and (5.55) as follows: 0

1 0 x0 1 Bx C B0 B 1C B x¼B C¼B @ x2 A @ 0 x3

0 1

0 0

0 0

1 0

0

10 1 0 1 0 t t C B C B 0 CB x C Bx C C CB C ¼ B C 0 A@ y A @y A 1

ð5:56Þ

z

z

In the Bjorken-Drell metric, the γ matrices with upper subscripts 0, 1, 2, 3 are used; referring to Eqs. (5.26), (5.38), and (5.39), we define γ0 ¼ γ4 ¼ β ¼ γ ¼ γ α ¼ iγ i ¼ i

0 i

^I

^0

^0 ^0

^I

^ σi

^0

σ^i

! ,

γ 0þ ¼ γ 0 ,

αi ¼ αi ,

! ,

ð5:57Þ γ

iþ

¼ γ , i

i ¼ 1, 2, 3

354

5

Polarization Theory of Relativistic Nuclear Reactions

γ ¼ iγ γ γ γ ¼ γ 5 ¼ 5

0 1 2 3

! ^0 ^I , ^I ^0

γ 5þ ¼ γ 5

ð5:58Þ

μ, ν ¼ 0,1,2,3

ð5:59Þ

They satisfy the following anti-commutative relation: 

f γ μ , γ ν g ¼ 2gμν ,

γ 5 , γ μ ¼ 0,

Introduce the following differential operators: ∂μ 

∂ ¼ ∂xμ



 ∂ ! ,∇ , ∂t

μ

∂ 

∂ ¼ ∂xμ



! ∂ , ∇ ∂t

 ð5:60Þ

The four-dimensional energy-momentum vector can be expressed as contravariant components and covariant components, respectively: 0

1 0 1 E p0 B p1 C B p C B C B xC p ¼ B 2 C ¼ B C, @ p A @ py A p3

0

1 0 1 E p0 B p C Bp C B 1C B xC p¼B C¼B C @ p2 A @py A

pz

ð5:61Þ

pz

p3

The following expressions can be obtained according to Eqs. (5.5), (5.56), (5.60), and (5.61): μ

pμ ¼ i∂ ,

pμ ¼ i∂μ

ð5:62Þ

And we can obtain *

*

a  b ¼ aμ bμ ¼ gμν aν bμ ¼ a0 b0  a  b ,

!

!

a  b ¼ ai bi ¼ ai bi

ð5:63Þ

The square of energy-momentum of the free particles with mass is *

p2 ¼ pμ pμ ¼ E2  p 2 ¼ m20

ð5:64Þ

The D’Alembert operator can be expressed as μ

2

!2

□  ∂μ ∂ ¼ ∂0  ∇

ð5:65Þ

From Eq. (5.57) we can know that γ 4 ¼ γ 0, γ i ¼  iγ i, so the Dirac equation given by Eq. (5.30) can be rewritten as

 iγ μ ∂μ  m0 ψ ðxÞ ¼ 0

where the ħ ¼ c ¼ 1 unit system is used.

ð5:66Þ

5.1

Basic Theory of Relativistic Quantum Mechanics

5.1.4

355

Plane Wave Solution of the Dirac Equation

Equation (5.19) has given that the relativistic Hamiltonian of free particles is *

*

H ¼ cα  p þ β m0 c2

ð5:67Þ

The commutative relation between the first component L1 ¼ x2p3  x3p2 of space * angular momentum L and Hamiltonian is h i * * ½H, L1 ¼ cα  p , x2 p3  x3 p2

ð5:68Þ

Using the Possion brackets given by Eq. (4.130) and Eq. (4.131) and according the way given by Eq. (4.174), Heisenberg introduced the following commutative relations:

qi , pj ¼ iħδij ,





q i , q j ¼ pi , p j ¼ 0

ð5:69Þ

The dimension of the momentum p is MeVs fm , where s represents time second, so the dimension of the multiplication of pq is MeVs; it is the same as the dimension of ħ. So the following result can be obtained from Eq. (5.68): ½H, L1 ¼ iħc

X

  * * αi ðδi2 p3  δi3 p2 Þ ¼ iħcðα2 p3  α3 p2 Þ ¼ iħc α  p 6¼ 0 1

i

It can be expressed by a three-dimensional vector as follows: h !i h i * * * * * H, L ¼ cα  p , L ¼ iħcα  p 6¼ 0

ð5:70Þ

*

Therefore, each component of orbital angular momentum L in the Dirac equation is not a conservation quantity. * In order to ensure the conservation of total angular momentum J , it is necessary * to introduce a four-dimensional spin angular momentum S to get the total angular * * * momentum J ¼ L þ S . Let ħ* ħ S¼ Σ¼ 2 2

*

! σ^ ^0 , 0^ σ^

We can prove the following relation:

*

Σ¼

σ^ ^ 0 ^ σ^ 0

! ð5:71Þ

356

5

^0 σ^  σ^ ^0 σ^  σ^

ħ2 SS¼ 4

*

Polarization Theory of Relativistic Nuclear Reactions

*

!

2iħ2 ¼ 4

σ^ ^ 0 ^ 0 σ^

!

*

¼ iħ S

ð5:72Þ

The following commutative relation can be obtained utilizing Eqs. (5.38) and (5.71): h

0 ^0 i h i * ħ * * * ħc B i H, S ¼ cα  p , Σ ¼ @ h * 2 2 σ^  p , σ^

h * σ^  p ,

σ^

^ 0

i1 C A

ð5:73aÞ

First of all, the first component of the commutative relation in the above formula is found as follows: h i * σ 3 p2 þ 2i^ σ 2 p3 σ^  p , σ^1 ¼ ½σ^1 p1 þ σ^2 p2 þ σ^3 p3 , σ^1 ¼ 2i^   * ¼ 2i σ^  p 1

ð5:73bÞ

So we can get h

  *i * * H, S ¼ iħc α  p 6¼ 0

ð5:74Þ

But the following result can be obtained from Eqs. (5.70) and (5.74): h

i h *i h *i * H, J ¼ H, L þ H, S ¼ 0

ð5:75Þ

*

So the component Jz of the total angular momentum J and J2 are the conserved quantities. S3 is a diagonal matrix as follows:

S3 ¼

ħ 2

σ^3 ^0 ^0 σ^3

!

0

1

B ħB0 ¼ B 2@0 0

The four eigenvectors of S3 are

0

0

0

1 0 0 1

0 0

0

0

1

1 C C C A

ð5:76Þ

5.1

Basic Theory of Relativistic Quantum Mechanics

357

0 1 0 1 0 0 B C B C B1C B C ħB1C B C S3 B C ¼  B C, 2B0C B0C @ A @ A 0 0 0 1 0 1 0 0 B C B C B0C B C ħB0C B C S3 B C ¼  B C 2B0C B0C @ A @ A 1 1

0 1 0 1 1 1 B C B C B0C B C ħB0C B C S3 B C ¼ B C, 2B0C B0C @ A @ A 0 0 0 1 0 1 0 0 B C B C B0C B C ħB0C B C S3 B C ¼ B C, 2B1C B1C @ A @ A 0 0

ð5:77Þ

ħ The corresponding eigenvalues are , respectively; they are double degeneracy. It 2 can be seen that the Dirac equation describes the particles (such as electrons and 1 nucleons) that have intrinsic spin . 2 From Eq. (5.21) and Eq. (5.19), it can be seen that the Dirac equation satisfied by the momentum p ¼ 0 stationary particles is as follows: iħ

∂ψ ¼ β m0 c2 ψ ∂t

ð5:78Þ

According to Eq. (5.38), four independent solutions of above formula can be written as 0 1 1 B C 2 m0 c B0C ei ħ t B C, @0A

0 1 0 B C 2 m0 c B1C ei ħ t B C, @0A

0

0 1 0 B C 2 m0 c B0C ei ħ t B C , @1A

0

0 1 0 B C 2 m0 c B0C ei ħ t B C @0A

0

ð5:79Þ

1

Referring to Eq. (5.5) it can be seen that the first two solutions correspond to the 1 E ¼ m0c2 solutions for spin ħ, and the latter two solutions are the E ¼  m0c2 2 1 solutions for spin ħ. 2 When p 6¼ 0 the plane wave solution of the free particle is generally expressed as  ψ¼

ψA ψB



 ¼

uA ð p Þ u B ð pÞ



eħð p  r E tÞ i **

ð5:80Þ

i ** In the above formula eħð p  r E tÞ is the general form of the plane wave, where the energy-related terms are given in Eq. (5.79). The stationary Dirac equation can be written from Eq. (5.20) and using Eqs. (5.67), (5.38), and (5.80) as follows:

358

5

" c

^0 σ^ σ^

Polarization Theory of Relativistic Nuclear Reactions

!

^0

*

 p þ m0 c

^0

^I

2

^0 ^I

!#

! u A ð pÞ u B ð pÞ

¼E

! uA ðpÞ uB ð p Þ

,

*

c^ σ  p uB ðpÞ þ m0 c2 uA ðpÞ ¼ EuA ðpÞ, *

c^ σ  p uA ðpÞ  m0 c2 uB ðpÞ ¼ EuB ðpÞ The above two formulas can be rewritten as c * σ^  p uB ðpÞ E  m0 c2 c * σ^  p uA ðpÞ uB ðpÞ ¼ E þ m 0 c2

uA ð pÞ ¼

ð5:81Þ ð5:82Þ

Substituting Eq.    (5.82) into Eq. (5.81) and noting that Eq. (2.16) gives * * σ^  p σ^  p ¼ p2^I, so we can get uA ð p Þ ¼

p2 c2 uA ð pÞ E  m20 c4

ð5:83Þ

2

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Above formula gives the energy relations E2 ¼ p2 c2 þ m20 c4 , E ¼ p2 c2 þ m20 c4, which can be called the on-shell relation of relativistic energy and momentum. And we can obtain the following relation: *

σ^  p ¼

 X σ^i pi ¼ i

When E > 0 taking uA ¼

  1 0

and

  0 1

p3 p1 þ ip2

p1  ip2 p3

 ð5:84Þ

, and using Eqs. (5.82) and (5.84), we can

obtain 0

1 0 1 p3 c ðp1  ip2 Þc 2 2 C B E þ m0 c C B uB ¼ @ and @ E þ m0 c A ðp1 þ ip2 Þc A p3 c E þ m 0 c2 E þ m 0 c2 Thus the obtained E > 0 solutions of the two plane waves are 0

1 1 B C 0 B C B C p3 c C uð1Þ ðpÞ ¼ N B B E þ m 0 c2 C, B C @ ðp þ ip Þc A 1 2 E þ m 0 c2

0

1 0 B C 1 B C B ðp1  ip2 Þc C C uð2Þ ðpÞ ¼ N B B C B E þ m 0 c2 C @ p c A 3 E þ m 0 c2

ð5:85Þ

5.1

Basic Theory of Relativistic Quantum Mechanics

359

where N is a normalized coefficient. For negative energy particles, refer to Eq. (5.80), the general representation of free particle plane waves should be changed to  ψ¼

ψA



ψB

 ¼

u A ð pÞ u B ð pÞ



eħð p  r EtÞ i **

ð5:86Þ

!

That is, the negative signs are added in front of E and p of the phase of the plane ! wave; here E and p are still taken positive values. In this case the stationary Dirac equation given by Eqs. (5.20) and (5.67) also becomes   * * cα  p þ β m0 c2 ψ ¼ Eψ

ð5:87Þ

For negative energy case the relations corresponding to Eqs. (5.81) and (5.82) should be changed to c * σ^  p uB ðpÞ ð5:88Þ E þ m0 c2 c * uB ðpÞ ¼ σ^  p uA ðpÞ ð5:89Þ E  m 0 c2     1 0 and , and using Eqs. (5.88) and (5.84), we can In this case taking uB ¼ 0 1 obtain uA ð p Þ ¼

1 1 0 p3 c ðp1  ip2 Þc 2 2 C B E þ m0 c C B and @ E þ m0 c A uA ¼ @ ðp1 þ ip2 Þ c A p3 c E þ m0 c2 E þ m 0 c2 0

So the obtained solutions of the two plane waves for negative energy case are 0

1 p3 c B E þ m 0 c2 C B C B ðp1 þ ip2 Þc C ð1Þ B v ð pÞ ¼ N B E þ m c 2 C C, 0 B C @ A 1 0

0

1 ðp1  ip2 Þc B E þ m 0 c2 C B C B p3 c C ð2Þ B v ð pÞ ¼ N B E þ m c 2 C C 0 B C @ A 0

ð5:90Þ

1

Based on Eq. (5.42), let uðsÞ ðpÞ ¼ uðsÞþ ðpÞγ 0 and vðsÞ ðpÞ ¼ vðsÞþ ðpÞγ 0 . The following relations can be verified by using Eqs. (5.85) and (5.90):

360

5

Polarization Theory of Relativistic Nuclear Reactions

uðsÞ ðpÞuðs Þ ðpÞ ¼ δss0 , vðsÞ ðpÞvðs Þ ðpÞ ¼ δss0 , 0

0

vðsÞ ðpÞuðs Þ ðpÞ ¼ uðsÞ ðpÞvðs Þ ðpÞ ¼ 0 0

0

ð5:91Þ

For example, we can verify the following relations: "

# p2 c 2 2m0 c2 u ðpÞu ðpÞ ¼ 1  N2, N2 ¼ 2 E þ m0 c2 ð E þ m 0 c2 Þ " # 2 2 c p 2m0 c2 2 ¼  N2 N vð1Þ ðpÞvð1Þ ðpÞ ¼  1  E þ m0 c2 ðE þ m0 c2 Þ2 ð1Þ

ð1Þ

Then the normalization coefficient can be obtained as N¼

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi E þ m 0 c2 2m0 c2

ð5:92Þ

We know that u( p) is a positive energy plane wave function constituted by u(1)( p) and u(2)( p), and v( p) is a negative energy plane wave function constituted by v(1)( p) and v(2)( p). χ is a two-dimensional spin wave function; it is an eigenfunction of the ! helicity operator σ^  ^p ¼ σ^  !pp with eigenvalue 1, for which the corresponding  j j   1 0 , respectively. The positive energy solution basis spin functions are and 0 1 can be obtained according to Eqs. (5.85), (5.90), and (5.92) as follows: 1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0 χ 2 E þ m0 c @ * A uð pÞ ¼ c^ σp 2m0 c2 χ 2 E þ m0 c

ð5:93Þ

The negative energy solution is 1 * rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0 σp E þ m0 c2 @ c^ χ v ð pÞ ¼ E þ m 0 c2 A 2m0 c2 χ

ð5:94Þ

It can be seen that u( p) and v( p) are all four-component vectors. In order to explain the negative energy solution of the Dirac equation, Dirac proposed the hole theory in 1930, arguing that the hole of the electron is a particle with the same mass as the electron and the opposite charge, i.e., a positron (positive electron). The electron which energy E > m0c2 is in a free state, i.e., the so-called positive energy state; and the electron which energy E <  m0c2 is in the Dirac Sea, that is, the so-called negative energy state. Since the electrons belonging to Fermions obey the Pauli incompatibility principle, all the negative states in the Dirac Sea are

5.1

Basic Theory of Relativistic Quantum Mechanics

361

filled with electrons. At this time, the positive energy electrons cannot fall into the negative states, which can explain the stability of positive energy electrons. The electrons bound by the Coulomb field in the atom are at the energy region m0c2 < E < m0c2. The so-called vacuum state refers to that the positive energy states are all without electrons, and the negative energy states are all filled with electrons, and this image explains the generation and annihilation of positive and negative electron pairs. In 1932, positrons were observed in cosmic rays; it confirmed the Dirac’s prophecy.

5.1.5

Lorentz Covariant of the Dirac Equation

In the followingdiscussion  the Pauli metric will be used. Linear transformation of * the vector xμ  r , ict in the four-dimensional Minkowski space is as follows: xμ ! x0μ ¼ αμν xν

ð5:95Þ

After the transformation it is required that the first three components of x0μ are still maintained as real, and the fourth component is imaginary. Therefore, αik (i, k ¼ 1, 2, 3) is required to be real, αi4 and α4i (i ¼ 1, 2, 3) are imaginary, and α44 is real. In order to ensure that the speed of light is equal in any coordinate system, the length of the space-time vector after transformation is required to remain unchanged, that is, xμ xμ ! x0μ x0μ ¼ αμν xν αμρ xρ ¼ xρ xρ

ð5:96Þ

Hence from above formula there is the requirement αμν αμρ ¼ δνρ

ð5:97Þ

The two four-dimensional vectors contraction products (such as xμxμ, ∂μ∂μ, xμ∂μ) are scalar, and Eq. (5.96) indicates that the scalar is unchanged under the Lorentz transformation. Equation (5.95) is multiplied by αμρ from the left and summed to μ, and using Eq. (5.97) we can obtain αμρ xμ0 ¼ αμρ αμν xν ¼ δρν xν ¼ xρ

ð5:98Þ

The inverse transformation of Eq. (5.95) is 0 xρ ¼ α1 ρμ xμ

ð5:99Þ

Comparing Eq. (5.99) and Eq. (5.98), the following relation can be obtained:

362

5

Polarization Theory of Relativistic Nuclear Reactions

α1 ρμ ¼ αμρ

ð5:100Þ

Therefore the product of positive and inverse matrices satisfies α1 νμ αμρ ¼ αμν αμρ ¼ δνρ

ð5:101Þ

For the reference systems that move at a uniform speed v along the z-axis, let β ¼ v/c, and then the corresponding Lorentz transformation matrix is 0

1 B0 B B B α¼B0 B B @ 0

0 1

0 0 0 0 iβ 1 0 pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi 1  β2 1  β2 1 iβ 0 pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1  β2 1β

1 C C C C C C C A

ð5:102Þ

Using transformation 0

x

1

ByC B C r¼B C @zA

r0 ¼ α r,

ð5:103Þ

ict we can obtain ðv=c2 Þz vt z t x 0 ¼ x, y 0 ¼ y, z0 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi  pffiffiffiffiffiffiffiffiffiffiffiffiffi , t 0 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi  pffiffiffiffiffiffiffiffiffiffiffiffiffi ð5:104Þ 1  β2 1  β2 1  β2 1  β2 where (x, y, z, t) are the quantities in the fixed coordinate system, and (x0, y0, z0, t0) are the quantities in the moving coordinate system. This transformation is called the true Lorentz transformation. The spatial reflection transformation matrix is 0

1

B 0 B α¼B @ 0 0

0

0

1

0

0 0

1 0

The time inversion transformation matrix is

0

1

0C C C 0A 1

ð5:105Þ

5.1

Basic Theory of Relativistic Quantum Mechanics

0

1 B0 B α¼B @0

0 0 1 0

0

0 0

0 1

363

1 0 0 C C C 0 A

ð5:106Þ

1

It can be seen that the transformation matrices given by Eqs. (5.102), (5.105), and (5.106) all satisfy Eq. (5.97) and Eq. (5.101). If we write a Dirac equation according to Eq. (5.33) 

 γ 0μ ∂μ þ κ ψ 0 ¼ 0

ð5:107Þ

o γ 0μ , γ 0ν ¼ 2δμν , the Pauli’s theorem is that if there are two sets n o of 4  4 matrices that satisfy { γ μ, γ ν} ¼ 2δμν and γ 0μ , γ 0ν ¼ 2δμν , respectively, where γ 0μ satisfies

n

there must be a non-singular and unique 4  4 matrix S, so that they satisfy the following transformation relations [1, 6, 7]: Sγ μ S1 ¼ γ 0μ ,

S1 γ 0μ S ¼ γ μ

ð5:108Þ

where the non-singular means that there must be an inverse matrix of S to satisfy SS1 ¼ 1. If γ 0μ is also Hermitian, from Eq. (5.108) we can prove that S is unitary, i.e., S+ ¼ S1. According to this theorem, Eq. (5.107) is rewritten as

 Sγ μ S1 ∂μ þ κ SS1 ψ 0 ¼ 0

ð5:109Þ

Multiply S1 from the left of the above formula and note that S is commutable with ∂μ, and then Eq. (5.109) becomes

 γ μ ∂μ þ κ S1 ψ 0 ¼ 0

ð5:110Þ

It can be seen that S1ψ 0 is the solution of the Dirac equation, so Eq. (5.107) is equivalent to the Dirac equation (5.33). The relations between ψ and ψ 0 are as follows: ψ 0 ¼ Sψ,

ψ 0 þ ¼ ψ þ S1

ð5:111Þ

Note that ανμ is only a matrix element; the following expressions can be written according to Eqs. (5.95) and (5.97): ∂0 ν ¼ ανμ ∂μ , where

0

∂μ ¼ ανμ ∂ν ,

0

0

0

γ μ ∂μ ¼ γ μ ανμ ∂ν ¼ ανμ γ μ ∂ν  γ ν ∂ν

ð5:112Þ

364

5

Polarization Theory of Relativistic Nuclear Reactions

γ ν  ανμ γ μ

ð5:113Þ

So the Dirac equation (5.33) can be rewritten according to Eq. (5.112) as follows:

 0 γ ν ∂ν þ κ ψ ¼ 0

ð5:114Þ

The following expressions can be proved utilizing Eqs. (5.29) and (5.101):

 γ μ γ ν þ γ ν γ μ ¼ αμρ γ ρ ανδ γ δ þ ανδ γ δ αμρ γ ρ ¼ αμρ ανδ γ ρ γ δ þ γ δ γ ρ ¼ 2δμν

ð5:115Þ

According to the Pauli theorem (5.108), there is a 4  4 matrix denoted as L, and the transformation of the γ matrix can be written as follows: γ μ ¼ αμν γ ν ¼ L1 γ μ L

ð5:116Þ 0

here L has no relationship with coordinates, and is commutable with ∂μ . Then substituting Eq. (5.116) into Eq. (5.114), we can obtain 

 0 L1 γ μ L∂μ þ κ ψ ¼ 0

Multiplying L from the left side of the above equation, the following equation can be got 

 0 γ μ ∂μ þ κ Lψ ¼ 0

ð5:117Þ

Therefore under the Lorentz transformation, the change of the wave function and the satisfied equation are, respectively ψ 0 ¼ Lψ,



 0 γ μ ∂μ þ κ ψ 0 ¼ 0

ð5:118Þ

This proves that the Dirac equation has covariance under the Lorentz transformation of four-dimensional space-time coordinates. The above transformation can also be discussed in another way. Let L be the transformation operator of the matrix element αμν, since this transformation only transforms the four-dimensional coordinates, its matrix elements are all independent of the coordinates, so the Dirac equation given by Eq. (5.33) can be converted into   0 γ μ ∂μ þ κ ψ 0 ðx0 Þ ¼ 0 and there is

ð5:119Þ

5.1

Basic Theory of Relativistic Quantum Mechanics

365

ψ 0 ðx0 Þ ¼ Lψ ðxÞ

ð5:120Þ

From Eq. (5.95) and Eq. (5.100), we can get 0

∂μ ¼ ανμ ∂ν

ð5:121Þ

Substituting Eq. (5.120) and Eq. (5.121) into Eq. (5.119) and multiplying L1 from the left side of the obtained formula, the following expression can be obtained:

1  L γ μ Lανμ ∂ν þ κ ψ ðxÞ ¼ 0

ð5:122Þ

L1 γ μ L ¼ αμν γ ν

ð5:123Þ

Only there is

Equation (5.122) may become Eq. (5.33). The conjugate equation of the Dirac equation (5.33) is 

∂μ ψ þ γ μ þ κψ þ ¼ 0

ð5:124Þ

Multiplying γ 4 from the right side of the above equation, we can obtain 

∂μ ψ þ γ μ γ 4 þ κψ þ γ 4 ¼ 0 



Noting γ i γ 4 ¼ γ 4 γ i , ∂i ¼ ∂i , ∂4 ¼ ∂4, based on the definition of Eq. (5.42), the following expression can be obtained from the above formula: ∂μ ψγ μ  κψ ¼ 0

ð5:125Þ

0

And using the inverse transformation ∂μ ¼ ανμ ∂ν and Eq. (5.116), from the above equation we can obtain 0

0

0

ανμ ∂ν ψγ μ  κψ ¼ ∂ν ψανμ γ μ  κψ ¼ ∂ν ψL1 γ ν L  κψ ¼ 0

ð5:126Þ

Multiplying L1 from the right side of the above equation, and let ψ 0 ¼ ψL1

ð5:127Þ

then we get the following equation satisfied by ψ 0 under the Lorentz transformation: 0

∂ν ψ 0 γ ν  κψ 0 ¼ 0

ð5:128Þ

366

5

Polarization Theory of Relativistic Nuclear Reactions

It can be seen that ψ after the transformation still satisfy the same equation with Eq. (5.125), so under the Lorentz transformation there is ψ ! ψ 0 ¼ ψL1

ð5:129Þ

Therefore using Eq. (5.129) and Eq. (5.118), we can get ψ 0 ψ 0 ¼ ψL1 Lψ ¼ ψψ

ð5:130Þ

The above formula shows that ψψ is an invariant under the Lorentz transformation, i.e., ψψ is a scalar under the Lorentz transformation. In the following we discuss the γ 5 behavior under the true Lorentz transformation given by Eq. (5.102). First we can find



 L1 γ 5 L ¼ L1 γ 1 LL1 γ 2 LL1 γ 3 LL1 γ 4 L ¼ α1μ γ μ ðα2ν γ ν Þ α3ρ γ ρ ðα4λ γ λ Þ ! ! iβ 1 iβ 1 pffiffiffiffiffiffiffiffiffiffiffiffiffi γ 3 þ pffiffiffiffiffiffiffiffiffiffiffiffiffi γ 4 ¼ γ 1 γ 2 pffiffiffiffiffiffiffiffiffiffiffiffiffi γ 3 þ pffiffiffiffiffiffiffiffiffiffiffiffiffi γ 4 1  β2 1  β2 1  β2 1  β2   β2 1 ¼ γ1γ2 γ γ þ γ4 γ3 ¼ γ1 γ2 γ3 γ4 ¼ γ5 2 3 4 1β 1  β2 The above formula indicates that L and γ 5 are commutable under the true Lorentz transformation, i.e., L1 γ 5 L ¼ γ 5 ,

γ 5 L ¼ Lγ 5

ð5:131Þ

Under the spatial reflection transformation, xi ! x0i ¼ xi , t ! t 0 ¼ t, it can be seen from Eq. (5.105) that αik ¼  δik, α44 ¼ 1. For γ matrices there are L1 γ i L ¼ γ i ,

γ i L ¼ Lγ i ,

i ¼ 1, 2, 3;

L1 γ 4 L ¼ γ 4 , γ 4 L ¼ Lγ 4 ð5:132Þ

The spatial reflection transformation matrix satisfying Eq. (5.132) can be expressed as L ¼ ξγ 4

ð5:133Þ

Substituting Eq. (5.133) into (5.132) can verify its validity. Here ξ represents the particle’s intrinsic parity, due to requirements LL ¼ I, L1 ¼ L, so there are ξ2 ¼ 1, ξ ¼ 1. The transformations of γ matrices during the spatial reflection are αμν γ ν ¼ L1 γ μ L ¼ ξ2 γ 4 γ μ γ 4 ¼ γ 4 γ μ γ 4

ð5:134Þ

Under the spatial reflection transformation, the transformation of Dirac spinor is

5.1

Basic Theory of Relativistic Quantum Mechanics

367

ψ ðxÞ ! Lψ ðxÞ ¼ γ 4 ψ ðxÞ

ð5:135Þ

The particle’s intrinsic parity may be þ1 or 1. We can still get L1 γ 5 L ¼ ξ2 γ 4 γ 5 γ 4 ¼ γ 5 γ 4 γ 4 ¼ γ 5

ð5:136Þ

So for the reflection of space there are γ 5 L ¼ Lγ 5 ,

L1 γ 5 ¼ γ 5 L1

ð5:137Þ

For the 4  4 matrices, there must be 16 linearly independent matrices, and all 4  4 matrices can be composed of their linear combinations. It is possible to select 16 linearly independent matrices, denoted by Γ A, composed by four γ matrices that satisfy the anti-commutative relations, and then any 4  4 matrix can be given with the linear combination of these linearly independent matrices Γ A. In the following all the possible multiple products of the γ matrices and the number of independent matrices composed by the various multiple products of the γ matrices are given. highest product

γ5 ¼ γ1γ2γ3 γ4

1

triple product double product

γ μ γ ν γ σ ðμ 6¼ v 6¼ σ Þ or iγ 5 γ ρ γ μ γ ν ðμ 6¼ νÞ

4 6

single product zero product

γμ I

4 1

We select 16 linearly independent 4  4 matrices Γ A as T Γ S ¼ 1, Γ V μ ¼ iγ μ , Γ μν ¼ σ μν 

 1

γ γ  γ ν γ μ , Γ P ¼ iγ 5 , Γ A μ ¼ iγ 5 γ μ 2i μ ν ð5:138Þ

We can verify, utilizing Eqs. (5.29), (5.39), and (5.40), that they have the following properties: þ 1 1. The 16 Γ A matrices all satisfy jΓ A j2 ¼ Γ þ A Γ A ¼ I, so Γ A ¼ Γ A . S 2. Except for the unit matrix Γ , for each Γ A there is another Γ B which is anticommutative with it, i.e., Γ AΓ B ¼  Γ BΓ A. 3. Except for the unit matrix Γ S, for each ΓA there is tr{Γ A} ¼ 0; when A 6¼ B there is tr{Γ AΓ B} ¼ 0. 4. The determinants of 16 Γ A matrices are all 1. Since the determinant has the following properties: if A and B are the square matrices of the same rank, the determinant jABj ¼ jAjjBj, and after the interchange of rows and columns (reflection about a diagonal) the determinant value does not change. Because Γ A is also a square 4  4 matrix, there is determinant det(Γ A) ¼ det (Γ A), and

368

5

Polarization Theory of Relativistic Nuclear Reactions

 using Eq. (5.27) it is possible to prove det Γ þ A ¼ detðΓ A Þ by Γ A given by Eq. (5.138). So by use of jΓ Aj2 ¼ I we can obtain  



þ 2 det jΓ A j2 ¼ det Γ þ A Γ A ¼ det Γ A  detðΓ A Þ ¼ ½detðΓ A Þ ¼ 1, detðΓ A Þ ¼ 1 However, in Γ A diagonal representation, except the unit matrix Γ S, their traces are still 0. At this time, the numbers of the value þ1 and 1 in the four diagonal elements must be equal, so it can only be that det(Γ A) ¼ 1. 5. The 16 Γ A matrices are linearly independent of each other. This conclusion can be 16 P CA Γ A ¼ 0, i.e., 16 Γ A are not independent, multiplying proved as follows: if A¼1

Γ B for the front of left side of the above formula one can obtain C B þ P C A Γ B Γ A ¼ 0; using the preceding 3. clause, tracing the above formula one A6¼B 16 P C A Γ A ¼ 0 are 0. That is, they can find CB ¼ 0. Since B is arbitrary, all of CA in A¼1

are linearly independent. 6. If a matrix is commutative with all 16 Γ A matrices, since the above 16 Γ A matrices are linearly independent of each other, this matrix must be a multiple of the unit matrix. In the theory of relativistic quantum mechanics, the physical quantity is a Dirac covariable composed by ψ, ψ and γ μ matrices. The Dirac covariables of all physical quantities may only have the following five types: ① Scalar S ¼ ψψ corresponds to zero product. ② Pseudo-scalar P ¼ iψγ 5 ψ corresponds to the highest product. ③ Vector V μ ¼ iψγ μ ψ corresponds to single product. ④ Axis-vector Aμ ¼ iψγ 5 γ μ ψ corresponds to triple product.

 ⑤ Anti-symmetric tensor T μν ¼ 2i1 ψ γ μ γ ν  γ ν γ μ ψ corresponds to double product. When defining these covariables, except the scalar, the factor i representing imaginary number are introduced in order to make them to meet the properties of vectors and tensors. For example, it is possible to ensure that the first three components of the vector are real, and the fourth component is imaginary, the sign of its Hermitian conjugate real part is invariant, and the sign of its Hermitian conjugate imaginary part is altered. These conclusions can be proved as follows:

þ ¼ iψ þ γ μ γ 4 ψ ¼ iψγ 4 γ μ γ 4 ψ ¼ ðiψγ i ψ, iψγ 4 ψ Þ Vþ μ ¼ iψγ μ ψ And keep that the pseudo-scalar is real: Pþ ¼ iψ þ γ 5 γ 4 ψ ¼ iψγ 5 ψ ¼ P.

5.1

Basic Theory of Relativistic Quantum Mechanics

369

Under the true Lorentz transformation, the scalar is invariant: S0 ¼ ψ L1 Lψ ¼ ψψ ¼ S

ð5:139Þ

Using Eq. (5.131), under the true Lorentz transformation, the pseudo-scalar is also invariant: P0 ¼ iψ 0 γ 5 ψ 0 ¼ iψL1 γ 5 Lψ ¼ iψγ 5 ψ ¼ P

ð5:140Þ

Under the spatial reflection transformation there is still S0 ¼ ψ L1 Lψ ¼ S

ð5:141Þ

However, due to Eq. (5.137) there is P0 ¼ iψ L1 γ 5 Lψ ¼ iψ L1 Lγ 5 ψ ¼ iψγ 5 ψ ¼ P

ð5:142Þ

Therefore, P is called a pseudo-scalar. Under the true Lorentz transformation, the transformation of the vector Vμ is V 0μ ¼ iψ 0 γ μ ψ 0 ¼ iψL1 γ μ Lψ ¼ αμν ðiψγ ν ψ Þ ¼ αμν V ν

ð5:143Þ

and the transformation of the axis vector Aμ is A0μ ¼ iψ 0 γ 5 γ μ ψ 0 ¼ iψL1 γ 5 γ μ Lψ ¼ iψγ 5 L1 γ μ Lψ ¼ αμν Aν

ð5:144Þ

It can be seen that this transformation is the same as the vector Vμ. However, under the spatial reflection transformation, for the vector Vμ and the axis vector Aμ, using Eqs. (5.105) and (5.137), the following results can be obtained, respectively: ( V 0μ ¼ iψ 0 γ μ ψ 0 ¼ iψL1 γ μ Lψ ¼ αμν ðiψγ ν ψ Þ ¼ αμν V ν ¼ ( A0μ ¼ iψ 0 γ 5 γ μ ψ 0 ¼ iψL1 γ 5 γ μ Lψ ¼ αμν ðiψγ 5 γ ν ψ Þ ¼

V i , i ¼ 1,2,3 V 4,

μ¼4 ð5:145Þ

Ai ,

i ¼ 1,2,3

A4 ,

μ¼4 ð5:146Þ

It can be seen that the spatial inversion transformation of the axis-vector Aμ differs by a negative sign from the transformation of the vector Vμ, so Aμ is called the axis vector.

370

5

Polarization Theory of Relativistic Nuclear Reactions

Table 5.1 Dirac covariables and their transformation properties under the true Lorentz transformation and the spatial reflection transformation

Scalar Pseudo-scalar Vector

Dirac covariable S ¼ ψψ P ¼ iψγ 5 ψ V μ ¼ iψγ μ ψ

True Lorentz transformation S0 ¼ S P0 ¼ P V 0μ ¼ αμv V v

Axis-vector

Aμ ¼ iψγ 5 γ μ ψ

A0μ ¼ αμν Aν

Anti-symmetric tensor

1 T μν ¼ ψ 2i

 γμ γν  γνγμ ψ

T 0μν ¼ αμρ ανσ T ρσ

Space reflection transformation S0 ¼ S P0 ¼  P V 0i ¼ V i , V 04 ¼ V 4 A0i ¼ Ai , A04 ¼ A4 T 0ij ¼ T ij , T 0i4 ¼ T i4 , T 04j ¼ T 4j , T 044 ¼ T 44

Number 1 1 4 4 6

Under the true Lorentz transformation, the transformation of the anti-symmetric tensor Tμν is 

 1 0

1 ψ γ μ γ ν  γ ν γ μ ψ 0 ¼ ψL1 γ μ γ ν  γ ν γ μ Lψ 2i 2i  1

¼ ψ αμρ γ ρ ανσ γ σ  ανσ γ σ αμρ γ ρ ψ 2i  1

¼ αμρ ανσ ψ γ ρ γ σ  γ σ γ ρ ψ ¼ αμρ ανσ T ρσ 2i

T 0μν ¼

ð5:147Þ

Under the spatial reflection transformation, using Eq. (5.105) the transformations of the anti-symmetric tensor Tμν are



 T 0 ij ¼ αiρ αjσ T ρσ ¼ δiρ δjσ T ρσ ¼ T ij , T 0 i4 ¼ αiρ α4σ T ρσ ¼ δiρ δ4σ T ρσ ¼ T i4 ,

T 0 4j ¼ T 4j ,

ð5:148Þ

T 0 44 ¼ α4ρ α4σ T ρσ ¼ T 44 Summarizing the above results, all possible Dirac covariables and their transformation properties under the true Lorentz transformation and spatial reflection transformation are listed in Table 5.1. Any other form of Dirac covariables can always be represented by these covariables.

5.1.6

The Trace Formulas for γ Matrix Products

The trace formulas for the γ matrix product of the Bjorken-Drell metric are given below. The trace of an odd number of γ μ matrix product is zero, and for even number of γ μ matrix products, the following trace formulas are found [1, 8]:

5.2

Transformation of Relativistic Coordinate Systems

371

trf γ μ γ ν g ¼ 4gμν μ ν ρ σ

μν ρσ

ð5:149Þ

μσ νρ

μρ νσ

trf γ γ γ γ g ¼ 4ðg g þ g g  g g Þ trf γ σ γ μ γ ν γ ρ g ¼ trf γ μ γ ν γ ρ γ σ g h

  tr γ μ γ ν γ ρ γ σ γ λ γ η ¼ 4 gμν gρσ gλη þ gρη gσλ  gρλ gση



 gμρ gνσ gλη þ gνη gσλ  gνλ gση þ gμσ gνρ gλη þ gνη gρλ  gνλ gρη

i gμλ ðgνρ gση þ gνη gρσ  gνσ gρη Þ þ gμη gνρ gσλ þ gνλ gρσ  gνσ gρλ

ð5:150Þ ð5:151Þ

ð5:152Þ

where gμν is the matrix element of the metric tensor given by Eq. (5.53).

5.2

Transformation of Relativistic Coordinate Systems *

In the Pauli  metric, the  momentum-energy 4 vector composed of momentum p and E * energy E is p , i , and let c v β¼ , c

γ ¼ 1=

qffiffiffiffiffiffiffiffiffiffiffiffiffi 1  β2

ð5:153Þ

The Lorentz transformation matrix between the two coordinate systems moving each other at a speed v in the z-axis direction has been given by Eq. (5.102) as 0

1

B0 B α¼B @0

0

0

1 0

0 γ

0

0

iβγ

γ

1 0

0

0

0

1

0 C C C iβγ A

ð5:154Þ

From Eq. (5.100) one can also know 0

B0 1 B α1 ¼ B @0 0 0 0

0 γ iβγ

1

0 C C C iβγ A γ

The Lorentz transformation relation of the momentum-energy 4 vector is

ð5:155Þ

372

0

5

px 0

1

0

1 0

B p0 B y B @ pz 0

C B0 1 C B C¼B A @0 0

iE0 =c

0 0

0

0

Polarization Theory of Relativistic Nuclear Reactions

10

px

1

0

iβγ

1

C C B C B

 C C ð5:156Þ C¼B A @ γ pz  βE=c A

 iE=c iγ βpz  E=c

B 0 C CB p y CB iβγ A@ pz

0 γ

px py

γ

Adding the square of the third row and the square of the fourth row of the right end of the above equation, we can obtain h i h i γ 2 p2z þ β2 ðE=cÞ2  2βpz E=c  γ 2 β2 p2z þ ðE=cÞ2  2βpz E=c



 ¼ γ 2 1  β2 p2z  γ 2 1  β2 ðE=cÞ2 ¼ p2z  ðE=cÞ2

ð5:157Þ

It can be seen that the following relation does guarantee p0 2x

þ

p0 2y

þ

p0 2z

 0 2  2 E E  ¼ p2x þ p2y þ p2z  c c

ð5:158Þ

Then the constant m can be introduced, let p2 

 2 E ¼ m2 c2 c

ð5:159Þ

And above formula can be rewritten as E 2 ¼ p2 c 2 þ m 2 c 4

ð5:160Þ

where m is the stationary mass of the particle. * In the S0 system set p 0 ¼ 0, which indicates that the particle is stationary in the S0 system. Then the following formula can be obtained by Eq. (5.160): E 0 ¼ mc2

ð5:161Þ

The inverse transformation of momentum-energy 4 vector can be done with Eq. (5.155): 0

1 0 px 1 B p C B0 B y C B B C¼B @ pz A @ 0 iE=c So we can get

0

0 1

0 0

0

γ

0

iβγ

1 0 10 1 0 0 0 B C B C 0 C CB 0 C B 0 C C¼B CB C iβγ A@ 0 A @ γmv A γ iγmc iE0 =c

ð5:162Þ

5.2

Transformation of Relativistic Coordinate Systems

373

pz ¼ γmv

ð5:163Þ

E ¼ γmc2

ð5:164Þ

v¼

p c2 pz ¼ z γm E

ð5:165Þ

The reference system with a total momentum of zero is called the center of momentum system, referred to as c.m. system. From Eq. (5.163) it can be seen that the momentum is not proportional to speed, so the center of momentum system and the center of mass system are different. Only if they degenerate to satisfy the non-relativistic conditions, they are consistent to each other. For two particle center of momentum system the requirement is *

*

*

p ¼ p1 þ p2 ¼ 0

ð5:166Þ

*

In the laboratory system, the momentum p 1 of the incident particle m1 follows the zaxis direction, and the target nucleus m2 is stationary. Using the symbol with “ 0 ” to represent the physical quantity in the center of momentum system, so we can write the following Lorentz transformations as 0

0

1

0

1

B C B B 0 C B0 B C B B 0 C¼B B p C B0 @ 1z A @ iE 01 =c 0 0

0 B 0 B B 0 @ p2z iE 02 =c

1

0

1 C B0 C B C¼B A @0 0

0

0

1

0

0

γ

0

iβγ

0 1

0 0

0 0

γ iβγ

0

10

0

1

0

CB B 0 C CB 0 CB B iβγ C A@ p1z

1

0

C B C C B C 0 C B C  C C¼B

C B γ p1z  βE 1 =c C A @ A 

iγ βp  E =c iE 1 =c 1 1z

γ

10

0 B 0 C CB CB iβγ A@ γ

0 0 0 iE 2 =c

1

0

1

0 C B C 0 C B C C¼B C A @ γβE 2 =c A

ð5:167Þ

ð5:168Þ

iγE 2 =c

From the above two formulas we may get   E p01z ¼ γ p1z  β 1 c

 0 E 1 ¼ γ βcp1z  E 1

ð5:169Þ ð5:170Þ

374

5

Polarization Theory of Relativistic Nuclear Reactions

p02z ¼ γβm2 c

ð5:171Þ

E02 ¼ γm2 c2

ð5:172Þ

Based on p01z þ p02z ¼ 0 we can find p1z  β

E1  βm2 c ¼ 0, c

β¼

p1z c E1 þ m2 c2

ð5:173Þ

Here v ¼ βc is the speed of the motion coordinate system relative to the stationary particle 2. From Eq. (5.160) we get E1 ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p21z c2 þ m21 c4

ð5:174Þ

The total energy of the c.m. system is E 0 ¼ E01 þ E 02

ð5:175Þ

The following expression can be obtained from Eqs. (5.170) and (5.172)–(5.175): h

i 2 1 m2 c2 þ E 1  β2 E1 þ m2 c2 2 1β





2 2 2 2 ¼ 1  β E 1 þ m2 c ¼ E 1 þ m2 c2  p21z c2

2 E0 2 ¼ γ 2 m2 c2 þ E1  βcp1z ¼

ð5:176Þ

¼ E 21 þ 2E1 m2 c2 þ m22 c4  E 21 þ m21 c4 ¼ 2E 1 m2 c2 þ m21 c4 þ m22 c4 The total energy E0 of the c.m. system is also called reaction effective energy. In ħ ¼ c ¼ 1 unit system, define    1=2 pffiffi ! ! 2 2 s ¼ ðE 1 þ E 2 Þ  p 1 þ p 2

ð5:177Þ

pffiffi Comparing this formula with Eq. (5.3), it can be seen that s corresponds to a ! ! stationary mass term. In the c.m. system of two particles, there is p 1 þ p 2 ¼ 0, then we can obtain pffiffi s ¼ E1 þ E2

ð5:178Þ

pffiffi So it can be seen that s is the total energy of the two particles in the c.m. system, which contains the energy of the stationary mass. It has been assumed that the S0 system is uniformly moving with speed v along the z-axis direction relative to the

5.2

Transformation of Relativistic Coordinate Systems

375

S system, and the physical quantity in the S0 system is represented by the symbol with “ 0 ”. Eq. (5.156) has given p0x ¼ px , p0y ¼ py ,

 p0z ¼ γ pz  βE ,

 E 0 ¼ γ βpz  E

ð5:179Þ

So we can first find

0 2



 2 E 1 þ E 02 ¼ γ 2 β p1z þ p2z  E 1 þ E 2 h

2

2



i ¼ γ 2 β2 p1z þ p2z þ E 1 þ E2  2β p1z þ p2z E 1 þ E2

ð5:180Þ







2 2

2 !0 !0 2 p 1 þ p 2 ¼ p1x þ p2x þ p1y þ p2y þ γ 2 p1z þ p2z  βðE 1 þ E2 Þ

2

2

2

2 ð5:181Þ ¼ p1x þ p2x þ p1y þ p2y þ γ 2 p1z þ p2z þ γ 2 β2 E 1 þ E2



 2γ 2 β p1z þ p2z E1 þ E 2

pffiffi Using the above two formulas and Eq. p (5.177), it can be proved that s is invariant ffiffi in different coordinate systems, i.e., s is Lorentz invariant. In the laboratory system where particle 1 is incident and particle 2 is stationary, from Eqs. (5.177) and (5.3) we can obtain 1=2 pffiffi 2 s ¼ m1 þ m22 þ 2m2 E 1

ð5:182Þ

This formula is consistent with Eq. (5.176). The wave function equation in the interaction representation is (see Eq. (3.14.6) in Ref. [4]) i

∂ j t i ¼ V I ðt Þ j t i ∂t

ð5:183Þ

where VI(t) is the residual interaction. Here we use the ħ ¼ c ¼ 1 unit system. For wave functions we have the relation jt i ¼ uðt, t 0 Þ jt 0 i

ð5:184Þ

S  uð1, 1Þ

ð5:185Þ

S matrix element is defined as

We discuss the reaction process shown in Fig. 5.1 [9]. Here use 1 and 10 to represent incident and outgoing particles, and use 2 and 20 to represent the target nucleus and the residual nucleus, respectively. Set j1, 2i as the t ¼ 1 initial state, j10, 20i is the t ¼ 1 final state, and its reaction amplitude is

376

5

Polarization Theory of Relativistic Nuclear Reactions

Fig. 5.1 Reaction process 1 þ 2 ! 10 þ 20

h10 , 20 jSj1, 2i

ð5:186Þ

The corresponding probability is W f ¼ jh10 , 20 jSj1, 2ij

2

ð5:187Þ

In the reaction process, the energy conservation and momentum conservation must be satisfied. From Eq. (5.80) we can see that the free nucleon plane wave function is !*  eið p  r EtÞ , it is required to have an entire number of cycles in a square box with ! volume Ω, so the momentum p which can be taken is !

p¼

2π ðl, m, nÞ, Ω1=3

l, m, n ¼ 0, 1, 2,⋯ ðexcept l ¼ m ¼ n ¼ 0 Þ ð5:188Þ

where l, m, n represent the integer values taken in the x, y, z directions, respectively, ! * so that p  r is an integer multiple of 2π at the square box boundary of the volume Ω. Each group l, m, n corresponds to a wave vector, i.e., p is a separate variable, whereas p becomes a continuous variable only when Ω ! 1. Using the periodic boundary condition Eq. (5.188), one can easily obtain 1 Ω

Z

d r eið p  p Þ r ¼ δ!p 0 ,!p *

Ω

!0

!

*

ð5:189Þ

There is also the following δ function formula: 1 ð2πÞ3

Z

  * !0 ! * ! ! d r ei ð p  p Þ  r ¼ δ p 0  p

ð5:190Þ

Comparing the above two formulas shows that there is the following correspondence:   Ω ! ! δ! ! , δ p 0  p , 3 p 0, p ð2πÞ For momentum-energy 4 vectors there is

when Ω ! 1

ð5:191Þ

5.2

Transformation of Relativistic Coordinate Systems

Ω δp0 ,p , δ4 ðp0  pÞ, ð2πÞ4

377

when Ω ! 1

ð5:192Þ

For the two-particle subsystem, Ω should become Ω2, and there is

 Ω2 δ0 0 , δ4 p01 þ p0 2  p1 þ p2 , 4 p1 þp2 , p1 þp2 ð2πÞ

when Ω ! 1

ð5:193Þ

So we can write Eq. (5.186) as [9] h10 , 20 jSj1, 2i ¼ M

 ð2πÞ4 4 0 δ p1 þ p02  p1  p2 2 Ω

ð5:194Þ

where p1 , p2 , p01 , and p02 are the 4 momentum of particles 1, 2, 10 , and 20 , respectively. Ω represents the system volume. M is transition matrix element and contains the four exterior lines and the t matrix representing the interaction at the vertex. According to

 the graphical rules of quantum field theory, ð2πÞ4 δ4 p01 þ p02  p1  p2 is contributed by the vertex. Substituting Eq. (5.194) into Eq. (5.187), we can obtain W f ¼ jM j2

2   ð2πÞ4 4 0 0 p p δ p þ p 1 2 1 2 Ω2

ð5:195Þ

There are the following relations: δ 4 ð pÞ ¼

1 ð2πÞ4

Z

Z d4 xeipx , Z

ð2πÞ4 δ4 ðpÞ ¼

d4 xeipx

h i2 d4 xeipx ¼ ð2πÞ4 δ4 ðpÞ

4 4

ð2πÞ δ ðpÞ

ð5:196Þ ð5:197Þ

Since there is δ4( p) at the left end of the above formula, one can make p ¼ 0 in the integrand, and the following relation can be obtained from the definition of spacetime: Z d4 x ¼ ΩT

ð5:198Þ

where T represents the interval of time, and at last one must let Ω and T ! 1. Then we can obtain the following expression by Eq. (5.197):

378

5

Polarization Theory of Relativistic Nuclear Reactions

Fig. 5.2 Two-particle nuclear reaction diagram in center of momentum system, if both particle 1 and particle 2 are in the cylinder vTdσ, the reaction 1 þ 2 ! 10 þ 20 will occur within time T

h

ð2πÞ4 δ4 ðpÞ

i2

¼ ð2πÞ4 δ4 ðpÞΩT

ð5:199Þ

Substituting Eq. (5.199) into Eq. (5.195) we get W f ¼ jM j2

 ð2πÞ4 4 0 δ p1 þ p02  p1  p2 ΩT 4 Ω

ð5:200Þ

We study the nuclear reaction between the incident particle labeled with 1 and the target nucleus labeled with 2. The motion direction of the incident particle 1 is selected as the z-axis. In the laboratory system, the particle 2 is stationary, and we only consider the coordinate system of inertial motion along the z-axis. In the two-particle center of momentum coordinate particles 1 and 2 move in the * system, *  direction of the z-axis at speed v ¼  v 1  v 2  ¼ v1  v2 , where v1 and v2 representing the velocities of particles 1 and 2 in the z-axis direction, and v1 > v2 is demanded. We assume that 1 þ 2 ! 10 þ 20 reaction cross section is dσ. Now let us consider the cylinder shown in Fig. 5.2. Since the initial state corresponds to the states of only two particles 1 and 2 in the volume Ω, whereas the states, where particles 1 and 2 leave far away, need not to be considered. So the probability that particles 1 and 2 are in the column is vTdσ Ω

ð5:201Þ

If particles 1 and 2 are all in the column, the 1 þ 2 ! 10 þ 20 reaction will occur in time T, so its probability is equal to  2   T  0 0 T  h1 , 2 ju ,  j1, 2i 2 2

ð5:202Þ

The u in the above equation is just the u in Eq. (5.184). At T ! 1 limit, Eq. (5.202) becomes Eq. (5.200), and let Eq. (5.200) and Eq. (5.201) be equivalent to each other, then we can obtain σ¼

X ð2πÞ4 Ω v 2

 jM j2 δ4 p01 þ p02  p1  p2

ð5:203Þ

5.2

Transformation of Relativistic Coordinate Systems

379

The summation symbol is added to the right side of the above formula, which *0 *0 indicates to spread throughout various three-dimensional momentum p 1 and p 2 of the end state, while dσ in the left side of the equation also becomes σ representing the integral cross section. According to the phase space theory, we know X !

pi

Ω ! ð2πÞ3

Z

*

ð5:204Þ

d3 p i

Then the following expression can be obtained by Eq. (5.203): σ¼

1 ð2πÞ2 ðv1  v2 Þ

Z

 *0 *0 d3 p 1 d3 p 2 jM j2 δ4 p01 þ p02  p1  p2

ð5:205Þ

Since the particles 1 and 2 only move in the z-axis direction, using Eq. (5.165) we can obtain *  *    *   p 1   p 2  E 2 * * p 1   E1  p 2  *    ¼ v ¼ v 1  v 2 ¼ v1  v 2 ¼ E1 E2 E1 E2

ð5:206Þ

So Eq. (5.205) can be rewritten as σ¼

E E   1 2 *  * 2  ð2πÞ E 2 p 1   E 1  p 2 

Z

 *0 *0 d3 p 1 d3 p 2 jM j2 δ4 p01 þ p02  p1  p2 ð5:207Þ

Note that Eq. (5.205) or Eq. (5.207) gives an integral cross section that holds in any coordinate system of inertial motion along the z-axis. Eq. (5.207) can be rewritten as Z σ¼

*0

*0

d3 p 1 d 3 p 2

d6 σ

*0

ð5:208Þ

*0

d3 p 1 d 3 p 2

where d6 σ

*0 *0 d3 p 1 d 3 p 2

¼

 E1 E2 2 4 0 0 *  *  jM j δ p1 þ p2  p1  p2 ð2πÞ E2  p 1   E 1  p 2  2



*0 *0

ð5:209Þ

is a 6-differential cross section of two emitted particles with definite p 1 p 2 . Assume that Eq. (5.209) is an expression given in the S system, and it is assumed that the S0 system moves at a uniform speed v in the z-axis direction relative to the S system. We make convention that the relative velocity between the two coordinate systems is less than the velocity of the incident particle 1 in the laboratory system. And assume

380

5 Polarization Theory of Relativistic Nuclear Reactions

that the momentum-energy 4 vector of the particles in the S0 system is



 q , ie .

*

According to Eq. (5.209) the following expression can be written in the S0 system: d6 σ0

!0 !0 d3 q 1 d3 q 2

¼

 e e 2 4 0 0   1 2 *  jN j δ q1 þ q2  q1  q2 * ð2πÞ e2  q 1   e1  q 2  2

ð5:210Þ

where N is a transition matrix element in the S0 system. There are the relations q01x ¼ p01x , q01y ¼ p01y , q02x ¼ p02x , q02y ¼ p02y

ð5:211Þ

And using the transformation relations given by Eq. (5.156), we can obtain q1 ¼ γp1  βγE 1 , e1 ¼ βγp1 þ γE1 , q2 ¼ γp2  βγE 2 , e2 ¼ βγp2 þ γE 2 , q01z ¼ γp01z  βγE 01 , e01 ¼ βγp01z þ γE 01 , q02z ¼ γp02z  βγE 02 , e02 ¼ βγp02z þ γE 02 ð5:212Þ Note that q1, q2, p1, p2 represent the momentum values of the particles in the above equations; they are not the momentum-energy 4 vector. First we find the factor ðe2 q1  e1 q2 Þ=γ 2 ¼ ðβp2 þ E 2 Þðp1  βE 1 Þ  ðβp1 þ E 1 Þðp2  βE2 Þ ¼ βp1 p2 þ β2 p2 E1 þ E 2 p1  βE 1 E 2 þ βp1 p2  β2 p1 E2

 E1 p2 þ βE 1 E 2 ¼ 1  β2 ðE 2 p1  E 1 p2 Þ ð5:213Þ Then we can get γ 2 ðE 1  βp1 ÞðE2  βp2 Þ e 1 e2 ¼ E 2 p1  E 1 p2 e 2 q1  e 1 q2

ð5:214Þ

We have known     δ q01x þ q02x  q1x  q2x ¼ δ p01x þ p02x  p1x  p2x ,     δ q01y þ q02y  q1y  q2y ¼ δ p01y þ p02y  p1y  p2y

ð5:215Þ

Because the incident particle 1 and target 2 move only in the z direction in the S and S0 systems before the collision, therefore there are actually p1x ¼ p2x ¼ p1y ¼ p2y ¼ 0 and q1x ¼ q2x ¼ q1y ¼ q2y ¼ 0 in Eq. (5.215). Using Eq. (5.212) we can obtain

5.2

Transformation of Relativistic Coordinate Systems

381

 1



 δ q01z þ q02z  q1  q2 ¼ δ p01z þ p02z  p1  p2  β E 01 þ E 02  E 1  E2 , γ

0  1 0 

 0 δ e1 þ e2  e1  e2 ¼ δ E 1 þ E02  E1  E 2  β p01z þ p02z  p1  p2 γ ð5:216Þ So we can get



 δ q01z þ q02z  q1  q2 δ e01 þ e02  e1  e2 

 1

¼ 2 δ p01z þ p02z  p1  p2 δ E01 þ E 02  E 1  E 2 γ

ð5:217Þ

The solution of the positive energy plane wave of the Dirac equation has been given by Eq. (5.93) as     * * us p ¼ u p , þ χ s

ð5:218Þ

rffiffiffiffiffiffiffiffiffiffiffiffiffi0 ^I 1 E þ m@ * * u p, þ ¼ σ^  p A 2m Eþm

ð5:219Þ

where 



where χ s is a nucleon spin wave function with a spin projection of s. In Eq.(5.218)  * the upper and lower components of the four-dimensional wave functions u p , þ are permitted to act on the nucleon spin wave functions χ s, respectively. Let ^t S and ^t S0 be the t matrix operators of the two-body NN interaction of the S system and the S’ system, respectively, and the nuclear force or t matrix generally belongs to be the Lorentz invariant. The two-body transition matrix elements appearing in Eqs. (5.209) and (5.210) are as follows, respectively:  0  0     * * * * M ¼ us02 p 2 us01 p 1 ^t S us1 p 1 us2 p 2  0  0     * * * * N ¼ us02 q 2 us01 q 1 ^t S0 us1 q 1 us2 q 2

ð5:220Þ ð5:221Þ

0 where us ¼ uþ s γ . Since the Dirac equation satisfies Lorentz invariance, its solutions satisfy the following relations:

 0  0  0  0 * * * * us01 q 1 ¼ us01 p 1 , us02 q 2 ¼ us02 p 2 ,         * * * * u s 1 q 1 ¼ u s 1 p 1 , us 2 q 2 ¼ u s 2 p 2 And since there is ^t S0 ¼ ^t S , so we can get

ð5:222Þ

382

5

Polarization Theory of Relativistic Nuclear Reactions

N ¼M

ð5:223Þ

The following expression can be obtained comparing Eq. (5.209) with Eq. (5.210) and using Eqs. (5.214), and (5.217), and (5.223): 6 0

d σ

!0 !0 d3 q 1 d3 q 2

¼

 *   *  E 1  β p 1  E 2  β p 2  E1 E2

d6 σ

*0 *0 d3 p 1 d3 p 2

ð5:224Þ

From the above formula it can be seen that when v ¼ β ¼ 0 the differential cross sections in the two coordinate systems are equal.

5.3

Relativistic Optical Model and Phenomenological Optical Potential

In 1979, Arnold and Clark [10, 11] developed a nucleon relativistic optical model based on the Dirac equation. And the elastic scattering data of the medium energy protons were analyzed with relativistic phenomenological optical potential (RPOP). In order to better meet the experimental data of the proton elastic scattering angular dσ distribution , the polarization analyzing power Ay(θ), and spin rotation function dΩ Q(θ) for some specific targets and energies, the expressions and parameters of RPOP were further studied [12–15]. Then the energy-related proton RPOP [16–19] and neutron RPOP [18, 19] for certain target nuclei were studied. Later the universal proton RPOP [20–22] and neutron RPOP [23, 24] that can be used in a variety of target nuclei and energy regions below 1000 MeV were also developed. The neutrons and protons in the nucleus are all in the bound states, they all belong to the positive energy nucleons of total energy smaller than Mtc2(t ¼ n, p), Mt is the stationary mass of the nucleon t. The nuclear density is close to constant in the center region of the nucleus. If we look at the nucleus as an infinite medium without a surface area, we can introduce the hypothetical concept of nuclear matter. The nucleons in the nuclear matter are generally the positive energy nucleons of total energy smaller than Mtc2. If we study one of the nucleons in the nuclear matter, this nucleon must be in the potential field generated by other nucleons around it. This potential field is generated by this nucleon exchanging mesons with the surrounding nucleons in the nuclear matter. We call this potential field the nucleon self-energy Σ t, and we can also call Σ t the mean field felt by this nucleon. The studied nucleon may be in the bound state of total energy smaller than Mtc2; and if it obtains enough kinetic energy, then it may also be in the scattering state with the energy greater than Mtc2. However, as long as it is in nuclear matter, it will still be affected by the selfenergy Σ t. According to the local density approximation, if a nucleon is in the nuclear surface area with smaller nuclear density, the nucleon is considered to be in the nuclear matter with smaller nuclear density. Only when a nucleon is far from

5.3

Relativistic Optical Model and Phenomenological Optical Potential

383

the nucleus and is no longer affected by the nuclear potential field, it does think that the nucleon has become a free nucleon. The total energy of a free nucleon must be greater than Mtc2. In order to solve the relativistic two-body motion equation, it is generally decomposed into a free motion equation that describes the system center of momentum motion without potential field and a one-body motion equation that uses the mean field method to describe the relative motion of two particles in the center of momentum system. Dirac equation is a one-body motion equation that describes the 1 particle with the target nucleus in the center of momentum reaction of spin 2 system. Under normal circumstances, there is no need to solve the motion equation of the system center of momentum. It is only necessary to solve the one-body equation describing the movement with the two-body interaction in the center of momentum system. If necessary, the calculation results can be transformed to the laboratory system. In the ħ ¼ c ¼ 1 unit system, the stationary Dirac equation for free nucleon can be written from Eqs. (5.20) and (5.19) as follows: 

 * α  k t þ β M t ψ t ¼ Et ψ t

*

ð5:225Þ

If the studied nucleon is in an infinite nuclear matter unrelated to space coordinate, it will feel the existence of self-energy (mean field) Σ t(kt), and Eq. (5.225) becomes [25] h

i ! α  k t þ βðM t þ Σ t ðk t ÞÞ ψ t ðkt Þ ¼ Et ψ t ðk t Þ

*

ð5:226Þ

In a static infinite nuclear matter that is independent of space coordinate, the selfenergy Σ t(kt) should satisfy translational and rotational invariance, and it is assumed that it also satisfies the spatial reflection invariance. As can be seen from Eq. (5.138) and Table 5.1, under the

Pauli metric, the 1 γ μ γ ν  γ ν γ μ are 16 linearly independent matrices I, iγ 5, iγ μ, iγ 5γ μ and σ μν ¼ 2i scalar, pseudo-scalar, vector, axis-vector, and tensor matrices, respectively. Under the Bjorken-Drell (BD) metric, the following 16 linear independent matrices can be represented as I, γ 5, γ μ, γ 5γ μ and σ μν ¼ 2i ðγ μ γ ν  γ ν γ μ Þ . Although known by Eq. (5.57), γ 4 ¼ γ 0, γ i ¼  iγ i, but known by Eqs. (5.32) and (5.56), x4 ¼ ix0, xi(Pauli metric) ¼  xi(BD metric), therefore it is available that γ μxμ(Pauli metric) ¼ iγ μxμ(BD metric). Notice σ μμ ¼ 0, the most common form of self-energy Σ t of the nucleon t under the BD metric is [2, 26, 27] μ μ,v 5 μ μ,pv þ σ 0i Σ 0i,T þ σ ij Σ ij,T Σ t ¼ Σ st þ γ 5 Σ ps t þ γ Σt þ γ γ Σt t t 0

ð5:227Þ 0

It can be seen from Eq. (5.61) that under spatial reflection k 0 ¼ k0 , ki ¼ k i , and using Eq. (5.132) it can be seen that k it γ i remains unchanged under spatial reflection.

384

5 Polarization Theory of Relativistic Nuclear Reactions

Note that there is a coefficient β at the front of Eq. (5.226), and it can be seen from Eq. (5.132) that β ¼ γ 4 ¼ γ 0 is also unchanged under the spatial reflection transformation. Eq. (5.58) has given γ 5 ¼  γ 5, and according to Eq. (5.136) or Eq. (5.137) it can be seen that the sign of γ 5 will change under spatial reflection, so that the second term representing the pseudo-scalar (ps) and the fourth term representing the axis-vector (pv) in Eq. (5.227) do not satisfy the parity conservation, i.e., they cannot be selected. Note that the following covariant 4-vectors can be given by Eq. (5.61): 0

1 0 1 k0 E B k C Bk C B 1 C B xC k¼B C¼B C, @ k2 A @kyA k3

0

1 0 1 ρ j0 B j C Bj C B 1 C B xC j¼B C¼B C @ j2 A @jyA

k z

ð5:228Þ

jz

j3

 *

where ρ and j jx , jy , jz represent the nucleon density and nucleon flow density, respectively. Since it is assumed that the nuclear matter has a uniform finite density and is in a stable state, so the nucleon flow density is 0, such that the tensor formed by kit jjt will not contribute, therefore the last item formed by σ ij in Eq. (5.227) does not contribute; however the term related to k it j0t ¼ kit ρt has contribution to the penultimate item in Eq. (5.227). And let 0 0 γ 0 Σ 0,v t  γ Σt ,

i i v γ i Σ i,v t  γ kt Σ t ,

T

σ 0i Σ 0i,T  iγ 0 γ i k it ρt Σ t  γ 0 γ i kit Σ Tt t

ð5:229Þ

so Eq. (5.227) can be rewritten as *

*

*

*

Σ t ¼ Σ st þ γ 0 Σ 0t þ γ  k t Σ vt þ γ 0 γ  k t Σ Tt *

ð5:230Þ

*

where γ  k t ¼ γ i kit . Σ st , Σ 0t , Σ vt , and Σ Tt are the scalar, 4-vector time-like component, 4-vector space-like components, and tensor terms, respectively. For the sake of simplicity, we omit the tensor term, so Eq. (5.230) is simplified to *

*

Σt ¼ Σst þ γ 0 Σ0t þ γ  k t Σ vt

ð5:231Þ

In the ħ ¼ c ¼ 1 unit system, substituting Eq. (5.231) into Eq. (5.226) and using Eq. (5.57), as well as noting that αi in the BD metric is equal to αi in the Pauli metric, so we can obtain h

i * *

 1 þ Σ vt α  p t þ β M t þ Σ st þ Σ 0t ψ t ¼ Et ψ t p2

ð5:232Þ

If we only study the on-shell nucleon, that is, E t ¼ 2Mt , then the self-energy t Σ vt , Σ st , Σ 0t and wave function ψ t are only related to pt and Fermi momentum ktF, where ktF is determined by the density of the nuclear matter of the nucleon t. If knF 6¼ kpF, it is an asymmetric nuclear matter that shows that the number of neutrons

5.3

Relativistic Optical Model and Phenomenological Optical Potential

385

and the number of protons are not equal. In general, ktF is not explicitly included in the expression.

