Quantum Theory of Angular Momentum: educational manual 9786010426733

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AL-FARABI KAZAKH NATIONAL UNIVERSITY

М. А. Zhusupov R. S. Kabatayeva K. A. Zhaksybekova

QUANTUM THEORY OF ANGULAR MOMENTUM Educational manual

Almaty «Qazaq university» 2017

UDC 539.1 (075.8) LBC 22.3 я 383 я 73 Zh 99 Recommended for publication by the decision of the Academic Council of the Faculty of Physics and Technology and Editorial and Publishing Council of Al-Farabi Kazakh National University (Protocol №4 dated 26.05.2017) Reviewers: doctor of Physics and Mathematics, Professor Yu.V. Arkhipov candidate of Technical Sciences, Associate Professor G.А. Abdraimova

Zhusupov М.А. Zh 99 Quantum Theory of Angular Momentum: educational manual / М.А. Zhusupov, R.S. Kabatayeva, K.A. Zhaksybekova. – Almaty: Qazaq university, 2017. – 120 p. ISBN 978-601-04-2673-3 In the educational manual written on the base of the special course being given during a number of years at the Department of Theoretical and Nuclear Physics of the Faculty of Physics and Technology of al-Farabi Kazakh National University the fundamentals of quantum theory of angular momentum are given, a problem for eigen functions and eigen values of angular momentum operator is formulated, quantization and matrix elements of the momentum operator, vector coupling of two, three and more momenta are given, construction of spin and orbital wave functions of particles, spin quantum theory, problems and examples are detailed. This manual is assigned for students, master students and PhD students attended a general course of quantum mechanics in a volume being given to students of the department of physics. The knowledge of nuclear physics in a volume of a course of general physics is assumed. Publishing in authorial release.

UDC 539.1 (075.8) LBC 22.3 я 383 я 73 ISBN 978-601-04-2673-3

© Zhusupov М.А., Kabatayeva R.S., Zhaksybekova K.A., 2017 © al-Farabi KazNU, 2017

2

INTRODUCTION

An angular momentum is one of the important quantum characteristic of micro-world objects. This term denotes the orbital momentum of relative motion, spin momentum of particles, total momentum (spin) of a system of particles and etc. By its properties the isotopic spin of strongly interacting particles could be referred to the term. The momenta are characterized by quantum rules of adding, unusual from the point of view of classical representations. The most important property is a quantization. These rules are from the commutation properties for momenta operators. But one does not need the explicit form of the momenta operator – it is enough to know their commutation properties. Only for the operator of orbital momentum on the base of the principle of correspondence one can find the explicit form in the coordinate representation. It is believed that because of these restrictions (necessity of the explicit form beside the commutation relations) the rules of quantization for orbital momentum are stronger. Indeed, the orbital quantum number takes only integer values  0,1,2,... , but the spin quantum number j in dependence on the type of particles can take either integer or half integer values. Note, that the harder rules of quantization can be connected with the fact that the spatial parity of relative motion of particles a and b is defined by the orbital momentum of their relative motion  1 . The special place in the theory is taken by quantum rules of momenta coupling. Knowledge of the rules is necessary for construction of wave functions with total angular momentum. In this case the special coefficients of vector addition or ab

3

Clebsch-Gordan coefficients are introduced. In the manual the general formulas for their calculation are given. The derivation for the case when one of the momenta j 

1 is given. The 2

numerous examples of use of coefficients of vector addition in quantum applications are given. In a case, when one needs to add three or four momenta, the Racah coefficients (6 j -symbols) or generalized Racah coefficients (9 j -symbols) are introduced respectively.

4

Chapter I

ORBITAL MOMENTUM AND ITS QUANTIZATION §1. Motion in centrally symmetrical field. Integrals of motion The centrally symmetrical field is a field potential of which depends only on the absolute value r of the radius-vector r , that is V  r   V  r  , where r  x 2  y 2  z 2  is the module of the radius-vector. Problems for such potentials are convenient to consider in the spherical coordinate system. Since the potential does not depend on the angles, then the variables are separated: x, y, z  r , , ,

  x, y, z, t     r , , , t     r , t 

A connection of the Cartesian coordinates with the spherical ones is realized through the expressions: x  r sin  cos  y  r sin  sin  z  r cos .

Let us consider the stationary problem. In this case the time dependence of the wave function is ei  Et :

 r ,t    r   e

i  Et

.

In this case the total energy is the integral of motion. So the values of the energy of the system are defined from the Schrodinger stationary equation: 5

Hˆ   r , ,   E  r , ,  .

For the complete description of the system it is necessary to define all the integrals of motion. It is obvious, that along with the total energy in the centrally symmetrical field there are also other integrals of motion. Let us define them. §2. Parity of quantum states The parity is the quantum mechanical notion which does not have any classical analogue. The parity of a state characterizes the behavior of the wave function of a system when inversion the coordinates that is when the change occurs x   x, y   y, z   z . 1. Introduce the operator Iˆ , describing the inversion of coordinates in the wave function, – inversion operator:

Iˆ  x, y, z,...     x,  y,  z,... .

(2.1)

From the other hand one can write the operator equality for the inversion operator

Iˆ  I  ,

(2.2)

where I – Eigen values of the operator Iˆ . It is easy to find them since the double application of the operator to the function  should again give the initial function:

Iˆ2   I 2    .

(2.3)

It is not difficult to show that the inversion operator is Hermitian. Hence, the Eigen values of the inversion operator are real:

I 2  1 and I  1 . 6

The state of a system described by the wave function which does not change its sign when inversion of the coordinates is called even parity state (the parity is positive). In the case of the wave function’s sign change to opposite sign one has odd parity state of the system (the parity is negative). Whether the Eigen values of the inversion operator, that is the parity of the system, are integrals of motion? Let us consider the nuclear Hamiltonian A ˆ2 p Hˆ   i  Vij . i 1 2m i j

(2.4)

In this expression the first term is the operator of total kinetic energy of all nucleons, and the second one – is the operator of the potential energy of interaction. When inversion of the coordinates both the terms of the Hamiltonian (2.4) do not change. The first term contains the second order derivatives with respect to coordinates, and the potential energy of particles is a function of only relative position of the particles. If Hˆ – is an invariant with respect to inversion that is

Hˆ (r )  Hˆ (r ) , ˆ ˆ (r )(r )  Hˆ (r )(r )  HI ˆ ˆ(r ) . then IH Hence it is easy to see that the inversion operator commutates with the Hamiltonian (2.4)

[ Iˆ, Hˆ ]  0 , and, hence, in accordance with the rules of quantum mechanics the parity is the integral of motion. The law of parity conservation is also multiplicative. 2. Let us find the parity of two non-interacting particles a and b . The wave function of a system can be written in the following form:

 a b   a  b  la  lb   a  b  lab , 7

(2.5)

here  a and  b – inner wave functions of particles a and b . The function  lab describes their relative motion. The parity of the total system is defined by the following expression:

I a b  I a  Ib  Ilab ,

(2.6)

I a and I b – are the so called inner parities of particles a and b . An experiment showed that the inner parity of the proton and neutron is the same: it is chosen to be positive

I p  I n  1 .

I lab – parity of relative motion. Let us define it. The wave function of the relative motion in the centrally symmetrical field is represented in the following form:

 nlm  r , ,    Rnl  r   Pl m  cos   eim .

(2.7)

Rnl  r  – radial wave function, Pl m  cos   – associated Legendre polynomial. When inversion of the coordinates r  r one has:

r  r;      ;      . At this

cos   cos       cos ; eim  eim     1 eim ; m

Pl m   cos     1

l m

8

 Pl m  cos   .

Thus,

 nlm  r   Rnl  r  1

l m

 Pl m  cos     1  eim  m

(2.8)

  1   nlm  r  l

and the parity of the relative motion is defined by the orbital momenta of relative motion

I lab   1 ab . l

(2.9)

3. For the arbitrary nuclear reaction

ab cd

(2.10)

one can formulate the law of parity conservation

Ii  I f ,

(2.11)

where I i and I f – parities in initial and final states respectively, defined by the formula (2.6). For the reaction (2.10) one can also write the laws of total angular momentum conservation:

Ji  J f , Ji  J a  Jb 

(2.12)

ab

,

where

(2.13)

J f  Jc  Jd 

cd

The laws (2.11) and (2.12) give a total series of selection rules with respect to spin and parity for nuclear reactions. 4. If an atomic nucleus is presented in a form of ensemble of nucleons moving independently, as it is, for example, in the nuclear 9

shells model, then the parity is equal to the product of the inner

parities of the nucleons by the multiplier  1i i , where li – orbital l

quantum number for certain nucleon. 5. The inner parity is the quantum number which is to be referred to particles if one wants to consider the conservation of parity in the processes of strong interaction. A certain sort of elementary particles and nuclei is characterized by the certain values of spin and parity, which definite type wave functions correspond to. In the table 1 there are possible types of wave functions and particles which the wave functions describe. Table 1 Types of wave functions Spin 0 0 1 1

1/ 2 3/ 2 2

Parity +1 -1 -1 +1 +1

Type of wave function scalar pseudoscalar vector pseudovector or axialvector spinor (tensor of half order) halfspinor of odd order second order tensor

Particles  -particles  -mesons photons deutrons

p, n

 -meson graviton

Since, for example,  -mesons have zero spin and negative parity that is I   0 , then they are described by the wave functions which when coordinates inversion will be transformed as the pseudoscalars. That is why such particles are called pseudoscalar. Note, that for fermions the parity of antiparticles is opposite to particles parity, but for bosons the parity of antiparticles coincides with the particles parity. §3. Operator of orbital angular momentum In classical mechanics the projections of orbital momenta are the integrals of motion. See, what the situation in quantum mechanics is. The projections of angular momenta in the Cartesian system have the following form: 10

    Lˆx  i  y  z  , y   z    Lˆ y  i  z  x  , z   x     Lˆz  i  x  y  . x   y

The operator of angular momentum squared Lˆ2  Lˆ2x  Lˆ2y  Lˆ2z . Let us transfer to the spherical coordinates system:

  r sin  x   cos  y   sin  z   cos 

x2  y 2   2 .   x  y  z      x  y  z      r cos cos   r cos sin   r sin  . x y z

  x  y  z      x  y  z  0

     r sin  sin   r sin  cos   y x . x y x y  Hence, Lˆz  i .  Find Lˆ . y

11

 xz  yz         x  y z        y x  y x y

(3.1)

 yz   cos  ,   2    sin   ctg .    

x

From the left side there should not be any r , and from the right   side one has in order to find the Lˆ y and – to find the Lˆ x . x y  sin   r  cos  cos       x 2 z  xyz   y 2 z  xyz   2  2 x  2  .  x  y z  x  2 y 2 2   z  x  y    cos   sin   ctg  x ; 2    x z     cos   sin   ctg z x .   x z

Comparing with the expression for the Lˆ y one sees that     . Lˆ y  i  cos   sin   ctg      

Find Lˆ x :  xz  yz         x  y z        y x   x y

 x 12



y

 , eliminate xz  2 

  cos   r cos     r 2 sin 2   xyz  y 2 z   xyz  x 2 z     2  2 y  2  2 z y .  x  y z  x  y y z sin 

    Lˆx  i  sin    cos   ctg .    

Hence it is easy to find the form of the momentum squared operator in the spherical coordinates system: Lˆ2  

2

 1      sin    sin   

1 2    .  2 2   sin   

Commutative relations between the component of the momentum Lˆi and the Hamiltonian Hˆ Show that the operators Lˆ commutate with the Hamiltonian Hˆ i

of the centrally symmetrical field. 1. Hˆ  Tˆ  V  r  Find Tˆ , Lˆx  .  pˆ i , pˆ k   0 ,  xˆi , xˆk   0 ,  pˆ i , xˆk   i ik . 2 2 2 pˆ 2 pˆ x  pˆ y  pˆ z ˆ T  2m 2m ˆ ˆˆ ˆ ˆ Lx  ypz  zp y .

а)  pˆ x2 , Lˆx   0 , it is obvious, since Lˆ x does not contain xˆ . b)  pˆ y2 , Lˆx   pˆ y2  yp ˆˆ z  zp ˆˆ z  zp ˆˆ z  pˆ y2 zp ˆˆ z pˆ y2  ˆ ˆ y    yp ˆ ˆ y  pˆ y2  pˆ y2 yp ˆ ˆ y  yp 



    ˆ ˆ z  yp ˆ ˆ y2 pˆ z   pˆ y2 , yˆ  pˆ z   pˆ y  pˆ y , yˆ    pˆ y , yˆ  pˆ y  pˆ z  ˆ ˆ 3y  pˆ 2 yp  zp    i  i  

 2i pˆ y pˆ z . 13

c)  pˆ z2 , Lˆx   pˆ z2  yp ˆˆ z  zp ˆˆ z  zp ˆˆ z  pˆ z2 zp ˆˆ z3  ˆ ˆ y    yp ˆ ˆ y  pˆ z2  pˆ z2 yp ˆ ˆ y  yp 



ˆ ˆ y pˆ    pˆ z2 , zˆ  pˆ y  2i pˆ z pˆ y .  zp 2 z

Thus, 1 1 Tˆ , Lˆx    ˆ2 ˆ2 ˆ2 ˆ  ˆ ˆ ˆ ˆ   2m  px  p y  pz , Lx   2m  2i p y pz  2i pz p y   0 .

Similarly, it is easy to show that  pˆ 2 , Lˆ y   0 ,  pˆ 2 , Lˆz   0 . Tˆ , Lˆx   0 ; Tˆ , Lˆ y   0 ; Tˆ , Lˆz   0 , then      

Tˆ , Lˆ2   0 .  

Now find the permutation relations between the Lˆx , Lˆ y , Lˆz and Vˆ  r   V  x, y, z  .

ˆ ˆ  i  y   z  V  x, y, z    Lˆx ,Vˆ   LˆxVˆ  VL   x   y   z        V  x, y, z   i  y  z   i  y  V  z  V  y  y  z  z V  y

Find y

   V  z . z y 

   V and z  V . z y

 V  V       V  ; z z z   V V V  ; z z z

14

Similarly   V . V V  y y y   V  LˆxV  VLˆx  i  yV  y  zV  z y  z  V V   V  z  Vy  Vz   i  y z   0. y z y  y   z

Thus, we obtained a certain operator which is not equal to zero.  V V  i  y z   i r  gradV x , y   z since i x V x

j y V y

k z . V z

For the centrally symmetrical field V  V  r  .  V r V r  LˆxV  r   V  r  Lˆx  i  y z  r y   r z V y   V z  i  y z   0. r r   r r

r  x2  y 2  z 2 , r x 

1  1 2 r y r z  ; x  y2  z2  2  2x  x r ;  ;  2 y r z r

15

Similarly  Lˆ y ,V   0 and  Lˆz ,V   0 ,  Lˆ2 ,V   0 .   Thus, the operators Lˆx , Lˆ y , Lˆz commutate with the Hamiltonian of a system and are the integrals of motion. Similarly, it is easy to show that  Lˆ2 , Hˆ   0 .  

Now find the commutative relations between the projections of the momentum  Lˆi , Lˆk  and  Lˆ2 , Lˆi  .  Lˆx , Lˆ y  1.   ˆ ˆ z  zp ˆˆ y Lˆx  yp ˆˆ z ˆ ˆ x  xp Lˆ y  zp ˆˆ y  yp ˆˆx. Lˆz  xp

 Lˆx , Lˆ y     ˆˆ z  zp ˆˆ z    zp ˆˆ z    yp ˆˆ z  zp ˆ ˆ y    zp ˆ ˆ x  xp ˆ ˆ x  xp ˆˆ y     yp

ˆˆ z  zp ˆˆ z  xp ˆˆ z  zp ˆˆ z  zp ˆˆ z  zp ˆ ˆ x  yp ˆ ˆ y  zp ˆ ˆ x  zp ˆ ˆ y  xp ˆ ˆ x  yp ˆ ˆ x  zp ˆˆ y   yp ˆˆ z  yp ˆ ˆ z  xp ˆˆ z  zp ˆ ˆ x  pˆ z zˆ  zp ˆˆ y  ˆ ˆ y  yp ˆ ˆ z   zp ˆ ˆ z  pˆ z zˆ  xp  xp  i

 i

ˆˆ x  xp ˆˆ y   i Lˆz .  i  yp

ˆˆ y  yp ˆˆ x . Lˆz  xp Similarly one can get for the other components through the cyclic permutation: 16

 Lˆx , Lˆ y   i    Lˆ y , Lˆ z   i    Lˆz , Lˆx   i  

2.

