Physics 582 General Field Theory

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Eduardo Fradkin

Physics 582 General Field Theory

University of Illinois at Urbana-Champaign 2020

1 Introduction to Field Theory

The purpose of this book, is twofold. Here I will to introduce field theory as a framework for the study of systems with a very large number of degrees of freedom, N → ∞. And I will also introduce and develop the tools that will allow us to treat such systems. Systems that involve a large (in fact, infinite) number of coupled degrees of freedom arise in many areas of Physics, notably in High Energy and in Condensed Matter Physics, among others. Although the physical meaning of these systems and their symmetries are quite different, they actually have much more in common than it may seem at first glance. Thus, we will discuss, on the same footing, the properties of relativistic quantum field theories, classical statistical mechanical systems and condensed matter systems at finite temperature. This is a very broad field of study and we will not be able to cover each area in great depth. Nevertheless, we will learn that it is often that case that what is clear in one context can be used to expand our knowledge in a different physical setting. We will focus on a few unifying themes, such as the construction of the ground state (the “vacuum”), the role of quantum fluctuations, collective behavior, and the response of these systems to weak external perturbations.

1.1 Examples of fields in physics 1.1.1 The electromagnetic field Let us consider a very large box of linear size L → ∞, and the electromagnetic field enclosed inside it. At each point in space x we can define a vector (which is a function of time as well) A(x, t) and a scalar A0 (x, t). These are the vector and scalar potentials. The physically observable electric field

2

Introduction to Field Theory

E(x, t) and the magnetic field B(x, t) are defined in the usual way 1 ∂A (x, t) − !A0 (x, t) (1.1) E(x, t) = − c ∂t The time evolution of this dynamical system is determined by a local Lagrangian density (which we will consider in section 2.6). The equations of motion are just the Maxwell equations. Let us define the 4-vector field B(x, t) = ! × A(x, t),

A (x) = (A (x), A(x)) µ

0

0

A ≡ A0

(1.2)

where µ = 0, 1, 2, 3 are the time and space components. Here x stands for the 4-vector x = (ct, x) µ

µ

(1.3)

To every point x of Minkowski spacetime M we associate a value of the µ vector potential A . The vector potentials are ordered sets of four real num4 bers and hence are elements of R . Thus a field configuration can be viewed 4 as a mapping of the Minkowski spacetime M onto R , µ

A ∶M↦R

4

(1.4)

Since spacetime is continuous we need an infinite number of 4-vectors to specify a configuration of the electromagnetic field, even if the box were finite (which is not). Thus we have a infinite number of degrees of freedom for two reasons: spacetime is both continuous and infinite.

1.1.2 The elastic field of a solid Consider a three-dimensional crystal. A configuration of the system can be described by the set of positions of its atoms relative to their equilibrium state, (i.e. the set of deformation vectors d at every time t). Lattices are labeled by ordered sets of three integers and are equivalent to the set 3

Z =Z×Z×Z

(1.5)

whereas deformations are given by sets of three real numbers, and are ele3 ments of R . Hence a crystal configuration is a mapping 3

d∶Z ×R↦R

3

(1.6)

At length scales ", which are large compared to the lattice spacing a but small 3 compared to the linear size L of the system, we can replace the lattice Z by a continuum description in which the crystal is replaced by a continuum 3 three-dimensional Euclidean space R . Thus the dynamics of the crystal

1.1 Examples of fields in physics 3

3

4

requires a four-dimensional spacetime R × R = R . Hence the configuration space becomes the set of continuous mappings 4

d∶R ↦R

3

(1.7)

In this continuum description, the dynamics of the crystal is specified in terms of the displacement vector field d(x, t) and its time derivatives, the velocities ∂d (x, t), which define the mechanical state of the system. This is ∂t the starting point of the theory of elasticity. The displacement field d is the elastic field of the crystal.

1.1.3 The order-parameter field of a ferromagnet Let us now consider a ferromagnet. This is a physical system, usually a solid, in which there is a local average magnetization field M (x) in the vicinity of a point x. The local magnetization is simply the sum of the local magnetic moments of each atom in the neighborhood of x. At scales long compared to microscopic distances (the interatomic spacing a), M (x) is a continuous real vector field. In some situations, of interest, the magnitude of the local moment does not fluctuate but its local orientation does. Hence, the local state of the system is specified locally by a three-component unit vector n. Since the set of unit vector is in one-to-one correspondence with the points 2 on a sphere S , the configuration space is equivalent (isomorphic) to the sets 2 of mappings of Euclidean three-dimensional space onto S , 3

n∶R ↦S

2

(1.8)

In an ordered state the individual magnetic moments become spontaneously oriented along some direction. For this reason, the field n is usually said to be an order parameter field. In the theory of phase transitions, the order parameter field represents the important degrees of freedom of the physical system,( i.e., the degrees of freedom that drive the phase transition).

1.1.4 Hydrodynamics of a charged fluid Charged fluids can be described in terms of hydrodynamics. In hydrodynamics, one specifies the charge density ρ(x, t) and the current density j(x, t) µ near a spacetime point x . The charge and current densities can be represented in terms of the 4-vector j (x) = (cρ(x, t), j(x, t)) µ

(1.9)

4

Introduction to Field Theory

where c is a suitably chosen speed (generally not the speed of light!). Clearly, the configuration space is the set of maps µ

4

j ∶R ↦R

4

(1.10)

In general we will be interested both in the dynamical evolution of such systems and in their large-scale (thermodynamic) properties. Thus, we will need to determine how a system that, at some time t0 is in some initial state, manages to evolve to some other state after time T . In Classical Mechanics, the dynamics of any physical system can be described in terms of a Lagrangian. The Lagrangian is a local functional of the field and of its space and time derivatives. “Local” here means that the equations of motion can be expressed in terms of partial differential equations. In other words, we do not allow for “action-at-a-distance,” but only for local evolution. Similarly, the thermodynamic properties of these systems are governed by a local energy functional, the Hamiltonian. That the dynamics is determined by a Lagrangian means that the field itself is regarded as a mechanical system to which the standard laws of Classical Mechanics apply. Here, the wave equations of the fluid are the equations of motion of the field. This point of view will also tell us how to quantize a field theory. 1.2 Why quantum field theory? From a historical point of view, quantum field theory (QFT) arose as an outgrowth of research in the fields of nuclear and particle physics. In particular, Dirac’s theory of electrons and positrons was, perhaps, the first QFT. Nowadays, QFT is used, both as a picture and as a tool, in a wide range of areas of physics. In this course, I will not follow the historical path of the way QFT was developed. By and large, it was a process of trial and error in which the results had to be reinterpreted a posteriori. The introduction of quantum field theory as the general framework of particle physics implied that the concept of particle had to be understood as an excitation of a field. Thus photons become the quantized excitations of the electromagnetic field with particle-like properties (such as momentum), as anticipated by Einstein’s 1905 paper on the photoelectric effect. Dirac’s theory of the electron implied that even such “conventional” particles should also be understood as the excitations of a field. The main motivation of these developments was the need to reconcile, or unify, Quantum Mechanics with Special Relativity. In addition, the experimental discoveries of the spin of the electron and of electron-positron creation by photons, showed that not only was the Schr¨odinger equation

1.2 Why quantum field theory?

5

inadequate to describe such physical phenomena, but the very notion of a particle itself had to be revised. Indeed, let us consider the Schr¨odinger equation HΨ = ih̵

∂Ψ ∂t

(1.11)

where H is the Hamiltonian 2

pˆ H= + V (x) (1.12) 2m and pˆ is the momentum represented as a differential operator h̵ (1.13) pˆ = ! i acting on the Hilbert space of wave functions Ψ(x). The Schr¨ odinger equation is invariant under Galilean transformations, provided the potential V (x) is constant, but not under general Lorentz transformations. Hence, Quantum Mechanics as described by the Schr¨odinger equation, is not compatible with the requirement that the description of physical phenomena must be identical for all inertial observers. In addition, it cannot describe pair-creation processes since in the non-relativistic Schr¨ odinger equation, the number of particles is strictly conserved. Back in the late 1920s, two apparently opposite approaches were proposed to solve these problems. We will see that these approaches actually do not exclude each other. The first approach was to stick to the basic structure of “particle” Quantum Mechanics and to write down a relativistically invariant version of the Schr¨ odinger equation. Since in Special Relativity the natural 2 2 2 Lorentz scalar involving the energy E of a particle of mass m is E − (p c + 2 4 m c ), it was proposed that the “wave functions” should be solutions of the equation (the “square” of the energy) [(ih̵

2 ̵ ∂ 2 hc 2 4 ) − (( !) + m c )] Ψ(x, t) = 0 i ∂t

(1.14)

This is the Klein-Gordon equation. This equation is invariant under the Lorentz transformations, µ

µ,ν ′ xν

x =Λ

x = (x0 , x) µ

(1.15)

provided that the “wave function” Ψ(x) is also a scalar (i.e. invariant) under Lorentz transformations Ψ(x) = Ψ (x ) ′