 The various terms in Eq. (5.232) are divided by 1 þ Σ vt and note that 1 þ Σ vt Σ vt Et Eψ, v ψt ¼ v Et ψ t  1 þ Σt 1 þ Σt 1 þ Σ vt t t

 1 þ Σ vt M t þ Σ st  Σ vt M t M t þ Σ st ¼ 1 þ Σ vt 1 þ Σ vt

ð5:233Þ

So we can obtain h

i * α  p þ βðM þ U s Þ þ U 0 ψ ¼ Eψ

*

ð5:234Þ

where Us ¼

Σs  ΣvM , 1 þ Σv

U0 ¼

Σ0 þ ΣvE 1 þ Σv

ð5:235Þ

For the sake of simplicity, in the above two formulas we omit the footnote t representing the nucleon type. Due to the propagation of electromagnetic fields, the Coulomb potential VC should be introduced. It is determined by the empirical charge distribution of the nucleus. Here we temporarily add the VC contribution according to the practice in the references, so Eq. (5.234) can be rewritten as [10] h

i * α  p þ βðM þ U s Þ þ U 0 þ V C ψ ¼ Eψ

*

ð5:236Þ

Equation (5.38) has given !

α¼

! σ^ , σ^ 0^

^0

β¼

^I ^ 0

^ 0 ^I

! ð5:237Þ

and let  ψ¼

ψu ψd

 ð5:238Þ

so based on the discussion in Sect. 5.1, it can be seen that ψ u corresponds to the large upper component with positive energy and ψ d corresponds to the little lower component with positive energy. Then the following equation can be obtained by Eq. (5.236):

386

5

0  1 * σ^  p ψ d B C @  * A þ σ^  p ψ u

Polarization Theory of Relativistic Nuclear Reactions

!

   M þ Us þ U0 þ V C ψ u ψu

 ¼E s 0  M þ U  U  VC ψd ψd

 

 * σ^  p ψ d þ M þ U s þ U 0 þ V C ψ u ¼ Eψ u  

 * σ^  p ψ u  M þ U s  U 0  V C ψ d ¼ Eψ d ψd ¼ h



*

σ^  p

*

σ^  p ψ D u

 

i 1 * σ^  p þ M þ U s þ U 0 þ V C ψ u ¼ Eψ u D

ð5:239Þ

ð5:240Þ ð5:241Þ ð5:242Þ ð5:243Þ

where D ¼ E þ M þ Us  U0  V C

ð5:244Þ

First let us analyze the following expressions:        h  i  1 * 1 * * * * 1 * A  σ^  p σ^  p ψ u ¼ σ^  p ψ u þ σ^  p σ^  p ψ u σ^  p D D D ð5:245Þ Here we only discuss the spherical nuclei, so D is just a function of r, and there is *

*

p ¼ i∇ ¼ i

*

r d r dr

ð5:246Þ

Then using Eq. (2.16) we can obtain    *  1 σ^  r d 1 2 * σ^  p ψ u p ψ u þ i A¼ r dr Dðr Þ D ðr Þ   *  dDðr Þ σ^  r 1 2 i * σ^  p ψ u ¼ p ψu þ 2 dr r Dðr Þ D ðr Þ

ð5:247Þ

Equation (13.4.35) in Ref. [28] has given   *  * * * * * σ^  a σ^  b ¼ a  b þ i^ σ ab So there is

ð5:248Þ

5.3



Relativistic Optical Model and Phenomenological Optical Potential

    * * * * * * * * σ  r  p ¼ r  p þ i^ σ  L, σ^  p ¼ r  p þ i^

*

σ^  r

*

387 *

*

L¼ r p

ð5:249Þ

From Eq. (5.247) we can get (  *)  * *  1 d σ^  L d rp 2 p þi A¼  ð ln Dðr ÞÞ ψu ð ln Dðr ÞÞ r dr r dr Dðr Þ

ð5:250Þ

Substituting Eq. (5.250) into Eq. (5.243), we can also get (  *)  * *  1 d σ^  L d rp 2 p þi  ð ln Dðr ÞÞ ψu ð ln Dðr ÞÞ r dr r dr Dðr Þ

 þ M þ U s þ U 0 þ V C ψ u ¼ Eψ u

ð5:251Þ

And using Eq. (5.244) the above formula can be rewritten as (

 * *   *) p2 i d rp 1 d σ^  L þ ð ln Dðr ÞÞ  ð ln Dðr ÞÞ ψu 2E 2E dr r 2E dr r 

 1

M þ Us þ U0 þ V C E þ M þ Us  U0  V C ψ u 2E  1

¼ E þ M þ Us  U0  V C ψ u 2 þ

ð5:252Þ

Note that there is the following relation:



 M þ Us þ U0 þ V C E þ M þ Us  U0  V C



2 ¼ E M þ U s þ U 0 þ V C þ M 2 þ 2MU s þ ðU s Þ2  U 0 þ V C

ð5:253Þ

Substituting above formula into Eq. (5.252) and moving the terms corresponding to the first two items at the right end of the above formula to the right side of the equal sign of Eq. (5.252), then we can obtain (

 * *   *) i d r p 1 d σ^  L p2 þ ð ln Dðr ÞÞ  ð ln Dðr ÞÞ ψu 2E 2E dr r 2E dr r

 h

0 2 i E 2  M 2  2E U 0 þ V C 1 s s 2 þ ψu ψu ¼ 2MU þ ðU Þ  U þ V C 2E 2E ð5:254Þ

The above formula can be further rewritten as

388

5



Polarization Theory of Relativistic Nuclear Reactions

 * * dDðr Þ * 1 p2 i 1 dDðr Þ r  p 0 σ^  L þ U þ VC þ  2E r 2E Dðr Þ dr 2ErDðr Þ dr h

2 io E2  M 2 1 þ U s ð2M þ U s Þ  U 0 þ V C ψu ψu ¼ 2E 2E

ð5:255Þ

Next transform the wave function as follows: pffiffiffiffiffiffiffiffiffiffi Dðr Þφðr Þ

ð5:256Þ

  *2 r d * * d 1 d 2d , r  p ¼ ir , p2 ¼ ∇ ¼  2 r r dr dr dr r dr

ð5:257Þ

ψ u ðr Þ ¼ One has known *

*

p ¼ i∇ ¼ i

*

So we can obtain rffiffiffiffi pffiffiffiffi dφ 1 1 dD d φ ð5:258Þ ψ ¼ D þ dr u dr 2 D dr !# " rffiffiffiffi p ffiffiffi ffi dφ 1 1 dD d 1 2 D þ φ r2 p2 ψ u ¼ ∇ ψ u ¼  2 dr 2 D dr r dr rffiffiffiffi rffiffiffiffi   pffiffiffiffi 1 d 1 1 dD dφ 1 1 dD dφ 2 dφ ð5:259Þ  ¼ D 2 r  dr 2 D dr dr 2 D dr dr r dr rffiffiffiffi rffiffiffiffiffiffi 2   1 1 dD 1 1 1 d 2 dD φ φ  þ r 4 D3 dr dr 2 D r 2 dr ! rffiffiffiffi rffiffiffiffi rffiffiffiffiffiffi 2 * * 1 dD dφ 1 1 dD 1 dD r  p 1 dD pffiffiffiffi dφ 1 1 dD i D þ φ ¼ þ φ ψu ¼ dr 2 D dr D dr dr 2 D3 dr D dr r D dr ð5:260Þ So we can get rffiffiffiffiffiffi 2    * * pffiffiffiffi 1 d dφ 3 1 dD 1 dD r  p ψu ¼  D 2 þ p2 þ i φ r2 dr 4 D3 dr D dr r r dr rffiffiffiffi   1 1 1 d 2 dD φ  r dr 2 D r 2 dr Substitute Eqs. (5.256) and (5.261) into Eq. (5.255) and eliminate obtain [10, 20, 23]

ð5:261Þ

pffiffiffiffi D, then we can

5.3

Relativistic Optical Model and Phenomenological Optical Potential



 p2 E2  M 2 * * ðuÞ ðuÞ þ U eff ðr Þ þ V C ðr Þ þ U SO ðr Þ σ  L φðr Þ ¼ φð r Þ 2E 2E

389

ð5:262Þ

where h

2 i 1 ðuÞ U s ðr Þ ð2M þ U s ðr ÞÞ  U 0 ðr Þ þ V C ðr Þ þ U D ðr Þ 2E ð5:263Þ "  # 2   dDðr Þ 1 3 1 dDðr Þ 1 d ðuÞ ð5:264Þ U D ðr Þ ¼  2 r2 dr 2E 4 Dðr Þ dr 2r Dðr Þ dr

ðuÞ

U eff ðr Þ ¼ U 0 ðr Þ þ

ðuÞ

U SO ðr Þ ¼ 

dDðr Þ 1 2ErDðr Þ dr

ð5:265Þ

Since D(r) always appears in both the molecule and the denominator of the formulas, Eq. (5.244) can be modified to [20] 

Dðr Þ ¼ E þ M þ U s ðr Þ  U 0 ðr Þ  V C ðr Þ =ðE þ M Þ

ð5:266Þ

where UD(r) is called the Darwin potential, and it is usually considered to be a small amount. Equation (5.262) is an equation that describes the upper component of the positive energy nucleon wave function. The equation for the lower component of the positive energy nucleon wave function is derived below. From Eq. (5.240) we can obtain ψu ¼

*

σ^  p ψd T

T ¼ E  M  Us  U0  V C h   

i * 1 * σ^  p σ^  p  M þ U s  U 0  V C ψ d ¼ Eψ d T

ð5:267Þ ð5:268Þ ð5:269Þ

For spherical nuclei, refer to Eqs. (5.245) and (5.250); Eq. (5.269) can be rewritten as (  * *   *) d r d σ^  L  p 1 p2 þ i ð ln T ðr ÞÞ  ð ln T ðr ÞÞ ψd dr r dr r T ðr Þ

  M þ U s  U 0  V C ψ d ¼ Eψ d Using Eq. (5.268) the above formula can be rewritten as

ð5:270Þ

390

5

(

Polarization Theory of Relativistic Nuclear Reactions

 * *   *) i d rp 1 d σ^  L p2 þ ð ln T ðr ÞÞ  ð ln T ðr ÞÞ ψd 2E 2E dr r 2E dr r  

1

 M þ Us  U0  V C E  M  Us  U0  V C ψ d 2E  1

¼ E  M  Us  U0  V C ψ d 2

ð5:271Þ

And there is the following relation:



  M þ Us  U0  V C E  M  Us  U0  V C



2 ¼ E M þ U s  U 0  V C þ M 2 þ 2MU s þ ðU s Þ2  U 0 þ V C

ð5:272Þ

Substituting Eq. (5.272) into Eq. (5.271) and moving the terms corresponding the first two terms of the right side of Eq. (5.272) to the right side of the equal sign of Eq. (5.271), we can obtain  * *   *) i d rp 1 d σ^  L p2 þ ð ln T ðr ÞÞ  ð ln T ðr ÞÞ ψd 2E 2E dr r 2E dr r

 h

0 2 i E2  M 2  2E U 0 þ V C 1 s s 2 ψd þ 2MU þ ðU Þ  U þ V C ψ d ¼ 2E 2E

(

ð5:273Þ

The above formula can be further rewritten as 

 * * dT ðr Þ * 1 i 1 dT ðr Þ r  p p2 0 σ^  L þ U þ VC þ  2E r 2E T ðr Þ dr 2ErT ðr Þ dr  h

2 i E2  M 2 1 ψd ¼ ψd þ U s ð2M þ U s Þ  U 0 þ V C 2E 2E

ð5:274Þ

The form of the above formula is exactly the same as Eq. (5.255) for the upper component. Then the following transform of the wave function is introduced: ψ d ðr Þ ¼

pffiffiffiffiffiffiffiffiffi T ð r Þ ϕð r Þ

ð5:275Þ

The same derivation method as the upper component equation can be used to obtain 

 E2  M 2 p2 * * ðdÞ ðdÞ þ U eff ðr Þ þ V C ðr Þ þ U SO ðr Þ σ  L ϕðr Þ ¼ ϕð r Þ 2E 2E

where

ð5:276Þ

5.3

Relativistic Optical Model and Phenomenological Optical Potential ðdÞ

391

h

2 i 1 ðdÞ þ U D ðr Þ U s ðr Þ ð2M þ U s ðr ÞÞ  U 0 ðr Þ þ V C ðr Þ 2E ð5:277Þ "  2  # dT ðr Þ 1 3 1 dT ðr Þ 1 d ðdÞ ð5:278Þ U D ðr Þ ¼  2 r2 dr 2E 4 Tðr Þ dr dr 2r T ðr Þ

U eff ðr Þ ¼ U 0 ðr Þ þ

dT ðr Þ 1 2ErT ðr Þ dr

 T ð r Þ ¼ E  M  U s ð r Þ  U 0 ð r Þ  V C ð r Þ = ðE  M Þ ðdÞ

U SO ðr Þ ¼ 

ð5:279Þ ð5:280Þ

The form of Eq. (5.276) for the lower component is exactly the same as Eq. (5.262) for the upper component; their difference is that the definitions of T(r) and D(r) are different. Noting that the p in the first item at the left end of Eqs. (5.262) and (5.276) are !

!

operator, and there is p ¼ i∇ . We know that E2  M2 ¼ k2 by Eq. (5.3) or Eq. (5.160). If the nuclear potentials Us(r) and U0(r) as well as the Coulomb potential VC(r) are all equal to 0, Eq. (5.262) and Eq. (5.276) degenerate, respectively, into

 ∇2 þ k2 φ ¼ 0,



 ∇2 þ k 2 ϕ ¼ 0

ð5:281Þ

The above two formulas are all spherical Bessel equations, so the spherical Bessel function jl(kr) and the spherical Neumann function nl(kr) are applicable to the upper component wave function and the lower component wave function in any positive energy situation. In areas outside the nucleus, the Coulomb potential is V C ðr Þ ¼

zZe2 r

ð5:282Þ

where e is the electron charge and z and Z are the charge numbers carried by the incident particle and the target nucleus, respectively. When the nuclear potential is 0 and only the Coulomb potential exists, the motion equation satisfied by the charged particles in the ħ ¼ c ¼ 1 unit system is 

     k2 1 2 * *  ∇ þ V C ðr Þ ψ C r ¼ ψ C r 2μ 2μ

ð5:283Þ

where μ¼

MM A M þ MA

ð5:284Þ

392

5

Polarization Theory of Relativistic Nuclear Reactions

where M and MA are the masses of the incident particle and target nucleus, respectively, and μ is called reduced mass. We can also rewrite Eq. (5.283) as  ∇2 þ k 2 

2ηk r



  ! ψC r ¼ 0

ð5:285Þ

where η¼

μzZe2 k

ð5:286Þ

We refer to Eq. (5.283) and Eq. (5.285) as the Coulomb equation, which contains the charge numbers z and Z of the incident particle and the target, and the masses M and MA and other physical quantities. The Coulomb equation is derived from the non-relativistic Schrodinger equation and is suitable for the lower energy region. 2ηk When the energy is quite high, there will be k 2 in the area outside the nucleus, r so it can be seen from Eq. (5.285) that the Coulomb potential is no longer important for very high energy. In this case this equation basically becomes the spherical Bessel equation. Previously, we added the Coulomb potential VC in the Dirac equation (5.236) using side-by-side with U0, which is currently consistent method in the literature. But this approach will bring about the following two problems: Firstly, when we make both the nuclear potential U0 and Us 0, because there is a non-constant VC(r) in ðρÞ ðρÞ D(r) and T(r), the nuclear potential U D ðr Þ and U SO ðr Þðρ ¼ u, dÞ will not disappear, and secondly, even if we ignore the VC(r) contained in D(r) and T(r), at this time the

2 1 terms V C ðr Þ   2E V C ðr ÞÞ are stillretained in Eq. (5.262) and Eq. (5.276). If the 1 2 V C ðr Þ second term of V C ðr Þ  2E

is ignored and Eqs. (5.262) and (5.276) are

rewritten according to the form of the Coulomb equation (5.283), when compared them with the Coulomb equation (5.283), it will be found that there is one more coefficient E/μ in front of VC(r) in the newly obtained equation. This coefficient may be close to 1 only if the energy is very low and the target nucleus is very heavy. It can be seen that the current general method of the adding Coulomb potential VC in Dirac equation (5.236) is somewhat unreasonable in physics. It can be seen from the positive energy nucleon plane waves given by Eqs. (5.80) and (5.93) that when the energy is very low, there is p  E þ M; the contribution of the lower component in the incident plane wave is very small. We also know that the Coulomb potential only plays a significant role in low energy conditions, and the influence of the Coulomb potential on the high energy reaction is negligible. Therefore, we propose that the Coulomb potential VC is not added using side-byμ side with U0 in the Dirac equation (5.236). Instead, the Coulomb potential EV C is ðuÞ only added using side-by-side with U eff in the upper component equation (5.262), and the influence of the Coulomb potential VC is ignored in the lower component equation (5.276). Since the Coulomb potential only works under low energy conditions, this method of adding Coulomb potential according to the non-relativistic

5.3

Relativistic Optical Model and Phenomenological Optical Potential

393

Schrodinger equation is reasonable. Thus Eqs. (5.262), (5.263), (5.266), (5.276), (5.277), and (5.280) will be replaced by the following equations based on our proposed approach, respectively: 

 p2 E2  M 2 μ * * ðuÞ ðuÞ þ U eff ðr Þ þ V C ðr Þ þ U SO ðr Þ σ  L φðr Þ ¼ φðr Þ 2E 2E E h

2 i 1 ðuÞ ðuÞ U eff ðr Þ ¼ U 0 ðr Þ þ þ U D ðr Þ U s ðr Þ ð2M þ U s ðr ÞÞ  U 0 ðr Þ 2E

 Dðr Þ ¼ E þ M þ U s ðr Þ  U 0 ðr Þ =ðE þ M Þ  2  E2  M 2 p * * ðdÞ ðdÞ þ U eff ðr Þ þ U SO ðr Þ σ  L ϕðr Þ ¼ ϕð r Þ 2E 2E h

2 i 1 ðdÞ ðdÞ U eff ðr Þ ¼ U 0 ðr Þ þ þ U D ðr Þ U s ðr Þ ð2M þ U s ðr ÞÞ  U 0 ðr Þ 2E

 T ð r Þ ¼ E  M  U s ð r Þ  U 0 ð r Þ =ð E  M Þ ðuÞ

ð5:287Þ ð5:288Þ ð5:289Þ ð5:290Þ ð5:291Þ ð5:292Þ

ðuÞ

where U D ðr Þ and U SO ðr Þ have been given by Eqs. (5.264) and (5.265), respectively; ðdÞ ðdÞ U D ðr Þ and U SO ðr Þ have been given by Eq. (5.278) and Eq. (5.279), respectively. In this way, when let the nuclear potential U0 and Us be all 0, the equation (5.290) automatically degenerates into the spherical Bessel equation, and the equation (5.287) automatically degenerates into the Coulomb equation (5.283), and they can be applied to any target nuclei. In fact, when the energy is quite high, the equation (5.287) is also equivalent to the spherical Bessel equation. If the following nuclear potential symbols are introduced U cðρÞ ðr Þ ¼

E ðρÞ U ðr Þ, μ eff

U ðsoρÞ ðr Þ ¼

E ðρÞ U ðr Þ, μ SO

ρ ¼ u,d

ð5:293Þ

Equations (5.287) and (5.290) can be rewritten as 

 ∇2 k2 * * ðuÞ ðuÞ þ U c ðr Þ þ V C ðr Þ þ U so ðr Þ σ  L φðr Þ ¼ φðr Þ  2μ 2μ   ∇2 k2 * *  þ U ðcdÞ ðr Þ þ U ðsodÞ ðr Þ σ  L ϕðr Þ ¼ ϕðr Þ 2μ 2μ

ð5:294Þ ð5:295Þ

Equations (5.287) and (5.290) are called the Schrodinger-like equations, whereas Eq. (5.294) and Eq. (5.295) have been completely transformed into the Schrodinger equation form, but the nuclear potentials defined by Eq. (5.293) need to be taken. When the energy is not very high, it is only necessary to solve the upper component Eq. (5.287) or Eq. (5.294), so they are also called the spherical nucleus relativistic optical model equations. If assume that both Us(r) and U0(r) contain real and imaginary parts and phenomenally introduce their relations with the coordinates

394

5 Polarization Theory of Relativistic Nuclear Reactions

r, the incident particle energy, and the A and Z of the target, then solve the relativistic optical model equation (5.287) or (5.294), and by adjusting the parameters to meet the experimental data, thus the relativistic phenomenological optical potential (RPOP) can be obtained. If the nuclear bound state is described by Eq. (5.287) or Eq. (5.294), the massenergy relation of the on-shell nucleons corresponding to Eq. (5.3) becomes E2 ¼  p2 þ M2, 0 < E < M. So the bound state nucleons still belong to the positive energy nucleons. The S matrix elements can be solved by boundary conditions of the spherical nucleus relativistic optical model Eq. (5.287) or (5.294) (discussed in the next section). The elastic scattering amplitude can be written according to Eqs. (2.70), (2.82), and (2.83) as follows: ^ ðθ, φÞ ¼ AðθÞ^I þ BðθÞ * ð5:296Þ F n  σ^   h   i i X lþ1 l1 AðθÞ ¼ f C ðθÞ þ ðl þ 1Þ 1  Sl 2 þ l 1  Sl 2 e2iσl Pl ð cos θÞ ð5:297Þ 2k l BðθÞ ¼

  1 X lþ12 l1 Sl  Sl 2 e2iσl P1l ð cos θÞ 2k l

ð5:298Þ

*

where n is a unit vector perpendicular to the reaction plane, generally taking the * y-axis in the n direction. Moreover, the following formulas for the polarization analyzing power Ay(θ) and the spin rotation function Q(θ) have been given by Eqs. (2.131) and (2.132) as follows: Ay ðθÞ ¼

2 Re ðAðθÞB ðθÞÞ I 0 ðθ Þ

ð5:299Þ

QðθÞ ¼

2ImðAðθÞB ðθÞÞ I 0 ðθ Þ

ð5:300Þ

I 0 ðθÞ ¼ jAðθÞj2 þ jBðθÞj2

ð5:301Þ

where I0(θ) is the differential cross section of the unpolarized incident nucleons. In the non-relativistic optical model, the spin-orbital coupling potential is added separately. But the spin-orbital coupling potential in the relativistic optical model equation is naturally generated by the Dirac equation, and it is closely related to the central potential Ueff(r). So it has obvious advantages to analyze the polarization observable quantities. Therefore, since the birth of the relativistic optical model, it dσ has been used to analyze the elastic scattering angular distribution , polarization dΩ analyzing power Ay(θ), and spin rotation function Q(θ) of the spherical nucleus at the same time, and the better results have been obtained [10–21]. From experimental data and theoretical calculation results, it can be seen that the polarization phenomena exist obviously until the incident nucleon energy 1000 MeV.

5.4

Dirac S Matrix Theory of the Relativistic Nuclear Reactions

5.4

395

Dirac S Matrix Theory of the Relativistic Nuclear Reactions

1 particles, so only the Dirac S 2 1 matrix theory of the two-body nuclear reactions of the incident spin particles 2 1 and various possible outgoing spin particles is studied here. 2 The two-body nuclear reaction can be expressed as a þ A ! b þ B, a and b 1 represent spin particles, and the reaction channels a þ A and b þ B are expressed 2 by α and α0, respectively. m and m0 represent the stationary masses of the particles a and b; MA and MB represent the stationary masses of the target A and the residual nucleus B. v and v0 represent the velocities of the incident particle a and the outgoing particle b. Next we let εαn and εα0 n0 represent the excitation energies for the nth exited state of the target A and the n0th exited state of the residual nucleus B, respectively. InMn and I 0n M 0n represent their corresponding angular momentum and its z component, ΦI n M n ðξÞ and ΦI 0n M 0n ðξ0 Þ represent their corresponding spin wave functions. ξ and ξ0 represent the internal freedom of A and B, respectively. We call the system, for which the sum of incident particle momentum and target momentum is equal to 0, as center of momentum (c.m.) system, and set E as the total energy of the whole system in c.m. system. In the ħ ¼ c ¼ 1 natural unit system we let The Dirac equation is only applicable to spin

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi E 2  m2 , h  i1=2 2 ¼ E2  m2 þ M 2A þ m2 þ ε2αn  M 2B  m0  ε2α0 n0

k  k αn,αn ¼ k 0  kα0 n0 ,αn

ð5:302Þ

Select the direction of incident particle as the z-axis, according to Eq. (5.80) and Eq. (3.1.22) given in Ref. [4], we can write the initial state total wave function of the positive energy stationary incident particles as   1 * ðiÞ Ψ αnνM n ¼ pffiffiffi u p eikz χ ν ΦI n M n ðξÞ v rffiffiffiffiffi   4π * X^ ¼ u p ljl ðkr Þ il Yl0 ðθ, 0Þχ ν ΦI n M n ðξÞ v l

ð5:303Þ

pffiffiffiffiffiffiffiffiffiffiffiffi where ^l  2l þ 1. v is the velocity of the incident particle relative to the target in the nth excited state, and χ ν is the incident particle spin wave function   with a * magnetic quantum number ν. jl is the spherical Bessel function. u p has been given by Eq. (5.93) as follows: 0 1 ^I   rffiffiffiffiffiffiffiffiffiffiffiffiffi E þ m * @ σ^  ! u p ¼ p A 2m Eþm

ð5:304Þ

396

5

Polarization Theory of Relativistic Nuclear Reactions *

where ^I is the unit 2  2 matrix and p is the momentum operator. Eq. (5.303)   ! denotes that the upper and lower components of u p act on the two-dimensional spin wave function χ ν at the same time. ffi When the incident particle energy is very qffiffiffiffiffiffiffi Eþm low, there are p  (E þ m) and 2m ! 1, then from Eq. (5.304) we can see that the contribution of the lower component in this case can be ignored, and Eq. (5.303) will automatically degenerate into non-relativistic theoretical formula. Only when the energy of the incident particle is quite high, the contribution of the lower component needs to be considered. Introduce the following symbols: ljν ðΩÞ ¼ Cjν l0

1 2ν

il Yl0 ðθ, 0Þχ ν ,

φJM αnlj ðΩ, ξÞ ¼

X νM n

il Yl0 ðθ, 0Þχ ν ¼

X jν C l0 j

1 2ν

ljν ðΩÞ

ð5:305Þ

CJjν MI n M n ljν ðΩÞ ΦI n M n ðξÞ,

ljν ðΩÞ ΦI n M n ðξÞ ¼

X C Jjν MI n M n φJM αnlj ðΩ, ξÞ

ð5:306Þ

JM

So Eq. (5.303) can be rewritten as ðiÞ Ψ αnνM n

rffiffiffiffiffi   4π * X^ ¼ u p ljl ðkr Þ C jl0ν 1ν CJjν MI n M n φJM αnlj ðΩ, ξÞ v 2 ljJM

ð5:307Þ

If the incident particle is a charged particle, there will be a long-range Coulomb potential VC(r). Since the Coulomb potential can be ignored at very high energy, we only consider the Coulomb potential and Coulomb scattering in the upper component. The discussion in the 5.3 section has pointed out that the Coulomb functions obtained by Coulomb equation can be used directly in the case of relativity. For the charged particles, the upper component of the initial state total wave function of the incident particles can be obtained from Eqs. (5.307) and (5.304) as ðiÞðuÞ Ψ αnνM n

rffiffiffiffiffirffiffiffiffiffiffiffiffiffiffiffiffiffi 4π E þ mX^ iσl F l ðkr Þ j ν J M Cl0 1ν C jν I n M n φJM ¼ le αnlj ðΩ, ξÞ kr v 2m ljJM 2

ð5:308Þ

where Fl(kr) is Coulomb wave function. Further we can rewrite the above formula as ðiÞðuÞ Ψ αnνM n

pffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffi i π E þ mX^ iσ l j ν J M ¼ pffiffiffi le C l0 1ν Cjν I n M n ½ðGl  iF l Þ  ðGl þ iF l Þ φJM αnlj ðΩ, ξÞ 2m ljJM 2 kr v ð5:309Þ

l l represents the spherical incident wave and Gl þiF represents the where Gl iF r r spherical outgoing wave. If the incident particles do not undergo reaction, the

5.4

Dirac S Matrix Theory of the Relativistic Nuclear Reactions

397

strengths of the spherical incident wave and the spherical outgoing wave are equal to each other. If the incident particles undergone elastic scattering or other two-body reactions with the target, after the S matrix elements are introduced into the upper component, the upper component of the total wave function of the system can be obtained as follows: ðtÞðuÞ Ψ αnνM n

pffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffi X 1 i π E þ mX^ iσ l j ν pffiffiffiffi ¼ le C l0 1ν C Jjν MI n M n kr 2m ljJM 2 v0 α0 n0 l0 j0 i h ðuÞJ ðGl  iF l Þδα0 n0 l0 j0 ,αnlj  Sα0 n0 l0 j0 ,αnlj ðGl0 þ iF l0 Þ φJM α0 n0 l0 j0 ðΩ, ξÞ

ð5:310Þ

There is a formula iðGc  iF c Þ δc0 c  iSc0 c ðGc0 þ iF c0 Þ ¼ 2F c δc0 c þ iðδc0 c  Sc0 c ÞðGc0 þ iF c0 Þ ð5:311Þ So Eq. (5.310) can be rewritten as ðtÞðuÞ Ψ αnνM n

pffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiX 4π E þ m ^ iσ l j ν J M ¼ le C l0 1ν C jν I n M n kr 2m ljJM 2 

 X i  1 pffiffiffi F l φJM pffiffiffiffi δα0 n0 l0 j0 ,αnlj  Sαðu0 nÞJ0 l0 j0 ,αnlj ðGl0 þ iF l0 ÞφJM αnlj ðΩ, ξÞþ α0 n0 l0 j0 ðΩ, ξÞ 0 v α0 n 0 l0 j0 2 v

#

ð5:312Þ The first term in the square bracket of the above formula is clearly the upper component of the initial state wave function of the incident particles given by Eq. (5.308). So according to the two terms in square bracket in Eq. (5.312), the upper component of the total wave function of the system given by Eq. (5.312) can be divided into two terms representing the initial state and the final state as ðtÞðuÞ

ðiÞðuÞ

ðf ÞðuÞ

Ψ αnνM n ¼ Ψ αnνM n þ Ψ αnνM n

ð5:313Þ

pffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffi X 1 i π E þ mX^ iσ l j ν pffiffiffiffi ¼ le C l0 1ν C Jjν MI n M n 0 kr 2m ljJM 2 α0 n0 l0 j0 v   ðuÞJ δα0 n0 l0 j0 ,αnlj  Sα0 n0 l0 j0 ,αnlj ðGl0 þ iF l0 Þ φJM α0 n0 l0 j0 ðΩ, ξÞ

ð5:314Þ

where ðf ÞðuÞ Ψ αnνM n

Equation (3.8.5) in Ref. [4] has given

398

5

Polarization Theory of Relativistic Nuclear Reactions 0

0

Gl0 þ iF l0 ! eik r ðiÞl eiσl0 r!1

ð5:315Þ

We can obtain the following expression substituting Eqs. (5.306), (5.305), and (5.315) into Eq. (5.314): pffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffi X 1 i π E þ mX^ iσl j ν pffiffiffiffi eiσ l0 le Cl0 1ν CJjν MI n M n k 2m ljJM 2 v0 0 0 0 0 αnlj   ik0 r X j0 m0 e ðuÞJ δα0 n0 l0 j0 ,αnlj  Sα0 n0 l0 j0 ,αnlj Cl0 m0 j 1ν0 CJj0 mM0 j I 0 n M 0 n Yl0 m0 l χ ν0 ΦI 0 n M 0 n l r ν0 M 0 m 0 m 0 2

ðf ÞðuÞ Ψ αnνM n ! r!1

n

l

j

ð5:316Þ And we can write the following formula according to Eqs. (5.312) and (5.313): ðiÞðuÞ Ψ αnνM n

pffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiX 1 4π E þ m ^ iσ l j ν ¼ le C l0 1ν C Jjν MI n M n pffiffiffi F l φJM αnlj ðΩ, ξÞ kr 2m ljJM 2 v

ð5:317Þ

Further the following formula can be obtained substituting Eqs. (5.306) and (5.305) into Eq. (5.317): ðiÞðuÞ Ψ αnνM n

rffiffiffiffiffirffiffiffiffiffiffiffiffiffiffiffiffiffi 4π E þ mX^ iσl F l ðkr Þ l i Yl0 ðθ, 0Þχ ν ΦI n M n ¼ le kr v 2m l

ð5:318Þ

When r ! 1 the following relation can be obtained when making comparison between Eq. (3.3.76) and Eq. (3.3.45) given in Ref. [4]: pffiffiffiffiffiX iσ F l ðkr Þ l eikr i Yl0 ðθ, 0Þ ! eikz þ f C ðθÞ 4π ^le l r!1 kr r l

ð5:319Þ

where fC(θ) has been given by Eqs. (3.3.41), (3.3.3), and (3.3.42) in Ref. [4]. The first term on the right end of Eq. (5.319) represents the undistorted incident plane wave, and the second term represents the spherical outgoing wave of Coulomb scattering induced by Coulomb potential. In Eq. (5.318) the term corresponding to the second term of Eq. (5.319) is ð2ÞðuÞ Ψ αnνM n ! r!1

1 pffiffiffi v

rffiffiffiffiffiffiffiffiffiffiffiffiffi Eþm eikr χ Φ f C ðθ Þ r ν In Mn 2m

ð5:320Þ

The upper component of the total reaction wave function of the system can be obtained from Eqs. (5.316) and (5.320) as

5.4

Dirac S Matrix Theory of the Relativistic Nuclear Reactions

ðrÞðuÞ Ψ αnνM n

399

rffiffiffiffiffiffiffiffiffiffiffiffiffi  Eþm X 1 pffiffiffiffi f C ðθÞδα0 n0 ,αn δν0 M 0n ,ν M n ¼ þ  ! r!1 2m 0 0 0 0 v0 α n ν Mn pffiffiffi X   i π ^leiðσl þσl0 Þ δα0 n0 l0 j0 ,αnlj  Sðu0 ÞJ0 0 0 þ α n l j ,αnlj k ðf ÞðuÞ Ψ αnνM n

ð2ÞðuÞ Ψ αnνM n

ljJM l0 j0 m0 l m0j j0 m0 0 C jl0 ν 1ν C Jjν MI n M n C l0 m0 j 1ν0 CJj0 M m0j I 0n M 0n Yl m0 l ðθ, φÞ l 2 2



0

eik r χ 0Φ 0 0 r ν In Mn ð5:321Þ

From the above formula, the reaction amplitude of the upper component can be written as ðuÞ

f α0 n0 ν0 M 0 ,αnνM n ¼ n

rffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi i π X ^ iðσl þσl0 Þ Eþm f C ðθÞδα0 n0 ,αn δν0 M 0n ,ν M n þ le k 2m

  j0 m0 ðuÞJ δα0 n0 l0 j0 ,αnlj  Sα0 n0 l0 j0 ,αnlj C jl0 ν 1ν C Jjν MI n M n C l0 m0 j l

2

ljJM l0 j0 m0 l m0j



J M 1 0 C j0 m0 I 0 M 0 Yl0 m0 l ðθ, φÞ ν n n j 2

ð5:322Þ The lower component of the initial state total wave function of the incident particles can be written from Eqs. (5.307) and (5.304) as ðiÞðdÞ Ψ αnνM n

rffiffiffiffiffirffiffiffiffiffiffiffiffiffiffiffiffiffi ! 4π E þ m σ^  p X^ ¼ lj ðkr ÞC jl0ν v 2m E þ m ljJM l

1 2ν

C Jjν MI n M n φJM αnlj ðΩ, ξÞ

ð5:323Þ

Further the above formula can be rewritten as ðiÞðdÞ Ψ αnνM n

pffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffi ! i π E þ m σ^  p X^ j ν ¼ pffiffiffi lC 2m E þ m ljJM l0 v ½ðnl  ijl Þ  ðnl þ ijl Þ

1 2ν

CJjν MI n M n

φJM αnlj ðΩ,

ð5:324Þ

ξÞ

where k(nl  ijl) represents spherical incident wave and k(nl þ ijl) represents spherical outgoing wave. If the incident particles undergo elastic scattering or other two-body reactions with the target, after the S matrix element is introduced into the lower component, the lower component of the total wave function of the system can be obtained as

400

5

ðtÞðdÞ Ψ αnνM n

Polarization Theory of Relativistic Nuclear Reactions

pffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffi ! X 1 i π E þ m σ^  p X^ j ν pffiffiffiffi ¼ lC l0 1ν C Jjν MI n M n k 2m E þ m ljJM 2 v0 α0 n0 l0 j0 i h ðdÞJ k ðnl  ijl Þ δα0 n0 l0 j0 ,αnlj  Sα0 n0 l0 j0 ,αnlj k 0 ðnl0 þ ijl0 Þ φJM α0 n0 l0 j0 ðΩ, ξÞ

ð5:325Þ

There is the formula iðnc  ijc Þ δc0 c  iSc0 c ðnc0 þ ijc0 Þ ¼ 2jc δc0 c þ iðδc0 c  Sc0 c Þðnc0 þ ijc0 Þ ð5:326Þ So Eq. (5.325) can be rewritten as ðtÞðdÞ Ψ αnνM n

pffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffi  ! k 4π E þ m σ^  p X^ j ν J M pffiffiffi j φJM ðΩ, ξÞ lC ¼ C k 2m E þ m ljJM l0 12ν jν I n M n v l αnlj   X i  ðdÞJ 0 JM 0 0 pffiffiffiffi δα0 n0 l j0 ,αnlj  Sα0 n0 l0 j0 ,αnlj k ðnl þ ijl0 Þφα0 n0 l0 j0 ðΩ, ξÞ þ 0 α0 n0 l0 j0 2 v ð5:327Þ

The first term in the square bracket of the above formula is the lower component of the initial state wave function of the incident particles given by Eq. (5.323). So according to the two terms in square bracket in Eq. (5.327), we can divide the lower component of the total wave function of the system into two terms representing the initial state and the final state as ðtÞðdÞ

ðiÞðdÞ

ðf ÞðdÞ

Ψ αnνM n ¼ Ψ αnνM n þ Ψ αnνM n

ð5:328Þ

pffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffi ! X 1 i π E þ m σ^  p X ^ j ν pffiffiffiffi lC l0 1ν C Jjν MI n M n ¼ k 2m E þ m ljJM 2 v0 α0 n0 l0 j0   ðdÞJ δα0 n0 l0 j0 ,αnlj  Sα0 n0 l0 j0 ,αnlj k0 ðnl0 þ ijl0 Þ φJM α0 n0 l0 j0 ðΩ, ξÞ

ð5:329Þ

where ðf ÞðdÞ Ψ αnνM n

Equation (3.1106) has given nl ðkr Þ þ i jl ðkr Þ ! ðiÞl r!1

eikr kr

ð5:330Þ

Then the following expression can be obtained substituting Eqs. (5.306), (5.305), and (5.330) into Eq. (5.329):

5.4

Dirac S Matrix Theory of the Relativistic Nuclear Reactions

401

pffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffi ! X 1 i π E þ m σ^  p X^ j ν pffiffiffiffi lC l0 1ν CJjν MI n M n k 2m E þ m ljJM 2 v0 α0 n0 l0 j0 ð5:331Þ 0   ik r X j0 m0j e ðdÞJ J M 0 0 0 0 0 0 δα0 n0 l j ,αnlj  Sα0 n0 l0 j0 ,αnlj C C0 0 0 0Y 0 χ Φ r 0 0 0 0 l0 m0 l 12ν0 j mj I n M n l m l ν I n M n

ðf ÞðdÞ Ψ αnν M n ! r!1

ν M n ml mj

There are the following relations: *

*

∇¼ n

∂ 1* þ ∇Ω , ∂r r

*  ∂ ∇Ω ¼ , ∂θ θ *

*

*

n¼

r * ¼ ^r ¼ e r r

ð5:332Þ

1 ∂ sin θ ∂φ

ð5:333Þ

*  ∇Ω ¼ φ

*

*

*

l ¼ r  p,

*

*

p ¼ i∇

ð5:334Þ

!

It can be seen that l is perpendicular to r , and we can obtain i ∂ ∂ , lφ ¼ i sin θ ∂φ ∂θ *  *  ∇Ω ¼ ilθ ∇Ω ¼ ilφ ,

lr ¼ 0,

lθ ¼

θ

*

ð5:335Þ ð5:336Þ

φ

*

∇¼ n

  ∂ i * *  e φ lθ  e θ lφ ∂r r

ð5:337Þ

Note that the three spherical coordinate axes are perpendicular to each other and we can obtain *

*

*

*

n  l ¼ e φ lθ  e θ lφ , *

*

∇¼ n !

*

*

 * * * ∇Ω ¼ i n  l

  ∂ 1 * * i n l r ∂r

*

ð5:338Þ ð5:339Þ

*

*

Let a ¼ n , b ¼ l in Eq. (5.248), and apparently there is n  l ¼ 0, so we can get    * * * * i^ σ  n  l ¼ σ^  n σ^  l The following formula can be obtained using Eqs. (5.339) and (5.340):

ð5:340Þ

402

5

Polarization Theory of Relativistic Nuclear Reactions !!

*  d σ^  l σ  p ¼ iσ  ∇ ¼ i σ  ^r  dr r

*

*

*

*

ð5:341Þ

The total angular momentum of the system is *

*

*

*

*

*

J ¼ j þ In ¼ l þ s þ In *

*

*

ð5:342Þ

*

*

*

J 2 ¼ l2 þ s2 þ I 2n þ 2 l  s þ 2 l  I n þ 2 s  I n

ð5:343Þ

Ignore the last two terms of the above formula, and let the above formula act on its eigenfunction; the following result can be obtained: * * * 3 K Jαnl  2 l  s ¼ σ^  l ¼ J ðJ þ 1Þ  lðl þ 1Þ   I n ðI n þ 1Þ 4

ð5:344Þ

So Eq. (5.341) can be rewritten as  

*  d K Jαnl  σ  p ¼ iσ  ∇ ¼ i σ  ^r dr r

*

*

*

*

ð5:345Þ

We will substitute Eq. (5.345) into Eq. (5.331). First note 

J

d K α0 n0 l0  r dr

 ik0 r    ik0 r   ik0 r  J 1 K 0 00 e e e ¼ ik0   α n l ik0 ! r!1 r r r r r

ð5:346Þ

and it is known that 1 ! 1X σ^  ^r ¼ σ^  r ¼ ð1Þη σ^η r η r η r

ð5:347Þ

Equation (2.21) has given σ^η χ ν ¼

pffiffiffi 1 2ð1Þ2ν C 11 2

η νþη

1 2

χ ν νþη

ð5:348Þ

So we can get ðσ^  ^r Þχ ν ¼

pffiffiffi 1 1X ð1Þη r η 2ð1Þ2ν C11 r η 2

Equation (3.942) has also given

η νþη

1 2

χ ν νþη

ð5:349Þ

5.4

Dirac S Matrix Theory of the Relativistic Nuclear Reactions

403

rffiffiffiffiffi 3 r rY1η ðΩÞ ¼ 4π η

ð5:350Þ

so there is rffiffiffiffiffi 1 8πX ð1Þ2νþη C 11 ðσ^  ^rÞχ ν ¼ 3 η 2

η νþη

1 2

ν

Y1

η ðΩÞχ νþη

ð5:351Þ

From Eq. (3.943) we have Ylml Y1

η

rffiffiffiffiffi 3 ^X 1 L ml η ¼ l CL 0 Y C ^ l ml 1 η l0 10 L ml η 4π L L rffiffiffiffiffi X 3 ð1Þ1η C l1 mηl L ml η C Ll0 0 10 YL ¼ 4π L

ð5:352Þ ml η

Then we can obtain the following formula substituting Eqs. (5.345), (5.346), (5.351), and (5.352) into (5.331): ðf ÞðdÞ Ψ αnνM n  ! r!1

pffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffi   i π Eþm X k0 1 X ^ ðdÞJ pffiffiffiffi l δα0 n0 l0 j0 ,αnlj  Sα0 n0 l0 j0 ,αnlj k 2m α0 n0 ν0 M 0 E þ m v0 n

j0 m0

C jl0ν 1ν C JjνMI n M n C l0 m0 j 2

C Ll0 00 1 0 YL

l

m0 l η ðθ,

1 0 2ν

CJj0 mM0 j

φÞ

e

ik 0 r

r

I0 n M0 n

ljJM l0 j0 m0 l m0j

pffiffiffiP 0 1 2 ð1Þ2þν C 11 νη0 þη η

2

P 1 2

ν0

L

l0 m0l L m0 l η

C1 η

χ ν0 þη ΦI 0 n M 0 n ð5:353Þ

Let ν0 ¼ ν0 þ η, then Eq. (5.353) can be rewritten as

ν0 ¼ ν0  η

ð5:354Þ

404

5

ðf ÞðdÞ Ψ αnνM n ! r!1

Polarization Theory of Relativistic Nuclear Reactions

pffiffiffi 0 pffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffi i π Eþm X 2k 1 X pffiffiffiffi k 2m α0 n0 ν0 M 0 E þ m v0

ljJM l0 j0 m0 l m0j

n

Cjl0 ν 1ν CJjν MI n M n CJj0 M m0j 2

X L

X 1 j0 m0 0 ð1Þ2þν η Cl0 m0 j

1 2

l

η

ν0 η

C11 2

η ν0

ν0 þη

1 2

0

l0 m0l

C1 η

I 0n M 0n

  ^l δα0 n0 l0 j0 ,αnlj  Sðd0 ÞJ0 0 0 α n l j ,αnlj

L

m0l η

C Ll0

0 0

10

YL

m0l η

ðθ, φÞ

eik r χ Φ0 0 r ν0 I n M n ð5:355Þ

From the C-G coefficients in the above formula, we can see ν þ M n ¼ m0j þ M 0n ,

m0j ¼ m0l þ ν 0  η

ð5:356Þ

and the following relations can be obtained: m0l  η ¼ ν þ M n  ν0  M 0n , ν0  η ¼ ν þ M n  m0l  M 0n , η ¼ m0l þ ν0 þ M 0n  ν  M n

ð5:357Þ

So Eq. (5.355) can be rewritten as ðf ÞðdÞ Ψ αnνM n ! r!1

pffiffiffi 0 pffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffi   i π Eþm X 2k 1 X ^ ðdÞJ pffiffiffiffi l δα0 n0 l0 j0 ,αnlj  Sα0 n0 l0 j0 ,αnlj k 2m α0 n0 ν0 M 0 E þ m v0 n ljJM l0 j0 m0 l m0j

C jl0ν 1ν C JjνMI n M n C Jj0 mM0 j X L

2

I 0n M 0n ð1Þ

1 0 0 2þνþM n ml M n

l0 m0l m0l þν0 þM 0n νM n L νþM n ν0 M 0n

C1

j0 m0

C l0 m0 j

1 l 2

νþM n m0l M 0n

1 m0l þν0 þM 0n νM n 0 1 νM n þm0 þM 0 n l 2 2 ν

C1

0

C Ll0 0 010 YL νþM n ν0 M 0n ðθ, φÞ

eik r χ Φ0 0 r ν0 I n M n ð5:358Þ

From the above formula the reaction amplitude of the lower component can be written as

5.4

Dirac S Matrix Theory of the Relativistic Nuclear Reactions

ðdÞ f α0 n0 ν0 M 0n , αnνM n

405

rffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi   E þ m 2 k0 i π X ^ ðdÞJ ¼ l δα0 n0 l0 j0 ,αnlj  Sα0 n0 l0 j0 ,αnlj C jl0ν1ν C JjνMI n M n 2m E þ m k 2 ljJM l0 j0 m0 l m0j

CJj0 M m0j

I 0n M 0n ð1Þ

X

l0

L

C1

1 0 0 2þνþM n ml M n

j0 m0

C l0 m0 j 1 l 2

m0l m0l þν0 þM 0n νM n

1 m0 þν0 þM 0 νM

νþM n m0l M 0n

n C 1 ν0 l1 νMn þm0 þM 0 2

2

n

l

L 0 L νþM n ν0 M 0n C l0 0 1 0 YL νþM n ν0 M 0n ðθ,

n

φÞ

ð5:359Þ Equation (2.253) has given rffiffiffiffiffiffiffiffiffiffiffiffiffiffisffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 0 2l0 þ 1 l0  m0l ! m0l

0  Pl0 ð cos θÞeiml φ Yl0 m0 l ðθ, φÞ ¼ ð1Þ 0 4π l þ ml ! m0l

ð5:360Þ

Introduce the following symbols for Eq. (5.322): 0 f νðu0 νÞll J

s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  l0  m0l !X j ν

0  C 1 CJ M l þ m0l ! jj0 l0 2ν jν I n M n   ðuÞJ 0 I 0n M 0n δα0 n0 l j0 ,αnlj  Sα0 n0 l0 j0 ,αnlj

0 0 ml ¼ ^l^l eiðσl þσl0 Þ j0 m0 C l0 m0 j 1ν0 C Jj0 mM0 j l 2

ð5:361Þ

and we can get the following relation based on the C-G coefficients in the above formula: m0l ¼ ν þ M n  ν0  M 0n

ð5:362Þ

So Eq. (5.322) can be rewritten as ðuÞ f α0 n0 ν0 M 0n ,αnνM n

rffiffiffiffiffiffiffiffiffiffiffiffiffih Eþm ¼ f C ðθÞδα0 n0 ,αn δν0 M 0n ,νM n 2m i  m0 0 X 0 0

i f ðνu0 νÞll J m0l Pl0 l ð cos θÞ eiml φ þ ð1Þml 2k 0 ll J

Introduce the following symbols for Eq. (5.359):

ð5:363Þ

406

5

s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 0 þ M 0 !X þ ν L  ν  M n n ^

 þ M n  ν0  M n ¼ ^lL Cj ν 1 CJ M L þ ν þ M n  ν0  M 0n ! jj0 l0 2ν jν I n M n   1 0 j0 m0j 0 ðdÞJ 2þνþM n ml M n C 0 j0 ,αnlj  S δ 0 0 0 0 0 0 α n l In Mn α0 n0 l j ,αnlj ð1Þ l0 m0 1 νþM m0 M 0

ðdÞll0 m0 LJ

f ν0 ν l ν

C Jj0 mM0 j 1

C1 2

Polarization Theory of Relativistic Nuclear Reactions

 0

l

m0l þν0 þM 0n νM n l0 m0 C l ν0 12 νM n þm0l þM 0n 1 m0l þν0 þM 0n νM n

n

2

l

n

L 0 L νþM n ν0 M 0n C l0 0 10

ð5:364Þ So Eq. (5.359) can be rewritten as rffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 0 E þ m 2 k0 i 0 ¼ ð1ÞνþM n ν M n 2m E þ m 2k X ðdÞll0 m0 LJ

 νþM ν0 M 0n 0 0 f ν0 ν l ν þ M n  ν0  M 0n PL n ðcos θÞ eiðνþM n ν M n Þφ

ðdÞ f α0 n0 ν0 M 0n ,αnνM n

ll0 m0l LJ

ð5:365Þ Referring to Eqs. (2.258)–(2.261), we introduce the following symbols according to Eqs. (5.363) and (5.365): ðuÞ AM 0n M n ðθÞ

rffiffiffiffiffiffiffiffiffiffiffiffiffi" Eþm ¼ f C ðθÞ δα0 n0 ,αn δM 0n ,M n 2m X ðuÞll0 J

 M M 0 0 i þ f 11 M n  M 0n Pl0 n n ð cos θÞ ð1ÞM n M n 2k 22 0

#

ð5:366Þ

ll J

rffiffiffiffiffiffiffiffiffiffiffiffiffi" 0 Eþm 1 ðuÞ  ð1ÞM n M n 1 BM 0n M n ðθÞ ¼ 2m 2k X ðuÞll0 J

 M M 0 1 f 1 1 M n  M 0n  1 Pl0 n n ð cos θÞ ll0 J

ðuÞ C M 0n M n ðθÞ

ð5:367Þ

#

ð5:368Þ

2

rffiffiffiffiffiffiffiffiffiffiffiffiffi" 0 Eþm 1 ¼ ð1ÞM n M n þ1 2m 2k X ll0 J

ðuÞ DM 0n M n ðθÞ

2

#

ðuÞll0 J

f 1 1 M n 2 2



M 0n

þ1



M M 0 þ1 Pl0 n n ð cos θÞ

rffiffiffiffiffiffiffiffiffiffiffiffiffi" Eþm ¼ f C ðθÞδα0 n0 ,αn δM 0n ,M n 2m X ðuÞll0 J

 M M 0 0 i þ ð1ÞM n M n f 1 1 M n  M 0n Pl0 n n ð cos θÞ 2k 2 2 0 ll J

#

ð5:369Þ

5.4

Dirac S Matrix Theory of the Relativistic Nuclear Reactions

ðdÞ AM 0n M n ðθÞ

rffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi  0 i E þ m 2k 0 ¼ ð1ÞM n M n 2m E þ m 2k   M M 0 P ðdÞll0 m0l LJ

f1 1 M n  M 0n PL n n ð cos θÞ ll0 m0l LJ

ðdÞ BM 0n M n ðθÞ

ðdÞ CM 0n M n ðθÞ

2

ðdÞ DM 0n M n ðθÞ

ð5:371Þ

2

rffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi  0 E þ m 2k 0 1 ¼ ð1ÞM n M n þ1 2m E þ m 2k  X ðdÞll0 m0 LJ

 M M 0 þ1 f 1 1 l M n  M 0n þ 1 PL n n ð cos θÞ ll0 m0l LJ

ð5:370Þ

2 2

rffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi  0 1 E þ m 2k 0  ¼ ð1ÞM n M n 1 2k 2m E þ m  X ðdÞll0 m0 LJ

 M n M 0n 1 l 0 f 1 1 M n  M n  1 PL ð cos θÞ ll0 m0l LJ

407

2

2

rffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi  0 i E þ m 2k 0 ¼ ð1ÞM n M n 2m E þ m 2k  X ðdÞll0 m0 LJ M n M 0n 00 l f 1 1 ðM n  M n ÞPL ð cos θÞ ll0 m0l LJ

ð5:372Þ

2

ð5:373Þ

2

pffiffiffi 0 2k in all matrix elements of the lower component Eþm of the reaction amplitude given by Eqs. (5.370)–(5.373). When the incident particle energy is very low, this coefficient is very small. So it can be seen that the contribution of the lower component can be ignored in very lowrenergy ffiffiffiffiffiffiffiffiffiffiffiffiffi case. Eþm ! 1, Since when the incident particle energy is very low, the coefficient 2m so the matrix elements of the upper component of the reaction amplitude given by Eqs. (5.366)–(5.369) will automatically degenerate into the non-relativistic reaction amplitude matrix elements given by Eqs. (2.258)–(2.261) in very low energy case. It can be seen that the upper and lower components in the spin space are incoherent when determining the polarization physical quantities based on the * expression form of the four-dimensional spin operator S given by Eq. (5.71). In the case of omitting the footnote Mn0Mn and the argument (θ) or (θ, φ), starting from Eqs. (5.363) and (5.365), the following matrices of reaction amplitude with definite Mn0Mn can be written for the upper component and the lower component, respectively: Note that there is a coefficient

408

5

Polarization Theory of Relativistic Nuclear Reactions

¼

AðuÞ iCðuÞ eiφ

^ ðdÞ ¼ F

AðdÞ iC ðdÞ eiφ

^ F

ðuÞ

! iBðuÞ eiφ iðM n M 0n Þφ e DðuÞ ! iBðdÞ eiφ iðM n M 0n Þφ e DðdÞ

ð5:374Þ

ð5:375Þ

*

If the y-axis is selected in the direction of the unit vector n perpendicular to the reaction plane, then there is φ ¼ 0, and the above two formulas can be simplified to AðuÞ  iBðuÞ iCðuÞ DðuÞ

^ ðuÞ ¼ F ^ F

ðdÞ

AðdÞ  iBðdÞ iCðdÞ DðdÞ

¼

! ð5:376Þ ! ð5:377Þ

The following relations can be proved by means of the method to prove Eqs. (2.295) and (2.296) in Sect. 2.7: ðuÞ

0

0

ðuÞ

0

0

ðuÞ

M n ðθ Þ

ð5:378Þ

ðuÞ

M n ðθ Þ

ð5:379Þ

DM 0n M n ðθÞ ¼ ð1ÞΠþΛþI n þI n þM n M n AM 0n C M 0n M n ðθÞ ¼ ð1ÞΠþΛþI n þI n þM n M n BM 0n

Note the following two parameters defined by Eqs. (2.290) and (2.291) are used in the above relations: ( Π¼ 

0 1

Λ¼

when π i π I n ¼ π 0i π 0I n

0

ð5:380Þ

when π i π I n ¼ π 0i π 0I n

1

when I n and I 0n are all integers when I n and I 0n are all semi‐odd numbers

ð5:381Þ

Below we analyze the C-G coefficients in the lower component of the reaction amplitude given by Eq. (5.359). First we get C jl0 ν 1ν ¼ ð1Þlþ2j Cjl0 ν1 1

2

2

C Jj0 M m0j

ν

, C Jjν MI n M n ¼ ð1ÞjþI n J C jJ 0

I 0n M 0n

0

¼ ð1Þj þI n J CjJ0

M ν I n M n ,

M m0j I 0n M 0n ,

ð1Þ2þνþM n ml M n ¼ ð1Þ2ðνþM n ml M n Þ ð1Þ2νM n þml þM n ¼ ð1Þ2νM n þml þM n 1

0

0

0

0

1

0

0

1

0

0

5.4

Dirac S Matrix Theory of the Relativistic Nuclear Reactions j0 m0j

C l0 m0

0

1 2

l

1

C1 2

νþM n m0l M 0n

m0l þν0 þM 0n νM n ν0 12 νM n þm0l þM 0n

l0 m0 C1 m0lþν0 þM 0 νM n L νþM n ν0 M 0 n n l

m0j m0l

j0

0

¼ ð1Þl þ2j C l0 1

1

¼ C1 2

¼ ð1Þ

409

νM n þm0l þM 0n

1 2

,

m0l ν0 M 0n þνþM n , 1 ν0 νþM n m0l M 0n 2

1þLl0

l0 m0l m0l ν0 M 0n þνþM n L νM n þν0 þM 0n

C1

YL νþM n ν0 M 0n ðθ, 0Þ ¼ ð1ÞνþM n ν M n YL νM n þν0 þM 0n ðθ, 0Þ 0

0

In this way we can see that if we change the sign of all the spin magnetic quantum numbers in Eq. (5.359) in φ ¼ 0 case, what phase factor compared to the original formula is going to be present 0

0

0

0

0

ð1Þlþ2jþjþI n Jþj þI n Jþ1þl þ2j þ1þLl þνþM n ν M n 1

0

1

0

0

0

¼ ð1ÞlþI n JþI n Jþ1þLþνþM n ν M n ¼ ð1Þ2J1þlþI n þI n þLþνþM n ν M n 0

0

0

0

It can be seen that for the lower component, the two parameters defined by Eqs. (2.290) and (2.291) can also be introduced, but it should be noted that the orbital angular momentum of the emitted particles is L instead of l0. Then the following relations can be obtained for the lower component: ðdÞ

0

0

ðdÞ

0

0

ðdÞ

ð5:382Þ

ðdÞ

ð5:383Þ

DM 0n M n ðθÞ ¼ ð1ÞΠþΛþI n þI n þM n M n AM 0n M n ðθÞ C M 0n M n ðθÞ ¼ ð1ÞΠþΛþI n þI n þM n M n BM 0n M n ðθÞ

They are similar with Eq. (5.378) and Eq. (5.379), respectively. When the incident particle, target, and the residual nucleus are all unpolarized, refer to Eqs. (2.271), (2.278), and (2.279), the following differential cross section of the outgoing particle can be obtained by using Eqs. (5.378), (5.379), (5.382), and (5.383) as follows: I 0α0 n0 ,αn ¼

o X 1 n ðρÞ X ðρÞ ðρÞþ 1 1 ^M ^ M0 M F I ¼ tr F 0M n n n n 2I n þ 1 ρM M 0 0,M 0n M n 2I n þ 1 ρM M 0 2 n

n

n

ðρÞ

I 0,M 0n M n

     ðρÞ 2  ðρÞ 2 ¼ AM 0n M n  þ BM 0n M n  ,

ð5:384Þ

n

ρ ¼ u, d

ð5:385Þ

The component of the polarization vector of the outgoing particles corresponding to the unpolarized incident particles is P0α0 ni 0 ,α n ¼

o X 1 n ðρÞþ 1 ^ ðMρ0ÞM F ^M , tr σ^i F 0M n n n n 2I n þ 1 ρM M 0 2 n

From Eq. (2.303) we obtain

n

i ¼ x,y,z

ð5:386Þ

410

5

Polarization Theory of Relativistic Nuclear Reactions

P0x α0 n0 ,αn ¼ 0, P2 ¼

1 I 0α0 n0 ,αn

P0y P0z α0 n0 ,αn ¼ P2 , α0 n0 ,αn ¼ 0   X 2 ðρÞ ðρÞ Re AM 0n M n CM 0n M n 2I n þ 1 ρM M 0 n

ð5:387Þ ð5:388Þ

n

In the case where the target and the residual nucleus are unpolarized and the incident particle is polarized, the analyzing power of the incident particles is Aiα0 n0 ,αn ¼

1 I 0α0 n0 ,αn

o X 1 n ðρÞ ðρÞþ 1 ^M ^ M 0 M σ^i F , tr F 0M n n n n 2I 0 þ 1 ρM M 0 2 n

i ¼ x,y,z

ð5:389Þ

n

So from Eq. (2.301) we can obtain Axα0 n0 ,αn ¼ 0, P1 ¼

1 I 0α0 n0 ,αn

Ayα0 n0 ,αn ¼ P1 , Azα0 n0 ,αn ¼ 0   X 2 ðρÞ ðρÞ Re AM 0n M n BM 0n M n 2I n þ 1 ρM M 0 n

ð5:390Þ ð5:391Þ

n

The differential cross section of the outgoing particles is

 I α0 n0 ,αn ¼ I 0α0 n0 ,αn 1 þ py P1

ð5:392Þ

where pi(i ¼ x, y, z) is the component of the initial polarization vector of the incident particles. The polarization transfer coefficient of the outgoing particles is defined as K ji ¼

1 I 0α0 n0 ,αn

o X 1 n ðρÞ ðρÞþ 1 ^M ^M F ^ , tr σ^j F 0M σ 0M i n n n n 2I 0 þ 1 ρM M 0 2 n

i, j ¼ x, y, z

ð5:393Þ

n

According to the results given in Sect. 2.7 we have K xx ¼ W 2 ,

K yx ¼ 0,

K xy ¼ 0, K xz ¼ Q2 ,

K yy ¼ J 0 , K yz ¼ 0, X 2

Q1 ¼

1

K zx ¼ Q1 , K zy ¼ 0, K zz ¼ W 1

ð5:394Þ

  ðρÞ ðρÞ Im AM 0n M n BM 0n M n

ð5:395Þ

 X  ðρÞ 2 ðρÞ Im AM 0n M n C M 0n M n 2I n þ 1 ρM M 0

ð5:396Þ

I 0α0 n0 ,αn 2I n þ 1 ρM n M 0

n

Q2 ¼

1 I 0α0 n0 ,αn

n

n

5.5

Dirac Coupling Channel Theory Including Elastic Scattering. . .

W1 ¼ W2 ¼ J0 ¼

1 I 0α0 n0 ,αn 1 I 0α0 n0 ,αn

1 I 0α0 n0 ,αn

X  ðρÞ 2  ðρÞ 2  1 AM 0n M n   BM 0n M n  2I n þ 1 ρM M 0 n

411

ð5:397Þ

n

  X 1 ðρÞ ðρÞ ðρÞ ðρÞ Re AM 0n M n DM 0n M n  BM 0n M n C M 0n M n 2I n þ 1 ρM M 0

ð5:398Þ

  X 1 ðρÞ ðρÞ ðρÞ ðρÞ Re AM 0n M n DM 0n M n þ BM 0n M n C M 0n M n 2I n þ 1 ρM M 0

ð5:399Þ

n

n

n

n

The components of the polarization vector of the outgoing particles are  I 0α0 n0 ,αn

p W þ pz Q2 I α0 n0 ,αn x 2

ð5:400Þ

 I 0α0 n0 ,αn

P þ py J 0 I α0 n0 ,αn 2

ð5:401Þ

 I 0α0 n0 ,αn

px Q1 þ pz W 1 I α0 n0 ,αn

ð5:402Þ

Pαx0 n0 ,αn ¼

Pαy0 n0 ,αn ¼ Pαz0 n0 ,αn ¼

This section develops the Dirac S matrix theory of the relativistic nuclear reactions. In the case of the unpolarized target and residual nucleus which spins are not equal to 0, this section also gives the formulas for calculating the differential cross section and the polarization physical quantities of the relativistic two-body nuclear 1 1 reactions of spin incident particles and various possible spin emitted particles. 2 2 The above relativistic theories and calculation formulas will automatically degenerate into corresponding non-relativistic theories and calculation formulas when the incident particle energy is very low.