Lˆz Lˆ x Lˆ y

 Lˆ2 , Lˆi   

 Lˆ2 , Lˆx    Lˆ2x  Lˆ2y  Lˆ2z , Lˆx     

а)  Lˆ2x , Lˆx   0. b)  Lˆ2y , Lˆx   Lˆ y  Lˆ y , Lˆx    Lˆ y , Lˆx  Lˆ y  i  Lˆ y Lˆz  Lˆz Lˆ y  .  i Lˆz

 i Lˆz

c)  Lˆ2z , Lˆx   Lˆz  Lˆz , Lˆx    Lˆz , Lˆx  Lˆz  i  Lˆz Lˆ y  Lˆ y Lˆz  .  i Lˆ y

 i Lˆ y

Thus,  Lˆ2 , Lˆx   0.  Lˆ2 , Lˆ x   0   2 ˆ ˆ  L , Ly   0    Lˆ2 , Lˆ z   0  

It is known, that if the operators commutate with each other then they have the common system of Eigen functions. We have operators Hˆ and Lˆ2 which commutate with each other. In addition the operators Lˆx , Lˆ y and Lˆ z also commutate with them. But the last operators do not commutate with each other. Thus, Lˆx , Lˆ y and Lˆ z do 17

not have a common system of Eigen functions, but each of them has a common system of Eigen functions with the operators Hˆ and Lˆ2 . There is no contradiction here. Let us prove the following theorem: If one has two conserving physical quantities F and K , operators of which do not commutate with each other then the energy levels of the system are degenerated. Let   is a wave function of the stationary state in which along with the energy there is a certain value of the quantity F ˆ exists. Then the K does not coincide with the  (within the accuracy of an arbitrary multiplier), otherwise the quantity K would also have a certain value what is not possible since the Fˆ and Kˆ do not commutate with each other and cannot be measured ˆ simultaneously. But from the other hand the K is the Eigen function of the Hamiltonian referring to the same E that the  does:













ˆ   E Kˆ  , Hˆ Kˆ   Kˆ Hˆ   KE

that is there are more than one Eigen function corresponding to the level E , that is the level is degenerated. Thus, in the centrally symmetrical field the levels should be degenerated. Let  is the Eigen function of the Hˆ and Lˆ z , that is Hˆ   E

Lˆz   Lz  . Consider   Lˆx  . Since  Lˆx , Lˆz   0 , then   const   . ˆ ˆ   Lˆ Hˆ   Lˆ E  E Lˆ   E , that is But Hˆ   HL x x x x



Hˆ   E . 18



Hence,   is the Eigen function of the Hˆ , corresponding to the same E that the  does. The degeneration is shown. Conclusion: thus, a problem about motion in the centrally symmetrical potential is reduced to obtaining of the Eigen functions of three operators: Hˆ , Lˆ2 and Lˆ z . Instead of the Lˆ z there would be any of the projections Lˆ and Lˆ . But the Lˆ has the simplest form y

x

z

in the spherical coordinate system. Note that in these states the Lˆ x and Lˆ y will not have definite values. §4. Eigen functions of operators Hˆ , Lˆ2 and Lˆ z Let us transfer to the spherical system Hˆ  

2 

1  2   1 r  r 2 r  r  r 2

2

2

2  V  r  .

 1     1 2  sin    .     sin 2   2   sin   

The operator ˆ L2  

2

 1     1 2   2 ;  sin   2   sin   2   sin    Lˆz  i

 ; 

Hence, Hˆ  

 1   2   Lˆ2   r   2 2   V r  . 2  r 2 r  r  r  2

19

One should find the general solution of the following equations: Hˆ   r , ,   E   r , ,  Lˆ2   r , ,   L2   r , ,  Lˆz   r , ,   Lz   r , , 

Let’s first consider the last equation:   Lz  , 

i

  ce

i

Lz



 c  eim , m 

Lz

.

The function  m   should be a single-valued function of the variable  . For that the m should be an integer m  0, 1, 2,... . eim  eim  2 

e2 mi  cos 2 m  i sin 2 m  1 ,

m  0, 1, 2,...

If only m  1 2 , then the  m would be a double-valued function   i    

1  ei and 2  e  2

. If m  1 3 , then 3 such roots and so on. The constant c is found from the condition: 2

2

c

2

 eim  eim d  1

0

c

1 . 2 20

Thus, we obtained a quantization of the Eigen values of the projections of the moment on the axis z : Lz  m , m  0, 1, 2,... 1  m     eim . 2

Let us now consider the equation ˆ L2   L2  .

In the spherical system the equation has the form:  1     1 2 L2  sin      ,   0 .    2   sin 2   2  sin    

Let us compare this equation with the known equation for the spherical functions:  1      1 2 sin    l  l  1   Ylm  ,   0 .    2 2   sin    sin    

This equation has single-valued, continuous and finitesimal solutions if only the numbers l  0,1,2,...  integer and positive. The equations coincide if L2  2l  l  1 . Thus, the Eigen values of the angular momentum squared operator are defined by the quantum numbers l  0,1,2,... , and the Eigen functions of this operator coincide with the spherical functions Ylm  ,  of order of l . At this to certain l the  2l  1 spherical functions Ylm correspond. They differ by values of quantum number m , taking at the given l the values m  0, 1, 2,..., l . in total  2 l 1

21

The spherical functions can be presented in a form: Ylm  ,   lm   

exp  im  2

.

That is why the spherical functions are also the Eigen functions of the operator Lˆ z : LˆzYlm  ,   mYlm  ,  . Thus, the second index of the function Ylm allows distinguishing the states differing by the projections of the angular momentum on the axis z : 1 Y00  ; 4 Y1,1  

Y1,0 

Y1,1  Yl ,0 

3 sin   ei 8 3 cos 4 3 sin   ei 8

2l  1 Pl  cos  , Pl  Legendre polynomials. 4

 P0  1   P1  x   x 2  Pl  x    P  x   3 x  1 2 2  3  5 x  3x  P3  x   2  22

Y

Y d    ll  mm , d   sin  d d .

lm l m

Since in the spherical coordinates system for the central field the operator V  r  does not depend on the angles, then the Schrodinger equation’s solution can be searched in a form of product of a function dependent on the angles by another function dependent on only module of the radius-vector:   r , ,   R  r   Ylm  , 

(4.1)

Find the R  r  from the Schrodinger equation. After substitution of the wave function in a form (4.1), acting by the operator Lˆ2 to Ylm  ,  and cancelling the equation for the Ylm  ,  x , the Schrodinger equation takes the form: 

 1 d  2 d  l  l  1     R  r   V  r   R  r   ER  r  . r  2  r 2 dr  dr  r2  2

This equation is also called the radial Schrodinger equation. It defines the Eigen values of energy of a system. Denote the Eigen value through the E nl , where n  1,2,3,...  enumerates the Eigen values. n  first quantum number, l  orbital quantum number, m  magnetic quantum number. The wave function of a system is denoted as:  nlm  r , ,   Rnl  r   Ylm  ,  .

At this the energy E  Enl and does not depend on the m . The

multiplicity of this degeneration is equal to  2l  1 . The minimal multiplicity of the degeneration in the central field equals  2l  1 . 23

And it connects physically with the fact that the energy of a system does not depend on the projection of momentum on axis z . ˆ Thus, we showed, that the Eigen values of the operators L2 and Lˆ z are quantized. One can get the same results from the commutative relations not considering the Schrodinger equation’s solution and the explicit form of the operators (operator equations).

ˆ §5. Quantization of operator L2 from elementary formulas of probabilities theory One can show that the formula L2    1 can be obtained from the elementary formulas of the probabilities theory if one supposes that the possible projections of the momentum on the arbitrary axes equal m ( m  ,  1,...,  ) and all these values of the momentum projections are equally probable and the axes are equivalent. Here  1 . Due to the equal rightness of the axes x, y, z one has L2  L2  L2x  L2y  L2z  3L2z . Since the different values of the

projections of the momentum are equally probable then one has L2z 

1 m 2 2 m 2 m   m . 2  1 m 2  1 m0

The obtained sum of the integer numbers squared can be calculated by different ways. 1. Let us define the following expression:  d2 2 m    2 m0  d

d 2 1  e  1 m  e    2  m0   0 d  1  e

,  0

since the sum in the square brackets represents the geometric progression with common ratio q  e . After double differentiation 24

and expansion of the obtained ambivalence with use of L’Hospital rule one gets:

m

2





 1 2  1 6

m0

.

Finally one has L2  3 

2  2 1



 1 2  1 6





 1 .

2. Induction method. Let us consider the following table:

   

2

2

1 1

2 5

3 14

4 30

5 55

6 91

1

3

6

10

15

21

5 3

7 3

9 3

11 3

13 3

1

3 3

It is seen, that Since





 



2



 1

2

2 1 . 3

.

Hence:



2





 1 2  1 6

that gives the same formula for the L2 . 3. Method of finite difference. Denote S 

   k 2 . S  0  0 . k 0

25

,

S   1  S  Let us search for the S  S

  a0  a1





 1 . 2

(5.1)

in a form of expansion:

 a2

2

 a3

3

 a4 4 , a0  0 .

Substitute into (5.1):

a1   1  a2   1  a3   1  a4   1  a1  a2 2

3

a4

4



4

2

2

 a3

3



 2 1

Or a1  a2  2  1  a3  3

2

 3  1  a4  4

3

 ...  1 

2

 2 1.

For the arbitrary the equality is true if the coefficients at the same degrees of are equal. Setting equal to each other, one gets the system of equations: at

0

: a1  a2  a3  a4  1 .

at

1

: 2a2  3a3  2 .

at

2

: 3a3  1 .

at

3

: a4  0 .

Solving the equations, one gets: 1 1 1 a1  , a2  , a3  . 6 2 3

Hence S





 1 2  1 6 26

.

Problem: using this method find the sums S 

  k3

and

k 0

S

  k4 . k 0

Problems to chapter I 1. Which values can the orbital momentum L of two alpha-particles relative motion take? 2. Can the excited nucleus 4 Be8 with spin equal to 1 decay by two alphaparticles channel? 3. One has an excited nucleus with spin 1 in the even parity state. The reaction A  B  2 He4 is energetically possible. The stable nucleus B , forming in this reaction, has the quantum numbers I   0 . Show on the base of momentum and parity conservation laws that such a reaction is not possible. 4. In which state (ortho- or para-) the positronium is if it can decay with three gamma quanta emission? Test questions to chapter I 1. Which field is called potential? Which potential field is called centrally symmetrical? 2. What the operator, describing a system, should be that it’s Eigen values are the integrals of motion? 3. What properties should the Hamiltonian of a system possess that the parity is the integral of motion? 4. How is the parity of particles relative motion wave function defined? 5. What is the form of the operator Lˆ z in the spherical coordinate system? 6. What is the form of the Eigen functions of the operators Lˆ2 and Lˆ z ? 7. Is the parity of a quantum system an integral of motion? Why? 8. What does the degeneracy of energy levels of a system mean? Whether the energy levels of a system in the central field are degenerated? 9. Which values do the Eigen values of the operator of angular momentum projection and momentum squared operator take?

27

Chapter II

GENERAL THEORY OF ANGULAR MOMENTUM OPERATOR §6. General theory of angular momentum operator. ˆ Quantization of operators J 2 and Jˆ from commutative z

relations Lˆ Introduce the momentum operator Jˆi  i . Or for the sake of

simplicity let us put  1 . In the base of their definition the following permutative relations are taken: 1 i  Jˆx , Jˆ y   2  Lˆx , Lˆ y   Lˆz  iJˆz .      Jˆ x , Jˆ y   iJˆ z     ˆ ˆ ˆ  J y , J z   iJ x      Jˆ z , Jˆ x   iJˆ y    

 Jˆ 2 , Jˆ x   0     2 ˆ ˆ  J , J y   0     Jˆ 2 , Jˆ z   0    

Introduce the operators 1 Jˆ  2 1 Jˆ   2

 Jˆ

x

 Jˆ

Jˆ  non-Hermitian operator. 28

x

 iJˆ y



 iJˆ y



    Jˆ  

(6.1)

 Jˆ 2 , Jˆ   0 , what is obvious, since the Jˆ 2 commutates with   ˆ the J x and Jˆ y .



 



1 ˆ ˆ  Jˆz , Jˆ   J z J x  iJˆ y  Jˆx  iJˆ y Jˆ z      2   1 ˆ ˆ  1 ˆ ˆ  ˆ  J z J x  Jˆ x Jˆ z  i Jˆ z Jˆ y  Jˆ y Jˆ z     J x  iJ y   J  ; 2 2   iJˆ y  iJˆ x  







 



1 ˆ ˆ  Jˆz , Jˆ   J z J x  iJˆ y  Jˆx  iJˆ y Jˆz      2

 1 ˆ ˆ  J z J x  Jˆ x Jˆ z  i Jˆ z Jˆ y  Jˆ y Jˆ z 2   iJˆ y  iJˆ x 





  1  ˆ ˆ  ˆ   2 iJ y  J x    J  ;  

 Jˆz , Jˆ    Jˆ .  



1  Jˆ , Jˆ    Jˆ x  iJˆ y   2

 Jˆ

x

 

 iJˆ y  Jˆx  iJˆ y

(6.2)

 Jˆ

x



 iJˆ y   

1   Jˆx2  iJˆx Jˆ y  iJˆ y Jˆ x  Jˆ y2  Jˆ x2  iJˆ x Jˆ y  iJˆ y Jˆ x  Jˆ y2   2  i  Jˆ y Jˆx  Jˆx Jˆ y   Jˆz ; Calculate also the quantities:

2 Jˆ Jˆ  2 



1 ˆ J x  iJˆ y 2

 Jˆ

x







 iJˆ y  Jˆ x2  i Jˆ x Jˆ y  Jˆ y Jˆ x  Jˆ y2  iJˆ z

ˆ  Jˆx2  Jˆ y2  Jˆz  Jˆz2  Jˆz2  J 2  Jˆz2  Jˆz ; ˆ 2 Jˆ Jˆ  J 2  Jˆz2  Jˆz ; 29

ˆ ˆ ˆ Hence, 2 Jˆ Jˆ  2 Jˆ Jˆ  2 J 2  2 Jˆz2 , J 2  Jˆ Jˆ  Jˆ Jˆ  J z2 ; ˆ The operators J 2 and Jˆ z commutate, that is why they have the common system of Eigen functions and their matrices can be simultaneously reduced to the diagonal form. ˆ Consider J 2  Jˆ 2  Jˆ 2  Jˆ 2 . z

x

y

The Eigen functions of the operators Jˆ 2 and Jˆ z are denoted as J 2, Jz . J 2 , J z Jˆ 2  Jˆ z2 J 2 , J z  J 2  J z2  J 2 , J z Jˆ x2  Jˆ y2 J 2 , J z  0 

operator of essentially positive quantity

Hence, J 2  J z2 , that is at a certain fixed value of the J 2 the values of the J z are restricted both from up and down:  J 2  Jz  J 2 .

Denote J z  m

 J z max  mmax  J z min  mmin

Jˆz Jˆ  Jˆ Jˆz   Jˆ . Let us operate with this operator on the  m – Eigen function of the operators Jˆ 2 and Jˆ :   J 2 , J . z

m

z

Jˆz  m  m m





Jˆz Jˆ  m  Jˆ Jˆz  Jˆ  m Jˆz Jˆ  m   m  1 J   m , 30

that is the function Jˆ  m is the Eigen function of the operator Jˆ z with the Eigen values  m  1 .

















Jˆz Jˆ  m   m  1 Jˆ  m

Jˆz Jˆ  m   m  1 Jˆ  m

That is why the following is true Jˆ  m  const   m1 .

Thus the operator Jˆ transfers the wave function  m   m1 , and the operator Jˆ transfers  m   m1 .  0 , since the Jˆ is a raising operator, Further consider Jˆ  



mmax

Jˆ  mmin  0 , since the Jˆ is a lowering operator. Operate by the operators 2Jˆ and 2Jˆ , then one gets:

 Jˆ 



 Jˆ z2  Jˆ z  mmax  0   Jˆ 2  Jˆ z2  Jˆ z  mmin  0   2



Transferring to the Eigen values one gets: 2 J 2  mmax  mmax  0   2 2 J  mmin  mmin  0  

2 2 mmax  mmax  mmin  mmin  0 .

2

1  1 1 1 1 1  2 mmax     mmin  mmin     mmin       mmin   2  2 2 4 2 2  31

mmax   mmin  1  this is not possible, mmax  mmin .

Thus, the solution is mmax  mmin  j .