′

(1.16)

However, it soon became clear that the Klein-Gordon equation was not

6

Introduction to Field Theory

compatible with a particle interpretation. In addition, it cannot describe particles with spin. In particular, the solutions of the Klein-Gordon equation have the (expected) dispersion law 2

2 2

2 4

(1.17)

E =p c +m c

which implies that there are positive and negative energy solutions √ E = ± p2 c2 + m2 c4

(1.18)

From a “particle” point of view, negative energy states are unacceptable since they would imply that there is no ground state. We will see in chapter 4 that in quantum field theory there is a natural and simple interpretation of these solutions, and that in no way make the system unstable. However, the meaning of the negative energy solutions was unclear in the early thirties. To satisfy the requirement from Special Relativity that energy and momentum must be treated equally, and to avoid the “negative energy solutions” that came from working with the “square” of the Hamiltonian H, Dirac proposed to look for an equation that was linear in derivatives (Dirac, 1928). In order to be compatible with Special Relativity, the equation must be covariant under Lorentz transformations,( i.e. it should have the same form in all reference frames). Dirac proposed a matrix equation that is linear in derivatives with a“wave function” Ψ(x) in the form of a four-component vector, a 4-spinor Ψa (x) (with a = 1, . . . , 4) ih̵

̵ 3 ab ∂Ψa hc 2 ∑ αj ∂j Ψb (x) + mc βab Ψb (x) = 0 (x) + i ∂t

(1.19)

j=1

where αj and β are four 4 × 4 matrices. For this equation to be covariant it is necessary that the 4-spinor field Ψ should transform as a spinor under Lorentz transformations Ψa (Λx) = Sab (Λ)Ψb (x) ′

(1.20)

where S(Λ) is a suitable matrix. The matrices αj and β have to be pure numbers independent of the reference frame. By further requiring that the iterated form of this equation (i.e. the “square”) satisfies the Klein-Gordon equation for each component separately, Dirac found that the matrices obey the (Clifford) algebra {αj , αk } = 2δjk 1,

{αj , β} = 0,

2

2

αj = β = 1

(1.21)

where 1 is the 4 × 4 identity matrix. The solutions are easily found to have √ the energy eigenvalues E = ± p2 c2 + m2 c4 . (We will come back to this in

1.2 Why quantum field theory?

7

chapter .2) It is also possible to show that the solutions are spin 1/2 particles and antiparticles (we will discuss this later on). However, the particle interpretation of both the Klein-Gordon and the Dirac equations was problematic. Although spin 1/2 appeared now in a natural way, the meaning of the negative energy states remained unclear. The resolution of all of these difficulties was the fundamental idea that these equations should not be regarded as the generalization of Schr¨odinger’s equation for relativistic particles but, instead, as the equations of motion of a field, whose excitations are the particles, much in the same way as the photons are the excitations of the electromagnetic field. In this picture particle number is not conserved but charge is. Thus, photons interacting with matter can create electron-positron pairs. Such processes do not violate charge conservation but the notion of a particle as an object that is a fundamental entity and has a distinct physical identity is lost. Instead, the field becomes the fundamental object and the particles become the excitations of the field. Thus, the relativistic generalization of Quantum Mechanics is Quantum Field Theory. This concept is the starting point of Quantum Field Theory. The basic strategy to seek a field theory with specific symmetry properties and whose equations of motion are Maxwell, Klein-Gordon and Dirac equations, respectively. Notice that if the particles are to be regarded as the excitations of a field, there can be as many particles as we wish. Thus, the Hilbert space of a Quantum Field Theory has an arbitrary (and indefinite) number of particles. Such a Hilbert space is called a Fock space. Therefore, in Quantum Field Theory the field is not the wave function of anything. Instead the field represents an infinite number of degrees of freedom. In fact, the wave function in a Quantum Field Theory is a functional of the field configurations which themselves specify the state of the system. We will see below that the states in Fock space are given either by specifying the number of particles and their quantum numbers or, alternatively, in terms of the amplitudes (or configurations) of some properly chosen fields. Different fields transform differently under Lorentz transformations and constitute different representations of the Lorentz group. Consequently, their excitations are particles with different quantum numbers that label the representation. Thus, 1) The Klein-Gordon field φ(x) represents charge-neutral scalar spin-0 particles. Its configuration space is the set of mappings of Minkowski space onto the real numbers φ ∶ M ↦ R, or complex numbers for charged spin-0 particles φ ∶ M ↦ C. 2) The Dirac field represents charged spin-1/2 particles. It is a complex 4-

8

Introduction to Field Theory

spinor Ψα (x) (α = 1, . . . , 4) and its configuration space is the set of maps 4 Ψα ∶ M ↦ C , while it is real for neutral spin-1/2 particles (such as neutrinos).

3) The gauge field A (x) for the electromagnetic field, and its non-abelian generalizations for gluons (and so forth). µ

The description of relativistic quantum mechanics in terms of relativistic quantum fields solved essentially all of the problems that originated its initial development. Moreover, Quantum Field Theory gives exceedingly accurate predictions of the behavior of quantized electromagnetic fields and charged particles, as described by Quantum Electrodynamics (QED). Quantum Field Theory also gives a detailed description of both the strong and weak interactions in terms of field theories known as Quantum Chromodynamics (QCD), based on Yang-Mills gauge field theories, and Unified and Grand Unified gauge theories. However, along with its successes, Quantum Field Theory also brought with it a completely new set of physical problems and questions. Essentially, any Quantum Field Theory of physical interest is necessarily a nonlinear theory as it has to describe interactions. So even though the quantum numbers of the excitations (i.e. the “particle” spectrum) may be quite straightforward in the absence of interactions, the intrinsic non-linearities of the theory may actually unravel much of this structure. Note that the equations of motion of Quantum Field Theory are nonlinear, as they also are in Quantum Mechanics. However, the wave functional of a Quantum Field Theory obeys a linear Schr¨ odinger equation just as the wave function does in non-relativistic Quantum Mechanics. In the early days of Quantum Field Theory, and indeed for some time thereafter, it was assumed that perturbation theory could be used in all cases to determine the actual spectrum. It was soon found out that while there are several cases of great physical interest in which some sort of perturbation theory yields an accurate description of the physics, in many more situations this is not the case. Early on it was found that, at every order in perturbation theory, there are singular contributions to many physical quantities. These singularities reflected the existence of an infinite number of degrees of freedom, both at short distances, since spacetime is a continuum (the ultraviolet (UV) domain), and at long distances, since spacetime is (essentially) infinite (the infrared (IR) domain). Qualitatively, divergent contributions in perturbation theory come about because degrees of freedom from a wide range of length scales (or wavelengths) and energy scales (or frequencies) contribute to the expectation values of physical observables.

1.2 Why quantum field theory?

9

Historically, the way these problems were dealt with was through the process of regularization (i.e. making the divergent contributions finite), and renormalization (i.e. defining a set of effective parameters which are functions of the energy and/or momentum scale at which the system is probed). Regularization required that the integrals to be cutoff at some high energy scale (in the UV). Renormalization was then thought of as the process by which these arbitrarily introduced cutoffs were removed from the expressions for physical quantities. This was a physically obscure procedure, but it worked brilliantly in QED and, to a lesser extent, in QCD. Theories for which such a procedure can be implemented with the definition of only a finite number of renormalized parameters (the actual input parameters to be taken from experiment) are said to be renormalizable quantum field theories. QED and QCD are the most important examples of renormalizable quantum field fheories, although there are many others. Renormalization implies that the connection between the physical observables and the parameters in the Lagrangian of a Quantum Field Theory is highly non-trivial, and that the spectrum of the theory may have little to do with the predictions of perturbation theory. This is the case for QCD whose “fundamental fields” involve quarks and gluons but the actual physical spectrum consists only of bound states whose quantum numbers are not those of either quarks or gluons. Renormalization also implies that the behavior of the physical observables depends of the scale at which the theory is probed. Moreover, a closer examination of these theories also revealed that they may exist in different phases in which the observables have different behaviors, with a specific particle spectrum in each phase. In this way, to understand what a given Quantum Field Theory predicted became very similar to the study of phases in problems in Statistical Physics. We will explore these connections in detail later in this book when we develop the machinery of the Renormalization Group in chapter 15. In this picture, the vacuum (or ground state) of a quantum field theory corresponds to a phase much in the same way as in Statistical (or Condensed Matter) Physics. While the requirement of renormalizability works for the Standard Model of particle physics it fails for Gravity. The problem of unifying Gravity with the rest of the forces of Nature remains a major problem in contemporary physics. A major program to solve this problem is String Theory. String Theory is the only known viable candidate to quantize Gravity in a consistent manner. However, in String Theory, Quantum Field Theory is seen as an effective low energy (hydrodynamic) description of Nature, and the Quantum Field Theory singularities are “regularized” by String Theory in a natural way (but at the price of locality).