5.5

Dirac Coupling Channel Theory Including Elastic Scattering and Collective Inelastic Scattering Channels

For collective deformation nuclei, refer to Eq. (5.234) not including the Coulomb potential VC, the Dirac equation describing the nucleon scattering process can be written as [29–33] h

i * α  p þ βðm þ U s Þ þ U 0 þ H T ψ ¼ Eψ

*

ð5:403Þ

where m is the nucleon stationary mass; Us and U0 are the complex nuclear potentials; HT is a non-relativistic internal motion Hamiltonian of the target nucleus

412

5

Polarization Theory of Relativistic Nuclear Reactions

H T Φ I n M n ¼ ε n ΦI n M n

ð5:404Þ

The scalar potential Us and the time-like component U0 of the 4 vector potential appearing in Eq. (5.403) can take the phenomenological potential of the WoodsSaxon form as



 U j ¼ V j f r, Rj1 , aj1 þ i W j f r, Rj2 , aj2 ,

j ¼ s, 0

ð5:405Þ

where

f r, Rjn , ajn







r  Rjn ¼ 1 þ exp ajn

1 ,

j ¼ s, 0

ð5:406Þ

In the above formulas, n ¼ 1 and 2 correspond to the optical potential real and imaginary parts, respectively. Of course, other forms of the phenomenological optical potential can also be included   in Eq. (5.405), for example, the surface type d optical potential obtained by dx 1þ1ex can also be included. The optical potential of the nucleus is the potential field generated by the nucleus itself. If the density distribution of the nucleus deviates from the sphere, the corresponding optical potential will also deviate from the spherical symmetry. Here we assume that Us and U0 are deformed potentials. The most common form of the deformed nucleus radius is " Rj ! Rj 1 þ

X

# 0

0

αλμ Yλμ ðθ , φ Þ

ð5:407Þ

λμ

where θ0 and φ0 are the direction angles in the coordinate system fixed on the target nucleus. Here we assume that the target nucleus is an axis-symmetric rotational nucleus, so from Eq. (5.407) we can get " Rjn ! Rjn

# X 0 1þ βλ Yλ0 ðθ ,0Þ ,

n ¼ 1, 2

ð5:408Þ

λ

where βλ is the λ order deformation parameter of the target nucleus. The nuclear potential appearing in the Dirac equation (5.234) or (5.403) is UD ¼ γ0Us þ U0

ð5:409Þ

When the target nucleus radius Rjn appearing in Eq. (5.405) is transformed based on Eq. (5.408), from Eq. (5.409) we can obtain

5.5

Dirac Coupling Channel Theory Including Elastic Scattering. . .

U D ! U D þ ΔU D

413

ð5:410Þ

where U D represents the spherical potential that does not contain βλ items, ΔUD represents the nuclear potential contributed by the deformation items containing βλ in Eq. (5.408), and we can write its expression according to Eq. (5.405) as follows: ΔU D ¼

! X

 βλ Yλ0 ðθ0 ,0Þ γ 0 F s ðr Þ þ F 0 ðr Þ

ð5:411Þ

λ

F j ðr Þ ¼ V j Rj1

  ∂

∂

f r, Rj1 , aj1 þ i W j Rj2 f r, Rj2 , aj2 , ∂Rj1 ∂Rj2

j ¼ s, 0 ð5:412Þ

Due to the existence of the deformation potential energy term ΔUD, it is possible for the incident nucleon to occur the inelastic scattering with the target nucleus. In the above theoretical formulas related to optical potential, the microscopic optical potential obtained from the theories of nuclear force or meson exchange can also be used. Substituting Eq. (5.410) into Eq. (5.403) and replacing U with a spherical symmetric potential U(r) we can obtain h

i * α  p þ βðm þ U s ðr ÞÞ þ U 0 ðr Þ þ H T þ ΔU D ψ ¼ Eψ

*

ð5:413Þ

Let  ψ¼

ψu



ψd

ð5:414Þ

then utilizing Eq. (5.38), Eq. (5.413) can be rewritten as 0  1 * σ^  p ψ d B C @ * A þ σ^  p ψ u

i ! ! μ ψu V C þ H T þ gð F 0 þ F s Þ ψ u E ¼E ψd ½m  U s þ U 0 þ H T þ gðF 0  F s Þ ψ d

h

m þ Us þ U0 þ

ð5:415Þ Note that in the upper component equation, refer to the discussion made on the Coulomb potential and Eq. (5.287) in the 5.3 section, the spherical nucleus Coulomb mM potential Eμ V C is added side-by-side with U0, where μ ¼ mþM , m and M are the masses of the incident particle and target nucleus, respectively, and μ is called the reduced mass. And g appeared in Eq. (5.415) is

414

5

g¼

Polarization Theory of Relativistic Nuclear Reactions

X λ

βλ Yλ0 ðθ0 , 0Þ

ð5:416Þ

Equation (5.341) has given *!

d σ^  l  r dr

*

σ^  p ¼ iðσ^  ^r Þ

ð5:417Þ

!

where ^r ¼ r =r. So from Eq. (5.415) we can obtain iðσ^  ^r Þ

*!   μ d σ^  l ψ d þ E  H T  m  U s  U 0  V C ψ u ¼ ðF 0 þ F s Þgψ u  E r dr

iðσ^  ^r Þ

ð5:418Þ ! * d σ^  l ψ u þ ðE  H T þ m þ U s  U 0 Þ ψ d ¼ ðF 0  F s Þgψ d  r dr ð5:419Þ

The wave functions ψ u and ψ d with the total angular momentum JM can be expressed, respectively, as   1X J * u ðr Þ φJM ψ u, JM r , ξ ¼ nlj ðΩ, ξÞ, r nlj nlj

  1X J * ψ d, JM r , ξ ¼ d ðr Þ φJM nlj ðΩ, ξÞ r nlj nlj ð5:420Þ

where n is the energy level symbol of the target nucleus and φJM nlj has been given by Eq. (5.306) and Eq. (5.305). The α' ¼ α scattering processes will be discussed in this section. Using Eq. (5.305), Eq. (5.351), and the first equation of Eq. (5.352), we can obtain iðσ^  ^r Þ l0 j0 m0 j ðΩÞ ¼ C11 2

η ν0 þη

pffiffiffi l0 þ1 0 X j0 m0j 2 i ^l C l0 m0 l

ν0 m0l

X1 1 2

ν0 L

^ L

L

m0l η

C l0 m0 l

1

1 0 2ν

X 1 0 ð1Þ2ν þη η

L0 η C l0 0

ð5:421Þ 10 YL

m0l η ðΩÞχ ν0 þη

Note that χþ ν χ ν0 þη ¼ δν

ν0 þη

ð5:422Þ

5.5

Dirac Coupling Channel Theory Including Elastic Scattering. . .

Z

Ylml YL

m0l η dΩ

¼ δlL δml

415

ð5:423Þ

m0l η

The following relations can be obtained by using Eqs. (5.421)–(5.423): η ¼ ν  ν0 ¼ m0l  ml ,

m0j ¼ m0l þ ν0 ¼ ml þ ν ¼ mj

ð5:424Þ

Then we get Z

^ þ ljml iðσ

X

X

 ^r Þ

X j0 m pffiffiffi l0 lþ1 ^l0 X j mj dΩ ¼ 2 i C lm 1ν δmj m0j C l 0 m 0 j 1ν 0 j l 2 l 2 ^l ν ml ν0 m0 l

1 0 2ν þη

ð1Þ

η

¼

l0 j0 m0

C 11 ν η 2

pffiffiffi l0 lþ1 ^l0 l 0 3i C0 ^l l 0

1 2

l m0 η C l C l0 0 ν0 l0 m0 l 1 η l 0 10

10 δmj m0j

X νν0 ml m0l

j0 mj

j mj

Clm

1 2ν

l

C l0 m0

l

1 0 2ν

X 1 ν0 C 21ν 1 η

η

2

C ll0 mm0 ll

1 η

ð5:425Þ Utilizing Eq. (1.63) we can obtain X l m C l0 m0 ll η

1

1

2 η C 1ν 2

ν0 1 η

X  1 1  t m ¼ ð1Þlþml^l ^t W ll0 ; 1t C l0 m0 t l 2 2 tm

1 1 2

ν0

C2l

ν ml t mt

t

ð5:426Þ When deriving the above formula Eq. (5.424) is used. We can also obtain X j0 m C l0 m0 j m0l ν0

X ml ν

ð1Þ

lþml

l

1 0 2ν

Ctl0 mm0t

l

1 0 2ν

1 ν j m C lm j 1ν C 2l ml j0 mj l 2

¼ δtj0 δmt mj ¼ ð1Þ

lþ12j

ð5:427Þ pffiffiffi 2 δ0 ^j jj

ð5:428Þ

Then we get X¼

pffiffiffi l0 lþ1 1 ð1Þlþ2j^l0 Cll0 00 6i

 10 W

ll0

 1 1 ; 1j δjj0 δmj m0j 2 2

ð5:429Þ

Using the orthogonal normalization properties of the wave functions, we can obtain

416

5

Z

Polarization Theory of Relativistic Nuclear Reactions

  1 J ! φJMþ nlj ðΩ, ξÞ ψ u,JM r , ξ dΩdξ ¼ r unlj ðr Þ, Z 1 J φJMþ nlj ðΩ, ξÞ ψ d,JM ðr, ξÞdΩdξ ¼ r d nlj ðr Þ

ð5:430Þ

The following equation can be obtained utilizing Eqs. (5.417), (5.420), (5.306), (5.425), (5.429), and (5.344): Z

*!   d σ^  l * ψ u,JM r , ξ dΩdξ ξÞ iðσ^  ^r Þ  r dr !   J X d K J 0 unl0 j0 ðr Þ X nl CJjmMj I n M n C Jj0 mMj I n M n X ¼  r dr r mj M n l0 j0 !   J X d K Jnl0 unl0 j ðr Þ Bll0 j ¼  r r dr 0

φJMþ nlj ðΩ,

ð5:431Þ

l

where Bll0 j ¼

pffiffiffi l0 lþ1 1 0 6i ð1Þlþ2j ^l C ll0 00

  01 1 W ll ; 1j 10 2 2

ð5:432Þ

In the same way we can obtain Z

*!   d σ ^  l * ^  ^r Þ φJMþ ψ d,JM r , ξ dΩdξ  nlj ðΩ, ξÞ iðσ r dr ! X  d K J 0  d Jnl0 j ðr Þ nl Bll0 j ¼  dr r r 0

ð5:433Þ

l

Equations (3.1215) and (3.1220) have given, respectively Yλ0 ðθ0 , 0Þ ¼

X ð1Þmλ Dλ mλ 0 ðα, β, γ ÞYλ

mλ ðΩÞ

ð5:434Þ

mλ

Z

And we can obtain

ΦI n M n Dλ mλ 0 ΦI 0n M 0n dξ ¼

^I 0n I M Cn n ^I n I 0n M 0n

In0 λmλ C I 0n 0

λ0

ð5:435Þ

5.5

Dirac Coupling Channel Theory Including Elastic Scattering. . .

Z

^λ ^l0 l ml 0 p ffiffiffiffiffi Cl0 m0 l Y dΩ ¼ 0 mλ l m l 4π^l

Ylml Yλ

417

ð5:436Þ

l0 λ mλ C l0 0 λ0

Then the following equation can be obtained utilizing Eqs. (5.420), (5.306), (5.416), and (5.434)–(5.436): Z r , ξÞdΩdξ φJMþ nlj ðΩ, ξÞ gψ u,JM ð^ ¼

X X X X 1 0 βλ C JjmMj I n M n CJj0 mM0 j uJn0 l0 j0 ðr Þ il l r 0 0 0 0 0 mM λ nlj

X

j0 m0j

C l0 m0

1 2ν

l

j

ð1Þ

^0 mλ I n

n

mj M n

In0 λmλ C I 0 0

^λ^l pffiffiffiffiffi C ll0 mm0 ll λ0 4π^l

j mj

C lm

ml m0l ν

l

1 2ν

ð5:437Þ

0

n C II n0 M M0

^I n

mλ

X I 0n M 0n

n

n

n

l 0 λ mλ C l0 0 λ0

First using Eq. (3.14) we find ð1Þmλ

X ml m0l ν

j0 m0j

j mj

Clm

l

1 2ν

C l0 m0

l

1 2ν

C ll0 mm0 ll

λ mλ

  ^j 0 0 0 0 1 0 0 mλ ¼ ð1Þlþl λj þ1þmj þmλ ^l^j Cλj m λl ; lj 0 m0 W j j j j ^λ 2 0   ^ ^ ^ 1 1 jj l mλ W jlj0 l0 ; λ ¼ ð1Þlþ2þmj C λj m 0 j j m0 j ^λ 2

ð5:438Þ

Then utilizing Eq. (1.63) we can get X mλ

Cλj

mλ In Mn mj j0 m0 j C I 0n M 0n λmλ 0

¼ ð1ÞI n þλI n

X 0

 ^λ^J W jj0 I n I 0 ; λJ 0 CJ00 M0 0 0 0 C I n M n n j m j I n M n j mj J0 M0

ð5:439Þ J0M0

We can also obtain X m0j M 0n

X mj M n

C Jj0 mM0 j

J0 M0 I 0n M 0n C j0 m0 j I 0n M 0n

n CJjmMj I n M n ð1Þmj C Ij n Mm j

J0M0

¼ δJJ 0 δMM 0

¼ ð1ÞI n J

^I n δJJ 0 δMM 0 J^

ð5:440Þ

ð5:441Þ

Thus the following equation can be obtained substituting Eqs. (5.438)–(5.441) into Eq. (5.437):

418

5

Z

Polarization Theory of Relativistic Nuclear Reactions

  * φJMþ nlj ðΩ, ξÞ gψ u,JM r , ξ dΩdξ ¼

X X 1 ^I 0 ^λ^l0 0 1 0 βλ uJn0 l0 j0 ðr Þil l n C II n0 0 0λ0 pffiffiffiffiffi C ll0 00 λ0 ð1Þl þ2 n ^ ^ r In 4π l 0 0 0 λ nlj

 ^j^j0^l  0 0 1  ^I 0 W jlj l ; λ ð1ÞI n þλI n ^λ^JW jj0 I n I 0n ; λJ ð1ÞI n J n ^λ ^ 2 J 1 X X 1 J λþl0 þI 0n þ12J l0 l ¼ pffiffiffiffiffi βλ u 0 0 ðr Þi ð1Þ r n0 l j 4π λ n0 l 0 j 0  

 ^λ^l0^j^j0^I 0 C l0 0 CI n0 0 W jj0 ll0 ; λ 1 W jj0 I n I 0 ; λJ n l 0 λ0 I n 0 λ0 n 2 X X 1 J ¼ βλ V λJ nlj,n0 l0 j0 r un0 l0 j0 ðr Þ 0 0 0 λ

ð5:442Þ

nlj

where 1 0 0 1 l0 l V λJ ð1ÞI n þ2J^l^l ^j^j ^I 0 n C λl0 00 nlj,n0 l0 j0 ¼ pffiffiffiffiffi i 4π  

 1 W jj0 ll0 ; λ W jj0 I n I 0n ; λJ 2

In 0 l0 C I 0n 0 λ0

ð5:443Þ

In the same way we can get Z

  X X 1 J * φJMþ βλ V λJ nlj ðΩ, ξÞ gψ d,JM r , ξ dΩdξ ¼ nlj,n0 l0 j0 r d n0 l0 j0 ðr Þ 0 0 0 λ

ð5:444Þ

nlj

From Eq. (5.418) and Eq. (5.419) we can also obtain the following equations utilizing Eqs. (5.430), (5.431), (5.433), (5.442), and (5.444): !   J   X d K Jnl0 d nl0 j ðr Þ μ Bll0 j r þ E  εn  m  U s  U 0  V C uJnlj ðr Þ  r r dr E l0 X X J ¼ ðF 0 þ F s Þ βλ V λJ nlj,n0 l0 j0 un0 l0 j0 ðr Þ n0 l0 j0

λ

X l0



!



J d K Jnl0 unl0 j ðr Þ þ ðE  εn þ m þ U s  U 0 Þ dJnlj ðr Þ  r r dr X X J ¼ ðF 0  F s Þ βλ V λJ nlj,n0 l0 j0 d n0 l0 j0 ðr Þ

ð5:445Þ

Bll0 j r

λ

n0 l0 j0

ð5:446Þ

Relativistic Collective Deformation RDWBA Method and Calculation. . .

5.6

419

Let r ¼ rm be the outer boundary to find numerical solution of the coupling channel equation of the above radial wave function, when r > rm the nuclear potential can be ignored. We stipulate that the symbol “ 0 ” represents the derivative for k0r. Notice here α0 ¼ α is no longer marked for scattering process. According to Eq. (5.310), we can see that the radial wave function of the upper component satisfies the following boundary condition: 

0  

0 0  ðuÞJ

uJn0 l0 j0 ðr Þ  Gl0 ðk r Þ  iF 0l0 ðk 0 r Þ δn0 l0 j0 ,nlj  Sn0 l0 j0 ,nlj G0l0 ðk0 r Þ þ iF 0l0 ðk 0 r Þ    ¼  ðuÞJ 0 0 0 0 uJn0 l0 j0 ðr Þ  0 ðk r Þ  iF l0 ðk r ÞÞδn0 l0 j0 ,nlj  S 0 0 0 ðk r Þ þ iF l0 ðk r ÞÞ  G ð ð G l l  n0 l j ,nlj rm rm

ð5:447Þ According to Eq. (5.325), the radial wave function of the lower component satisfies the following boundary condition: 

0   0 0 ðdÞJ 0 0  dJn0 l0 j0 ðrÞ  ðk0 rnl0 ðk 0 rÞÞ  iðk 0 rjl0 ðk0 r ÞÞ δn0 l0 j0 ,nlj  Sn0 l0 j0 ,nlj ðk 0 rnl0 ðk0 r ÞÞ þ iðk0 rjl0 ðk 0 rÞÞ    ¼  ðdÞJ d Jn0 l0 j0 ðr Þ  ½ðk0 rnl0 ðk0 r ÞÞ  iðk 0 rjl0 ðk 0 rÞÞ δn0 l0 j0 ,nlj  Sn0 l0 j0 ,nlj ½ðk0 rnl0 ðk 0 rÞÞ þ iðk 0 rjl0 ðk 0 r ÞÞ  rm rm

ð5:448Þ From the similarity of the lower component equation (5.295) and the upper component equation (5.294), the similarity between the lower component boundary condition (5.448) and the upper component boundary condition (5.447) can be ðρÞJ understood. Thus using the various S matrix elements Sn0 l0 j0 ,nlj ðρ ¼ u, dÞ obtained from the above boundary conditions and the calculation formulas given in the 5.4 section, the elastic and inelastic scattering cross sections, differential cross sections, 1 and various polarization physical quantities of spin particles can be calculated. 2

5.6

Relativistic Collective Deformation RDWBA Method and Calculation of Nucleon Polarization Quantities

Equations (3.12.57), (3.12.56), and (7.2.40) in Ref. [4] give the following differential cross section formulas in the distorted wave Born approximation (DWBA) theory for a þ A ! b þ B reaction: X  dσ  dσ 1 ¼ 2 2 dΩ ^i ^I ma mA mb mB dΩ βα a A

ð5:449Þ

420

5



dσ dΩ

 βα

Polarization Theory of Relativistic Nuclear Reactions

 μα μβ k β  T βα 2 ¼

 2 2 kα 2πħ

ð5:450Þ

Z    ðÞ * * k β , r β Φβ ξβ T βα ¼ χ β h    i *  * * * * U β r β , ξβ  U β r β Φα ðξα Þχ αðþÞ k α , r α d r β dξβ

ð5:451Þ

where α and β represent the incident channel and the outgoing channel, respectively. For the scattering process the target nucleus is the same as the residual nucleus, so * * * r α ¼ r β ¼ r can be taking. For the inelastic scattering reaction of the rotational nucleus, ξα and ξβ are the intrinsic coordinate Euler angles, and there are dξ ¼ sin βdβdαdγ and     * * U β r β , ξβ  U β r β ¼ ΔU D

ð5:452Þ

For the axis-symmetric rotational nuclei, ΔUD has been given by Eq. (5.411). Assume that the target nucleus and the residual nucleus are in n and n0 states of the ground state rotation band, respectively, their angular momentum are In and I 0n , respectively, then the corresponding rotational nucleus eigenfunctions are, respectively ΦI n M n ΦI 0n M 0n

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2I n þ 1 I n  ¼ DM n 0 8π2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2I 0n þ 1 I 0n  ¼ DM 0n 0 8π2

ð5:453Þ ð5:454Þ

The above two formulas show that the theory discussed in this section only applies to even A nuclei, which In and I 0n are integers. Eqs. (3.1215), (3.1219), and (3.1220) have given Yλ0 ðθ0 , 0Þ ¼ Z

X

ð1ÞM λ Dλ M λ 0 ðα, β, γ ÞYλ

M λ ðΩr Þ

ð5:455Þ

Mλ I0

In DMn 0n 0 Dλ M λ 0 DM n 0 sin βdβdαdγ ¼

Z

ðÞ

ΦI 0n M 0n Dλ M λ 0 ΦI n M n dξ ¼

8π2 I 0n M 0n CIn Mn 2 ^I n0 ^I n I 0n M 0n C ^I 0n I n M n

I 0n 0 λM λ C I n 0 λ0

I 0n 0 λM λ C I n 0 λ0

ð5:456Þ ð5:457Þ

where χ ðαþÞ and χ β in Eq. (5.451) take the solutions of a spherical relativistic optical model with a target nucleus with spin 0; they can be expressed using an upper component and a lower component, respectively, as follows:

5.6

Relativistic Collective Deformation RDWBA Method and Calculation. . .

χ αðþÞ

ð Þ

χβ

0 *  1 * *  u k,r * B ν C ! ψ ν k , r ¼ @ *  A * dν k , r

0 *  1 * *  u k 0, r C B ν0 0 * ! ψ ν0 k , r ¼ @ *  A 0 * d ν0 k , r

421

ð5:458Þ

ð5:459Þ

where ν and ν0 are the spin magnetic quantum numbers of incident nucleon and * * outgoing nucleon, respectively; k and k 0 are the momentum wave vectors of incident nucleon and outgoing nucleon, respectively. According to the discussion in Sect. * * * * 5.3, uν ð k , r Þ and uν0 ð k 0 , r Þ are the solutions of the upper component of the spherical * * relativistic optical model in the presence of the Coulomb potential; dν ð k , r Þ and * * dν0 ð k 0 , r Þ are the solutions of the lower component of the spherical relativistic optical model in the absence of the Coulomb potential. Note that the incident nucleon is along the z-axis direction, then using Eqs. (5.411), (5.57), and (5.452)– (5.459), the transition matrix elements of the collective inelastic scattering can be obtained from Eq. (5.451) as follows: ðρÞ

T n0 ν0 M 0n ,nνM n ðθÞ ¼

ðuÞ

qn0 ν0 M 0n ,nνM n

ðdÞ

^I n X I0 0 βCn ^I 0n λ λ I n 0

λ0 ð1Þ

M 0n M n

I0 M0

C I nn M nn

ρ ¼ u, d Z *0 * ¼ uþ ν0 ð k , r Þ ðF 0 ðr Þ þ F s ðr ÞÞYλ

qn0 ν0 M 0n ,nνM n ¼

Z

λ

ðρÞ M 0n M n qn0 ν0 M 0n ,nνM n ,

ð5:460Þ

* * 2 M n M 0n ðΩr Þuν ð k , r Þ r drdΩr

ð5:461Þ *

*

0 dþ ν0 ð k , r Þ ðF 0 ðr Þ  F s ðr ÞÞYλ

* * 2 M n M 0n ðΩr Þd ν ð k , r Þ r drdΩr

ð5:462Þ The above formulas give the transition matrix elements contributed by the upper and lower components of the relativistic wave functions. Note that (F0(r) þ Fs(r)) appear in the formula of the upper component; and (F0(r)  Fs(r)) appear in the formula of the lower component. The following expressions can be obtained according to Eqs. (5.239), (5.241), and (5.446): * *

*

* *

σ  p Þ uν ð k , r Þ dν ð k , r Þ ¼ Dn ð^

ð5:463Þ

422

5

Polarization Theory of Relativistic Nuclear Reactions

1 D n ¼ E  εn þ m þ U s  U 0

ð5:464Þ

where m is the nucleon stationary mass. In the case of a spherical target nucleus with spin 0, based on Eq. (5.345) one can obtain

*  σ  p ¼ iσ  ∇ ¼ i σ  ^r

*

*

*

*



d Slj  r dr

 ð5:465Þ

where Slj ¼ jðj þ 1Þ  lðl þ 1Þ 

3 4

ð5:466Þ

*

When the direction of the wave vector k of the incident nucleon is selected as the zaxis and in the case of the target nucleus with spin 0, the general form of the coefficient of the front side of the spin wave function χ v is as (see Eqs. (3.5.2) and (3.4.6) in Ref. [4]): *  pffiffiffiffiffirffiffiffiffiffiffiffiffiffiffiffiffiffiX E þ m ^ j ν iσl F lj ðkr Þ * e uν k , r ¼ 4π lC ljν 2m lj l0 12ν kr

ð5:467Þ

ljν ¼ Cjl0ν 1ν il Yl0 ðΩr Þ

ð5:468Þ

2

where Flj(kr) is a radial wave function obtained by relativistic optical model equation. For the Schrodinger-like equation (5.287), in the ħ ¼ c ¼ 1 unit system, there is k¼p¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi E 2  m2

ð5:469Þ

qffiffiffiffiffiffiffiffi Multiply the coefficient

Eþm 2m

in Eq. (5.467) so that when there is no nuclear

potential and Coulomb potential, this formula will automatically degenerate to a planar wave that is consistent with Eq. (5.93), and the coefficient is close to 1 when the energy is very low. In the DWBA theory, the outgoing wave is also a spherical *0 distorted wave, but the k direction is not along the z-axis, set θ to be the angle *0 * *0 * between k and the z-axis (i.e., k direction), γ is the angle between k and r , so the coefficient of the front side of the outgoing distorted wave χ v0 is *  pffiffiffiffiffi * uν0 k 0 , r ¼ 4π

rffiffiffiffiffiffiffiffiffiffiffiffiffiffi E0 þ mX^0 j0 ν0 l C l0 0 2m 0 0 lj

1 0 2ν

eiσl0

F l0 j0 ðk0 r Þ l0 j0 ν0 k0 r

ð5:470Þ

Relativistic Collective Deformation RDWBA Method and Calculation. . .

5.6

l0 j0 ν0 ¼ where E0 ¼

0 0 0 C jl0 0ν 1ν0 il Yl0 0 ðγ, 2

0Þ ¼

0 0 0 C jl0 0ν 1ν0 il 2

423

pffiffiffiffiffi X 4π Yl0 m0 ðΩr Þ Yl0 m0 ðΩk0 Þ ^l0 m0

ð5:471Þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi E 2 þ ε2n  εn20 . And note 0

Yl0 m0 ðΩr Þ Yλ

M n M 0n ðΩr Þ

0 1 X ^l ^λ L1 M 1 ¼ ð1Þm pffiffiffiffiffi C ^1 l0 m0 4π L1 M 1 L

λ

L1 0 M n M 0n C l0 0 λ0 YL1 M 1 ðΩr Þ

ð5:472Þ

Z YL1 M 1 ðΩr Þ dφ ¼ 2π YL1 0 ðθ, 0Þ δM 1 0

ð5:473Þ

From the above two formulas, m0 ¼ M n  M 0n can be got. Further we can obtain Z YL1 0 ðθ, 0ÞYl0 ðθ, 0Þd cos θ ¼

1 δ 2π L1 l

ð5:474Þ

Then we can get Z

Yl0

M n M 0n

Yλ

M n M 0n Yl0 dΩr

^l0 ^λ 0 0 ¼ ð1ÞM n M n pffiffiffiffiffi Cll0 M 0 n þM n 4π ^l

l 0 λ M n M 0n C l0 0 λ0

ð5:475Þ And there is the relation ðσ^  ^r Þ ðσ^  ^r Þ ¼ ^I

ð5:476Þ

*0

Noting that k is axis-symmetric relative to the z-axis, the following result can be obtained substituting Eqs. (5.463), (5.465), (5.467), and (5.470) into Eqs. (5.460)– (5.462) and using Eqs. (5.468), (5.471), (5.475), and (5.476): T

ðρÞ n0 ν0 M 0n ,nνM n ðθ Þ

C jl0ν

j0 ν0 1 C 0 l 0 12ν0 2ν

Yl0

M n M 0n ðθ,

1 ¼ 0 kk

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðE þ mÞ ðE þ εn  εn0 þ mÞ ^I n X^0 ll0 iðσl σl0 Þ l i e 4m2 ^I 0n ljl0 j0

X I0 0 βλ^λCI nn 0 λ

I 0n M 0n λ0 C I n M n

l 0 l 0 λ M 0n M n C l0 0 λ0 C l0 M n þM 0n

λ

!

M n M 0n

ðρÞ

0Þ T n0 l0 j0 ,nlj , ρ ¼ u,d ð5:477Þ

424

5 ðuÞ

Z

T n0 l0 j0 ,nlj ¼

Polarization Theory of Relativistic Nuclear Reactions

F l0 j0 ðk0 r ÞðF 0 ðr Þ þ F s ðr ÞÞF lj ðkr Þ dr

 0  d Sl0 j0 F l0 j0 ðk r Þ Dn0 ðF 0 ðr Þ  F s ðr ÞÞ  dr r r    d Slj F lj ðkr Þ 2 r dr Dn  r r dr

ð5:478Þ

Z 

ðdÞ

T n0 l0 j0 ,nlj ¼

ð5:479Þ

The definition of Dn has been given by Eq. (5.464). And we know that when the energy of incident particles is very low, the contribution of the lower component can be ignored. The theory described above is called the relativistic collective deformation RDWBA method. In the expression of the above transition operator, I n þ λ  I 0n and l0 þ λ  l are required to be even. When substituting the transition matrix elements of ρ ¼ u and d given by Eq. (5.477) into Eqs. (5.450) and (5.449), the contributions of the upper component and the lower component to the collective excitation inelastic scattering differential cross section can be obtained. The optical potential parameters can be determined by fitting the elastic scattering experimental data, while the deformation parameters βλ can be determined by fitting the inelastic scattering experimental data. The λ value takes the smallest one or two positive integers greater than or equal to 2 that satisfy the angular momentum coupling *

*

*0

relation I n þ λ  I n as well as that I n þ λ  I 0n is even number. In Refs. [34–40] the collective inelastic scattering problem of even-even target nuclei with spin 0 is studied. It can be seen according to Eqs. (5.449) and (5.450) that the reaction amplitude corresponding to Eqs. (2.250) and (2.251) is as follows: ðρÞ f n0 ν0 M 0n , nνM n ðθÞ

μ ¼ 2πħ2

rffiffiffiffi k0 ðρÞ ðθÞ, T k n0 ν0 M 0n , nνM n

ρ ¼ u, d

ð5:480Þ

where μ is the nucleon reduced mass. The RDWBA method is an approximation theory. The distorted wave Born approximation is that the wave function of the outgoing particle is replaced with the distorted wave function calculated by the spherical optical model. Of course, in this case the coupling of the incident nucleon * * angular momentum j and the target nucleus spin I n is not considered. It can be seen that there is no C ‐ G coefficient to represent the coupling of nucleon angular * * momentum j and target spin I n in the transition matrix element expression given by Eq. (5.477). When the coupling between inelastic scattering channel and elastic scattering channel is relatively weak, the calculation of inelastic scattering reaction by RDWBA method still has a certain degree of credibility. Refer to Eqs. (5.376) and (5.377), and let

Relativistic Impulse Approximation of Elastic Scattering. . .

5.7

ðρÞ

ðρÞ

ðρÞ

AM 0n M n ðθÞ ¼ f n0 1M 0 , n1M ðθÞ, 2

ðρÞ

n

2

12M 0n , n12M n

ðρÞ

BM 0n M n ðθÞ ¼ if n0 1M 0 ,n

n

ðρÞ

CM 0n M n ðθÞ ¼ if n0

425

2

ðθÞ,

ðρÞ

n

12M n

ðθÞ,

ðρÞ

DM 0n M n ðθÞ ¼ f n0

12M 0n , n 12M n

ðθÞ, ρ ¼ u,d ð5:481Þ

Thus the cross sections, differential cross sections, and various polarization physical quantities of the nucleon collective inelastic scattering with target can be calculated using the formulas given in Sect. 5.4.

5.7

Relativistic Impulse Approximation of Elastic Scattering and Calculation of Nucleon Polarization Quantities

If the incident nucleon energy is relatively high, its wavelength is smaller than the distance between the nucleons in the nucleus. The interaction between the incident nucleon and the nucleon in the nucleus can be regarded as quasi-free nucleonnucleon (NN) scattering and can be handled by multiple scattering theories. At this time, we need to know the free NN scattering amplitude and the density distribution of the nucleus. When the incident nucleon energy is high enough, the correction of the NN scattering amplitude by the nuclear density is negligible. For the elastic scattering process, there is no need to consider changes in the internal states of the nucleus. So in 1983, the theory of relativistic impulse approximation (RIA) of the medium energy nucleon elastic scattering was proposed by McNeil et al. [41–43], and this theory was soon applied by others and further studied [44–47]. Subsequently, a number of studies on the nucleon elastic scattering with target using RIA theory were carried out [48–57]. In the stationary Dirac equation (5.226), the Bjurken-Drell metric is used, theself-energy operator Σ is expressed p0 ¼ p0 ¼ E is taken, and   as  an elastic * ð þÞ * ^ scattering optical potential U r . And then the wave function ψ * r describing k ,s

the positive energy nucleon is used, where s represents the nucleon spin magnetic quantum number. Thus in the ħ ¼ c ¼ 1 unit system using Eq. (5.61) and Eq. (5.57), the following equation can be obtained by multiplying γ 0 from left side of Eq. (5.226) with definite energy h

 i   ð þÞ * ^ * r ψ* r ¼ 0 γ μ pμ  m  U k ,s

ð5:482Þ

mass, the on-shell mass-energy relation is where m represents the nucleon stationary   pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi * * 2 μ 2 E ¼ k þ m , and there are p ¼ E, i∇ , pμ ¼ E, i∇ .

426

5

Polarization Theory of Relativistic Nuclear Reactions

* E  We use  k , sðþÞ to represent the positive energy free nucleon wave function in   * ^ * momentum space when the potential U r ¼ 0, the momentum vector is k , and the spin projection is s. Then from Eq. (5.482) we can obtain E  * γ μ pμ  m  k , sðþÞ ¼ 0

ð5:483Þ

* E *    k , sðþÞ ¼ u k , þ jχ s i

ð5:484Þ



And there is

where jχ si is the nucleon spin wave function. Equation (5.93) has given 0 1 ^I *  rffiffiffiffiffiffiffiffiffiffiffiffiffi E þ m@ * A u k, þ ¼ σ^  k 2m Eþm

ð5:485Þ

Equation *  (5.484) indicates that both the upper component and lower component of u k , þ all act on the two-dimensional spin wave function jχ si, respectively. For the negative energy states there are E  * γ μ pμ  m  k , sðÞ ¼ 0

ð5:486Þ

* *  E   k , sðÞ ¼ u k ,  jχ s i

ð5:487Þ



Equation (5.94) has given 1 0 * *  rffiffiffiffiffiffiffiffiffiffiffiffiffi σ^  k E þ mB C u k,  ¼ @E þ mA 2m ^I

ð5:488Þ

  * ^ * When the potential U r ¼ 0 and the momentum vector is k , using Eqs. (5.484), (5.487), (5.80), and (5.86), the stationary wave function of the positive energy (þ) and negative energy () free nucleons with the spin projection s can be written as follows:    * * * r ¼ e i k  r u k , jχ s i

ð Þ *

φ*

k ,s

Their conjugate wave functions are

ð5:489Þ

5.7

Relativistic Impulse Approximation of Elastic Scattering. . .

427

  *  ** 0 r ¼ hχ s0 juþ k 0 , e i k  r

ð Þþ *

φ *0

k ,s0

ð5:490Þ

  ð þÞ * From the above results it can be seen that φ* r is the positive energy solution k ,s   ^ * of Eq. (5.482) in U r ¼ 0 case. It can also be seen that the general solution of Eq. (5.482) is     ð þÞ * r ¼ φ* r þ

ðþÞ *

ψ*

k ,s

k ,s

    1 ðþÞ * ^ * r U r ψ * γ μ pμ  m þ iδ k ,s

ð5:491Þ

The introduction of iδ is to deal with the poles and one finally makes δ ! 0+. Here (γ μpμ  m þ iδ)1 is a propagator. Let L  γ μ pμ  m

ð5:492Þ

so Eq. (5.491) can be rewritten as       ð þÞ ð þÞ * ^ * r ψ* r r ¼ φ* þ L1 U

ðþÞ *

ψ*

k ,s

k ,s

k ,s

ð5:493Þ

Since     Z       * * ^ *0 * ðþÞ * ðþÞ * ^ * r ψ* r 0 d r 0 U r ψ* r ¼ δ r  r 0 U k ,s

k ,s

ð5:494Þ

so there is     Z       * * ^ *0 * ðþÞ * ð þÞ * ^ * r ψ* r 0 d r 0 L1 U r ψ * r ¼ L1 δ r  r 0 U k ,s k ,s Z       * * * ð þÞ * ^ * r 0 ψ* r 0 d r 0  G r, r0 U

ð5:495Þ

k ,s

where     * * * * G r , r 0 ¼ L1 δ r  r 0 ,

    * * * * LG r , r 0 ¼ δ r  r 0

ð5:496Þ

Note here we only study the stationary state problem. Substituting Eq. (5.495) into Eq. (5.493), one can get     Z       * * * ðþÞ * ð þÞ * ^ * r 0 ψ* r 0 d r 0 r ¼ φ* r þ G r , r 0 U

ð þÞ *

ψ*

k ,s

k ,s

There is the following δ function formula:

k ,s

ð5:497Þ

428

5

  * * δ r  r0 ¼

Polarization Theory of Relativistic Nuclear Reactions

1 ð2πÞ3

Z

* * * * 0 ei q ð r  r Þ dq

ð5:498Þ

Thus the following expression can be obtained from Eqs. (5.496), (5.492), and (5.498):   * * G r, r0 ¼

1 1 γ μ pμ  m ð2πÞ3

Z

* * * * 0 ei q  ð r  r Þ d q

ð5:499Þ

We have γ μ pμ þ m 1 ¼

2 γ μ pμ  m γ μ p  m2

ð5:500Þ

μ

Substituting Eq. (5.500) into Eq. (5.499) and letting (γ μpμ)2 act on the function * * * r 0 ÞiE ðtt 0 Þ iqμ ðrr0 Þμ ei q  ð r  , and using Eqs. (5.59) and (5.53) as well as noting   ¼e *

pμ ¼ E, i∇ , then one can obtain

γ μ pμ

2

1 ¼ γ μ pμ γ ν pν ! q μ qν γ μ γ ν ¼ qμ qν ð γ μ γ ν þ γ ν γ μ Þ ¼ E 2  q2 2

Further there is

γ μ pμ

2

 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  m2 ¼ E 2  m2  q2 ¼  q  E 2  m2 q þ E2  m2 ð5:501Þ

The following expression can be obtained substituting Eqs. (5.500) and (5.501) into Eq. (5.499): 

* *0

G r, r



¼

γ μ pμ þ m ð2πÞ3

Z

* * * 0 ei q ð r  r Þ *  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d q 2 2 2 2 q E m qþ E m

ð5:502Þ

  * *  * * * In spherical coordinate system q  r  r 0 ¼ q r  r 0  cos θ is hold, and there is *

d q ¼ sin θ dθ dφ q2 dq

ð5:503Þ

* *  Then when letting x ¼  r  r 0  the following expression can be obtained by Eq. (5.502):

Relativistic Impulse Approximation of Elastic Scattering. . .

5.7

 G

* *0 r, r



¼

γ μ pμ þ m

Z



429

eiqx cos θ 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dφ d cos θ q dq q  E 2  m2 q þ E2  m2

ð2πÞ3 Z

iqx  γ μ pμ þ m 1 q iqx  dq ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e  e 2 4π ix 0 q  E2  m2 q þ E 2  m2

ð5:504Þ Therefore using the residual theorem, the following result can be obtained from Eq. (5.504) [4]: Z   γ μ pμ þ m 1 qeiqx * *  G r, r0 ¼  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dq 2 4π ix 1 q  E 2  m2 q þ E 2  m2 ! !0    γ μ pμ þ m eik r  r   ¼ ! !0  4π r  r 

ð5:505Þ

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Here k ¼ E2  m2 is the momentum wave vector value of the incident or outgoing nucleon.   ^ * r 0 given in Eq. (5.497) is not We know that only when r0 is relatively smaller U   ^ * 0, so when r ! 1 there must be r0  r in U r 0 6¼ 0 region, then one can get   *   * *  r * ! ! 2 1=2  kr 1  2  r 0 k  r  r 0  ¼ k r 2  2 r  r 0 þ ðr 0 Þ r *0

ð5:506Þ

*0

¼ kr  k  r

*0

k ¼k

*

r r

ð5:507Þ

*  *   If the z-axis is taken along the k 0 direction, because  k 0  ¼ k so there are k 0x ¼ k0y ¼ 0, k 0z ¼ k, then from Eqs. (5.484) and (5.485) we can obtain

430

5

Polarization Theory of Relativistic Nuclear Reactions

 ED* X*0  0  k , s0 ðþÞ k , s0 ðþÞ γ 0 s

20

1

0

1

0

3

1

6B 7 B C 6B 0 C 7  B 1 C C 6 7 B C k E þ m 6B k C 7 C 0 1 0 ¼ 0 þB B k C 1 0  6 B C Eþm 7 2m 6B Eþm C 7 B 0 C @ A 4 Eþm 5 @ k A  0 Eþm 1 0 20 13 k 0 0 0 0 1 0  0 C B 6B Eþm C7 k C B0 6B C7 1 0 C 6B 0 B C7 0 0 0 E þ m B C 6 E þ m 6B B C7 C   ¼ þ B0 C7 2 B C 6 0 0 0 2m 6B k B C7 k C C7 0C B   6B E þ m 0  E þ m 2 A7 A @ 4@ 5 k k 0  0  Eþm Eþm 0 0 0 0 1 0 k 1 0  0 Eþm C B C B k C B 1 0 C B 0 Eþm C B E þ mB C   2 ð5:508Þ ¼ C B k 2m B k C 0 0  C BE þ m E þ m C B B  2 C A @ k k 0  0  Eþm Eþm Utilizing Eq. (5.57) we can get Eγ 0  γ i k 0i þ m ¼

^0 ðE þ mÞ^I ^0 ðE  mÞ^I

! 

^ 0

σ^i k0i 0 ^ σ i k 0i ^

! ð5:509Þ

We can still get k2 E 2  m2 ¼ Eþm Eþm ! ! i 0 1 0 ky þ k0z ¼ 0 0 1

Em¼ σ^i k0i

¼

0 1 1 0

! k0x

þ

0 i

ð5:510Þ k 0z

k0x  ik0y

k0x þ ik 0y

k 0z

!

ð5:511Þ For a plane wave which selects the direction of the outgoing particle beam as the z-axis, so there are k0z ¼ k, k0x ¼ k 0y ¼ 0. In this case substituting Eqs. (5.510) and (5.511) into Eq. (5.509) and comparing them with Eq. (5.508), we can obtain

5.7

Relativistic Impulse Approximation of Elastic Scattering. . . *

*

Eγ 0  γ  k 0 þ m ¼ 2m

431

 ED* X*0  0  k , s0 ðþÞ k , s0 ðþÞ

ð5:512Þ

s

where  D*  D*   k 0 , s0 ðþÞ ¼ k 0 , s0 ðþÞγ 0

ð5:513Þ

!

The above simplified proof of taking k 0 direction as z-axis does not lose its universality. The following result can be obtained substituting Eqs. (5.505), (5.512), and (5.489) into Eq. (5.497):   E ED*   * * * m X  *0 0   ψ * r ¼ ei k  r  k , sðþÞ   k , s ðþ Þ k , s0 ðþ Þ 2π s0 k ,s Z ikj*r *r 0 j     e * ð þÞ * ^ * * *  U r 0 ψ* r 0 d r 0  r  r 0 k ,s

!

ð þÞ *

ð5:514Þ

Using Eq. (5.506) and Eq. (5.507), Eq. (5.514) can be rewritten as follows: ð þÞ * 

ψ*

k ,s

r

 !   X *0   * * m ¼ eik r  k ; sðþÞ  jk s0 ðþÞi k 0 ; s0 ðþÞ 2π s0 Z  * *

 ðþÞ * 0  * 0 eikr ik 0 r 0 ^ * 0 e U r ψ * r dr r k ,s * *

ð5:515Þ

From the above formula, we can extract the scattering amplitude as [41] Z * *  D     * * 0 m *0 0  * ðþÞ * ^ * r 0 ψ* r 0 d r 0 F s0 ,s k 0 , k ; E ¼  k , s ðþÞ ei k  r 0 U 2π k ,s

ð5:516Þ

We introduce the Dirac T matrix in the following way:         * ð þÞ * ð þÞ * ^ * U r ψ * r ¼ T^ r φ* r k ,s

k ,s

ð5:517Þ

and introduce the following symbol: D*  *E Z * *   ** 0   * * k 0 T^  k ¼ ei k  r T^ r ei k  r d r So Eq. (5.516) can be rewritten as

ð5:518Þ

432

5

Polarization Theory of Relativistic Nuclear Reactions

D   E * *  D E m *0 0  *  *  * F s0 ,s k 0 , k ; E ¼  k , s ðþÞ k 0 T^  k  k , sðþÞ 2π

ð5:519Þ

If assumed that the incident nucleon energy is relatively high, it only interacts with each nucleon in the target nucleus once, i.e., ignoring the influence of the density of the nucleus on the NN scattering, then the T matrix element can be approximated as follows: D*  *E D*  ð1Þ *E     k 0 T^  k ffi k 0 T^  k ¼

 + *  X A D   E *0  *   k ^t i  k 0 0   i¼1

ð5:520Þ

where j0i represents the ground state of the spin saturation two-full shell even-even nucleus, A is the mass number of the target nucleus, and ^t i is the t matrix of scattering of the incident nucleon the ith nucleon in the target nucleus. From it *E with   (5.518) D*Eq. E **  0  * ik r is the non-relativistic plane wave, so k ^t i  k is the can be seen that  k ¼ e non-relativistic t matrix element. Substituting Eq. (5.520) into Eq. (5.519) one can obtain  + *   X A D   E * * *  D E *0  *  m *0 0    0 k , s ð þ Þ  0 k ^t i  k 0  k , sðþÞ F s0 ,s k , k ; E ¼  2π  i¼1 

ð5:521Þ

We study in the NA center of momentum system composed of the incident * * nucleon and target nucleus. k and k 0 appearing in Eq. (5.521) are the wave vectors of the incident nucleon and the outgoing nucleon in the NA center of momentum system, respectively. And let *

ka 

  1 * *0 kþk , 2

*

*

*

q  k  k0

ð5:522Þ

be the average momentum wave vector and momentum transfer wave vector of the incident nucleon and the outgoing nucleon, respectively. Apparently there are *

* 1* k ¼ k a þ q, 2

*0

* 1* k ¼ ka  q 2

ð5:523Þ

*

And from k 0 ¼ k^r given by Eq. (5.507) we know *

*

ka  q ¼ 0

ð5:524Þ

For the sake of simplicity, we assume that each nucleon in the target nucleus is stationary relative to the center of the target nucleus, i.e., ignoring the nucleon Fermi movement in the nucleus [2, 42, 48, 49]. In the NA center of momentum system, the

5.7

Relativistic Impulse Approximation of Elastic Scattering. . .

433

total momentum wave vectors of the target nucleus and the recoil nucleus should be * *  k and  k 0, respectively; the average momentum wave vectors of each nucleon of * * the target nucleus and the recoil nucleus are  k =A and  k 0 =A, respectively. This coordinate system is usually called the Breit coordinate system. In RIA theory, assuming that the nucleons not collided are not affected, we set the momentum * wave vector of the nucleon which will be collided before the colliding is p ¼ * *  k =A þ x , and each nucleon momentum wave vector of the other A  1 nucleons, * * which will not be collided, is  k =A  x =ðA  1Þ. At this time, the total momentum * * wave vector of the target nucleus is still  k , but x is to be determined. After the * * * collision, the momentum wave vector of the emitted nucleon is k 0 ¼ k  q . The momentum wave vector of the nucleon, which has been collided, changed to be * *0 * * p ¼  k =A þ x þ q . The momentum wave vectors of the other A  1 nucleons, which were not collided, do not change. And the total momentum wave vector of the * * * recoil nucleus is  k þ q ¼  k 0. Since we study elastic scattering, it is required that the momentum value of the collided nucleon in the target nucleus should not change before and after the collision, that is,  *   *    * * *  k =A þ x  ¼  k =A þ x þ q 

ð5:525Þ

 *  * * * q  q þ 2  k =A þ x  q ¼ 0

ð5:526Þ

So we can get *

Using Eq. (5.524) the above formula can be rewritten as  *  * * * * q  q þ 2  k =A þ k a =A þ x  q ¼ 0

*

ð5:527Þ

Therefore we can obtain [2, 42, 48, 49] *

*

*

*

x ¼ k =A  k a =A  q =2

*

*

p ¼ k a =A  q =2,

*0

*

ð5:528Þ *

p ¼  k a =A þ q =2

ð5:529Þ

The total energies of the incident nucleon and the collided nucleon before being collided can be written according to Eqs. (5.523), (5.529), and (5.524) as follows, respectively:

1=2 E ð1Þ ¼ k 2a þ q2 =4 þ m2

ð5:530Þ

434

5

Polarization Theory of Relativistic Nuclear Reactions

h i1=2 E ð2Þ ¼ ðka =AÞ2 þ q2 =4 þ m2

ð5:531Þ

From Eq. (5.523) and Eq. (5.524) we can get

1=2 ka ¼ k2  q2 =4

ð5:532Þ

The physical quantity s introduced according to Eq. (5.177) can be expressed using Eqs. (5.523) and (5.529) as follows:

2 ! !2

2 s ¼ Eð1Þ þ Eð2Þ  k þ p ¼ Eð1Þ þ Eð2Þ  k2a ð1  1=AÞ2

ð5:533Þ

If A ¼ 1, Eq. (5.533) degenerates into a formula for a two-nucleon system. It can be seen that s is the square of the total energy of the two particle system in the center of momentum system and is the Lorentz invariant. And let ð5:534Þ

t ¼ q2

Here t and q are not Lorentz invariants. k is determined by the kinetic energy TL of the incident nucleon in the laboratory system. If TL and q2 are known, s and t can be calculated out. However, when TL is determined, the k values obtained in different coordinate systems are different, so the q values are also different. The general form of NN scattering amplitude that satisfies Lorentz invariance is [41–49, 55] F NN ¼ F S I ð1ÞI ð2Þ þ F P γ 5 ð1Þγ 5þ ð2Þ þ F V γ μ ð1Þγ μþ ð2Þ þF A γ 5 ð1Þγ 5þ ð2Þγ μ ð1Þγ μþ ð2Þ þ F T σ μν ð1Þσ μνþ ð2Þ

ð5:535Þ

where FS, FP, FV, FA, and FT represent scalar, pseudo-scalar, vector, axial-vector, and anti-symmetric tensor items, respectively. Here (1) and (2) represent the incident nucleon and the being collided nucleon in the target nucleus, respectively. For the sake of simplicity, there is no distinction between neutrons and protons here. In order to rewrite Eq. (5.535) into another form, firstly the following structural matrices are introduced [55, 58]: 0 Γ1 ¼ I ¼ @

^I ^0

^0 0 ^0 5 @ Γ3 ¼ γ ¼ ^I

^I

1

0

A,

Γ2 ¼ γ ¼ @

^I ^0

1 A,

^I

^ 0

1

A, ^ ^ 0 I 0 1 ^ 0 ^I 0 5 A Γ4 ¼ γ γ ¼ @ ^ ^ I 0 0

It is easy to write the following relations using Eqs. (5.57)–(5.59):

ð5:536Þ

5.7

Relativistic Impulse Approximation of Elastic Scattering. . .

435

γ 5 ð1Þγ 5þ ð2Þ ¼ Γ 3 ð1ÞΓ 3 ð2Þ, !

!

γ μ ð1Þγ μþ ð2Þ ¼ Γ 2 ð1ÞΓ 2 ð2Þ  Γ 4 ð1ÞΓ 4 ð2ÞΣ ð1Þ  Σ ð2Þ, μ

ð5:537Þ

!

μþ

!

γ ð1Þγ ð2Þγ ð1Þγ ð2Þ ¼ Γ 4 ð1ÞΓ 4 ð2Þ  Γ 2 ð1ÞΓ 2 ð2ÞΣ ð1Þ  Σ ð2Þ 5

5þ

*

The definition of the 4  4 matrix Σ has been given by Eq. (5.71). The tensor matrices are defined as i σ μν ¼ ðγ μ γ ν  γ ν γ μ Þ, 2

i σ μνþ ¼  ðγ νþ γ μþ  γ μþ γ νþ Þ 2

ð5:538Þ

We can further find the following expression using σ^i σ^j ¼ δij^I þ iεijk σ^k , i, j, k ¼ 1, 2, 3 1 σ μν ð1Þσ μνþ ð2Þ ¼ ½γ μ ð1Þγ ν ð1Þγ νþ ð2Þγ μþ ð2Þ  γ ν ð1Þγ μ ð1Þγ νþ ð2Þγ μþ ð2Þ 4 γ μ ð1Þγ ν ð1Þγ μþ ð2Þγ νþ ð2Þ þ γ ν ð1Þγ μ ð1Þγ μþ ð2Þγ νþ ð2Þ 1 ¼ ½γ μ ð1Þγ ν ð1Þγ νþ ð2Þγ μþ ð2Þ  γ ν ð1Þγ μ ð1Þγ νþ ð2Þγ μþ ð2Þ 2 1 0 ¼  γ ð1Þγ i ð1Þγ i ð2Þγ 0 ð2Þ  γ i ð1Þγ 0 ð1Þγ i ð2Þγ 0 ð2Þ 2

þγ i ð1Þγ 0 ð1Þγ 0 ð2Þγ i ð2Þ  γ 0 ð1Þγ i ð1Þγ 0 ð2Þγ i ð2Þ

þ γ i ð1Þγ j ð1Þγ j ð2Þγ i ð2Þ  γ j ð1Þγ i ð1Þγ j ð2Þγ i ð2Þ  ¼ 2γ 0 ð1Þγ 0 ð2Þγ i ð1Þγ i ð2Þ  γ i ð1Þγ j ð1Þγ i ð2Þγ j ð2Þ i6¼j  From γ i ¼ Γ 4Σ i and σ^i σ^j i6¼j ¼ iεijk σ^k we can get γ i ð1Þγ j ð1Þγ i ð2Þγ j ð2Þji6¼j ¼ ðΓ 4 ð1ÞΓ 4 ð2ÞÞ2

X

i6¼j ! 2

Σ i ð1ÞΣ j ð1ÞΣ i ð2ÞΣ j ð2Þ !

¼ 2ðΓ 4 ð1ÞΓ 4 ð2ÞÞ Σ ð1Þ  Σ ð2Þ Then we obtain σ μv ð1Þ σ μvþ ð2Þ !

!

!

!

¼ 2Γ 2 ð1ÞΓ 2 ð2ÞΓ 4 ð1ÞΓ 4 ð2ÞΣ ð1Þ  Σ ð2Þ þ 2ðΓ 4 ð1ÞÞ2 ðΓ 4 ð2ÞÞ2 Σ ð1Þ  Σ ð2Þ ð5:539Þ !

!

¼ 2½Γ 1 ð1ÞΓ 1 ð2Þ þ Γ 3 ð1ÞΓ 3 ð2Þ Σ ð1Þ  Σ ð2Þ

436

5

Polarization Theory of Relativistic Nuclear Reactions

The following expression can be obtained substituting Eqs. (5.537) and (5.539) into Eq. (5.535): F NN ¼

4   X ! ! an þ bn Σ ð1Þ  Σ ð2Þ Γ n ð1ÞΓ n ð2Þ

ð5:540Þ

n¼1

where a1 ¼ F S ,

a2 ¼ F V ,

a3 ¼ F P ,

a4 ¼ F A ,

b1 ¼ 2F T ,

b2 ¼ F A ,

b3 ¼ 2F T ,

b4 ¼ F V

ð5:541Þ

The coefficients Fi for each item in Eq. (5.535) are all functions of s and t given by Eqs. (5.533) and (5.534), usually s is not labeled, and they can be expressed only as F NN(q). Set fNN as the non-relativistic NN scattering amplitude in the NA center of momentum system; its matrix element of the spin state is þ M s01 s02 ,s1 s2 ¼ χ þ s0 χ s0 f NN χ s1 χ s2 2

1

ð5:542Þ

Using the positive energy plane wave of the Dirac equation given by Eq. (5.218) and using Eq. (5.523) and Eq. (5.529), the matrix element of the plane wave state of the relativistic NN scattering amplitude F NN can be written as follows [42]:   *  *    1* 1* 1* 1* 1 * e s0 s0 , s1 s2 ¼ us0 1 * 0 M  k  u k þ   u F u q k q q q k a a a a NN s s s1 1 2 1 2 2 2 A 2 2 2 A ð5:543Þ If we make a non-relativistic approximation, i.e., we ignore the contribution of the small *lower component of the positive energy plane wave, at this time there is us k  χ s, but the difference between the non-relativistic and relativistic cases is a qffiffiffiffiffiffiffiffi normalized coefficient Eþm 2m , so we let *  * B k a, q h    *   *      i1=2 1* 1 * 1* 1* 1* 1 * E q  ka þ m E ka  q þ m E ka þ q þ m E  q  ka þ m 2 A 2 2 A 2 ¼ 4m2

ð5:544Þ

5.7

Relativistic Impulse Approximation of Elastic Scattering. . .

437

*  * where E k a þ 12 q is the energy corresponding to the nucleon momentum *  * * * k a þ 12 q and k a and q have been given by Eq. (5.522). Since the impulse approximation *  is only applicable to medium-high energy nuclear reactions, the value * of B k a , q given by the above formula should be significantly greater than 1. Here we give a convention *  e s0 s0 ,s1 s2 ¼ B k a , * q M M s01 s02 ,s1 s2 1 2

ð5:545aÞ

The corresponding convention given in the early literature of RIA theory [41–47] is as e s0 s0 ,s1 s2 ¼ 1 M s0 s0 ,s1 s2 M 1 2 2ik 1 2

ð5:545bÞ

This convention will make the dimensions of the two matrix elements in Eq. (5.545b) different. The nuclear force parameters corresponding to the above two kind of convention methods are different. **

When the incident plane wave is taken as ei k  r i (there is no coefficient 1/(2π)3/2 in front), Eq. (3.12.50) of Ref. [4] and Eq. (3.76) of Ref. [49] all give the relation between the non-relativistic NN scattering amplitude fNN and the t matrix element as follows: f NN ðqÞ ¼ 

  μNN D*0  *E k ^t NN  k 2πħ2

ð5:546Þ

where μNN is the NN reduced mass and ^t NN is the t matrix in the NA center of mass system. The usual nuclear force parameters are given in the NN center of mass system. In the case of relativity, since the general convention is that t matrix is Lorentz invariant, so the t matrix in the NN center of momentum system can be used directly in the NA center of momentum system. Note the following relation: D*    E Z   ** **  ! * * ! * * k 0 δ r  r i  k ¼ δ r  r i ei q  r d r ¼ ei q  r i so under the zero-range force approximation and refer to Eq. (5.546), the NN scattering amplitudes in Eq. (5.521) for the above two convention methods can be expressed, respectively, as

438

5



Polarization Theory of Relativistic Nuclear Reactions

  D   E 1 m *0 ^ * * *  F NN ðqÞ exp i q  r i k t i  k ¼  * * 2π B k ,q

ð5:547aÞ

a

D   E   m *0 ^ * * * ð5:547bÞ k t i  k ¼ 2ikF NN ðqÞ exp i q  r i 2π D*  *E   As mentioned earlier, k 0 ^t i  k is the non-relativistic t matrix element, and F NN(q) 

is the relativistic scattering amplitude, so the convention corrections are introduced in the above two equations. And we can rewrite the above two formulas as D*  *E   k 0 ^t i  k ¼ 

  2π * * *  F NN ðqÞ exp i q  r i * mB k a , q

ð5:548aÞ

  D*  *E 4πik   * * F NN ðqÞ exp i q  r i k 0 ^t i  k ¼  m

ð5:548bÞ

  ^ * r We know that the optical potential is equivalent to the mass operator. The U appearing in the equations is the   equation  previous   mass operator Σ in theDDirac *0  *E * * ^ r is converted to T^ r by using Eq. (5.517), so k T^  k given (5.226), and U by Eq. (5.520) can be regarded as the optical potential Uopt(q) in the momentum space. Introduce a coefficient symbol

C ð qÞ ¼

8 > > >

> > :  4πik , m

ð5:549Þ

For convention given by Eq:ð5:545bÞ

So substituting Eq. (5.548a) and Eq. (5.548b) into Eq. (5.520), we can obtain  + *  X A    * *  U opt ðqÞ ¼ C ðqÞ 0 F ðqÞ exp i q  r i 0   i¼1 NN

ð5:550Þ

In the independent particle shell model, the anti-symmetric wave function of the double-full shell target nucleus can be written as [55] A   Y 1 * j0i ¼ pffiffiffiffiffi det unljμ r i A! i¼1

where

ð5:551Þ

5.7

Relativistic Impulse Approximation of Elastic Scattering. . .

   * unljμ r ¼

439

 ϕnlj ðr Þ Y μlj ð^r Þ i^ σ  ^r λnlj ðr Þ

ð5:552Þ

We have the following expressions according to Eqs. (17.3.21) and (17.3.22) in Ref. [28]: Y μlj ð^r Þ ¼

X j Cl

μ μν

ν

1 2ν

il Y l

r Þχ 12ν μν ð^

ð5:553Þ

ϕnlj ðr Þ ¼ unlj ðr Þ=r

ð5:554Þ

where unlj(r) is a radial wave function of the nuclear bound state, and neutrons and protons are not distinguished. The Dirac equation can be written as h

i     * * * α  p þ βðm þ U S Þ þ U 0 u r ¼ Eu r

*

ð5:555Þ

And the following expression can be obtained from Eqs. (5.239), (5.241), and (5.345): 

 d Slj ϕnlj ðr Þ  λnlj ðr Þ ¼ D dr r

ð5:556Þ

3 4

ð5:557Þ

D ¼ ðE þ m þ U S  U 0 Þ1

ð5:558Þ

where Slj ¼ jðj þ 1Þ  lðl þ 1Þ 

The effect of Coulomb potential is not considered in the lower component of the nucleon wave function. The following expression can be obtained from Eqs. (5.553), (5.351), and (5.352): Z μlj ð^r Þ  iðσ^  ^r ÞY μlj ð^r Þ ¼ X1 L

^ L

C lL

pffiffiffi lþ3 X j 2 i ^l C l

μνη μν 1 η

ν

CLl0

μ μν

X 1 2ν

0 rÞ 10 YL μνη ð^

ð1Þ2νþη C11

η

1

2

η νþη

1 2

ν

χ νþη ð5:559Þ

So Eq. (5.552) can be rewritten as   * unljμ r ¼

ϕnlj ðr ÞY μlj ð^r Þ λnlj ðr ÞZ μlj ð^r Þ

! ð5:560Þ

440

5

Polarization Theory of Relativistic Nuclear Reactions

Introduce the following momentum distribution shape factors according to Eqs. (5.540) and (5.550):  + *  X  **  iq r i  ρn ðqÞ ¼ 0 Γ n ðiÞ e 0  i   + *   X **  ! !  j n ðqÞ ¼ 0 Γ n ðiÞΣ ðiÞ ei q  r i 0   i

ð5:561Þ

ð5:562Þ

*

where ρn(q) and j n ðqÞ are determined by the wave function |0i of the target nucleus and have no relationship with the nuclear force parameters an and bn. So Eq. (5.550) can be rewritten as U opt ðqÞ ¼ CðqÞ

4 X

  ! ! Γ n an ðqÞρn ðqÞ þ bn ðqÞ j n ðqÞ  Σ

ð5:563Þ

n¼1

Using the Fourier transformation, the optical potential of the coordinate space can be obtained as [2, 46–48] U opt ðr Þ ¼

1 ð2πÞ3

Z

**

!

ei q  r U opt ðqÞ d q

ð5:564Þ

Substituting Eq. (5.563) into Eq. (5.564), we can obtain U opt ðr Þ ¼

4 X

  ! ! Γ n U n ðr Þ þ W n ðr Þ  Σ

ð5:565Þ

n¼1

where U n ðr Þ ¼

1 ð2πÞ3

*

1 ð2πÞ3

W n ðr Þ ¼ !

Z Z

**

!

ð5:566Þ

!

ð5:567Þ

ei q  r CðqÞan ðqÞρn ðqÞdq **

!

ei q  r C ðqÞbn ðqÞ j n ðqÞd q

Notice that Un(r) and W n ðr Þ in Eq. (5.565) correspond to an and bn in Eq. (5.541), respectively, for example, U1(r) is equivalent to US(r), U2(r) is equivalent to UV(r). It is also noted that the optical potential Uopt(r) contains terms obviously appearing ! spin operator Σ , which are obtained from the space-like components of the 4 vector terms and the 4 axis-vector terms as well as the tensor terms in the nuclear force. There are 8 items in the NN scattering amplitude given by Eq. (5.535), which correspond to scalar, pseudo-scalar, time-like, and space-like components of the

5.8

Relativistic Impulse Approximation of Inelastic Scattering. . .