Then J 2  j 2  j  j  j  1 ,  mmax   j . Operate now by the lowering operator on the wave functions: Jˆ  j   j 1 Jˆ  j 1   j  2 

One will get the values of m until mmin   j , where the neighboring values of m differs for 1: m  j, j  1, j  2,...,  j . This is possible only if j  is an integer or half-integer. Thus, not considering the explicit form of the operators Jˆ 2 , Jˆ , z

we showed, that the Eigen values are quantized: 1 3 5 J 2  j  j  1 , where j  0,1,2,3,... or j  , , ,... 2 2 2

In addition, the Eigen values of the operator Jˆ z is m , where mmax  j , and mmin   j , that is m  j, j  1, j  2,...  j  take in

total  2 j  1 different values at any, integer or half-integer, value of j . §7. Matrix elements of operators Jˆ , Jˆ , Jˆx and Jˆ y in diagonal representation of operators Jˆ 2 and Jˆ z General properties of matrix elements of Hermitian operators 1) Fˆ m   Fnmn n

32

A set of all matrix elements of operators forms the matrix of operators. 2) n Fˆ  m  m Fˆ n



 for an arbitrary matrix element.

F  F  – for Hermitian operators. 

n Fˆ m  m Fˆ n , that is the diagonal elements are real.

If the matrix element is real then n Fˆ m  m Fˆ n . 3) In a representation of Eigen functions of an operator it’s matrix elements are diagonal: n Fˆ m   n Fˆ m d  Fn  nm d  Fn mn .

4) Matrix elements ˆ ˆ FG   Fnl Glm .

 

nm

of

two

operators’

product:

l

j m Jˆ Jˆ jm , here

Find the matrix element

jm  is an

Eigen function of the operators Jˆ 2 and Jˆ z . And in the representation of own Eigen functions the matrix element of operators have the diagonal form. 1 Jˆ Jˆ  Jˆ 2  Jˆ z2  Jˆ z 2





j m Jˆ 2 jm  j  j  1   jj   mm j m Jˆz2 jm  m2   jj   mm

j m Jˆz jm  m   jj   mm

1 j m Jˆ Jˆ jm   j  j  1  m2  m   jj    mm 2

– different from zero only diagonal elements. 33

(7.1)

From the other hand let’s use the rule of obtaining of matrix elements of two matrices product: j m Jˆ Jˆ jm   j m Jˆ j m j m

j m Jˆ jm

The operators Jˆ do not change the quantum number j , but change the m . (7.1) = j 2  j  m2  m   j  m  j  m   j  m   j  m  j  m  1 . That is why j   j , and m  m  1 , since for the Jˆ the elements with

Jˆ 

j m  1 Jˆ jm are different from zero, and for the

jm Jˆ j m  1 , that is m  m . jm Jˆ Jˆ jm  jm Jˆ j m  1  Jˆ  Jˆ ,  jm Jˆ j m  1

j m  1 Jˆ jm 

j m  1 Jˆ jm  jm Jˆ j m  1







jm Jˆ j m  1  jm Jˆ j m  1

2



1  j  m  j  m  1. 2  j  m  j  m  1 

j , m Jˆ j , m  1 

2

(7.2)

The expression (7.2) – is real number. Here a certain choice of the phase is done. Similarly j, m Jˆ j, m  1  j, m Jˆ j, m  1

 j, m  1 Jˆ jm



34

 j, m  1 Jˆ j, m .

Hence, at once one has:

 j  m  j  m  1

j , m  1 Jˆ j , m 

2

(7.3)

Changing m  m  1 , one can obtain also other matrix elements. The wave functions of the angular momentum operator in the general case have the form of matrix (vector-column for s  1 2 ). That is why the operators which operate on them should have the form of matrices that is the operators should be written in a matrix form. Find Jˆ jm  M  j , m  1 . Certainly, that the M   is not the Eigen value of the operator

Jˆ : M   j, m  1 Jˆ j, m . Similarly Jˆ jm  M  j, m  1 , M   j, m  1 Jˆ j, m . M  and M  are easy to find using the formulas (7.1) and (7.2): Change in (7.1) m  m  1 :

Jˆ j , m 

 j  m  1 j  m  2

 j  m  1 j  m  ,

j , m  1 Jˆ j , m 

Jˆ jm 

j, m  1 ,

2

 j  m  1 j  m  2

(7.4)

j, m  1 .

Example: Consider the Eigen functions of the operators Lˆ2 , Lˆ z :

Ylm  ,  .









1 ˆ 1 ˆ Lˆ  Lx  iLˆ y , Lˆ  Lx  iLˆ y . 2 2 35

LˆYlm  ,  

LˆYlm  ,  

 l  m  1 l  m  2

Yl ,m 1  ,  ,

 l  m  1 l  m  2

Yl ,m 1  , 

Knowing the matrix elements of the operators Jˆ and Jˆ , it is easy to find the matrix elements of the operators Jˆ x and Jˆ y .



1 ˆ Jˆ  J x  iJˆ y 2







1 ˆ Jˆ  J x  iJˆ y  Jˆ 2

Jˆ  Jˆ .

Calculate: 1  j, m Jˆ j , m  1  j , m Jˆ x j , m  1  i j , m Jˆ y j , m  1   2  1  j, m Jˆ j , m  1  0  j , m Jˆx j , m  1  i j , m Jˆ y j , m  1  (7.5)  2 

Hence, summing up, one gets: j , m Jˆx j , m  1 

 j  m  j  m  1 2

Interchanging m  m  1 , one gets: j , m  1 Jˆx j, m  36

 j  m  1 j  m  2

(7.6)

j , m  1 Jˆx j , m 

 j  m  1 j  m  2

(7.7)

Subtracting in (7.1), one gets: 2  i j , m Jˆ y j , m  1  j , m Jˆ j , m  1     j, m Jˆ j, m  1  j , m  1 Jˆ jm  .  

and from (7.1): i j, m Jˆ y j, m  1   2

 j  m  j  m  1 .

(7.8)

From (7.8), since the matrix element Jˆ y  is purely imaginary: i j, m  1 Jˆ y j, m  2

 j  m  j  m  1 .

(7.8)

Interchanging m  m  1 : i j, m  1 Jˆ y j , m   2

 j  m  1 j  m  .

(7.9)

From (7.6): 

j, m Jˆx j, m  1  j, m  1 Jˆx j , m ,

since the matrix element Jˆ x  is real: j, m  1 Jˆx j, m 

1 2

37

 j  m  j  m  1 .

(7.6)

The formulas (7.6 – 7.9) can be reduced to the following form: j , m  1 Jˆ x j , m 

1 2

j , m  1 Jˆ y j , m 

j i 2

j

m  j  m  1 m  j  m  1

(7.10)

Consider examples. The momentum operators in the matrix form: 1. j  0 . In this case the Jˆ 2 and all components of the momentum are presented as first-order zero matrices. 1 1 2. j  . Then m   . Find the matrix elements of operators 2 2 Jˆx , Jˆ y , Jˆz and Jˆ 2 . The matrices of operators in this case have the dimension  2 j  1   2 j  1  2  2 . The rows and columns of the matrices are enumerated by the values of projections:

jm Jˆi jm .

We chose a representation in which the operators Jˆ 2 and Jˆ z are diagonal, that is a representation of the Eigen functions of the operators Jˆ 2 and Jˆ z . It is clear, that the matrices of the operators Jˆ 2 and Jˆ are diagonal: z

1 1 0  ˆ2 3 1 0 . Jˆ z   , J    2  0 1 4 0 1 1 1 1 1 1 1 1 1 1 1 j  , m  Jˆz j  , m   j  ,m  j  ,m   . 2 2 2 2 2 2 2 2 2 2

The element for the operator Jˆ 2 : 1 1 1 1 3 1 1 1 1 3 j  , m  Jˆ 2 j  , m   j  ,m  j  ,m   . 2 2 2 2 4 2 2 2 2 4

38

Calculate the matrix of the operator Jˆ x : 1 2 Jˆ x   . 3 4

From the formula (7.10): 1 1 1 1  0. The first element: j  , m  Jˆx j  , m  2 2 2 2 The second element: 1 1 1 1 1  1 1  1 1  1 j  , m  Jˆx j  , m         1  ; 2 2 2 2 2  2 2  2 2  2 1 1 1 1 1 j  , m   Jˆx j  , m   ; 2 2 2 2 2 The fourth element equals zero.

The third element:

1  0 1 ˆ 1 2 Finally one has: Jˆ x   .  , Jy   2 1 0 3 4

The elements 1 and 4 equal zero. The second element: 1 1 1 1 i i j  , m  Jˆ y j  m     1   ; 2 2 2 2 2 2

The third element:

1 1 1 1 i i j  , m   Jˆ y j  m   1 ; 2 2 2 2 2 2

1  0 i  Jˆ y   . 2 i 0  39

Within the accuracy of the multiplier

1 the matrices of Jˆ x , Jˆ y 2

and Jˆ z are the spin Pauli matrices:  0 i  0 1 1 0   , ˆ y    , ˆ z   . i 0  1 0  0 1

ˆ x  

3. j  1 , m  0, 1 (photon). 1 0 0  1 0 0     2 Jˆ z   0 0 0  , Jˆ  2  0 1 0   0 0 1 0 0 1    

1 1 j  1, m  0 Jˆx j  1, m  1  2 ; 2 2 1   0   0 2   0 1 0   1  1 1  ˆ Jx   0 1 0 1 .   2 2   2 0 1 0 1   0   0 2       Jˆ y      

0



i 2

i 2

0

0

i 2

 0    0 i 0  i  1    i 0 i  .   2 2  0 i 0   0  

3 15 3 1 1 3 4. j  : j  j  1  , m  , ,  ,  . 2 4 2 2 2 2 40

1  15  0 2 ˆ J  4 0  0

0 1 0 0

0 0 1 0

3  1 0 ˆ Jz  20  0

0  0 , 0  1

0 0 0  1 0 0 0 1 0   0 0 3 

Thus, we obtained a representation for the matrices of the ˆ J x , Jˆ y , Jˆz and Jˆ 2 for the different j .





1 ˆ Jˆ Jˆ  J 2  Jˆz2  Jˆ z . 2

§8. Geometric interpretation of momentum projection operator Let us consider the problem about centrally symmetrical potential:  nlm  r , ,   Rnl  r   Ylm  ,  . Let a particle is in a state with quantum numbers  lm  , the number l characterizes the Eigen values of the operator Lˆ2 in this state, and the numbers m characterize the Eigen values of the operator of projection on the axis z . 1) What can be said about the momentum projections on other axes ( x and y ) in the state  lm  ? If Lˆz  m , then Lx  Ly  0 . This is clear from the above shown consideration since Lx  lm Lˆx lm , Ly  lm Lˆ y lm .

These integrals equal zero, since m1  m2  1 . 41

lm1 Lˆx lm2  0 only when

Thus, the angular momentum seems to precessing around the axis z in a space (fig. 1), so that the average values of projections on the axes x and y equal zero.

Figure 1. Precession of angular momentum around

z axis

2) Consider the projection on the arbitrary axis oz  in the state  l , m  . The unit vector in a direction of z  with the normal vector n  sin   cos , sin   sin  , cos  .





Lˆ  Lˆx , Lˆ y , Lˆz . In classical mechanics Lz  L  n .

Figure 2. Spherical coordinates system

In quantum mechanics:

Lˆz  Lˆx sin  cos  Lˆ y sin  sin   Lˆz cos . 42

Find the average value of the operator Lˆ z  : lm Lˆz lm  cos  lm Lˆz lm  m cos .

Hence, one has

Lz  m cos . Problems to chapter II 1. Explain how a motion of a particle with orbital momentum  0 occurs. 2. If the orbital momentum of a system  0 , then in which states in the Hamiltonian there appears the so called centrifugal energy 2

Ecentrifug. 



 1

2

. Estimate its value for a problem about a deuteron

R  1 and 2.

and for the 3. Show, that the electron spin cannot be connected with its rotation around own axis, since in this case we are in contradiction with the relativity theory.

ˆˆ ˆ . 4. Find the matrix element Fmk of a product of three operators Fˆ  KLM Test questions to chapter II 1. How do the Eigen values of angular momentum operator and momentum squared operator quantized in quantum mechanics? 2. What properties do the operators Jˆ and Jˆ possess? 3. What properties do the matrix elements of Hermitian operators posses? 4. Which are the matrix elements of the momentum projection operator and momentum squared operator? 5. What properties do the operators’ matrix elements in quantum mechanics possess? 6. Which operators have a common system of Eigen functions? Which are the matrix elements of such operators? 7. What is the geometric interpretation of the angular momentum projections?

43

Chapter III

VECTOR ADDITION OF TWO MOMENTA. APPLICATION OF THEORY §9. Vector addition of two momenta Consider a system consisting of two parts (subsystems), state of each part is defined by the momenta Jˆ 1 and Jˆ  2  respectively. These can be 2 particles with orbital momenta l1 and l2 , or a particle with orbital momentum l and spin s . If there is no interaction between the two parts of the system then the operators of projections of these momenta commutate that is:  Jˆi 1 , Jˆk  2   0 ,  

i, k  x, y, z .

In this case the total system can be in states in which there simultaneously exist the definite values of squared momenta of the subsystems: J 2 1 

2

j1  j1  1 ,

J 2  2 

2

j2  j2  1 ,

further the Eigen values of the operators Jˆ 2 and Jˆ z will be presented in units of 2 and , respectively, and their projections on one of the coordinate axes which is taken as z : J z 1  m1 J z  2   m2 44

Such states are described by wave functions which are the product of the Eigen functions for every of the momentum operators in separate:    1  2 or (9.1) j1m1 j2 m2  j1m1  j2 m2 . If one fixes j1 and j2 , then there exist  2 j1  1 2 j2  1 different functions distinguishing by the values of the numbers pair m1 and m2 . Define now the operator of the total momentum of the total system Jˆ : ˆ ˆ ˆ J  J 1  J  2  . ˆ ˆ Operators of projections for the J 1 and J  2  satisfy the commutative relations:

 Jˆ x , Jˆ y   i    Jˆ y , Jˆ z   i    Jˆ z , Jˆ x   i  

Jˆ z    ˆ Jx   Jˆ y  

That is why, it is easy to see, that the components of the operator ˆ J also satisfy these conditions. It is easy t see, that the wave functions (9.1) are the Eigen functions of the operator of total momentum projection Jˆz  Jˆz 1  Jˆz  2  , corresponding to the Eigen value

Jz  m 

 m1  m2  , since

Jˆz j1m1 j2 m2  ˆ Consider the operator J 2 :

 m1  m2  45

j1m1 j2 m2 .

ˆ ˆ ˆ ˆ ˆ ˆ ˆ J 2  J 2 1  J 2  2   J 1 J  2   J  2  J 1  ˆ ˆ ˆ ˆ  J 2 1  J 2  2   2 J 1 J  2  . ˆ ˆ This operator commutates with the operators J 2 1 and J 2  2  , hence, the total momentum squared can have a value defined simultaneously with the squared momenta of each of the subsystems. But the functions (9.1) are not the Eigen functions of the ˆ operator J 2 because of that the third term in the expression for the ˆ J 2 mixes the states differing by the m1 and m2 for a unit. One can show that from the functions (9.1) it is possible to form the linear combinations which are the Eigen functions of the operator ˆ J 2 as well. Since the operator Jˆ is linear, then these functions will z

be its Eigen functions as well. In this case the wave function of the system, described by a definite value of the total momentum squared, its projection and the subsystems momenta squared Jˆ 2 1 and Jˆ 2  2  , can be written in a form: (9.2) j1 j2 jm    j1m1 j2 m2 jm   j1m1 j2 m2 . m1m2

Here  j1m1 j2 m2 jm   coefficients, defining the contribution of the different functions (9.1) into the expression (9.2). These coefficients are called Clebsch-Gordan coefficients or coefficients of vector addition. The phase multipliers before the wave functions are chosen so that the vector addition coefficients are real. Further we will get the explicit expression in a case when one of the 1 momenta j1 or j2 is . 2 In literature there exist different denotations for these coefficients:

jm j m 1

1 2

2

jm    j1 j2 m1m2 jm    j1 j2 m1m2 j1 j2 jm   C jjm . 1m1 j2 m2 46

From the formula (9.2) it follows that the Clebsch-Gordan coefficients are the matrices of transformation from the representation where the subsystems momenta projections are given to a representation where the system total momentum and its projection are given. §10. General formula for Clebsch-Gordan coefficients through factorials. Basic properties of vector addition coefficients The general formula for calculation of the vector addition coefficients was derived by Wigner using theoretical group methods. Racah found the general formula for these coefficients from the recurrent relations. These formulas have enough enormous form. For example, the Racah formula is written in the following form:

jm j m 1

1 2

2

jm     2 j  1 j1  j2  j ! j1  j2  j ! j  j2  j1 !     j1  j2  j  1!   12

  m,m1  m2

n   1  j1  m1 ! j1  m1 ! j2  m2 !    n  n ! j1  j2  j  n ! j1  m1  n ! j2  m2  n !   j2  m2 ! j  m ! j  m !   .  j  j2  m  n ! j  j1  m2  n !