2 Classical Field Theory

In what follows we will consider rather general field theories. The only guiding principles that we will use in constructing these theories are (a) symmetries and (b) a generalized Least Action Principle.

2.1 Relativistic invariance Before we saw three examples of relativistic wave equations. They are the Maxwell equations for classical electromagnetism, the Klein-Gordon equation and the Dirac equation. Maxwell’s equations govern the dynamics of a µ 0 vector field, the vector potentials A (x) = (A (x), A(x)), whereas the KleinGordon equation describes excitations of a scalar field φ(x) and the Dirac equation governs the behavior of the four-component spinor field ψα (x), (α = 0, 1, 2, 3). Each one of these fields transforms in a very definite way under the group of Lorentz transformations, the Lorentz group. The Lorentz group is defined as a group of linear transformations Λ of Minkowski spacetime M onto itself Λ ∶ M ↦ M such that the new coordinates are related to the old ones by a linear (Lorentz) transformation ′µ

µ ν

x = Λν x

(2.1)

0

The space-time components of a Lorentz transformation, Λi , are the Lorentz boosts. Lorentz boosts relate inertial reference frames moving at rel1 ative velocity v with respect to each other. Lorentz boosts along the x -axis

2.1 Relativistic invariance

have the familiar form

11

x + vx /c 0′ x =√ 1 − v 2 /c2 0

1

x + vx /c 1′ x =√ 1 − v 2 /c2 1

2′

2

3′

3

x =x x =x

0

(2.2) 0

1

2

3

where x = ct, x = x, x = y and x = z (note: the superscripts indicate 2 2 −1/2 components, not powers!). If we use the notation γ = (1 − v /c ) ≡ cosh α, we can write the Lorentz boost as a matrix: ⎛x ⎞ ⎛cosh α sinh α 0 0⎞ ⎛x ⎞ 1 1′ ⎜ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ x ⎟ sinh α cosh α 0 0⎟ x ⎟ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ = ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 2 2′ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 0 1 0 x x ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 3 3′ ⎠ ⎝ 0 0 0 1 ⎝x ⎠ ⎝x ⎠ 0

0′

(2.3)

i

The space components of Λj are conventional rotations R of three-dimensional Euclidean space. Infinitesimal Lorentz transformations are generated by the hermitian operators

where ∂µ = the algebra

∂ ∂xµ

Lµν = i(xµ ∂ν − xν ∂µ )

(2.4)

and µ, ν = 0, 1, 2, 3. The infinitesimal generators Lµν satisfy

[Lµν , Lρσ ] = igνρ Lµσ − igµρ Lνσ − igνσ Lµρ + igµσ Lνρ

(2.5)

where gµν is the metric tensor for flat Minkowski space-time (see below). This is the algebra of the Lie group SO(3, 1). Actually, any operator of the form Mµν = Lµν + Sµν

(2.6)

where Sµν are 4 × 4 matrices satisfying the algebra of Eq.(2.5) are also generators of SO(3, 1). Below we will discuss explicit examples. Lorentz transformations form a group, since (a) the product of two Lorentz transformations is a Lorentz transformation, (b) there exists an identity

12

Classical Field Theory

transformation, and (c) Lorentz transformations are invertible. Notice, however, that in general two transformations do not commute with each other. Hence, the Lorentz group is non-Abelian. The Lorentz group has the defining property of leaving invariant the relativistic interval 2

2

2

2 2

x ≡ x0 − x = c t − x

2

(2.7) 2

The group of Euclidean rotations leave invariant the Euclidean distance x and it is a subgroup of the Lorentz group. The rotation group is denoted by SO(3), and the Lorentz group is denoted by SO(3, 1). This notation makes manifest the fact that the signature of the metric has one + sign and three − signs. The group SO(3, 1) of linear transformations is non-compact in the following sense. Let us consider first the group of rotations in three-dimensional space, SO(3). The linear transformations in SO(3) leave the Euclidean dis2 2 2 2 tance (squared) R = x1 + x2 + x3 invariant. The set of points with a fixed value of R is the two-dimensional surface of a sphere of radius R in three dimensions, that we will denote by S2 . The elements of the group of rotations SO(3) are in one-to-one correspondence with the points on S2 . The area of a 2-sphere S2 of unit radius is 4π. Then we will say that the “volume” of the group SO(3) is finite and equal to 4π. A group of linear transformations with finite volume is said to be compact. In contrast, the Lorentz group is the set of linear transformations, denoted by SO(3, 1), that leave the relativisµ tic interval xµ x invariant, which is not positive definite. As we well know, Lorentz boosts which map points along hyperbolas of Minkowski space time. In this sense, the Lorentz group is non-compact since its “volume” is infinite. We will adopt the following conventions and definitions: 1) Metric Tensor: We will use the standard (“Bjorken and Drell”) metric for Minkowski space-time in which the metric tensor gµν is

gµν = g

µν

⎛ ⎜ ⎜ =⎜ ⎜ ⎜ ⎜ ⎝

1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

(2.8)

With this notation the infinitesimal relativistic interval is 2

µ

µ

ν

2

2

2

2

ds = dx dxµ = gµν dx dx = dx0 − dx = c dt − dx 2) 4-vectors:

i) x is a contravariant 4-vector, x = (ct, x) µ

µ

2

(2.9)

2.1 Relativistic invariance

13

ii) xµ is a covariant 4-vector xµ = (ct, −x) iii) Covariant and contravariant vectors (and tensors) are related through the metric tensor gµν µ

µν

(2.10)

A = g Aν iv) x is a vector in R

3

v) p = ( Ec , p) is the energy-momentum 4-vector . Hence, pµ p = is a Lorentz scalar. µ

µ

2

E c2

−p

2

3) Scalar Product:

µ

p ⋅ q = pµ q = p0 q 0 − p ⋅ q ≡ pµ q ν g 4) Gradients: ∂µ ≡

∂ ∂xµ

µ

∂ . ∂xµ

and ∂ ≡

µν

(2.11)

We define the D’Alambertian ∂

2

1 2 2 ∂t − & (2.12) 2 c which is a Lorentz scalar. From now on we will use units of time [T ] and length [L] such that h̵ = c = 1. Thus, [T ] = [L] and we will use units like centimeters (or any other unit of length). 2 5) Interval: The interval in Minkowski space is x , 2

µ

∂ ≡ ∂ ∂µ ≡

2

2

µ

x = xµ x = x0 − x

2

2

(2.13) 2

Time-like intervals have x > 0 while space-like intervals have x < 0. Since a field is a function (or mapping) of Minkowski space onto some other (properly chosen) space, it is natural to require that the fields should have simple transformation properties under Lorentz transformations. For µ example, the vector potential A (x) transforms like 4-vector under Lorentz ′µ µ ν ′µ ′ µ ν transformations, i.e. if x = Λν x , then A (x ) = Λν A (x). In other words, µ µ A transforms like x . Thus, it is a vector. All vector fields have this property. A scalar field Φ(x), on the other hand, remains invariant under Lorentz transformations, Φ (x ) = Φ(x) ′

′

(2.14)

A 4-spinor ψα (x) transforms under Lorentz transformations. Namely, there exists an induced 4 × 4 linear transformation matrix S(Λ) such that S(Λ ) = S −1

and

−1

(Λ)

Ψ (Λx) = S(Λ)Ψ(x) ′

Below we will give an explicit expression for S(Λ).

(2.15)

(2.16)

14

Classical Field Theory < = 0 > 1 2 ? 3 @ ;4 5 6 /7 8 9 :.A -B + ,C *# D $ % )"E (F !& G 'K JH L M IN ~ O }P |Q m u w v{zyxR tsln ko p q S rjT U g h iV a b d ce W f_ `^]X \Y Z [

x0

time-like

x2 > 0

space-like x2 < 0

x1

Figure 2.1 The Minkowski space-time and its light cone. Events at a rel2 2 2 ativistic interval with x = x0 − x > 0 are time-like (and are causally 2 2 2 connected with the origin), while events with x = x0 − x < 0 are spacelike and are not causally connected with the origin.