441

4-vector and the 4-axis-vector, and the corresponding tensors with σ 0i and σ ij. Due to the requirement to satisfy the conservation of parity, it is generally believed that the pseudo-scalar and axis-vector terms do not contribute. And it is believed that in the case of the stationary state of the spherical symmetric system, the 4-vector space-like component terms and tensor terms (b3) of σ 0i do not contribute [2, 46–48, 53, 57, 59]. In addition, the calculation results also show that the contribution of tensor items (b1) of σ ij is also very small [46, 47]. Therefore it can be seen that the relativistic optical potential is mainly contributed by scalar term US and 4-vector time-like component term U0. The spherical optical potential Uopt(r) can be calculated using Eqs. (5.565)– (5.567), (5.561), (5.562), (5.551), (5.560), and (5.541) and nuclear force parameters. Then substituting Uopt(r) into Eq. (5.482) and making a transformation similar to the 5.3 section, and introducing the Coulomb potential, the relativistic optical model calculation can be carried out. In the specific calculation, the selection of NN nuclear force should be considered. For example, someone used a simple Yukawa potential [43]; and some people use the RIA method to study the elastic scattering process of protons and non-double full shell nucleus (such as 13C) [55]. When the elastic scattering amplitudes are obtained, the differential cross section, analyzing power, and spin rotation function of the elastic scattering can be calculated using Eqs. (5.296)–(5.301) [2, 43–49, 51–57].

5.8

5.8.1

Relativistic Impulse Approximation of Inelastic Scattering and Calculation of Nucleon Polarization Quantities Relativistic Distorted Wave Impulse Approximation of Single Particle State Inelastic Scattering [58, 60, 61]

We consider the process by which the nucleus in the initial state Ψ J i M i is excited to the final state Ψ J f M f . Its transition amplitude can be written as * Tf i ¼

A X j¼1

ð þÞ

ðÞ

k ,s

k ,s0

where ψ * and ψ *0

+ ðÞ

ð þÞ

k ,s0

k ,s

ψ *0 Ψ J f M f ^t ð0, jÞ ψ * Ψ J i M i

ð5:568Þ

are the wave functions representing the incident nucleon and

the outgoing nucleon with the positive energy, respectively. The incident and outgoing nucleons have a label number of 0. Ψ JM is a wave function of the nucleus composed of A nucleons ( j ¼ 1, 2, . . ., A). ^t ð0, jÞ is a t matrix describing the interaction between the label 0 nucleon and the jth nucleon in the nucleus. And ψ *k ,s and Ψ JM satisfy the following Dirac equations, respectively:

442

5



Polarization Theory of Relativistic Nuclear Reactions

 * α  p þ β m þ V 0 ψ *k ,s ¼ Ek ψ *k ,s " # A   X * * α j  p j þ βj m þ V t Ψ JM ¼ EΨ JM *

ð5:569Þ ð5:570Þ

j¼1

where Vt in Eq. (5.570) is the potential felt by the nucleon in the target nucleus, it is selected to solve the bound state equation (5.570), so that Ei and Ψ J i M i for the initial state and Ef and Ψ J f M f for the final state can be obtained. The final state f is a single particle excited state. Equation (5.569) is an equation describing scattering state of nucleon labeled with 0. In higher energy condition, it can be assumed that the incident nucleon only interacts with one nucleon in the target nucleus and then runs out of the nucleus. Based on the nuclear wave function Ψ J i M i the optical potential V0 can be obtained by the RIA method described in the 5.7 section, and substituting the obtained optical potential V0 into Eq. (5.569), then the distorted ðþÞ wave ψ * can be solved. If starting from the nuclear wave function Ψ J f M f , the k ,s

ðÞ

distorted wave ψ *0

k ,s0

can be found by the same method. The t matrix ^t ð0, jÞ in

Eq. (5.568) can be given with the method described in the 5.7 section. If all wave functions and t matrices are given in the spatial coordinate, the transition amplitude Tf i can be obtained from Eq. (5.568). Then the differential cross section and various polarization physical quantities of the single particle state inelastic scattering can be calculated using Eq. (5.480) and Eq. (5.481) as well as the calculation formulas given in Sect. 5.4. The above method is called the relativistic distorted wave impulse approximation of the single particle state inelastic scattering.

5.8.2

Relativistic Distorted Wave Impulse Approximation of Collective State Inelastic Scattering [62–64]

We know that when calculating the target nucleus ground state wave function j0i appeared in Eqs. (5.550) and (5.551), we must first give the radius R of the nucleus, so the RIA optical potential Uopt given by Eq. (5.565) is actually a function of the nuclear radius R. We assume that the nucleus is not spherical and its radius is deformed similar to Eq. (5.407), so we introduce the following deformation optical potential according to Eq. (5.565): ! X ∂U n ðr Þ sn 0 Γn βλ Yλ0 ðθ , 0Þ Rsn ΔU ¼ ∂Rsn n¼1 λ ! # ! X ∂W n ðr Þ ! 0n 0 þ βλ Yλ0 ðθ , 0Þ R0n Σ ∂R0n λ 4 X

"

ð5:571Þ

5.9

Relativistic Impulse Approximation of (p, n) Reactions and Calculation. . .

443

where βsn and βλ0n ðn ¼ 1, 2, 3, 4Þ are the adjustable deformation parameters. λ And let ΔU D ¼ γ 0 ΔU

ð5:572Þ

Then it is possible to perform various calculations using the methods described in Sect. 5.6. Since the optical potential used here is calculated using the RIA method, the theory is said to be a relativistic distorted wave impulse approximation of the collective state inelastic scattering.

5.9

Relativistic Impulse Approximation of (p, n) Reactions and Calculation of Nucleon Polarization Quantities

In the real part of the nucleon universal optical potential, it usually contains a term NZ that is proportional to α  NZ A . The coefficient signs at the front edge of the A terms of the neutron and proton optical potentials are opposed; this form of optical potential indicates that the optical potential should be isospin related. Lane proposed the following form of optical potential [65]: *

V ¼ V0 þ

*

t T V1 A

ð5:573Þ

where V0 and V1 are the isospin scalar part and isospin vector part of the optical * * potential, respectively. t and T are the isospin operators of the incident nucleon and the target nucleus, respectively, and they are Their components in the  all vectors.

rectangular coordinate system are ^t x , ^t y , ^t z and T^ x , T^ y , T^ z , respectively, and their components in the spherical basis coordinate system are ð^t þ , ^t 0 , ^t  Þ and

 T^ þ , T^ 0 , T^  , respectively, and there are ^t 0 ¼ ^t z , T^ 0 ¼ T^ z . The total isospin value of the nucleon is t ¼ 12 , and its isospin z component value τ ¼ 12 represents the neutron and τ ¼  12 represents the proton. The total isospin value of the target nucleus is T ¼ A2 , and if the nucleus is composed entirely of neutrons, there is T z ¼ T 0 ¼ A2 ; if the nucleus is composed entirely of protons, there is T z ¼ T 0 ¼  A2 , then there is T z ¼ T 0 ¼ NZ 2 for the nucleus composed of A ¼ N þ Z. If a proton in the nucleus becomes a neutron, N increases by 1, Z decreases by 1, and then Tz ¼ T0 increases by 1. In spherical basis coordinate there is *

*

t T ¼

1 X μ¼1

ð1Þμ^t μ T^ μ

ð5:574Þ

444

5

Polarization Theory of Relativistic Nuclear Reactions

and there is the following angular momentum formula: J 0 ψ jm ¼ J z ψ jm ¼ mψ jm

ð5:575Þ

Introducing the angular momentum operators  1

J  pffiffiffi J x iJ y 2

ð5:576Þ

and using the commutative relation [Ji, Jk] ¼ iεiklJl of the angular momentum operators, it can be proved that  1  1

½J z , J ¼ pffiffiffi iJ y iðiÞJ x ¼  pffiffiffi J x iJ y ¼ J

2 2

ð5:577Þ

Let ð Þ

φjm ¼ J ψ jm

ð5:578Þ

so we can obtain



 J z J ψ jm ¼ fJ J z þ ½J z , J g ψ jm ¼ J ðJ z 1Þ ψ jm ¼ ðm 1Þ J ψ jm ð5:579Þ and ð Þ

ð Þ

J z φjm ¼ ðm 1Þ φjm

ð5:580Þ

ð Þ

ð5:581Þ

And let J ψ jm ¼ φjm ¼ Γ ψ j m 1 and from Eq. (5.576) we know ðJ Þþ ¼ J

ð5:582Þ

so from the above two formulas we can obtain     jΓ j2 ¼ J ψ jm , J ψ jm ¼  ψ jm jJ J jψ jm We can also obtain

ð5:583Þ

5.9

Relativistic Impulse Approximation of (p, n) Reactions and Calculation. . .

445

  

 1

1 J x i J y J x i J y ¼  J 2x þ J 2y i J x J y i J y J x 2 2 



 1 1 2 2 ¼  Jx þ Jy i Jx, Jy ¼  J 2  J z ð J z 1Þ 2 2

ð5:584Þ

1 1 jΓ j2 ¼ ½J ðJ þ 1Þ  mðm 1Þ ¼ ðJ mÞðJ m þ 1Þ 2 2

ð5:585Þ

J J ¼ 

Then we can get 1 J ψ jm ¼ pffiffiffi ½ðj mÞðj m þ 1Þ 1=2 ψ j 2

m 1

ð5:586Þ

Let φtτ and ΦTT 0 be the isospin wave functions of the nucleon and the target nucleus, respectively, so that from Eq. (5.575) we can obtain ^t 0 φtτ ¼ τφtτ ,

T^ 0 ΦTT 0 ¼ T 0 ΦTT 0

ð5:587Þ

From Eq. (5.586) we know ^t φtτ ¼ p1ffiffiffi ½ðt τÞ ðt τ þ 1Þ 1=2 φt 2

ð5:588Þ

τ 1

If τ ¼ 12 (representing neutron), there will be ^t þ φ11 ¼ 0, 22

^t  φ11 ¼ p1ffiffiffi φ1 22 2 2

12

ð5:589Þ

¼0

ð5:590Þ

If τ ¼  12 (representing proton), there will be ^t þ φ1 2

12

1 ¼ pffiffiffi φ12 12 , 2

^t  φ1 2

12

From Eq. (5.586) we can also write 1 T^ ΦTT 0 ¼ pffiffiffi ½ðT T 0 Þ ðT T 0 þ 1Þ 1=2 ΦT 2

T 0 1

ð5:591Þ

So we have T^ þ ΦA2

NZ 2

1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffi Z ðN þ 1ÞΦA2 2

NZþ2 2

,

T^  ΦA2

NZ 2

1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffi N ðZ þ 1ÞΦA2 2

NZ2 2

ð5:592Þ Using the formulas given above, for the p þ A reaction there is

446

5

* * t  T φ12

12 ΦA2

NZ 2

¼

Polarization Theory of Relativistic Nuclear Reactions

N Z φ12 4

12 ΦA2

NZ 2



1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi N ðZ þ 1Þφ12 12 ΦA2 2

NZ2 2

ð5:593Þ The first item in above formula represents the (p, p) elastic scattering channel, and the second one represents the (p, n) reaction channel. For the n þ A reaction there is * * t  T φ12 12 ΦA2

NZ 2

¼

NZ φ12 12 ΦA2 4



NZ 2

1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Z ðN þ 1Þφ12 2

12 ΦA2

NZþ2 2

ð5:594Þ

The first item in above formula represents the (n, n) elastic scattering channel, and the second one represents the (n, p) reaction channel. It is assumed that the target nucleus is a spherical nucleus with spin 0, regardless of the internal freedom and excited state of the target nucleus, i.e., only its isospin state is considered. For incident and outgoing nucleons, their spatial coordinates and spins must be considered to be same as the spherical optical model. The isospin wave , but the nucleon wave function of the target nucleus is still represented by ΦA2 NZ 2 function is rewritten as φ12τ ¼

X uτl0 j0 r

l0 j0 m0 j

m0

l0 1j j0 ðΩÞχ 12τ

ð5:595Þ

2

where m0

l0 1j j0 ðΩÞ ¼ 2

X j0 m0 C l0 m0 j l

m0l ν0

0

1 0 2ν

il Yl0 m0 l ðΩÞχ 12ν0

ð5:596Þ

where χ 12τ is a nucleon isospin wave function. Using the new φ12τ given by Eq. (5.595), based on Eqs. (5.589) and (5.590), we agree 2 X u l0 j0 ðr Þ 1

^t φ1 2

1 2

¼

l0 j0 m0 j

r

m0 1 l0 1j j0 ðΩÞ pffiffiffi χ 12 2 2

12

ð5:597Þ

1

^t þ φ1 2

12

¼

X u2l0 j0 ðr Þ l0 j0 m0 j

r

m0 1 l0 1j j0 ðΩÞ pffiffiffi χ 12 2 2

1 2

ð5:598Þ

This means that under the action of the isospin operator ^t , when the proton is turned into a neutron or the neutron is turned into a proton, the corresponding nucleon space wave function is turned into a space wave function that represents the new nucleon. This is physically reasonable. This agreement is to ensure that τ in the radial wave

5.9

Relativistic Impulse Approximation of (p, n) Reactions and Calculation. . .

447

function uτl0 j0 ðr Þ is consistent with τ in χ 12τ . The total wave function of the system can be expressed as ψ τ ¼ φ12τ ΦA2

ð5:599Þ

NZ 2

In the case of non-relativistic theory, the stationary Schrodinger equation satisfied by ψ τ is "

ħ2 * 2 ∇ þ  2μτ

*

*

t T V0 þ V1 A

!# ψ τ ¼ Eτ ψ τ

ð5:600Þ

where μτ and Eτ are the reduced mass and the energy of the nucleon τ, respectively. 1 For τ ¼  protons, while substituting Eqs. (5.599), (5.595), (5.593), and (5.598) 2 P þ þ R mþ ΦA T 0 χ 1τ0 r dΩ l1jj ðΩÞ from the left side of into Eq. (5.600) and multiplying τ0 T 0 0

2

0

2

2

the obtained expression, the following equation can be obtained: 

     NZ ħ 2 d2 l ð l þ 1 Þ  V0 þ VC   V 1 þ Ep uplj ðr Þ 4A 2μp dr 2 r2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi N ðZ þ 1Þ þ V 1 unlj ðr Þ ¼ 0 2A

ð5:601Þ

1 neutrons, 2 similar methods are used, but Eqs. (5.594) and (5.597) need to be used, and the following equation can be obtained: In the above equation the Coulomb potential VC is added. For τ ¼



     ħ2 d2 l ð l þ 1 Þ NZ  V0 þ  V 1 þ E n unlj ðr Þ 2μn dr 2 4A r2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Z ð N þ 1Þ þ V 1 uplj ðr Þ ¼ 0 2A

ð5:602Þ

Whether proton is the incident particle or neutron is the incident particle, Eqs. (5.601) and (5.602) must be solved together. The S matrix elements Sljp,p , Sljn,p , Sljn,n , Sljp,n can be obtained through boundary conditions, and the relevant cross sections, differential cross sections, and polarization quantities can also be calculated. Through (p, n) or (n, p) reaction the target nucleus will become its own isospin analogous state (IAS). In the case of relativity, the following form of optical potential can be taken [66]:



* * U opt ¼ U 0s þ γ 0 U 00 þ U 1s þ γ 0 U 10 t  T =A

ð5:603Þ

448

5

Polarization Theory of Relativistic Nuclear Reactions

The Dirac equation given by Eq. (5.226) is h

i

 * α  p þ β mτ þ U opt  Eτ ψ τ ¼ 0

*

ð5:604Þ

Substituting Eq. (5.603) into Eq. (5.604) and using Eqs. (5.593), (5.594), (5.597), and (5.598), the following equations can be obtained: h

i

  N Z 1 * βU s þ U 10  E p ψ p α  p þ β mp þ U 0s þ U 00  4A pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 N ðZ þ 1Þ  βU 1s þ U 10 ψ n ¼ 0 2A h i

  N Z 1 * * α  p þ β mn þ U 0s þ U 00 þ βU s þ U 10  E n ψ n 4A pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  Z ð N þ 1Þ 1  βU s þ U 10 ψ p ¼ 0 2A *

ð5:605Þ

ð5:606Þ

where ψ p and ψ n are the 4-component wave functions and they can be divided into upper and lower components. Referring to Eq. (5.287) the proton upper component μ equation needs to be added to the Coulomb potential Epp V C; μp is the proton reduced mass. Eqs. (5.605) and (5.606) should be solved simultaneously. The optical potential appeared in the equations may use the phenomenological potential, or the microscopic potential calculated by the RIA method; it means that the RIA method can be used to study (p, n) and (n, p) reactions. The above equations can also be solved by the coupling channel method described in Sect. 5.5. In the spherical ρÞlj ðρÞlj ðρÞlj ðρÞlj , Sn,p , Sn,n , Sp,n are introduced, where nucleus case, the S matrix elements Sðp,p ρ ¼ u, d represent the upper component and the lower component, respectively. Finally, the relevant cross sections, differential cross sections, and polarization quantities can be calculated by using the calculation formulas given in Sect. 5.4.

5.10 5.10.1

Relativistic Classical Field Theory and Lagrangian Density in Quantum Hadron Dynamics Relativistic Classical Field Theory [2, 67, 68]

Under the ħ ¼ c ¼ 1 unit system and the Bjorken-Drell metric, the physical quantity of the field is set to be φλ ðxÞ, where

λ ¼ 1,2,⋯, n

ð5:607Þ

5.10

Relativistic Classical Field Theory and Lagrangian Density. . .

 

  * * xμ ¼ x0 , x1 , x2 , x3 ¼ t, r , xμ ¼ ðx0 , x1 , x2 , x3 Þ ¼ t,  r

449

ð5:608Þ

and n represents the number of physical quantities of the field at each space-time point. Assume that the Lagrangian density L is a function of φλ and ∂μφλ, ∂μ  ∂x∂μ has been given by Eq. (5.60). The Lagrange function is equal to Z L¼

 d4 xL φλ , ∂μ φλ

ð5:609Þ

Make a slight change for the physical quantity of the field as follows: φλ ðxÞ ! φλ ðxÞ þ δφλ ðxÞ

ð5:610Þ

and it is required that δφλ(x) is equal to zero at the boundary of the Lagrange function integral; at the same time assume that the realistic field physical quantity φλ makes the Lagrange function taking the extreme value, i.e., any δφλ does not change the Lagrange function value. So the partial integral can be used to obtain (

)

 ∂L ∂L  δ ∂μ φλ δφ þ

δ L ¼ d4 x ∂φλ λ ∂ ∂μ φλ ( !) Z ∂L ∂L 4   ∂μ

¼0 ¼ d xδφλ ∂φλ ∂ ∂μ φ λ Z

ð5:611Þ

Thus the Euler-Lagrange equation is obtained as ∂L ∂L   ∂μ

∂φλ ∂ ∂ μ φλ

! ¼ 0,

λ ¼ 1, 2, ⋯, n

ð5:612Þ

The energy-momentum tensor is defined as ∂L  ∂ν φ λ T μν  gμν L þ

∂ ∂ μ φλ

ð5:613Þ

The metric matrix element gμν has been given by Eq. (5.53). The following expressions can be obtained using Eqs. (5.613) and (5.612): ∂μ T

μν

μν

¼ g ∂μ L þ ∂μ ¼ gμν ∂μ L þ

! ∂L ∂L

 ∂ν φλ þ

 ∂μ ∂ν φλ ∂ ∂ μ φλ ∂ ∂μ φλ

∂L ν ∂L  ∂μ ∂ν φλ , ∂ φλ þ

∂φλ ∂ ∂ μ φλ

450

5

Polarization Theory of Relativistic Nuclear Reactions

∂L ∂L  ∂ μ ∂ μ0 φ λ , ∂ μ φλ þ

∂φλ ∂ ∂μ0 φλ 8 ∂L 0 ∂L > >  ∂ μ0 ∂ 0 φ λ ,  ∂ φλ 

> > < ∂φλ ∂ ∂ μ0 φ λ gμν ∂μ L ¼ δμν > > ∂L i ∂L > >  ∂μ0 ∂i φλ , ∂ φλ 

: ∂φλ ∂ ∂ μ0 φ λ ∂μ L ¼

ν¼0 ν ¼ i ¼ 1, 2, 3

Then we get the continuous equation as ∂μ T μν ¼ 0

ð5:614Þ

The 4-momentum is defined as Pν ¼

Z

*

ð5:615Þ

d r T 0ν

And the conjugate physical quantity π λ corresponding to the generalized coordinate φλ is defined as πλ 

∂L ∂φ_ λ

ð5:616Þ

λ , π λ is the generalized momentum. Then we introduce the Hamilwhere φ_ λ ¼ ∂φ ∂t tonian density ℋ and from Eq. (5.613) we can get

ℋ  π λ φ_ λ  L ¼ T 00

ð5:617Þ

The total energy of the field is Z E ¼ P0 ¼ H ¼

*

drℋ

ð5:618Þ

In Eq. (5.617), it is understood that L and ℋ are the functions of φσ, π σ, and φ_ σ ; moreover φ_ λ is a function of φσ and π σ. From Eq. (5.617) we can obtain ∂φ_ ∂ℋ ∂L ∂φ_ λ ¼ φ_ σ þ π λ λ  ¼ φ_ σ ∂π σ ∂π σ ∂φ_ λ ∂π σ Utilizing Eq. (5.612) we also have

ð5:619Þ

5.10

Relativistic Classical Field Theory and Lagrangian Density. . .

∂φ_ λ ∂L ∂L ∂φ_ λ ∂L ∂ℋ ¼ πλ   ¼ ∂φσ ∂φ_ λ ∂π σ ∂φσ ∂φσ ∂φσ 2 3 " # 7 ∂L ∂ 6 6 ∂L 7  ¼ π_ σ  ¼ ∂μ

4 ∂xi ∂φσ 5 ∂ ∂ μ φσ ∂ ∂xi ∂φ_ ∂φ_ ∂ℋ ∂L ∂L ∂L   ¼ πλ  λ      λ ¼   ∂φ_ λ ∂φσ ∂φσ ∂φσ ∂φσ ∂φσ ∂ ∂ ∂ ∂ ∂ ∂xi ∂xi ∂xi ∂xi ∂xi

451

ð5:620Þ

ð5:621Þ

Substituting Eq. (5.621) into Eq. (5.620) we can get 2 π_ σ ¼ 

3

7 ∂ℋ ∂ 6 6 ∂ℋ 7 þ 4 ∂φσ ∂xi ∂φσ 5 ∂ ∂xi

ð5:622Þ

Equations (5.619) and (5.622) are the Hamiltonian canonical motion equations of the field.

5.10.2

Lagrangian Density in Quantum Hadron Dynamics [2, 68–74]

In order to study high-temperature and high-density systems, it is necessary to introduce strong interacting particles on the basis of relativistic field theory and develop quantum Hadron dynamics models. In the Walecka model [69] it is believed that nucleon-nucleon interaction is achieved by exchanging one boson, and there are many kinds of mesons and photons to be changed between the nucleons. Introduce the following Lagrangian density: L ¼ LN þ LM þ Lint

ð5:623Þ

where LN and LM represent the Lagrangian densities of free nucleons and free mesons, respectively, and Lint represents the Lagrangian density of interaction between nucleons and mesons. The Lagrangian density of free nucleons is taken as

 LN ¼ ψ i iγ μ ∂μ  M ψ i

ð5:624Þ

where ψ i is the Dirac spinor field describing the ith nucleon in the nucleus, and M is the nucleon stationary mass. The sum of i is included in Eq. (5.624). Using partial integral we have

452

5

Polarization Theory of Relativistic Nuclear Reactions

 ψ i iγ μ ∂μ ψ i ! iγ μ ∂μ ψ i ψ i In Eq. (5.612) when φλ is replaced with ψ i we can obtain ∂LN ¼ Mψ i ∂ψ i " # ∂LN  ¼ iγ μ ∂μ ψ i ∂μ

∂ ∂μ ψ i

ð5:625Þ ð5:626Þ

Using the Lagrange equation (5.612) we can get

 iγ μ ∂μ  M ψ i ¼ 0

ð5:627Þ

This is the Dirac equation (5.66) of free nucleons. σ meson is a scalar meson whose spin is equal to zero in charge neutrality (isospin scalar) and is used to simulate the attraction between nucleons. However, the existence of σ mesons has not been confirmed experimentally so far. In Eq. (5.612) φλ is replaced with σ, and the Lagrangian density of the free σ meson is taken as 1 μ Lσ ¼ ∂μ σ ∂ σ  U ðσ Þ 2 1 1 1 U ðσ Þ ¼ m2σ σ 2 þ g2 σ 3 þ g3 σ 4 3 4 2

ð5:628Þ ð5:629Þ

where mσ is σ meson stationary mass, the second and third terms of U(σ) are the non-linear potential of σ meson, and g2 and g3 are the non-linear term coupling constants of σ meson self-interaction. Substituting the scalar meson Lσ without the non-linear terms by taking g2 ¼ g3 ¼ 0 into Eq. (5.617), the following expression can be obtained:   1 1 * 2 1 ℋσ ¼ π 2σ  π 2σ þ ∇σ þ m2σ σ 2 2 2 2    * 2 1 ¼ π 2σ þ ∇σ þ m2σ σ 2 2

ð5:630Þ

This proves that the Hamiltonian density of scalar mesons without non-linear terms is always positive. We know ∂μ∂μ ¼ ∂μ∂μ according to Eqs. (5.55) and (5.60), and then the following expressions can be obtained from Eqs. (5.628) and (5.629): ∂Lσ ¼ m2σ σ  g2 σ 2  g3 σ 3 ∂σ

ð5:631Þ

5.10

Relativistic Classical Field Theory and Lagrangian Density. . .

# ∂Lσ  ¼ ∂μ ∂μ σ ∂μ

∂ ∂μ σ

453

"

ð5:632Þ

Using the Lagrange equation (5.612), the free σ meson equation is obtained as

 μ ∂μ ∂ þ m2σ σ þ g2 σ 2 þ g3 σ 3 ¼ 0

ð5:633Þ

Substituting Eq. (5.65) into Eq. (5.6) we can get the Klein-Gordon (KG) equation of free particles as

 μ ∂μ ∂ þ m20 ψ ¼ 0

ð5:634Þ

In g2 ¼ g3 ¼ 0 case, Eq. (5.633) is the same as Eq. (5.634). The ω meson is a vector meson whose spin is equal to 1 in charge neutrality (isospin scalar), and the field physical quantity ωμ is 4-vector. The ω mesons are used to describe short-range repulsion between nucleons. The Lagrangian density of the free ω mesons is taken as 2 1 1 1

Lω ¼  Ωμν Ωμν þ m2ω ωμ ωμ þ C3 ωμ ωμ 4 2 4

ð5:635Þ

The anti-symmetric field tensor of the ω mesons is μ

ν

Ωμν ¼ ∂ ων  ∂ ωμ

ð5:636Þ

where mω is ω meson stationary mass and C3 is a non-linear term constant. The following relations can be obtained from Eqs. (5.55) and (5.60): ωμ ωμ ¼ ωμ ωμ ,

Ωμv Ωμv ¼ Ωμv Ωμv

ð5:637Þ

We can get the following expressions from Eqs. (5.635) and (5.636):

 ∂Lω ¼ m2ω ων þ C 3 ωμ ωμ ων ∂ων " # ∂Lω  ¼ ∂μ Ωμν ∂μ

∂ ∂μ ων

ð5:638Þ ð5:639Þ

In the derivation of Eq. (5.639), a number 2 will come out when making the square of Ωμν, and the other 2 can be understood by the following formula:

454

5

Polarization Theory of Relativistic Nuclear Reactions



    ∂Ωμν ∂ð∂1 ω2 ∂2 ω1 Þ  1 2 2 μν Ω ¼ ∂ ω ∂ ω1 ∂ð∂1 ω2 Þ ∂ð∂1 ω2 Þ        ∂ð∂2 ω1 ∂1 ω2 Þ 2 1 1 2 2 1 ∂ ω1 ∂ ω2 ¼ ∂ ω2 ∂ ω1  ∂ ω1 ∂ ω2 ¼2Ω12 þ ∂ ð ∂ 1 ω2 Þ So the equation of the free ω vector meson obtained by using Eq. (5.612) is

 ∂μ Ωμν þ m2ω ων þ C 3 ωμ ωμ ων ¼ 0

ð5:640Þ

If C3 ¼ 0 above formula becomes ∂μ Ωμν þ m2ω ων ¼ 0

ð5:641Þ

This formula is called the Proca equation of vector particles. Taking divergence of Eq. (5.641) we can obtain [9]

 μ ν ∂ν ∂μ Ωμν þ m2ω ων ¼ ∂ν ∂μ ∂ ων  ∂ν ∂μ ∂ ωμ þ m2ω ∂ν ων ¼ 0 Since the sums to μ, ν should be done simultaneously, it can be proved that μ

ν

μ

μ

∂ν ∂μ ∂ ων  ∂ν ∂μ ∂ ωμ ¼ ∂ν ∂μ ∂ ων  ∂μ ∂ν ∂ ων ¼ 0 So for the free vector particles with mass m 6¼ 0, there is the following relation: ∂ ν ων ¼ 0

ð5:642Þ

This formula is called the Lorentz gauge of vector particles. This formula does not necessarily hold when there is an interaction potential. The following equation can be obtained substituting Eq. (5.642) into Eqs. (5.640) and (5.641):



 μ ∂μ ∂ þ m2ω ων þ C 3 ωμ ωμ ων ¼ 0

 μ ∂μ ∂ þ m2ω ων ¼ 0

ð5:643Þ ð5:644Þ

Equation (5.644) is the KG equation satisfied by the vector meson. The ρ meson is a vector meson whose spin is equal to 1 with charge (isospin !μ vector), its field physical quantity is represented by ρ , where the arrow represents a three-dimensional vector in the isospin space, and it can be positively charged (+), uncharged (0), and negatively charged (). The isospin space of the nucleon is two-dimensional, and the ρ meson isospin space is three-dimensional. The inclusion of ρ mesons in the Lagrangian density is to better describe the isospin effect. The Lagrangian density of the ρ mesons is taken as [2, 70–72]

5.10

Relativistic Classical Field Theory and Lagrangian Density. . .

455

1 * *μν 1 * *μ Lρ ¼  R μν  R þ m2ρ ρ μ  ρ 4 2

ð5:645Þ

where *μν

R

 μ  * *ν μ *ν ν *μ ¼ ∂ ρ  ∂ ρ  gρ ρ  ρ

ð5:646Þ

and mρ is ρ meson stationary mass. In order to make the ρ meson coupled with the isospin vector flow, a non-linear ρ meson interaction term is added in Eq. (5.646), and gρ is a ρ meson coupling constant. If this item is not added, the field equation of the ρ meson can be derived by the same way with the derivation of the ω meson equation. Their difference is that the ρ meson is an isospin vector meson. Therefore let [2] *μν

L

μ *ν

ν *μ

¼∂ ρ ∂ ρ

ð5:647Þ

1 * *μν 1 * *μ L0ρ ¼  L μν  L þ m2ρ ρ μ  ρ 4 2

ð5:648Þ

then we can obtain ΔLρ  Lρ  L0ρ h  !μν  μ  *    μ i 1 * * * *ν * * * *ν ¼ gρ ρ μ  ρ ν  L þ ρ  ρ  L μν  gρ ρ μ  ρ ν  ρ  ρ 4 h   *μν  μ  * i 1 * * * *ν ¼ gρ  ρ ν  ρ μ  R þ ρ  ρ  L μν 4 ð5:649Þ  Using the vector formula

*  * * * * a  b  c ¼ a  b  c , from above formula the

*

following expression can be obtained: h    μ  * i !μν 1 * * * *ν þ ρ  ρ  L μν ΔLρ ¼ gρ  ρ ν  ρ μ  R 4

ð5:650Þ

When substituting ΔLρ into the Lagrange equation (5.612), we ignore the contributions of the following items: "

# ∂ðρ  ρ Þ *  L μ 0 ν0 , * ∂ρν *μ

0

So we can get

*ν 0

2 *

ρ ν0  4

0

*μ ν 0

*

∂ðρ μ0  R *

∂ρ ν

Þ

3 5,

2 3 0 * μ ν0 * ∂ðρ  R Þ 0 * μ  5 ρ ν0  4 * ∂ ∂μ ρ ν

456

5

Polarization Theory of Relativistic Nuclear Reactions

  !μν 1 * ¼  gρ ρ μ  R 4

ð5:651Þ

 μ  ∂ΔLρ 1 *ν   ¼ gρ * ρ  ρ * 2 ∂ ∂μ ρ ν

ð5:652Þ

∂ΔLρ *

∂ρν

Referring to Eq. (5.640) the field equation of the ρ meson can be written as *μν

∂μ L

*ν

þ m2ρ ρ ¼ gρ

h    μ i !μν 1 1 * * *ν þ ∂μ ρ  ρ ρμ  R 2 4

ð5:653Þ

If the non-linear term is not included in Eq. (5.646), i.e., let gρ ¼ 0 in Eq. (5.653), *ν and using the Lorentz gauge ∂ν ρ ¼ 0 of the vector particle, Eq. (5.653) can be rewritten as   ν ! μ ∂μ ∂ þ m2ρ ρ ¼ 0

ð5:654Þ

This is the KG equation satisfied by ρ meson. π meson is a pseudo-scalar meson whose spin is equal to zero with charge (isospin * vector); the field physical quantity is represented by symbol π , where the arrow indicates that π meson is a three-dimensional vector in the isospin space. Similarly, in order to consider the coupling problem with the isospin vector flow, the following form of Lagrangian density is introduced for π meson [2, 70, 71, 74]: 1 ! !μ 1 * * Lπ ¼ P μ  P  m2π π  π 2 2 *μ

μ!

!μ

!

ð5:655Þ

P ¼ ∂ π  gρ ρ  π

ð5:656Þ

1 * μ* 1 * * L0π ¼ ∂μ π  ∂ π  m2π π  π 2 2

ð5:657Þ

Let

then we can obtain ΔLπ  Lπ  L0π h   μ     μ i 1 * * * * * * * * * μ* ¼  gρ ρ μ  π  ∂ π þ ρ  π  ∂ μ π  g ρ ρ μ  π  ρ  π 2 h   *μ  μ  i 1 * * * * * ¼  gρ  π  ρ μ  P þ ρ  π  ∂ μ π 2 h    μ  i *μ 1 * * * * * ¼  gρ π  ρ μ  P þ ρ  π  ∂μ π 2 ð5:658Þ

5.10

Relativistic Classical Field Theory and Lagrangian Density. . .

457

When substituting ΔLπ into the Lagrange equation (5.612), we ignore the contributions of the following items: 3 2  0 !μ * ∂ ρ μ0  P 7 * 6 π 6  7 4 ∂ ∂ * 5 μπ

2 *μ *3 ∂ ρ π 4 5  ∂μ * π, * ∂π So we can get

  *μ ∂ΔLπ 1 * ¼ g ρ  P ρ μ * 2 ∂π  μ  1 ∂ΔLπ * *

* ¼  gρ ρ  π 2 ∂ ∂μ π

ð5:659Þ ð5:660Þ

Refer to Eqs. (5.634) and (5.653) and noting the difference between the m2 term signs in Eq. (5.648) and Eq. (5.657), the field equation of the π meson can be written as

h   μ i * *μ  1 1 * * * μ ρ μ  P þ ∂μ ρ  π ∂μ ∂ þ m2π π ¼ gρ 2 2

ð5:661Þ

If the non-linear term is not included in Eq. (5.656), i.e., let gρ ¼ 0 in Eq. (5.661), and then Eq. (5.661) can be rewritten as

* μ ∂μ ∂ þ m2π π ¼ 0

ð5:662Þ

This is the KG equation satisfied by π meson. The electromagnetic field is a vector field with a spin equalto 1 and a mass equal * * to zero, and it is expressed as Aμ, A ¼ (A0, A1, A2, A3) ¼ ϕ, A , A is a threedimensional vector potential and ϕ represents a scalar potential. The Lagrangian density of the free electromagnetic field is taken as 1 LA ¼  F μν F μν 4 μ

ν

F μν ¼ ∂ Aν  ∂ Aμ

ð5:663Þ ð5:664Þ

From the Lagrange equation (5.612) we can obtain ∂μ F μν ¼ 0

ð5:665Þ

The free electromagnetic field satisfies the following Lorentz gauge given by Eq. (5.642):

458

5

Polarization Theory of Relativistic Nuclear Reactions

∂ν Aν ¼ 0

ð5:666Þ

So Eq. (5.665) can be rewritten as μ

∂μ ∂ Aν ¼ 0

ð5:667Þ

The Lagrangian densities of the interaction between the nucleon and σ, ω, ρ mesons can be taken as Lint, σ ¼ gσ ψ i σψ i

ð5:668Þ

fω ψ σ μν Ωμν ψ i 4M i fρ * * * * ¼ gρ ψ i γ μ τ  ρ μ ψ i  ψ i σ μν τ  L μν ψ i 4M

Lint, ω ¼ gω ψ i γ μ ωμ ψ i  Lint, ρ

ð5:669Þ ð5:670Þ

*

where σ μν has been given by Eq. (5.538) and Ωμν and L μν have been given by * Eq. (5.636) and Eq. (5.647), respectively. τ is a nucleon Pauli isospin matrix. The first items of Eq. (5.669) and Eq. (5.670) represent vector coupling, and the second items represent tensor coupling and fω and fρ are tensor coupling constants. For π meson there is *

*

Lint, π ¼ igπ ψ i γ 5 τ  π ψ i 

* fπ * ψ i γ 5 γ μ τ  ∂μ π ψ i mπ

ð5:671Þ

In the above formula the first term represents the pseudo-scalar (ps) coupling, and the second term represents the axis-vector (pv) coupling. For the electromagnetic field there is Lint, A ¼ eψ i γ μ

1  τ3 Aμ ψ i 2

ð5:672Þ

In the above formula τ3 ¼ 1 represents the neutrons, and τ3 ¼ 1 represents the protons. And e is a proton charge.

5.10.3

Relativistic Mean Field Equations

Substituting the interaction Lagrangian density introduced earlier into the Lagrange equation (5.612) and merging the obtained result into the corresponding free field equation, the nucleon Dirac equation (5.627) becomes μ



γ i∂μ  V μ  ðM þ SÞ  σ μν T μν ψ i ¼ 0

ð5:673Þ

5.10

Relativistic Classical Field Theory and Lagrangian Density. . .

459

where *

*

SðxÞ ¼ gσ σ ðxÞ þ igπ γ 5 τ  π ðxÞ

ð5:674Þ

The first item on the right end of the above formula is the scalar field, and the second item is the pseudo-scalar field related to the isospin; x represents the position of the space-time vertex where the interaction occurs. *

*

V μ ð x Þ ¼ g ω ωμ ð x Þ þ g ρ τ  ρ μ ð x Þ þ

f π 5* 1  τ3 * γ τ  ∂ μ π ð xÞ þ e Aμ ðxÞ mπ 2

ð5:675Þ

The first item at the right end of the above formula is the vector term, the second item is the vector field related to the isospin, the third item is the axis-vector field related to the isospin, and the fourth item is the Coulomb potential. T μν ðxÞ ¼

fρ * fω G ð xÞ þ τ  Lμν ðxÞ 4M μν 4M

ð5:676Þ

The first item on the right end of the above formula is the tensor field, and the second item is the tensor field related to the isospin. The σ meson equation (5.633) becomes

 μ ∂μ ∂ þ m2σ σ ¼ gσ ρS  g2 σ 2  g3 σ 3

ð5:677Þ

where A X

ρ S ð xÞ ¼

ψ i ð xÞ ψ i ð xÞ

ð5:678Þ

i¼1

ρS(x) is a nucleon scalar density. The ω meson equation (5.640) becomes ∂μ Ωμν þ m2ω ων ¼ gω jν 

fω ν t  C 3 ðmω ωμ Þων 2M

ð5:679Þ

where j ν ð xÞ ¼

A X

ψ i ðxÞγ ν ψ i ðxÞ

ð5:680Þ

∂μ ðψ i ðxÞσ μν ψ i ðxÞÞ

ð5:681Þ

i¼1

t ν ð xÞ ¼

A X i¼1

460

5

Polarization Theory of Relativistic Nuclear Reactions



  * Here jν is the nucleon flow density, and there is j ¼ j0 , j1 , j2 , j3 ¼ ρB , j , where A X

ρB ð xÞ ¼

ψþ i ðxÞ ψ i ðxÞ

ð5:682Þ

i¼1

ρB is the usual nucleon density, which is different from the nucleon scalar density ρS given by Eq. (5.678). tν is called nucleon tensor flow density. The ρ meson equation (5.653) becomes *μν

∂μ L

h*ν    μ i f *ν *μν 1 1 ! *ν * *ν ρ þ m2ρ ρ ¼ gρ j τ þ ð5:683Þ þ ∂μ ρ  ρ  t ρμ  R 2 2M τ 4

where !ν j τ ð xÞ

¼

A X

*

ψ i ðxÞγ ν τ ψ i ðxÞ

ð5:684Þ

 * ∂μ ψ i ðxÞσ μν τ ψ i ðxÞ

ð5:685Þ

i¼1 !ν t τ ð xÞ

¼

A X i¼1

*ν

*ν

Here j τ is a nucleon isospin vector flow density vector, and t τ is a nucleon isospin tensor flow density vector. In the case of taking fρ ¼ 0 and not considering π mesons, if the non-linear term related to gρ is not added in Eq. (5.646), the second and third items related to gρ will not appear at the right end of Eq. (5.683). If the initial state !ν !ν isospin vector flow j τ is 0, at this time, a ρ meson field ρ ¼ 0 is obtained. Then in the nucleon equations (5.673) and (5.675), the gρ term also has no contribution, so it cannot cause the nucleons to produce an isospin vector flow, which is equivalent to that the ρ meson field does not work. This is why the non-linear term related to gρ is added in Eq. (5.646). The π meson equation (5.661) becomes

h    μ i * *μ f * 1 1 ! * ! * μ ∂μ ∂ þ m2π π ¼ igπ ρ ps þ π ρ pv þ gρ ρ μ  P þ ∂μ ρ  π mπ 2 2 ð5:686Þ

where !

ρ ps ðxÞ ¼

A X i¼1

*

ψ i ðxÞγ 5 τ ψ i ðxÞ

ð5:687Þ

5.11

Relativistic Green Function Theory at Zero Temperature

!

ρ pv ðxÞ ¼

A X

461

 * ∂μ ψ i ðxÞγ 5 γ μ τ ψ i ðxÞ

ð5:688Þ

i¼1 *

*

Here ρ ps is a nucleon pseudo-scalar density vector, and ρ pv is a nucleon axis-vector density vector. The gρ terms in Eq. (5.686) are generated by non-linear terms, and it can be seen that π mesons and ρ mesons are coupled each other. The equation of electromagnetic field (5.665) becomes ∂μ F μν ¼ ejpv

ð5:689Þ

where jνp ðxÞ ¼

A X i¼1

ψ i ðxÞγ ν

1  τ3 ψ i ð xÞ 2

ð5:690Þ

here jνp is a proton flow density. In general, we only study the static properties of the nucleus. It can be assumed that the meson field and the photon field are static classical fields, and the nucleons perform independent movements in the classical field, then the relativistic mean field theory can be formed. The A nucleons in the nucleus are all the positive nucleons above the Dirac sea, so it can be said that the above theory was established without considering the negative energy nucleons in the Dirac sea. If we want to consider the Dirac sea effect, i.e., considering vacuum polarization, the theory must be renormalized. When considering vacuum polarization, Lagrangian density will change somewhat [74–77]. However, the introduced theory previously is an effective theory containing multiple parameters. When determining the theoretical parameters by meeting the experimental data, most of the vacuum polarization effects can be considered. Therefore, the relativistic mean field theory by using the approximation without considering the Dirac sea effect still has considerable practical value.

5.11

Relativistic Green Function Theory at Zero Temperature

Let jΨ 0i represent the ground state of a non-interacting fermions system at finite density. The ground state jΨ 0i contains positive energy baryon levels filled to some Fermi wave number kF and no antiparticles and free mesons.

462

5.11.1

5

Polarization Theory of Relativistic Nuclear Reactions

The Propagator of Neutral Scalar Bosons with Spinless [2, 67]

  * The Klein-Gorden equation satisfied by the wave function σ x , t of the free scalar meson σ can be obtained by substituting Eq. (5.65) into Eq. (5.6) or by Eq. (5.634) as follows:

 *  μ ∂μ ∂ þ m2σ σ x , t ¼ 0

ð5:691Þ

In g2 ¼ g3 ¼ 0 case, from Eq. (5.628) and (5.629), we can obtain 1 1 μ Lσ ¼ ∂μ σ ∂ σ  m2σ σ 2 2 2

ð5:692Þ

where σ is called the generalized coordinate. The generalized momentum can be obtained by Eq. (5.616). π

∂Lσ ¼ σ_ ∂σ_

ð5:693Þ

Using Eq. (5.617) and Eq. (5.692), we can write 1 1 1 μ ℋσ ¼ π 2  Lσ ¼ π 2  ∂μ σ ∂ σ þ m2σ σ 2 ¼ 2 2 2

h *   *  i ∇σ  ∇σ þ π 2 þ m2σ σ 2 ð5:694Þ

The following canonical motion equations can be obtained according to the Eqs. (5.619) and (5.622): ∂σ ¼ π, ∂t

∂π * 2 ¼ ∇ σ  m2σ σ ∂t

ð5:695Þ

In t ¼ t0 case, introduce the following equal-time commutation relation: h    0 i   * * * *0 σ x , t , π x , t ¼ iδ x  x

ð5:696Þ

This formula is similar to Eq. (4.175), which is the usual boson quantification rule. To facilitate the discussion, we assume that the field is limited to a large cubic box of volume V ¼ L3 and impose the following periodic boundary conditions: ki ¼

2π ni , L

i ¼ 1, 2, 3;

ni ¼ 0, 1, 2,⋯

ð5:697Þ

5.11

Relativistic Green Function Theory at Zero Temperature

463

The simplest eigenfunction of Eq. (5.691) is the plane wave. We can write σ and π as the following superposed forms of the plane waves: 1 X i*k *x q* e , σ ¼ pffiffiffiffi V * k k

1 X i*k *x π ¼ pffiffiffiffi p* e V * k

ð5:698Þ

k

Since σ and π are the Hermitian operators, so there are q*k ¼ qþ *,

p*k ¼ pþ *

k

ð5:699Þ

k

where q*k and p*k are all time functions. We can rewrite Eq. (5.698) as 1 q*k ¼ pffiffiffiffi V

Z

*

dx σ e

**

i k  x

,

1 p*k ¼ pffiffiffiffi V

Z

*

**

d x π ei k  x

ð5:700Þ

Then we can obtain the following commutation relation from Eqs. (5.696) and (5.700): h

i q*k , p*k 0 ¼ iδ*k *k 0

ð5:701Þ

Substituting Eq. (5.698) into Eq. (5.695), then comparing it with Eq. (5.698), and noting that σ and π are all real numbers, thus we can obtain dq*k ¼ pþ *, dt k

dp*k ¼ ω2k qþ * dt k

ω2k  m2σ þ k2

ð5:702Þ ð5:703Þ

Equation (5.703) means that the σ meson is on-shell. It can also be seen that the * movements of different degrees of freedom marked with different k can be completely separated. þ From Eq. (5.702) we can see that q*k , qþ * , p* , p* are all the superpositions of k k

k

eiωk t and eiωk t . We will quantify q*k in the following way:

  1 q*k ¼ pffiffiffiffiffiffiffiffi c*k eiωk t þ cþ* eiωk t k 2ωk

ð5:704Þ

where cþ* and c*k are the σ meson creation operator and annihilation operator, k

respectively. The following expression can be obtained by Eqs. (5.702), (5.699), and (5.704):

464

5

Polarization Theory of Relativistic Nuclear Reactions

rffiffiffiffiffiffi  ωk iωk t c*k eiωk t  cþ p*k ¼ i *e 2 k

ð5:705Þ

Further we can obtain the following relations from Eqs. (5.704) and (5.705): c*k ¼

rffiffiffiffiffiffi  ωk 1 q*k þ i pffiffiffiffiffiffiffiffi p*k eiωk t , 2 2ωk

cþ * ¼ k

rffiffiffiffiffiffi  ωk 1 q*k  i pffiffiffiffiffiffiffiffi p*k eiωk t 2 2ωk ð5:706Þ

Substituting Eq. (5.706) into Eq. (5.701), we can obtain the following commutation relation: h

i c*k , cþ ¼ δ*k *k 0 *

ð5:707Þ

k

The remaining pairs of operators are all commutated. The selection of the front edge coefficient of the right end of Eq. (5.704) ensures that the commutative relation Eq. (5.707) can be obtained. Since σ is a real number, from Eqs. (5.698) and (5.704) we can obtain   ** 1 X 1 pffiffiffiffiffiffiffiffi c* eiωk t þ cþ* eiωk t ei k  x σ ðxÞ ¼ pffiffiffiffi k V * 2ωk k k   μ 1 X 1 ikμ xμ pffiffiffiffiffiffiffiffi c* eikμ x þ cþ ¼ pffiffiffiffi *e k V * 2ωk k

ð5:708Þ

k

1=2 * * . where kμ xμ ¼ k 0 t  k  x . In the on-shell case there is k 0 ¼ ωk  k2 þ m2σ Note that |Ψ 0i is the vacuum state of the σ meson, and the zero order Green function of the neutral free scalar meson σ is defined as iΔ0 ðx, x0 Þ ¼ hΨ 0 jTfσ ðxÞσ ðx0 Þg jΨ 0 i

ð5:709Þ

where T{⋯} represents the time ordered product of {⋯}. So there is iΔ0 ðx, x0 Þ ¼ hΨ 0 jσ ðxÞσ ðx0 ÞjΨ 0 iθðt  t 0 Þ þ hΨ 0 jσ ðx0 Þσ ðxÞjΨ 0 iθðt 0  t Þ !

ð5:710Þ

where θ(t  t0) is the Heaviside step function. Note that for all k there are c!k jΨ 0 i ¼ 0 and hΨ 0 jcþ ! ¼ 0 ; therefore we can obtain the following expression k

using Eq. (5.708):

5.11

Relativistic Green Function Theory at Zero Temperature

hΨ 0 jσ ðxÞσ ðx0 ÞjΨ 0 i ¼

465

  μ 1 1X ik μ xμ pffiffiffiffiffiffiffiffiffiffiffiffi hΨ 0 j c* eikμ x þ cþ *e k V * * 2 ωk ωk 0 k k k0

  0 0μ 1 X 1 ikμ ðxx0 Þμ ik 0μ x0 μ i ¼ e c*k 0 eikμ x þ cþ jΨ * e 0 V * 2ωk k0

ð5:711Þ

k

The following expression can also be obtained in the same way: hΨ 0 jσ ðx0 Þσ ðxÞjΨ 0 i ¼

1 X 1 ikμ ðxx0 Þμ e V * 2ωk

ð5:712Þ

k

From Eq. (5.697) we know Δki ¼

2π Δn , L i

Δni ¼ 1;

i ¼ 1, 2, 3

ð5:713Þ

So there is 1 1 L3 1 ¼ Δn Δn Δn ¼ Σ Δk1 Δk 2 Δk 3 Σ Σ 1 2 3 V ð2πÞ3 *k V *k V *k

ð5:714Þ

In the limit of an infinitely large volume, that is, V ! 1 and Δki ! 0, there is [9] 1 1 Σ! V *k ð2πÞ3

Z

*

dk

ð5:715Þ

Then the following expression can be obtained substituting Eqs. (5.711), (5.712), and (5.715) into (5.710): iΔ0 ðx, x0 Þ ¼ 1 ¼ ð2πÞ3

Z

h i 0 μ 0 μ 1X 1 eikμ ðxx Þ θðt  t 0 Þ þ eikμ ðxx Þ θðt 0  t Þ V * 2ωk k



* *0 * i* k xx

dke



h i 0 0 1 eiωk ðtt Þ θðt  t 0 Þ þ eiωk ðtt Þ θðt 0  t Þ 2ωk

ð5:716Þ

And because θðt  t 0 Þ þ θðt 0  t Þ ¼ 1 thus we can define temporarily the following quantity using Eq. (5.703):

ð5:717Þ

466

5

Polarization Theory of Relativistic Nuclear Reactions

Fig. 5.3 Integral poles

I

1 2πi

1 ¼ 2πi

Z

Z

0

eik0 ðtt Þ 1 dk ¼ kμ kμ  m2σ þ iε 0 2πi

Z

0

eik0 ðtt Þ dk0 2 k0  ω2k þ iε

0

eik0 ðtt Þ ½θðt  t 0 Þ þ θðt 0  t Þ dk 0 ðk0  ωk þ iεÞ ðk 0 þ ωk  iεÞ

ð5:718Þ

where k0 is equivalent to off-shell energy and ε is a positive infinitesimal. In the integral given by Eq. (5.718), there is a pole in both upper and lower half planes (see Fig. 5.3). The integrals in Eq. (5.718) for θ(t  t0) and θ(t0  t) terms are doing using the residual theorem in the lower half plane and the upper half plane, respectively, and one obtains I¼

h i 0 0 1 eiωk ðtt Þ θðt  t 0 Þ þ eiωk ðtt Þ θðt 0  t Þ 2ωk

ð5:719Þ

Equation (5.716) can be rewritten as follows using Eqs. (5.718) and (5.719): Z

0

iΔ ðx, x Þ ¼ i 0

0 μ

eikμ ðxx Þ d4 k μ ð2πÞ4 k μ k  m2σ þ iε

ð5:720Þ

In the above formula, only the integral on k0 has an extreme value. From the Fourier transform relation, it can be seen that Δ 0 ðk Þ ¼

1 kμ k  m2σ þ iε μ

ð5:721Þ

This is the propagator of a neutral scalar boson with spinless in the momentum space.

5.11.2 The Propagator of Fermions with Spin 12 [2, 6–8] Equation (5.66) or (5.627) has given the Dirac equation of the free spin 12 particle as

 iγ μ ∂μ  M ψ ¼ 0

ð5:722Þ

5.11

Relativistic Green Function Theory at Zero Temperature

467

1 particle stationary mass. The particles with semi-odd number 2 spin are called fermions, and the particles with integer spin are called bosons. Baryons are the heavier strong interaction fermions. The nucleon belongs to baryon. All mesons are bosons. The hadron is the collective name for all the particles involved in the strong interaction, such as baryon and meson. Eq. (5.624) has given the corresponding Lagrangian density as where M is the spin

 L ¼ ψ iγ μ ∂μ  M ψ

ð5:723Þ

Since the spinor field ψ is a complex, we may consider ψ(x) and ψ ðxÞ as independent variables. Taking ψ as the generalized coordinates, then the generalized momentum can be obtained by Eq. (5.616) as Π¼

∂L ¼ iψ þ ∂ψ_

ð5:724Þ

Utilizing Eq. (5.617) the Hamiltonian density can be obtained as   ∂ ! * ℋ ¼ ψ þ iα  ∇ þ β M ψ ¼ ψ þ i ψ ∂t

ð5:725Þ

In t ¼ t0 case, we introduce the following equal-time anti-commutation relation: n

   0 o   * * * *0 ψα x , t , ψþ ¼ δαβ δ x  x β x , t

ð5:726Þ

and the other pairs of spinor fields are all anti-commutated each other. Both the positive- and negative-energy solutions of the free nucleon Dirac equation can be written as follows by Eqs. (5.80), (5.86), (5.93), and (5.94): μ

ψ ðþÞ ðxÞ ¼ eikμ x uðk Þ 1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffi0 χ E þ M@ * A uð k Þ ¼ σ^  k 2M χ EþM μ

ψ ðÞ ðxÞ ¼ eikμ x vðk Þ 1 * rffiffiffiffiffiffiffiffiffiffiffiffiffiffi0 E þ M @ σ^  k χ A vð k Þ ¼ EþM 2M χ

ð5:727Þ ð5:728Þ ð5:729Þ ð5:730Þ

    1 0 where χ 1 ¼ , χ2 ¼ . u(k) and v(k) are all four-component vectors and 0 1 satisfy the orthonormalization relations given by Eq. (5.91)

468

5

Polarization Theory of Relativistic Nuclear Reactions

uðsÞ ðk Þ uðs Þ ðk Þ ¼ δss0 , vðsÞ ðkÞvðs Þ ðkÞ ¼ δss0 , 0 0 vðsÞ ðk Þ uðs Þ ðk Þ ¼ uðsÞ ðkÞvðs Þ ðkÞ ¼ 0 0

0

ð5:731Þ

where s and s0 stand for nucleon spins. Substituting Eq. (5.727) into Eq.(5.722) there is

h i



! ! iγ 0 ðiE Þ þ iγ j ikj  M uðkÞ ¼ γ 0 E  γ  k  M uðkÞ ¼ 0

Then from the above formula we can get

μ  γ kμ  M uðkÞ ¼ 0,

!

!

γ μ kμ  γ0 E  γ  k

ð5:732Þ

Note that k μ is a newly introduced symbol. Substituting Eq. (5.729) into Eq. (5.722) there is

 γ μ k μ þ M vð k Þ ¼ 0

ð5:733Þ

The following relation can be obtained by Eqs. (5.57) and (5.59): ðγ μ Þ þ ¼ γ 0 γ μ γ 0

ð5:734Þ

Taking complex conjugate to Eq. (5.732) and using Eq. (5.734), we can obtain

 uþ ðkÞ γ 0 γ μ γ 0 kμ  γ 0 Mγ 0 ¼ 0 That is,

 uð k Þ γ μ k μ  M ¼ 0

ð5:735Þ

 vð k Þ γ μ k μ þ M ¼ 0

ð5:736Þ

In the same way we can get

In the particle stationary system, there is k μ ¼ ðM, 0Þ ; thus Eqs. (5.732) and (5.733) are converted into [8]

 γ 0  1 uðM, 0Þ ¼ 0,



 γ 0 þ 1 vðM, 0Þ ¼ 0

ð5:737Þ

We can directly get the following relations from Eqs. (5.85), (5.90), and (5.92):

5.11

Relativistic Green Function Theory at Zero Temperature

469

0 1 1 B C B0C B C uð1Þ ðM, 0Þ ¼ B C, B0C @ A

0 1 0 B C B1C B C uð2Þ ðM, 0Þ ¼ B C, B0C @ A

0 0 1 0 B C B0C B C ð1Þ v ðM, 0Þ ¼ B C, B1C @ A 0

0 0 1 0 B C B0C B C ð2Þ v ðM, 0Þ ¼ B C B0C @ A 1

ð5:738Þ

Substituting Eq. (5.738) into Eq. (5.737), we can confirm that Eq. (5.738) is indeed the solution of Eq. (5.737). And let ð1Þ

φ ðM, 0Þ ¼

1 0

! ð2Þ

, φ ðM, 0Þ ¼

1

! , !

ð5:739Þ

1 0 rffiffiffiffiffiffiffiffiffiffiffiffiffiffi φðsÞ ðM, 0Þ E þ MB C * uðsÞ ðkÞ ¼ A, @ 2M σ^  k ðsÞ φ ðM, 0Þ EþM 0 1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffi σ^  * k ðsÞ ϕ ðM, 0Þ C E þ MB vðsÞ ðkÞ ¼ @E þ M A 2M ðsÞ ϕ ðM, 0Þ

ð5:740Þ

ϕð1Þ ðM, 0Þ ¼

1 0

!

0

, ϕð2Þ ðM, 0Þ ¼

0 1

in this case, Eqs. (5.728) and (5.730) can be rewritten as

We can also obtain γ μ kμ þ M ¼ γ 0 E  γi ki þ M ¼

*

EþM

^ σk

σ^  k

ðE  M Þ

*

! ð5:741Þ

0 1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffi φðsÞ ðM, 0Þ γ kμ þ M E þ MB C uðsÞ ðM, 0Þ ¼ @ σ^  * A ¼ uðsÞ ðk Þ ð5:742Þ k ðsÞ 2M ½2M ðE þ M Þ 1=2 φ ðM, 0Þ EþM μ

470

5

Polarization Theory of Relativistic Nuclear Reactions

0 γ μ k μ þ M ¼ γ 0 E þ γ i ki þ M ¼

1 * σ^  k A EþM 1

@ ðE  M Þ *

^ σk

ð5:743Þ

0 rffiffiffiffiffiffiffiffiffiffiffiffiffiffi σ^  * k ðsÞ γ kμ þ M E þ MB ϕ ðM, 0Þ C¼ vðsÞ ðk Þ ð5:744Þ ðsÞ v ð M, 0 Þ ¼ @ A E þ M 1=2 2M ½2M ðE þ M Þ ϕðsÞ ðM, 0Þ μ

We have known uðsÞ ðk Þ ¼ uðsÞþ ðkÞγ 0 , vðsÞ ðk Þ ¼ vðsÞþ ðkÞγ 0; by using Eqs. (5.57) and (5.59), we can obtain the following relations from Eqs. (5.742) and (5.744): γμ kμ þ M

uðsÞ ðk Þ ¼ uðsÞ ðM, 0Þ vðsÞ ðk Þ ¼ vðsÞ ðM, 0Þ

ð5:745Þ

½2M ðE þ M Þ 1=2 γ μ k μ þ M

ð5:746Þ

½2M ðE þ M Þ 1=2

First the following formula can be found: 0 1 20 1 3 1 0 B1C 6B 0 C 7 X B C 6B C 7 uðsÞ ðM, 0ÞuðsÞ ðM, 0Þ ¼ 6B Cð1 0 0 0Þ þ B Cð0 1 0 0Þ7 @ A @ A 4 5 0 0 s¼1,2 0 1 1 0 0 0 B0 1 0 0C B C ¼B C¼ @0 0 0 0A 0

0 ^I ^ 0 ^ 0 ^ 0

ð5:747Þ ! ¼

1 þ γ0 2

0 0 0 0 So there is Λ þ ðk Þ 

X

uðsÞ ðkÞ uðsÞ ðkÞ

s¼1,2

μ  1 ¼ γ kμ þ M 2M ðE þ M Þ

"

X

ðsÞ

ðsÞ

u ðM, 0Þ u ðM, 0Þ

#

γ μ kμ þ M



s¼1, 2



 1 þ γ0 μ  1 ¼ γ μ kμ þ M γ kμ þ M 2 2M ðE þ M Þ ð5:748Þ Then the following relation can be obtained by using Eqs. (5.728) and (5.731):

5.11

Relativistic Green Function Theory at Zero Temperature *

uðsÞ ðk ÞuðsÞ ðkÞ ¼

σ^  k ðsÞþ EþM χ χ ðsÞþ EþM 2M

2

!

 *2 3 σ^  k EþM6 1 7 ¼ ¼ 41  25 2M 2M ð E þ MÞ ðE þ M Þ

^I

471

^ 0

!