Here n  0,1,2,... , but when summing there are taken only such values when the arguments of the factorials in the denominator are not negative. Basic properties of Clebsch-Gordan coefficients 1. Coefficients are different from zero only if m  m1  m2 , that is why if there is given the total projection m , then the 47

summation by one of the indices takes the formal character, since m1  m  m2 . 2. Define the possible values of the quantum number j  0 when j1 and j2 are fixed. Firstly define the possible values of the quantum number m . Since m  m1  m2 ,  m1 max  j1 ,  m2 max  j2

and mmax  j1  j2 . Only one state defined by the function j1 , m1  j1  j2 , m2  j2

and j  j1  j2 corresponds to this value.

The next value m  j1  j2  1 can be realized by the linear combination

of

two

functions

j1 , j1  1  j2 , j2

like

and

j1 , j1  j2 , j2  1 . One of these linear combinations will correspond to j  j1  j2 , and the another j  j1  j2  1 . To the value m  j1  j2  2 there correspond three linear combinations of three

j1 , j1  2 j2 , j2 ,

functions:

j1 , j1  1 j2 , j2  1

and

j1 , j1 j2 , j2  1 , three values of the quantum number j correspond to these functions: j1  j2 , j1  j2  1, j1  j2  2 . One can see, that decreasing the value of m for a unit, one always gets the new value of j until it reaches the values when either m1   j1 or m2   j2 . Thus the minimal value of j : jmin  j1  j2 . So, at the given values of j1 and j2 the quantum number j takes the sequence of values differing for a unit, and these values satisfy the inequality j1  j2  j  j1  j2 .

(10.1)

At that to every j the  2 j  1 values of m   j,   j  1 ,... correspond. The total number of states with all possible j will be equal to j1  j2

  2 j  1   2 j

1

j  j1  j2

48

 1 2 j2  1 .

For example, if j2  j1 , then

 2

j1  j2

j1  j2

j1  j2

j1  j2

2 j





 j1  j2  j1  j2  2 j2  1  2

 2 j2  1   2 j1  1 2 j2  1 .

The inequality (10.1) is called a triangle relation and in brief it is denoted as   j1 j2 j  : j1  j2  j  j1  j2 j  j1  j2  j  j1 j  j2  j1  j  j2 .

If one of these conditions is not satisfied then the ClebschGordan coefficients automatically equal zero. Examples. 2.1. One has j1  1 , j2  1 . What is the quantum number of total momentum j equal to? 0 j 2,

that is the possible values of the total momentum j  0,1,2 . Whether the following combinations of momenta satisfy the triangle rule: 5 1 а) j1  , j2  2 , j  (answer: yes). 2 2 5 1 b) j1  1 , j2  , j  (answer: no). 2 2 The triangle rule is not satisfied then the corresponding ClebschGordan coefficients equal zero. 49

3. Clebsch-Gordan coefficients satisfy the orthonormality conditions:

 j m j m 1

jm

1 2

jm  j1m1 j2 m2 jm    m1m1   m2m2

2

 j m j m 1

1 2

2

jm  j1m1 j2 m2 j m    jj    mm .

m1m2

4. From the orthonormality relation using the fact that the Clebsch-Gordan coefficients are real one gets the following property. Direct transformation: j1 j2 jm 

 j m j m 1

1 2

2

jm   j1m1 j2 m2 ,

m1m2

multiplied by the  j1m1 j2 m2 jm  and summing up

: jm

  j m j m 1

1 2

jm



  j m j m 1

1 2

2

2

jm  j1 j2 jm 

jm  j1m1 j2 m2 jm   j1m1 j2 m2 

m1m2 jm

=   m1m1   m2 m2  j1m1 j2 m2 . m1m2

Thus, j1m1  j2 m2    j1m1 j2 m2 jm  j1 j2 jm jm

– inverse transformation is realized using the same ClebschGordan coefficients. 5. Note the property of symmetry of Clebsch-Gordan coefficients:

jm j m 1

1 2

2

jm    1 1

j  j2  j

50

j m 2

j m1 jm  .

2 1

Hence the most important property of wave function regarding the momenta place permutation follows:

j1 j2 jm   1 1

j  j2  j

j2 j1 jm ,

that is in quantum mechanics the order of momenta addition is important. In dependence on the order the phases of wave functions are changed. 6.  j1m1 j2 m2 jm    1 1

j  j2  j

j

1

 m1 j2  m2 jm  .

§ 11. Use of Clebsch-Gordan coefficients. Comparison of electrons angular distributions with j  3 2 and different projections m j

1,

A special interest presents a comparison of the total cross sections of nuclear reactions when one can use the principle of isotopic invariance of strong interactions. The total cross section of a reaction characterizes the probability of the process. In this case the ratio of the total cross sections of the processes is defined by the ratio of the corresponding Clebsch-Gordan coefficients by isotopic spin since the other quantum numbers of interacting particles are the same. In literature such relations are called Shmushkevich rule. As an example consider the relation of cross sections of pion scattering on nucleons:

   p    p

(11.1)

0    n   p  N       p

(11.2)

and

in the energy range of 33-resonance. At that the scattering occurs mainly through the intermediate state of  N -system with the total isotopic spin T  3 2 (non-resonance interaction in the state with T  1 2 is negligibly small). The first resonances were discovered by 51

Fermi when scattering of  -mesons on nucleons:       p    p,   p      p , where  -resonance has the following quantum numbers of spin, parity and isospin  j , T  3 2 ,3 2 . In literature it is called 33-resonance. Find the mass of the  -resonance. 2 2 M 2 c 4   E  E p    p  p p  c 2   

(11.3)

Considering, that 2 4 2 2 2 2  m c  E  p c , E p  m p c ,  2 E  T  m c , p  0     p 

one gets M  c 2  E2  m2p c 4  2 E m p c 2  p2 c 2   m2 c 4  m2p c 4  2m p c 2 T  m c 2   1236 MeV,

where T  180 MeV,

mp c2  940 MeV,

m c2  140 MeV (11.4)

Thus, the mass of 33-resonance equals 1236 MeV. Find the relation between the differential cross sections of these three reactions at the same relative energies, scattering angles and spins orientation: (11.5) I     p    p

 II     p   0  n  III     p     p

(11.6)

The reactions (II) and (III) represents the two channels of one (with reference to isospin) reaction   p   N , distinguishing by different charge states of pion-nucleon system in the final state, 52

isospin of which is equal to T  3 2 . Compare  d  II   d  III   with d  I  . Under the d cross sections selection rule the reactions being considered are completely identical in the view of coordinate and spin degrees of freedom and according to the isotopic invariance (with account of the dominating role of states with T  3 2 in  N -interaction) the ratio of the cross sections

 d  II   d  III  : d  I 

equals the ratio of «weights» of the

necessary isotopic state with T  3 2 in the initial states of the

  p - and   p -pion-nucleon systems. The pointed weights are defined through the corresponding Clebsch-Gordan coefficients by isospin.

Figure 3. Excited states of nucleons observed at elastic scattering of pions on proton:   total effective cross section in millibarns, Т  kinetic energy of -mesons in laboratory system

Clebsch-Gordan coefficients by isotopic spin are different, in difference from other momenta, hence the Shmushkevich rule is true: 2

 1 T1  .  2 T2 2

53

(11.7)

For the reaction    p in the range of 33-resonance the Clebsch-Gordan coefficient T1  111 21 2 3 23 2   1 .

(11.8)

For the reaction    p in the range of 33-resonance the Clebsch-Gordan coefficient T2  1  11 21 2

3 2 1 2  1

3.

(11.9)

Hence, d1 d 2  3 ,

(11.10)

that is enough good coincides with the given experimental data (fig. 3). At the resonance energy 190 MeV one has d 1 d 2  215 mb

75 mb

 2,9 .

(11.10')

Thus, the physical side of the theory of Clebsch-Gordan coefficients is quite essential. Find now the ratio d  II  : d  III  . It equals the ratio of «weights» of charge states  0 n and   p of the pion-nucleon system in the state with T  3 2 (and T3  1 2 ), which are defined by Clebsch-Gordan coefficients. For the reaction    p     p in the range of 33-resonance the Clebsch-Gordan coefficient equals: T2  1  11 21 2

3 2 1 2  1

3.

(11.11)

For the reaction    p   0  n in the range of 33-resonance the Clebsch-Gordan coefficient equals: 54

T3  101 2  1 2



3 2 1 2  2 3 ,

d  II 

d  III 



T2 T3

(11.12)

2 2

 2.

(11.13)

From (11.10) and (11.13) it follows d  I  : d  II  : d  III   9 : 2 :1 .

Referring to ∆ (1236) resonance with spin j  3 2 can be confirmed by angular correlations of elastic   p -scattering in the range of this resonance. The angular correlations of pion in elastic   p scattering can be found using the formulas

 ( )      * , 2

(11.14)

where ( j, m j ) 

 ( m sm

s

/ jm j ) ( , ml )  ( s, ms )

m ms



 1, m   0, 1 , s p  1 2, ms p  1 2 ,

j  p  1 2,3 2; m j  =  1 2,  3 2 .  p

1) Consider j  3 2, m j  1 2 . In this case one has: 3 1  1 1 31 1 1   ,   10  1,0    ,   2 2  2 2 2 2 2 2  1 1 3 1 1 1  11   1,1   ,     2 2 2 2 2 2 55

(11.15)



1 1 1 1  1  1 1   2 2   1,0   ,  2 2   1,1   1 ,  1       2 1  1 2 1  1 2 2 2 2

1

=

2 1 1 1 1 1  1,0    ,    1,1   ,   . 3 3 2 2 2 2

Spherical functions:

 1,1  

3 ei sin  4 2

(11.16)

3  1,0   cos  4

 ( )        2

       

2 3 3 ei 1 1 1  1 1   cos    ,    sin      ,     3 4 3 4 2  2 2  2 2 2 3 3 ei 1 1 1  1 1   cos    ,    sin      ,     3 4 3 4 2 2 2  2 2 

 =     =

1 1 1 1  1 1  cos    ,   sin   ei    ,     2 8 2 2  2 2  1 1 1 1  1 1  cos    ,   sin   ei    ,     2 8 2 2  2 2 

1 1 1 1 1 1 1 1 1 1 cos2     ,     ,   sin  cos  ei    ,      ,   2 2 2 2 2 4      2 2 2 2

1 1 1 1 1 cos sin   ei    ,     ,    4 2 2 2 2 1 1 1 1 1  sin 2     ,      ,    8 2 2 2 2 1 1 1 1  cos2   sin 2   sin 2   4cos 2     1  3cos2   . 2 8 8 8 

56

In calculations we consider the normality condition for the spin functions:

  s, ms     s, m' s    m m '

(11.17)

1 1  3cos2   . 8

(11.18)

s

s

Hence, one gets

   

3 3 2) Consider j  , m j  . 2 2 In this case one has: 3 3  ,   2 2  11  11  22

3 3 1 1   1,1   ,   2 2 2 2

3 1  2 2  1,1   1 , 1     2 1  1 2 2

1

3 ei 1 1 1 1   1,1   ,    sin      , . 4 2 2 2 2 2

          2

3 1 1 sin 2    sin 2  . 4 2 8

Hence, one gets

   

1 sin 2  . 8

1 1 3) Consider j  , m j  . 2 2

In this case one has: 1 1  ,   2 2

 1 1 11   2 2

1 1 1 1  1,1   ,    2 2 2 2 57

(11.19)

 11  10  22

1 =

1 1 1 1  1,0    ,   2 2 2 2

1 1 1 1  1  1 1   2 2   1,1  ,   2 2   1,0    1 , 1       3 3 2 2 2 2

2 1 1 1 1 1   1,1   ,      1,0    ,   3 3 2 2 2 2 2 1 3 1 1 1 1 3 1 1   sin   ei    ,     cos    ,   3 2 2 3 2  2 2 2 2 1 1 1 1 1 1  sin   ei    ,    cos    ,  . 4 4 2 2 2 2 

          2

1 1 1 . sin 2   cos2   4 4 4

Hence, one gets

   

1 = const. 4

(11.20)

3 3 1 One can see, that the states with j  , m j  and m j  2 2 2 1 1 have the larger probability than the state with j  , m j  (fig. 4). 2 2 One can consider a number of other such examples of nuclear reactions. Consider reactions which are inhibited from the point of view of isotopic spin conservation law. The reaction d  d     0 is inhibited under the law of isospin conservation law since Td  T  0 , T  1 , hence, Td  d  0 , and T  0  1 . The

experiment confirms the inhibition for this reaction in the sense that this reaction is observed only with very small effective cross section which can be referred to only electromagnetic interaction since it does not obey the law of isospin conservation. 58

Figure 4. Angular correlations of pion scattered in elastic 

   

  p -scattering: curve 1

1 1 1 = 1  3cos 2   , 2      sin 2  , 3       8 8 4 const, 4  experimental dots

The isotopic invariance supposes that different particles relating to the same isotopic multiplet are to be considered as identical particles in different states distinguishing by the value of Т3-component of isospin (and respectively the charge). At that the quantum-mechanical principle of identical particle indistinguishability, requiring the certain symmetry of the wave function regarding to permutation of any two such particles’ variables, should be distributed for different particles of the same isomultiplet as well. If one considers two-pion system then a permutation of the pions’ coordinates is equivalent to their inversion with respect to the mass center and hence the symmetry of the coordinate wave function of the state with the given value of relative motion momentum L (or as well – total momentum J  L ) coincides with its parity  1 . The wave function of two pions – identical bosons – should be symmetric with respect to permutation of their variables: coordinates and isotopic variables. Hence, in the state with the given L the isospin part of the wave function is to be symmetric for the even L

59

values of L and antisymmetric for the odd values of L . Considering the value of pion isospin T  1 and the formal similarity of momentum and isospin properties one can conclude that the isospin part of two pion system with respect to their permutation is symmetric when the total isospin T  0;2 and antisymmetric when T  1 . Hence in the states with even L only the values T  0;2 are possible, and with odd L  only T  1 . Find a probability  T  of different values of the total isotopic spin T of the pion-nucleon system and the average values of Tˆ 2 in the following isotopic states:

  p,   n,  0 p,  0 n,   p,   n . In  N -system states being considered the t3 -components of the nucleon and pion isospins have certain values (note: (t3 ) p  1 , (t3 ) n   1 , (t3 )   1, (t3 ) 0  0, (t3 )   1 ). 2 2 Let the pion and nucleon are weak-interacting systems with quantum numbers  t , t 3  and  t N , t3 N  of isospin and its third component. The possible values of the isospin T of the total system satisfy the conditions:

max  t N  t , t N 3  t 3   T  t N  t The operator of total isospin Tˆ  tˆN  tˆ .

(11.21)

2 Tˆ 2   tˆN  tˆ   tˆN2  tˆ2  2tˆN tˆ ,

(11.22)

Hence,

and the average value

60

1 1 Tˆ 2  t N (t N  1)  t (t  1)  2t N 3t 3  (  1)  1  (1  1)  2t N 3t 3  2 2 3 11   2  2t N 3t 3   2t N 3t 3 . 4 4 11 Tˆ 2   2t N 3t 3 . 4

(11.23)

From the other hand, for the  N -system only two values of the 1 3 1 total isospin are possible: T  and T  , so, denoting    as 2 2 2 1 the probability of the value T  in states being considered (at that 2 3 1     1     ) one has: 2   2 1 1   1 3 3   3 T 2   T T  1  T     1       1     2 2   2 2 2   2 T 3  1  15  3  3  1  15   1   15 1              1        3   . 4 2 4 2 4 2 4   2  4 2 T2 

15 1  3   . 4 2

( 11.24)

Substituting into (11.23) the different values for the third component of isospin of nucleon and pion ( (t3 ) p  1 , 2 1 (t3 ) 0  0, (t3 )   1 ), one finds the (t3 )n   , (t3 )   1, 2 average value of the isospin squared. Using (11.24), one gets: а) for the   p  and   n  systems: T 2  15 ,  3  1; 4 2   b) for the  n  and  p  systems:

 

61

 

 

T2 7 ,  1 2 ,  3  1 ; 2 2 3 3 4

c) for the  0 p  and  0 n  systems:

 

 

T 2  11 ,  1  1 ,  3  2 . 2 2 3 3 4

Thus, one can see, that for the systems   p and   n there exists only one state with the isospin T  3 2 . For the systems   n and   p the weight of T  1 2 is larger, than the weight of the state with T  3 2 . For the  0 p and  0 n – vice versa, the probability of the state with T  3 2 is larger, than of the state with T  1 2 . At that the total probability of the states in all three cases is equal to 1. Consider a neutral particle f 0 with isotopic spin T  0 , which has the probability of decay by two-pion channel: f 0  2 . The possible decay channels: f 0     , f 0  2 0 . Find the relation between the probabilities of decay of particle f 0 by the mentioned channels ( t =1, (t3 )   1, (t3 ) 0  0, (t3 )    1 ). The wave function of the states is presented in a form:  TT3  

 t t  t t  t

t3  t

1

3

1

2

3

2

 

 

TT3   t1  t3    t 2  t3  1

2

 (11.25)

The possible values of the third component of the total isospin T3  0 , since T  0 .  2  0,0   111  1

00     1,1    1, 1 

 1010

+ 1  111

00    0 1,0    0 1,0  

00     1, 1    1,1 .