2.2 The Lagrangian, the action, and the Least Action Principle The evolution of any dynamical system is determined by its Lagrangian. In the Classical Mechanics of systems of particles described by the generalized coordinates q, the Lagrangian L is a differentiable function of the coordinates q and their time derivatives. L must be differentiable since, otherwise, the equations of motion would not be local in time, i.e. could not be written in terms of differential equations. An argument `a-la Landau and Lifshitz (Landau and Lifshitz, 1959a) enables us to “derive” the Lagrangian. For example, for a particle in free space, the homogeneity, uniformity and isotropy of space and time require that L be only a function of the absolute value of the velocity ∣v∣. Since ∣v∣ is not a differentiable function of v, the La2 2 grangian must be a function of v . Thus, L = L(v ). In principle there is no reason to assume that L cannot be a function of the acceleration a (or 2 rather a ) or of its higher derivatives. Experiment tells us that in Classical Mechanics it is sufficient to specify the initial position x(0) of a particle and its initial velocity v(0) in order to determine the time evolution of the

2.2 The Lagrangian, the action, and the Least Action Principle

15

system. Thus we have to choose 1 2 2 L(v ) = const + mv (2.17) 2 The additive constant is irrelevant in classical physics. Naturally, the coef2 ficient of v is just one-half of the inertial mass. However, in Special Relativity, the natural invariant quantity to consider is not the Lagrangian but the action S. For a free particle the relativistic invariant (i.e. Lorentz invariant √ ) action must involve the invariant interval, v2 c2

the proper length ds = c 1 − relativistic massive particle as S = −mc ∫

sf si

dt. Hence one writes the action for a

ds = −mc ∫

tf

2

The relativistic Lagrangian then is 2

L = −mc

√

dt ti

1−

√

1−

v2 c2

v2 c2

(2.18)

(2.19)

As a power series expansion, it contains all powers of v /c . It is elementary to see that, as expected, the canonical momentum p is 2

p=

2

mv ∂L =√ ∂v 2 v 1− 2 c

(2.20)

from which it follows that the Hamiltonian (or energy) is given by 2

mc H=√

= 2

v 1− 2 c

√

p2 c2 + m2 c4 ,

(2.21)

as it should be. Once the Lagrangian is found, the classical equations of motion are determined by the Least Action Principle. Thus, we construct the action S S = ∫ dt L (q, q) ˙

(2.22)

where q˙ = dq , and demand that the physical trajectories q(t) leave the action dt S stationary, i.e. δS = 0. The variation of S is δS = ∫

tf ti

dt (

∂L ∂L δq + δq) ˙ ∂q ∂ q˙

(2.23)

16

Classical Field Theory q

i

t

Figure 2.2 The Least Action Principle: the dark curve is the classical trajectory and extremizes the classical action. The curve with a broken trace represents a variation.

Integrating by parts, we get δS = ∫ Hence, we get

tf ti

dt q˙ (

tf ∂L d ∂L ∂L δq) + ∫ dt δq [ − ( )] ∂ q˙ ∂q dt ∂ q˙ ti

(2.24)

1tf tf ∂L ∂L 111 d ∂L δS = δq 111 + ∫ dt δq [ − ( )] (2.25) ∂ q˙ 11 ∂q dt ∂ q˙ ti 1ti If we assume that the variation δq is an arbitrary function of time that vanishes at the initial and final times ti and tf , i.e. δq(ti ) = δq(tf ) = 0, we find that δS = 0 if and only if the integrand of Eq.(2.25) vanishes identically. Thus, ∂L d ∂L − ( )=0 (2.26) ∂q dt ∂ q˙

These are the equations of motion or Newton’s equations. In general the equation that determine the trajectories that leave the action stationary is called the Euler-Lagrange equation. 2.3 Scalar field theory For the case of a field theory, we can proceed very much in the same way. Let us consider first the case of a scalar field Φ(x). The action S must be invariant under Lorentz transformations. Since we want to construct local

2.3 Scalar field theory

17

theories it is natural to assume that S is given in terms of a Lagrangian density L S=∫ d xL 4

(2.27)

where L is a local differentiable function of the field and its derivatives. These assumptions in turn guarantee that the resulting equations of motion have the form of partial differential equations. In other terms, the dynamics does not allow for action-at-a-distance. 4 Since the volume element of Minkowski space d x is invariant under Lorentz transformations, the action S is invariant if L is a local, differentiable function of Lorentz invariants that can be constructed out of the field Φ(x). Simple invariants are Φ(x) itself and all of its powers. The graµ 2 ∂Φ dient ∂ Φ ≡ ∂x is not an invariant but the D’alambertian ∂ Φ is. Bilinears µ

µ

such as ∂µ Φ∂ Φ are also Lorentz invariant as well as under a change of the sign of Φ. So, we can write the following simple expression for L: 1 µ L = ∂µ Φ∂ Φ − V (Φ) 2

(2.28)

where V (Φ) is some potential, which we can assume is a polynomial function of the field Φ. Let us consider the simple choice 1 2 2 V (Φ) = m ¯ Φ 2

̵ Thus, where m ¯ = mc/h. L=

1 1 2 2 µ ∂µ Φ∂ Φ − m ¯ Φ 2 2

(2.29)

(2.30)

This is the Lagrangian density for a free scalar field. We will discuss later on in what sense this field is “free”. Notice, in passing, that we could have 2 added a term like ∂ Φ. However this term, in addition of being odd under Φ → −Φ, is a total divergence and, as such, it has an effect only on the boundary conditions but it does not affect the equations of motion. In what follows, unless stated to the contrary, will will not consider surface terms. The Least Action Principle requires that S be stationary under arbitrary variations of the field Φ and of its derivatives ∂µ Φ. Thus, we get δS = ∫ d x [ 4

δL δL δΦ + δ∂ Φ] δΦ δ∂µ Φ µ

(2.31)

Notice that since L is a functional of Φ, we have to use functional derivatives, i.e. partial derivatives at each point of space-time. Upon integrating by parts,

18

Classical Field Theory

we get δS = ∫ d x ∂µ ( 4

δL δL δL 4 )] δΦ) + ∫ d x δΦ [ − ∂µ ( δ∂µ Φ δΦ δ∂µ Φ

(2.32)

Instead of considering initial and final conditions, we now have to imagine that the field Φ is contained inside some very large box of space-time. The term with the total divergence yields a surface contribution. We will consider field configurations such that δΦ = 0 on that surface. Thus, the EulerLagrange equations are

More explicitly, we find

δL δL )=0 − ∂µ ( δΦ δ∂µ Φ

(2.33)

∂V δL =− δΦ ∂Φ

(2.34)

∂L µ = ∂ Φ, δ∂µ Φ

(2.35)

and, since

Then ∂µ

δL µ 2 = ∂µ ∂ Φ ≡ ∂ Φ δ∂µ Φ

(2.36)

By direct substitution we get the equation of motion (or field equation) 2

∂ Φ+

∂V =0 ∂Φ

(2.37)

For the choice 2

the field equation is

m ¯ 2 Φ , V (Φ) = 2

(∂ + m ¯ )Φ = 0 2

2

2

∂V 2 =m ¯ Φ ∂Φ

2

2

(2.38)

(2.39)

∂ where ∂ = c12 ∂t 2 − & . Thus, we find that the equation of motion for the free massive scalar field Φ is 2

1 ∂ Φ 2 2 −& Φ+m ¯ Φ=0 (2.40) 2 2 c ∂t This is precisely the Klein-Gordon equation if the constant m ¯ is identified . Indeed, the plane-wave solutions of these equations are with mc h̵ i(p0 x0 −p⋅x)/h̵

Φ = Φ0 e

(2.41)

2.4 Classical field theory in the canonical formalism

19

where p0 and p are related through the dispersion law 2

2 2

2 4

(2.42)

p0 = p c + m c

This means that, for each momentum p, there are two solutions, one with positive frequency and one with negative frequency. We will see below that, in the quantized theory, the energy of the excitation is indeed equal to 1 h̵ ∣p0 ∣. Notice that m = mc has units of length and is equal to the Compton ¯ wavelength for a particle of mass m. From now on (unless it stated the contrary) I will use units in which h̵ = c = 1 in which m = m. ¯ 2.4 Classical field theory in the canonical formalism In Classical Mechanics it is often convenient to use the canonical formulation in terms of a Hamiltonian instead of the Lagrangian approach. For the case of a system of particles, the canonical formalism proceeds as follows. Given a Lagrangian L(q, q), ˙ a canonical momentum p is defined to be ∂L =p ∂ q˙

(2.43)

The classical Hamiltonian H(p, q) is defined by the Legendre transformation H(p, q) = pq˙ − L(q, q) ˙

(2.44)

If the Lagrangian L is quadratic in the velocities q˙ and separable, e.g. 1 2 L = mq˙ − V (q) 2 then, H(pq) ˙ is simply given by

where

H(p, q) = pq˙ − (

2

(2.45)

2

mq˙ p − V (q)) = + V (q) 2 2m

p=

∂L = mq˙ ∂ q˙

(2.46)

(2.47)

The (conserved) quantity H is then identified with the total energy of the system. In this language, the Least Action Principle becomes

Hence

δS = δ ∫ L dt = δ ∫ [pq˙ − H(p, q)] dt = 0 ∫ dt (δp q˙ + p δq˙ − δp

∂H ∂H )=0 − δq ∂p ∂q

(2.48)

(2.49)

20

Classical Field Theory

Upon an integration by parts we get ∫ dt [δp (q˙ −

∂H ∂H ) + δq (− − p)] ˙ =0 ∂p ∂q

(2.50)

which can only be satisfied for arbitrary variations δq(t) and δp(t) if q˙ =

∂H ∂p

p˙ = −

∂H ∂q

(2.51)

These are Hamilton’s equations. Let us introduce the Poisson Bracket {A, B}qp of two functions A and B of q and p by ∂A ∂B ∂A ∂B {A, B}qp ≡ − (2.52) ∂q ∂p ∂p ∂q