0

1

C * A σ^  k ðsÞ χ EþM   *2  2 ¼1 ðE þ M Þ  σ^  k ^ 0  ^I

B @

χ ðsÞ

Further we can get 

*2

σ^  k

¼ ðE þ M Þ2  2M ðE þ M Þ ¼ ðE þ M ÞðE  M Þ

ð5:749Þ

And we can obtain the following formula using Eqs. (5.741) and (5.749):

 1 þ γ0 μ  γ μ kμ þ M γ kμ þ M 2 0 1 1 !0 ! ! ^I ^0 EþM ^ σk EþM ^ σk A @ A ¼@ ! ! ^0 ^0 σ^  k ðE  M Þ σ^  k  ðE  M Þ 0 1 * *! EþM ^ σk E þ M ^ σ k @ A ¼ * ^0 ^0 σ^  k ðE  M Þ 1 0 0 1 * ! ðE þ M Þ2 ðE þ M Þ^ σk EþM ^ σk C B A ¼@  *2 A ¼ ðE þ M Þ@ * ! ðE þ M Þ^ σk  σ^  k σ^  k ðE  M Þ

μ  ¼ ðE þ M Þ γ kμ þ M ð5:750Þ

Then the following result can be obtained by Eq. (5.748): X

Λþ ðkÞ 

uðsÞ ðkÞ uðsÞ ðk Þ ¼

γμkμ þ M 2M

ð5:751Þ

vðsÞ ðkÞ vðsÞ ðk Þ ¼

γ μ k μ þ M 2M

ð5:752Þ

s¼1,2

In the same way we can prove Λ ðkÞ  

X

s¼1,2

So from the above two formulas, the following relation can be seen

472

5

Polarization Theory of Relativistic Nuclear Reactions

Λþ ðk Þ þ Λ ðk Þ ¼ I

ð5:753Þ

Therefore we call Λ+(k) and Λ(k) as the projection operators of the positive- and negative-energy states, respectively. We still use the periodic boundary conditions given by Eq. (5.697), so the spinor fields appeared in Eqs. (5.722), (5.723), and (5.726) can be expanded, respectively [2, 6, 7]: rffiffiffiffiffih i μ μ 1 X M ðsÞ ðkÞ eikμ x , a*k s uðsÞ ðk Þ eikμ x þ bþ ψ x , t ¼ pffiffiffiffi * v E ks V * ks r ffiffiffiffi ffi   h i μ μ 1 X M þ ðsÞ * ψ x , t ¼ pffiffiffiffi a* u ðk Þ eikμ x þ b*k s vðsÞ ðkÞ eikμ x E ks V * 



*

ð5:754Þ

ks

In above formulas, a+ and a, b+ and b represent the creation and annihilation operators of nucleon and anti-nucleon, respectively. Next we will prove that the expansion coefficients selected in Eq. (5.754) will make the introduced creation and annihilation operators satisfy the following anti-commutation relations: n

a*k s , aþ * 0

k s0

o

* *  ¼ δss0 δ k  k 0 ,

n

b*k s , bþ * 0

k s0

o

* *  ¼ δss0 δ k  k 0

ð5:755Þ

The other pairs of these operators are all anti-commutated each other. First we assume that Eq. (5.755) and other anti-commutated relations are valid, and then substituting Eq. (5.754) into Eq. (5.726) and using Eqs. (5.751), (5.752), and (5.715), as well as noting t0 ¼ t, thus we can obtain [6, 7] n

   0 o * *  1X M X * * p ffiffiffiffiffiffiffiffiffiffiffi ψα x , t , ψþ x , t ¼ δ δ k  k0 0 ss β V ** Ek Ek0 ss0 k k0 



 * * *0 * * *0 ð s0 Þ ðs0 Þ uðαsÞ ðk Þuβ ðk 0 Þγ 0 ei k  x  x þ vðαsÞ ðkÞvβ ðk 0 Þγ 0 ei k  x  x

 nh i o   1 X 1 i*k  *x *x 0 * * * * γ 0 k 0  γ  k þ M  γ 0 k 0  γ  k þ M γ 0 ¼ e V * 2E αβ k Z

* *0  * *  * i* 1 k xx δ d k ¼ e ¼ δ δ k  k0 αβ αβ ð2πÞ3

ð5:756Þ    0  * * Note that in the on-shell case k0 ¼ E. This proves that ψ α x , t and ψ x , t expressed by Eq. (5.754) may satisfy Eq. (5.726) under the condition that Eq. (5.755) is satisfied. Since the nucleon is fermion, the zero order Green function of the nucleon is defined as

5.11

Relativistic Green Function Theory at Zero Temperature

473

   iG0αβ ðx  x0 Þ ¼ hΨ 0 T ψ α ðxÞ ψ β ðx0 Þ jΨ 0   ¼ hΨ 0 ψ α ðxÞ ψ β ðx0 ÞjΨ 0 θðt  t 0 Þ  hΨ 0 jψ β ðx0 Þ ψ α ðxÞjΨ 0 iθðt 0  t Þ

ð5:757Þ

Note the following relations: *

aþ * jΨ 0 i ¼ 0, ks *    a*k s jΨ 0 i ¼ 0,  k  > k F

b*k s jΨ 0 i ¼ 0, for all k ;

*    k  < kF ;

ð5:758Þ

Also note that only the positive energy nucleon can be created above the Fermi surface; using Eqs. (5.754), (5.758), and (5.715), we can get firstly E   1 X M XD  pffiffiffiffiffiffiffiffiffiffiffi Ψ 0 a*k s aþ hΨ 0 ψ α ðxÞ ψ β ðx0 ÞjΨ 0 ¼ * jΨ 0 V ** E k E k0 ss0 k 0 s0 k k0 h i * * *0 *0  0 0 ð5:759Þ uðsÞ ðkÞuðs Þ ðk0 Þ ei k  x  k  x eiðEk tEk0 t Þ αβ Z *i ! ! !0    h * 1

0 1   μ d k ¼ k þ M 1  θ k  γ  k  ei k  x  x eiEðtt Þ μ F αβ 2E ð2πÞ3 There are the relations Z

1

dω eiωt 1 2π ω iε Z 0 0 dω eiðωþEÞðtt Þ eiEðtt Þ θðt  t 0 Þ ¼ i ω þ iε 2π

θð t Þ ¼ i

ð5:760Þ ð5:761Þ

Let k0 ¼ ω þ E, ω ¼ k0  E, dω ¼ dk0, E is the on-shell energy, and k0 is equivalent to the off-shell energy. Then Eq. (5.761) may be rewritten as 0

eiEðtt Þ θðt  t 0 Þ ¼ i

Z

0

dk0 eik0 ðtt Þ 2π k0  E þ iε

ð5:762Þ

So the following expression can be obtained utilizing Eqs. (5.759) and (5.762):   hΨ 0 ψ α ðxÞ ψ β ðx0 Þ jΨ 0 θðt  t 0 Þ Z ¼i

*    1  θ k  k F  0 μ d k 1 μ eikμ ðxx Þ γ k μ þ M αβ 4 2E k  E þ iε 0 ð2πÞ 4

ð5:763Þ

And we can write the following expression by using Eqs. (5.754) and (5.758):

474

5

Polarization Theory of Relativistic Nuclear Reactions

  hΨ 0 ψ β ðx0 Þψ α ðxÞjΨ 0 0 0μ 1X M X ðs0 Þ 0 ik 0μ x0 μ ðs0 Þ pffiffiffiffiffiffiffiffiffiffiffi hΨ 0 j aþ þ b*0 0 vβ ðk0 Þeikμ x *0 uβ ðk Þe ¼ V k s 0 0 E E s k k k ss0 **0

kk

! ð5:764Þ

  μ μ a*k s uαðsÞ ðkÞeikμ x þ b*k s vαðsÞ ðk Þeikμ x jΨ 0 i  I þ II

where  0  * * *0 *0  0 1X M X ðs Þ ðsÞ pffiffiffiffiffiffiffiffiffiffiffi hΨ 0 jaþ ei k  x  k  x eiðEk tEk0 t Þ * a* jΨ 0 i uβ ðk Þuα ðk Þ V ** E k E k0 ss0 k 0 s0 k s k k0 ! * ! ! !0   0 1 X M X ðsÞ   ðsÞ ¼ uβ ðk Þuα ðkÞ θ k F   k  ei k  x  x eiEðtt Þ V * E s

I¼

k

ð5:765Þ Since uβ and uα are only the components of the vector, there are uβ uα ¼ uα uβ ¼ ðuuÞαβ

ð5:766Þ

We may see the following expression from Eqs. (5.760)–(5.762): e

iE ðtt 0 Þ

Z

0

θðt  t Þ ¼ i

0

dω eiðωþEÞðtt Þ ¼i 2π ω  iε

Z

0

dk0 eik0 ðtt Þ 2π k0  E  iε

ð5:767Þ

Then from Eq. (5.765) and Eq. (5.767), the following expression can be obtained using Eqs. (5.715) and (5.751):

I  θðt0  t Þ ¼ i

Z

*    θ k  k F

 0 μ d k 1 μ k þ M eikμ ðxx Þ γ μ 4 2E αβ k  E  iε 0 ð2πÞ 4

ð5:768Þ

According to Eq. (5.764), in the similar way we can prove  0  * * *0 *0  0 1X M X ðs Þ ðsÞ pffiffiffiffiffiffiffiffiffiffiffi hΨ 0 jb*k 0 s0 bþ i v ð k Þv ð k Þ ei k  x  k  x eiðEk tEk0 t Þ jΨ * 0 α β V ** Ek Ek0 ss0 ks k k0 !

 ! ! !0 0 1 X M X ðsÞ ðsÞ ¼ vβ ðk Þvα ðkÞ ei k  x  x eiEðtt Þ V * E s k

 ! ! !0  0 1X 1 μ ¼ γ k μ þ M αβ ei k  x  x eiEðtt Þ V * 2E

II ¼

k

ð5:769Þ

5.11

Relativistic Green Function Theory at Zero Temperature

475

This term represents the generation of the negative-energy nucleon in vacuum. Introduce the new symbol !

!

k μ ¼ γ 0 E  γ  k γ μe *

ð5:770Þ

*

Letting k !  k after the summation symbol of Eq. (5.769), we can get II ¼ 

   * * *0 0 1X 1 k μ þ M eiEðtt Þ ei k  x  x γ μe V * 2E αβ

ð5:771Þ

k

From Eq. (5.760)–(5.762) we know Z

Z 0 0 dk 0 eik0 ðtt Þ dω eiðωEÞðtt Þ ¼i ð5:772Þ ω  iε 2π k 0 þ E  iε 2π Z   0 μ d4 k 1 1 μe þ M eikμ ðxx Þ ð5:773Þ k γ II  θðt 0  t Þ ¼ i μ 4 2E αβ k 0 þ E  iε ð2πÞ 0

eiEðtt Þ θðt 0  t Þ ¼ i

It can be seen that the zero order Green function (propagator) of the free nucleon in momentum space is, according to Eqs. (5.757), (5.763), (5.764), (5.768), and (5.773), as follows: ( 1 G0αβ ðkÞ ¼ 2E



γμ kμ þ M

"



*    1  θ kF   k 

αβ

k 0  E þ iε )

  kμ þ M  γ μe

þ

*#    θ kF   k  k 0  E  iε

ð5:774Þ

1 αβ k 0 þ E  iε

where *

*

γ μ kμ ¼ γ0 E  γ  k ,

*

*

k μ ¼ γ 0 E  γ  k γ μe

ð5:775Þ

The first item in the first line of the right end of Eq. (5.774) represents the positive energy nucleon propagator above the Fermi surface, the second item represents the positive energy nucleon-hole propagator below the Fermi surface, and the last item appeared in the second line on the right end represents the propagator of negativeenergy nucleon-hole (called anti-nucleon) in the infinite Dirac sea. In the *   non-relativistic (NR) approximation,  k   M, E ffi M þ k 2 =2M, the lower component of the positive energy nucleon will tend to have no contribution. Notice ðγ 0 E þ M Þαβ !2Mδαβ and ignore the last item representing the anti-nucleon in NR

Eq. (5.774), we obtain

476

5

" G0αβ ðkÞ!δαβ NR

Polarization Theory of Relativistic Nuclear Reactions

*    1  θ kF   k  k0  E þ iε

þ

*#    θ kF   k  k 0  E  iε

ð5:776Þ

This is just the non-relativistic nucleon zero order Green function [28]. Equation (5.774) can be rewritten as    *

 1 1    θ kF   k  2EG0αβ ðkÞ ¼ γ μ k μ þ M αβ k 0  E  iε k0  E þ iε   3 2

μ  μe γ þ M k μ γ kμ þ M αβ αβ 5 þ4  k0  E þ iε k 0 þ E  iε

ð5:777Þ

Using the principal value integral formula   1 1

iπδðk0  E Þ ¼P k0  E k0  E iε

ð5:778Þ

*    in the first item containing θ k F   k  of Eq. (5.777), the real part of the principal value integral is just eliminated. The corresponding usual symbol is *

*

γ μ kμ ¼ γ 0 k0  γ  k

ð5:779Þ

γ μ k μ δ ð k 0  E Þ ¼ γ μ k μ δð k 0  E Þ

ð5:780Þ

then we can get the relation

In order to derive the second item of Eq. (5.777), we first seek ðk 0  E þ iεÞðk 0 þ E  iεÞ ¼ k 20  E 2 þ iε ¼ kμ k μ  M 2 þ iε,   * * ðk0 þ EÞγ μ kμ  ðk0  E Þγ μe k μ ¼ 2E γ 0 k 0  γ  k ¼ 2Eγ μ k μ , ðk0 þ EÞM  ðk 0  E ÞM ¼ 2EM Then by Eq. (5.777) we can obtain *  iπ 1   ð  E Þ θ k  þ δ k k 0 F k μ kμ  M 2 þ iε E ð5:781Þ  G0F ðkÞαβ þ G0D ðk Þαβ

 G0αβ ðkÞ ¼ γ μ kμ þ M αβ



The two propagators appeared in the second line of the above formula correspond to the first term and the second term in brace {. . .} of the above formula, respectively.

5.11

Relativistic Green Function Theory at Zero Temperature

477

Dirac propagator G0D does not contribute when kF ! 0; δ(k0  E) represents that is on-shell. In the case of kF > 0 finite nuclear density, one should carry out the study in the ground state coordinate system in which total three-dimensional momentum is equal to 0. G0F ðkÞ is called the Feynman propagator of the free nucleon and anti-nucleon without interaction in the vacuum state. It is off-shell. One can also derive G0F ðk Þ in another way. The equation satisfied by the propagator of a free nucleus in vacuum is [6, 7] G0D



 iγ μ ∂μ  M GF ðx, x0 Þ ¼ δ4 ðx  x0 Þ

ð5:782Þ

We can see from the above equation that GF(x, x0) is only related to (x  x0), and we may obtain the following expression: GF ðx, x0 Þ ¼ GF ðx  x0 Þ ¼

Z

0 μ

d4 k eikμ ðxx Þ 4 γ μ k  M þ iε μ ð2πÞ

ð5:783Þ

Then we get GF ð k Þ ¼

5.11.3

γ μ kμ þ M 1 ¼ γ μ k μ  M þ iε kμ k μ  M 2 þ iε

ð5:784Þ

The Propagator of Neutral Vector Bosons with Spin 1

When the nonlinear term constant C3 ¼ 0, the Lagrangian density of the free vector meson given by Eq. (5.635) is 1 1 L ¼  Ωμν Ωμν þ m2 V μ V μ 2 4

ð5:785Þ

where μ

ν

Ωμν ¼ ∂ V ν  ∂ V μ

ð5:786Þ

where m is meson stationary mass. The Proca equation corresponding with L has been given by Eq. (5.641) ∂μ Ωμν þ m2 V ν ¼ 0

ð5:787Þ

Free vector meson satisfies the Lorentz gauge given by Eq. (5.642): ∂ν V ν ¼ 0

ð5:788Þ

478

5

Polarization Theory of Relativistic Nuclear Reactions

Then the vector meson satisfies the Klein-Gorden equation given by Eq. (5.644):



 μ ∂μ ∂ þ m2 V ν ¼ □ þ m2 V ν ¼ 0

ð5:789Þ

Since the 4-dimensional vector Vν satisfies the Lorentz gauge condition given by Eq. (5.788), one of the four degrees of freedom is not independent. Since Ωμν is antisymmetric for μν and noting Eq. (5.60), so we can write j Ω0j ¼ V_ þ ∇j V 0 ¼ Ωj0

Note ∂j ¼ ∂ j,

ð5:790Þ

Vj ¼ V j, we can also write Ω0j ¼ V_ j  ∇j V 0 ¼ Ωj0

ð5:791Þ

Ω0j ¼ Ωj0 ¼ Ω0j ¼ Ωj0

ð5:792Þ

0 It can be seen that V_ ¼ V_ 0 is not included in L; therefore V0 ¼ V0 is not an independent variable [9]. And we can also write

Ωij ¼ ∇i V j þ ∇j V i ¼ Ωji ,

Ωij ¼ ∇i V j  ∇j V i ¼ Ωji ,

Ω ¼ Ω ¼ Ωij ¼ Ωji ij

ji

ð5:793Þ

Noting Vj ¼ V j, so we can get      j 1 _j 1 1 * * 2 V þ ∇j V 0 V_ þ ∇j V 0  ∇  V  Ωμν Ωμν ¼ 2 4 2

ð5:794Þ

The conjugate momentum can be obtained by using Eqs. (5.616) and (5.794): πj ¼

∂L 0 _j j ¼ V þ ∇j V _ ∂V

ð5:795Þ

Since V0 is not an independent variable, we take V0 ¼ 0 here, and then from Eq. (5.795) we get j π j ¼ V_

ð5:796Þ

*

For the certain k , we take the unit vectors as *

k e !k 3 ¼ * ,   k

*

*

*

e *k i  e *k j ¼ δij ,

i, j ¼ 1, 2, 3

ð5:797Þ

5.11

Relativistic Green Function Theory at Zero Temperature

479

In above formulas, i ¼ 1, 2 represents a transverse wave, and i ¼ 3 represents a ! ! longitudinal wave. Since V is the Hermitian operator, so V can be expanded with the plane waves and quantized as follows: 3 i   h ** 1 X 1 X* pffiffiffiffiffiffiffiffi e *k i a*k i eiωk tþi k  x þ h:c: V x , t ¼ pffiffiffiffi V * 2ωk i¼1

! *

ð5:798Þ

k

where h.c. represents the complex conjugate term of the previous term and a and a+ are the particle annihilation and creation operators, respectively. Substituting Vi given by Eq. (5.798) into Eq. (5.789), we can obtain ω2k ¼ k2 þ m2

ð5:799Þ

We can obtain the following expression according to Eqs. (5.796) and (5.798): 



1 X ðiÞ π x , t ¼ pffiffiffiffi V * *

k

rffiffiffiffiffiffi 3 h i ** ωk X * e *k i a*k i eiωk tþi k  x  h:c: 2 i¼1

ð5:800Þ

Assume that the annihilation operator and the creation operator satisfy the following commutation relations:   ¼ δij δ*k *k 0 , a*k i , aþ * 0

h

k j

 i  þ ¼0 a*k i , a*k 0 j ¼ aþ * , a* 0 ki

k j

ð5:801Þ

We can prove the following equal-time commutation relation by use of Eqs. (5.798), (5.800), and (5.801):

 h    0 i 1 X i*k  *x *x 0 * * e V i x , t , π j x , t ¼ iδij V * k

¼ iδij

1 ð2πÞ3

Z

*

*

d k ei k 

*

*0

xx



  * *0 ¼ iδij δ x  x

ð5:802Þ

And it can also be proved that h    0 i h    0 i * * * * V i x , t , V j x , t ¼ πi x , t , πj x , t ¼ 0

ð5:803Þ

These are just the equal-time commutation relations required by the quantization rule. * Set e *k 0 be the unit vector along the time axis and require to satisfy the following conditions in the 4-dimensional space-time coordinate system:

480

5 *

Polarization Theory of Relativistic Nuclear Reactions

*

e *k μ  e *k ν ¼ δμν ,

μ,ν ¼ 0,1,2,3

ð5:804Þ

Be careful not to attempt to express the four unit vector axes of the 4-dimensional space-time coordinate in 3-dimensional space. Since V0 is not an independent variable, refer to Eq. (5.798), assume that we can expand the field quantity V in the 4-dimensional space-time coordinate as follows: 3   h i ρ 1 X 1 X* * pffiffiffiffiffiffiffiffi e *k μ a*k μ eikρ x þ h:c: V x , t ¼ pffiffiffiffi V * 2ωk μ¼0

ð5:805Þ

k

And assume that there are the following communication relations: h i a*k μ , aþ ¼ δμν δ*k *k 0 , * 0 k ν

h

i

a*k μ , a*k ν ¼



aþ * , kμ

aþ * kν

 ¼0

ð5:806Þ

The zero order Green function of a free vector meson is defined as  iD0μν ðx, x0 Þ ¼ hΨ 0 jT V μ ðxÞ V ν ðx0 Þ jΨ 0 i

ð5:807Þ

When the study work is done only in 3-dimensional coordinate space, the above formula is reduced to  iD0ij ðx, x0 Þ ¼ hΨ 0 jT V i ðxÞ V j ðx0 Þ jΨ 0 i

ð5:808Þ

This is the definition of the usual propagator. Eq. (5.807) can be rewritten as iD0μν ðx, x0 Þ ¼ hΨ 0 jV μ ðxÞ V ν ðx0 Þ jΨ 0 iθðt  t 0 Þ  hΨ 0 jV ν ðx0 Þ V μ ðxÞ jΨ 0 iθðt 0  t Þ ð5:809Þ Noting Vi ¼  Vi, then we obtain i gμν X 1 h ikρ ðxx0 Þρ 0 ρ e θðt  t 0 Þ þ eikρ ðxx Þ θðt 0  t Þ V ! 2ωk k Z

! !0  h i * i* gμν 1 k xx iωk ðtt 0 Þ 0 iωk ðtt 0 Þ 0 ¼ d k e e θ ð t  t Þ þ e θ ð t  t Þ 2ωk ð2π Þ3

iD0ij ðx, x0 Þ ¼

ð5:810Þ Using the derivation method of Eq. (5.720), we can obtain iD0μν ðx, x0 Þ ¼ i

gμν ð2πÞ4

Z

0 ρ

d4 k

eikρ ðxx Þ k λ kλ  m2 þ iε

ð5:811Þ

5.11

Relativistic Green Function Theory at Zero Temperature

481

So the propagator of the free vector mesons in momentum space is D0μν ðkÞ ¼

λ

gμν

k λ k  m2 þ iε

ð5:812Þ

Another derivation method of Eq. (5.812) is given below. In the Eq. (5.795), no longer taking non-independent variable V0 ¼ 0, but in Eq. (5.787) taking ν ¼ 0, then by using Eqs. (5.790)–(5.792) and (5.795) we can obtain V0 ¼

1 * * ∇ π m2

ð5:813Þ

Substituting Eq. (5.813) into Eq. (5.795), we can also obtain [9]   !_ 1 * * * π ¼V þ 2∇ ∇ π m

*

*

ð5:814Þ

*

Since V and π are the Hermitian operators, the following plane wave expansion can be done: 3   1 X 1 X * iωk tþi*k *x pffiffiffiffiffiffiffiffi q *k i e þ h:c: V ¼ pffiffiffiffi V * 2ωk i¼1

ð5:815Þ

3   1 X 1 X ! iωk tþi*k *x pffiffiffiffiffiffiffiffi π ¼ pffiffiffiffi p *k i e þ h:c: V * 2ωk i¼1

ð5:816Þ

*

k

*

k

Further from Eqs. (5.815) and (5.816), we can obtain 3 h i ** 1 X 1 X * pffiffiffiffiffiffiffiffi V ¼ pffiffiffiffi ðiωk Þ q *k i eiωk tþi k  x þ h:c: V * 2ωk i¼1

ð5:817Þ

3 h  i  ** * * 1 X 1 X * * pffiffiffiffiffiffiffiffi ∇  π ¼ pffiffiffiffi ik i e *k i  p *k i eiωk tþi k  x þ h:c: V * 2ωk i¼1

ð5:818Þ

*_

k

k

3   1 X 1 X* 1 1 * * * ffiffiffi p p ffi ffiffiffiffiffiffiffiffi ∇ ∇  π e* ¼ m2 V * 2ωk i¼1 k i m2 k " # 3   ** X * * ðiki Þ ik j e *k j  p *k j eiωk tþi k  x þ h:c: j¼1

ð5:819Þ

482

5

Polarization Theory of Relativistic Nuclear Reactions

The following relations can be obtained substituting Eqs. (5.816), (5.817), and (5.819) into Eq. (5.814): *

*

*

p *k i ¼ iωk q *k i  e *k i

*þ p *k i

¼

*þ iωk q *k i

3   ki X * * *  p* k e , j kj kj m2 j¼1

3   k X * **þ  e *k i i2 kj e *k j  p k j m j¼1

ð5:820Þ

*

Then rewrite the above formulas as *

q *k i

*þ q *k i

" # 3   ki X * * i * * ¼ p *k i þ e *k i 2 k e *  p *k j , ωk m j¼1 j k j "  # 3 i **þ * k i X * *þ ¼ p k i þ e *k i 2 k e *k j  p ! ωk kj m j¼1 j

ð5:821Þ

Introduce the quantization in the following way: *þ p *k i

*

*

p *k i ¼ e *k i ðiωk Þa*k i ,

*

¼ e *k i ðiωk Þaþ *

ki

ð5:822Þ

Einstein’s convention is still used below. In the KD metric, two identical Greek letter subscripts represent sums from 0 to 3, and two identical English letter subscripts represent sums from 1 to 3. So the Eq. (5.821) can be rewritten as   ki kj * q *k i ¼ e *k i a*k i þ 2 a*k j , m

*

*þ q *k i

  ki kj þ * ¼ e *k i aþ þ a * * ki m2 k j

ð5:823Þ

Substituting Eq. (5.823) and Eq. (5.822) into Eq. (5.815) and Eq. (5.816), respectively, we can get the following relations:    3 ** ki kj 1 X 1 X* iωk tþi k  x * * * p ffiffiffiffiffiffiffiffi p ffiffiffi ffi e aki þ 2 akj e þ h:c: V¼ m V * 2ωk i¼1 k i

*

ð5:824Þ

k

1 X ðiÞ π ¼ pffiffiffiffi V *

*

k

rffiffiffiffiffiffi 3 h i ** ωk X * e *k i a*k i eiωk tþi k  x  h:c: 2 i¼1

ð5:825Þ

The equal-time commutation relation can be obtained, by using Eqs. (5.824), (5.825), and Eq. (5.801), as follows:

5.11

Relativistic Green Function Theory at Zero Temperature

  X *

 h    0 i * *0 ki kl 1 i * j * V x , t , π x , t ¼ i δij þ 2 δlj ei k  x  x V * m k     kk * *0 ¼ i δij þ i 2l δlj δ x  x m

483

ð5:826Þ

ki kl δ than Eq. (5.802). Of course it can be prove that m2 lj h    0 i h    0 i * * * * V i x , t , V j x , t ¼ πi x , t , πj x , t ¼ 0

This formula is more the term

Equation (5.805) can be rewritten as follows referring to Eq. (5.824):    3   kμ kv ρ 1 X 1 X* * pffiffiffiffiffiffiffiffi V x , t ¼ pffiffiffiffi e *k μ a*k μ þ 2 a*k v eikρ x þ h:c: m V * 2ωk μ¼0

ð5:827Þ

k

Using the similar method, from Eq. (5.807) the following results can be obtained by using Eq. (5.827):

iD0μν ðx,

0

Z

xÞ¼i

kμ kν μν d4 k g þ m2 ikρ ðxx0 Þρ e ð2πÞ4 kλ kλ  m2 þ iε

D0μν ðkÞ

kμ kν m2 ¼ k λ kλ  m2 þ iε gμν þ

ð5:828Þ

ð5:829Þ

This is exactly the result given in Refs. [8, 26, 78]. Since the vector mesons must be coupled with the conserved nucleon current in practical applications, the conserved nucleon current jν(x) must satisfy ∂νjν(x) ¼ 0; therefore kνjν(k) ¼ 0 must be satisfied in momentum space. When D0μν ðkÞ given by Eq. (5.829) acting on jν(k), the second term will give kμkνjν ¼ 0, so the kμkν term of D0μν ðk Þ given by Eq. (5.829) does not contribute to the physical quantity [8, 69, 79]. At the same time, the term kmi k2l δlj appeared in Eq. (5.826) violates the usual quantization rule, which also indicates that the kμkν term should not be retained in Eq. (5.829). Therefore, in practical applications, the D0μν ðk Þ given by Eq. (5.812) is used. Although the time-like component V0 of the free vector meson field is not an independent variable, but the zero order propagators obtained here are generally recognized results. In addition, this problem was studied by introducing Lagrange multiplier λ in the Refs. [8, 79]. The selection of λ value is uncertain, but in the end it is recommended to use D0μν ðkÞ given by Eq. (5.812).

484

5.11.4

5

Polarization Theory of Relativistic Nuclear Reactions

Feynman Rules of Nucleon-Meson Interaction in Momentum Representation

The Feynman rules of the non-relativistic many-body theory are summarized by introducing secondary quantification, Wick theorem, expressions of various Green functions, corresponding Feynman diagrams, and so on in Ref. [28]. In the relativistic many-body theory, it is also necessary to go through a series of theoretical preparations, in particular to use the Wick theorem of bosons and fermions, and then to summarize the Feynman rules of nucleon-meson interaction through a large number of examples [2, 8, 67, 78]. The relativistic many-body theory will automatically degenerate to the non-relativistic many-body theory under very low energy condition. Let jΨ i represent the ground state of a system of interacting many nucleons at finite density (jΨ i is vacuum state for mesons); the corresponding nucleon propagator is defined as  iGαβ ðx, x0 Þ ¼ hΨ jT ψ α ðxÞ ψ β ðx0 Þ jΨ i

ð5:830Þ

where ψ(x) is the nucleon field in Heisenberg representation. Gαβ is a matrix element and G represents a matrix. The interacting nucleon propagator G and the non-interacting nucleon propagator G0 satisfy the following Dyson equation: GðkÞ ¼ G0 ðkÞ þ G0 ðkÞΣ ðkÞGðkÞ

ð5:831Þ

where Σ represents the self-energy operator. The above Dyson equation, the HartreeFock equation of the self-energy operator Σ, and the interacting meson propagator can be represented by Fig. 5.4 when only σ mesons exist. Fig. 5.4 Graphical representation of the selfconsistent relativistic Hartree-Fock equation (a) Nucleon Dyson equation; (b) Hartree-Fock equation of Σ; (c) Interacting σ meson propagator

5.11

Relativistic Green Function Theory at Zero Temperature

485

In the non-relativistic theory, the nucleon-nucleon interaction occurs simultaneously for two nucleons, whereas in the relativistic theory, the nucleon-nucleon interaction occurs by exchanging mesons. Since the propagation of mesons needs time, so the relativistic nucleon-nucleon interaction does not occur at the same time for two nucleons. In the Feynman diagram, the position point where two nucleon lines meet a meson line is called vertex, a meson has two vertexes and the number of the vertexes is even in a Feynman diagram. Therefore, the non-relativistic first order graph corresponds to relativistic second order graph, and the non-relativistic second order graph corresponds to relativistic fourth order graph. The main advantage of calculating the interacting propagators is that they allow one to evaluate observable quantities. For example, using the definition of the nucleon Green function given by Eq. (5.830) and the anti-commutation property of the nucleon wave function, the expectation value of the operator Γ can be written as 

 hΨ jψ ðxÞΓψ ðxÞ jΨ i ¼ !lim! 0lim0þ hΨ T ψ α ðxÞ ψ β ðyÞ jΨ Γ βα y ! x y !x

¼ !lim! lim tr½Γ iGðx, yÞ

ð5:832Þ

y ! x y0 !x0þ

 P ^B ^ ¼ Aβα Bαβ is used, y0 ! x0+ represents that y0 is less where the relation tr A αβ

than x0+ but infinitely closer to x0+. For a two-line closed ring graph (loop) of the operator G interacting with σ meson, in this case there is x' ¼ x, refer to above formula we can write    lim hΨ T ψ α ðxÞ ψ β ðyÞ jΨ δβα hΨ jψ ðxÞ ψ ðxÞjΨ i ¼  lim ! ! 0 0þ y ! x y !x

¼ !lim! lim tr½iGðx, yÞ

ð5:833Þ

y ! x y0 !x0þ

First the physical process of the interaction between the nucleon and the meson is expressed by the connected Feynman diagram with two external propagator lines and n vertexes connected by internal propagator lines. Assign a direction to each line. The rule that points out what each part of the diagram corresponds to what each part in the physical formula is called the Feynman rule. We will summarize some of the main elements of the Feynman rule of the nucleon-meson interaction in the momentum representation using the Bjorken-Drell metric as follows: 1. Each directed line gets a factor, such as: A single solid line represents a free nucleon propagator ! iG0. The double solid lines represent interacting nucleon propagator ! iG. The dotted line represents the free σ meson propagator ! iΔ0. The ripple line represents the free ω meson propagator ! iD0μν . Similar methods can be used for isospin vector meson ρ and π.

486

5

Polarization Theory of Relativistic Nuclear Reactions

2. The vertexes of σ and ω mesons are assigned a factor igσ and igωγ μ, respectively. Conserve the energy and momentum at each vertex. In general, spin and isospin are summed at the vertex, but if the spin and isospin of the two nucleons at the vertex are the same, there is no need to sum them. Similar methods can be used for ρ and π mesons. 3. Sum over all repeated indices. R d4 q over all independent four-momentum q. 4. Integrate ð2πÞ4 5. Include a factor of (1) for each closed fermions loop, and trace the corresponding physical quantity, such as the following correspondence:

Require placing the contribution of the vertex of ω meson line and the nucleon loop in the trace.

Require placing the contributions of the two vertexes of ω meson lines and the nucleon loop in the trace. If the contribution of the vertex located on the nucleon loop contains the isospin matrix, it should also be placed in the trace. Then we should make tracing of the two kinds of matrices, respectively. Other cases could be dealt with in a similar manner.

5.12 5.12.1

Real Part of Nucleon Relativistic Microscopic Optical Potential and Relativistic Nuclear Matter Properties Self-Energy Operator and Green Function in Nuclear Matter [26, 80]

When the tensor term is omitted, the self-energy operator in nuclear matter has been given by Eq. (5.231) as *

*

Σ ðk Þ ¼ Σ s ðk Þ þ γ 0 Σ 0 ðk Þ þ γ  k Σ v ðk Þ

ð5:834Þ

For the no-interacting state jΨ 0i, refer to Eqs. (5.66) and (5.62), we can write the definition of the zero order Green function G0 as

5.12

Real Part of Nucleon Relativistic Microscopic Optical Potential. . .



 γ μ k μ  M G0 ¼ I

487

ð5:835Þ

The above formula can be formally written as

G0

1

¼ γ μ kμ  M

ð5:836Þ

The Dyson equation (5.831) can also be rewritten as

 1  G0 Σ G ¼ G0

ð5:837Þ

For the above equation, multiplying (G0)1 from left end and G1 from right end, and using Eqs. (5.836) and (5.834), we can obtain

1 ! *  Σ ¼ γ μ kμ  M  Σ s  γ0 Σ 0  γ  k Σ v G1 ¼ G0

ð5:838Þ

If we let k 0 ¼ k 0  Σ 0 ,

*

!

k ¼ k ð1 þ Σ v Þ,

M k ¼ M þ Σ s

ð5:839Þ

Equation (5.838) can be rewritten as G1 ¼ γ μ kμ  M k

ð5:840Þ

1=2 Ek ¼ k2 þ M 2 k

ð5:841Þ

We also define

Formally Ek is equivalent to on-shell energy when existence of interactions. We know that in no-interacting case the on-shell single particle energy should be k0 ¼ Ek ¼ (k2 þ M2)1/2, and k0 ¼ Ek corresponds to k 0 ¼ E k , i.e., when k0 ¼ Ek there is k 0 ¼ E k . In Σ ! 0 case we can easily see that the above statement is correct. So when k0 ¼ Ek from the first equation of Eq. (5.839) and Eq. (5.841), we can obtain Ek ¼

E k

 h * i2 h * i2 1=2 2 v   s   þ Σ ¼ k 1 þ Σ  k , Σ k þ M þ Σ  k , E k *    þ Σ 0  k , E k 0

ð5:842Þ

In the above formula, Ek is a transcendent function that requires * to be solved by self  consistent iteration way. Obviously Ek is only related to  k  and nuclear matter

1=2 . density expressed by kF. And from Eq. (5.842) we can see E k ! k2 þ M 2 Σ!0

488

5

Polarization Theory of Relativistic Nuclear Reactions

Since Eq. (5.840) and Eq. (5.836) are the same in form, so refer to Eq. (5.781), we can write GðkÞ ¼ GF ðkÞ þ GD ðk Þ   1 GF ðkÞ ¼ γ μ kμ þ M k  μ kμ k  M 2 k þ iε *    iπ   GD ðk Þ ¼ γ μ kμ þ M k  δðk 0  E k Þ θ kF   k  Ek

ð5:843Þ ð5:844Þ ð5:845Þ

where Ek in Eq. (5.845) is the on-shell single particle energy given by Eq. (5.842).

5.12.2

Real Part of Nucleon Relativistic Microscopic Optical Potential

1. σ Meson Figure 5.5 shows the Feynman diagram of Hartree-Fock equation of exchange σ mesons. According to the Feynman rules described in the 5.11 section, the expression corresponding to the abovementioned Feynman diagram can be written as iG0 ðk ÞΣ σ ðkÞGðkÞ

Z

¼ iG ðkÞðigσ Þðigσ Þ iΔ ð0Þ 0

0

Z þiG0 ðk Þðigσ Þðigσ Þ

d4 q trðiGðqÞÞ iGðkÞ ð2πÞ4

ð5:846Þ

4

dq 0 iΔ ðk  qÞiGðqÞ iGðkÞ ð2πÞ4

where G0Σ σG on the left side of the above equation is regarded as a total propagator, so only one imaginary unit i is multiplied in front. Or according to the Feynman rules, interaction V, effective interaction T, and self-energy Σ contribute iV,  iT,  iΣ, respectively, and there is (i)3 ¼ i. Then noting Eq. (5.721) we can obtain Z

Fig. 5.5 Feynman diagram of Hartree-Fock equation of exchange σ mesons

d4 q trGðqÞ þ ig2σ ð2πÞ4

Z

d4 q 0 Δ ðk  qÞGðqÞ ð2πÞ4 ( ) Z 4 G ð q Þ q 1 d trGðqÞ þ ¼ ig2σ ðk  qÞμ ðk  qÞμ  m2σ þ iε ð2πÞ4 m2σ

Σ σ ðk Þ ¼

ig2σ Δ0 ð0Þ

ð5:847Þ

5.12

Real Part of Nucleon Relativistic Microscopic Optical Potential. . .

489

where the first term is the Hartree term, which is independent with momentum k ; and the second term is the Fock term, which is related to momentum k. We have known G ¼ GF þ GD, where the Feynman propagator GF represents the contribution of virtual nucleons and anti-nucleons in vacuum, and the GF term in the integral given by Eq. (5.847) are divergent. In principle, through studying vacuum fluctuation correction, the contribution of the GF item can be studied using the renormalization method [75]. However, this is a subject that needs to be further studied. Here we only study the real nucleons in the Fermi sea, i.e., we neglect the contribution of the GF item. When taking G ¼ GD, using Eq. (5.845) we can rewrite Eq. (5.847) as Z Σ σ ðk Þ ¼

ig2σ

( )   γ μ qμ þ M q 1 d4 q μ   tr γ qμ þ M q þ ðk  qÞμ ðk  qÞμ  m2σ þ iε ð2πÞ4 m2σ !   iπ

  ¼ Σ H ðk Þ þ Σ F ðk Þ  δ q0  E q θ k F  q σ σ Eq ð5:848Þ

Here we assume that there is only one kind of nucleon in the nuclear matter, i.e., without considering the nucleon isospin. Then we can find the Hartre term as 2 ΣH σ ðk Þ ¼ gσ

π ð2πÞ4

Z

!

dq

 2 Z k F  * 4M q  Mq 2 gσ 1   θ k  q q dq ¼  F  2 2 mσ π 0 E q mσ Eq ð5:849Þ

It can be seen that the Hartree item is independent with momentum k. In the following we will find the Fock item by using Eq. (5.848). Noting when q0 ¼ Eq there is q0 ¼ E q , then we can obtain Z

* γ μ qμ þ M q   iπ

d4 q δ q0  E q θ k F   q  4 ðk  qÞ ðk  qÞμ  m2 þ iε E ð2πÞ q σ μ  * q   μ  2 Z θ k  F γ q þ M πgσ * q μ q ¼E ¼ dq q 0 Eq ðk  qÞμ ðk  qÞμ  m2σ ð2πÞ4   *   Z * *  θ kF   q  πg2σ γq * Mq 0 ¼ dq þγ  jq ¼E E q E q ðk  qÞμ ðk  qÞμ  m2σ 0 q ð2πÞ4

Σ Fσ ðkÞ ¼ ig2σ

ð5:850Þ *

*

*

Taking z-axis along the k direction and setting that the angle between γ and k is * ! θ1 in the φ ¼ 0 plane, i.e., γ ¼ ðγ, θ1 , 0Þ, and letting q ¼ ðq, θ, φÞ, then in the rectangular coordinate system we can write

490

5

Polarization Theory of Relativistic Nuclear Reactions

q1 ¼ q sin θ cos φ q2 ¼ q sin θ sin φ

γ 1 ¼ γ sin θ1 γ2 ¼ 0

q3 ¼ q cos θ

γ 3 ¼ γ cos θ1

*

*

From Eq. (5.839) we know q ¼ ð1 þ Σ v Þ q , and we can obtain *

*

γ  q ¼ γ sin θ1 q sin θ cos φ þ γ cos θ1 q cos θ

ð5:851Þ

Substituting this formula into Eq. (5.850), when making integral to φ, we can know that the first item in Eq. (5.851) does not contribute. When only retaining the items that have contributions to the integral, from Eq. (5.851) we can get *

*

γq ¼



*

*!

γk

*

kq

 =k 2

ð5:852Þ

We can also obtain

2 * *2 ðk  qÞμ ðk  qÞμ jq0 ¼Eq ¼ E  Eq  k  q

2 ¼ E  E q  k2  q2 þ 2kq cos θ

ð5:853Þ

Let

2 Ai ðk, qÞ ¼ k 2 þ q2 þ m2i  E  E q ,

i ¼ σ, ω, π, ρ

ð5:854Þ

so there is  1 1 q ¼E ¼ ðk  qÞμ ðk  qÞμ  m2i 0 q Ai ðk, qÞ þ 2kq cos θ

ð5:855Þ

And there are the following indefinite integral formulas: Z

du 1 ¼ ln ða þ buÞ þ C, a þ bu b

Z

udu 1 ¼ ½a þ bu  a ln ða þ buÞ þ C a þ bu b2 ð5:856Þ

Let us find the following integral:   2π Ai ðk, qÞ  2kq 1 d cos θ ¼  ln 2π 2kq Ai ðk, qÞ þ 2kq 1 Ai ðk, qÞ þ 2kq cos θ   π π A ðk, qÞ þ 2kq ¼  ln  i ¼  Θi ðk, qÞ kq kq Ai ðk, qÞ  2kq Z

1

ð5:857Þ

Real Part of Nucleon Relativistic Microscopic Optical Potential. . .

5.12

491

where   A ðk, qÞ þ 2kq , Θi ðk, qÞ ¼ ln  i Ai ðk, qÞ  2kq

i ¼ σ,ω,π,ρ

ð5:858Þ

We can also get the following integral: Z

1

2π 1

2πkq kq cos θ d cos θ ¼ 2 2 ½4kq  Ai ðk, qÞ Θi ðk, qÞ Ai ðk, qÞ þ 2kq cos θ 4k q ð5:859Þ   A ðk, qÞ Θi ðk, qÞ ¼ 2πΦi ðk, qÞ ¼ 2π 1  i 4kq

where Φi ðk, qÞ ¼

Ai ðk, qÞ Θi ðk, qÞ  1, 4kq

i ¼ σ,ω,π,ρ

ð5:860Þ

Therefore we can obtain the following result according to Eqs. (5.850), (5.852), (5.855), (5.857), and (5.859): ΣFσ ðk,

 Z kF  Mq π EÞ ¼ Θ ðk, qÞqdq 4 k Eq σ ð2πÞ 0  Z Z kF  π kF q * * 2π þγ 0 Θσ ðk, qÞqdq  γ  k 2 Φ ð k, q Þqdq  σ k 0 k 0 Eq  Z Z kF M  kF g2 q 0 Θσ ðk, qÞqdq ¼ σ2  Θσ ðk, qÞqdq þ γ Eq 16π k 0 0 ) * *Z 2 γ  k k F q   Φσ ðk, qÞqdq k 0 Eq πg2σ

ð5:861Þ

The coefficients of the constant term which is not related to γ matrix, γ 0 term, and * * F0 Fv γ  k term in the above equation are V Fs σ ðk Þ, V σ ðk Þ, and V σ ðk Þ, respectively. 2. ω Meson In Fig. 5.5 we replace the dotted line with a corrugated line to represent the exchange of the ω meson. According to Feynman rules, we can write iG0 ðkÞΣ ω ðkÞ Gðk Þ ¼ iG0 ðk Þðigω γ μ Þ iD0μν ð0Þ Z d4 q tr½ðigω γ ν ÞðiGðqÞÞ iGðkÞ þ iG0 ðk Þðigω γ μ Þ ð2πÞ4 Z d4 q iD0μν ðk  qÞ iGðqÞðigω γ ν Þ iGðkÞ ð2πÞ4

ð5:862Þ

492

5

Polarization Theory of Relativistic Nuclear Reactions

And using Eq. (5.812) we can further obtain Z d4 q d4 q 0 ν 2 μ Σ ω ðk Þ ¼ trðγ GðqÞÞ þ igω γ D ðk  qÞGðqÞγ ν 4 4 μν ð2πÞ ð2πÞ " # Z 4 μ μ tr ð γ G ð q Þ Þ G ð q Þγ q d ¼ ig2ω γ μ gμμ þ m2ω ð2πÞ4 ðk  qÞλ ðk  qÞλ  m2ω þ iε ig2ω γ μ

gμν m2ω

Z

F ¼ ΣH ω ðk Þ þ Σ ω ðk Þ

ð5:863Þ The above formula implicit summation to μ. It can be seen that when the sum to μ is not carried out, according Eq. (5.149) we know 

 gμμ tr γ μ γ ν qν þ M k ¼ 4qμ

ð5:864Þ

Still taking G ¼ GD, so using Eqs. (5.845) and (5.864), we can obtain the following formula from Eq. (5.863): 2 μ ΣH ω ðk Þ ¼ igω γ

Z

 *   d4 q 4qμ iπ

q δ q  E  θ k q F  0 ð2πÞ4 m2ω E q

ð5:865Þ

Utilizing Eq. (5.852) we also have 

 γ μ gμ  !

q0 ¼E q

¼

γ 0 Eq

*

*

kq γk k2 *

*

ð5:866Þ

!

Since there is k  q ¼ kq cos θ, when making integral to cosθ it can be known that the second item in Eq. (5.866) has no contribution, so the following result can be obtained by Eq. (5.865). 0 ΣH ω ðk Þ ¼ γ

g2ω 2 π m2ω

Z

kF

 q2 dq ¼ γ 0

0

gω mω

2

k3F 3π2

ð5:867Þ

And the following formula can be obtained according to Eqs. (5.863) and (5.845): Z Σ Fω ðk Þ ¼ ig2ω

  gμμ γ μ γ ν qν þ M q γ μ

*   iπ

dq δ q0  E q θ k F   q   4 λ ð2πÞ ðk  qÞλ ðk  qÞ  m2ω þ iε E q 4

ð5:868Þ

5.12

Real Part of Nucleon Relativistic Microscopic Optical Potential. . .

493

Notice X μ

gμμ γ μ γ ν γ μ ¼

X μ

gμμ γ μ ðγ μ γ ν þ 2gμν Þ ¼ 4γ ν þ 2γ ν ¼ 2γ ν

ð5:869Þ

so there is   X gμμ γ μ γ ν qν þ M q γ μ ¼ 2γ ν qν þ 4M q

ð5:870Þ

μ

Then Eq. (5.868) can be rewritten as Z Σ Fω ðkÞ

¼

πg2ω

* 2γ ν qν  4M q   d4 q 1

q δ q  E  θ k q F  0 ð2πÞ4 ðk  qÞλ ðk  qÞλ  m2ω þ iε E q ð5:871Þ

Comparing this formula with the first equation of Eq. (5.850) and then according to Eq. (5.861), we can obtain ( Z kF  Mq g2ω EÞ ¼ Θ ðk, qÞqdq  4 2 E q ω 16π k 0 ) Z kF * * Z kF   k q 4 γ þ2γ 0 Θω ðk, qÞ qdq  Φω ðk, qÞqdq k 0 0 Eq

Σ Fω ðk,

ð5:872Þ

The contribution of ω meson tensor term is not considered here. 3. π Meson Equation (5.671) gives the interaction Lagrangian density of the π meson with the nucleon. The first and second terms represent the pseudo-scalar (ps) and axis-vector (pv) coupling, respectively. Replace the σ meson with the π meson in Fig. 5.5, and compare Eq. (5.671) with Eq. (5.669) describing the ω meson; according to the Feynman rules we can write 0 5 0 iG0 ðkÞΣ ps π ðk ÞGðk Þ ¼ iG ðk Þgπ γ τi iΔij ð0Þ Z

d4 q 5  tr gπ γ τj ðiGðqÞÞ iGðkÞ þ iG0 ðkÞgπ γ 5 τi 4 ð2πÞ Z d4 q 0 iΔij ðk  qÞiGðqÞgπ γ 5 τj iGðkÞ ð2πÞ4

ð5:873Þ

494

5

Polarization Theory of Relativistic Nuclear Reactions

And then get Z Σ ps π ðk Þ

¼

ig2π

h i

 d4 q 5 0 γ τi Δij ð0Þ tr γ 5 τj GðgÞ  γ 5 τi Δ0ij ðk  qÞGðqÞγ 5 τj ð5:874Þ 4 ð2πÞ

In the above formula the first item is Hartree one, and the second item is Fock one. π meson is a pseudo-scalar meson, whose propagator similar to σ meson is taken as Δ0ij ðqÞ ¼

qμ

qμ

δij  m2π þ iε

ð5:875Þ

where i and j are isospin footnotes and δij means that the charge exchange between the π meson and the nucleon is not considered, i.e., the nuclear force is approximately independent of the charge. When G(q) in the first item of Eq. (5.874) is replaced by GD(q) given by Eq. (5.845), because there are trγ 5 ¼ 0 and tr(γ 5γ μ) ¼ 0, so we can get ðk Þ ¼ 0 Σ psH π

ð5:876Þ

5 μ That is, the Hartree item of Σ ps π has no contribution. Noting τ iτ i ¼ 3 and { γ , γ } ¼ 0, the following result can be obtained from Eqs. (5.874) and (5.875):

Z

γ 5 τi GðqÞγ 5 τi d4 q 4 ðk  qÞ ðk  qÞμ  m2 þ iε ð2πÞ π μ    μ  Z 3 γ qμ  M q *   iπ

d4 q ¼ ig2π δ q0  E q θ k F   q  4 ðk  qÞ ðk  qÞμ  m2 þ iε E  ð2πÞ μ π μ

psF ðkÞ ¼ ig2π Σ ps π ðk Þ ¼ Σ π

ð5:877Þ Comparing this formula with Σ Fσ ðk Þ in Eq. (5.848), here an extra 3 is multiplied, and in the front of M q there is a negative sign, so according to Eq. (5.861) we can write Σ ps π ðk,

3g2π EÞ ¼ 16π2 k

Z

kF 0

"

*

* M q 2 γ  k q qdq   Θπ ðk, qÞ þ γ 0 Θπ ðk, qÞ  Φ ðk, qÞ Eq k Eq π

#

ð5:878Þ It is generally believed that the pseudo-scalar coupling of π mesons cannot give reasonable results, and the pseudo-scalar coupling of π mesons can be converted into axis-vector coupling, so the contribution of Σ ps π term is generally not considered [26, 80].

5.12

Real Part of Nucleon Relativistic Microscopic Optical Potential. . .

495

Referring to Eq. (5.708) the π meson wave function of the isospin i component can be written as   μ 1 X 1 ik μ xμ pffiffiffiffiffiffiffiffi a* eikμ x þ aþ π i ðxÞ ¼ pffiffiffiffi * e ki V * 2ωk k i

ð5:879Þ

k

And we can also write the following expression according to Eqs. (5.720) and (5.721):  iΔ0ij ðx, x0 Þ ¼ hΨ 0 jT π i ðxÞπ j ðx0 Þ jΨ 0 i ¼ i

Z

0 μ d4 k 0 Δ ðkÞ eikμ ðxx Þ 4 ij ð2πÞ

ð5:880Þ

In the Fock term at the right end of Fig. 5.5, there are two vertices of the upper annihilation meson x and the lower generation meson x0. Since ∂μ  ∂x∂μ , according ! * to Eq. (5.879) it is known that ∂μx π contributes ikμ and ∂μx0 π contributes ikμ in Eq. (5.671). According to Eq. (5.671), in Eq. (5.873) let the corresponding vertex x term gπ γ 5 ! i mf ππ γ 5 γ μ ipμ and let the corresponding vertex x0 term gπ γ 5 ! i mf ππ γ 5 γ ν ðipν Þ, where p ¼ 0+ or k  q, so for the axis-vector (pv) coupling term we can get    f π 5 μ γ γ i0þ iG ¼ iG ðk Þ i τi iΔ0ij ð0Þ μ mπ    Z f π 5 ν þ d4 q γ γ i0ν τj ðiGðqÞÞ iGðkÞ tr i mπ ð2πÞ4   Z   f d4 q μ 0 þiG0 ðkÞ i π γ 5 γ i ð k  q Þ μ τ i iΔij ðk  qÞiGðqÞ mπ ð2πÞ4   f i π γ 5 γ ν iðk  qÞν τj iGðkÞ mπ 0

ðk ÞΣ pv π ðk ÞGðk Þ

0

ð5:881Þ

According to the above formula, Σ pv π is divided into Hartree and Fock terms: pvH ðk Þ þ Σ pvF ðk Þ Σ pv π ðk Þ ¼ Σ π π

ð5:882Þ

In Eq. (5.881) G(q) is replaced with GD(q). By using trγ 5 ¼ 0, tr(γ 5γ ν) ¼ 0, tr (γ 5γ νγ μ) ¼ 0 or by 0þ ν ! 0 we can always get ðk Þ ¼ 0 Σ pvH π So we can obtain

ð5:883Þ

496

5

Polarization Theory of Relativistic Nuclear Reactions

pvF Σ pv ðk Þ π ðk Þ ¼ Σ π  2 Z f d4 q 5 μ ¼i π γ γ ðk  qÞμ τi Δ0ij ðk  qÞ GðqÞ γ 5 γ ν ðk  qÞν τj mπ ð2πÞ4  2 Z 5 μ 5 ν f d4 q γ γ ðk  qÞμ GðqÞγ γ ðk  qÞν ¼ 3i π η mπ ð2πÞ4 ðk  qÞη ðk  qÞ  m2π þ iε    2 Z  5 μ λ  γ γ ð k  q Þ γ q þ M γ 5 γ ν ðk  qÞν 4 q λ μ fπ dq ¼ 3i mπ ðk  qÞη ðk  qÞη  m2π þ iε ð2πÞ4  *  iπ

   δ q0  E q θ k F  q Eq

ð5:884Þ

Note that γ λγ ν ¼  γ νγ λ þ 2gλν, gλν(k  q)ν ¼ δλν(k  q)ν, and when the γ μ(k  q)μγ ν(k  q)ν item is making summation to μ, ν, the μ 6¼ ν items will cancel each other, and there is γ 5γ 5 ¼ 1. When the sum to μ is not carried out, γ μγ μ ¼ gμμ can be obtained by Eq. (5.59). Then using Eq. (5.852) we can obtain  

 I  γ 5 γ μ ðk  qÞμ γ λ qλ þ M q γ 5 γ ν ðk  qÞν δ q0  Eq  

 ¼ γ μ ðk  qÞμ γ λ qλ  M q γ ν ðk  qÞν δ q0  Eq h   i

 ¼ ðk  qÞμ ðk  qÞμ γ λ qλ þ M q þ 2γ μ ðk  qÞμ qν ðk  qÞν δ q0  E q ( 

2 * *2 M q ¼  Ek  Eq þ k  q  * 2 2

* * * * E k  E q E q þ k  q E q  2 Ek  Eq q  k  q "*    *

2 * *2 * ! k  q þγ  k E  E  k  q k q k2 #) * *! * 



  kq * * 2 1  2 E q E k  E q  q  k  q δ q0  E q k þγ 0



ð5:885Þ

Simplify some of the items in the above formula as follows: ! * *! *2 * 2 * *   *  kq k  q * * * q * * þ2 1 2 q  k  q ¼ k  q 1 þ 2  2qq  kq k2 k k ð5:886Þ So Eq. (5.885) can be rewritten as

Real Part of Nucleon Relativistic Microscopic Optical Potential. . .

5.12

497

nh

h

2 2 * *i  E k  E q þ k 2 þ q2  2 k  q M q þ γ 0 Eq Ek  E q 

 i * * * *

þ k2 þ q2  2 k  q E q þ 2qq Ek  E q  2 k  q Ek  Eq " * * 

  2k  *   k * 2 ð5:887Þ q

q

* ! E  E þ E  E þ γ  k 2 E k  Eq Eq þ E k q k q q k2 k2   

 * * q2 þ k  q 1 þ 2  2qq δ q0  E q k

I¼

The following expression can be got utilizing Eqs. (5.732) and (5.852): γ μ qμ

þ

M q

¼

M q

þ

γ 0 E q

*

*

kq γk k2 *

*

ð5:888Þ

Comparing the first equation of Eq. (5.850) with Eq. (5.861) and noting Eq. (5.888), it can be seen that after removing the same coefficient, 1 corresponds to Θσ, and * * k  q corresponds to 2kqΦσ. And comparing the first equation of Eq. (5.850) with Eq. (5.884) and using the result given by Eq. (5.887), referring to Eq. (5.861), we can obtain 8  2 Z k F < M  

2 2 2  f 3 q pv π qdq  E E þk þq Θπ ðk, qÞ Σ π ðkÞ¼ k q  16π2 k mπ : Eq 0 2

  

2 2 2 2qq Ek E q 04 Θπ ðk, qÞ E k E q þk þq þ  4kqΦπ ðk, qÞ þγ E q 3 2

  

 2qq * 4kq E k Eq * 5 4 Φπ ðk, qÞ  γ  k  4kqþ 2 E k E q þ  Θπ ðk, qÞ E q Eq 1 0 39   =  2

 2kq

2 q 4kq 2kq @  1þ 2 þ 2 Ek Eq þ 2  E k E q AΦπ ðk, qÞ5 Eq ; k k Eq k 

ð5:889Þ Using Eq. (5.860) and Eq. (5.854) to simplify the first item on the right side of the above formula as follows h

i 2  E k  E q þ k 2 þ q2 Θπ  4kqΦπ ¼ 4kq  m2π Θπ

ð5:890Þ

then simplify some of the items in the second term on the right side of Eq. (5.889) as follows:

498

5



Polarization Theory of Relativistic Nuclear Reactions

h

i  2 k2 þ q2 Θπ  4kqΦπ ¼ E k  E q  m2π Θπ þ 4kq

ð5:891Þ

So Eq. (5.889) can be rewritten as Σ pv π ðk, E Þ ¼

 2 Z k F   Mq

fπ 3 2 qdq  4kq  mπ Θπ ðk, qÞ 2 Eq 16π k mπ 0

  

2 2qq E k  E q 0 2 Θπ ðk, qÞ þγ 4kq þ Ek  Eq  mπ þ Eq

   * *  4kq Ek  E q 2γ  k qq Φπ ðk, qÞ  k E k  E q þ  Θπ ðk, qÞ  Eq k Eq     

2

 q 2 2   k þ q þ Ek  Eq þ 2q E k  E q Φπ ðk, qÞ Eq ð5:892Þ

Because the contribution of the π meson pseudo-scalar terms is generally not considered, the coupling terms with one vertex as the pseudo-scalar and the other vertex as the axis-vector are not studied here. 4. ρ Meson Equation (5.670) gives the interaction Lagrangian density of the ρ meson with the nucleon. The first and second terms represent vector coupling and tensor coupling, respectively. Referring to Eqs. (5.812) and (5.875), we can write the propagators of the ρ mesons without interaction as follows: D0μν,ab ðqÞ ¼

qλ

qλ

gμν δab  m2ρ þ iε

ð5:893Þ

where δab represents that the charge exchange between the ρ meson and the nucleon is not considered. First studying the vector items, referring to Eq. (5.862) we can write

 0 μ iG0 ðk ÞΣ V τa iD0μν,ab ð0Þ ρ ðk ÞGðk Þ ¼ iG ðk Þ igρ γ Z 

 d4 q

tr τb igρ γ ν ðiGðqÞÞ iGðkÞ þ iG0 ðk Þ igρ γ μ τa 4 ð2πÞ Z

 d4 q iD0μν,ab ðk  qÞ iGðqÞ igρ γ ν τa iGðkÞ 4 ð2πÞ and get

ð5:894Þ

5.12

Real Part of Nucleon Relativistic Microscopic Optical Potential. . .

ΣV ρ ðk Þ

¼

ig2ρ γ μ τa

Z Z

499

d4 q 0 Dμν,ab ð0Þ trðτb γ ν GðqÞÞ ð2πÞ4

d4 q 0 Dμν,ab ðk  qÞ GðqÞγ ν τb ð2πÞ4 " # Z trðτa γ μ GðqÞÞ G ð qÞ γ μ τ a d4 q 2 μ ¼ igρ γ τa gμμ þ m2ρ ð2πÞ4 ðk  qÞλ ðk  qÞλ  m2ρ þ iε þig2ρ γ μ τa

ð5:895Þ Notice trτa ¼ 0 (a ¼ 1, 2, 3), so we can get ðk Þ ¼ 0 Σ VH ρ

ð5:896Þ

Then using Eq. (5.845) and noting τaτa ¼ 3, from Eq. (5.895) we can obtain VF ðk Þ ΣV ρ ðk Þ ¼ Σ ρ    μ ν  Z 3g γ γ q þ M γ μ iπ

4 *   μμ q ν d q q θ k δ q  E  ¼ ig2ρ q F  0 ð2πÞ4 ðk  qÞλ ðk  qÞλ  m2ρ þ iε E q

ð5:897Þ When including summation to μ there is gμμγ μγ μ ¼ (gμμ)2 ¼ 4, so we get   gμμ γ μ γ ν qν þ M q γ μ ¼ γ ν qν gμμ γ μ γ μ þ 2gμμ gμν qν γ μ þ M q gμμ γ μ γ μ   ¼ 2 γ ν qν þ 2M q

ð5:898Þ

Comparing the first equation of Eq. (5.850) with Eq. (5.897) and using the result given by Eq. (5.861), and comparing Eq. (5.888) with Eq. (5.898), as well as using the conclusion obtained through the analyzing after Eq. (5.888): 1 corresponds to Θ * * and k  q corresponds to 2kqΦ, then we can obtain ΣV ρ ðk,

3g2ρ EÞ ¼ 2 8π k

Z

kF 0

"

*

* 2M q 2 γ  k q qdq   Θρ ðk, qÞ þ γ 0 Θρ ðk, qÞ  Φ ðk, qÞ Eq k E q ρ

#

ð5:899Þ In the tensor term of the Lagrangian density of the interaction between ρ meson * and nucleon given by Eq. (5.670), σ μν and L μν have been given by Eq. (5.538) and Eq. (5.647), respectively. And noting that ∂μx contributes ikμ, ∂μx0 contributes ikμ, then at the vertex x we can find

500

5

Polarization Theory of Relativistic Nuclear Reactions

   i μ ν * * γ γ  γ ν γ μ ∂μ ρ ν  ∂ν ρ μ 2   i μ ν * * * * ¼ γ γ ∂μ ρ ν  γ ν γ μ ∂μ ρ ν  γ μ γ ν ∂ν ρ μ þ γ ν γ μ ∂ν ρ μ 2  

!

σ μν L μν ¼

*

*

*

*

¼ i γ μ γ ν ∂μ ρ ν  γ ν γ μ ∂μ ρ ν ¼ i 2γ μ γ ν ∂μ ρ ν  2gμν ∂μ ρ ν     * * * ν* ¼ 2i γ μ γ ν ∂μ ρ ν  ∂ ρ ν ¼ 2 γ μ qμ γ ν ρ ν  qν ρ ν



ð5:900Þ

*

where σ μν L μν changes its sign at the vertex x0. The momentum propagated in the Hartree term is 0, i.e., in Eq. (5.900), there are qμ ! 0 and qμ ! 0, so we can get ðk Þ ¼ 0 Σ TH ρ

ð5:901Þ

Comparing Eq. (5.670) with the second item of Eq. (5.898), noting that the multiplication of two imaginary unit i from two σ μν will come out (1), and using the result given by Eq. (5.900), referring to the first formula of Eq. (5.884), we can write  ðk Þ ¼ 4i Σ Tρ ðk Þ ¼ Σ TF ρ

fρ 4M

2 Z

d4 q ð2πÞ4

h

 γ ξ ð k  q Þ ξ γ μ  ðk  q Þμ τ a

 2 Z   i fρ d4 q D0μν,ab ðk  qÞGðqÞ γ ζ ðk  qÞζ γ ν þ ðk  qÞν τb ¼ 12i 4M ð2πÞ4 h i h i γ ξ ðk  qÞξ γ μ þ ðk  qÞμ γ η qη þ M q γ ζ ðk  qÞζ gμμ γ μ þ ðk  qÞμ ðk  qÞλ ðk  qÞλ  m2ρ þ iε *   iπ

   δ q0  E q θ k F  q Eq

ð5:902Þ

And we can find i h i h I  γ ξ ðk  qÞξ γ μ þ ðk  qÞμ γ η qη þ M q γ ζ ðk  qÞζ gμμ γ μ þ ðk  qÞμ h ih ¼ γ ξ ðk  qÞξ γ μ þ ðk  qÞμ γ η qη γ ζ ðk  qÞζ gμμ γ μ þ γ η qη ðk  qÞμ i γ ζ ðk  qÞζ gμμ γ μ M q þ ðk  qÞμ M q ¼ γ ξ ðk  qÞξ γ μ γ η qη γ ζ ðk  qÞζ gμμ γ μ  γ ξ ðk  qÞξ γ μ γ η qη ðk  qÞμ þγ ξ ðk  qÞξ γ μ γ ζ ðk  qÞζ gμμ γ μ M q  γ ξ ðk  qÞξ γ μ ðk  qÞμ M q γ η qη γ ζ ðk  qÞζ gμμ γ μ ðk  qÞμ þ γ η qη ðk  qÞμ ðk  qÞμ γ ζ ðk  qÞζ gμμ γ μ ðk  qÞμ M q þ ðk  qÞμ ðk  qÞμ M q ð5:903Þ

5.12

Real Part of Nucleon Relativistic Microscopic Optical Potential. . .

501

Simplify some of the items in the above formula as follows: γ ξ ðk  qÞξ γ μ γ η qη γ ζ ðk  qÞζ gμμ γ μ ¼ γ ξ ðk  qÞξ γ η qη γ μ γ ζ ðk  qÞζ gμμ γ μ þ 2γ ξ ðk  qÞξ gμη qη γ ζ ðk  qÞζ gμμ γ μ ¼ γ ξ ðk  qÞξ γ η qη γ ζ ðk  qÞζ γ μ gμμ γ μ  2γ ξ ðk  qÞξ γ η qη gμζ ðk  qÞζ gμμ γ μ þ2γ ξ ðk  qÞξ γ ζ ðk  qÞζ γ η qη ¼ 2γ ξ ðk  qÞξ γ η qη γ ζ ðk  qÞζ þ 2γ ξ ðk  qÞξ γ ζ ðk  qÞζ γ η qη ¼ 2γ ξ ðk  qÞξ γ ζ ðk  qÞζ γ η qη þ 4γ ξ ðk  qÞξ qη gηζ ðk  qÞζ þ2γ ξ ðk  qÞξ γ ζ ðk  qÞζ γ η qη ¼ 4γ ξ ðk  qÞξ qη ðk  qÞη ð5:904Þ γ ξ ðk  qÞξ γ μ γ ζ ðk  qÞζ gμμ γ μ M q ¼ γ ξ ðk  qÞξ γ ζ ðk  qÞζ γ μ gμμ γ μ M q þ 2γ ξ ðk  qÞξ gμζ ðk  qÞζ gμμ γ μ M q ¼ 2γ ξ ðk  qÞξ γ ζ ðk  qÞζ M q ð5:905Þ γ η qη γ ζ ðk  qÞζ gμμ γ μ ðk  qÞμ ¼ γ η qη γ ζ ðk  qÞζ γ μ ðk  qÞμ ¼ γ ζ ðk  qÞζ γ η qη γ μ ðk  qÞμ  2gηζ qη ðk  qÞζ γ μ ðk  qÞμ ¼ γ ζ ðk  qÞζ γ μ ðk  qÞμ γ η qη þ 2γ ζ ðk  qÞζ gημ qη ðk  qÞμ

ð5:906Þ

2qη ðk  qÞη γ μ ðk  qÞμ ¼ γ ζ ðk  qÞζ γ μ ðk  qÞμ γ η qη And notice γ ξ ðk  qÞξ γ ζ ðk  qÞζ ¼ ðk  qÞξ γ ζ γ ξ ðk  qÞζ þ 2gξζ ðk  qÞξ ðk  qÞζ ¼ ðk  qÞξ ðk  qÞξ

ð5:907Þ

Also note that the ξ 6¼ ζ items in the first term after the first equal sign of the above formula will disappear when the sum to ξ and ζ is proceeded. From Eq. (5.903) the following result can be obtained by using Eqs. (5.904)–(5.907), (5.854), and (5.852):

502

5

Polarization Theory of Relativistic Nuclear Reactions

I jq0 ¼Eq ¼ 4γ ξ ðk  qÞξ qη ðk  qÞη  γ ξ ðk  qÞξ γ μ ðk  qÞμ γ η qη 2γ ξ ðk  qÞξ γ ζ ðk  qÞζ M q  γ ξ ðk  qÞξ γ μ ðk  qÞμ M q  γ ζ ðk  qÞζ γ μ ðk  qÞμ γ η qη þ ðk  qÞμ ðk  qÞμ γ η qη γ ζ ðk qÞζ γ μ ðk  qÞμ M q þ ðk  qÞμ ðk  qÞμ M q ¼ ðk  qÞμ ðk  qÞμ γ η qη  3M q ðk  qÞμ ðk  qÞμ þ 4γ ξ ðk  qÞξ qη ðk  qÞη   

2 * *2 * * 0  ¼  γ Eq  γ  q Ek  Eq  k  q  

2 * *2   3M q E k  E q  k  q h

 * * *ih 

 * * *i þ4 γ 0 E k  Eq  γ  k  q Eq Ek  Eq  q  k  q * *   * * * * * * k  q 2 m  A þ 2 kq ¼ γ 0 E q m2ρ  Aρ þ 2 k  q þ γ  k ρ ρ k2   * * 3M q m2ρ  Aρ þ 2 k  q h

 2 * *



i þ4γ 0 E q E k  Eq  k  q E k  E q þ qq E k  E q * *!h i

 * * kq * * 4 γ  k 1  2 E q E k  E q  k  q þ qq k h  

2 * * * * ¼ 3M q m2ρ  Aρ þ 2 k  q  γ 0 E q m2ρ  Aρ þ 2 k  q  4Eq E k  E q 

i * *

þ4 k  q E k  Eq  4qq E k  Eq "*  *  

 * * k  q m2ρ  Aρ þ 2qq cos 2 θ  4Eq Ek  E q þγ  k 2 k # * *  * * * * qq k  q 

  2 þ4 2 E q E k  E q þ 4 k  q  4qq  4qq cos θ þ 4 k  q 2 k k

ð5:908Þ There is an indefinite integral formula as Z

i h u2 du 1 1 ¼ 3 ða þ buÞ2  2aða þ buÞ þ a2 log ða þ buÞ þ C a þ bu b 2

Referring to Eq. (5.859) and using Eqs. (5.858) and (5.860), we can obtain

ð5:909Þ

5.12

Real Part of Nucleon Relativistic Microscopic Optical Potential. . .