The Clebsch-Gordan coefficients: 62

111  1 1010 1  111

00  

1 , 3

00   

1 , 3

00    1

11 0

111  1

00  

1 . 3

Hence,  2  0,0  

1 1    1,1    1, 1   0 1,0    0 1,0   3 3 

1   1, 1    1,1 . 3  1  2  0,0     1,1   1, 1   0 1,0   0 1,0   3 







   1, 1   1,1 .

(11.26)

here  2  0,0  – isospin wave function of two-pion state with T  0 ,   t , t3  – isospin wave functions of pions with a certain

value of the component t3 . The probabilities of decay of f 0 into different charge states of two-pion system (    , 2 0 ) are proportional to the probabilities of finding of pions in such states at T  0 (under the conservation of isospin in the decay). The probability of the charge state is defined by the Clebsch-Gordan coefficient squared, that is the probability of 2

the state of two  0 equals T 0  2 0    1   1 . Respectively 3 3 

T 0      1  T 0  2 0   1   . 1 3

63

2 3

Hence,

  f 0        f 0  2 0 

2.

(11.27)

Consider another way of finding of decay probability. Represent a decay of some number of N particles f 0 . As a result of their

decay there appears the number N  f 0        -mesons, the

same number of   -mesons and 2 N  f 0  2 0   0 -mesons (in the decay f 0  2 0 there appear at once two  0 ). The initial system of f 0 is, obviously, isotopic symmetric (since T f 0  0 ). The final state, containing the decay pions, should be isotopic symmetric as well. The reflection of this symmetry is the same number of pions in different charge states   ,  0 ,   , what immediately leads to the relation (11.27) found above. The consideration given is represented to be very obvious from the physical point of view. Consider two reactions p  p  d    and n  p  d   0 and find the ratio of differential cross sections of these reactions taken at the same relative energies, scattering angles and spins mutual orientation. Since Td  0, T  1 , then the final states of both the reactions are different isospin states of the same physical system (pion + deutron) with isospin T  1 , distinguishing only by the value T3 -component of the isospin. Under the isospin conservation the reactions being considered occur only in states of initial nucleonnucleon system with the isospin T  1 . At that in the reaction pp  d  both the nucleon are just in the required isotopic state with T  1 (and T3  1 ), while in the reaction pn  d 0 the required state of two nucleons with T  1 (and T3  0 ) is presented, obviously, with the probability 1 2 (with the same probability 1 2 there is presented the state with T  0, T3  0 ). Under the d cross sections selection rule both the reactions are completely identical in the sense of coordinate and spin degrees of freedom, so that under the isotopic invariance the ratio of their cross sections equals the 64

ratio of probabilities of the necessary isotopic state with T  1 in the initial states, that is to two: d  p  p  d     d  n  p  d   0 

2.

Show now, that d  p  d  d  n    

d  p  d  d  p   0 

 2.

Since Td  0 , then in the sense of isospin in the reactions being considered the deuteron plays a role of «accelerator» in the process of proton «dissociation» into a nucleon and a pion:



pN T  1

2

   T  1

In the initial stage of the process T  1 2, T3  1 2 , and under the isospin conservation the same values of T and T3 the pionnucleon system in the final state has. Under the d cross sections selection rule both the reactions are completely identical in the sense of coordinate and spin degrees of freedom so that under the isotopic invariance the ration of their cross sections equal the ratio of «weights» of the charge states of the n  and p 0 pion-nucleon system with T  1 2, T3  1 2 . The last ratio equals 2, what proves our statement. § 12. Operator of electron total momentum. Construction of wave function of a particle with spin 1 2 Remind of relations from quantum mechanics: ˆ L2Yl ,m  65

2

l  l  1 Yl ,m ,

LˆzYl ,m  mYl ,m ,

Lˆ  Lˆx  iLˆ y , Lˆ  Lˆx  iLˆ y .

LˆYl ,m  ,  

 l  m  1l

1

0

m Yl ,m1  , 

 1 1    ,  1 1    ,  1  1   m m . , , ,m , m 0 1 2 2 2 2 2 2 s

s

s

s

ˆ ˆ ˆ Introduce an operator of total momentum J  L  S . It is easy to see that it obeys the commutative relations usual for momentum operators:

 Jˆ x , Jˆ y   i    Jˆ y , Jˆ z   i   ˆ ˆ Jz , Jx   i    Jˆ 2 , Jˆ   0 i  

Jˆ z Jˆ x

(12.1)

Jˆ y

   Lˆ  Sˆ  2 Lˆ Sˆ  Lˆ Sˆ  Lˆ Sˆ 

ˆ ˆ ˆ J2  L  S

2

2

2

x

x

y

y

z

z

(12.2)

Jˆz  Lˆz  Sˆz

Determine the complete set of commuting operators. It is ˆ ˆ ˆ ˆ obvious, that  J 2 , L2   0 and  J 2 , S 2   0 . It is seen from the     ˆ expression (12.2). For example, L2 commutate with Lˆ x , Lˆ y , Lˆ z and ˆ Sˆ x , Sˆ y , Sˆ z and S 2 . 66

ˆ ˆ ˆ Hence, there are J 2 , S 2 and L2 in the complete set. ˆ ˆ Also it easy to see, that  L2 , Jˆi   0 and  S 2 , Jˆi   0 .     For example,  Lˆ2 , Jˆz    Lˆ2 , Lˆz  Sˆz   0 .     However from the expression (12.2) on can see, that ˆ  J 2 , Lˆ   0 and  Jˆ 2 , Sˆ   0 . i i     Thus, it is possible to measure simultaneously the quantities ˆ ˆ ˆ corresponding to the operators J 2 , L2 , S 2 and any of the projections ˆ of the operator J , for example Jˆ , that is one can find the common z

system of Eigen functions of these operators. In the central field the Hamiltonian Hˆ  Tˆ  Vˆ  r 

does not depend on spin variables and the following quantities are included in the complete set: ˆ ˆ ˆ S 2 , Jˆz , L2 , J 2 , Hˆ .

(12.3)

Problem: construct the common Eigen function of these operators (it is necessary to find the common function of these operators). For construction of the wave function with spin 1 2 let us ˆ choose the Eigen functions of operators S 2 and Sˆ z as basis function:

1

1 , ms  2 2

 0 1    and  1 1   , ms  1 0 2 2

 this is a complete set.

1 1 2

, ms

2

, ms

67

  ms ms .

(12.4)

Let us expand the required common Eigen function for the set (12.3) by the functions (12.4):   1  r , ,   1 1   2  r , ,   1

1 , 2 2

, 2 2

   r , ,     1 .   2  r , ,   

The operators, which commutate with the operator Sˆ z , will be ˆ presented as a diagonal matrices Jˆz , L2 , Lˆx , Lˆ y , Lˆz . In the





representation of the Eigen functions of the operator Sˆ z the operators have the following form: ˆ  Lz  2 ˆ ˆ ˆ ˆ ˆ J z  Lz  I  S z  I    0  

 ˆ2 ˆ2 ˆ  L LI 0 

1 2

, ms

ˆ L2  1 2

   ˆ Lz   2 0

0  ˆ2  L 

, ms

ˆ  L2  I .

 0 i  Sˆ y   , 2 i 0 

0 1 Sˆx   , 2 1 0

1 0  ˆ2 3 Sˆz   , S   4 2  0 1

2

1 0  . 0 1

The matrices are equal, if the corresponding matrix elements are equals. 68

Consider the equation for the Eigen values of the operator Jˆ z :

Jˆz   m j  .     Jˆ z  1   m j  1  ,  2   2 

   i   2   0  

Lˆz  i

 . 

        1   m j   1  .     2  i    2  2  0

      1    i  2    m j 1    .     m j 2     i   2    2   

The matrices are equal, if the corresponding elements of each matrix are equal to each other. 1  1  m j 1 ,  2  2 i  2  m j2 ,  2

i

or 1  1   m j   1 ,   2  2  1 i   m j   2.   2

i

A solution will be written in the following form: 69

1  i  m j   2

1  f1  r ,  e 

1  i  m j   2

 2  f 2  r ,  e 

1   i  m j     f1  r ,  e  2      – Eigen function. 1  i  m j     2    f 2  r ,  e 

ˆ Further the  should be the Eigen function of the operator L2 : ˆ L2  

2

l  l  1  .

ˆ It is obvious, that the  is the Eigen function of both the L2 and Jˆ z one has to write: 1  R1  r  Y

l ,m j 

 2  R2  r  Y

1 2

l ,m j 

1 2

 ,   c1  R  r   Yl ,m  1  ,  , j

2

 ,   c2  R  r   Yl ,m  1  ,  . j

2

Here one used the fact that since the angular operators do not operate on the radial part of the wave function, then it should be the same for both the components of the wave function. So,  c1Y 1  ,    l ,m j  2   1    R r      c Y  ,    2 1  l ,m j    2 2   ˆ ˆ is now the Eigen function of the operators S 2 , Jˆ z and L2 . It is ˆ necessary to make it the Eigen function of the operator J 2 as well: 70





ˆ ˆ ˆ J 2  L2  S 2  2 Lˆx Sˆx  Lˆ y Sˆ y  Lˆz Sˆz . ˆ J 2 

2

j  j  1  .

Let us write this equation in a matrix form:  Lˆ2  0 

0   1  3 2  1 0   1     ˆ2    2  4  0 1    2  L   0 1 ˆ  0 i  ˆ  1 0    1   2  Lˆx    Ly    Lz      2i 0  2  0 1    2   2  1 0    = 2 j  j  1  1  .  2 

or  ˆ2 3 2 ˆ  L  4  Lz   Lˆx  iLˆ y  





  ˆ2 3 2 ˆ  L  4  Lz  Lˆ   Lˆx  iLˆ y  





 Lˆ

x

ˆ 3 L2  4

 Lˆ

x

Lˆ

 iLˆ y

ˆ 3 L2  4

2





   1     2  2  Lˆz      c Y    1 l ,m j  12      c2Yl ,m  1  j 2   Lˆz  

 iLˆ y



2

2

  j  j  1  1  ,  2 

 c1Y 1   l ,m j  2  j  j  1  .  c2Yl ,m j  1  2  

ˆ Operating by the operators L2 , Lz and Lˆ , one gets 2 algebraic equations:  3 2 1   2  2  m j    c1 Y 1 + l  l  1  4 2   l ,m j  2   2



2

 1 2  l    m j c2 Yl , m j  1   2 2 71

2

j  j  1 c1 Y

l ,m j 

1 2

2

2

  

 1 2  l    m j  c1 Yl ,m j 1   2 2

l  l  1 

3 4

2



2

1    m j    c2 Yl , m j  1  2   2 2 = j  j  1 c2 Y 1 . l ,m j 

2

2  1    1 2  l l  1  j j  1   m c  l   m  c  0     j 1 j 2     4     2  2  1 1     2  l    m j  c1  l  l  1  j  j  1   m j  c2  0  4  2   

(12.4)

One has a system of homogeneous linear algebraic equations with respect to unknown coefficients c1 and c2 . In order that these equations are compatible it is necessary that the determinant composed of the coefficients at the unknowns equals zero. 1 Denote q  l  l  1  j  j  1  . 4 2

 1 2 l    mj  2

q  mj 2

 1 2 l    mj 2  

q  mj 2

 1 q 2  m2j   l    m2j  0 ,  2  1 q  l   .  2 1 1 1 1. q  l  : l  l  1  j  j  1   l  . 2 4 2 72

 0.

l2  l  j2  j  j2  j  l2 

1 1 l , 4 2

1 0 4

1 1 2 1 1 j   l     l; 2 4 4 2 1 j1  l  , 2 1 j2  l  – this solution should be omitted, since it should be 2 that j  0 , and the quantum number l  0 .

1 1 2 1 : l  l  j 2  j   l  . 4 2 2 3 j 2  j  l 2  2l   0 , 4 1 1 2 3 1 j   l  2l      l  1 2 4 4 2 1 j1  l  , 2 3 j2  l  – this solution should be omitted as well. 2

2. q  l 

Thus, one found that j  l  1 2 , hence one proved the vector model of momenta addition. Find the wave function. 1 Let j  l  . 2    1  3  1  1 2 l  l  1   l  2  l  2   4  m j  c1   l  2   m j  c2  0 ;        2

2

3 1 2   1 2 2 l  l  l  2l  4  4  m j  c1   l  2   m j  c2  0 ;     73

2

1    1 2  l   m j  c1    l    m j  c2 ; 2    2 1 l   mj 2 c2   c1 2  1 2 l    mj 2  

 1    1   1 l   mj  l  2   m j    l  2   m j        c  2 c2    c1 . 1 1  1    1   l   mj  l  2   m j   l  2   m j  2      

For the j  l 

1 : 2 2

1 1 2   1 2 2 l  l  l  4  4  m j  c1   l  2   m j  c2  0 ;     2

 1   1 2 l  2  m j  c1    l  2   m j  c2    

 1   l   mj  2   c  c2   1 2  1 2 l    mj  2

 1   l   mj   2  c . 1  1  l   m  j   2 

For finding of the coefficients theirselves one has to require the normality condition: c12  c22  1 .

For the j  l 

1 one has: 2 74

1 1 1 l   mj  l   mj  mj 2 2 2 c12  c12  1 ; c12  c12  1 1 1 l   mj l   mj 2 2 1 1 1 1 l   mj l   mj l   mj l   mj 2 2 2 2 ; c2  ; c1    1 2l  1 2 l  1 2 l 1 l   mj 2   1  l   mj  2   Y 1  ,   l ,m j  2l  1 1     2   j  l  , m j ,l   R  r    . 2    l 1 m  j   2  Y 1  ,    l , m  j 2l  1 2   l  0,1,2... 1 For the j  l  one has: 2 1  1  1   c12  l   m j  l   m j  l   mj  2 2  2  c2  1 ;  1. c12   1  1  1   l   mj  l   mj   2  2   l

1 1 1 1  mj l   mj l   mj l   mj 2 2 2 2 c1    ; c2   ; 1 2l  1 2 l  1 2 l  1 l   mj 2   1  l   mj  2   Y 1  ,   l ,m j  2l  1 1     2   j  l  , m j ,l   R  r     2    l 1 m  j   2  Y 1  ,    l , m  j 2l  1 2   l  1,2,3... l

75

These functions are orthonormal. The radial dependence of the wave function R  r  is defined by a form of the interaction potential V  r  . The coefficients coefficients. 1 1 s  , ms   . 2 2  j ,m j ,l  R  r 

obtained

are

the

Clebsch-Gordan

  1  lml ms jm j   Yl ,ml   s ,ms 2 ml  ms  m j  



1

0

1 1    , 1 1    . , , 0 1 2 2 2 2   1 jm j  lml ms jm j   Clml 1 ms – Clebsch-Gordan coefficients 2   2

j l

ms 

1 2

ms  

1 2

1 2

1  mj 2 2l  1

l

1  mj 2 2l  1

l

j l



1 2

1  mj 2 2l  1

l

1  mj 2 2l  1

l

In the second column the signs at the coefficients are changed for the opposite signs (in comparison with the obtained ones). Exactly such phases are obtained when deriving of the coefficients using theoretical group methods. In this form they are given in tables. The coefficients squared are the probabilities of finding of a system with momentum j , m j in the states with certain values of orbital momentum , m and spin s, ms . 76

§13. Construction of spin and orbital functions of two electrons s1 

1 1 and s2  . 2 2

 s m  1  and  s m  2  . 1 s1

Denote:

2

 1 1  1    1 ,  1

1  2 2

22

 1 1  2     2  ,  1

1  2 2

22

s

2

1     1 ,

 2      2  .

The spin function of a total system: s1s2 SM S  1 2   SM S  

 s m

ms1  ms2  M S

1



s ms2 SM S  1

s1 2

2

 1   ms 1

1 ms 2 2

 2  .

In accordance with the triangle rule the total spin of the entire system takes two values: S  0 and 1 (singlet and triplet). Calculating the corresponding coefficients of vector addition using the tables one gets for the triplet state of the system: S  1 , M S  1,0, 1 .