Let F (q, p, t) be some differentiable function of q, p and t. Then the total time variation of F is ∂F ∂F dq ∂F dp dF = + + dt ∂t ∂q dt ∂p dt

(2.53)

Using Hamilton’s Equations we get the result dF ∂F ∂F ∂H ∂F ∂H = + − dt ∂t ∂q ∂p ∂p ∂q

(2.54)

or, in terms of Poisson Brackets, dF ∂F = + {F, H}qp dt ∂t

In particular,

dq ∂H ∂q ∂H ∂q ∂H = = − = {q, H}qp dt ∂p ∂q ∂p ∂p ∂q

since

∂q =0 ∂p

and

∂q =1 ∂q

(2.55)

(2.56)

(2.57)

Also the total rate of change of the canonical momentum p is dp ∂p ∂H ∂p ∂H ∂H = − ≡− dt ∂q ∂p ∂p ∂q ∂q since

∂p ∂q

= 0 and

∂p ∂p

(2.58)

= 1. Thus, dp = {p, H}qp dt

(2.59)

2.4 Classical field theory in the canonical formalism

21

Notice that, for an isolated system, H is time-independent. So, ∂H =0 ∂t and

since

∂H dH = + {H, H}qp = 0 dt ∂t {H, H}qp = 0

(2.60)

(2.61)

(2.62)

Therefore, H can be regarded as the generator of infinitesimal time translations. Since it is conserved for an isolated system, for which ∂H = 0, we can ∂t indeed identify H with the total energy. In passing, let us also notice that the above definition of the Poisson Bracket implies that q and p satisfy {q, p}qp = 1

(2.63)

This relation is fundamental for the quantization of these systems. Much of this formulation can be generalized to the case of fields. Let us first discuss the canonical formalism for the case of a scalar field Φ with Lagrangian density L(Φ, ∂µ , Φ). We will choose Φ(x) to be the (infinite) set of canonical coordinates. The canonical momentum Π(x) is defined by Π(x) =

δL δ∂0 Φ(x)

(2.64)

If the Lagrangian is quadratic in ∂µ Φ, the canonical momentum Π(x) is simply given by ˙ Π(x) = ∂0 Φ(x) ≡ Φ(x) (2.65) The Hamiltonian density H(Φ, Π) is a local function of Φ(x) and Π(x) given by H(Φ, Π) = Π(x) ∂0 Φ(x) − L(Φ, ∂0 Φ)

(2.66)

If the Lagrangian density L has the simple form 1 2 L = (∂µ Φ) − V (Φ) 2 then, the Hamiltonian density H(Φ, Π) is

(2.67)

1 1 2 2 Π (x) + (!Φ(x)) + V (Φ(x)) (2.68) 2 2 which is explicitly a positive definite quantity. Thus, the energy of a plane wave solution of a massive scalar field theory, i.e. a solution of the KleinGordon equation, is always positive, no matter the sign of the frequency. ˙ − L(Φ, Φ, ˙ ∂j Φ) ≡ H = ΠΦ

22

Classical Field Theory

In fact, the lowest energy state is simply Φ = constant. A solution made of linear superpositions of plane waves (i.e. a wave packet) has positive energy. Therefore, in field theory, the energy is always positive. We will see that, in the quantized theory, in the case of a complex field, the negative frequency solutions are identified with antiparticle states and their existence do not signal a possible instability of the theory. The canonical field Φ(x) and the canonical momentum Π(x) satisfy the equal-time Poisson Bracket (PB) relations {Φ(x, x0 ), Π(y, x0 )}P B = δ(x − y)

(2.69)

where δ(x) is the Dirac δ-function and the Poisson Bracket {A, B}P B is defined to be {A, B}P B = ∫ d x [ 3

δA δB δA δB ] (2.70) − δΦ(x, x0 ) δΠ(x, x0 ) δΠ(x, x0 ) δΦ(x, x0 )

for any two functionals A and B of Φ(x) and Π(x). This approach can be extended to theories other than that of a scalar field without too much difficulty. We will come back to these issues when we consider the problem of quantization. Finally we should note that while Lorentz invariance is apparent in the Lagrangian formulation, it is not so in the Hamiltonian formulation of a classical field.

2.5 Field theory of the Dirac equation We now turn to the problem of a field theory for spinors. We will discuss the theory of spinors as a classical field theory. We will find that this theory is not consistent unless it is properly quantized as a quantum field theory of spinors. We will return to this problem in chapter 7. Let us rewrite the Dirac equation ih̵

̵ ∂Ψ hc 2 α ⋅ !Ψ + βmc Ψ ≡ HDiracΨ = i ∂t

(2.71)

in a manner in which relativistic covariance is apparent. The operator HDirac is the Dirac Hamiltonian. We first recall that the 4 × 4 hermitian matrices α and β should satisfy the algebra {αi , αj } = 2δij I,

{αi , β} = 0,

where I is the 4 × 4 identity matrix.

2

2

αi = β = I

(2.72)

2.5 Field theory of the Dirac equation

23

A simple representation of this algebra are the 2×2 block (Dirac) matrices β=(

0 σ α =( i ) σ 0 i

i

i

I 0 ) 0 −I

(2.73)

where the σ matrices are the three 2 × 2 Pauli matrices σ =( 1

0 1 ) 1 0

σ =(

0 −i ) i 0

2

σ =( 3

1 0 ) 0 −1

(2.74)

and I is the 2 × 2 identity matrix. This is the Dirac representation of the Dirac algebra. It is now convenient to introduce the Dirac gamma matrices, 0

i

γ =β

i

(2.75)

γ = βα

µ

The Dirac gamma matrices γ have the block form γ =β=( 0

an obey the Dirac algebra

I 0 ), 0 −I {γ , γ } = 2g µ

ν

γ =( i

µν

0 σ ) i −σ 0 i

I

(2.76)

(2.77)

where I is the 4 × 4 identity matrix. In terms of the gamma matrices, the Dirac equation takes the much simpler, covariant, form mc µ (iγ ∂µ − ̵ ) Ψ = 0 (2.78) h

where Ψ is a 4-spinor. It is also customary to introduce the notation (known as Feynman’s slash) µ a/ ≡ aµ γ

(2.79)

Using Feynman’s slash, we can write the Dirac equation in the form mc (i∂/ − ̵ ) Ψ = 0 h

(2.80)

From now on I will use units in which h̵ = c = 1. In these units energy has −1 units of (length) and time has units of length. Notice that, if Ψ satisfies the Dirac equation, then it also satisfies (i∂/ + m)(i∂/ − m)Ψ = 0

(2.81)

24

Classical Field Theory

Also, we find 1 µ ν 1 µ ν µ ν ∂/ ⋅ ∂/ =∂µ ∂ν γ γ = ∂µ ∂ν ( {γ , γ } + [γ γ ]) 2 2 =∂µ ∂ν g

µν

=∂

2

(2.82)

where we used that the commutator [γ , γ ] is antisymmetric in the indices µ and ν. As a result, we find that each component of the 4-spinor Ψ must also satisfy the Klein-Gordon equation µ

ν

(∂ + m ) Ψ = 0 2

2

(2.83)

2.5.1 Solutions of the Dirac equation Let us briefly discuss the properties of the solutions of the Dirac equation. Let us first consider solutions representing particles at rest. Thus Ψ must be constant in space and all its space derivatives must vanish. The Dirac equation becomes 0 ∂Ψ = mΨ (2.84) iγ ∂t where t = x0 (c = 1). Let us introduce the bispinors φ and χ Ψ=(

φ ) χ

(2.85)

We find that the Dirac equation reduces to a simple system of two 2 × 2 equations ∂φ ∂χ i = +mφ, i = −mχ (2.86) ∂t ∂t The four linearly independent solutions are −imt

φ1 = e and

imt

χ1 = e

(

1 ) 0

(

1 ) 0

−imt

φ2 = e

imt

χ2 = e

(

(

0 ) 1

0 ) 1

(2.87)

(2.88)

Thus, the upper component φ represents the solutions with positive energy +m, while χ represents the solutions with negative energy −m. The additional two-fold degeneracy of the solutions is related to the spin of the particle, as we will see below.

2.5 Field theory of the Dirac equation

25

More generally, in terms of the bispinors φ and χ the Dirac Equation takes the form, ∂φ 1 i (2.89) = mφ + σ ⋅ !χ i ∂t ∂χ 1 (2.90) = −mχ + σ ⋅ !φ i ∂t Furthermore, in the non-relativistic limit, taking formally c → ∞, it should reduce to the Schr¨ odinger-Pauli equation. To see this, we define the slowly varying amplitudes φ˜ and χ ˜ i

−imt ˜

φ=e

φ

−imt

χ=e

χ ˜

(2.91)

The field χ ˜ is small and nearly static. We will now see that the field φ˜ describes solutions with positive energies close to +m. In terms of φ˜ and χ, ˜ the Dirac equation becomes ∂ φ˜ 1 = σ ⋅ !χ ˜ i ∂t

(2.92)

1 ∂ χ˜ = −2mχ ˜ + σ ⋅ !φ˜ i ∂t

(2.93)

i i

Indeed, in this limit, the l.h.s of Eq. (2.93) is much smaller than its r.h.s. Thus we can approximate 2mχ ˜≈

1 σ ⋅ !φ˜ i

(2.94)

We can now eliminate the “small component” χ ˜ from Eq. (2.92) to find that ˜ φ satisfies ∂ φ˜ 1 2 i =− & φ˜ (2.95) 2m ∂t which is indeed the Schr¨odinger-Pauli equation.