Z 2π

503

1

k2 q2 cos 2 θ d cos θ 1 Ai þ 2kq cos θ n h i o 2πk 2 q2 1 ð2kq  Ai Þ2  ð2kq þ Ai Þ2 þ 8kqAi  A2i Θi ¼ 3 2 ð2kqÞ    π

AΘ ¼ 4kqAi þ 8kqAi  A2i Θi ¼ πAi 1  i i ¼ πAi Φi 4kq 4kq

ð5:910Þ

*

*

The previous summary of the correspondence is that 1 corresponds to Θi and k  q corresponds to 2kqΦi. The comparison between Eq. (5.910) and Eq. (5.859) shows * 2 * i that k  q ¼ k2 q2 cos 2 θ should correspond to 2kqΦ 2π πAi ¼ kqAi Φi . Referring to Eq. (5.889) and using Eq. (5.908), from Eq. (5.902) we can obtain Σ Tρ ðkÞ ¼

 2 Z kF   i 3M q h 2 fρ 12 qdq  mρ  Aρ Θρ þ 4kqΦρ  2 Eq 16π k 4M 0  

2  8kq

γ 0 m2ρ  Aρ Θρ þ 4kqΦρ  4 Ek  Eq Θρ þ  E k  E q Φρ Eq

     

 4qq Ek  Eq * * 2q 2  Θρ þ γ  k  mρ  Aρ Φρ  4 E k  E q Θρ  Eq kE q    8q

8kq 4qq 8q2 q 2q þ Φ AΦ E k  E q Φ ρ þ  Φρ   Θ ρ þ Eq Eq k kEq ρ kE q ρ ρ ð5:911Þ

From Eq. (5.859) we know 4kqΦρ ¼ Aρ Θρ  4kq Merge the coefficients of the containing

ð5:912Þ

4q * *  Φρ items of γ  k items in Eq. (5.911) as kEq

follows:

2 1 1 2k2 þ 2q2 þ m2ρ  Aρ ¼ k2 þ q2 þ E k  E q  m2ρ 2 2 So Eq. (5.911) can be rewritten as

ð5:913Þ

504

5

Σ Tρ ðk, E Þ ¼

Polarization Theory of Relativistic Nuclear Reactions

  2 Z kF  h 3M q  2 fρ 3 0 qdq  m Θ  4kq  γ m2ρ Θρ ρ ρ E q 4π2 k 4M 0



 

2 4qq Ek  Eq 8kq Ek  Eq 4kq  4 Ek  Eq Θρ  Θρ þ Φρ Eq E q 

  qq 2q

* * þ4 γ  k   Θρ  E k  E q Θρ þ E k  E q Φρ Eq k  

2 1 2  q 2 2 ð5:914Þ þ  k þ q þ E k  E q  m ρ Φρ 2 kE q

For ρ mesons, if the tensor coupling exists at the vertex x and the vector coupling exists at the vertex x0; or the vector coupling exists at the vertex x and the tensor coupling exists at the vertex x0, so the vector-tensor coupling terms will appear. From the previous discussion it can be seen that the vertex on the loop of the Hartree item does not contribute for either vector or tensor coupling, so we can get ðk Þ ¼ 0 Σ VTH ρ

ð5:915Þ *

Also note that the negative sign needs to be added when σ μν L μν given by Eq. (5.900) acting on the vertex x0. Referring to the second term of the first equation of Eq. (5.895) and the first equation of Eq. (5.902), we can write Z

nh i d4 q μ ξ μ γ ð k  q Þ γ  ð k  q Þ ξ ð2πÞ4 h i o τa D0μν,ab ðk  qÞGðqÞγ ν τb  γ μ τa D0μν,ab ðk  qÞGðqÞ γ ξ ðk  qÞξ γ ν  ðk  qÞν τb i 8h Z 4 < γ ξ ðk  qÞ γ μ  ðk  qÞμ g GðqÞγ ν μν fρ ξ d q ¼ 6igρ 4 λ 2 4M ð2πÞ : ðk  qÞλ ðk  qÞ  mρ þ iε h i9 γ μ gμν GðqÞ γ ξ ðk  qÞξ γ ν  ðk  qÞν =  ; ðk  qÞλ ðk  qÞλ  m2ρ þ iε i  8h Z 4 < γ ξ ð k  qÞ γ μ  ð k  qÞ μ γ η q þ M  g γ μ μμ fρ η q ξ d q ¼ 6igρ 4 λ 2 4M ð2πÞ : ðk  qÞλ ðk  qÞ  mρ þ iε  h i9 γ μ gμμ γ η qη þ M q γ ξ ðk  qÞξ γ μ  ðk  qÞμ = iπ

*    δ q0  E q θ k F   q   λ ; Eq ðk  qÞ ðk  qÞ  m2 þ iε

VTF ðk Þ ¼ 2igρ Σ VT ρ ðk Þ ¼ Σ ρ

λ

fρ 4M

ρ

ð5:916Þ We can also find

5.12

Real Part of Nucleon Relativistic Microscopic Optical Potential. . .

505

h i  I  γ ξ ðk  qÞξ γ μ  ðk  qÞμ γ η qη þ M q gμμ γ μ  h i γ μ gμμ γ η qη þ M q γ ξ ðk  qÞξ γ μ  ðk  qÞμ ¼ γ ξ ðk  qÞξ γ μ γ η qη gμμ γ μ þ γ ξ ðk  qÞξ γ μ gμμ γ μ M q ðk  qÞμ γ η qη gμμ γ μ  ðk  qÞμ gμμ γ μ M q

γ μ gμμ γ η qη γ ξ ðk  qÞξ γ μ þ γ μ gμμ γ η qη ðk  qÞμ γ μ gμμ γ ξ ðk  qÞξ γ μ M q þ γ μ gμμ ðk  qÞμ M q ¼ γ ξ ðk  qÞξ γ η qη γ μ gμμ γ μ þ 2γ ξ ðk  qÞξ gμη qη gμμ γ μ þ4γ ξ ðk  qÞξ M q  γ η qη γ μ ðk  qÞμ þ γ μ gμμ γ η qη γ μ γ ξ ðk  qÞξ 2γ μ gμμ γ η qη gμξ ðk  qÞξ þ γ μ ðk  qÞμ γ η qη þγ μ gμμ γ μ γ ξ ðk  qÞξ M q  2γ μ gμμ gμξ ðk  qÞξ M q ¼ 4γ ξ ðk  qÞξ γ η qη þ 2γ ξ ðk  qÞξ γ η qη þ 4γ ξ ðk  qÞξ M q þγ μ ðk  qÞμ γ η qη  2qη gημ ðk  qÞμ  γ μ gμμ γ μ γ η qη γ ξ ðk  qÞξ þ2γ μ gμμ gμη qη γ ξ ðk  qÞξ  2γ μ ðk  qÞμ γ η qη þγ μ ðk  qÞμ γ η qη þ 4γ ξ ðk  qÞξ M q  2γ ξ ðk  qÞξ M q ¼ 2γ ξ ðk  qÞξ γ η qη  2ðk  qÞμ qμ  2γ η qη γ ξ ðk  qÞξ þ 6γ ξ ðk  qÞξ M q ¼ 4gξη ðk  qÞξ qη  2ðk  qÞμ qμ þ 6γ ξ ðk  qÞξ M q h i ¼ 6 ðk  qÞμ qμ þ γ ξ ðk  qÞξ M q ð5:917Þ Further from above formula we can get " I jq0 ¼Eq ¼ 6  2





Ek  E q E q þ

*    

 * * * * * k  q  q þ γ 0 Ek  E q M q  γ  k  q M q

#

3

* *!



 * * kq ** 6

7 ¼ 64  E k  Eq Eq þ k  q  qq þ γ 0 E k  Eq M q  γ  k 1  2 M q 5 k ð5:918Þ Then the following expression can be obtained by Eqs. (5.916) and (5.918):

506

Σ VT ρ ðk,

5

Polarization Theory of Relativistic Nuclear Reactions

Z fρ d4 q 1 E Þ ¼ 36igρ 4M ð2πÞ4 ðk  qÞ ðk  qÞλ  m2 þ iε ρ λ h

  * *

  0  E k  E q E q þ k  q  qq þ γ Ek  E q M q # * *! *

  kq * *  iπ  γ  k 1  2 M q  δ q0  E q θ k F   q  Eq k  Z kF

 fρ 9 qq 2kq ¼ 2 gρ qdq E k  Eq Θρ þ  Θρ   Φρ Eq Eq 4π k 4M 0

    ) 2qM q Ek  Eq M q * * Mq 0 Θρ þ γ  k Θ  Φ γ Eq E q ρ kE q ρ

ð5:919Þ

For a nuclear matter containing only neutrons or protons, the previously obtained self-energy formula can be used directly. For N ¼ Z symmetric nuclear matter, it is necessary to multiply the isospin number λ ¼ 2 in front of the self-energy formula 2 and using ρ ¼ 2 k 3F to determine Fermi momentum. For N 6¼ Z asymmetric nuclear 3π matter, neutrons and protons have Fermi momentum kFn and kFp, respectively, so their self-energies obtained separately need to be added. Finally the optical potential of the finite nucleus is obtained by the local density approximation. For the nucleons describing the Feynman diagram intermediate state, because Dirac propagator GD containing δ function is used, so the intermediate state nucleons are on-shell. In the Ai expression given by Eq. (5.854), if k and E are not related, i.e., both are independent variables, then the real part of the relativistic microscopic

1=2 þ Σ 0 ðkÞ or potential given above is off-shell; if taking E ¼ k2 þ M 2 k 2 2 E ¼ k þ M , it is on-shell.

5.12.3

Relativistic Nuclear Matter Properties

The Dirac equation for free nucleon is 

 *  *  * α  k þ βM u k , ξ ¼ Ek u k , ξ

*

ð5:920Þ

where ξ represents spin. When there is an interaction in the nuclear matter, Eq. (5.920) can be rewritten as 

*   *  * α  k þ βM k u k , ξ ¼ E k u k , ξ

*

ð5:921Þ

And referring to Eq. (5.93), the plane wave solution of Eq. (5.921) can be written as

5.12

Real Part of Nucleon Relativistic Microscopic Optical Potential. . .

*  u k, ξ ¼

1 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0 χξ   Ek þ M k B  C A @ σ^  * k 2M k χ   ξ Ek þ M k

507

ð5:922Þ

Moreover utilizing Eq. (5.839), Eq. (5.921) can be rewritten as h

*  i *  * * * α  k þ β M þ β Σ s þ Σ 0 þ α  k Σ v u k , ξ ¼ Ek u k , ξ

*

*

ð5:923Þ

*

In the above formula, α  k þ β M are the kinetic energy terms, and the sum of the three terms related to self-energy Σ are potential energy terms.   * In the 5.11 section, the nucleon wave function ψ x , t given by Eq. (5.754) and Eq. (5.755) has been proved to satisfy the anti-commutative relation given by Eq. (5.726). Therefore, in the case with interaction, the plane wave function of the positive energy nucleons is taken as ψ *k ξ T





1 x , t ¼ pffiffiffiffi V

*

sffiffiffiffiffiffiffi M k *  ikμ xμ u k, ξ e χT Ek

ð5:924Þ

where T stands for isospin. The wave function so defined is orthogonal normalized by using ψ +ψ. In the N ¼ Z symmetric nuclear matter, using Eq. (5.678) a scalar density can be given as ρS ¼

XZ

kF

ξT

Z

kF

*

Vd k ψ* ψ* ð2πÞ3 k ξ T k ξ T

ð5:925Þ

*

Vd k where is the number of states in the phase space. Since 3 *  *ð2πÞ u k , ξ u k , ξ ¼ 1, using Eq. (5.715) the following expression can be obtained from Eqs. (5.924) and (5.925) as Z ρS ¼ 4

kF

*

d k M k 2 ¼ 2 3 E π ð2πÞ k

Z 0

kF

M k 2 k dk E k

ð5:926Þ

where E k has been given by Eq. (5.841). And using Eq. (5.682) the usual nucleon density can also be given as ρB ¼

XZ ξT

Since

M k Ek

kF

*

Vd k þ ψ* ψ* ð2πÞ3 k ξ T k ξ T

*  *  uþ k , ξ u k , ξ ¼ 1, so we have

ð5:927Þ

508

5

Polarization Theory of Relativistic Nuclear Reactions

ρB ¼

2 3 k 3π2 F

ð5:928Þ

Equation (5.617) gives the Hamiltonian density as ℋ ¼ π λ φ_ λ  L ¼ T 00

ð5:929Þ

∂φ

where φ_ λ ¼ ∂tλ , π λ ¼ ∂∂L . The energy density of the system is φ_ λ     ε ¼ Ψ T 00 Ψ

ð5:930Þ

Since L contains the contributions of nucleon and various mesons, the energy density calculated from the above formula must also be related to nucleon and various mesons. Some people have studied Eq. (5.930) when only considering σ and ω mesons [26, 69]. It is believed that in N ¼ Z symmetric nuclear matter, only the smaller items need to be ignored to write Eq. (5.930) in the following familiar form:

λ ¼ 2 π

Z

Z

kF

Z

dk T ðkÞ þ 2λ ð2πÞ3

kF

λ T ðk Þ k 2 dk þ 2 2π

ε ¼ 2λ

0

*

kF

Z

kF

*

dk 1 V ðk Þ ð2πÞ3 2

ð5:931Þ

V ðkÞ k2 dk

0

where T(k) and V(k) are the distributions of nucleon kinetic energy density and nucleon potential energy density, respectively. The 2 in 2λ represents two spins, and λ ¼ 2 taken here represents two isospins. The kinetic energy density distribution is  *  M k þ * * * k, ξ α  k þ β M u k, ξ  u Ek 0 1 I *! * !   M σ^  k B E þM σ^  k * C ¼ k  k χþ I @ σ^  k Aχ ξ ξ * 2E k E k þ M k σ^  k  M E k þ M k 0 1  kk M þ !    * Ek þ M k B C E þ M  σ^  k B C ¼ k  k χþ I B Cχ ξ  ξ * 2E k E k þ M k @ * σ^  k A σ^  k  M  Ek þ M k     E þM Ek  M k kk  kk  ¼ k  k Mþ  þ  M 2E k E k þ M k E k þ M k Ek þ M k   MM k þ kk ¼ E k

T ðk Þ ¼

ð5:932Þ

5.12

Real Part of Nucleon Relativistic Microscopic Optical Potential. . .

509

The distribution of potential energy density is  *  M k þ *  s * * k , ξ βΣ þ Σ 0 þ α  k Σ v u k , ξ  u Ek 1 0 10 I * * ! s 0 v   Σ þ Σ σ ^  k Σ E þM σ^  k * C @ AB ¼ k  k χþ I @ σ^  k Aχ ξ ξ * 2E k E k þ M k v s 0 σ^  k Σ Σ þΣ E k þ M k 0 1 kk  Σ v s 0 Σ þ Σ þ ! * E k þ M k B C E þ M  σ^  k B C ¼ k  k χþ I  Cχ ξ   B ξ * 2E k Ek þ M k @ *

 σ^  k A σ^  k Σ v  Σ s  Σ 0  Ek þ M k   

s   E þ M  kk  Σ v kk Σ v 0 Ek  M k ¼ k  k Σs þ Σ0 þ  þ  Σ  Σ 2E k E k þ M k E k þ M k Ek þ M k Σ s M k þ Σ v kk  þ Σ0 ¼ E k

V ðk Þ ¼

ð5:933Þ If only σ and ω mesons are considered, the energy density is formally divided into kinetic energy terms, Hartree potential energy terms, and Fock potential energy terms, namely, ε ¼ εT þ εH þ ε F

ð5:934Þ

For N ¼ Z symmetric nuclear matter, from Eq. (5.931) and Eq. (5.932), we can obtain 2 εT ¼ 2 π

Z 0

kF

MM k þ kk  2 k dk Ek

ð5:935Þ

In fact the contribution of the self-energy operator Σ is also included in the εT expression given above. And there is εT ! Σ!0 or ρB !0

2 π2

Z

kF

E k k2 dk

ð5:936Þ

0

Equations (5.849), (5.861), (5.867), and (5.872) give the λ ¼ 1 self-energy expressions, respectively, as follows:

510

5

Polarization Theory of Relativistic Nuclear Reactions

 Σ sHσ ¼  Σ sFσ

gσ mσ

2

g2σ ¼ 16π2 k

1 π2

Z 0

kF

Z 0

kF

M q 2 q dq Eq

M q Θ qdq E q σ

Z kF g2σ Θσ qdq 16π2 k 0 Z kF  g2 q Σ vFσ ¼  2σ 2  Φσ qdq 8π k 0 E q  2 3 kF gω 0 Σ Hω ¼ mω 3π2 Z kF  Mq g2 Σ sFω ¼  ω2 Θ qdq 4π k 0 E q ω Z kF g2 Θω qdq Σ 0Fω ¼ ω2 8π k 0 Z kF  g2ω q v Σ Fω ¼  2 2  Φω qdq 4π k 0 E q Σ 0Fσ ¼

ð5:937Þ ð5:938Þ ð5:939Þ ð5:940Þ ð5:941Þ ð5:942Þ ð5:943Þ ð5:944Þ

When λ ¼ 2 the following expression can be found using Eqs. (5.931), (5.933), (5.937), and (5.941): "   # Z  2 M k gσ λ kF M q 2  q dq k2 dk Ek mσ π2 0 E q 0  2  2 Z kF " 2 3 # gω λkF 2 1 gσ 1 gω λ 2 dk ¼  ρ þ ρ2B þ 2 k S 2 mσ 2 mω mω 3π2 2π 0

λ εH ¼ 2 2π

Z

kF

ð5:945Þ

We can also find Z

(



 λ s Σ Fσ þ Σ sFω M k þ Σ vFσ þ Σ vFω kk   Ek 0 )

0  2 0 þ λ Σ Fσ þ Σ Fω k dk

λ εF ¼ 2 2π

Z

kF

(



 1 s Σ Fσ þ Σ sFω M k þ Σ vFσ þ Σ vFω kk   E k 0 )

0  2 0 þ Σ Fσ þ Σ Fω k dk

2 ¼ 2 π

kF

ð5:946Þ

5.12

Real Part of Nucleon Relativistic Microscopic Optical Potential. . .

511

The binding energy of a single nucleon in nuclear matter is ε  M ¼ a, ρB

a  16 MeV

ð5:947Þ

According to the law of thermodynamics, the formula for calculating the pressure p is p¼

ρ2B

  ∂ ε ∂ε ¼ ε þ ρB ∂ρB ρB ∂ρB

ð5:948Þ

Let p ¼ 0 in Eq. (5.948) then ρ0  ρB can be solved. At this time, the system is the most stable, p ¼ 0 is called the saturation condition, and ρ0 is called the saturation density. Generally ρ0  0.15~0.16 fm3, kF can be obtained by ρ0. The chemical potential under zero temperature is defined as  μ¼

∂E ∂N



 ¼

V

∂ε ∂ρB

 ð5:949Þ V

For * the ideal gas there is μ ¼ EF. Fermi energy EF is the solution of Eq. (5.842) when    k  ¼ k F . So the saturation condition can be written as p ¼ ε þ EF ρ0 ¼ 0

ð5:950Þ

If the effective nuclear force has only two free parameters gσ and gω, they can be determined by the above nuclear matter property parameters a and ρ0. The incompressible coefficient is [28]      ðε=ρB Þ ε 1 ∂ε ∂ 2  K¼ ¼ 9ρB  2þ  2 ∂ρB ρB ρB ∂ρB ρB ¼ρ0 ∂ρB ρB ¼ρ0    2  2  2 ∂ε ∂ ε  ∂ ε  þ 2  ¼ 9ρB  2 ε þ ρB ¼ 9ρB 2  ∂ρB ρB ∂ρB ρB ¼ρ0 ∂ρB ρB ¼ρ0 ∂ 9ρ2B

2

ð5:951Þ

The saturation condition is used in the above formula derivation. If the effective nuclear force has more free parameters, they can also be constrained by other properties of nuclear matter such as the incompressible coefficient K and the nucleon effective mass M k =M. For the asymmetric nuclear matter, the symmetric energy Esym and symmetric energy slope L can also be used to constrain the effective nuclear force parameters. If the theory is used for finite nuclei, they can also be constrained by experimental data on the binding energies and charge radii of some specific nuclei.

512

5

5.13

Polarization Theory of Relativistic Nuclear Reactions

Imaginary Part of Nucleon Relativistic Microscopic Optical Potential

We only study the lowest order Feynman diagrams that have contributions to the imaginary part of the self-energy Σ [25, 80]. For the case where the incident nucleon energy is above the Fermi surface, only the 2p‐1h polarization diagrams are considered for the time being. When only σ and ω mesons are included, Fig. 5.6 gives the fourth order Feynman diagrams that need to be considered. Each graph in Fig. 5.6 can be regarded as a second order correction graph of meson propagators. Figure 5.7 shows the Feynman diagram corresponding to the second order propagators of σ mesons.

5.13.1

Contribution of σ 2 σ Meson Exchange Processes to the Imaginary Part of the Optical Potential

Using the Feynman rules, the expression corresponding to Fig. 5.7 describing the exchange σ  σ mesons can be written as Z

ð2Þ

iΔ ðqÞ ¼ iΔ ðqÞðiqσ Þ 0

d4 p tr½iGðpÞ iGðp þ qÞ ðiqσ Þ iΔ0 ðqÞ ð2πÞ4

ð5:952Þ

 iΔ ðqÞ π σ ðqÞΔ ðqÞ 0

0

And we can get Z π σ ðqÞ ¼ ig2σ

d4 p tr½GðpÞGðp þ qÞ ð2πÞ4

ð5:953Þ

Fig. 5.6 σ and ω mesons forth order polarization diagrams (------------ σ meson,  ω meson) (a) exchange σ  σ mesons (b) exchange ω  ω mesons (c) exchange σ  ω mesons

Fig. 5.7 Feynman diagram corresponding to second order propagators of σ mesons

5.13

Imaginary Part of Nucleon Relativistic Microscopic Optical Potential

513

Note that π σ(q) is in the nuclear matter, so we use G instead of G0. Eq. (5.721) has given Δ0 ðqÞ ¼

1 q2  q20 þ m2σ

ð5:954Þ

Let us change the Fock term self-energy operator diagram on the right side of Fig. 5.5 to the following form:

In the above figure,

q represents a positive energy σ meson. According to the

second term of the first equation of Eq. (5.847), one can write the expression corresponding to the above graph as Z Σ 0σ ðkÞ ¼ ig2σ

d4 q 0 Δ ðqÞ Gðk  qÞ ð2πÞ4

ð5:955Þ

If the second order propagator given in Fig. 5.7 is embedded in the above figure, it becomes

As long as the replacement Δ0(q) with Δ0(q) π σ(q)Δ0(q) in Eq. (5.955) is used, the expression corresponding to the above figure can be obtained as Z Σ σ ðk Þ ¼

ig2σ

d4 q 0  2 Δ ðqÞ π σ ðqÞ Gðk  qÞ ð2πÞ4

ð5:956Þ

where π σ(q) has been given by Eq. (5.953). The self-energy operator given by Eq. (5.834) is !

!

Σ ðk, EÞ ¼ Σ s ðk, E Þ þ γ 0 Σ 0 ðk, EÞ þ γ  k Σ v ðk, E Þ

ð5:957Þ

And there are Σ s ðk, EÞ ¼ V s ðk, EÞ þ iW s ðk, EÞ,

Σ 0 ðk, EÞ ¼ V 0 ðk, E Þ þ iW 0 ðk, E Þ,

Σ v ðk, EÞ ¼ V v ðk, E Þ þ iW v ðk, E Þ

ð5:958Þ

514

5

Polarization Theory of Relativistic Nuclear Reactions

where k and E are the momentum and energy of the nucleon, respectively, and in the on-shell case there is k0 ¼ E. The following expression can be written according to Eqs. (5.843)–(5.845) as h

i

(

1 ðk   qÞμ ðk   qÞμ  M 2 kq þ iε *     iπ  * þ  δ k0  q0  E kq θ kF   k  q  Ekq μ



Gðk  qÞ ¼ γ ðk  qÞμ þ

M kq

ð5:959Þ

The practical Hartree-Fock calculation shows that the Vv value is very small and can be ignored. From the discussion in the 5.12 section, it can be seen that for Vs and V0 their Hartree terms are independent with momentum. Since Vs and V0 values are of the same order of magnitude as the nucleon stationary mass M, the total calculation result of Hartree-Fock term is also relatively weak in relation with momentum, so here we assume that Vs and V0 are constants. Then when only the real part of Σ is taken in Eq. (5.839), one can obtain !

k 0 ¼ k0  V 0 ¼ E  V 0 ,

!

k ¼ k,

M ¼ M þ V s

ð5:960Þ

Define and find * 2 * 2 * * E 2kq  k  q þ M 2 ¼ k  q þ M 2 ¼ E2 kq

ð5:961Þ

so Eq. (5.959) can be rewritten as h

i

(

1 ðk   qÞμ ðk  qÞμ  M 2 þ iε ) *    iπ

 * þ δ k0  V 0  q0  Ekq θ kF   k  q  E kq μ



G ð k  qÞ ¼ γ ðk  qÞμ þ M



ð5:962Þ

Substitute Eq. (5.962) into Eq. (5.956), and we first make the integral to q0. The second item in Eq. (5.962) has only imaginary part, and for the first term when using the residual theorem, only the imaginary part is retained. Firstly we analyze the following expression: * 2 * ðk  qÞμ ðk   qÞμ  M 2 þ iε ¼ ðk0  V 0  q0 Þ2  k  q  M 2 þ iε







¼ q0  k 0  V 0 þ E kq þ iε q0  k 0  V 0  Ekq  iε ð5:963Þ

5.13

Imaginary Part of Nucleon Relativistic Microscopic Optical Potential

515

The pole in the second medium bracket [ ] part of the above equation is in the upper half plane. This pole requires q0 ¼ k0  V 0  E kq

ð5:964Þ

This formula is the same as the requirement of the δ function in the second item of Eq. (5.962). The residual theorem can be used to obtain πi 2πi

¼ Ekq k0  V 0  E kq  k 0  V 0 þ Ekq *     *    * * And noting that 1  θ kF   k  q  ¼ θ  k  q   kF , then the imaginary part of the optical potential contributed by the σ meson can be obtained by Eq. (5.956) as Z W σ ðk Þ ¼

g2σ

 μ  * d q 0 2 γ ðk  qÞμ þ M Δ ð q Þ Imπ σ ðqÞ 2E kq ð2πÞ3   *    * θ  k  q   kF 

ð5:965Þ

q0 ¼k 0 V 0 E kq

where k0  V0  q0 ¼ Ekq represents that Wσ(k) is on-shell. Moreover, the contribution of the pole of meson propagator to the imaginary part of the optical potential is ignored in the above formula. ! Select the z-axis along the k direction; the entire reaction system is axissymmetric relative to the z-axis. The following relation can be got from Eqs. (5.964) and (5.961):

1=2 q0 ¼ E  V 0  k2 þ q2  2kq cos θ þ M 2

ð5:966Þ

!

We know dq ¼ q2 dqd cos θ dφ, cosθ can be replaced with q0 in integral variables, so the following relations can be obtained by Eq. (5.966): dq0 ¼

kq d cos θ, E kq

d cos θ ¼

E kq dq0 kq

ð5:967Þ

E kq ð5:968Þ dq0 qdqdφ k  *  *  R   * * In Eq. (5.965) there is dφ ¼ 2π. θ  k  q   kF requests  k  q   kF , and !

dq ¼

using Eq. (5.966) we can get

516

5

Polarization Theory of Relativistic Nuclear Reactions

1=2 ðq0 Þ max ¼ E  V 0  k 2F þ M 2

ð5:969Þ

 e F ¼ k 2F þ M 2 1=2 þ V 0 E

ð5:970Þ

Define

then there are eF, ðq0 Þ max ¼ E  E

ðq0 Þ min ¼ 0

ð5:971Þ

From Eq. (5.964) and Eq. (5.961) we can get  1=2  * *2 , q0 ¼ E  V 0   k  q  þ M 2

*  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  * 2  k  q  ¼ ðE  V 0  q0 Þ  M 2  P ð5:972Þ

The condition that the second equation in the above equations has a solution is ðE  V 0  q0 Þ2  M 2

ð5:973Þ

It can be seen from Eq. (5.969) that this condition is satisfied. Then from Eq. (5.966) we can obtain cos θ ¼

k 2 þ q2 þ M 2  ðE  V 0  q0 Þ2 2kq

ð5:974Þ

Since jcos θj  1 we can also obtain h i2 X  k 2 þ q2 þ M 2  ðE  V 0  q0 Þ2  4k2 q2 h h i

i2 ¼ q4  2 ðE  V 0  q0 Þ2 þ k 2  M 2 q2 þ ðE  V 0  q0 Þ2  k2 þ M 2 0 ð5:975Þ When X ¼ 0 the solution of q2 obtained from the above formula and Eq. (5.972) is q2r  ðE  V 0  q0 Þ2 þ k2  M 2 2k ¼ k2 þ P2 2kP

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðE  V 0  q0 Þ2  M 2

ð5:976Þ

The definition of P is given by Eq. (5.972). The two solutions obtained from the above formula are

5.13

Imaginary Part of Nucleon Relativistic Microscopic Optical Potential

qr1 ¼ k þ P,

qr2 ¼ jk  Pj

517

ð5:977Þ

Because X ! 1 when q ! 1, so the area satisfying X  0 is qr2  q  qr1

ð5:978Þ

Then we get q max ¼ k þ P,

q min ¼ jk  Pj

ð5:979Þ

According to Eq. (5.957) we can write *

*

W ¼ W s þ γ0 W 0 þ γ  k W v *

*

ð5:980Þ

*

*

Noting k  k ¼ ki k i ¼ ki ki ¼ k2 , γ  k ¼ γ i ki, then using Eq. (5.149) we can obtain the following results by Eq. (5.980) [81]:  1 1

W s ¼ trW, W 0 ¼ tr γ 0 W , 4 4

Wv ¼ 

 *  1 * tr γ  kW 4k 2

ð5:981Þ

And notice  1 0 0  tr γ γ k 0  q0 ¼ E  V 0  q0 ¼ E kq 4 n h io * * 1 * * tr γ  k γ μ ðk  qÞμ ¼ k2  k  q 4

ð5:982Þ ð5:983Þ

so from Eq. (5.966) and Eq. (5.972) we find *

*

2 k  q ¼ k2 þ q2  ðE  V 0  q0 Þ2 þ M 2 ¼ k2 þ q2  P2 ,  * * 1

k 2  k  q ¼ k 2  q2 þ P 2 2

ð5:984Þ

Here we introduce an abbreviated symbol representing a double definite integral: ZZ Z  

Z

~F EE

dq0 0

kþP

qdq jkPj

ð5:985Þ

We can write the following expressions according to Eqs. (5.965), (5.968), (5.971), (5.979), and (5.981)–(5.984): W sσ ðk,

ZZ g2σ M  0 2 E Þ ¼ 2  Δ ðqÞ Imπ σ ðqÞ 8π k

ð5:986Þ

518

5

Polarization Theory of Relativistic Nuclear Reactions

ZZ

2 g2σ EÞ ¼ 2  ðE  V 0  q0 Þ Δ0 ðqÞ Imπ σ ðqÞ 8π k ZZ 2



g2σ v W σ ðk, EÞ ¼   k2  q2 þ P2 Δ0 ðqÞ Imπ σ ðqÞ 3 2 16π k W 0σ ðk,

ð5:987Þ ð5:988Þ

When carrying out the practical calculation, pay attention to whether (Δ0(q))2 has !1 singularities. We use the form of the Green function given by Eq. (5.774), the last item of which describes the propagation of anti-nucleons with negative energy. Due to the very high energy required to generate anti-nucleon particle-hole pair, the contribution of this item can be ignored in the energy areas interested by us. So G( p) can be expressed in nuclear matter as

G ð pÞ ¼



2

   * θ  p   kF

 * θ kF   p 

3 þM 4 5 þ 2E p p0  V 0  Ep þ iε p0  V 0  E p  iε

γ μ pμ

ð5:989Þ

where E 2p ¼ p2 þ M 2

ð5:990Þ

2 2 We know E 2 ¼ E 2p , so according to Eq. (5.732) we have p ¼p þM !

!

γ μ pμ ¼ γ 0 E p  γ  p

ð5:991Þ

And we can write γ μ ð p þ qÞ μ þ M  Gðp þ qÞ ¼ 2E pþq    2 3 ! ! ! ! θ  p þ q   kF θ kF   p þ q  4 5 þ p0  V 0 þ q0  E pþq þ iε p0  V 0 þ q0  Epþq  iε

ð5:992Þ

where   ! ! 2 E 2pþq ¼ p þ q þ M 2

 ! ! ! γ μ ð p þ qÞ μ ¼ γ 0 E p þ q0  γ  p þ q

ð5:993Þ ð5:994Þ

Substituting Eq. (5.989) and Eq. (5.992) into Eq. (5.953), and we first study the integral to p0. Note that for functions with two poles, if both poles are in the upper

5.13

Imaginary Part of Nucleon Relativistic Microscopic Optical Potential

519

half plane or both poles in the lower half plane, the contribution of the loop integration of the half plane without pole is zero, so we first remove these items without contribution. Then the loop integration for lower half plane or upper half plane can give h 

i   μ  ν  p þ M ð p þ q Þ þ M γ tr γ μ ν dp0 dp π σ ðqÞ ¼ ig2σ 4E p E pþq 2π ð2πÞ3   8 * * * < θ kF   p  θ  p þ q   kF 



: p0  V 0  E p  iε p0  V 0 þ q0  Epþq þ iε     9 * * * = θ  p   kF θ kF   p þ q   

þ

p0  V 0  E p þ iε p0  V 0 þ q0  E pþq  iε ; h 

i   μ  ν  Z ! tr γ p þ M ð p þ q Þ þ M γ μ ν dp ¼ g2σ 3 4E E p pþq ð2πÞ      8  * * * * *9 * 0, using Eq. (5.1255) we can obtain   !2 2 δðP0  E Þ ¼ 2E δ s  E þ P

ð5:1258Þ

and Eq. (5.1251) can be rewritten as     E þ E k ð1Þ 1 * * ð2Þ 1 * * Gðk, PÞ  Gðk, sÞ ¼ 4π2 M 2 kþ Λþ P þ k Λþ Pk E kþ E k 2 2     2 * 1 δ k 0  ðE kþ  E k Þ δ s  ðE kþ þ E k Þ2 þ P 2 ð5:1259Þ Let z  α ¼ ρ eiφ, there is the following loop integral on the full plane I

dz ¼ zα

I

 Z 2π d α þ ρ eiφ iρ eiφ dφ ¼ ¼ 2πi iφ ρe ρ eiφ 0

If we make a half-loop integration of the half plane corresponding to the positive real axis, and there is no singularity on the longitudinal complex axis, so there is . Then for the function f(s) using the residual theorem of the positive real axis half plane, we can write f ðsÞ ¼

1 πi

Z

1 4M 2

s0

f ðs0 Þ ds0  s  iε

ð5:1260Þ

In the above formula, 4M2 is the lower limit of s. Replace f(s) with G(k, s) in Eq. (5.1260) and use Eq. (5.1259), then we can obtain 2M 2 ðEkþ þ E k Þ iGðk, PÞ ¼ iGðk, sÞ ¼ 2π E kþ E k  * ð2Þ 1 * * ð1Þ 1 * i Λþ P þ k Λþ Pk h 1 2 2 δ k0  ðEkþ  Ek Þ 2 * 2 ðEkþ þ E k Þ2  P  s  iε Define

ð5:1261Þ

5.15

Relativistic Bethe-Salpeter Equation

569

  * 2 1=2 1 ! E q ¼ E12P q ¼ M 2 þ P q 2

ð5:1262Þ

and since s is a conservation quantity, in Eq. (5.1261) s can be taken as

2 *2 s ¼ Eqþ þ Eq  P

ð5:1263Þ

So Eq. (5.1261) can be rewritten as    1 * * ð2Þ 1 * * þ k  k Λ P P þ 2M ðEkþ þ Ek Þ 2 2 iGðk, PÞ ¼ 2π

2 Ekþ Ek Eqþ þ Eq  ðEkþ þ E k Þ2 þ iε h i 1 δ k0  ðEkþ  E k Þ 2 2

ð1Þ

Λþ



ð5:1264Þ

Substituting Eq. (5.1264) into the BS equation given by Eq. (5.1243) and noting that there is the δ function containing k0 in Eq. (5.1264), we can obtain     Z * * * * * * t q 0, q ; P ¼ v q 0, q ; P þ

*  * * * * *  dk * * * v q 0, k ; P g k , P t k , q ; P 3 ð2πÞ

ð5:1265Þ where  * *  * * ð1Þ ð2Þ * * 2M 2 ðE þ E Þ Λþ 12 P þ k Λþ 12 P  k kþ k g k, P ¼

2 E kþ E k E qþ þ E q  ðE kþ þ E k Þ2 þ iε

ð5:1266Þ

! ! Then, G(k, P) was transformed into a three-dimensional propagator g k , P ; we call Eq. (5.1260) and Eq. (5.1264) as Blankenbecler-Sugar (BbS) selection [90]. It can be seen from Eq. (5.1264) that the relative momentum time-like component k0 of the intermediate state is not an independent variable. Since the two-nucleon interaction also contains elastic scattering process, it is possible to infer that the time-like components q0 and q00 of the initial and final states are not independent variables, so Eqs. (5.1265) and (5.1266) are equations that only contain three-dimensional * * * * momentum q , k , q 0 , and P .  * *2 *2 *2 Using the mean approximation of angles, there are 12 P þ k  14 P þ k  * *2 *2 *2 and 12 P  k  14 P þ k , and then Eq. (5.1264) and Eq. (5.1266) can be simplified to

570

5

iGðk, PÞ ¼ 2π

2

M E kþ



Polarization Theory of Relativistic Nuclear Reactions

ð1Þ 1 * 2P

Λþ

* * M2 g k, P ¼ E kþ

* ð2Þ  * * þ k Λþ 12 P  k

δð k 0 Þ E2qþ  E 2kþ þ iε  * *  * * ð1Þ ð2Þ Λþ 12 P þ k Λþ 12 P  k E 2qþ  E2kþ þ iε

ð5:1267Þ

ð5:1268Þ

It can be seen from δ(k0) that in this case there is k0 ¼ q0 ¼ q00 ¼ 0. The  positive  ! energy Dirac spinor that appears in Eq. (5.1252) can be expressed by u p when  *   *  * * without considering nucleon spin and isospin. Using u1 12 P þ q 0 u2 12 P  q 0  *   *  * * and u1 12 P þ q u2 12 P  q to act on Eq. (5.1265) from left and right sides,     ! * * * respectively, since u p and v q 0 , q ; P are both Lorentz invariants, so we can define  u1

  *     *   *  1 * *0 1 1 1 * * * * * * P þ q u 2 P  q 0 v q 0 , q ; P u 1 P þ q u2 P  q 2 2      2    2   *0 *0 *0 * * * * * ¼ u 1 q u 2  q v q , q u1 q u 2  q  V q 0 , q

ð5:1269Þ          1* * 1* * 1* * 1* * * * * u 1 P þ q 0 u 2 P  q 0 t q 0 , q ; P u 1 P þ q u2 P  q 2 2 2           2   *0 *0 *0 * * * * * * * ¼ u1 q u 2  q t q , q ; P u 1 q u 2  q  T q 0 , q ; P 

ð5:1270Þ   * * * Strictly speaking, the t matrix t q 0 , q ; P is not a Lorentz invariant. Using Eq. (5.1268) and the projection operator expression without summation to λ given by Eq. (5.1252), we can obtain     Z * * * * * T q 0, q ; P ¼ V q 0, q þ

*  * 2 !  dk M 1 *0 * * V q , k , q ; P T k Ekþ E 2qþ  E 2kþ þ iε ð2πÞ3

ð5:1271Þ The equation composed of Eqs. (5.1265) and (5.1266) and the equation given by Eq. (5.1271) are all called BbS equation. * In the two-nucleon center of momentum system (P ¼ 0 ), Eqs. (5.1267) and (5.1271) are simplified to

5.15

Relativistic Bethe-Salpeter Equation

iGðk, sÞ ¼ 2π 

*0 *





*0 *

Z



T q, q ¼V q, q þ

571

   * ð1Þ * ð2Þ k Λþ  k

Λþ

2

M E kþ

s 4

 E2k þ iε

δðk 0 Þ

ð5:1272Þ

*  * 2 *  dk M 1 *0 * T k, q V q , k 2 3 * 2 E k * ð2πÞ q  k þ iε ð5:1273Þ

where 

!2

1=2

Ek ¼ M þ k 2

ð5:1274Þ

According to the residual theorem of the positive real axis half plane, it can be written that 1 f ðP0 Þ ¼ πi

Z

1 2M

 f P00 dP0 P00  P0  iε 0

ð5:1275Þ

Utilizing G(k, P0) to replace f(P0) in Eq. (5.1275) and using Eq. (5.1251), we can obtain  * *  * 1 ð2Þ 1 * þ k  k Λ P P þ M 2 2 iGðk, PÞ ¼ iGðk, P0 Þ ¼ 2π E kþ E k ðEkþ þ Ek Þ  P0  iε h i 1 δ k0  ðEkþ  Ek Þ 2  * ð2Þ 1 * * ð1Þ 1 * i Λ þ k  k h Λ P P 2 þ þ 1 M 2 2 ¼ 2π δ k0  ðE kþ  E k Þ 2 E kþ E k E qþ þ E q  Ekþ  Ek þ iε 2

ð1Þ

Λþ

ð5:1276Þ And substituting Eq. (5.1276) into Eq. (5.1243), we can obtain Eq. (5.1265), where * * g k, P ¼

2



ð1Þ 1 * 2P

Λþ

* ð2Þ  * * þ k Λþ 12 P  k

M Ekþ Ek E qþ þ Eq  Ekþ  E k þ iε

ð5:1277Þ

As before, we use the mean approximation of angles and the matrix element of the positive energy Dirac spinor, the following equation can be obtained:

572

5





*0 * *





*0 *

Z

T q , q; P ¼ V q , q þ

Polarization Theory of Relativistic Nuclear Reactions

*   * 2 * dk M 1 * * * V q 0, k T k , q; P 3 2 2E E kþ qþ  2E kþ þ iε ð2πÞ ð5:1278Þ

The equation composed of Eq. (5.1265) and Eq. (5.1277) and the equation given by Eq. (5.1278) are all called Thompson equation [91]. In the two-nucleon center of * momentum system (P ¼ 0), Eq. (5.1278) is simplified to 



*0 *





*0 *

T q, q ¼V q, q þ

Z

*  * 2 *  dk M 1 * * V q 0, k T k, q 3 2 2E  2E þ iε E q k ð2πÞ k ð5:1279Þ

Since the BbS equation and the Thompson equation are derived from the BS equation but a certain approximation was made, it can only be said that they approximately satisfy  Lorentz  invariance, and at the same time it can be assumed *0 * * that the t matrix t q , q ; P also approximately satisfies Lorentz invariance.

5.16 Bonn One Boson Exchange Potential For the description of strong interacting nuclear forces, the meson theory based on perturbation theory is usually applied. The contribution to the lowest order of NN scattering is the one boson exchange graph. This section introduces the Bonn one boson exchange potential (OBEP) [92–94]. When the distance between the two nucleons is very short, the meson exchange graph will no longer be applicable due to the internal structure of the hadron. The usual method is to introduce the vertex shape factor phenomenally. Since it is a repulsive force in a very short range, it is equivalent to establishing a repulsion wall. Use (Jπ, I ) to represent the spin, parity, and isospin of the meson. In the one boson exchange (OBE), three isospin scalar mesons σ(0+, 0), ω(1, 0), and η(0, 0) and three isospin vector mesons δ(0+, 1), ρ(1, 1), and π(0, 1) are included. Since the existence of σ meson has not been confirmed experimentally, people usually regard σ exchange as an effective 2π mesons exchange. Refer to Eqs. (5.668)–(5.671), we can write the Lagrangian densities of the scalar s(σ), vector v(ω), pseudo-scalar ps(η), and axial vector pv(η) mesons, which all belong to the isospin scalar mesons coupled to the nucleons, as follows [89, 92–94]: L s ¼ gs ψψφðsÞ

ð5:1280Þ

5.16

Bonn One Boson Exchange Potential

L v ¼ gv ψγ μ ψφμðvÞ 

573

  fv ψσ μν ψ ∂μ φðνvÞ  ∂ν φμðvÞ 4M

L ps ¼ gps ψ iγ 5 ψφðpsÞ L pv ¼ 

ð5:1281Þ ð5:1282Þ

f ps 5 μ ψγ γ ψ ∂μ φðpsÞ mps

ð5:1283Þ

ðαÞ

where ψ is a nucleon field and φðμÞ is a meson field. For the corresponding isospin !

!ðαÞ

ðαÞ

vector meson δ, ρ, π, we use τ  φ ðμÞ instead of φðμÞ . The study of the interaction between two bare nucleons is usually carried out in the two-nucleon center of momentum (c.m.) system. Figure 5.11 gives a Feynman diagram of the contribution of one boson exchange to NN scattering in the c.m. system. From Eq. (5.721) we can write the scalar meson propagator corresponding to this graph as Δ0s ðq0  qÞ ¼

q00

 q0

2

1   ! ! 2  q 0  q  m2s

ð5:1284Þ

Equation (5.1267) indicates that the two-nucleon propagator contains δ(k0) in the case by using the mean approximation of angles; therefore there is k0 ¼ q0 ¼ q00 ¼ 0, so the following three-dimensional static propagator can be used [92]: Δ0s ðq0  qÞ ¼  

1

!

 ! 2

q0  q

ð5:1285Þ þ m2s

OBEP is defined as vOBEP ¼

X

vOBE α

ð5:1286Þ

α¼σ,δ,ω,ρ,η,π

For the isospin vector mesons δ, ρ, π, above formula needs to multiply by a factor * * τ 1  τ 2 . The on-shell positive energy free nucleon plane wave spinor is Fig. 5.11 Feynman diagram of the contribution of one boson exchange to NN scattering in c. m. system

574

5

Polarization Theory of Relativistic Nuclear Reactions

1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffi0 χλ E þ M@ * ! A u q, λ ¼ σ^  q 2M χλ EþM 



ð5:1287Þ

 2 1=2 ! where E ¼ q þ M 2 . In addition to the Feynman rules described in Sect. 5.11, there are the following Feynman rules:

  ! 1. The initial state positive energy nucleon outside line is expressed by u q , λ , and   ! the final state positive energy nucleon outside line is expressed by u q , λ .

2. A v line describing nucleon-nucleon interaction is expressed by iv. E ! 3. In c.m. system, the two-nucleon initial state is expressed by  q λ1 λ2 , and the D  ! corresponding final state is expressed by q 0 λ01 λ02 . Using the Feynman rules, we can first write the OBE amplitude of the scalar meson as D

*0 0 0  OBE * q λ 1 λ 2 vs q λ1 λ2

E

        * ! * ! g2s u q 0 , λ01 u q , λ1 u  q 0 , λ02 u  q , λ2 ¼   *0 ! 2 q  q þ m2s ð5:1288Þ

According to Eq. (5.812) the static propagator of the vector meson is D0μν ðq0  qÞ ¼ 

*0

gμν  ! 2

q q

ð5:1289Þ þ m2v

Here introduce the following γ matrix expressions with the lower footnote in the BD metric: γ μ  gμμ γ μ σ μν 

 i i

γ γ  γ ν γ μ ¼ gμμ gνν ðγ μ γ ν  γ ν γ μ Þ ¼ gμμ gνν σ μν 2 μ ν 2

ð5:1290Þ ð5:1291Þ

Note that the γ matrices with the lower footnote when using the BD metric in the above two formulas are not the same as the γ matrices with the lower footnote when using the Pauli metric introduced by Eq. (5.26). From Eq. (5.880) we can see that the meson propagator in space-time coordinate is

5.16

Bonn One Boson Exchange Potential

Z

0

iΔ ðx, x Þ ¼ i 0

575 0 μ d4 k 0 Δ ðk Þ eikμ ðxx Þ 4 ð2πÞ

ð5:1292Þ

The contribution of ∂μx at the vertex x is ikμ; the contribution of ∂μx0 at the vertex x0 is ikμ. In the second item of Eq. (5.1281), σ μν is anti-symmetric; therefore ∂μ φνðvÞ and ∂ν φðμvÞ have the same contribution. So we take the second item and multiply it by 2, and make the change gμν ! gμμ in Eq. (5.1289) to be used to the first vertex; also note that in Fig. 5.11 the 4-momentum of the meson propagator is q0  q, and then the OBE amplitude of the vector meson can be obtained as 

D

*0 0 0  OBE * q λ 1 λ 2  vμ j q λ 1 λ 2



E

   f     * *0 * ν 0 v 0 u q , λ1 σ μν iðq  qÞ u q , λ1 ¼ gv u q , γ μ u q , λ1 þ 2M      f     * * * * gv u  q 0 , λ02 γ μ u  q , λ2  v u  q 0 , λ02 σ μν iðq0  qÞν u  q 0 , λ2 2M 1   *0 ! 2 q  q þ m2v 

*0

λ01

ð5:1293Þ We can also write         * *0 0 * 5 5 0 , λ0 E g2ps u * q u q , λ , λ u  q , λ iγ u  q iγ 1 2   1 2 *0 0 0  OBE * q λ1 λ2 vps q λ1 λ2 ¼   *0 ! 2 q  q þ m2ps

D

ð5:1294Þ D



*0 0 0  OBE * q λ1 λ2 vpv jq λ1 λ2

E

¼

    f 2ps * 0  5 μ *0 0 0 , λ γ γ i ð q 0  qÞ u * u q q , λ , λ u  q 1 1 2 μ m2ps   1 * γ 5 γ μ i ð q 0  qÞ μ u  q , λ 2   *0 ! 2 q  q þ m2ps ð5:1295Þ

The shape factor used for each meson-nucleon vertex is  Fα

*0

2 

*

q q

where Λα is called truncation mass.

2 6 ¼4

3nα Λ2α Λ2α



 *0

m2α

7

 5 * 2

þ q q

ð5:1296Þ

576

5

Polarization Theory of Relativistic Nuclear Reactions

Since the multiple meson exchanges occur between two bare nucleons, it is necessary to use Brueckner theory to obtain an effective interaction that considers the effect of multiple meson exchanges. Now use symbol v instead of vOBEP, and let   D E  * * * * V q 0 λ01 λ02 , q λ1 λ2 ¼ q 0 λ01 λ02 v q λ1 λ2

ð5:1297Þ

so the Thompson equation in the c.m. system given by Eq. (5.1279) can be rewritten as     XZ *   dk * * * * *0 0 0 !e e V q λ λ , k T q 0 λ01 λ02 , q λ1 λ2 ¼ V q 0 λ01 λ02 , q λ1 λ2 þ λ λ 1 2 1 2 3 eλ1eλ2 ð2πÞ !  M2 1 * e e T k , q λ λ λ λ 1 2 1 2 E2k 2E q  2E k ð5:1298Þ In the c.m. system we choose the direction of the initial state incident particles as the z-axis, and q0 of the final state particle is located on the xz plane (see Fig. 5.12); it ! ! is equivalent that q 0 is obtained by rotating q an angle θ around the y-axis. A point P(x, y, z) in the original coordinate system becomes P(x0 , y0 , z0 ) in the new coordinate system after rotation (see Fig. 5.13), and there are x0 ¼ x cos θ  z sin θ y0 ¼ y 0

z ¼ x sin θ þ z cos θ

Fig. 5.12 Scattering schematic diagram of two particles in c.m. system

Fig. 5.13 Two coordinate systems that one of them rotates an angle θ around the y-axis

ð5:1299Þ

5.16

Bonn One Boson Exchange Potential

577

According to the above formula, we can find   0  ∂x ∂ ∂z0 ∂  ∂ ∂ ∂  ¼ þ ¼ z þ x ∂x ∂z ∂θ θ¼0 ∂θ ∂x0 ∂θ ∂z0 θ¼0  0 ! The wave function ψ r can be written as  0     ! ! ^yψ ! r ¼ ψ0 r ψ r ¼R

ð5:1300Þ

 0 ^ y can be obtained from the following expansion of ψ ! r The operator R  0 ! ψ r ¼ ψ ðx cos θ  z sin θ, y, x sin θ þ z cos θÞ  2     0  θ2 ∂ ! !0 ∂ ψ r þ ψ r þ ⋯⋯ ¼ ψ ðx, y, zÞ þ θ ∂θ 2 2! ∂θ θ¼0 θ¼0 # 2    2   ∂ ∂ θ ∂ ∂ ! ¼ 1θ z x þ þ ⋯⋯ ψ r z x 2! ∂x ∂z ∂x ∂z ∂ ∂ ¼ eθðz∂xx∂zÞ ψ ðr Þ ð5:1301Þ ^ y ¼ eiθ L^y R

ð5:1302Þ

Let χ λ1 , χ λ2 , χ λ01 , χ λ02 be the two-dimensional helicity wave functions of the four     particles taking their own motion direction as the axis, and jλ1i, jλ2i, λ01 , λ02 be their two-dimensional spin wave functions with the previously selected common ! z-axis as the axis; also note that  q of the λ2 particle is in the opposite direction with ! ! ! q of the λ1 particle, and q 0 is rotated relative to q , so we can write jλ1 i ¼ χ λ1 ,  0 λ ¼ e2i θ σ^y χ λ0 , 1

1

jλ2 i ¼ χ λ2  0 λ ¼ e2i θ σ^y χ λ0 2 2

ð5:1303Þ ð5:1304Þ

 ð0 Þ pi ¼ 12 σ^izð0Þ of where jλi i ði ¼ 1, 2Þ is the eigenstate of the helicity operator 12 σ^i  ^ the ith nucleon with unit momentum ^pi , (0) represents whether there is a apostrophe or no apostrophe, so there is  0E  E 1  ðÞ ð0 Þ  ð0 Þ σ^i  ^pi λi ¼ λi λi , i ¼ 1,2 2 ð0 Þ

ð5:1305Þ

where λi represents the nucleon helicity value. The definition of a reduced rotation matrix element is

578

5

Polarization Theory of Relativistic Nuclear Reactions

 E D  ^   dJλ0 λ ðθÞ ¼ Jλ0 eiθ J y Jλ

ð5:1306Þ

And it has the following properties: 0

dJλ0 λ ðθÞ ¼ dJλ λ0 ðθÞ ¼ ð1Þλ λ dJλλ0 ðθÞ Z 0 2 d Jλ0 λ ðθÞdJλ0 λ ðθÞ d cos θ ¼ δ 0 2J þ 1 JJ

ð5:1307Þ

d J00 ðθÞ ¼ PJ ðcos θÞ

ð5:1309Þ

ð5:1308Þ

Let  0 0 λ λ ¼ χ λ0 χ λ0

jλ1 λ2 i ¼ χ λ1 χ λ2 ,

1 2

1

2

ð5:1310Þ

then we can write the following two-nucleon spin coupled state wave functions as φ λ1 λ2 ¼

X S

CS1λ1 2

λ1 λ2 1 2

λ2

  λ1 λ2 ,

φλ01 λ02 ¼

X S0 C 1 λ0 S0

2 1

λ01 λ02  0 0  0 λ1 λ2 1 2 λ2

ð5:1311Þ

Because the total spin of the two nucleon system is conserved, S0 ¼ S should be maintained when the above two formulas are expanded at the same time; their desirable values are 0 and 1. Therefore, according to Eq. (5.1306) we can write the following expansion [93–95]: *

*

hq 0 λ01 λ02 jvj q λ1 λ2 i ¼

1 X ð2S þ 1Þ d Sλ0 λ0 1 2 4π S

λ1 λ2 ðθ Þ

hλ01 λ02 jvS ðq0 , qÞjλ1 λ2 i ð5:1312Þ

Utilizing Eq. (5.1308), from above formula we can get hλ01 λ02 jvS ðq0 , qÞjλ1 λ2 i ¼ 2π

Z

1

1

d Sλ0 λ0 1

2

λ1 λ2 ðθ Þ

*

*

hq 0 λ01 λ02 jvj q λ1 λ2 id cos θ

ð5:1313Þ

Then one can carry out the following expansion:   * * * * T q 0 λ01 λ0 , q λ1 λ2  hλ01 λ02 jTðq 0 , q Þ jλ1 λ2 i  0

1X 0 ð2L þ 1Þ T L q0 λ01 λ02 , qλ1 λ2 PL0 ð cos θÞ ¼ 2 L0

ð5:1314Þ

Substituting Eq. (5.1312) and Eq. (5.1314) into Eq. (5.1298), multiplying PL(cosθ) * * * * * * from the left and integral over cosθ, and noting that q 0  k and q 0  q with k  q satisfy the additive angle formula, so it can be converted into an integral equation

5.16

Bonn One Boson Exchange Potential

579

containing only variable k; each item of these is a 4  4 matrix. Taking into account that the parity is conserved, time inversion is not changed, and the total spin is  * *  conserved, so only the following 6 of the 16 matrix elements λ01 λ02 T q 0 , q jλ1 λ2 i are independent [95–98]: hþþjT jþþi,

hþþjT ji,

hþjT jþi

hþjT jþi,

hþþjT jþi,

hþjT jþþi

1 1 where þ and  represent the helicity value and  . In the on-shell case, from the 2 2 time inversion invariant, we can get hþþjTjþi ¼  hþjTjþþi, so only 5 matrix elements are independent. It is also necessary to consider the nucleon isospin, distinguish between neutrons and protons, and include Coulomb interaction in p-p interactions. There is a relation between the S matrix and the T matrix as [92, 93] 

   

 p01 p02 Sjp1 p2 i ¼ p01 p02 jp1 p2  2πi δð4Þ p01 þ p02  p1  p2 p01 p02 T jp1 p2 i ð5:1315Þ

After the S matrix elements are obtained, the cross sections, angular distributions, and polarization data of the NN scattering can be calculated. At the same time, OBEP can also be used to solve the homogeneous equation of the deuterium bound state to meet the properties of the deuterium ground state [92–95]. In the Ref. [92] three groups of OBEP (Table 5.2) of the relativistic momentum space obtained by the Thompson equation and the pv coupling of π, η mesons are given, where gps ¼ f ps m2Mps, M is the average nucleon mass. The above Bonn potential parameters are obtained by the Thompson equation calculation through fitting the NN scattering experimental data and the properties of the deuteron ground state. But the Thompson equation only allows the projection of the intermediate state to the positive energy scattering state, i.e., in the process of determining the nuclear force parameters by fitting to the experimental data, the influence of the intermediate state

Table 5.2 The relativistic momentum space OBEP obtained by Thompson equation and pv coupling of π,η mesons [92] α σ δ ω ρ η π

mα

Bonn A

ðMeVÞ

g2α =4π 8.3143 0.7709 20 0.99 7 14.9

550 983 782.6 769 548.8 138.03

Bonn B Λα(GeV) 2.0 2.0 1.5 1.3 1.5 1.05

Note: gps ¼ f ps m2Mps , fω/gω ¼ 0, fρ/gρ ¼ 6.1,

g2α =4π 8.0769 3.1155 20 0.95 5 14.6 nα ¼ 1

Bonn C Λα(GeV) 2.0 1.5 1.5 1.3 1.5 1.2

g2α =4π 8.0279 5.0742 20 0.95 3 14.6

Λα(GeV) 1.8 1.5 1.5 1.3 1.5 1.3

580

5 Polarization Theory of Relativistic Nuclear Reactions

as a negative energy state has been ignored. In fact, this kind of determining nuclear force parameters is equivalent to considering the vacuum polarization effect to a certain extent. That is, in the nuclear matter calculation by using the Thompson equation or the BbS equation, the intermediate state as a negative energy state is not considered, it is consistent with that of determining the nuclear force parameter, and it is also equivalent to considering the vacuum polarization effect to a certain extent. In recent years, a new breakthrough has been made in the study of nuclear forces with the chiral effective field theory [99]. This is currently the forefront of the study of nuclear force.

5.17 5.17.1

Nucleon Relativistic Microscopic Optical Potential Based on the Dirac-Brueckner-Hartree-Fock Theory Relativistic Brueckner Theory

The Dirac equation satisfied by a nucleon with momentum k in nuclear medium has been given by Eq. (5.482) as

 *  γ μ k μ  M  Σ ðk Þ u k , λ ¼ 0

ð5:1316Þ

where λ represents the nucleon spin projection; self-energy Σ(k) is equivalent to the mean field. Eq. (5.158) has given *

*

Σ ðk Þ ¼ Σ s ðk Þ þ γ 0 Σ 0 ðk Þ þ γ  k Σ v ðk Þ

ð5:1317Þ

The components of self-energy can be written according to Eq. (5.981) as   *  1 1

1 * Σ s ðk Þ ¼ trΣ ðk Þ, Σ 0 ðkÞ ¼ tr γ 0 Σ ðk Þ , Σ v ðkÞ ¼  2 tr γ  k Σ ðk Þ 4 4 4k ð5:1318Þ Σ(k) and its three components are all complex numbers. Substituting Eq. (5.1317) into Eq. (5.1316) we can obtain h

i !   * * γ 0 k 0  Σ 0 ð k Þ  γ  k ð 1 þ Σ v ð k ÞÞ  ð M þ Σ s ð k ÞÞ u k , λ ¼ 0

ð5:1319Þ

Let M  ðk Þ ¼

M þ Σ s ðk Þ , 1 þ Σ v ðk Þ

k 0 ¼

k 0  Σ 0 ðk Þ , 1 þ Σ v ðk Þ

*

*

k ¼k

ð5:1320Þ

5.17

Nucleon Relativistic Microscopic Optical Potential Based. . .

581

Fig. 5.14 Relationship between valence nucleon self-energy Σ, effective interaction Τ, and selfconsistent propagator G

The definition here is somewhat different from Eq. (5.839). Dividing the right end of Eq. (5.839) by [1 þ Σ v(k)] will give Eq. (5.1320). The advantage of Eq. (5.1320) is *

*

k ¼ k . So Eq. (5.1319) can be rewritten as h

i *  γ μ k μ  M  ðk Þ u k , λ ¼ 0

ð5:1321Þ

The above formula is formally equivalent to the Dirac equation of a free nucleon. We call this kind of nucleons an equivalent free nucleon in nuclear matter, or simply a valence nucleon. Let Τ(k) be the effective interaction or reaction matrix of the two nucleons in nuclear matter discussed in the 5.16 section. The relationship between the valence nucleon self-energy Σ(k), effective interaction Τ(k), and self-consistent propagator G can be expressed by Fig. 5.14 [87], where the first item on the right represents the Hartree direct item, and the second item represents the Fock exchange item. In Feynman rules, the interaction line V or Τ is expressed by iV or iT, so from Fig. 5.14 we can write Z iGðkÞΣ ðk ÞGðkÞ ¼ iGðkÞ Z þiGðkÞ

d4 q ½ðiGðqÞÞðiT ðkq,kqÞÞ iGðk Þ ð2πÞ4 d4 q iGðqÞðiT ðkq,qk ÞÞ iGðkÞ ð2πÞ4

ð5:1322Þ

and get [100] Z Σ ðkÞ ¼ i

d4 q ftr½GðqÞT ðkq,kqÞ  GðqÞT ðkq, qk Þg ð2πÞ4

ð5:1323Þ

The specific numerical calculation of the relativistic Hartree-Fock (HF) theory in the nuclear matter (n.m.) stationary coordinate system shows that the relation of Σ s and Σ 0 with momentum in the interesting nuclear matter density range is very weak, and there is Σ v  1, so the following approximation is used for Eq. (5.1317): Σ ð k Þ ffi ðM   M Þ þ γ 0 Σ 0

ð5:1324Þ

582

5

Polarization Theory of Relativistic Nuclear Reactions

where M ffi M þ Σ s is consistent with Eq. (5.1320). In the specific calculation, we first take M and Σ 0 as the real constants, i.e., ignoring their imaginary parts, it is called the quasi particle approximation. This is similar to the fact that the mean field contains the momentum independent Lorentz scalar Σ s and time-like vector Σ 0. In the calculation process, the new M and Σ 0 are obtained, and their values can be determined by repeated iterations. The approximation expression (5.1324) is obtained in the relativistic HF calculation in the n.m. system, and Eq. (5.1324) can be generalized in any coordinate system as Σ ð k Þ ffi ðM   M Þ þ γ μ Σ μ

ð5:1325Þ

!