1 1 1 1   10    222 2 

1 1  0 1 2 2  ; 1 2 2  1 2

1 1 1 1   222 2

1 1  0 1 2 2  ; 1 2 2  1 2

 00   

77

1 1  1 2 2 1; 1 2  1 2

1 1 1 1  11   2222 

1 1  1 2 2 1. 1 2  1 2 1,1   1   2  ,

1 1 1 1   1  1    2 22 2 

1,0 

1   1    2     2    1 , 2 1, 1    1    2  ,

0,0 

1   1    2     2    1  . 2

Symmetry with respect to permutations: SM S 1 2    1

1 S

SM S  21  .

The triplet states S  1 – symmetric, and the singlet ones S  0 – antisymmetric. Construction of orbital function of two electrons l1l2 LM 

 l m l m

m1  m1  M

1

1 2

l1m1  Yl1m1  ,  и

78

2

LM  l1m1 l2 m2 ,

l2 m2  Yl2 m2  ,  .

When l1  l2  l  electrons are equivalent. The symmetry is defined as:

l1l2 LM   1

2l  L

l2l1LM   1 l2l1LM , L

that is depends on the parity of L . §14. Quantum theory of spin. Spin Pauli matrices. A complete set of two-column matrices The spin of electron – is a particular case of angular momentum ˆ J.

ˆ Like other operators of angular momentum the spin operator S is a vector, projections of which (as for any momentum) satisfy the following commutative relations:

  Sˆx , Sˆ y   i Sˆ z     ˆ ˆ ˆ S y , Sz   i Sx      Sˆz , Sˆ x   i Sˆ y      Sˆ 2 , Sˆ   0, i  x, y, z  i   

(14.1)

ˆ S 2  Sˆx2  Sˆ y2  Sˆz2 . Earlier it was shown that the operators satisfying the commutative relations (14.1), are quantized by the following rules: ˆ S2 

1 3 , 1, ,... 2 2 ˆ S z  mS ; mS  S , S  1,...,  S  in total  2S +1 values  . 2

S  S  1 ; S  0,

79

The spin of electron (and nucleon) – is a particular case of the momentum when S  1 2 . Then the projections of the spin mS  1 2; 1 2 take in total only two values, that is S z  2; и S z   2 . Earlier we considered the form of momentum operators for the case when j  0,1 2,1 and etc. In a case j  1 2  S : 0 1 Sˆx   ; 2 1 0

 0 i  Sˆ y   ; 2 i 0 

1 0  Sˆz   . 2  0 1

(14.2)

These operators are Hermitian and was obtained in a ˆ representation of the Eigen functions of the operators S 2 and Sˆ z . 1 0  Find the Eigen functions of the operator Sˆz    . Find 2  0 1 a the required wave function in a form of column  1    . Then , mS b 2 a a Sˆz    S z   ; b b

1 0  a a       Sz  . 2  0 1  b  b

The wave functions satisfy the condition of orthonormality

    1 , or  1

 1

2

2

, mS

, mS

  mS mS .

Since a a Sˆz    S z   ; b b 1 0  a a       Sz   2  0 1  b  b

then 80

   2  a   Sz a    ,    b   Szb     2 

hence S z  ; и S z   . 2 2 The condition of normality: a 2  b2  1 . 1. S z 

2

; 1 0  a     2  0 1  b 

a    2b

 a  a     ;  b   b 

a  a; b  b  0 .

Solution: a  1, b  0   1

1 , mS  2 2

1   .  0

2. S z   ; 2 1 0  a a         2  0 1  b  2b  a   a       b   b 

a  a  0; b  1 .

Then  1

1 , mS  2 2

 0  . 1 81

The wave functions

1 2

are called spinors and form a

, mS

complete set. It is easy to show that Sˆz  1 2

, mS

 mS  1 2

. , mS

Pauli spin matrices In quantum theory the Pauli spin matrices are widely used. They are introduced by the relation: ˆ (14.3) S  ˆ , 2 Sˆx  ˆ x , Sˆ y  ˆ y , Sˆz  ˆ z . 2 2 2

From (14.2) one has: 0 1

 0 i  1 0   ; ˆ z   ; 0  0 1

ˆ x    ; ˆ y   1 0 i

(14.4)

1 0

ˆ ˆ 2  ˆ x2  ˆ y2  ˆ z2  3    3I ; 0 1

1 0 Iˆ     unitary matrix. 0 1

The operators (14.4) are Hermitian and unitary. Further

ˆ x2  ˆ y2  ˆ z2  Iˆ .

(14.5)

A complete set of two-column matrices One can show that the system of four two-column matrices 1 0 ˆ x , ˆ y , ˆ z and Iˆ    from a complete set. 0 1 82

In the complete set all elements are to be linearly independent that is it is not possible to get any matrix through a multiplication of any other one by a constant number. This condition is easy to be tested and in addition any two-column matrix given in advance can be expanded by the complete set: A  a0 I  axˆ x  ayˆ y  azˆ z .

(14.6)

It is necessary only to determine the coefficients of expansion. This can be done using spurs – traces of matrices. The spur А equals the sum of diagonal elements: Sp A   Ann

(14.7)

n

1 It is obvious, a0  Sp A . The spur of all Pauli matrices equals 2 zero. The coefficients ax , a y , az have a form:

1 1 1 ax  Sp  Aˆ x  , a y  Sp  Aˆ y  , az  Sp  Aˆ z  . 2 2 2

§15. Isotopic spin for nucleons and nuclei The proton and neutron possess the very close properties. The small difference in their masses giving the difference in their average life time is the consequence of a difference of their electromagnet characteristics. That is why it is convenient to consider the proton and the neutron as two charge states of the same particle – nucleon. For a description of the charge state of a nucleon one has to introduce in addition to the simple space and spin variables the new, charge variable taking two values – one, corresponding to a proton state of the nucleon, and another – to a neutron one. The similar charge characteristics are introduced for a description of different states of a nucleus in total as well. 1. In accordance with two possible values of the charge variable the wave function is to be two-component: 83

 1  x     x   ,    x    2 

(15.1)

where x  r ,  z – is a set of spatial and spin coordinates of a nucleon. The meaning of the functions 1  x  and  2  x  will be clear, of one introduces a charge function of the proton and neutron states: 1 

p    0

and

 

0

n    . 1  

(15.2)

Then the expression (15.1) is written in a form of the following expansion: 1   0   x   1  x      2  x    .  0 1 

(15.3)

It is known that all linear operators operating on twocomponent wave functions can be expressed through the Pauli matrices: 0 1 , 1 0

ˆ1  

 0 i  , 0

ˆ2   i

1 0    0 1

ˆ3  

(15.4)

and the unitary matrix 1 0 1ˆ   . 0 1

(15.5)

It is easy to see that the operator ˆ1 plays a role of operator transferring the proton into the neutron and vice versa:

84

 0 1  1   0         n ,  1 0  0   1 

ˆ1 p  

(15.6)  0 1  0   1         p .  1 0  1   0 

ˆ1n   Introduce an operator

ˆp 

1 0 1 1  ˆ3    . 2 0 0

(15.7)

It can be considered as a projecting proton operator since it selects only the proton state from arbitrary states:  1 0  1   1         p ,  0 0  0   0 

ˆ p p  

(15.8)  1 0  0 

ˆ pn      0.  0 0  1  Using the operator ˆ p it is easy to get an operator of electric charge of a nucleon: q  e  ˆp 

e  1  ˆ3  , 2

(15.9) qˆ p  e   p ,

qˆ  n  0.

The operators ˆ1 ,ˆ2 ,ˆ3 possess all mathematical properties of components of operators of simple spin ˆ x ,ˆ y ,ˆ z . The last ones, as it is known, form a vector operator of the simple spin sˆ  1 2ˆ . Absolutely similarly one can unite the three components ˆ1 ,ˆ2 and 85

ˆ3 in a vector operator of isotopic spin tˆ  1 2ˆ , operating in a certain abstract space, called a charge one. It is worth to remember that the isotopic spin corresponds to a rotation in a certain space «1,2,3», which does not have any relation to the axes of simple space with which the direction of the simple spin can be connected. Consider an operation of the operator tˆ3  1 2ˆ3 on the charge nucleon functions:  1 0  1  tˆ3 p  1 2     1 2  p ,  0 1 0 

(15.10)  1 0  0  tˆ3n  1 2     1 2 n .  0 1 1 

Thus, the Eigen values of the operator tˆ3 can be considered as the additional variables of the nucleon characterizing its charge state: t3  1 2

for a proton,

t3  1 2

for a neutron.

(15.11)

2. For a nucleus, consisting of A nucleons, one can introduce an operator of the total isotopic spin: A

ˆ T   tˆ  i 

(15.12)

i 1

and an operator of total electric charge: A e A Qˆ   qˆ  i    1  ˆ3  i   . 2 i 1 i 1

86

(15.13)

It is obvious that a projection of the total isotopic spin of a nucleus is directly defined by a number of protons and neutrons in the nucleus: A

Tˆ3   tˆ3  i 

and

i 1

T3 

ZN . 2

(15.14)

ˆ

3. The operators T 2 and Tˆ3 commutate and that is why the quantities corresponding to them can be determined simultaneously and can characterize the nuclear states. Whether the Eigen values of these operators are integrals of motion? It is obvious that the T3 – is an integral of motion, since the number of nucleons and the charge of a nucleus are conserving quantities. For determination whether the Eigen value of the operator

ˆ T 2 is an integral of motion it is necessary to calculate the ˆ commutator of the operator T 2 with the total Hamiltonian of a nucleus. However, here one can use another, simpler receipt as well. From quantum mechanics it is known that the total momentum of a system of particles is an integral of motion if the total Hamiltonian of this system is invariant with respect to the coordinate system rotation. In a case of the isotopic spin it is necessary to determine whether the nuclear Hamiltonian is invariant with respect to coordinate system rotation in an isotopic space. At such rotations an orientation of a vector of the isotopic spin of a nucleus will be changed with respect to axes of this space, in particular, the projection of the isospin on axis «3». According to the formula (15.14), an increase of T3 for a unit corresponds to a change of one neutron by a proton, and a decrease of T3 for a unit – to a change of one proton by a neutron. So the rotation operation in the abstract isotopic space is imitated by the change of one of the protons of the nucleus by a neutron and vice versa. The general expression for the nuclear Hamiltonian has the following form: 87

Hˆ  Eˆ kin.  Wˆ  VˆCoul.  Vˆnucl.

(15.15)

Here Eˆ kin. – is an operator of the total kinetic energy of nucleons, W  is a total eigenenergy of nucleons, VCoul. – is energy of Coulomb interaction between nucleons, Vnucl. – is energy of nuclear interaction between nucleons. Because of the mass difference of proton and neutron the Ekin. and W are changed when interchanging of a proton by a neutron. When changing the number of protons the VCoul. is changed as well. How will the Vnucl. behave when such a replacement occurs? All collections of available at present data on a nucleus say that the nuclear interaction between two nucleons does not depend on their electric charge. This fundamental statement of the nuclear physics is called the hypothesis of charge independence of nuclear forces. Thus, in approach of charge independence of nuclear forces from the four terms of the Hamiltonian (15.15) the three – Ekin. , W and VCoul. are changed when interchanging a proton by a neutron (and vice versa) and the one – Vnucl. remains unchanged. That is why, speaking generally, the isotopic spin T is not an integral of motion. However it is easy to be convinced that the Coulomb interaction gives the largest contribution into the isotopically non-invariant part of the Hamiltonian. That is why for light nuclei where the ratio of Coulomb interaction is small in comparison with the nuclear interaction between nucleons one can neglect the isotopically noninvariant part of the Hamiltonian in comparison with the invariant one. Hence, in light nuclei one can approximately consider the isotopic spin to be an integral of motion. In heavy nuclei the relative role of the Coulomb interaction increases and the isotopic spin is not even approximately to be considered as an integral of motion. For them each state is to be considered as a mixture of the large number of states with different T . 4. With the concept of the isotopic spin the conception of generalized Pauli principle is closely connected. Since in an 88

approximation of the charge independence of the nuclear forces a proton and a neutron can be considered as identical particles then a wave function of a nucleus should be now antisymmetric not only by spatial and spin variables of nucleons but also by the variables of the isotopic spin:

  x1 , x2 , x3 ,...    x2 , x1 , x3 ,... ,

(15.16)

where x   r , sz , t3  , t3 – is a projection of the isotopic spin of the nucleon. 5. For nuclear reactions one van formulate the law of isotopic spin conservation. In a case, for example, of arbitrary two-particle reaction

ab cd the equality is true

Ti  T f

(15.17)

here Ti and T f – are isotopic spins of initial and final states of a system:

Ti  Ta  Tb , (15.18)

T f  Tc  Td . Similarly the conservation of isospin projection is expressed. 6. As an experiment shows in a case of light nuclei the following statement is quite true: the states with the least isotopic spin are energetically more favorable. Since it is obvious that a vector cannot be less than its projection then

Tmin  T3 89

(15.19)

and the isotopic spin of the ground states of light nuclei is expressed by an expression:

T

ZN . 2

(15.20)

§ 16. Construction of charge functions of two nucleons Construct a charge function of two nucleons if the charge functions of a proton and a nucleon are known –  p and  n (the indices p and n refer to values of projections of isotopic spin

t3  1 2 and 1 2 , respectively). The required function satisfies the operator equation

ˆ T 2T , T3  T T  1 T , T3 .

(16.1)

Suppose, that one has a system of two particles with momenta j1 , m1 and j2 , m2 . Denote their wave functions through the functions

 j m and  j m . As it is known, a wave function of a system with 2

1 1

2

total momentum I and its projection M is constructed by the following way:

  j1 j2 ; IM  



j1m1 j2 m2 IM  j1m1   j2m2 .

(16.2)

m1m2

Here

j1m1 j2 m2 IM

– is Clebsch-Gordan coefficient or a

vector addition coefficient. Using the formula (16.2) and finding the Clebsch-Gordan coefficients values from the corresponding tables it is easy to construct the following expressions:

0,0 

1  p 1n  2   p  2n 1 . 2 90

(16.3)

The given function describes the np-system in an antisymmetric isospin state with T  0 and T3  0 .

1,0 

1  p 1n  2   p  2n 1 . 2

(16.4)

This function describes the np-system in a symmetric state with T  1.

1,1  n 1 n  2  , (16.5)

1,1   p 1  p  2  . These functions describe the nn - and pp -systems in symmetric states respectively. § 17. Possible states of two nucleons system The basic knowledge about properties of nuclear forces is obtained from a study of two nucleons system. At that the primary problem is a classification of the possible states of this system. 1. The total wave function of a system of two nucleons can be written in a form of product of separate multipliers:

  spatial  spin  isospin .

(17.1)

The possibility of such a representation of the total wave function is based on the fact that there is no connection between the spin, isospin and spatial variables of nucleons. According to the generalized Pauli principle a wave function of a system of two identical nucleons is to be antisymmetric with respect to permutation of all coordinates. Since each of the multipliers in the expression (17.1) is defined by the corresponding momentum – orbital, spin and isospin, then it is necessary to clear 91

out how the symmetry of the wave function of a system consisting of two subsystems with the definite values of subsystems’ momenta is defined. As it has already been said in the previous paragraph if one has a system of two particles with momenta j1 , m1 and j2 , m2 , denote theirs wave functions through  j1m1 and  j2 m2 , then the wave function of a system with total momentum I and its projection M is constructed by the following way:

  j1 j2 ; IM  



j1m1 j2 m2 IM  j1m1   j2m2 .

(17.2)

m1m2

– is a Clebsch-Gordan coefficient or a

j1m1 j2 m2 IM

Here

vector addition coefficient. At that the total momentum I and its projection are defined by the following rules:

I  j1  j2 , j1  j2  1 ,... j1  j2 ,

(17.3)

M  m1  m2 . Permutating two particles, one gets a new function:

  j2 j1; IM  



j2 m2 j1m1 IM  j1m1   j2m2 .

m1m2

The symmetry of the wave function  is defined by the known properties of symmetry of Clebsch-Gordan coefficients:

j1m1 j2 m2 IM   1 1

j  j2  I

If

one

 j2 m2 j1m1 IM .

(17.4)

considers the spin function of two nucleons ( s1  s2  1 2 ), then the total spin of a system of two nucleons takes 92

two values S  0 and S  1 . According the (17.4) the symmetry of the spin function is defined by the expression  1

1 S

. Similarly the

symmetry of the isospin function is defined by the multiplier

 1

1T

. If the orbital momenta of nucleons are identical l1  l2

(such nucleons are called equivalent), then the symmetry of orbital function is defined by the multiplier

 1

L

, where the L – is an

orbital momentum of the system. Thus, the symmetry of the total wave function of two equivalent nucleons is defined by the expression:

 1  1  1 L

1 S

1T

  1

L  S T

.

In accordance with the requirements of antisymmetricity of the total wave function the combination of the quantum numbers L  S  T for a system of two identical nucleons is to be odd:

L  S  T = odd.