2.5.2 Conserved current ¯ by Let us introduce one last bit of useful notation. Let us define Ψ ¯ = Ψ† γ 0 Ψ in terms of which we can write down the 4-vector j µ

µ

¯ Ψ j = Ψγ

(2.96) µ

(2.97)

26

Classical Field Theory

which is conserved, i.e. µ

∂µ j = 0

(2.98)

µ

Notice that the time component of j is the density 0 ¯ 0 Ψ ≡ Ψ† Ψ j = Ψγ

(2.99)

µ

and that the space components of j are † 0 † ¯ j = ΨγΨ = Ψ γ γΨ = Ψ αΨ

(2.100)

Thus the Dirac equation has an associated 4-vector field, j (x), which is conserved and obeys a local continuity equation µ

∂0 j0 + ! ⋅ j = 0

(2.101)

However it is easy to see that in general the density j0 can be positive or negative. Hence this current cannot be associated with a probability current (as in non-relativistic quantum mechanics). Instead we will see that that it is associated with the charge density and current.

2.5.3 Relativistic covariance Let Λ be a Lorentz transformation. Let Ψ(x) be a spinor field in an inertial ′ ′ frame and Ψ (x ) be the same Dirac spinor field in the transformed frame. The Dirac equation is covariant if the Lorentz transformation ′

ν

xµ = Λµ xν

(2.102)

induces a linear transformation S(Λ) in spinor space Ψα (x ) = S(Λ)αβ Ψβ (x) ′

′

(2.103)

such that the transformed Dirac equation has the same form as the original equation in the original frame, i.e. we will require that if (iγ

∂ − m) Ψβ (x) = 0, ∂xµ αβ

(iγ

∂ ′ ′ Ψβ (x ) = 0 ′µ − m) ∂x αβ (2.104) Notice two important facts: (1) both the field Ψ and the coordinate x change under the action of the Lorentz transformation, and (2) the gamma matrices and the mass m do not change under a Lorentz transformation. Thus, the gamma matrices are independent of the choice of a reference frame. However, they do depend on the choice of the basis in spinor space. µ

then

µ

2.5 Field theory of the Dirac equation

27

What properties should the representation matrices S(Λ) have? Let us ′µ µ ν first observe that if x = Λν x , then ∂ ∂x ∂ −1 ν ∂ ′µ = ′µ ∂xν ≡ (Λ )µ ∂xν ∂x ∂x ν

(2.105)

Thus, ∂x∂ µ is a covariant vector. By substituting this transformation law back into the Dirac equation, we find ∂ ∂ ′ ′ µ −1 ν Ψ (x ) = iγ (Λ ) µ ν (S(Λ)Ψ(x)) ′µ ∂x ∂x Thus, the Dirac equation now reads iγ

µ

iγ (Λ )µ S(Λ) −1

µ

Or, equivalently

ν

(2.106)

∂Ψ − mS (Λ) Ψ = 0 ∂xν

(2.107)

(Λ) iγ (Λ ) µ S (Λ)

∂Ψ − mΨ = 0 ∂xν Therefore, covariance holds provided S (Λ) satisfies the identity S

−1

−1

µ

S

−1

ν

(Λ) γ S (Λ) (Λ )µ = γ −1

µ

ν

(2.108)

ν

(2.109)

Since the set of Lorentz transformations form a group, the representation matrices S(Λ) should also form a group. In particular, it must be true that the property S

−1

(Λ) = S (Λ ) −1

(2.110) 2

holds. We now recall that the invariance of the relativistic interval x = xµ x implies that Λ must obey ν

λ

λ

Λ µ Λν = gµ ≡ δµ Hence,

Λµ = (Λ ) ν −1

ν

λ

µ

(2.111)

µ

(2.112)

So we can rewrite Eq.(2.109) as

S (Λ) γ S (Λ) µ

−1

= (Λ )

−1 µ ν

γ

ν

(2.113)

Eq.(2.113) shows that a Lorentz transformation induces a similarity transformation on the gamma matrices which is equivalent to (the inverse of) a Lorentz transformation. From this equation it follows that, for the case of Lorentz boosts, Eq.(2.113) shows that the matrices S(Λ) are hermitian. Instead, for the subgroup SO(3) of rotations about a fixed origin, the matrices S(Λ) are unitary. These different properties follow from the fact that the

28

Classical Field Theory

matrices S(Λ) are representation of the Lorentz group SO(3, 1) which is a non-compact Lie group. We will now find the form of S (Λ) for an infinitesimal Lorentz transformaµ µ tion. Since the identity transformation is Λν = gν , a Lorentz transformation infinitesimally close to the identity should have the form µ

µ

µ

Λν = gν + ων , where ω

µν

and

(Λ )ν = gν − ων −1 µ

µ

µ

(2.114)

is infinitesimal and antisymmetric in its space-time indices ω

µν

= −ω

νµ

(2.115)

Let us parameterize S (Λ) in terms of a 4 × 4 matrix σµν which is also antisymmetric in its indices, i.e. σµν = −σνµ . Then, we can write i µν S (Λ) = I − σµν ω + . . . 4 i µν −1 S (Λ) = I + σµν ω + . . . 4

(2.116)

where I stands for the 4 × 4 identity matrix. If we substitute back into Eq.(2.113), we get i i µν λ αβ λ λ ν (I − σµν ω + . . .)γ (I + σαβ ω + . . .) = γ − ων γ + . . . 4 4 Collecting all the terms linear in ω, we obtain i λ µν λ ν [γ , σµν ] ω = ων γ 4 Or, what is the same, the matrices σµν must obey

[γ , σνλ ] = 2i(gν γλ − gλ γν ) µ

µ

µ

(2.117)

(2.118)

(2.119)

This matrix equation has the solution σµν =

i [γ , γ ] 2 µ ν

(2.120)

′

Under a finite Lorentz transformation x = Λx, the 4-spinors transform as Ψ (x ) = S (Λ) Ψ ′

with

′

i µν S (Λ) = exp (− σµν ω ) 4

(2.121)

(2.122)

The matrices σµν are the generators of the group of Lorentz transformations in the spinor representation. From this solution we see that the space

2.5 Field theory of the Dirac equation

29

components σjk are hermitean matrices, while the space-time components σ0j are antihermitean. This feature is telling us that the Lorentz group is not a compact unitary group, since in that case all of its generators would be hermitian matrices. Instead, this result tells us that the Lorentz group is isomorphic to the non-compact group SO(3, 1). Thus, the representation matrices S(Λ) are unitary only under space rotations with fixed origin. The linear operator S (Λ) gives the field in the transformed frame in terms of the coordinates of the transformed frame. However, we may also wish to ask for the transformation U (Λ) that compensates the effect of the coordinate transformation. In other words we seek for a matrix U (Λ) such that Ψ (x) = U (Λ) Ψ(x) = S (Λ) Ψ(Λ x) ′

−1

(2.123)

For an infinitesimal Lorentz transformation, we seek a matrix U (Λ) of the form i µν U (Λ) = I − Jµν ω + . . . (2.124) 2 and we wish to find an expression for Jµν . We find i i µν µν ρ ρ ν (I − Jµν ω + . . .) Ψ =(I − σµν ω + . . .)Ψ (x − ων x + . . .) 2 4 i µν ρ ν ≅ (I − σµν ω + . . .) (Ψ − ∂ρ Ψ ων x + . . .) 4 (2.125)

Hence, i ′ µν µν Ψ (x) ≅ (I − σµν ω + xµ ω ∂ν + . . .) Ψ(x) 4

(2.126)

From this expression we see that Jµν is given by the operator 1 σ + i(xµ ∂ν − xν ∂µ ) (2.127) 2 µν We easily recognize the second term as the orbital angular momentum operator (we will come back to this issue shortly). The first term is then interpreted as the spin. In fact, let us consider purely spacial rotations, whose infinitesimal generator are the space components of Jµν , i.e. Jµν =

1 Jjk = i(xj ∂k − xk ∂j ) + σjk (2.128) 2 We can also define a three component vector J( as the 3-dimensional dual of Jjk Jjk = -jkl J(

(2.129)

30

Classical Field Theory

where -

ijk

-

ijk

is the third-rank Levi-Civita tensor: ⎧ ⎪ 1 ⎪ ⎪ ⎪ ⎪ =⎨ −1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩0

if (ijk) is an even permutation of (123) if (ijk) is an odd permutation of (123)

(2.130)

otherwise

̵ Thus, we get (after restoring the factors of h)

ih̵ h̵ -(jk (xj ∂k − xk ∂j ) + -(jk σjk 2 4 h̵ 1 = i h̵ -(jk xj ∂k + ( -(jk σjk ) 2 2 ̵h ˆ ( + σ( J( ≡ (x × p) 2

J( =

(2.131)

The first term is clearly the orbital angular momentum and the second term can be regarded as the spin. With this definition, it is straightforward to check that the spinors (φ, χ), which are solutions of the Dirac equation, carry spin one-half. 2.5.4 Transformation properties of field bilinears in the Dirac theory We will now consider the transformation properties of several physical observables of the Dirac theory under Lorentz transformations. Let Λ be a general Lorentz transformation, and S(Λ) be the induced transformation for the Dirac spinors Ψa (x) (with a = 1, . . . , 4): Ψa (x ) = S(Λ)ab Ψb (x) ′

′

(2.132)

Using the properties of the induced Lorentz transformation S(Λ) and of the Dirac gamma matrices, is straightforward to verify that the following Dirac bilinears obey the following transformation laws: Scalar: ¯ ′ (x′ ) Ψ′ (x′ ) = Ψ(x) ¯ Ψ Ψ(x)

which transforms as a scalar.