Note in the n.m. system Σ μ ¼ δμ0Σ 0, Σ ¼ 0. In the n.m. system, using the symbol introduced by Eq. (5.1320) and letting E k ¼ E  ðk Þ, according to Eq. (5.781) the propagator of the valence nucleon in the nuclear matter can be written as   GðkÞ ¼ γ μ kμ þ M 

 iπ

1 þ  δ k0  E k θðkF  kÞ  ν 2 k ν k  M þ iε E k

 ð5:1326Þ

*   where k ¼  k  . The δ function in the above formula indicates that the valence nucleon is on-shell, and there is

1=2 E k ¼ k2 þ M 2

ð5:1327Þ

E k is obtained through the nuclear matter correction to the definition of Ek. Utilizing Eq. (5.1325) we can obtain 

 ν  

 1   2 θ k 0  δþ kν k ν  M 2  δ k0  Ek ¼ δ kν k  M 2E k

ð5:1328Þ

The above formula has given the definition of the δ+ function, and it is a Lorentz scalar. According to Eq. (5.93), the plane wave spinor of the positive energy valence nucleon in the nuclear matter satisfying Eq. (5.1321) can be written as *  u k, λ ¼

0 1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi χλ   Ek þ M B C * @ σ^  k A 2M  χ E k þ M  λ

The baryon flow density 4-vector is defined as

ð5:1329Þ

5.17

Nucleon Relativistic Microscopic Optical Potential Based. . .

583

Bμ ¼ ψγ μ ψ

ð5:1330Þ

where ψ is a baryon Dirac spinor. The baryon density is ρB ¼ ψ þ ψ ¼ ψγ 0 ψ ¼ B0

ð5:1331Þ

 *

where B B1 , B2 , B3 is a baryon flow density. In the n.m. system (baryons do not flow) and in the case of zero temperature, there is

2 Bμ Bμ ¼ B0 ¼ ρ2B

ð5:1332Þ

For the symmetric nuclear matter Eq. (5.928) has given ρB ¼ 3π22 k3F . When we do ! observing in any new coordinate system, which moves at a constant speed v relative ! to n.m. system, baryons are flowing at a speed  v . Therefore it can be seen that the baryon velocity in the new observation system is !

β

!

!

v B ¼ 0, c B

!

!

B ¼ β B0

ð5:1333Þ

Since BμBμ is a Lorentz invariant, using Eqs. (5.1332) and (5.1333) in any coordinate system, we can obtain

2 ! 2

 2 Bμ Bμ ¼ B0  B ¼ 1  β2 B0 ¼ ρ2B

ð5:1334Þ

1=2 B0 ¼ γ ¼ 1  β2 ρB

ð5:1335Þ

!

*

! v ¼ 0, there will be a baryon flow density B ¼ 0 and a selfc ! energy flow density Σ ¼ 0 . When the total three-dimensional momentum of the nuclear matter is 0, i.e., in the n.m. system, for the symmetric nuclear matter the distribution of the momentum density can be represented by θðkF  kÞ, which is * called the Fermi sphere. Here k represents the speed with which the nucleons do

In the n.m. system β ¼

!

!

irregular Fermi movement in nuclear matter; by paying attention k ¼ k we can obtain  ! !2   ν 2 k0 B0  k  B B k 2 k2F þ kν kν  ν μ ¼ k 2F þ k 2 0 k  Bμ B ρ2B

 2 2 2 2  ¼ k2F þ k2 0  k  k 0 þ k β þ 2β k 0 k

! k2F  k2 ¼ ðkF  kÞðk F þ kÞ β!0

ð5:1336Þ

584

5

Polarization Theory of Relativistic Nuclear Reactions

Because of θðk F þ k Þ ¼ 1, the Lorentz invariant n(k; B) representing the distribution of nucleon momentum density can be taken as " 

nðk ; BÞ ¼ θ

k 2F

þ

kν k ν

 ν 2 # k B  ν μ  ! θðkF  kÞ n:m:system Bμ B

ð5:1337Þ

Equation (5.1326) can be rewritten, using Eqs. (5.1328) and (5.1337), as follows:   Gðk; BÞ ¼ γ μ k μ þ M 



   1 þ  ν 2 k k  M ð k ; B Þ þ 2πi δ n ν kν kν  M 2 þ iε ¼ GF ðk; BÞ þ GD ðk; BÞ ð5:1338Þ

At this time G(k, B) becomes Lorentz invariant. Since GF represents the propagation between virtual nucleon and anti-nucleon with effective mass M, if only the actual valence nucleons are considered when calculating self-energy Σ, so while making calculation for the expression of Σ(k) given by Eq. (5.1323), G(q) can be replaced by GD(q, B) [87, 100]. Then we get Z Σ ðkÞ ¼ i

5.17.2

d4 q ftr½GD ðq; BÞT ðkq, kqÞ  GD ðq; BÞT ðkq, qk Þ g ð2πÞ4

ð5:1339Þ

Relativistic Pauli Incompatibility Operators

The two intermediate state nucleons shown in Fig. 5.10 can be represented by 1  1  According to Eq. (5.1337), their density 2 P þ k and 2 P  k in nuclear  matter.

 distributions are n 12 P þ k; B and n 12 P  k; B , respectively. From the discussion in the 5.16 section, we can see that in the case of angle averaging, there is k0 ¼ 0. Since it is required that one of the two intermediate state nucleons cannot be scattered to the occupying state of another nucleon, the following Pauli incompatibility operator can be introduced in the n.m. system as h  i h  i 1 1 1  n P  k; B Qðk, P ; BÞ ¼ 1  n P þ k; B 2 2

ð5:1340Þ

where Q(k, P; B) represents the part that is not occupied by any one of the two intermediate state nucleons. Note that the above Pauli incompatibility operator does not exclude the occupying state of the nucleons in the nuclear matter. Therefore it is not convenient for practical utilization.

5.17

Nucleon Relativistic Microscopic Optical Potential Based. . .

585

Note that there is a total invariant in any coordinate system as !2

s ¼ Pμ Pμ ¼ P2 0 P

ð5:1341Þ

!

In the n.m. system there is B ¼ 0, i.e., the medium does not flow. In the two particle !

center of momentum (c.m.) system, there is P c ¼ 0, i.e., the coordinate origin is fixed at the center of momentum of two interacting particles, and the entire nuclear matter is flowing relative to the center of momentum of the two particles. It can be seen from Eq. (5.165) that the velocity of baryon flow is !

!

β¼

P P0

ð5:1342Þ

!

where β represents the velocity of the center of momentum of the two particles relative to the n.m. system, and when making observation in the c.m. system the ! velocity of the baryon flow is β . And there is

1=2 P0 ¼ pffiffiffiffi γ ¼ 1  β2 s

ð5:1343Þ

The Lorentz matrix of the Pauli metric given by Eq. (5.102) or Eq. (5.450) should become in the BD metric as follows: 0

γ

B B α¼B B @  For space-time coordinate

0 0

0

1 0

0

0 1

cγβ

0 0

γ 1  β c C 0 C C C 0 A

ð5:1344Þ

γ



!

t, r

using the transformation given by the above

matrix, the result given by Eq. (5.104) can also be obtained. If the components of ! the speed v on (x, y, z) three axes  are all non-zero, in the ħ ¼ c ¼ 1 unit system, the ! space-time coordinate a0 , a in the original coordinate system can be changed to   ! a0c , a c in the new coordinate system represented by footnote c by using the following formulas [87]:  ! ! a0c ¼ γ a0  β  a

ð5:1345Þ

586

5

Polarization Theory of Relativistic Nuclear Reactions



γ * ! ac ¼ a þ βγ β  a  a0 γþ1

!

!

!

 ð5:1346Þ

From above two formulas, the corresponding Lorentz transformation matrix can be written as 0

γ

B B B γβx B B α¼B B B γβy B B @ γβz

γβx 1þ

γβy

γ 2 β2x γþ1

γ 2 βy β x γþ1 γ 2 β z βx γþ1

γβz

γ 2 βx βy γþ1 γ 2 β2y 1þ γþ1 γ 2 βz βy γþ1

1

C C C C C γ 2 β y βz C C C γþ1 C C γ 2 β2z A 1þ γþ1 γ 2 β x βz γþ1

ð5:1347Þ

And notice γ 2 β 2 ð γ  1 Þ γ 2 β 2 ð γ  1Þ γ 2 β2 ¼ 2 ¼γ1 ¼ γþ1 γ2  1 γ ð1  γ 2 Þ then when βx ¼ βy ¼ 0 and βz ¼ β, Eq. (5.1347) can be automatically reduced to Eq. (5.1344). We take n.m. system as the original coordinate system, and take c.m. system with

! Pc

¼ 0 as the new coordinate system c, so there are 



 ! * * 1 ð5:1348Þ P  k ¼ k 2 c c *   In the following, k will be used to represent the k ¼  k  used previously. In the

 0 n.m. system the particles are on-shell, there is p1 ¼ E  ðp1 Þ, in the c.m. system,

0 there is p1 c ¼ E  ðp1c Þ ¼ E  ðkÞ, and we can obtain ! p1 c

¼



 * 1 ! * P þ k ¼ k, 2 c

*2

P2 0 P



! p2



¼

2 ¼ s ¼ P0 c ¼ ½E  ðp1c Þ þ E  ðp2c Þ 2

 ¼ ½E  ðkÞ þ E  ðkÞ 2 ¼ 4 k 2 þ M 2

ð5:1349Þ

and k¼

 1=2 1  s  M 2 4

ð5:1350Þ

In the following we study the Pauli incompatibility operator in the c.m. system. First let

5.17

Nucleon Relativistic Microscopic Optical Potential Based. . .

1=2 E F ¼ k2F þ M 2

587

ð5:1351Þ

where kF is defined in the n.m. system. From Eq. (5.1336) the following expression can be obtained in the c.m. system by using Eqs. (5.1349)–(5.1351) and (5.1334): "

 #  ν 2 B k ν nðk  ; BÞ ¼ θ E 2 F  ρ2B      k  Bν k  Bν k  Bν E F þ ν ¼ θ E F  ν ¼ θ E F  ν ρB ρB ρB

ð5:1352Þ

  ! kv Bv 1  0 ! ! ¼ k0 B  k  B , the observed B =ρB is negative number flow in the ρB ρB c.m. system, so the second factor after the second equal sign of Eq. (5.1352) is always positive. Note that the term after the last equal sign in Eq. (5.1352) contains * * k  B ¼ kB cos θk item, so that the θ function given by this formula forms a Fermi ellipsoid. Using Eq. (5.1333) and Eq. (5.1335), from Eq. (5.1352) we can see that in the c.m. system the condition, which guarantee that the intermediate state particles are outside the Fermi ellipsoid formed by the on-shell particle occupancy states, is as follows:

Since

 E F 

*

*

Ek B0  k  B ρB



 * ! ¼ γ Ek þ k  β

ð5:1353Þ

Then it can be obtained that the Pauli incompatibility operator satisfied by the two * * intermediate state particles with k and  k in the c.m. system is h  i h  i * ! * ! Qðk  ; BÞ ¼ θ γ Ek þ k  β  E F θ γ E k  k  β  E F

ð5:1354Þ

In the above formula the multiplied two terms on the right side of the equal sign * * correspond to k and  k two intermediate state particles, respectively. We choose ! * ! the β direction as the z-axis and the angle between k and β as θk. From Eq. (5.1353) we can see that the equation satisfied by the above Fermi ellipsoid surface is

 γ Ek þ kβ cos θk  E F ¼ 0

ð5:1355Þ

Taking cos θk ¼ 0 in Eq. (5.1355), so we can get

 γ 2 k 2 þ M 2 ¼ k2F þ M 2 , Let its solution be

k 2  k2F þ β2 E2 F ¼ 0

ð5:1356Þ

588

5

ku ¼

Polarization Theory of Relativistic Nuclear Reactions

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k2F  β2 E2 F

ð5:1357Þ

Taking cos θk ¼ 1 in Eq. (5.1355), so we can get 1 2kβ  2 2 Ek ¼ E F  kβ, k2 ¼ k 2F  β2 E 2 F þ k β  γ EF , γ k 1 2 2kβ  2 k þ E þ β2 E 2 þ βEF ¼ k F F ¼ kF , γ F γ γ2

ð5:1358Þ

Let its solution be

 k v ¼ γ kF  βE F

ð5:1359Þ

Taking cos θk ¼  1 in Eq. (5.1355), then we can get k  βE F ¼ kF γ

ð5:1360Þ

 kþ ¼ γ k F þ βE F

ð5:1361Þ

Let its solution be

From Eq. (5.1359) and Eq. (5.1361), it can be seen that the total length of the long axis of the Fermi ellipsoid is 2γkF. Since the short half axis of the Fermi ellipsoid is * perpendicular to the β direction, the short half axis corresponding to the center of the major long axis is still to be kF, but for the short radius ku corresponding to the center of momentum there is ku < kF. And let k ¼ min ðk u , k v Þ cos α ¼

γ

E k

 γβk

ð5:1362Þ

E F

ð5:1363Þ

so Eq. (5.1354) can be rewritten as 8 >

: 1

k < k k  k  kþ ð5:1364Þ k > kþ

The relativistic Pauli incompatibility operator after averaging for the direction angle ! of k can be obtained by Eq. (5.1364) as

5.17

Nucleon Relativistic Microscopic Optical Potential Based. . .

Qðk  ; BÞ ¼

1 4π

Z

dΩk Qðk  ; BÞ

589

ð5:1365Þ

In the k  k  k+ case of Eq. (5.1364), the condition, which guarantee that the two θ functions are equal to 1 at the same time, is  cos α  cos θk  cos α

ð5:1366Þ

So the following expression can be got from Eqs. (5.1363)–(5.1366):

Qðk ; BÞ ¼

8 0 > > < γE   E k

γβk 1

> > :

F

k < k k  k  kþ

ð5:1367Þ

k > kþ

In Refs. [87, 101] the more detailed derivations of the above relativistic Pauli incompatibility operator in the c.m. system are given, and Refs. [88, 102–104] also quoted its final result. In the following we study the Pauli incompatibility operator in the n.m. system. According to Eq. (5.1337) we can write  n

     ν h  i2

 1  1  1 1  P k  P k; B ¼ θ k 2F þ P k P k Bν = Bμ Bμ 2 2 2 ν 2 ν ð5:1368Þ

The following expressions can be obtained using Eqs. (5.1333)–(5.1335), and (5.1342): 

 ν  2   1  1  1  1 ! ! 2 ð5:1369Þ P k ¼ P k P0 Ek  P0 β k 2 2 2 ν 2 h  i2

   i  h1  1 * * * 2 1  P0 Ek B0  P0 β k  B =ρ2B P k Bν = Bμ Bμ ¼ 2 2 2 ν h   * *i2 1 1 ¼ γ2 P0 E k þ P0 β2 k  β 2 2 ð5:1370Þ Let 1 ¼ P0 E k 2  1 1

ð Þ ¼ Dk þ P0 β2 ¼ P0 1 þ β2 E k 2 2 ð Þ

Dk

ð Þ

Rk

ð5:1371Þ ð5:1372Þ

590

5

Polarization Theory of Relativistic Nuclear Reactions

From Eq. (5.1368) we can obtain that the equation satisfied by the Fermi ellipsoid surface in the n.m. system is  ν h  i 2 

 1  1  Bμ Bμ P k  P k Bν 2 ν 2 ν    2 * *2 1 ! ! ð Þ2 ð Þ ¼ k2F þ Dk  P0 β k  γ 2 Rk k  β 2 1 2 2 ð Þ2 2 ¼ kF þ Dk  P0 β  k2 P0 βk cos θk 4 h i k2F þ



1  P k 2

ð Þ2

γ 2 Rk

ð5:1373Þ

ð Þ

þ k2 β2 cos 2 θk 2kβRk cos θk ¼ 0

Here introduce the following symbols: ð Þ

Fk ð Þ

Ck

ð Þ2

¼ γ 2 Rk

ð Þ

¼ γ Rk

ð Þ2

 Dk

þ

1  P 2γ 0

ð5:1374Þ

1 2 2 þ β2 P2 0 þ k  kF 4

ð Þ

ð5:1375Þ

ð Þ

so it can be seen that when k  kF, F k and Ck are all positive. For 12 P k from Eq. (5.1373) we can get the following equations, respectively: ð þÞ

ð þÞ

¼0

ð5:1376Þ

ð Þ

ð Þ

¼0

ð5:1377Þ

γ 2 k2 β2 cos 2 θk þ 2γ kβF k cos θk þ Ck γ 2 k2 β2 cos 2 θk  2γ kβF k cos θk þ Ck

The above two formulas are equivalent to Eq. (5.1373) to be multiplied by (-1), so that if the left end of the above two formulas is greater than 0, it means that it is outside the Fermi ellipsoid. The solution of Eq. (5.1376) is ( ð þÞ cos θk

1 ¼ 2 2 2 γ k β

ð þÞ γkβF k

 1=2 ) ð þÞ  k ð þÞ 2 2 2 2 ð þÞ

γkβF k  γ k β Ck 

k ð5:1378Þ

where ðþÞ k

  1=2  1 ð þÞ ðþÞ2 ðþÞ ¼ F k F k  C k γβ

The solution of Eq. (5.1377) is

ð5:1379Þ

5.17

Nucleon Relativistic Microscopic Optical Potential Based. . .

ð Þ

591

ðÞ

cos θk 

k

k

ð5:1380Þ

where ðÞ k

  1=2  1 ðÞ ðÞ2 ðÞ ¼ F Fk  Ck γβ k

ð5:1381Þ

From Eq. (5.1374) and Eq. (5.1375), we can obtain ð Þ

Hk

1 2 1 ð Þ ð Þ2 2 2 P0 þ P0 Rk þ Dk  β2 P2 0  k þ kF 2 4 4γ ð5:1382Þ  1

ð Þ ð Þ2 2 2 ¼ P0 Rk þ Dk þ 1  2β2 P2  k þ k 0 F 4 ð Þ2

 Fk

ð Þ

 Ck

¼

ð Þ

It can be seen that H k  0 are the conditions that Eqs. (5.1376) and (5.1377) have ð Þ solutions, respectively, and H k < 0 represent that Eqs. (5.1376) and (5.1377) have no solutions, respectively. That is, if the left end of the corresponding equation is greater than 0, it is outside the Fermi ellipsoid. Also notice k ðþÞ < 0. We can express the Pauli incompatibility operator in the n.m. system as Qðk  , P ; BÞ ¼ QðþÞ ðk , P ; BÞQðÞ ðk , P ; BÞ

ð5:1383Þ

where Qð Þ ðk , P ; BÞ 8 ð Þ ð Þ ð Þ > 0, k < kþ and k < k and H k  0 > > > ! > > ð Þ   > > k ð Þ ð Þ ð Þ > þ >  , k θ cos θ and k < kð Þ and H k  0 , k  max k > k þ þ > k < ¼  ð Þ  >

 > k ð Þ ð Þ > > θ  cos θk , k < k þ and k  max k ð Þ , k ð Þ and H k  0 > > k > > > >   > > ð Þ ð Þ ð Þ : 1, k < k þ and k < k ð Þ and H k  0 or H k < 0 ð5:1384Þ ð Þ

ð Þ

Note cos θk  cos θkþ and cos θk  cos θk cannot be satisfied at the same time. The relativistic Pauli incompatibility operator after the angle averaging in the n.m. system is

592

5

Qðk , P ; BÞ ¼

Polarization Theory of Relativistic Nuclear Reactions

1 2

Z

1

1

Qðk  , P ; BÞ d cos θk

ð5:1385Þ

Referring to the Pauli incompatibility operator in the c.m. system described previously, according to Eqs. (5.1383)–(5.1385), their analytical expressions may be further found.

5.17.3

Calculation Formulas for T Matrix Elements and Nuclear Matter Properties in Symmetric Nuclear Matter

According to Eq. (5.1279) the Thompson equation in nuclear matter can be written in the c.m. system as     Z * * * * T q 0, q ¼ V q 0, q þ

*  ! 2 Qðk ; BÞ !  dk M ! ! V q 0, k ð5:1386Þ   T k, q 3 2E  2E E 2 ð2πÞ q k k

  * * where Qðk ; BÞ is the angle averaging Pauli incompatibility operator. T q 0 , q is *

the T matrix element in the c.m. system. The speed β of the center of momentum * * * relative to the n.m. system has been given by Eq. (5.1342). Using the J ¼ L þ S angular momentum coupling mode, the method used previously to study the Bonn one meson exchange potential can also be used to convert Eq. (5.1386) into a one-dimensional equation of relative momentum k, and then we can use the inverse matrix method to solve the obtained equation. Let I ¼ 0, 1 be the total isospin of the two-nucleon system, and the anti-symmetry requires ð1ÞLþSþI ¼ 1

ð5:1387Þ

where L is the orbital angular momentum. As mentioned previously, for symmetric nuclear matter and in the on-shell case, there are only five linear independent helicity T matrix elements, so that the T matrix elements can be expanded with five linearly independent covariables. The following 5 Fermi covariables are usually used [87, 103, 105, 106]: S ¼ 1,

V ¼ γ μ1 γ 2μ ,

T ¼ σ μν 1 σ 2μν ,

A ¼ γ 51 γ μ1 γ 52 γ 2μ ,

P ¼ γ 51 γ 52

ð5:1388Þ

We also define the following five exchange Fermi covariables: e e ¼e S¼e SS, V SV,

e¼e T ST,

e¼e A SA,

e¼e P SP

ð5:1389Þ

5.17

Nucleon Relativistic Microscopic Optical Potential Based. . .

593

where e S is the exchange covariable of S, whose role is to exchange the Dirac footmarks of particles 1 and 2, for example, e Suð1Þσ uð2Þτ ¼ uð1Þτ uð2Þσ . The Fierz transform relations satisfied between 5 exchange Fermi covariables and 5 original Fermi covariables are as follows [105–107]: 0 1 0 1 e 1 S S B 1 BeC B C B BV C BV C B 4 2 B C B C B BT C ¼ F B T C ¼ 1 B 12 e 0 B C B C 4B BeC B C B @AA @AA B 4 2 @ e P P 1 1 0

1 2 0

1

2

0

0 1 2

2

2

1

1 0 1 1 C S CB C 4 CB V C CB C B C 12 C CB T C CB C 4 C@ A A A P 1

ð5:1390Þ

Equation (5.1388) has given the specific form of the Fermi covariant operators Γ i ¼ {S, V, T, A, P}, and Eq. (5.1389) can be verified from Eq. (5.1390) using the γ matrix commutative relations. According to the Hartree-Fock theory, the t matrix whose isospin is I should be written as t I ðq, θÞ ¼ t I,dir ðq, θÞ  t I,exc ðq, θÞ *

ð5:1391Þ

!

where θ represents the angle between q and q 0. The direct item can be expanded as follows: t I,dir ðq,θÞ ¼

1 I F ðq,θÞS þ F IV ðq,θÞ V þ F IT ðq,θÞT þ F IA ðq,θÞA þ F IP ðq,θÞP 2 S ð5:1392Þ

!

!

If q corresponds to an angle θ, then  q corresponds to an angle π  θ, and the two-nucleon system with isospin I ¼ 1 is symmetric and that with isospin I ¼ 0 is anti-symmetric. We can write [103, 106] h 1 I e S þ F IV ðq, π  θÞ V F ðq, π  θÞe 2 S i e þ F I ðq, π  θÞP e þ F IA ðq, π  θÞA e þF IT ðq, π  θÞT P

t I,exc ðq, θÞ ¼ ð1ÞIþ1

ð5:1393Þ

The five Fermi covariables given by Eq. (5.1388) are not unique. For example, P is a pseudo-scalar quantity, but P can also be replaced by a normal axis vector or a complete axis vector form [103, 106]. In the actual calculation, it was found that if all meson exchange potentials V were converted to effective interaction T, one could not obtain the satisfactory results, so it was proposed to divide T into two parts [96, 97]:

594

5

Polarization Theory of Relativistic Nuclear Reactions

T ¼ V þ ΔT

ð5:1394Þ

where V is the bare nucleon-nucleon interaction. Since the one π exchange plays a major role in the NN scattering process, in order to obtain satisfactory calculation results, the subtracted T matrix method is proposed [103, 106]. For the π and η mesons the bare NN interaction represented by a complete pseudo-vector is used, while for other mesons the effective interaction represented by a pseudo-scalar is used, namely, T ¼ T sub þ V π,η

ð5:1395Þ

The Pauli operator Q is related to speed β, so the T matrix element derived from Eq. (5.1386) is also related to speed β. Section 5.15 has clearly pointed out that the T matrix approximately satisfies the Lorentz invariance. Therefore, in the projection method discussed earlier in this section, the expansion of the T matrix element with five linear independent covariables can only be regarded as an approximate method. Strictly speaking, the results obtained by this method may lose the non-Lorentz invariant component of the T matrix element, maybe which will cause the wrong calculation result to some extent. The above method for solving the T matrix is performed in the c.m. system. The bare nucleon-nucleon interaction V obtained by the method described in Sect. 5.16 is the Lorentz invariant, but the entire process from the interaction V to the T matrix cannot be guaranteed to be Lorentz invariant. Therefore, in general the T matrix found in the c.m. system is not applicable directly in the n.m. system. The coordinate system transformation of the calculation results can be considered according to the method described in Sect. 5.2. Of course, it can also be done that the Thompson equation is solved directly in the n.m. system like Ref. [89], but the Pauli incompatibility operator of the n.m. system given earlier in this section should be used. After the T matrix is solved in the nuclear matter, the nucleon self energy Σ(k) in the nuclear matter can be calculated by Eqs. (5.1337)–(5.1339). Note that we use the uu ¼ 1 plane wave normalization method, the nucleon spinor field given by qffiffiffi Eq. (5.754) or (5.924) contains the factor

M E,

so the relativistic single particle

potential in nuclear matter is U ðk Þ ¼

X M 2 M Re hkqjT jkq  qk i  Re hk jΣ ðk Þjk i ¼ Ek Ek Eq qk

ð5:1396Þ

F

If Σ v(k) is ignored, there is Σ(k) ffi Σ s(k) þ γ 0Σ 0(k); in this case the Σ 0(k) item corresponds to the uþu ¼ 1 normalization method, so from Eq. (5.1396) we can obtain

5.17

Nucleon Relativistic Microscopic Optical Potential Based. . .

U ðk Þ ¼

M s Σ ðk Þ þ Σ 0 ðk Þ E k

595

ð5:1397Þ

The energy density in nuclear matter comes from kinetic energy and half potential energy, and there is ε¼

i *  X M  *  h! ! 1 u k , λ γ  k þ M þ Σ ð k Þ u k, λ Ek 2 kλ

ð5:1398Þ

The energy of each nucleon in nuclear matter is called binding energy, and its expression is Eb ¼

E ε M M ¼ A ρB

ð5:1399Þ

In the on-shell case, the single particle energy without static mass contribution in nuclear matter can be obtained by Eq. (5.1320) as

1=2 þ Σ 0 ðk Þ  M εsp ¼ ð1 þ Σ v ðkÞÞ k 2 þ M  ðkÞ

ð5:1400Þ

At the same time, the pressure p can be calculated using Eq. (5.948), and the incompressible coefficient K can be calculated using Eq. (5.951).

5.17.4

Nucleon Self-Energy in Asymmetric Nuclear Matter [103, 104]

The Fermi momentum of neutrons and protons in symmetric nuclear matter are equal. However, in asymmetric nuclear matter, neutrons and protons occupy the Fermi spheres of different sizes. Therefore they also have different Pauli incompatibility operators and correspond to different effective masses and self-energies. For asymmetric nuclear matter, it is necessary to distinguish between nn, pp, np three reaction channels. For the np reaction channel, taking into account the antisymmetry effect, the following decompositions on the interaction V and effective interaction T can be done: exc V np ¼ V dir np  V np ,

exc T np ¼ T dir np  T np

ð5:1401Þ

V exc np ¼ hnpjvjpni

ð5:1402Þ

where V dir np ¼ hnpjvjnpi, and there are the similar expressions for T.

596

5

Polarization Theory of Relativistic Nuclear Reactions

In the BS equation of symmetric nuclear matter given by Eq. (5.1243), iG represents the propagator of two free nucleons, which can be formally represented by iGG. For nuclear matter, it is also necessary to add a Pauli operator Q, v and t both need to take their helicity matrix elements, and then Eq. (5.1243) can be written in a simplified form as Z T ¼V þi

VQGGT

ð5:1403Þ

For asymmetric nuclear matter, the following BS equations need to be processed simultaneously: Z T nn ¼ V nn þ i

V nn Qnn Gn Gn T nn

ð5:1404Þ

V pp Qpp Gp Gp T pp

ð5:1405Þ

Z T pp ¼ V pp þ i Z dir T dir np ¼ V np þ i

Z dir V dir np Qnp Gn Gp T np þ i

Z T exc np

¼

V exc np

þi

exc V exc np Qpn Gp Gn T np

ð5:1406Þ

exc V dir np Qnp Gn Gp T np

ð5:1407Þ

Z dir V exc np Qpn Gp Gn T np

þi

where Vij(i, j ¼ n, p) represent the one boson exchange potential between bare nucleons. Assuming that in the np reaction channel neutrons and protons have the same dir average effective mass, from Eq. (5.1393) it can be seen that the relations of T dir np , V np exc and T exc are established through the Fierz transformation given by np , V np Eq. (5.1390), so Eq. (5.1406) and Eq. (5.1407) can be reduced to Z T np ¼ V np þ i

V np Qnp Gn Gp T np

ð5:1408Þ

According to Eq. (5.1279) it can be written that the effective Thompson propagator in nuclear matter is iGi Gj ¼

 M i M j 1   pffiffiffiffi   Ei Ej s  Ei  Ej þ iε

ð5:1409Þ

The Fermi momentum kFn and k Fp of neutrons and protons in asymmetric nuclear matter are different, so two Fermi ellipsoids of different sizes are considered for the Pauli incompatibility operator of the np reaction channel. ! For neutrons and protons, their baryon flow speed β and the corresponding γ are the same, and it is also noted that the effective masses of neutrons and protons are

5.17

Nucleon Relativistic Microscopic Optical Potential Based. . .

597

assumed to be equal for the np reaction channel. So the following relations can be introduced for i ¼ n, p according to Eqs. (5.1357), (5.1359), (5.1361)–(5.1364): qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2Fi  β2 E2 Fi   k vi ¼ γ kFi  βE Fi   kþi ¼ γ kFi þ βEFi

ð5:1411Þ

ki ¼ min ðkui , kvi Þ

ð5:1413Þ

γE k  E Fi γβk

ð5:1414Þ

ð5:1410Þ

kui ¼

cos αi ¼

8 > < 0,  Qi ðk ; BÞ ¼ θð cos αi  cos θk Þ θð cos θk þ cos αi Þ, > : 1,

ð5:1412Þ

k < ki ki  k  kþi ð5:1415Þ k > kþi

The angle averaging Pauli incompatibility operator of the np reaction channel in the c.m. system can be approximated as Qnp ðk  ; BÞ ¼

1 2

Z

1

1

Qn ðk  ; BÞQp ðk ; BÞdcosθk

ð5:1416Þ

According to Eq. (5.1323), the neutron self-energy and proton self-energy can be formally written as follows, respectively: Z

Z



½trðGn T nn Þ  Gn T nn  i tr Gp T np  Gp T np

Σ n ¼ i Z

Fn

Σ p ¼ i

ð5:1417Þ

Fp





tr Gp T pp  Gp T pp  i

Fp

Z



 tr Gn T np  Gn T np

ð5:1418Þ

Fn

The asymmetry degree of nuclear matter is defined as β¼

ρn  ρ p ρ

ð5:1419Þ

where ρn, ρp, ρ are neutron, proton, and total nuclear density in nuclear matter, respectively. The binding energy Εb(ρ, β) is a function of ρ and β, and it can be expanded as

 Ε b ðρ, βÞ ¼ Ε b ðρÞ þ Εsym ðρÞβ2 þ O β4

ð5:1420Þ

598

5

Polarization Theory of Relativistic Nuclear Reactions

From the above formula, the symmetric energy can be obtained as  2  1 ∂ Eb ðρ, βÞ  Ε sym ðρÞ ¼  2  ∂β2

ð5:1421Þ β¼0

If the β4 term and the more high order terms of β are ignored in Eq. (5.1420), we can obtain Ε sym ðbÞ ffi Ε b ðρ, β ¼ 1Þ  Εb ðρ, β ¼ 0Þ

5.17.5

ð5:1422Þ

Nucleon Relativistic Microscopic Optical Potential Based on DBHF Theory

The Dirac-Brueckner-Hartree-Fock (DBHF) theory described above can be used to calculate the properties of nuclear matter and the nucleon self-energy in nuclear matter. The real and imaginary parts of the nucleon self-energy given by Eq. (5.1318) correspond to the real and imaginary parts of the nucleon optical potential in the nuclear matter. The nuclear matter saturation density ρ0~0.16 fm3, the DBHF numerical calculation at low density will encounter divergence problems, in Ref. [103] only the calculation results of the ρ  0.08 fm3 range are given. If we want to use the local density approximation to calculate the microscopic optical potential of finite nuclei, we will encounter the problem of how to select the self-energy of the low-density region in nuclear matter. In order to solve this problem, people usually use a so-called relativistic mean field method with density-related coupling constants [108–114]. The general approach is to adjust the coupling constants of the σ, ω, etc. mesons in the relativistic mean field theory to fit the self-energy obtained by the microscopic DBHF calculation at this density. Then using the density-related coupling constants obtained above and the local density approximation, the relativistic mean field method can be used to calculate the ground state properties [108–113] or the microscopic optical potential [114] of the finite nucleus. After the density- related meson coupling constants are determined, the finite nucleus equation can be solved directly in the relativistic mean field theory. In the Ref. [114] the density-related meson coupling constants are obtained only from the real part of the microscopic optical potential calculated by DBHF theory. In fact, the density-related meson coupling constants can be obtained simultaneously from the real and imaginary parts of the microscopic optical potential in the asymmetric nuclear matter. How to solve finite nucleus problems directly by DBHF theory is a theory to be further developed in the future. In addition, in Ref. [115] the off-shell self-energy of asymmetric nuclear matter was also studied using DBHF theory.

5.18

Proca Relativistic Dynamics Equation of Spin 1 Particle. . .

5.18

599

Proca Relativistic Dynamics Equation of Spin 1 Particle and Its Application in Elastic Scattering Calculations

Deuteron is a weakly bound state vector particle with spin 1 consisting of neutron and proton. In order to calculate the cross sections, differential cross sections, and various polarization physical quantities of the elastic scattering and other reactions induced by the deuterons, we first need to study the relativistic dynamics equation of the particles with spin 1. We do not consider the limited size and internal structure of the deuterons for the time being; we only think of the deuteron as a point particle. In addition, the W boson is also a particle with spin 1 and non-zero mass, while the photon is a particle with spin 1 and mass 0. This section discusses the Proca relativistic dynamics equation that describes the particles with spin 1.

5.18.1

Proca Equation of Free Particles

The physical quantity of the vector particle field with spin 1 can be described by ψ μ(x), and the corresponding free vector particle Lagrangian density can be expected to be [116, 117] 1 1 L ¼  F μν F μν þ m2 ψ μ ψ μ 2 4

ð5:1423Þ

where the field strength tensor is μ

ν

F μν ¼ ∂ ψ ν  ∂ ψ μ

ð5:1424Þ

Substituting Eq. (5.1423) into the Lagrange equation (5.612), we can get ∂μ F μν þ m2 ψ ν ¼ 0

ð5:1425Þ

This is the Proca equation satisfied by the free particles with spin 1 [116–118]. In order to eliminate the additional components of the particle field, the following Lorentz gauge of the vector particles given by Eq. (5.642) can be used: ∂ν ψ ν ¼ 0

ð5:1426Þ

At this time, Eq. (5.1425) becomes the following Klein-Gordon equation:

 μ ∂μ ∂ þ m2 ψ ν ¼ 0

ð5:1427Þ

However, when there is an interaction potential, Eq. (5.1426) and Eq. (5.1427) do not necessarily hold.

600

5

5.18.2

Polarization Theory of Relativistic Nuclear Reactions

Proca Equation with Interaction Potential

We assume that the interaction potential of the point deuteron and the nucleus consists of the scalar potential S and the time-like component V of the vector potential, and the interaction potential is added in the following form [118]: ∂μ ! Dμ ¼ ∂μ þ iVδμ0

ð5:1428Þ

e ¼mþS m!m

ð5:1429Þ

∂ , so Eq. (5.1428) is equivalent to E ! E  V. In Equation (5.5) has given iE ! ∂t this way, the Proca equation (5.1425) can be rewritten as

e 2ψ ν ¼ 0 Dμ ðDμ ψ ν  Dν ψ μ Þ þ m

ð5:1430Þ

Referring to the electromagnetic field theory, the wave function ψ can be expressed as ψ¼

φ

!

!

A

ð5:1431Þ

!

where φ and A are the scalar component and vector component of the electromagnetic field, respectively. For ν ¼ 0 there is ψ 0 ¼ φ, and note that the μ ¼ ν item in Eq. (5.1430) has no contribution. Let us assume that S and V are both spherical symmetrical, and note that 

 ! ! ∂ dV ! þ iV ψ i ¼ i∂i ðE  V Þ ψ i ¼ i ^r  A þ iðE  V Þ∇  A Di D ψ ¼ ∂i dr ∂t 0

i

Let ω¼EV

ð5:1432Þ

1 dV ω dr

ð5:1433Þ

Θ¼ then from Eq. (5.1430) we can get 

!2

e ∇ m

 2

 * * * φ ¼ iω Θ ^r  A  ∇  A

For ν ¼ i ¼ 1, 2, 3 items, first note

ð5:1434Þ

5.18

Proca Relativistic Dynamics Equation of Spin 1 Particle. . . μ

Dμ D ψ



i

μ6¼i





μ

¼ ∂μ þ iVδμ0 ∂ þ iVδμ0





ψ i μ6¼i

 ¼

601

 !2 ω  ∇ ψ i þ ∇i ψ i , 2

!2

X* *  Dμ Di ψ μ μ6¼i ¼ iω ∂i ϕ þ ∇j ∇i ψ j , X

!2

∇i ψ i þ

X! ! ∇j ∇i ψ j

j6¼i

! ¼

j6¼i

i

X! ! !  ! * ∇i ∇j ψ j ¼ ∇ ∇  A ij

then from Eq. (5.1430) we can obtain 

 *2 *  * * * * e 2 A ¼ ∇ ∇  A  iω∇φ ∇ þ ω2  m

ð5:1435Þ

*

In order to eliminate the ^r  A item in Eq. (5.1434), let *

A¼

E* A ω 1

ð5:1436Þ

then we can obtain *

*

∇A ¼ **

* E* * E dV * E* * E ^r  A 1 ¼ ∇  A 1 þ Θ^r  A 1 , ∇  A1 þ 2 dr ω ω ω ω

∇A ¼

E ** E dV * E ** E * ^r A ¼ ∇ A 1 þ Θ^r A 1 ∇A1 þ 2 ω ω ω dr 1 ω

So Eq. (5.1434) can be rewritten as  2  * * * e 2 φ ¼ iE ∇  A 1 ∇ m

ð5:1437Þ

There are the following formulas: * 1

∇

r

¼

*

r , r3

*

*

∇ r ¼3

ð5:1438Þ

so we can get *

*

∇  ^r ¼ ∇  *2

*

*

* !  1 * r 1* * 2 ¼∇  r þ ∇ r ¼ r r r r

Notice ∇ ¼ ∇  ∇; we can also find

ð5:1439Þ

602

5

Polarization Theory of Relativistic Nuclear Reactions

 * 2* *  * * * E * * ** E * E E * 2* ∇ A ¼ ∇  ∇A ¼ ∇  ∇ A 1 þ Θ^r A 1 ¼ ∇ A 1 þ Θ^r  ∇ A 1 ω ω ω ω þ

** E 2* E * E2 * E Θ A 1 þ Θ0 A 1 þ ΘA 1 þ Θ^r  ∇ A 1 ω ω ωr ω

ð5:1440Þ

where Θ0 ¼

dΘ dr

ð5:1441Þ

And notice *

*

∇ ^r  A 1



*

*

r  A1 ¼∇ r *

! ¼

    r * * 1* * * r  A r  A þ ∇ 1 1 r r3 *

ð5:1442Þ

so we can obtain   * *  *  * * * E * * *  E* * * E E ∇ ∇  A ¼ ∇ ∇  A 1 þ Θ^r  A 1 ¼ ∇ ∇  A 1 þ Θ^r ∇  A 1 ω ω ω ω  *   *   *  * E E 1 E * E þ Θ2^r ^r  A 1 þ Θ0^r ^r  A 1  Θ ^r ^r  A 1 þ Θ∇ ^r  A 1 ω ω r ω ω ð5:1443Þ Substituting Eq. (5.1440) and Eq. (5.1443) into Eq. (5.1435), we can obtain  2   * *   * ** *  * 2 * e 2 A 1 ¼  Θ0 þ Θ2 þ Θ A 1  2Θ ^r  ∇ A 1 þ ∇ ∇  A 1 ∇ þ ω2  m r    * *  * * *  Θ Θ ** * þ r ∇  A 1 þ ∇ r  A 1 þ Θ0 þ Θ2  ^r ^r  A 1 þ R r r ð5:1444Þ *

R ¼ i

ω2 * ∇φ E

*

ð5:1445Þ *

If A 1 is known, φ can be solved from Eq. (5.1437), and R can be obtained using Eq. (5.1445). Introduce the following operator: *2

e2 Z^  ∇ þ ω2  m

ð5:1446Þ

5.18

Proca Relativistic Dynamics Equation of Spin 1 Particle. . .

603

e 2 will offset Since in the above formula, E2 contained in ω2 and m2 contained in m 2 ^ each other, Z can be approximated as much less than ω when the energy is not very high. Eq. (5.1445) can be formally rewritten, according to Eq. (5.1437), as *

*

R ¼ ω2 ∇

5.18.3



1 * * ∇  A1 ω2  Z^

 ð5:1447Þ

Some Expressions Related to S=1 Spin Operators [119]

Equation (3.51) has given that in the rectangular basis representation, the matrix element of the spin operator of particle with spin 1 in the rectangular coordinate system is

^Sk

ij

¼ iεkij ¼ iεijk

ð5:1448Þ

1.7). ^ S is where εijk is a three-dimensional anti-symmetry unit tensor given by a  Eq. ( * * * * ! S r A S r ^ 3  3 matrix, and r and A are three-dimensional vectors; therefore ^ is also a three-dimensional vector, and using Eq. (5.1448) we can write its vector element as h   i * * ^S  * r ^S  r A ¼ εijk εjlm r k r m Al i

ð5:1449Þ

There is the following equation [120]: εrmn εrst ¼ δms δnt  δmt δns

ð5:1450Þ

The left end of the above formula implies summation to r. So Eq. (5.1449) can be rewritten as h

^S  * r



*i ^S  * r A i ¼ r k r k Ai  r k r i Ak

ð5:1451Þ

Turn the above formula into vector form as  *    * * * * r r  A ¼ r 2 A  ^S  r ^ S r A

* *

ð5:1452Þ

604

5

Polarization Theory of Relativistic Nuclear Reactions

There is also the following vector cross multiplication formula: * * C  D ¼ εijk C j Dk i

ð5:1453Þ

Using the above formula we can write h*  *i * ∇ r A ¼ εijk ∇j εklm r l Am i

ð5:1454Þ

Noting pj ¼  i∇j and using Eq. (5.1448), the right end of the above formula can be rewritten as εijk ∇j εlkm r l Am ¼ i Sj ik pj Sl km r l Am Substituting above formula into Eq. (5.1454), the following vector relation can be obtained:     * * * ! * * S r A ∇  r  A ¼ i ^S  p ^

ð5:1455Þ

Using the relation pixj ¼ xjpi  iδij, we can find

** * * ** *! p r ¼ pi e i xj e j ¼ xj pi  iδij e i e j ¼ r p  i e i e i

!*

2 And noting ^S ¼ 2, we can write



     * * ! ^S  ! S  p  2i p ^S  r ¼ ^S  r ^

ð5:1456Þ

In this way, Eq. (5.1455) can be rewritten as     * * * * * ! * S  p A  2A ∇  r  A ¼ i ^S  r ^

ð5:1457Þ

 ! * * Noting the orbital angular momentum L ¼ i r  ∇ , so we can obtain 

* * * * r  ∇  A ¼ iL  A

*

ð5:1458Þ

and we can write * * i L  A ¼ iεijk Lj Ak ¼ iεjik Lj Ak i

ð5:1459Þ

5.18

Proca Relativistic Dynamics Equation of Spin 1 Particle. . .

605

So the following relation can be obtained according to Eqs. (5.1448), (5.1458), and (5.1459): 

 ** * * r  ∇  A ¼ ^S  L A

*

ð5:1460Þ

There is the following vector identity: * *

 * * * * * * * * * * * * * C ∇D ¼ C∇ Dþ∇ CD  D∇ CþD ∇C ð5:1461Þ

And notice * *  * * * d * * ** * xk e k ¼ A i e i  e j e j ¼ A A  ∇ r ¼ Ai e i  e j dxj

ð5:1462Þ

using the above formula from Eq. (5.1461), we can get  * *    * * * * * * * r ∇  A ¼ 2A þ r  ∇ A þ ∇  r  A

*

ð5:1463Þ

Substituting Eq. (5.1457) into the above formula, we have  * *      * * * * ! * S p A r ∇  A ¼ r  ∇ A  i ^S  r ^

*

ð5:1464Þ

From Eq. (5.1453) we can write h

 * *i r  ∇A ¼ εijk r j εklm ∇l Am ¼ εkij εklm r j ∇l Am

!

i

ð5:1465Þ

Using Eq. (5.1450) we can also write εkij εklm ¼ δil δjm  δim δjl þ δml δij  δml δij ¼ εikm εkjl þ δml δij  δim δjl So Eq. (5.1465) can be rewritten as h

 * *i

 ¼ εkjl εikm þ δij δml  δim δjl rj ∇l Am r  ∇A

*

i

ð5:1466Þ

Write the above formula by vector form as  * *      * * * * * * * * * r  ∇A ¼ r ∇ Aþ r ∇A  r ∇ A

*

ð5:1467Þ

606

5

Polarization Theory of Relativistic Nuclear Reactions

Utilizing Eq. (5.1460) and Eq. (5.1464), from Eq. (5.1467), we can get  * *  **    * ! * r  ∇  A ¼ ^S  L A  i ^S  r ^ S p A

*

ð5:1468Þ

There is also the following vector identity: * * * * * * *  * * *  * * * * *  ∇ CD ¼ D∇ Cþ C∇ DþD ∇C þC ∇D ð5:1469Þ *

*

Noting there is ∇  r ¼ 0, from the above formula we can write * *     * * * * * * * * * * ∇ r A ¼ A∇ r þ r ∇ Aþ r  ∇A

ð5:1470Þ

From the above formula, the following relation can be obtained using Eqs. (5.1462), (5.1467), and (5.1463) as     * * * * * * * * * * * * * ∇ r  A ¼ 3A þ r  ∇ A þ r  ∇  A þ ∇  r  A

ð5:1471Þ

Substituting Eq. (5.1457) and Eq. (5.1460) into the above formula, we can obtain    **    * * * * * * * * ! * ∇ r  A ¼ A þ r  ∇ A þ ^S  L A  i ^ S r ^ S p A

5.18.4

ð5:1472Þ

Proca Equation in the Form of Schrodinger-like Equation

Equation (3.927) gives the following expression:  *  *  * *    * 1 * * * * * * * T2 a, b  T2 c , d ¼ a  c b  d  ab  cd 2    1 * * * *  ab cd 3 *

*

*

*

*

ð5:1473Þ

*

If let a ¼ b ¼ ^S, c ¼ A , d ¼ B in the above equation, we can get  * * * *   i 2 * * T AB ¼ ^S  A ^S  B  ^S  A  B  AB 2 3

ð5:1474Þ

Proca Relativistic Dynamics Equation of Spin 1 Particle. . .

5.18

*

*

607

*

If taking A ¼ B ¼ r , we can define the following spatial-spatial tensor operator: h

 2 2i T RR ¼ r 2 ^S  ^r  3 *

*

*

ð5:1475Þ

*

If taking A ¼ r , B ¼ p , we can define the following spatial-momentum tensor operator:      i * 2 * * * * rp T RP ¼ ^S  r ^S  p  ^S  L  3 2      * i 2i * ! * * ¼ ^S  r ^S  p  ^S  L þ r ∇ 2 3

ð5:1476Þ

k 2 ¼ E 2  m2

ð5:1477Þ

Let

*2

and note ∇ ¼ p2 ; dividing Eq. (5.1444) by (2E) we can obtain 

 h  i* p2 1 2 2 2 2 0 e þk  Θ þΘ Θþ A1 þ ω þ m 2E 2E r     h * * * i 1 2Θ * * * 1 * * * 1 Θ ** *  r  ∇ A1 þ ∇ ∇  A1 þ r ∇  A1 þ ∇ r  A1 2E r 2E 2E r h  i  *  * k2 * 1 R 1 ^r ^r  A 1 þ ¼ A þ Θ0 þ Θ Θ  2E 2E 1 2E r ð5:1478Þ

k2 is equivalent to non-relativistic kinetic energy. 2E The following three items appeared in Eq. (5.1478) are first combined by using Eqs. (5.1464), (5.1472), and (5.1476):

When the energy is very low,

       * * *  * * * * * * * * * * * S  p A1 þ A1 S r ^ r ∇  A1 þ ∇ r  A1  2 r  ∇ A1 ¼ r  ∇ A1  i ^    **      * * * * * * * * * þ r  ∇ A 1 þ ^S  L A 1  i ^S  r ^S  p A 1  2 r  ∇ A 1  **    * * * * ¼ A 1 þ ^S  L A 1  2i ^S  r ^S  p A 1   **     * 2i * * * i ^ * ^ ¼ A 1 þ S  L A 1  2i T RP þ S  L  r  ∇ A1 3 2     * * * 4 * * * ¼ A 1 þ 2 ^S  L A 1  r  ∇ A 1  2iT RP 3 ð5:1479Þ

**

608

5

Polarization Theory of Relativistic Nuclear Reactions

According to Eq. (5.1452) and Eq. (5.1475), we have  * h

2 i* T * 1* ^r ^r  A 1 ¼ 1  ^S  ^r A 1 ¼  RR A1 þ A1 2 3 r

ð5:1480Þ

Then merge some of the central potentials appeared previously as follows: h  i h  i h  i 2 Θ 1 1 2 2  Θ0 þ Θ Θ þ þ þ Θ0 þ Θ Θ  ¼  Θ0 þ Θ Θ þ r r 3 r 3 r ð5:1481Þ Using the result given by Eq. (5.1479)–(5.1481), Eq. (5.1478) can be rewritten as 

     * * T RR p2 * * ^ þ U c þ U SO S  L þ U D r  ∇ þ U RR 2 þ iU RP T RP A 1 2E r   1 * * k2 * þ QþR ¼ A 2E 2E 1

ð5:1482Þ

where Uc, USO, UD, URR, URP are the central potential, spin-orbit coupling potential, Darwin potential, spatial-spatial tensor potential, and spatial-momentum tensor potential, respectively. Their expressions are as follows: Uc ¼

n h  io 1 2 2 e 2 þ k 2  Θ0 þ Θ Θ þ ω2 þ m 2E 3 r 1 2Θ 2E r 1 4Θ UD ¼  2E 3r h  i 1 1 ¼ Θ0 þ Θ Θ  2E r U SO ¼

U RR

U RP ¼ 

1 2Θ 2E r

ð5:1483Þ ð5:1484Þ ð5:1485Þ ð5:1486Þ ð5:1487Þ

And there is ** *  * Q ¼ ∇ ∇  A1 *

ð5:1488Þ

where R has been given by Eq. (5.1445). Equation (5.1482) is called the Proca equation in the form of Schrodinger-like equation. In addition, it is also necessary to artificially add Coulomb potential. The deuteron Coulomb potential VC can also be 1 discussed like the discussion to the spin particle Coulomb potential in Sect. 5.3. 2 Then one can let U c ! U c þ Eμ V C in equation (5.1482), where μ is the deuteron

5.18

Proca Relativistic Dynamics Equation of Spin 1 Particle. . .

609

reduced mass. If the other components of the optical potential are added in addition to the scalar potential S and the time-like component V of the vector potential in the nuclear potential, the resulting Proca equation also needs to be changed accordingly.

5.18.5

Discussion on the Application of the Proca Equation to Calculate the Elastic Scatterings Between Deuteron and Nucleus

When the Proca equation is used to calculate the elastic scattering of the deuteron and the nucleus, the deuteron phenomenological optical potential can be directly used, and the deuteron optical potential can also be obtained from the nucleonnucleon interaction potential using the folding model. The contribution of the deuteron breakup channel can be or cannot be considered in the folding model. It is also possible to use the deuteron microscopic optical potential obtained from effective nuclear force, realistic nuclear force, or meson exchange coupling constants. Equation (5.1482) Eq. (5.1437) constitute a set of simultaneous equations for  and *  * * the 4-vector field φ, A 1 . If we make Q ¼ R ¼ 0 in Eq. (5.1482), it is equivalent to ** *  * * ignore ∇ ∇  A 1 items and φ items. In fact, as long as we make ∇  A 1 ¼ 0, we *

*

will get Q ¼ R ¼ 0 simultaneously. At this time, Eq. (5.1482) gives three equations * of the three components of A 1 ; they are not coupled to each other. If the incident particle direction is selected as the z-axis, the entire system is axis-symmetric, so * only the two equations of the radial component and the polar angle component of A 1 need to be considered in the spherical coordinate system. For the two variables we can use mathematical methods such as transforming wave functions and separating variables as we solve the Schrodinger equation in the non-relativistic case. When solving the radial equation, the S matrix can be introduced to give the scattering amplitude, and then the differential cross sections and various polarization physical quantities of the elastic scattering of the deuteron with nucleus can be calculated. Sometimes it is thought that in Eq. (5.1482) the Darwin term and two tensor terms dσ contribute less and can be ignored. In the Ref. [118] the differential cross section , dΩ the vector analyzing power Ay, and the tensor analyzing power Ayy of the elastic scattering of the 400 MeV d þ 58Ni and 700 MeV d þ 40Ca in the laboratory system were calculated by using Proca equation. 58Ni and 40Ca are the target nuclei with spinless. According to the theory introduced in Chap. 3 of this book, the target nucleus with non-zero spin can also be calculated. *

*

When ∇  A 1 6¼ 0, it can be seen from Eqs. (5.1482) and (5.1437) that the three * components of A 1 and the scalar wave function φ satisfy the coupling equations. In the axis-symmetric system, there are also three wave function components to be

610

5 Polarization Theory of Relativistic Nuclear Reactions

coupled and solved. And there is the radial variable in each wave function component. In this case, how to solve the numerical problems and how to calculate the differential cross sections and various polarization physical quantities of the nuclear reaction are the topics that need to be further studied. In the Proca equation, the wave function describing the field physical quantity is * represented by a 4-vector field consisting of a scalar field φ and a vector field A . There is also a Kemmer-Duffin-Petiau (KDP) equation [118, 119, 121] that can be used to describe spin 1 particles, where the wave function describing the field physical quantity  is* represented  by a 10-dimensional vector formed by a column * * * constituted by φ, A , E , B . We know that in electromagnetic field theory E and *

*

B can be derived from φ and A . According to convention, all kinds of physical * * quantities should be able to be derived from the wave function, so putting E and B belong to the derived quantities into the wave function as a basic physical quantity is not consistent with the convention. In fact, when solving the KDP equation, it will be * ultimately converted into the equations of φ and A to be solved.

5.19

Weinberg Relativistic Dynamics Equation with Spin 1 Particles and Discussion on its Application

5.19.1

Weinberg Equation of Free Particles

1 nucleon are represented by 2  2 matrices. In the 2 Dirac equation, the wave function is divided into upper and lower components, so the entire spin space is described by 4  4 γ matrices. And in the case of low energy approximation, the contribution of the lower component can be ignored. Then the obtained equation will automatically degenerate into 2  2 spin space with only the upper component. The spin operators of the spin 1 particle are represented by 3  3 matrices. In the Weinberg equation [122, 123] describing the spin 1 particles, the wave function is also expressed as The spin operators of the spin

*

ψ¼ *

!

A * B

ð5:1489Þ

*

where A and B are the upper and lower components of the wave function, and both are three-dimensional vectors. The Weinberg equation can be written as [116–118]

 pμ pν γ μν  m2 ψ ¼ 0

ð5:1490Þ

5.19

Weinberg Relativistic Dynamics Equation with Spin 1 Particles. . .

611

where m is the particle stationary mass. γ μν is 6  6 matrix and their specific expressions are γ

00

^I 3 ^0 0^ ^I 3

¼

γ 0i ¼ γ i0 ¼ γ ¼ ij

^Si ^Sj þ ^Sj ^Si  δij^I 3 0^

! ð5:1491Þ

^0 ^Si

^Si ^0

! ð5:1492Þ

^0

  ^Si ^Sj þ S^j ^ Si  δij^I 3

! ð5:1493Þ

where ^I k is a unit k  k matrix, ^Si is spin operator of spin 1 particle. Let us do the following matrix decomposition: pμ pν γ μν ¼



Γ 11

Γ 12

Γ 21

Γ 22

 ð5:1494Þ

where Γ ij(i, j ¼ 1, 2) is 3  3 matrix. For the stationary systems, according to Eq. (5.61) we can see   ! ðp0 , p1 , p2 , p3 Þ ¼ E,  p

ð5:1495Þ

The following expressions can be obtained using Eqs. (5.1491)–(5.1495):     ! 2 !2 ! 2 Γ 11 ¼ E 2 þ 2 ^S  p  p ¼ 2 ^ S  p þ m2

ð5:1496Þ

! Γ 12 ¼ 2E ^S  p

ð5:1497Þ

!

Γ 21 ¼ 2E ^S  p        ! 2 !2 ! 2 2 2 ^ ^ ¼ E þ2 S p  p ¼ 2 S p þm 

Γ 22

ð5:1498Þ ð5:1499Þ

Substituting Eq. (5.1494) into Eq. (5.1490) we can get 

Γ 11  m2

Γ 12

Γ 21

Γ 22  m2

We can also rewrite the above formula as



*

A * B

! ¼0

ð5:1500Þ

612

5

1 2



Polarization Theory of Relativistic Nuclear Reactions

Γ 12 ðΓ 11  m2 Þ Γ 21 Γ 22  m2



*

!

A

*

¼0

ð5:1501Þ

B

Then the following equation can be obtained substituting Eqs. (5.1496)–(5.1499) into Eq. (5.1501): 0  2 ^ ! B S  p B   @ ! E ^S  p

1   ! ! E S^  p C * A  C * ¼ 0  A ! 2 2 ^ B  S p þm

ð5:1502Þ

This is the Weinberg equation for free particles with spin 1.

5.19.2

Weinberg Equation with Interaction Potential

We assume that the particle with spin 1 only has the scalar potential S and the timelike component V of the vector potential, and the interaction potentials are still introduced by the method given Eq. (5.1428) and Eq. (5.1429) [118]. And noting Eq. (5.62) we can rewrite Eq. (5.1490) as

 e2 ψ ¼ 0 i∂μ  Vδμ0 ði∂ν  Vδν0 Þγ μν  m

ð5:1503Þ

Let us do the following matrix decomposition:



i∂μ  Vδμ0 ði∂ν  Vδν0 Þγ

μν

 ¼

Λ11

Λ12

Λ21

Λ22

 ð5:1504Þ

where Λij(i, j ¼ 1, 2) is 33 matrix. For the stationary systems, using ω ¼ E  V introduced by Eq. (5.1432), the following expressions can be obtained using Eqs. (5.1491)–(5.1493) and (5.1504):   *2 ! 2 Λ11 ¼ ω2 þ ∇ þ 2 ^S  p   ! Λ12 ¼ 2ω2 ^S  p   ! Λ21 ¼ 2ω ^S  p     *2 ! 2 S p Λ22 ¼  ω2 þ ∇ þ 2 ^

ð5:1505Þ ð5:1506Þ ð5:1507Þ ð5:1508Þ

5.19

Weinberg Relativistic Dynamics Equation with Spin 1 Particles. . .

613

So the following equation can be obtained using Eqs. (5.1503)–(5.1508): 0

  *2 ! 2 2 e 2 þ ∇ þ 2 ^S  p Bω  m B   @ ! 2ω ^S  p

1   ! ! 2ω ^ S p C * A    C * ¼ 0 *2 ! 2 A B 2 2 ^ e þ∇ þ2 S p  ω þm ð5:1509Þ

And we can further get  2      * ! 2 ! * 2 * 2 ^ e A  2ω ^ S p B ¼0 ∇ þ2 S p þω m

ð5:1510Þ

 2     * ! * ! 2 2 * 2 ^ ^ e B¼0 2ω S  p A  ∇ þ 2 S  p þ ω þ m

ð5:1511Þ



Equation (5.1511) can be formally rewritten as   ! 2ω S^  p * B ¼ *2 A  2 ! e2 ∇ þ 2 ^S  p þ ω2 þ m

*

ð5:1512Þ

Since ω ¼ E  V, the molecule of Eq. (5.1512) ~E, and the denominator ~E2, so when the incident particle energy is quite low there is p < 1, > : - 1,

when s =

1 2

when s = -

1 2

ð6:235Þ

6.5

Polarized Particle Transport Equation

699

In the previous expressions, the relations between (θ1, φ1) and (θ0, φ0, θ, φ) can be derived from Eq. (6.218) and Eq. (6.219). So the function related to angle  *  * * θ1 , φ1 , Ω 0 after being integrated to angle Ω 0 becomes just related to angle Ω . The neutron polarization vector components given by Eq. (6.234) are just that used in Eq. (6.221) or Eq. (6.222) in O0 system. When the initial conditions and boundary conditions of the four functions to be solved are determined, after substituting Eq. (6.233) into Eq. (6.213), the obtained equation combined with Eqs. (6.234) can be solved jointly. In fact, three neutron polarization vector components can be calculated conveniently in the process of calculating neutron flux function. From the point of view of calculation mathematics, the calculation of neutron polarization vector components will not bring much difficulty. It can be predicted that when the required neutron normal and polarization microscopic nuclear data are available, the method of solving the usual neutron transport equation can be extended to solve the polarized neutron transport equation. Next we study the polarized deuteron transport equation, and consider only one * polarized reaction channel. At the time t and the space point r , the direction of the * incident deuterons with energy E0 in O0 system is Ω 0. Now we are going to study in O1 system; in this case the incident deuterons are along the z1-axis direction. After the incident deuteron energy E0 in the laboratory system is determined, the 42 kinds of the differential physical quantities in the laboratory system I0(θ1), Ay(θ1), Ai(θ1) (i = 1~3), P y(θ1), Pi(θ1) (i = 1~3), Ri(θ1) (i = 1~16), Qi(θ1)(i = 1~15), Wi(θ1) * (i = 1, 2) can be obtained from the polarization nuclear database for the 1 þ * A → 1 þ B reaction. Then the various polarization quantities related to angles (θ1, φ1) can be calculated according to Eqs. (6.63), (6.70), (6.71), (6.76), (6.81), (6.86), (6.91), ), and(6.101).  (6.96 * v * 0 Let Pi r , E , Ω 0 , s0 , t , i = x, y, z be the deuteron polarization vector compo*

0 nents in O0 system with E0, motion  energy  direction Ω , and spin magnetic quantum * 0 t * 0 0 0 number s , and let Pi r , E , Ω , s , t , i = xy, xz, yz, xx - yy, zz be the deuteron *

polarization tensor components in O0 system with energy E0, motion direction Ω 0 , rate functions are and spin magnetic quantum number s0. The eight polarization  * * solved in conjunction with the deuteron flux function ψ r , E, Ω, s, t . If we look at   * * Pvi r , E0 , Ω 0 , s0 , t ,i = x,y,z in O1 system, they become 

*v * p1 r ,

   * * *v * E 0 , Ω 0 , s0 , t = ^bðθ0 , φ0 ÞP r , E0 , Ω 0 , s0 , t

ð6:236Þ

where matrix ^bðθ0 , φ0 Þ has been given by Eq. (2.204) or Eq. (6.43). If we look at  the * * Pti r , E0 , Ω 0 , s0 , t ,i = xy,xz,yz,xx - yy,zz in O1 system, they become

700

6 Basis of the Polarized Particle Transport Theory

0 rffiffiffi 0 rffiffiffi  1  1 * 1 t 1 t * * 0 *0 0 0 0 0 B 6 p1 xx - yy r , E , Ω , s , t C B 6 Pxx - yy r , E , Ω , s , t C C C B rffiffiffi B rffiffiffi B B   C   C C C B B * * 2 t 2 t * 0 0 0 * B B p1 xy r , E0 , Ω 0 , s0 , t C Pxy r , E , Ω , s , t C C C B B C C B r3ffiffiffi B r3ffiffiffi C C B B     C C B 2 t 2 0 0 B * 0 *0 0 * 0 *0 0 t C = ^γ ðθ , φ ÞB C B p1 xz r , E , Ω , s , t Pxz r , E , Ω , s , t C C B B 3 3 C C B rffiffiffi B rffiffiffi B B   C   C * * 2 t 2 t * 0 0 0 C C B B * 0 0 0 r,E ,Ω ,s ,t p P r,E ,Ω ,s ,t C C B B 3 1 yz 3 yz C C B B C C B rffiffiffi B rffiffiffi   A   A @ @ * * 1 t 1 t * 0 0 0 * 0 0 0 r,E ,Ω ,s ,t p P r,E ,Ω ,s ,t 2 1 zz 2 zz ð6:237Þ where the matrix ^γ ðθ0 , φ0 Þ has been given by Eq. (6.118). And according to Eq. (6.103) we define     * * 3 * * pv1i r , E 0 , Ω 0 , s0 , t = pv1i r , E0 , Ω 0 , s0 , t , 2 8   2 t * 0 *0 0 > > p1i r , E , Ω , s , t , > >3 <   >   * 1 t * 0 *0 0 0 0 t * 0 p1i r , E , Ω , s , t = p1i r , E , Ω , s , t , > 6 > >   > >1 t * 0 *0 0 : p1i r , E , Ω , s , t , 2

i = x,y,z i = xy, xz, yz i = xx - yy

ð6:238Þ

i = zz

Then the Stokes vector of the incident deuterons in O1 system can be written out as 0

1 1  * B pv * C 0 0 0 B C 1x r , E , Ω , s , t B   C B C * 0 0 0 B pv * C B C 1y r , E , Ω , s , t B   C * B C * B pv1z r , E 0 , Ω 0 , s0 , t C B C B C   * B C   * 0 t 0 0 * p * r , E , Ω , s , t B C SðiÞ r , E0 , E, θ1 , Ω 0 , s0 , t = I 0 ðE 0 , E, θ1 Þ B 1xy C ð6:239Þ   B C * B pt * C 0 0 0 B C 1xz r , E , Ω , s , t B C   B C * 0 0 0 B pt * C r , E , Ω , s , t 1yz B C B C   B t C * 0 *0 0 B p1 xx - yy r , E , Ω , s , t C B C   A @ * 0 *0 0 t p1zz r , E , Ω , s , t 

The expression of Stokes vector for outgoing deuterons is

6.5

Polarized Particle Transport Equation

  * * Sðf Þ r , E0 , E, θ1 , φ1 , Ω 0 , s0 , t

701

0 

1

1 

C B * C B Pv1x * r , E 0 , E, θ1 , φ1 , Ω 0 , s0 , t C B B  C  C B * 0 0 0 C B Pv * 1y r , E , E, θ 1 , φ1 , Ω , s , t C B C B   C B * C B Pv * 0 0 0 C B 1z r , E , E, θ 1 , φ1 , Ω , s , t C B C B   * C ð6:240Þ B   * * * C B Pt1xy r , E 0 , E, θ1 , φ1 , Ω 0 , s0 , t 0 0 0 = I r , E , E, θ1 , φ1 , Ω , s , t B C C B   * C B 0 0 0 C B Pt1xz * r , E , E, θ , φ , Ω , s , t 1 1 C B C B   C B * 0 0 0 C B Pt * r , E , E, θ , φ , Ω , s , t 1 1 1yz C B C B   C B * * 0 C B Pt 0 0 B 1 xx - yy r , E , E, θ1 , φ1 , Ω , s , t C C B   A @ * * Pt1zz r , E 0 , E, θ1 , φ1 , Ω 0 , s0 , t The above two formulas satisfy the following relation: Sðf Þ = Z 1 SðiÞ

ð6:241Þ

where the expression of the Mueller matrix Z1 has been given by Eq. (6.106) as 0

Ay 0 Axy Axz Ayz 1 Ax B Px x x x x x Kx Ky Kz K xy K xz K xyz B B B Py K yx K yy K yz K yxy K yxz K yyz B B B 0 K zx K zy K zz K zxy K zxz K zyz B B xy P K xy K xy K xy K xy K xy K xy Z1 = B x y z xy xz yz B B xz xz xz xz xz xz xz P K K K K K K B x y z xy xz yz B yz B P yz yz yz yz yz Kx Ky Kz K xy K xz K yz B yz B xx - yy xx - yy xx - yy xx - yy xx - yy xx - yy xx - yy BP Kx Ky Kz K xy K xz K yz @ zz zz zz zz zz P Kx Ky 0 K xy K xz K zz yz

Axx - yy K xxx -yy K yxx -yy K zxx -yy K xy xx -yy K xz xx -yy K yz xx -yy -yy K xx xx -yy

K zz xx -yy

1 Azz K xzz C C C y K zz C C C 0 C C C K xy zz C C C K xz zz C C C K yz zz C C - yy C K xx A zz zz K zz 1 ð6:242Þ

' Here  the subscript 1 represents  the energy E and the angles (θ1, φ1). The function * I r , E0 , E, θ1 , φ1 , Ω0 , s0 , t is the microscopic differential cross section of the

outgoing deuterons, and it is also a function of (θ0, φ0, θ, φ) according to

702

6 Basis of the Polarized Particle Transport Theory

Eq. (6.218) and Eq. (6.219). Let number of the outgoing  the spin magnetic quantum  * deuterons be s; the function I r , E 0 , E, θ1 , φ1 , Ω0 , s0 , t in Eq. (6.240) contains the contributions of three parts s = ± 1, 0. 