(17.5)

2. For denotation of the spin states of nucleons the terminology accepted in atomic spectroscopy is borrowed. If the total spin of a system is equal to S , then the quantity  2S  1 is called the multiplicity of the state. At S  0 one has the singlet states, and at S  1 – triplet states. Similar definitions are accepted for a classification of isospin states as well. Along with the orbital momentum L of a system, the value of total momentum is given which are defined as the following vector sum:

I  LS .

(17.6)

The states of two nucleons with total momentum equal to I , orbital momentum L , isospin T and spin S are denoted by the following way:  2T 1 2 S 1

93

LI .

(17.7)

Instead of the numerical value of L the Latin alphabet letters are usually put. At that to the values of L , equal to 0, 1, 2, 3,…, the letters S , P, D, F ,... are put in correspondence respectively. Using the above-pointed rules it is easy for example to determine the possible states of a system of P -nucleons, that is with l1  l2  1 : 13

S1 , 31S0 , 11P1 , 33P0,1,2 , 13D1,2,3 , 31D2 .

(17.8)

If the nucleons are equivalent, that is l1  l2 , then all values of

S , T and L are possible, defined by the rules of vector addition:

S  0,1

T  0,1

L  l1  l2 .

An antisymmetrization of the wave function is to be done additionally. 3. The simplest nuclear system – deuteron, presents a bound state of neutron-proton pair. A study of a deuteron is free of difficulties of quantum-mechanical problem of many bodies, nonavoidable when considering other nuclei. It gives direct information about properties of nuclear forces – forces between nucleons. The following properties of a deuteron are known. The binding energy is   2, 23 MeV. The smallness of the binding energy is one of the typical properties of a deuteron. The total momentum (spin) is I  1 . The magnetic momentum is d  0,860 . This value is very close to the algebraic sum of the magnetic momenta of a neutron and a proton:  p  n  2,79  (1,91)  0,880 . Hence one can conclude that the orbital momentum of relative motion of nucleons of a deuteron is equal to zero L  0 , and spins of the neutron and proton are parallel S  1 . Thus, in the spectroscopic denotations the deuteron state is written as 13S1 . The quadrupole momentum of a deuteron is equal to Q  (2,74  0,01) 101 fm2. An existence of the electric quadrupole momentum at a deuteron points out directly to the 94

violation of the spherical symmetry of charge distribution in this nucleus that is to contribution of higher, non-zero, orbital momenta of relative motion of proton and neutron. That is why the real function of a deuteron corresponds to a mixture of the states 13 S1  13 D1 , at that, as it will be shown below, the contribution of

D -state is not large. Thus, a problem about a deuteron consists in a solving of the Schrodinger equation for a potential V  r  of nucleon-nucleon interaction

 2 2    2   V  r     r   E   r  .  

(17.9)

Here r  rn  rp – are coordinates of relative motion of a neutron and a proton in a deuteron,  

mn m p mn  m p

– is a reduced

mass of a deuteron. In place of V  r  the different forms of interaction potentials were chosen. It turned out, however, that the main characters of the problem is weakly dependent on the specific properties of the potential V  r  . That is why for the sake of simplicity of the V  r  chose the rectangular potential well:

V , r  a, V r    0 0 , r  a.

(17.10)

The given well is characterized by two parameters – a depth V0 and a width a . The Schrodinger equation for the radial wave function has a form:

  1 d  2 d  2   2 r   2 E V r   r dr dr      95

l  l  1      Rl  r   0 . (17.11) 2 2 r   

2

Let us rewrite the radial equation, introducing a function U l  Rl  r ,

d 2U l 2   2 E V r   dr 2 

l  l  1   Ul  r   0 . 2 r 2 

2

(17.12)

Find from this equation the energy of the ground state of a deuteron For that let us introduce the binding energy which is equal to the required energy by value but opposite by sign:    E . For the S -state of mutual motion of deuterons the equation (17.12) takes the following form:

d 2U  r  2  2    V  r   U  r   0 . dr 2

(17.13)

Solve this equation for two parts of a space – inside and out of the potential well:

where  2 

2 2

d 2U1  r    2 U1  r   0 , dr 2

(17.14)

d 2U 2  r    2 U 2  r   0 , 2 dr

(17.15)

V0    and  2 

2 2

.

The solutions of the equations (17.14) and (17.15) have the form:

U1  r   A  sin  r  B  cos  r ,

(17.16)

U 2  r   C  e  r  D  e  r .

(17.17)

96

From the boundary condition in zero U1  0   0 one has B  0 . The coefficient D is equal to zero as well, since the function U  r  is to be finite at r   . Finally one has

 A  sin  r , при r  a, U r     r при r  a. C  e ,

(17.18)

The equality of logarithmic derivatives in point r  a

gives the condition

U1  a  U 2  a   U1  a  U 2  a 

(17.19)

 ctg  a    .

(17.20)

Introduce the quantities x   a and y   a . For them one has two equations  xctg(x)  y ,

x2  y 2 

2 2

V0 a 2 .

(17.21)

Figure 5. Bound state of deuteron in intersection points of concentric circumferences with line 97

Hence, the intersections of the concentric circumferences of radii a 

2V0 2

with the lines y   xctg(x) define the values of

V0 , at which the bound state of a deuteron appears (fig. 5). Because of the negative sign in the equation (17.20) the cotangent argument is to be in the even quarters. Now find the condition of existence of the bound state which has zero energy. At that y  0, ctg( a)  0 and

a 

 2

.

(17.22)

The required condition connects the parameters V0 and a 2 :

V0 a 2 

2 2 . 8

(17.23)

If one accepts that the quantity a  2 1015 м – average radius of nuclear forces action, then from the (17.23) one gets the minimal depth of the potential well where zero energy bound state exists: V 0 min  25 MeV. Since the binding energy of the ground state of a deuteron equals   2, 23 MeV, then the real well for that value should be deeper

V0  25 MeV. Simple calculations show that for the acceptable values of the parameter a the depth of the well is a quantity of order of

V0 =(30 ÷ 40) MeV.

(17.24)

In some specified sense the well describing the interaction of a proton and a neutron turns out to be shallow, the interaction between the nucleons are hardly enough to keep the system in a bound state with very small (in comparison with the depth of the well) binding 98

energy. One can show that in such a well there are no other bound states. Even a small positive addition at the expense of centrifugal energy 2l  l  1 2 r 2 at l  0 turns out to be enough in order to break the bounds between the nucleons. It is easy to be convinced that a deuteron cannot have excited states with l  0 as well, since the depth of the potential well in this case is essentially exceeds the obtainable values given by the formula (17.24). Consider now the wave function of a deuteron. In figure 6 the solid curve shows the exact function (17.18). It is seen, that approximately 40% of time the nucleons in the deuteron are at a distance exceeding the radius of nuclear forces action, that is the size of a deuteron exceeds the potential well width a . In this sense it is usually said that the deuteron presents a «porous» system. A quantity, characterizing the size of a deuteron, is R  1/  – a distance at which the wave function of a deuteron decreases “ e ” times

R

1





2

 4,3 fm.

Figure 6. Deuteron wave function

In the majority of calculations, where the deuteron wave function is included, one neglects the range of nuclear forces action, 99

which is small in comparison with the average sizes of the deuteron and uses the asymptotic expressions at all values of r :

U  r   C  e  r .

(17.25)

This function is shown in the figure 6 by dashed line. The constant C can be found from the condition of normality of the wave function   r   C



2

e  r : r

dV  4 C

 2

e

2  r

dr 

2 C 2

0



 1.

Hence

C

 . 2

Thus, the normal approximate wave function of the deuteron has a form:

U r  

  r e . 2

(17.26)

Note one more time, that the results, obtained here for the rectangular well are practically do not changed when using other short-range potentials with parameters chosen correctly, such as exponential potential:

V  V0  e r / a , or Yukawa potential

V  r   V0 

100

er a , r/a

V  r   V0  er

Gauss potential

2

b2

and other.

Problems to chapter III 1. Write the wave function of electron in the central field in the state with

l  3 , j  7 2 and m j  1 2 . Solution:

l , j ,m j 

  lm sm l

ml  ms 1 2

jm j  Y3ml  1

s

2

ms

 1 1 7 1   30  Y30  1 1   2 2 2 2 22

 1 1 7 1 4 3   31   Y30  1 1   Y31  1 1 .  Y31  1  1   2 2 2 2 7 7   2 2 22 2 2 2. Write the orbital function of two electrons with l1  2 and l2  3 in the state with total momentum L  4 and M  3 (without calculation of ClebschGordan coefficients). Solution: ml1  2,1, 0, 1, 2 , ml2  3, 2,1,0, 1, 2, 3

M  3  ml1  2,1, 0 , ml2  3, 2,1 

 l m l m

ml1  ml2  M

1

l1 2

l2



LM Yl1ml Yl2 ml  1

2

  2231 43 Y22Y31   2122 43 Y21Y22   2033 43 Y20Y33 . 3. An electron in an atom has an orbital momentum l  2 , spin s  1 2 and a total momentum j  5 2 . Construct the wave function of the electron  jm at

m j  3 2 using spin functions 1 2, ms and functions of orbital motion Ylml . Solution: ml  1 , ms  1 2 ;

ml  2 , ms  1 2 ;

 

5 3 j  ,m j  2 2

5 3 j  ,m j  2 2

 

 1 5 3  2ml ms Y2 m  1 ; 2 2 2  l 2 ms ml  ms  m j 



4 1  Y21  1 1   Y22  1 1 .  5 5 22 2 2 101

4. Construct the spin function of two electrons. At what values of the total spin S it will be symmetric relatively to coordinates permutation and at what values – antisymmetric? Solution: s1  s2  1 2 .

1

12 

 2m

1 ms 2 2

 SM S  1  1 ;  2 ms1 2 ms2

1

1 ms 2 1

 SM S  1  1 ;  2 ms 2 2 ms1

ms1 ms2

s1



 2m

 21 

ms1 ms2

1 1  ms2 ms1 2 2



s2

 1 1 S  1 SM S    1  ms1 ms2 2  2

 SM S  ; 

S  0,1 ; At S  0  the function is antisymmetric; At S  1  the function is symmetric. 5. The same should be done for the orbital function of two electrons with

l1  l2  l .

l1l2    lmlm LM l1 l2 .

6. According to the Pauli principle the total function of two electrons is to be antisymmetric with respect to permutation of all coordinates:  anti     . A system of two electrons with l1  l2  3 is given. What values of the total momentum L are consistent with S  0 and S  1 ? Solution:  anti   symm  anti or  anti   anti 

symm.

S  0 , L  0, 2, 4,6 S  1 , L  1,3,5 . 7. The same for a system of p  and f  electrons. What is the difference between these cases? L  2,3, 4 for S  0 and 1. The electrons are not equivalent. 8. Find the angular distribution for a proton in the state with j 

mj 

1 , l  1. 2 102

3 and 2

Y11 

3 sin   ei , 8

Y11 

 1 1  3 1  11   2 2 22

3 sin   ei , 8

Y10 

3 1  11  Y11  1  1  10 2 2  22 2 2

31  22

2

3 cos  . 4

3 1  Y10  1 1 ; 2 2 22

1 2  Y11  1 1   Y10  1 1 ;  3 3 2 2 22

1 2 1 1  3cos 2   .  Y112  Y102  3 3 8 

P  3 1 ,

2 2

ˆ

9. Find the explicit form of the operator S 2  Sˆx2  Sˆ y2  Sˆz2 and show that its 2

Eigen values have the form

S  S  1 , where S  1 2 .

Solution: 2 0 1  0 1  1 0         4 1 0 1 0 4  0 1 2

Sˆx2 

Sˆ y2 

2  0 i   0 i  1 0      4  i 0   i 0  4 0 1

Sˆz2 

2 1 0  1 0  1 0      4  0 1  0 1 4  0 1 

2

2

Then

3 ˆ S 2  Sˆx2  Sˆ y2  Sˆz2  4   Sˆ 2  1   ,m 2

S

1 2

2

1 0 3   0 1 4

ˆ 1 3  S2     0 4

2

From the other hand, since S  1 2 , then

103

2

  I  

1 0 1 3      0 1  0 4

2

1  .  0

ˆ S 21

1 , mS  2 2



ˆ 1  S2    0 2

1 S  S  1    0 1  1  1 3 2 1  .   1   0   22    4 0 2

For the function 1 2, mS 1 2  similarly. 10. Show that for the matrices (14.2) the commutative relations (14.1) are true. Solution:

Sx S y  S y Sx 

 0 1  0 i   0 i  0 1         4  1 0  i 0   i 0  1 0   2 2  i 0   i 0   i 0           4  0 i   0 i   2  0 i  2

1 0     i Sz . 2  0 1

i and etc.

11. Find the commutative relations for the ˆ x , ˆ y , ˆ z :

ˆ xˆ y  ˆ yˆ x  2iˆ z  

ˆ yˆ z  ˆ zˆ y  2iˆ x  ˆ zˆ x  ˆ xˆ z  2iˆ y  Solution: from the (14.1) one has: S x S y  S y S x  i S z or 2

2

ˆ xˆ y  ˆ yˆ x   i ˆ z , 4  2 Hence

ˆ xˆ y  ˆ yˆ x  2iˆ z and etc. 12. Find ˆ xˆ y  ˆ yˆ x . Solution: 104

(*)

2i ˆ xˆ y  ˆ yˆ x   2iˆ xˆ y  ˆ y  2iˆ x    ˆ yˆ z  ˆ zˆ y  ˆ y  ˆ y ˆ yˆ z  ˆ zˆ y    ˆ yˆ zˆ y  ˆ zˆ y2  ˆ y2ˆ z  ˆ yˆ zˆ y  0. Thus, the matrices ˆ i are anticommutative, that is ˆ x , ˆ y   0 and etc. 

ˆ xˆ y  ˆ yˆ x ˆ yˆ z  ˆ zˆ y

(**)

ˆ zˆ x  ˆ xˆ z Combining the expressions (*) and (**), one gets:

ˆ xˆ y  ˆ yˆ x  iˆ z ; ˆ yˆ z  ˆ zˆ y  iˆ x ; ˆ zˆ x  ˆ xˆ z  iˆ y . 13. Find ˆ xˆ yˆ z in two ways: 1) using the explicit form of Pauli matrices; 2) using the commutative relations.



14. Find ˆ , ˆ z



2

, using only the commutative relations.

  , where a  is a certain vector.

15. Calculate ˆ a

2

16. Whether electron spin projections squared on the axes x, y, z have values defined simultaneously? Solution: the problem is reduced to a calculation of the commutator:

ˆ x2 , ˆ y2   ˆ xˆ xˆ yˆ y  ˆ yˆ yˆ xˆ x

(  )

Transform ˆ xˆ xˆ yˆ y in a following way:

ˆ xˆ xˆ yˆ y  ˆ xˆ yˆ xˆ y  ˆ yˆ xˆ xˆ y  ˆ yˆ xˆ yˆ x  ˆ yˆ yˆ xˆ x . In the result, substituting the obtained expression into the expression (    ), one gets:

105

ˆ x2 , ˆ y2   ˆ yˆ yˆ xˆ x  ˆ yˆ yˆ xˆ x  0 . The same result can be obtained by other way as well:

ˆ x2 , ˆ y2    I , I   0 . Thus, the electron spin projections squared can have definite values simultaneously.

    





17. Prove the equality: ˆ a ˆ b  ab I  i  a  b  ˆ .



18. Expand by two-column matrices ˆ and I the following matrices:

1 1  1 0   3 1   i 0  ,  ,   и  . 1 2  1 0   0 2   2 1 19. For the following nuclei define the isotopic spin of the ground and states

T and its projection: 1

H2 , 1 H3 , 2 He3 , 2 He4 , 4 Be9 , 7 N15 , 8 O15 , 8 O16 .

20. Suppose that a certain nucleus decay occurred with emission of alphaparticle. What can be said about the isotopic spin of the state formed? 21. Find the form of the projecting neutron operator. 22. Can the total spin of a system of two protons in s -state be equal to 1? 23. Estimate the minimal depth of the potential well for the second excited state in a deuteron with l  0 . 24. Estimate the order of value of the centrifugal energy of a deuteron for the orbital momentum l  1 . 25. Two electrons are in the f -shell of an atom. Determine, which values of the total orbital momentum L are consistent with the total spin S  0 and S  1 . Consider in a similar way the possible states of the configurations pf ( l1  1, l2  3 ). What is the difference between these two cases? 26. A nucleon in the 8 O17 nucleus has an orbital momentum l  2 , spin

s  1/ 2 , total momentum j  3 / 2 . Construct the wave function if the nucleon  jm at m  3 / 2 using spin function 1 2, ms and orbital function Y2, ml . Use the below-given reduced table of Clebsch-Gordan coefficients.