(2.133)

2.5 Field theory of the Dirac equation

31

Pseudoscalar: Let let us define the Dirac matrix γ5 = iγ0 γ1 γ2 γ3 . Then the bilinear ¯ ′ (x′ ) γ5 Ψ′ (x′ ) = det Λ Ψ(x) ¯ Ψ γ5 Ψ(x)

(2.134)

transforms as a pseudo-scalar. Vector: Likewise

ν ¯ ′ (x′ ) γ µ Ψ′ (x′ ) = Λµν Ψ(x) ¯ Ψ γ Ψ(x)

(2.135)

transforms as a vector, and Pseudovector:

ν ¯ ′ (x′ ) γ5 γ µ Ψ′ (x′ ) = det Λ Λµν Ψ(x) ¯ Ψ γ5 γ Ψ(x)

(2.136)

transforms as a pseudo-vector. Finally, Tensor:

αβ ¯ ′ (x′ ) σ µν Ψ′ (x′ ) = Λµα Λνβ Ψ(x) ¯ Ψ σ Ψ(x)

(2.137)

transforms as a tensor µ Above we have denoted by Λν a Lorentz transformation and det Λ is its determinant. We have also used that S

−1

(Λ)γ5 S(Λ) = det Λ γ5

(2.138)

Finally we note that the Dirac algebra provides for a natural basis of the space of 4 × 4 matrices, which we will denote by S

Γ ≡ I,

V

Γµ ≡ γµ ,

T

Γµν ≡ σµν ,

A

Γµ ≡ γ5 γµ ,

P

Γ = γ5

(2.139)

where S, V , T , A and P stand for scalar, vector, tensor, axial vector (or pseudo-vector) and parity respectively. For future reference we will note here the following useful trace identities obeyed by products of Dirac gamma matrices trI = 4,

trγµ = trγ5 = 0,

trγµ γν = 4gµν

(2.140)

Also, if we denote by aµ and bµ two arbitrary 4-vectors, then a / /b = aµ b − iσµν a b , µ

µ ν

and

tr (a / /b ) = 4a ⋅ b

(2.141)

32

Classical Field Theory

2.5.5 The Dirac Lagrangian We now seek a Lagrangian density L for the Dirac theory. It should be a local differentiable Lorentz-invariant functional of the spinor field Ψ. Since the Dirac equation is first order in derivatives and it is Lorentz covariant, the Lagrangian should be Lorentz invariant and first order in derivatives. A simple choice is ← → ¯ ∂/ − m)Ψ ≡ 1 Ψi ¯ ∂/ Ψ − mΨΨ ¯ L = Ψ(i (2.142) 2 ← → ¯ ∂/ Ψ ≡ Ψ( ¯ ∂/Ψ)−(∂µ Ψ)γ ¯ µ Ψ. This choice satisfies all the requirements. where Ψ The equations of motion are derived in the usual manner, i.e. by demand4 ing that the action S = ∫ d x L be stationary δS = 0 = ∫ d x [ 4

δL δL ¯ δΨ + δ∂ Ψ + (Ψ ↔ Ψ)] δΨα α δ∂µ Ψα µ α

(2.143)

The equations of motion are

δL δL − ∂µ =0 δΨα δ∂µ Ψα δL δL − ∂µ ¯ ¯α = 0 δ Ψα δ∂µ Ψ

(2.144)

By direct substitution we find (i∂/ − m)Ψ = 0,

← − ¯ ∂/ + m) = 0 and Ψ(i

(2.145)

← − Here, ∂/ indicates that the derivatives are acting on the left. Finally, we can also write down the Hamiltonian density that follows from the Lagrangian of Eq.(2.142). As usual we need to determine the canonical momentum conjugate to the field Ψ, i.e. Π(x) =

δL 0 † ¯ = iΨ(x)γ ≡ iΨ (x) δ∂0 Ψ(x)

(2.146)

Thus the Hamiltonian density is

¯ 0 ∂0 Ψ − L H =Π(x)∂0 Ψ(x) − L = iΨγ ¯ ⋅ !Ψ + mΨΨ ¯ =Ψiγ =Ψ (iα ⋅ ! + mβ) Ψ †

(2.147)

Thus we find that the “one-particle” Dirac Hamiltonian HDirac of Eq.(2.71) appears naturally in the field theory as well. Since the Hamiltonian of Eq.(2.147) is first order in derivatives, unlike its

2.6 Classical electromagnetism as a field theory

33

Klein-Gordon relative, it is not manifestly positive. Thus, there is a question of the stability of this theory. We will see below that the proper quantization of this theory as a quantum field theory of fermions solves this problem. In other words, it will be necessary to impose the Pauli Principle for this theory to describe a stable system with an energy spectrum that is bounded from below. In this way we will see that there is natural connection between the spin of the field and the statistics. This connection, which actually is an axiom of Quantum Field Theory, is known as the Spin-Statistics Theorem.

2.6 Classical electromagnetism as a field theory We now turn to the problem of the electromagnetic field generated by a set of sources. Let ρ(x) and j(x) represent the charge and current densities at a point x of space-time. Charge conservation requires that a continuity equation has to be obeyed, ∂ρ +!⋅j =0 ∂t

(2.148)

Given an initial condition, that specifies the values of the electric field E(x) and the magnetic field B(x) at some t0 in the past, the time evolution is governed by the Maxwell equations ! ⋅ E =ρ 1 ∂E =j !×B− c ∂t

! ⋅ B =0 1 ∂B !×E+ c =0 ∂t

(2.149) (2.150)

It is possible to reformulate classical electrodynamics in a manner in which (a) relativistic covariance is apparent and (b) that the Maxwell equations follow from a Least Action Principle. A convenient way to see the above is µν to define the electromagnetic field tensor F , which is the (contravariant) antisymmetric real tensor whose components are given by ⎛ 0 −E −E −E ⎞ 1 3 2 ⎜ ⎟ ⎜ E 0 −B B ⎟ ⎟ ⎜ ⎜ =⎜ ⎟ 2 3 1⎟ ⎜ B 0 −B ⎟ ⎟ ⎜ ⎟ ⎜E ⎝E 3 −B 2 B 1 0 ⎠ 1

F

µν

= −F

νµ

Or, equivalently F F

0i ij

= −F

i0

= −E

= −F

ji

=-

ijk

2

3

(2.151)

i

B

k

(2.152)

34

Classical Field Theory

̃µν is defined as follows The dual tensor F ̃µν = −F ̃νµ = F

1 µνρσ Fρσ (2.153) 2 µνρσ where is the fourth rank Levi-Civita tensor, defined similarly to the third rank Levi-Civita tensor of Eq.(2.130). In particular ̃µν F

⎛ ⎜ ⎜ ⎜ =⎜ ⎜ ⎜ ⎜ ⎜ ⎝

0 −B −B −B ⎞ 1 3 2 ⎟ B 0 E −E ⎟ ⎟ ⎟ 2 3 2 ⎟ B −E 0 E ⎟ ⎟ ⎟ 3 2 2 B E −E 0 ⎠ 1

2

3

(2.154)

With these notations, we can rewrite the left column of Maxwell equations, Eq.(2.150), in the more compact and manifestly covariant form ∂µ F

µν

=j

ν

(2.155)

which we will interpret as the equations of motion of the electromagnetic field. The right column of the Maxwell equations, Eq.(2.150), become the constraint ̃µν = 0 ∂µ F (2.156) which is known as the Bianchi identity. Then, consistency of the Maxwell equations requires that the continuity equation µ