*v

The three components Pv1x , Pv1y , Pv1z of the polarization vector P 1 and the five   *t components Pt1xy , Pt1xz , Pt1yz , Pt1 xx - yy , Pt1zz of the polarization tensor P 1 of the

outgoing deuterons in Eq. (6.240) are given in O1 system. If we look at them in O0 *v system, P 1 becomes *v

*v

^ ðθ0 , φ0 ÞP 1 p =B

ð6:243Þ *t

^ ðθ0 , φ0 Þ has been given by Eq. (2.203) or Eq. (6.42); P 1 becomes where matrix B 1 1 0 rffiffiffi 0 rffiffiffi 1 t 1 t B 6 pxx - yy C B 6 P1 xx - yy C C C B rffiffiffi B rffiffiffi C C B B C C B B 2 t C 2 t C B B p P 1 xy xy C C B B C C B r3ffiffiffi B r3ffiffiffi C C B B C B 2 t C=Γ 2 0 0 B t ^ ðθ , φ ÞB B pxz C P1 xz C C C B B 3 3 C C B rffiffiffi B rffiffiffi C C B B 2 t C 2 t C B B pyz C P1 yz C B B 3 3 C C B B C C B rffiffiffi B rffiffiffi A A @ @ 1 t 1 t pzz P1 zz 2 2

ð6:244Þ

v ^ ðθ0 , φ0 Þ has been given by Eq. (6.117). The polarization vector * where matrix Γ p *t and the polarization tensor p of the outgoing deuterons in O0 system are along the deuteron polarization direction. In the Cartesian coordinate system XYZ taking the deuteron polarization direction as the Z-axis, only the component pvZ of the polari*v zation vector p and the component ptZZ of the polarization tensor are not equal to *v 0. The absolute value of the outgoing deuteron polarization vector p can be found to be

pvZ

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2   2  2 = pvx þ pvy þ pvz

ð6:245Þ

Assume that the polarization direction of the outgoing deuterons is (θ2, φ2) in O0 *t system, and then the relations between the components of p and ptZZ can be given by Eq. (6.124) as follows:

6.5

Polarized Particle Transport Equation

703

3 3 sin 2 θ2 cos ð2φ2 ÞptZZ , ptxy = sin 2 θ2 sin ð2φ2 ÞptZZ 2 4 3 3 t t t pxz = sin ð2θ2 Þ cos φ2 pZZ , pyz = sin ð2θ2 Þ sin φ2 ptZZ 4 4  1 ptzz = 3 cos 2 θ2 - 1 ptZZ 2 ptxx - yy =

ð6:246Þ

It can be derived from the above formulas that 

 2  2 9 þ 4 ptxy = sin 4 ðθ2 Þ ptZZ 4  t 2  t  2 9  2 pxz þ pyz = sin 2 ðθ2 Þ cos 2 ðθ2 Þ ptZZ 4   t 2 9  2 1  t 2 pzz = cos 4 θ2 - cos 2 θ2 þ pZZ 4 3 9



2  2   2    2 4  t 2 1 2 t t þ þ ptzz = ptZZ pxz þ ptyz pxx - yy þ 4 pxy 3 3  

2  2   2   1=2 4  t 2 1 2 ptZZ = þ þ ptzz pxz þ ptyz ptxx - yy þ 4 ptxy 3 3 ptxx - yy

2

ð6:247Þ

ð6:248Þ ð6:249Þ

In fact, in the five-dimensional orthogonal Cartesian coordinate system, using the notations introduced by Eq. (6.120), we can write the following relation directly: 

htxx - yy

2

 2   2   2 2 2 þ htxy þ htxz þ htyz þ htzz = htZZ

ð6:250Þ

According to Eq. (6.120), we can get Eq. (6.248) from Eq. (6.250) immediately. This clearly shows that the use of five-dimensional orthogonal Cartesian coordinate system will bring greater convenience. *v *t The components of the polarization vector p and the polarization tensor p of the outgoing deuterons in O0 system are calculated results numerically, and pvZ and ptZZ can be obtained by Eq. (6.245) and Eq. (6.249), respectively. The shares N1, N0, N-1 of the three spin projections of the deuterons along the polarization direction satisfy the following relations given by Eqs. (3.163) and (3.165): N 1 þ N 0 þ N - 1 = 1,

pvZ = N 1 - N - 1 ,

ptZZ = N 1 - 2N 0 þ N - 1 ð6:251Þ

From the above three formulas we can obtain 1 2 þ 3pvZ þ ptZZ , 6 1 N - 1 = 2 - 3pvZ þ ptZZ 6

N1 =

N0 =

1 1 - ptZZ , 3

ð6:252Þ

704

6 Basis of the Polarized Particle Transport Theory

  * * Since pvZ and ptZZ are the functions of r , E0 , E, θ1 , φ1 , Ω 0 , s0 , t , the Ns(s = ± 1, 0) derived from the above   * * 0 r , E , E, θ1 , φ1 , Ω 0 , s0 , t .

equations

are

also

the

functions

of

Using the above results we can obtain that the polarization macroscopic double differential cross section in Eq. (6.213) is  * *  * Σ p r , t; E 0 , Ω 0 , s0 → E, Ω, s       * * * * * = ρ r , t N s r , E 0 , E, θ1 , φ1 , Ω 0 , s0 , t I r , E0 , E, θ1 , φ1 , Ω 0 , s0 , t

ð6:253Þ

  * where ρ r , t is the density of the nuclear medium, Ns has been given by   * * Eq. (6.252), and I r , E 0 , E, θ1 , φ1 , Ω 0 , s0 , t appearing in Eq. (6.240) can be solved by Eq. (6.241). Furthermore, the polarization vector components of the outgoing * deuterons with energy E, motion direction Ω , and spin quantum number s in O0 system can also be obtained as   * * Pvi r , E, Ω, s, t ZZ Xh  i   * * * * * = dvs N s r , E 0 , E, θ1 , φ1 , Ω 0 , s0 , t pvi r , E 0 , E, θ1 , φ1 , Ω 0 , s0 , t dE 0 dΩ 0 s0

s = ± 1,0;

i = x,y,z ð6:254Þ

  * * where pvi r , E 0 , E, θ1 , φ1 , Ω 0 , s0 , t , i = x,y,z have been given by Eq. (6.243). In O0 system the polarization tensor components of the outgoing deuterons with energy * E, motion direction Ω, and spin quantum number s are   * * Pti r , E, Ω, s, t ZZ Xh    i * * * * * = dts N s r , E 0 , E, θ1 , φ1 , Ω 0 , s0 , t pti r , E 0 , E, θ1 , φ1 , Ω 0 , s0 , t dE 0 dΩ 0 s0

s = ± 1,0;

i = xx - yy,xy,xz,yz,zz ð6:255Þ

  * * where pti r , E 0 , E, θ1 , φ1 , Ω 0 , s0 , t , i = xx - yy,xy,xz,yz,zz have been given by Eq. (6.244). The deuteron vector direction factor d vs is defined as

6.6

Prospect of the Polarized Particle Transport Theory

8 > < 1, dvs = 0, > : - 1,

705

when

s=1

when

s=0

when

s= -1

ð6:256Þ

The deuteron tensor direction factor d ts is defined as 8 > < 1, t ds = - 2, > : 1,

when

s=1

when

s=0

when

s= -1

ð6:257Þ

The components of the polarization vector and the polarization tensor of deuterons given by Eq. (6.254) and Eq. (6.255) are just that used in Eq. (6.236) and Eq. (6.237) in O0 system, respectively. When the initial conditions and boundary conditions of the nine functions to be solved are determined, after substituting Eq. (6.253) into Eq. (6.213), the obtained equation combined with Eqs. (6.254) and (6.255) can be solved jointly. For the transport problem of the spin 1 deuterons, in addition that the ionization loss effect needs to be considered, due to the presence of (d, n), (n, d), and * so on reactions, it is necessary to study the conjunction transport problem of 1 þ * * * * * * * 1 1 1 1 A → 1 þ B, 1 þ A → þ B, þ A → 1 þ B, þ A → þ B reaction channels. 2 2 2 2 In other words, it is necessary to solve the polarized particle transport equation (6.213) by using the source term Q to combine the neutron transport with the charged particle transport. So this problem becomes quite complicated.

6.6 6.6.1

Prospect of the Polarized Particle Transport Theory To Perfect the Polarization Theory of Nuclear Reactions

The theory of calculating various polarization physical quantities from the nuclear reaction amplitude is called polarization theory of nuclear reactions, and it still belongs to the category of nuclear reaction theory. S matrix theory is the basic theory describing the nuclear reactions. The expressions of the reaction amplitude given by S matrix theory satisfy the basic physical requirements, such as angular momentum conservation and parity conservation. The polarization theory of the two-body 1 reaction of the spin or 1 incident particle with a target and residual nucleus with 2 arbitrary spins was studied in this book. Some internal relations of the matrix elements of the nuclear reaction amplitude described by the S matrix elements were found. The clear expressions of various polarization quantities of the corresponding nuclear reactions were obtained. The optical model, coupled channel

706

6 Basis of the Polarized Particle Transport Theory

optical model, distorted wave Born approximation, R matrix theory, phase-shift analysis method, and even some microscopic nuclear reaction theory can all be used to calculate the S matrix elements of the nuclear reactions. The above theories belong to the usual nuclear reaction theory. Therefore, the publication of this book 1 makes that the polarization theory of the incident particles of spin or 1 and the 2 unpolarized targets and residual nuclei with arbitrary spins have reached the stage 1 where the actual calculations can be done. The particle n, p, t, 3Heof S = and d of 2 S = 1 are the most concerned particles in the particle transport theory. In the future, it is necessary to study the universal optical potential of deuteron containing tensor term, and the polarization experimental data should be included in the process of adjusting the theoretical model parameters to fit the experimental data. The spin of the particle α is equal to 0, and there is no polarization problem for itself. However, there are the polarization problems in the incident channel or outgoing channel corresponding to (n, α), (d, α), (α, n), (α, d) and other nuclear ! ! → 1 1 þ A→0 þ B, 1 þ A→0 þ B, 0 þ A→ þ B, 0 þ reactions indicated by 2 2 → A → 1 þ B . It is relatively easy to study the polarization phenomena of the two-body direct reaction where only the incident channel or only the outgoing channel is polarized. 3 The spin of the particles such as 5He, 7Li, 7Be, 9Be and so on is ; these particles 2 are relatively heavy. And for the sake of the integrity of the nuclear reaction 3 polarization theory, the spin particle polarization theory should also be studied. 2 3 This book begins by studying the polarization theory of the spin particles in 2 spherical basis coordinates, then based on the studying on the five-dimensional rank2 tensors required for spin 1 particles, the unitary transformation relations of the rank-2 and higher-rank tensors between the spherical basis coordinates system and the orthogonal Cartesian coordinate system are found, and the theoretical methods 3 for studying the spin ≥ particle polarization phenomena are given. 2 It is necessary to study the polarization effect of nuclear reactions in the energy region with the anisotropic differential cross sections. For the few-body nuclear reaction, there is a clear anisotropy phenomenon even in the keV energy region. If the target is a medium-weight nuclide or a heavy nuclide, the anisotropy is only apparent when the incident particle energy approaches the MeV energy region. This book also studied some nuclear reactions for which both the incident particle and the target are polarized at the same time. They can be expressed as symbols * * * 1 1 1 * * * þ , þ 1 , 1 þ 1 : Of course, their theoretical formulas are different corres2 2 2 ponding to different outgoing channels. This book only studies part of the outgoing * * channels for above reactions. For example, we have studied n þ d → n þ d reac* * tion, whereas we have not specifically studied d þ t → n þ α reaction. But it is not very difficult to study it using the methods described in this book. This book also

6.6

Prospect of the Polarized Particle Transport Theory *

*

707

studied d þ d → p þ t reaction, but because there are too many items in the formulas, we recommend using numerical calculation method including trace to deal with this problem. For the case where the target and the residual nucleus are a medium-weight nuclide or a heavy nuclide, it is usually assumed that the target and the residual nucleus are unpolarized. For the target and the residual nucleus which are also polarized, when the polarization rates of the target and/or residual nucleus with arbitrary spin are known, the methods described in Sect. 2.8, 3.7, 3.9, 3.10 sections can also be used to calculate various polarization quantities, but this method cannot calculate the polarization rates of the residual nucleus. If the spins of the heavier 1 polarized target and the residual nucleus are or 1, various polarization physical 2 quantities can also be calculated using the methods described in Sects. 2.9, 3.11, and 3.12. Although the theory itself is relatively complex, however, compared with the previous method in Sects. 2.8, 3.7, 3.9, 3.10, it is not necessary to know the polarization rates of the residual nucleus in advance; conversely the polarization rates of the residual nucleus can be given through theoretical calculations. In this book the light beam polarization theory has only made preliminary exploration, and the further research work on it still needs to be carried out. In addition, the polarization theory of the three-body reaction also needs to be further explored. In fact, some research works have been carried out on the polarization problem of three-body reaction in some papers studying the few-body reactions [21–26]. Since the pre-equilibrium and equilibrium compound nuclear reactions belong to the nuclear reaction process in which many nucleons are involved. Even if the incident nucleon is polarized, after the reaction process in which many nucleons participate, taking into account the average effect, the outgoing nucleons should be unpolarized. Since the spin-orbit potential can be automatically generated in the relativistic Dirac equation, it is also a hot topic to study the polarization phenomena using relativistic nuclear reaction theory. In the relativistic optical model, the Dirac equation is converted into a Schrodinger-like equation, and naturally it becomes that using non-relativistic formulas to calculate the corresponding polarization physical quantities. This book develops the Dirac S matrix theory of relativistic nuclear reactions. When the energy is quite low, this theory will automatically 1 degenerate into a non-relativistic S matrix theory for spin particles. In this way, 2 as long as the relativistic reaction amplitude is solved, using the Dirac equation, the 1 Dirac S matrix theory can be used to calculate various physical quantities of spin 2 particles, including an amount of polarization physical quantities. The relativistic polarization theory of the deuteron induced reactions needs to be studied further. When the incident nucleon energy is high enough to need to be described by relativistic theory, the intra-nuclear cascade reaction becomes one of the very important reaction mechanisms. The quantum molecular dynamics (QMD) model

708

6 Basis of the Polarized Particle Transport Theory

is just a theory to describe the intra-nuclear cascade reaction. In the intra-nuclear cascade process, it is possible to impact the nucleon going outside the nucleus after only a few collisions, so the intra-nuclear cascade reaction belongs to a nuclear reaction in which a small number of nucleons participate. Assume that the incident nucleon is polarized and the nucleons in the target are unpolarized, so the first * * * 1 1 1 1 collision belongs to þ → þ reaction. The following collision generally 2 2 2 2 also belongs to this kind of reaction, but the collision between the two nucleons, * * * * 1 1 1 1 which have been collided before, belongs to þ → þ reaction. The prob2 2 2 2 ability of such a collision is relatively small. In the QMD model, after simulating a large number of collision events, considering the axial symmetry of the system, the average double differential cross sections of the neutrons and protons being collided out by an incident nucleon can finally be obtained as Nn(E, θ) and Np(E, θ). Making the integrals to them on E and θ we can get the numbers of the neutrons and protons which are collided out. If the nucleon polarization rates are also passed during the nucleon-nucleon collision in the nucleus, the average nucleon polarization rates Pni(E, θ) and Ppi(E, θ) (i = x, y, z) of the neutrons and protons being collided out will be obtained. In order to realize the above idea, it is necessary to develop a polarization QMD model.

6.6.2

To Establish a Polarization Nuclear Database

The aim of the establishment of polarization nuclear database is to provide the basic nuclear data for the calculation of polarized nucleus transport theory. At present, we only consider the unpolarized target and residual nucleus. With regard to polarization nuclear databases, we propose to use the classification based on the kinds of the incident particles, such as neutron polarization nuclear databases, proton polarization nuclear databases, deuteron polarization nuclear databases, and so on. In each particle polarization nuclear database, there are many kinds of targets, and for each target there are many possible reaction channels. For example, for the neutron induced nuclear reactions, the elastic scattering channel only needs to contain the shape elastic scattering part, the discrete inelastic scattering channel only needs to contain the corresponding direct inelastic scattering part, and the direct reaction parts of (n, p), (n, d), (n, α) and so on reactions can also be included. The polarization problem is not considered for the parts contributed by the pre-equilibrium and equilibrium compound nuclear reaction, as well as for the parts contributed by the second emission and the third emission. For each reaction channel to consider the polarization effect, many sets of polarization differential physical quantities only related to angle θ for multi-energy points of the laboratory ! 1 þ system are contained. It has been pointed out in 6.2–6.4 sections that for 2

6.6

Prospect of the Polarized Particle Transport Theory

709

! 1 þ B reaction there are 8 sets of polarization differential physical quantities, 2 ! ! → → 1 1 for 1 þ A → þ B and þ A → 1 þ B reactions there are 18 sets, respectively, 2 2 → → and for 1 þ A → 1 þ B reaction there are 42 sets. It can also be inferred that there ! ! → 1 1 are 2 sets for þ A → 0 þ B and 0 þ A → þ B reactions and 5 sets for 1 þ 2 2 → A → 0 þ B and 0 þ A → 1 þ B reactions, respectively. The establishment of a polarization nuclear database should be combined with the calculation and evaluation of a complete set of nuclear data, including cross sections, angular distributions, double differential cross sections, and polarization data. Only when the used theories and model parameters can better fit the multiple experimental data of the reaction at the same time, the polarization nuclear data predicted theoretically will possess a certain degree of credibility. The establishment of the polarization nuclear database is a gradual accumulation process. For example, at the beginning it can only contain few targets and neutron induced data, and can only contain the data of the shape elastic scattering and the direct inelastic scattering of one or two discrete levels. In the initial stage, it needs only that we can give the basic data needed to carry out the research on the polarized nucleus transport theory for the individual macroscopic experiment inspection device. In this book, for the polarization theory of spin 1 particles, the five- dimensional ^ i ði = xy, xz, yz, xx - yy, zzÞ given independent Cartesian coordinate system taking Q by Eq. (3.21) as the basis functions is used, which is not yet a five-dimensional orthogonal Cartesian coordinate system. In the future, if someone wishes using the five dimensional orthogonal Cartesian coordinate system taking ^ i ði = xx - yy, xy, xz, yz, zzÞ given by Eq. (3.142), or Eq. (6.111), or Eq. (6.113) H as the basis functions and using the generalized Pauli matrix introduced in this book to derive a set of spin 1 particle polarization theory formulas, so that the order and coefficients of the terms contained in the new formulas will be different from those given in this book. But I suggest that you still use the polarization physical quantities, which are only relevant to polar angles, in the polarization nuclear database as defined in this book; thus two sets of spin 1 particle polarization theoretical formulas can use the same spin 1 particle polarization nuclear database. ^ and ^γ given by Eq. (6.117) and Eq. (6.118) In addition, the rotation matrices Γ should be used only for five-dimensional orthogonal Cartesian coordinate system, so for the spin 1 theoretical formulas given in this book, when making rotation using these matrices, appropriate changes should be made to the relevant physical quantities, such as doing in Eq. (6.130) (or Eq. (6.211), or Eq. (6.237), or Eq. (6.244)). A→

710

6.6.3

6

Basis of the Polarized Particle Transport Theory

To Research the Polarized Particle Transport Theory

In the initial stage, we can choose to explore the polarized nucleus transport theory for macroscopic experiment inspection devices with incident neutron energy in the MeV region and containing only one or two nuclides in the nuclear medium. The Monte-Carlo method or the polarized particle transport equation method can all be used. It is recommended that the Cartesian coordinate system of the laboratory system is used and the initial incident neutron direction is taken as the z-axis. Then assume that the initial incident neutrons, targets, and residual nuclei are all unpolarized. When doing the calculation, the usual nuclear database and polarization nuclear database are used at the time. First, in the usual nuclear database, we deduct the reaction channels that are already included in the polarization nuclear database. For example, in the polarization nuclear database, there are the shape elastic scattering channel and the first two level direct inelastic scattering channels, and then in the normal nuclear database, the elastic scattering channel can only include the contribution of the compound elastic scattering; in the first two level inelastic scattering channels, the contribution of direct inelastic scattering should be deducted. The above approach is equivalent to divide the original reaction channel into two reaction channels in form. The reaction channel, for which the polarization effect is not considered, is still placed in the normal nuclear database, and the reaction channel, for which the polarization effect is considered, is placed in the polarization nuclear database. When using the Monte-Carlo method, assume that the initial incident neutrons are unpolarized, after sampling, the obtained secondary neutrons generated through the unpolarization channel are still unpolarized, and secondary neutrons generated through the polarization channel are polarized. The polarization channels of the second reaction can process for the unpolarized and polarized secondary neutrons of the first reaction, and their corresponding outgoing particles are all polarized. For unpolarization channels, polarized incident neutrons are still treated as unpolarized neutrons, and the corresponding outgoing particles are also unpolarized. The ionization loss effect should be considered for the charged particle transport process. Generally, in MeV energy region, the contribution of the elastic scattering channels basically comes from the shape elastic scattering; the main contribution of the low-energy discrete level inelastic scattering channels comes from the direct inelastic scattering. Although the polarization effect is not considered in the continuous inelastic scattering and (n, 2n), (n, 3n), and so on reaction channels, the contribution given by the reaction channels considering the polarization effect are still accounts for a considerable share. In the energy region that needs to be studied with relativistic theory, for the elastic scattering channel the polarization effect can be considered. At this time, since the incident particle energy is much larger than the heights of the discrete energy levels, the discrete level direct inelastic scattering and other direct reaction channels are not very important, but the intra-nuclear cascade mechanism does have a greater contribution. Therefore, the development of the

References

711

polarization QMD model is very valuable for the study of the polarization problem of the relativistic nuclear reactions. Both the experiments and the theories have proved that the nuclear potential of 1 the spin nucleon contains spin-orbital potential, and the nuclear potential of the 2 spin 1 deuteron contains tensor potential in addition to the spin-orbital potential. Therefore, the nuclear forces felt by the particles with different spin magnetic quantum numbers when moving in the nuclear potential are different. Therefore their motion behavior will also be different. This is the polarization phenomenon of nuclear reaction that exists objectively in nature. The polarization phenomenon is not considered in the existing transport theory that describes the movement of the heavy particles in nuclear media. It means that all particles are unpolarized. From a microscopic physical point of view, this is only an approximate approach. In order to be able to describe the physical process of the natural world in a more realistic way and to study the influence of polarization phenomenon, the research on polarized nucleus transport theory should be carried out. This shows that it is necessary to carry out research on polarized nucleus transport theory. We know that the spins of most nuclei are not equal to 0. However, before writing this book, there was no polarization theory system that described the nuclear reaction 1 of spin and 1 particle with target and residual nucleus with non-zero spin, so that a 2 full set of polarization nuclear data could not be obtained. Of course, it was impossible to carry out polarization nucleus transport theory research in this case. However, in this book, the theory formulas describing the polarization reactions of 1 the incident particles of spin and 1 with the unpolarized target and residual nucleus 2 with arbitrary spins are clearly given. So it is feasible to establish a polarization nuclear database and carry out research on polarized nucleus transport theory. For the first time, this book puts forward the research topic of establishing polarization nuclear database and carrying out polarized nucleus transport theory. I think that this topic will encounter many difficulties and challenges in the future development. In the future, how the subject will develop, and what will be the results and conclusions, these questions can only be answered by scientific researchers through their future scientific research work.

References 1. Xiao XB, Li XM, Zhou GH. Properties of electron spin polarization transport of the quantum lines under the electromagnetic wave irradiation. J Phys. 2007;56:1649. (in Chinese) 2. Johnson M. Spin accumulation in gold films. Phys Rev Lett. 1993;70:2142. 3. Hershfield S, Zhao HL. Charge and spin transport through a metallic ferromagneticparamagnetic-ferromagnetic junction. Phys Rev. 1997;B56:3296. 4. Moroz AV, Barnes CHW. Effect of the spin-orbit interaction on the band structure and conductance of quasi-one-dimensional systems. Phys Rev. 1999;B60:14272.

712

6

Basis of the Polarized Particle Transport Theory

5. Brataas A, Nazarov YV, bauer G E W. Finite-element theory of transport in ferromagnet-normal metal systems. Phys Rev Lett. 2000;84:2481. 6. Moroz AV, Samokhin KV, Barnes CHW. Spin-orbit coupling in interacting quasi-one-dimensional electron systems. Phys Rev Lett. 2000;84:4164. 7. Nadgorny B, Soulen RJ, Osofsky MS, et al. Transport spin polarization of NixFe1-x: Electronic kinematics and band structure. Phys Rev. 2000;B61:R3788. 8. Tang HX, Monzon FG, Lifshitz R, et al. Ballistic spin transport in a two-dimensional electron gas. Phys Rev. 2000;B61:4437. 9. Mireles F, Kirczenow G. Ballistic spin-polarized transport and Rashba spin precession in semiconductor nanowires. Phys Rev. 2001;B64:024426. 10. Wang XF. Spin transport of electrons through quantum wires with a spatially modulated Rashba spin-orbit interaction. Phys Rev. 2004;B69:035302. 11. Xu YL, Guo HR, Han YL, et al. New Skyrme interaction parameters for a unified description of the nuclear properties. J Phys G Nucl Part Phys. 2014;41:015101. 12. Kulsrud RM, Furth HP, Valeo EJ, Goldhaber M. Fusion reactor plasmas with polarized nuclei. Phys Rev Lett. 1982;49:1248. 13. More RM. Nuclear spin-polarized fuel in inertial fusion. Phys Rev Lett. 1983;51:396. 14. Fletcher KA, Ayer Z, Black TC, et al. Tensor analyzing power for 2H(d,p)3H and 2H(d,n)3He at deuteron energies of 25,40, and 80 keV. Phys Rev. 1994;C49:2305. 15. Zhang JS, Liu KF, Shuy GW. Neutron suppression in polarized dd fusion reaction. Phys Rev. 1999;C60:054614. 16. Schieck H, Paetz gen. The status of “polarized fusion”. Eur Phys J. 2010;A44:321. 17. Schieck H, Paetz gen. “Polarized fusion”: new aspects of an old project. Few-Body Syst. 2013;54:2159. 18. Liu KF, Chao AW. Accelerator based fusion reactor. Nucl Fusion. 2017;57:084002. 19. Robson BA. The theory of polarization phenomena. Oxford: Clarendon Press; 1974. 20. Varshalovich DA, Moskalev AN, Khersonskii VK. Quantum theory of angular momentum. Singapore/New Jersey/Hong Kong: World Scientific; 1988. 21. Deltuva A. Spin Observables in three-body direct nuclear reactions. Nucl Phys. 2009;A821:72. →

22. Felsher PD, Howell CR, Tornow W, et al. Analyzing power measurement for the d þ d → d þ p þ n breakup reaction at 12 MeV. Phys Rev. 1997;C56:38. 23. Kievsky A, Viviani M, Rosati S. Cross section, polarization observables, and phase-shift parameters in p-d and n-d elastic scattering. Phys Rev. 1995;C52:R15. 24. Kievsky A, Rosati S, Viviani M. Proton-deuteron elastic scattering above the deuteron breakup. Phys Rev Lett. 1999;82:3759. 25. Kievsky A, Viviani M, Rosati S. Polarization observables in p-d scattering below 30 MeV. Phys Rev. 2001;C64:024002. 26. Kievsky A. Polarization observables in p-d and p-3He scattering. Nucl Phys. 2001;A689:361c.

Appendix: Clebsch-Gordan Coefficients, Racah Coefficients, and 9j Symbols

j m

In the Clebsch-Gordan (C-G) coefficient Cj31 m31 j2 m2, the angular momentum magnetic quantum numbers satisfy the relation expression m1 + m2 = m3. Only a, b, c are all integers and a + b + c is even; the C-G coefficient C c0 a0 b0 with the magnetic quantum numbers to be zero may not be zero. The Racah coefficient is also known as the W coefficient. The following abbreviated symbols are introduced: ^j


pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi 2j + 1 , ^a
2a + 1, ⋯⋯

ðA1Þ

Here are some common formulas related to the C-G coefficients, the Racah coefficients, and the 9j symbols [1, 2]: j -m

j m

Cj31 m31

j2 m2

= ð - 1Þj1 + j2 - j3 Cj31 -m31 j2 -m2 j m = ð - 1Þj1 + j2 - j3 Cj32 m32 j1 m1

ðA2Þ

^j j -m = ð - 1Þj1-m1 3 Cj21 m1 2 j3 -m3 ^j 2 ^j j m = ð - 1Þj1-m1 3 Cj23 m23 j1 -m1 ^j 2 ^j j -m = ð - 1Þj2 + m2 3 Cj13 -m13 j2 m2 ^j 1 ^j j m = ð - 1Þj2 + m2 3 Cj12 -m1 2 j3 m3 ^j 1

ðA3Þ

j m

Cj31 m31

j2 m2

j m

C j21 m21

00

= δj1 j2 δm1 m2

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 Q.-B. Shen, Polarization Theory of Nuclear Reactions, https://doi.org/10.1007/978-3-031-11878-4

ðA4Þ

713

714

Appendix: Clebsch-Gordan Coefficients, Racah Coefficients, and 9j Symbols

X m1 m2

j0 m0 j2 m2 C j1 m1 j2 m2

C jm j1 m1

X jm C j1 m1 jm

jm j2 m2 C j1 m01 j2 m02

= δjj0 δmm0

ðA5Þ

= δm1 m01 δm2 m02

ðA6Þ

W ðabcd; ef Þ = W ðbadc; ef Þ = W ðcdab; ef Þ = W ðdcba; ef Þ = W ðacbd; feÞ

ðA7Þ

W ðabcd; ef Þ = ð - 1Þe + f - a - d W ðebcf ; ad Þ = ð - 1Þe + f - b - c W ðaefd; bcÞ ðA8Þ X 2 ^e2^f W ðabcd; ef ÞW ðabcd; egÞ = δfg ðA9Þ e

X ^e2 ð - Þa + b - e W ðabcd; ef ÞW ðbacd; egÞ = W ðafgb; cdÞ

ðA10Þ

e

X

C eε aα

αβδεφ

X

cγ fφ c0 γ 0 bβ C eε dδ C bβ dδ C aα f φ cγ fφ bβ C eε dδ C bβ dδ

Ceε aα

βδε

X C eε aα ε

cγ bβ C eε dδ

=

 b0 W

abcd; e

= ^e^f Ccγ aα

fφ

fφ

cγ dδ C aα

ðA12Þ ðA13Þ

fφ

ð - Þf - b - d δab δcd ^bd^

 ð - Þc - d - a + b e 0 1 C c -1 =2 2 ^a^b

ðA14Þ ðA15Þ

d12

9 c> = X 2 ^k W ðaidh; kgÞW ðbfhd; keÞW ðaibf ; kcÞ f = > ; k h i 9 8 a b c> > = < X ^c2 W ðaibf ; kcÞ d e f = W ðaidh; kgÞW ðbfhd; keÞ > > ; : c g h i 9 8 > =

^c X cγ gρ fφ iv iv δcc0 δγγ0 d e f = C aα bβ C hη bβ eε C c0 γ 0 f φ C aα dδ C dδ eε C gρ 2 > > ^ ^ ^ ; f ^ghi αβδε : g h i 8 >

: g

ðA11Þ

W ðabcd; ef Þ

X ^e^f W ðabcd; ef ÞCfbβφ

W ðabcd; 0f Þ = C e0 a0

= δcc0 δγγ0 ^e^f W ðabcd; ef Þ

b e

φρην

ðA16Þ

ðA17Þ

hη

ðA18Þ

Appendix: Clebsch-Gordan Coefficients, Racah Coefficients, and 9j Symbols

C caαγ

X C ivcγ δφρν

X δρ

715

8 >

: g

9 b c> = ^c X h η e f = C bβ eε Cicγv f φ C gaαρ dδ Cfdδφ eε Cigρv 2 > ; ^f ^g^h^i δεφ h i

gρ f φ C aα

X ^b ^f ^g^i cγ Caα ^c^h

fφ dδ C dδ

ðA19Þ

ρην

fφ dδ C dδ

2

iv eε C gρ

hη

=

2

bβ

Cgρ aα

hη

iv eε C gρ

hη =

X ^b2^c^f ^g cγ Caα ^h bβcγ

8 >

: g

hη bβ C bβ

9 8 > =

hη ðA20Þ d e f C bβ bβ eε > > ; : g h i 9 8 > =

iv ðA21Þ C d e f eε cγ f φ > > ; : g h i

9 b c> = ð - 1Þ c + g - a - e δ δ cf gh e f = W ðabde; cgÞ ^ ^ c g > ; h 0

ðA22Þ

9 8

= a-b-c+d+1 e c = ð - 1Þpffiffiffi Ce0 Cc1 c0 a12 b12 > 1 6 ^c2 d^ ; 1 2 9 8 a b c > > > > > > = < d e c+1 0 d0 > > > > 1 > > :1 1 ; 2 2 ^2 + c + 1 c 0 1 ðd - aÞ^a2 + ðe - bÞb = ð - 1Þa - 2 - c + d + e pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 Ca 1 2 6ðc + 1Þð2c + 3Þ ^c d^

ðA23Þ

8 a > < d d0 > :1 2

C ec +0 1

C ec -0 1

0

8 a > > < d d0 > > :1 2

ðA24Þ

ðA25Þ b - 12

9 c > > = e c-1 > > 1 1 ; 2 b

ðA26Þ

ðd - aÞ^a2 + ðe - bÞ^ b -c c 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 Ca 1 2 6cð2c - 1Þ ^c d^ 2

= ð - 1Þ a - 2 - c + d + e 1

b - 12

716

Appendix: Clebsch-Gordan Coefficients, Racah Coefficients, and 9j Symbols

8 98 9 a b c> a b c> > > < = < = X 2 2 ^c2^f ^g2 ^h d e f d e f = δgj δhk > > > : ;> : ; cf g h i j k i 9 8 98 8 a e j> a b c> > >

< = < X 2 2 d b k = d e e d f ð - 1Þ2e + k - f - h ^j ^k > > > : ; > : ;> : jk g h g h i j k i

ðA27Þ 9 c> = f > ; i

ðA28Þ

In addition, the 3j symbol is 

a α

b β

b c

e f

c γ

 =

ð - 1Þb - a + γ c -γ Caα bβ ^c

ðA29Þ

The 6j symbol is 

a d



= ð - 1Þa + b + c + d W ðabcd; ef Þ

ðA30Þ

9 8 > =

is invariant under interchange of rows and columns The 9j symbol d e f > > ; : g h i (reflection about a diagonal) and is multiplied by (-1)p (where p = a + b + c + d + e + f + g + h + i ) upon interchange of two adjacent rows or columns.

References 1. Rose M E. Elementary Theory of Angular Momentum. New York: John Wiley and Sons, Inc., 1957 2. Brink D M and Satchler G R. Angular Momentum. Oxford: Clarendon Press, 1962.

Index

A Accelerator-driven sub-critical device, 629 Analyzing power, viii, ix, 2, 40, 41, 45, 67, 73, 84, 98, 143, 154, 156, 158, 163, 171, 182, 187, 196, 199, 203, 212, 214, 225, 229, 240, 245, 315, 316, 318, 320, 382, 394, 410, 441, 550, 609, 614, 616, 618, 627–629, 632, 635, 640, 642–645, 658, 680, 689 Angular distribution, vii, xiii, 40, 47, 171, 187, 199, 204, 269–276, 294–297, 382, 394, 550, 579, 628, 629, 635, 709 Angular momentum coupling, 64, 77, 159, 179, 192, 204, 286, 424, 592 Angular momentum operator, 10, 286, 444 Anti-commutation relation, 15, 25, 252, 467, 472 Anti-nucleon, 472, 475, 477, 489, 518, 584 Anti-symmetric wave function, 438 Asymmetric nuclear matter, 384, 506, 511, 595–598 Axis-symmetric rotational nuclei, v, 420 Axis-vector, 368, 369, 383, 384, 440, 441, 458, 459, 461, 493–495, 498, 593 Azimuthal angles, xi, xii, 9, 29, 38, 39, 47, 51, 53, 57, 60, 74, 100, 114, 131, 312, 525, 630–634, 670–686, 688

B Basis spin functions, 6, 7, 9, 13, 18, 23, 25, 26, 31, 61, 73, 74, 101, 102, 107, 108, 110, 117, 305, 306, 341, 360, 668 Binding energy, 249, 277, 511, 595, 597

Bjorken-Drell metric, 352, 353, 370, 448, 485 Bonn one boson exchange potential, xi, 572 Boson quantification rule, 462 Boundary conditions, 248, 272, 273, 276, 295, 297, 298, 332, 333, 376, 394, 419, 447, 462, 472, 614, 699, 705 Bound states, 250, 257–263, 268–270, 272, 277, 285–287, 382, 394, 439, 442, 579, 599, 614, 615 Breakup reactions, v, x, xiii, 258, 285, 618 Brueckner theory, 576, 580–584

C Canonical transformation, 331 Cartesian coordinate system, 4, 5, 7, 8, 32, 37, 56, 102, 105–108, 110–120, 125, 126, 128, 129, 131, 139, 252, 307, 325, 637, 660, 661, 666–668, 690, 693, 694, 702, 709, 710 Center-of-mass system, 44, 259, 260, 269, 275, 277, 279, 283–285, 297–299, 373, 437, 695 C–G coefficient, 3, 15, 20, 34, 63, 68, 78, 103, 105, 110, 134, 172, 188, 200, 205, 226, 244, 267, 306, 309, 310, 404, 405, 408 Charge conservation, 327 Charge density, 327 Charged particles, 204, 225, 391, 396, 629, 705, 710 Charge radii, 511 Clebsch-Gordan (C-G) coefficient, 3, 15, 20, 24, 34, 63, 68, 78, 103–105, 110, 134,

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 Q.-B. Shen, Polarization Theory of Nuclear Reactions, https://doi.org/10.1007/978-3-031-11878-4

717

718 172, 188, 200, 205, 226, 244, 267, 306, 309, 310, 404, 405, 408, 713–716 Collective deformation nuclei, 411 Collective inelastic scattering, xi, 411, 421, 424, 425 Commutation relation, 5, 10, 13, 15, 463, 464, 479 Completeness condition, 7, 10, 31, 102 Compton wavelength, 347 Conjugate equation, 349, 351, 365 Continuity equation, 352 Continuum discretized coupled channels (CDCC) theory, v, x Contravariant components, 11, 27, 73, 74, 112, 352–354 Coordinate system, xi, xii, 4, 9, 10, 12, 28, 29, 36, 37, 43–48, 55, 56, 74, 112, 116, 131–133, 249, 276, 278, 279, 284, 289, 290, 312, 332, 341, 345, 361, 362, 371, 374, 375, 378, 379, 382, 412, 433, 434, 443, 477, 479, 489, 576, 581–583, 585, 586, 594, 603, 635, 638, 639, 641, 658– 660, 667, 680, 689, 690, 692 Coordinate transformation relation, 47, 325 Correlation analyzing power, 84, 98, 213, 225, 241 Coulomb equation, 392, 393, 396 Coulomb field, 361 Coulomb gauge, 329 Coulomb potential, xiii, 244, 253, 259, 264, 288, 385, 391, 392, 396, 398, 411, 413, 421, 422, 439, 441, 447, 448, 459, 608, 613 Coulomb scattering amplitude, 34 Coulomb wave function, 269, 396 Coupled channel optical model, ix, 3, 705 Coupling channel equation, 274, 419 Covariant components, 113, 353, 354

D Darwin potential, 389, 608 Deformation parameters, 412, 424, 443 Density matrix, x, 16–21, 28, 29, 33, 35, 55, 60, 123–131, 139, 143, 154, 158, 311, 321, 340–342, 668 Deuteron induced reactions, 259, 707 D function, 9, 31, 74, 290 Diagonal matrix, 17, 28, 131, 356 Differential cross sections, 19, 33, 36, 39, 41, 42, 44–47, 61, 62, 67, 81, 98, 136, 146, 156, 158, 171, 187, 199, 225, 228, 229, 276–284, 297–300, 312, 320, 382, 394, 409–411, 419, 424, 425, 441, 442, 447,

Index 448, 599, 609, 610, 614, 616, 626, 627, 629, 631, 632, 635, 636, 638–641, 645, 658, 659, 669, 671, 674, 680, 681, 683, 689, 691, 693, 697, 698, 701, 704, 706, 708, 709 Differential operators, 354 Dirac-Brueckner-Hartree-Fock (DBHF), xi, 598 Dirac coupling channel theory, xi, 411 Dirac equation, viii, xiii, 2, 348–352, 354, 355, 357, 359–361, 363–365, 381–383, 392, 394, 395, 411, 412, 425, 436, 438, 439, 441, 448, 452, 458, 466, 467, 506, 580, 581, 610, 613, 707 Dirac propagator, 477, 506, 550 Dirac sea, 360, 461, 475 Dirac S matrix theory, v, xi, xiii, 395, 411, 707 Dirac spinor, 366, 451, 567, 570, 571, 583, 616 Dirac T matrix, 431 Direct inelastic scattering, 708–710 Direction factor, 698, 704, 705 Direct product, 12, 80, 207, 350 Distorted wave Born approximation (DWBA), ix, xi, 419, 422, 424, 706 Double definite integral, 517 Double-full shell target nucleus, 438 DWBA method, 3, 203 Dyson equation, 484, 487

E Effective interaction, 488, 565, 576, 581, 593–595 Effective mass, 511, 584, 595, 596 Effective nuclear force, 511, 609 Eigenfunctions, 6, 9, 10, 31, 291, 305, 360, 402, 420, 463 Eigenvalues, 357, 360 Einstein’s agreement, 350 Elastic scattering, viii, x–xii, 2, 3, 33–60, 69, 77, 84, 139, 143, 156, 158, 173, 179, 192, 204, 258, 259, 269–276, 285, 294, 295, 297, 311, 382, 394, 397, 399, 424, 425, 433, 441, 446, 550, 569, 599, 609, 610, 614, 618, 626–630, 635, 636, 708, 710 Electric current density, 327 Electromagnetic field theory, 326–329, 600, 610 Emitted particles, xii, xiii, 4, 38–40, 42, 47, 48, 51, 57, 60, 81, 98, 100, 187, 199, 204, 379, 409, 411, 635, 638–640, 659, 669, 689, 691 Energy-momentum tensor, 449

Index Equal-time commutation relation, 462, 479, 482 Equilibrium compound nuclear reactions, 707, 708 Euler angles, 9, 55, 290, 420, 660 Euler-Lagrange equation, 449 Even-even nucleus, 432 Expectation value, 19, 20, 28, 60, 62, 114, 127, 128, 230, 485

F Faddeev equation, 241, 616, 618 Fermi covariables, 592, 593 Fermi energy, 511 Fermi momentum, 384, 506, 595, 596 Fermi movement, 432, 583 Fermions system, 461 Fermi sea, 489 Fermi surface, 473, 475, 512, 551 Few-body nuclear reaction, 706 Feynman diagrams, 484, 485, 488, 506, 512, 550, 551, 573 Feynman propagator, 477, 489, 615 Feynman rules, 484–486, 488, 491, 493, 512, 522, 552, 565, 574, 581, 617 Final state, 18, 35, 50, 52, 56, 60, 62, 79, 272, 274, 275, 295, 296, 375, 397, 400, 441, 442, 565, 569, 574, 576 Four-dimensional energy-momentum vector, 354 Fourier transformation, 440

G Gamma function, 34 Gauge invariance, 328 Gauge transformation, 328 Generalized coordinate, 450, 462, 467 Generalized momentum, 330, 450, 462, 467 Generalized Pauli matrix, xi, 321, 323, 325, 668, 709 Generation and annihilation, 337, 361 Ground states, 251, 253, 255, 256, 258, 259, 285, 291, 305, 420, 432, 442, 461, 477, 484, 564, 579, 598

H Hamilton canonical equations, 329–331 Hamiltonian density, 450, 452, 467, 508 Hamiltonian operator, 348 Hamilton quantity, 330

719 Hartree-Fock, 484, 488, 514, 581, 593 Heisenberg representation, 484 Helicity basis functions, 9, 10, 31, 32, 117 Hermitian conjugate functions, 6, 73 Hermitian operator, 463, 479, 481

I Impulse approximation, 437, 441–443, 616 Incident particles, v, viii, x, xiii, 2, 4, 29, 33–37, 39–52, 57–63, 67–69, 74, 80, 98–100, 130, 131, 134, 139–141, 143, 146, 153, 154, 156, 158, 159, 171, 173, 179, 184, 185, 187, 192, 196, 197, 199, 214, 242, 258, 311–313, 320, 373, 378–380, 391, 392, 394–397, 399, 400, 407, 409–411, 413, 424, 447, 576, 609, 613, 626–629, 631, 632, 634–640, 645, 657–659, 668– 670, 674, 680, 683, 688–694, 705, 706, 708, 710, 711 Incident plane wave, 392, 398, 437 Incompressible coefficient, 511, 595 Indefinite integral formulas, 490, 502, 555 Independent particle shell model, 438 Inelastic scattering, xi, xiii, 3, 259, 285, 286, 294–298, 413, 419, 420, 424, 441–443, 708, 710 Initial state, 18, 20, 35, 52, 56, 62, 79, 285, 321, 375, 378, 395–397, 399, 400, 441, 442, 460, 565, 574, 576 Inverse transformation relation, 8, 324 Irreducible tensor, x, 9–13, 20, 114, 125, 254 Isospin analogous state (IAS), 447 Isospin operators, 443, 446 Isospin scalar, 443, 452, 453, 572 Isospin value, 443 Isospin vector, 443, 454–456, 485, 572, 573 Isospin vector flow, 455, 456, 460 Isospin wave functions, 445, 446

J j–j coupling, 3, 62, 159, 178, 179, 192 3j symbol, 8 6j symbol, 716 9j symbol, 292

K Kinetic energy, 48, 259, 260, 278, 282, 285, 346, 382, 434, 507–509, 595, 607, 625 Klein-Gordon (KG) equation, 453

720 L Laboratory system, 49, 53, 60, 276, 278, 279, 282, 284, 297, 299, 300, 373, 375, 378, 379, 383, 434, 609, 635–637, 640, 658, 659, 669, 670, 680, 689, 692, 693, 695, 699, 708, 710 Ladder approximation, 565, 614 Ladder function, 521 Lagrange equation, 330, 452, 453, 455, 457, 458, 599 Lagrange function, 329, 449 Lagrangian density, xi, 449, 451–458, 461, 467, 477, 493, 498, 499, 572, 599 Laplace operator, 346 Legendre function, 3, 34 Light beam polarization theory, 707 Light complex particle, x, xiii, 258, 285, 626, 629 Linear independent components, 13, 103, 309 Local density approximation, 382, 506, 550, 598 Lorentz covariant, 345, 346, 352, 361, 565 Lorentz force, 326 Lorentz gauge, 328, 454, 456, 457, 477, 478, 599 Lorentz invariant, 375, 381, 434, 437, 570, 583, 584, 594

M Magnetic field, 26, 28, 131 Magnetic quantum number, vii, 2, 9, 26, 161, 180, 194, 230, 270, 395 Many-body theory, xxv, 484, 565 Mass-energy relation, 346, 348, 394, 425 Mass operator, 438 Matrix element, ix, xiv, 3, 7, 9, 14, 17, 19, 27, 34, 38, 55, 63, 78, 79, 108, 110, 111, 116, 120, 129, 134, 136, 139, 158, 161, 170, 180, 186, 194, 198, 203–205, 227, 241, 246–248, 272–274, 276, 290, 294, 307, 311, 312, 363, 364, 371, 375, 394, 397, 399, 407, 419, 436, 437, 447–449, 484, 571, 577, 579, 596, 603, 614, 705, 706 Maxwell equations, xi, 326, 328 Mean field theory, 461, 564, 565, 598 Meson coupling constant, 455, 598 Meson exchange, 413, 512–550, 572, 576, 592, 593, 609 Meson field, 347, 460, 461, 483, 573 Minkowski space, 352, 353, 361 Momentum-energy 4 vector, 371, 372, 380

Index Momentum space, 426, 438, 466, 475, 481, 483, 579 Momentum wave vector, 278, 283, 421, 429, 432, 433 Monte-Carlo method, v, xii, xiii, 636, 710

N Negative energy nucleons, 461, 475, 616 Non-relativistic approximation, 436 Non-relativistic theory, 411, 447, 485 Normalization coefficient, 360 Normalization condition, 6, 16, 27, 28, 31, 61, 73, 74, 102, 123, 253, 264, 288, 698 Nuclear devices, 305, 629, 693 Nuclear force, vii, 1, 4, 244, 249, 251, 258, 262, 284, 346, 347, 381, 413, 437, 440, 441, 494, 572, 579, 580, 629, 711 Nuclear matter, xi, 382–384, 486, 487, 489, 506–511, 513, 518, 564, 567, 580–585, 592, 594–598 Nuclear potential, vii, 1, 269, 271–274, 288, 383, 391–393, 411–413, 419, 422, 609, 613, 711 Nuclear reaction amplitudes, viii, ix, 2, 705 Nuclear transport process, 4 Nucleon flow density, 384, 460 Nucleon-meson interaction, 484–486 Nucleon-nucleon elastic scattering, 63, 78 Nucleon-nucleon interaction, 256, 258, 451, 485, 594, 609 Nucleon scalar density, 459, 460 Nucleon tensor flow density, 460

O Off-shell, 466, 473, 477, 506, 550, 598 On-shell relation, 358, 521 Orthogonal Cartesian coordinate system, v, xi, xiii, 127, 322, 324, 325, 661, 665, 666, 668, 703, 706, 709 Orthogonal complete basis, 17 Orthonormality condition, 7, 10 Orthonormal property, 17, 60 Outgoing particles, vii, ix, 36, 37, 44–46, 61– 63, 68, 69, 74, 81, 99, 130, 131, 142, 143, 146, 153, 158, 159, 162, 171–173, 179, 181, 184, 185, 192, 194–197, 200, 214, 313, 315, 320, 375, 395, 409–411, 424, 430, 630–634, 636, 637, 639, 640, 645, 657–659, 668–670, 674, 679, 680, 683, 688, 689, 691, 692, 694, 695, 710

Index P Parity conservation, ix, 37, 78, 178, 205, 226, 240, 242, 246, 262, 384, 705 Partial integral, 330, 449, 451 Pauli incompatibility principle, 360, 565 Pauli matrix, viii, x, 2, 13, 23–26, 32, 37, 252, 321, 322, 350 Pauli metric, 352, 361, 371, 383, 384, 574, 585 Phase shift analysis method, ix, 3, 78, 204, 241, 706 Phase shift factor, 34 Phase space theory, 379 Phenomenological optical potential, x, xi, 242, 244, 245, 256, 382, 394, 412, 609, 627 Photon beams, xi, 4, 326, 339–342 Photon field, 461 Plane wave, 332, 340, 342, 355, 357–360, 376, 392, 430, 432, 436, 463, 479, 481, 506, 507, 573, 582, 594 Poisson bracket, 330, 331, 337 Polar angles, xii, xiii, 29, 47, 51, 53, 57, 609, 635, 658, 680, 688, 709 Polarization degree, 33, 39, 42, 697 Polarization density matrix, 18, 27, 29, 32, 36, 39, 80, 207, 228 Polarization direction, 28, 29, 37, 48, 131, 342, 667, 696, 698, 702, 703 Polarization nuclear database, v, xii–xiv, 4, 60, 635, 636, 638–640, 658, 659, 669, 680, 688, 689, 691, 692, 696, 699, 708–711 Polarization observables, viii, 2, 3, 48, 60, 394 Polarization operator, x, 12–15, 19, 20, 55, 101–103, 105–130, 307, 325 Polarization phenomenon, v, vii, viii, x, xi, xiii, xiv, 1, 2, 4, 242, 394, 626, 706, 707, 711 Polarization rate, viii, ix, xii, 2, 28, 44, 68, 73, 74, 77, 80, 131, 214, 311, 342, 627, 632–636, 639, 640, 645, 657, 659, 668, 679, 688–690, 692, 696, 699, 707, 708 Polarization tensor, 113, 125–127, 131, 140, 142, 147, 153, 162, 163, 171, 172, 182, 185, 195, 197, 199, 200, 208, 322, 324, 661, 666–668, 699, 702–705 Polarization theory, v, viii–xiv, 1–4, 13, 33–60, 62–90, 92, 95, 98–103, 105–134, 139, 143, 159, 161, 172, 178, 192, 194, 204, 225, 258–276, 305–316, 318, 320–326, 339–342, 629–634, 668, 705–709, 711 Polarization transfer coefficient, ix, 45, 66, 68, 73, 98–100, 146, 147, 153, 166, 171, 184, 185, 187, 196, 197, 199, 213, 214, 320, 410, 634, 635, 640, 649, 658, 680, 689

721 Polarization vector, 1, 28, 33, 35, 39, 40, 42, 44–46, 53, 55, 56, 61, 62, 81, 100, 123, 131, 140, 142, 146, 147, 153, 162, 163, 171, 172, 181, 182, 184, 185, 187, 194, 196, 197, 199, 200, 208, 313, 315, 322, 409–411, 633, 634, 636, 638, 668, 696– 699, 702–705 Polarized ion source, 28 Polarized nucleus transport theory, xi–xiv, 4, 60, 626, 629, 630, 708–711 Polarized particle beams, 26 Polarized particle transport theory, xi, xii, 625– 630, 636–640, 659–662, 664–670, 689– 692, 705–711 Positive electron, 360 Positive energy nucleon, 382, 389, 392, 394, 425, 473, 475, 507, 574 Positive energy plane wave, 360, 381, 436 Positive energy solution, 360, 427 Pre-equilibrium reaction theory, 3 Principal value integral, 476, 519 Probability conservation, 349, 352 Probability density, 352 Probability flow density, 352 Proca equation, 454, 477, 599–603, 606–610 Propagator, 427, 462–485, 488, 494, 498, 512, 513, 515, 528, 531, 533, 535, 537, 546, 550, 565, 566, 569, 573–575, 581, 582, 596 Pseudo-scalar, 368, 369, 383, 384, 434, 440, 441, 456, 458, 459, 461, 493, 494, 498, 533, 572, 593, 594

Q Quantum effect, 626 Quantum hadron dynamics, xi, 448, 451–458 Quantum number, 1, 62, 246, 270, 697, 704

R Racah coefficient, 15 Reaction amplitudes, ix, xi, xiii, xiv, 3, 18, 62– 65, 69, 74, 77, 159, 161, 173, 178–180, 192, 193, 204, 226, 227, 241, 276, 277, 298, 375, 399, 404, 407, 408, 424, 630, 640, 670, 680, 705, 707 Reaction mechanism, vii Realistic nuclear force, 3, 258, 565, 609 Recursion method, xi, 324, 325 Reduced mass, 34, 49, 245, 259, 392, 413, 424, 437, 447, 448, 609 Relativistic Bethe-Salpeter (BS) equation, 565

722 Relativistic classical field theory, xi, 448–451 Relativistic Green function theory, xi, 461 Relativistic impulse approximation, xi, 425, 441, 443 Relativistic mean field, 458–461, 564, 565, 598 Relativistic microscopic optical potential, xi, 488–506, 550, 551, 556, 580, 598 Relativistic nuclear reaction, v, xi, xiii, 4, 395, 411, 614, 616, 707, 711 Relativistic optical model, viii, xi, 2, 382, 393, 394, 420–422, 441, 707 Residual nucleus, viii–x, xii–xiv, 2–4, 62, 63, 65, 67, 69, 73–78, 143, 161, 170–173, 178, 181, 186, 188, 192, 194, 198–200, 285, 300, 320, 375, 395, 409–411, 420, 630, 631, 635, 640, 670, 671, 680, 681, 705, 707, 708, 711 Residual theorem, 429, 466, 514, 515, 568, 571 Resonance group theory, 241 RIA method, 441–443, 448 R matrix theory, ix, 3, 158, 241, 706 Rotation band, 420 Rotation matrix, 55, 116, 577, 665, 709 Rotation operator, 9, 116

S Scalar boson, 462–466 Scalar meson, 452, 462, 464, 572–574 Scalar potential, 327, 328, 334, 346, 412, 457, 600, 609, 612, 613 Scalar product, 11, 20, 125, 126, 254 Scattering amplitudes, x, 3, 33, 35, 37, 39, 51, 60, 134, 154, 272, 275, 297, 311, 312, 394, 425, 431, 434, 436–438, 440, 441, 609 Schrödinger equation, 2, 245, 249, 251, 261, 262, 272, 392, 393, 447, 609 Schrödinger-like equation, viii, 393, 422, 606– 609, 707 Self-energy operator, 425, 484, 486, 509, 513 Self-rotation motion, 626 Semi-odd number, vii, 10, 11, 69, 173, 189, 201, 285, 293, 467, 626 Separation variable method, 332 Shape elastic scattering, 635, 636, 658, 659, 695, 708–710 Single particle energy, 487, 488, 595 Single particle level, 1 Skyrme force, xxv, 627 S-L coupling, 3, 63, 68, 159, 172, 179, 188, 192, 200

Index S matrix, ix, xiii, 3, 34, 63, 78, 134, 158, 203, 204, 241, 248, 258, 271–274, 276, 294, 311, 375, 394, 397, 399, 419, 447, 448, 579, 609, 614, 705–707 Space-time coordinate, 345, 364, 479, 480, 574, 585 Space-time vector, 352, 353, 361 Speed of light, 326, 332, 345, 361 Spherical basis coordinate system, v, x, xi, xiii, 2, 4, 7, 8, 102, 103, 105, 107, 108, 110, 111, 113, 114, 117, 119, 125, 126, 134, 136, 139, 154, 156, 325, 443, 661 Spherical Bessel equations, 391–393 Spherical coordinate system, 4, 252–254, 428, 609 Spherical harmonic function, 3, 10, 34, 243 Spherical Neumann function, 391 Spherical optical model, 3, 34, 158, 260, 424, 446 Spherical symmetric system, 441 Spinless target, viii, x, xiii, 2, 3, 33–46, 60, 62, 134, 139, 143, 156, 158, 242, 311 Spin magnetic quantum number, vii, ix, xii, 1, 28, 29, 61, 62, 69, 74, 78, 131, 132, 173, 188, 200, 205, 226, 409, 421, 425, 626, 630, 640, 670, 680, 694, 696, 698, 699, 702, 711 Spin observables, 12, 13 Spin operator, x, 4–8, 13, 23, 24, 55, 101–103, 105–130, 266, 305–307, 325, 341, 342, 407, 440, 603–606, 610, 611, 668 Spin-orbital coupling potential, vii, 1, 242, 253, 255, 394, 614, 626 Spinor field, 451, 467, 472, 594 Spin projection, 4, 6, 9, 27, 31, 381, 426, 567, 580, 698, 703 Spin rotation function, 44, 382, 394, 441, 550, 635 Spin space, 18, 19, 21, 125, 126, 128, 129, 242, 309, 322, 323, 325, 407, 610, 666 Spin wave function, 18, 26, 27, 29, 60, 61, 73, 74, 116, 117, 123–130, 230, 261, 285, 360, 381, 395, 422, 426, 577 Stationary mass, 338, 346, 372, 374, 382, 395, 411, 422, 425, 451–453, 455, 467, 477, 514, 566, 611 Stationary particles, 357, 374 Statistical average, 3, 18 Stokes vector, 46, 48–50, 52, 56, 60, 637, 639, 659, 669, 689, 691, 696, 700 Symmetric nuclear matter, 506–509, 583, 592–596 Symmetric tensor, 106, 130

Index T Target nuclei, v, 278, 382, 393, 424, 609, 628 Tensor polarization rate, x, 55, 130, 133, 134, 143, 207, 658, 689 Thompson equation, 572, 576, 579, 580, 592, 594, 618 Three-body reaction, 707 Three-body reaction theory, 3 T matrix elements, 3, 204, 432, 437, 438, 592–595 Trace formula, 13 Transition matrix elements, 339, 377, 380, 381, 421, 424 Transverse field, 327 Transverse gauge, 329 Transverse potential, 329 Two-body direct reaction, ix, x, xii, 3, 630, 635, 636, 658, 659, 680, 688, 689, 695, 706 Two-body nuclear reaction, 3, 395, 411 Two-nucleon center of momentum system, 570, 572

U Unitary transformation relation, v, x, xi, xiii, 324, 661, 706 Unit matrix, 17, 37, 55, 80, 306, 367, 368, 666 Unpolarized incident particles, vii, ix, 40, 42, 44, 45, 61, 62, 67, 68, 73, 136, 142, 162, 171, 181, 187, 194, 195, 199, 315, 409, 633, 635, 640, 641, 658, 671, 680, 681, 689 Unpolarized particle beams, 26

V Vacuum region, 329 Vacuum state, 361, 464, 477, 484

723 Vector meson, 453–455, 477, 478, 480, 481, 483, 485, 572–575 4-vector momentum, 565 Vector polarization rate, x, 130, 143, 207, 680, 689 Vector-tensor coupling terms, 504 Vertex, 377, 459, 485, 486, 495, 498–500, 504, 531, 533, 537, 550, 572, 575

W Walecka model, 451 Wave function, viii, 2, 6, 9, 10, 12, 17, 27, 28, 30, 123, 171, 186, 199, 245, 246, 248–251, 253, 256, 258, 261–264, 269–275, 286–288, 294–296, 341, 346, 348, 351, 360, 364, 375, 376, 381, 384, 388–391, 395–400, 414, 415, 419, 421, 422, 424–426, 439–442, 446–448, 462, 485, 495, 507, 577, 578, 600, 609, 610, 614, 616 Weinberg equation, 610–614 Wick theorem, 484 Wigner D function, 10, 12, 116 Woods-Saxon potential, 412

Y Yukawa potential, 347, 441

Z Zero-range force approximation, 437