106

 j m 1/ 2 m 1

j1  1 2

k 

2

jm 

m2  1 2

m2  1 2

j1  m  1 2 2 j1  1

j1  m  1 2 2 j1  1

g 1 (2 ) 2 k 2   2

j1  m  1 2 2 j1  1

j

j1  1 2

1

27. Find the relation between the cross sections of reactions (cross sections are defined by squares of the corresponding Clebsch-Gordan coefficients):

p  p  d  , p  n  d   0. 28. Show, that Clebsch-Gordan coefficients

l 0l 0 1

2

L0   0 ,

if l1  l2  L  odd number. Use the property of symmetry for the coefficients:

 j m 1/ 2 m 1

1

2

jm    1 1

j  j2  j

  j1  m1 1/ 2  m2 j  m  .

29. Show that the reaction    d  n  n is inhibited for the scalar (that is zero spin and positive parity) π-meson in an assumption that it is captured from the s -level of «mesodeuterium» (π-meson stopped). 30. Which of the given particles

f 0  I  , T   2 ,0;  0  I  , T   1 ,0;  0  I  , T   0 ,0 can decay by two-pion channel? 31. Which of the given below systems can exist in the state with T  1 :

 0 0 ,    ,    , 0 0 ,  0 . When solving use the following table:

107

T T3

0





0



1 0

1 +1

1 -1

1 0

0 0

32. Determine the ratios of the reaction total cross sections

p  d 3 He   0 , p  d 3 He    . Test questions to chapter III 1. What are the coefficients of vector addition? 2. What are the properties of the Clebsch-Gordan coefficients? 3. How can the Clebsch-Gordan coefficients be used in nuclear reactions theory? 4. How are the spin and orbital wave functions of two electrons constructed? 5. Which system of matrices is called a complete one? 6. What is the generalized Pauli principle? 7. What are the possible states of a system of two nucleons? 8. How is the vector addition of momenta defined? 9. What is the main point of equivalent and non-equivalent nucleons momenta addition? 10. Which system is called the 33-resonanse? 11. What is the main point of the Shmushkevich isotopic rule?

108

Chapter IV

VECTOR ADDITION OF THREE AND FOUR MOMENTA §18. Vector addition of three momenta. Racah coefficients. 6j-symbols If one adds three angular momenta j1 , j2 , j3 , then one can firstly add j1 and j2

j  j 1

2



 J12 , then adds j3 to the first sum

and obtains a state of a system of three angular momentum with total momentum J and its projection М on axis z J12  j3  J . The





vector of state, constructed in such a way, is denoted as | j1 j2  J12  , j3 : JM . It is obvious, that j1 j2  J12  , j3 : JM 



( j1m1 j2 m2 | J12 M12 ) 

m1m2 m3 M12

 ( J12 M12 j3m3 | JM ) j1m1 j2 m2 j3m3 . Note, that two other vectors of state j3 , j1 j2  J12  : JM

(18.1)

j1 j2  J12  , j3 : JM ,

distinguishing from the (18.1) by an order of

angular momentum addition j1 , j2 and J12 , j3 respectively, coincide with the (18.1) within the accuracy of the phase: j1 j2  J12  , j3 : JM   1 1

j  j2  J12

  1 12

J  j3  J

j2 j1  J12  , j3 : JM  j3 , j1 j2  J12  : JM .

Addition of the same momenta can be done in other way: firstly





add j2 and j3 j2  j3  J 23 , and the obtained sum is added to the 109





j1 j1  J 23  J . Then one gets a state of system described by a

vector j1 , j2 j3  J 23  : JM 



( j2 m2 j3m3 | J 23 M 23 ) 

m1m2 m3 M 23

 ( j1m1 J 23 M 23 | JM ) j1m1 j2 m2 j3m3 .

(18.2)

The states (18.1) and (18.2) are connected with each other by the unitary transformation: j1 j2  J12  , j3 : JM    j1 j2  J12  , j3 : J | j1 , j2 j3  J 23  : J

j1 , j2 j3  J 23  : JM ;

J 23

j1 , j2 j3  J 23  : JM    j1 j2  J12  , j3 : J | j1 , j2 j3  J 23  : J

j1 j2  J12  , j3 : JM .

J12

The coefficients of transfer from one coupling scheme to another are called Racah coefficients and are denoted by a symbol in short U  j1 j2 Jj3 : J12 J 23   j1 j2  J12  , j3 : J | j1 , j2 j3  J 23  : J .

The Racah coefficients U are diagonal by total momentum and its projection and do not depend on this projection. From (18.1) and (18.2) it is seen, that the Racah coefficients are expressed through the Clebsch-Gordan coefficients by the following way: U  j1 j2 Jj3 : J12 J 23   



( j1m1 j2 m2 | J12 M 12 )( J12 M 12 j3 m3 | JM ) 

m1m2 m3 M12 M 23

 ( j2 m2 j3 m3 | J 23 M 23 )( j1m1 J 23 M 23 | JM ).

The relations of orthonormality are true for them: 110

U  j j Jj 1 2

3

: J12 J 23 U  j1 j2 Jj3 : J12 J 23    J12 J12 .

J 23

The coefficients U within the multiplier coincide with Wigner  j1 j2 J12  coefficients W  j1 j2 Jj3 : J12 J 23  and with 6j-symbols  :  j3 J J 23 

ab  e  , d : c | a, bd  f  : c  U  abcd : ef   

 2e  1 2 f

  1

a b  c  d

 1 W  abcd : ef  

 2e  1 2 f

a  1  d

b e . (18.3) c f

The Racah coefficients possess quite simple properties of symmetry with respect to permutation of momenta which they depend on. However the 6j-symbols have the largest symmetry: they do not change when permutating any columns:  j1 j2  l1 l2

j3   j1 j3 j2   j2   l3  l1 l3 l2  l2

j3 l3

j1    ... l1 

and when interchanging two any momenta in the upper row by the corresponding momenta of the lower row:  j1 j2  l1 l2

j3  l1 l2  l3   j1 j2

j3    ... l3 

The additional properties of symmetry of 6j-symbols are also known which were stated by Regge. There is a formula below for calculation of 6j-symbol, when one of the momentum equals zero:  j1 j2  0 l2

j3     l2 j2  l3 j3  l3  111

 1

j1  j2  j3

 2 j2  1 2 j3  1

.

(18.4)

§19. Vector addition of four momenta. Generalised Racah coefficients. 9j-symbols If one adds four angular momenta j1 , j2 , j3 , j4 , then one can

j  j

firstly add j1 and j2

1

2



 J12 , then j3 and j4

j

3



 j4  J 34 ,

and after that the intermediate momenta J12 and J 34 are added into the total momentum J . The obtained in a result wave function is denoted as j1 j2  J12  , j3 j4  J 34  : JM . One can also change the scheme

of

J13  J 24  J .

momenta At

coupling: that

j2  j4  J 24 ,

j1  j3  J13 ,

one

obtains

the

function

j1 j3  J13  , j2 j4  J 24  : JM . The coefficients of transfer from the first

coupling of four momenta to another one

j1 j2  J12  , j3 j4  J 34  : J | j1 j3  J13  , j2 j4  J 24  : J are called the generalized Racah coefficients. The more symmetric quantity is the so called 9j-symbol:  j1   j3 J  13

J12   J 34    2 J12  1  J 

j2 j4 J 24

1/ 2

  2 J 34  1 2 J13  1 2 J 24  1    j1 j2  J12  , j3 j4  J 34  : J | j1 j3  J13  , j2 j4  J 24  : J .

Using the 9j-symbols one can write: j1 j2  J12  , j3 j4  J 34  : JM  

  2 J

12

 1 2 J 34  1 2 J13  1 

J13 J 24

  2 J 24  1 

1/2

 j1   j3 J  13

j2 j4 J 24

J12   J 34  j1 j3  J13  , j2 j4  J 24  : JM . J  112

The coefficients of transfer between any other two coupling schemes of four momenta can be reduced to the 9j-symbols, which are tabled quantities. The 9j-symbol is easy to express through the 6j-symbol or 3jsymbols: a b  d e g h 

a d g b e h c f    h i k d k f k a j3  3   ji  mi  li  ni  ki  qi  l3     1  i 1 m1 m2 m3  k3  n1 n2 n3

 j1 j2   l1 l2 k k  1 2

q1

j  1  m1

c 2k  f     1  2k  1  k i 

j2 m2

q2

i ; b

q3

j3  l1 l2  m3  n1 n2

l3  k1 k2  n3  q1 q2  j  2  m2

l2 n2

k3  j1 l1 k1    q3   m1 n1 q1  k2   j2 l2 k2   . q2   m2 n2 q2 

When even permutation of the columns or rows and also when inversion with respect to both the diagonals the 9j-symbol remains unchanged. When odd permutation of columns or rows it multiplied by the  1 : s

 j1   l1 k  1

j2 l2 k2

j3   j3   l3    l3  k3    k3

j2   j2 s   l2    1  l2 k k2    2

j1 l1 k1

3

and etc., where s    ji  li  ki  . i 1

113

j1 l1 k1

j3   l3  k3  

The property of orthonormality a b  2e  1 2 f  1 2 g  1 2h  1 d e  gh g h 

c  a b  f d e i   g h

c   f    cc ff  . i  

When one of the momenta in the 9j-symbol equals zero, it reduces to the 6j-symbol: a  c f 

b d f

e  e  0 

 1

bc e f

a   2e  1 2 f  1 d

b e . c f

The 9j-symbols are very convenient when transferring from LSto jj-coupling (or vice versa): l1l2  L  , s1s2  S  : JM    2 L  1 2S  1 2 j1  1  j1 j2

  2 j2  1 

1/2

 l1   s1 j  1

L  S  l1s1  j1  , l2 s2  j2  : JM ; J 

l2 s2 j2

l1s1  j1  , l2 s2  j2  : JM    2 L  1 2S  1 2 j1  1  LS

  2 j2  1 

1/2

 l1   s1 j  1

l2 s2 j2

L  S  l1l2  L  , s1s2  S  : JM . J 

Problems to chapter IV 1. Expand by the spherical harmonics the cos 4  , cos6  , cos   Ylm  ,   ,

dYlm  ,   / d .

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2. Express the wave function  nlm  r  of three-dimensional harmonic oscillator, corresponding to the energy level En    n  3 / 2  and given in the spherical coordinates, through the oscillatory wave functions  nx ny nz  r  in the Cartesian coordinates. A calculation should be done for the cases: а)  200  r  , b)  220  r  . 3. Two particles with spins s1=1 and s2=2 are in a state when projections of both the spins on the axis z equal 0. Show, that in this case the total spin of two particles S = s1 + s2 cannot take the value S=2. 4. Two electrons are on the f-shell of an atom (l1=l2=3). Determine, which values of the total orbital momentum are consistent with the total spin S=0 and S=1. Consider similarly the possible states of configuration pf (l1=1, l2=3). What is the difference between these two cases? 5. In nuclear physics various multiplicity (   2, 3,... ) collective oscillations of the nuclear surface are considered. The quadrupole oscillations   2 correspond to ellipsoidal, octopole   3 – to pear-shaped deformations of spherical (in equilibrium state) nuclei and etc. Each quanta of these oscillations (it can be considered as a certain quasiparticle – phonon which is a boson) carries an angular momentum  and has a parity  1 . Determine which values of total momentum 

L are admitted in a system of two quadrupole phonons. Which values of L are possible in a system of two octopole phonons? 6. Show that the wave function of a system of three particles in a form

jj

1 2 j3

 j1  m1 , m2 , m3  m1

1, 2,3  

j2 m2

j3   j m 1 j2 m2  2  j3 m3  3 m3  1 1

describes a

state of these three particles with total momentum J  j1  j2  j3 , equal zero. Using this show that the system of three octopole phonons cannot possess zero total momentum. 7. Using properties of Regge symmetry show that 3j-symbol

3 5 / 2 7 / 2    equals zero.  0 3 / 2 3 / 2  8. Using Racah coefficients or 6j-symbols express the wave function in a form

j1 j2  J12  , j3 : J

through the wave functions, corresponding the following

coupling schemes: а) j1   j2  j3  ; b) j3   j1  j2  ; c) j2   j3  j1  ; d) j1   j3  j2  ; e) j2   j1  j3  ; f) j3   j2  j1  ; 115

 j3  j1   j2 ; j)  j2  j1   j3 . g)

h)

 j2  j3   j1 ;

(in some cases the transfer coefficients are reduced to the phase multipliers). 9. Prove the relation:

 a b | e  e d | c  b d | f      

 2e  1 2 f

 1  W  abcd ; ef  a f  | c  .

10. а) Transfer from jj-coupling to LS-coupling in the wave function s1/ 2 d3/ 2 : J  2 of two electrons; b) Transfer from LS -coupling to jj –coupling in the wave functions of two f-electrons

f 2 :1 G4 .

Test questions to chapter IV 1. 2. 3. 4. 5.

What are the Racah coefficients necessary for? What are the 6j-symbols introduced for? What are the generalized Racah coefficients? What are the 9j-symbols introduced for? What are the LS- and jj-couplings of momenta?

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REFERENCES 1. D.A. Varshalovich, A.N. Moskalev, V.K. Khersonskii. Quantum Theory of Angular Momentum. Translation. ISBN 9971-50-107-4. 2. Julian Schwinger. On Angular Momentum. Printed in the USA by Courier Corporation, ISBN-13: 978-0-486-78810-4. 3. M.E. Rose. Elementary Theory of Angular Momentum. Dover Publications, Inc. − New York. − ISBN 0-486-68480-6. 4. A.R. Edmonds. Angular Momentum in Quantum Mechanics. − Princeton University Press, Princeton, New Jersey. − ISBN 0-691-07912-9. 5. Ishkhanov B.S., Kapitonov I.M., Yudin N.P. Particles and atomic nuclei. – M.: LKI, 2007. – 584 p.

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CONTENT

INTRODUCTION .............................................................................................. 3 Chapter I. ORBITAL MOMENTUM AND ITS QUANTIZATION .................. 5 §1. Motion in centrally symmetrical field. Integrals of motion .................................................................................... 5 §2. Parity of quantum states ............................................................................ 6 §3. Operator of orbital angular momentum .................................................... 10

Hˆ , Lˆ2

Lˆ z .......................................... 19

§4.

Eigen functions of operators

§5.

Quantization of operator L from elementary formulas of probabilities theory ............................................................................... 24

and

ˆ2

Chapter II. GENERAL THEORY OF ANGULAR MOMENTUM OPERATOR .............................................................................. 28 §6. General theory of angular momentum operator.

ˆ2

Quantization of operators J and Jˆ z from commutative relations ............................................................................... 28

Jˆ , Jˆ , Jˆx and Jˆ y in diagonal

§7.

Matrix elements of operators

§8.

representation of operators Jˆ and Jˆ z .................................................... 32 Geometric interpretation of momentum projection operator ..................... 41 2

Chapter III. VECTOR ADDITION OF TWO MOMENTA. APPLICATION OF THEORY .......................................................................... 44 §9. Vector addition of two momenta .............................................................. 44 §10. General formula for Clebsch-Gordan coefficients through factorials. Basic properties of vector addition coefficients .................................................................................. 47 §11. Use of Clebsch-Gordan coefficients. Comparison of electrons angular distributions with  1 , j  3 2 and different projections m j ..................................................................... 51 §12. Operator of electron total momentum. Construction of wave function of a particle with spin 1 2 ....................... 65

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§13. Construction of spin and orbital functions of two electrons ........................................................................................ 77 §14. Quantum theory of spin. Spin Pauli matrices. A complete set of two-column matrices .................................................... 79 §15. Isotopic spin for nucleons and nuclei ....................................................... 83 §16. Construction of charge functions of two nucleons ................................... 90 §17. Possible states of two nucleons system .................................................... 91 Chapter IV. VECTOR ADDITION OF THREE AND FOUR MOMENTA.......... 109 §18. Vector addition of three momenta. Racah coefficients. 6j-symbols.................................................................. 109 §19. Vector addition of four momenta. Generalised Racah coefficients. 9j-symbols ........................................................................... 112 REFERENCES ................................................................................................... 117

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Educational edition Zhusupov Marat Abzhanovich Kabatayeva Raushan Sarsembekovna Zhaksybekova Kulyan Aitmagambetovna

QUANTUM THEORY OF ANGULAR MOMENTUM Educational manual Computer page makeup and cover designer N. Bazarbaeva The website used for front-page designing http://bakaraban.ru IS No.10989 Signed for publishing 09.06.17. Format 60x84 1/16. Offset paper. Digital printing. Volume 7,5 printer’s sheet. Edition 80. Order No.3495 Publishing house «Qazaq university» Al-Farabi Kazakh National University, 71 Al-Farabi, 050040, Almaty Printed in the printing office of the «Qazaq university» publishing house

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