∂µ j = 0

(2.157)

be satisfied. µν By inspection we see that the field tensor F and the dual field tensor µν ̃ map into each other upon exchanging the electric and magnetic fields. F This electro-magnetic duality would be an exact property of electrodynamics µ if in addition to the electric charge current j the Bianchi Identity Eq.(2.156) included a magnetic charge current (of magnetic monopoles). µ At this point it is convenient to introduce the vector potential A whose contravariant components are 0

The current 4-vector j (x) µ

A µ A (x) = ( c , A)

j (x) = (ρc, j) ≡ (j , j) µ

0

(2.158)

(2.159)

The electric field strength E and the magnetic field B are defined to be 1 ∂A 1 0 , E = − c !A − c ∂t

B = !×A

(2.160)

2.6 Classical electromagnetism as a field theory

35

In a more compact, relativistically covariant, notation we write F

µν

µ

ν

ν

µ

(2.161)

=∂ A −∂ A µ

In terms of the vector potential A , the Maxwell equations remain unchanged under the (local) gauge transformation A (x) ↦ A (x) + ∂ Φ(x) µ

µ

µ

(2.162)

where Φ(x) is an arbitrary smooth function of the space-time coordinates µ x . It is easy to check that, under the transformation of Eq.(2.162), the field µν µν strength remains invariant i.e. F ↦ F . This property is called Gauge Invariance and it plays a fundamental role in modern physics. By directly substituting the definitions of the magnetic field B and the µ electric field E in terms of the 4-vector A into the Maxwell equations, we obtain the wave equation. Indeed, the equation of motion ∂µ F

µν

=j

ν

(2.163)

yields the equation for the vector potential

∂ A − ∂ (∂µ A ) = j 2

ν

ν

µ

ν

(2.164)

which is the wave equation. We can now use gauge invariance to further restrict the vector potential µ A , which is not completely determined. These restrictions are known as the procedure of fixing a gauge. The choice µ

∂µ A = 0

(2.165)

known as the Lorentz gauge, yields the simpler and standard form of the wave equation 2

µ

∂ A =j

µ

(2.166)

Notice that the Lorentz gauge preserves Lorentz covariance. Another popular choice is the radiation (or Coulomb) gauge !⋅A=0

(2.167)

which yields (in units with c = 1)

∂ A − ∂ (∂0 A ) = j 2

ν

ν

0

ν

(2.168) ν

which is not Lorentz covariant. In the absence of external sources, j = 0, 0 we can further impose the restriction that A = 0. This choice reduces the set of three equations, one for each spacial component of A, which satisfy 2

∂ A = 0,

provided

!⋅A=0

(2.169)

36

Classical Field Theory

The solutions are, as we well know, plane waves of the form i(p0 x0 −p⋅x)

(2.170)

A(x) = A e 2

2

which are only consistent if p0 − p = 0 and p ⋅ A = 0. For the last reason, this choice is also known as the transverse gauge. We can also regard the electromagnetic field as a dynamical system and construct a Lagrangian picture for it. Since the Maxwell equations are local, gauge invariant and Lorentz covariant, we should demand that the Lagrangian density should be local, gauge invariant and Lorentz invariant. Since, in terms of the vector potential Aµ , the Maxwell equations are second order in derivatives, we seek a local Lagrangian density which is also second order in derivatives of the vector potential. A simple choice that satisties all the requirements is 1 µν µ L = − Fµν F − jµ A 4

(2.171)

This Lagrangian density is manifestly Lorentz invariant. Gauge invariance µ is satisfied if and only if jµ is a conserved current, i.e. if ∂µ j = 0, since, under a gauge transformation Aµ ↦ Aµ + ∂µ Φ(x) the field strength does not change. However, the source term changes as follows ∫ d x jµ A ↦ ∫ d x [jµ A + jµ ∂ Φ] 4

4

µ

µ

µ

= ∫ d x jµ A + ∫ d x ∂ (jµ Φ) − ∫ d x ∂ jµ Φ 4

µ

4

µ

4

µ

(2.172)

If the sources vanish at infinity, lim∣x∣→∞ jµ = 0, the surface term can be 4 dropped. Thus the action S = ∫ d xL is gauge-invariant if and only if the µ current j is locally conserved, µ

∂µ j = 0

(2.173)

which is the continuity equation. We can now derive the equations of motion by demanding that the action S be stationary, i.e. δS = ∫ d x [ 4

δL δL µ ν µ δA + ν µ δ∂ A ] = 0 δAµ δ∂ A

(2.174)

Much as we did before, we can now proceed to integrate by parts to get δS = ∫ d x ∂ [ 4

ν

δL δL δL µ 4 µ ν )] (2.175) ν µ δA ] + ∫ d x δA [ µ −∂ ( δ∂ A δA δ∂ ν Aµ

2.7 The Landau theory of phase transitions as a field theory

37

µ

By demanding that at the surface the variation vanishes, δA = 0, we get

Explicitly, we find

δL δL ν ) µ =∂ ( δA δ∂ ν Aµ δL = −jµ , δAµ

and

δL µν =F δ∂ ν Aµ

(2.176)

(2.177)

Thus, we obtain µ

j = −∂ν F

µν

(2.178)

or, equivalently ν

j = ∂µ F

µν

(2.179)

Therefore, the Least Action Principle implies the Maxwell equations. 2.7 The Landau theory of phase transitions as a field theory We now turn to the problem of the statistical mechanics of a magnet. In order to be a little more specific, we will consider the simplest model of a ferromagnet: the classical Ising model. In this model, one considers an array of atoms on some lattice (say cubic). Each site is assumed to have a net spin magnetic moment S(i). From elementary quantum mechanics we know that the simplest interaction among the spins is the Heisenberg exchange Hamiltonian H = − ∑ Jij S(i) ⋅ S(j)

(2.180)

where < i, j > are nearest neighboring sites on the lattice. In many situations, in which there is magnetic anisotropy, only the z-component of the spin operators plays a role. The Hamiltonian now reduces to that of the Ising model HI = −J ∑ σ(i)σ(j) ≡ E[σ]

(2.181)

where [σ] denotes a configuration of spins with σ(i) being the z-projection of the spin at each site i. The equilibrium properties of the system are determined by the partition function Z which is the sum over all spin configurations [σ] of the Boltzmann weight for each state, E[σ] ) Z = ∑ exp (− (2.182) T {σ}

38

Classical Field Theory

! S(i)

! S(j) i

j

Figure 2.3 Spins on a lattice.

where T is the temperature and {σ} is the set of all spin configurations. In the 1950’s, Landau developed a picture to study these type of problems which in general are very difficult. Landau first proposed to work not with the microscopic spins but with a set of coarse-grained configurations (Landau, 1937). One way to do this, using the more modern version of the argument due to Kadanoff (Kadanoff, 1966) and Wilson (Wilson, 1971), is to partition a large system of linear size L into regions or blocks of smaller linear size / such that a0 ≪ / ≪ L, where a0 is the lattice spacing. Each one of these regions will be centered around a site, say x. We will denote such a region by A(x). The idea is now to perform the sum, i.e. the partition function Z, while keeping the total magnetization of each region A(x) fixed to the values M (x) =

1 N [A]

∑ σ(y)

(2.183)

y∈A(x)

where N [A] is the number of sites in A(x). The restricted partition function

2.7 The Landau theory of phase transitions as a field theory

is now a functional of the coarse-grained local magnetizations M (x), Z[M ] = ∑ exp {− {σ}

⎛ ⎞ 1 E[σ] } ∏ δ ⎜M (x) − ∑ σ(y)⎟ T N (A) ⎝ ⎠ x y∈A(x)

39

(2.184)

The variables M (x) have the property that, for N (A) very large, they effectively take values on the real numbers. Also, the coarse-grained configurations {M (x)} are smoother than the microscopic configurations {σ}. At very high temperatures the average magnetization ⟨M ⟩ = 0 and the system is in a paramagnetic phase. On the other hand, at low temperatures the average magnetization may be non-zero and the system may be in a ferromagnetic phase. Thus, at high temperatures the partition function Z is dominated by configurations which have ⟨M ⟩ = 0 while at very low temperatures, the most frequent configurations have ⟨M ⟩ ≠ 0. Landau proceeded to write down an approximate form for the partition function in terms of sums over smooth, continuous, configurations M (x) which, formally, can be represented in the form Z ≈ ∫ DM (x) exp ( −

E [M (x), T ] ) T

(2.185)

where DM (x) is an integration measure which means “sum over all configurations.” We will define this more properly later on, in chapter 15. We will assume that for the dominant configurations in the sum (integral!) shown in Eq.(2.185) the local averages M (x) are smooth functions of x and are small. Then, the energy functional E[M ] can be written as an expansion in powers of M (x) and of its space derivatives. With these assumptions the free energy of the magnet in D dimensions can be approximated by the Ginzburg-Landau form (Landau and Lifshitz, 1959b) 1 1 1 D 2 2 4 E(M ) = ∫ d x ( K(T )∣!M (x)∣ + a(T )M (x) + b(T )M (x) + . . . ) 2 2 4! (2.186) Ginzburg and Landau made the additional (and drastic) assumption that there is a single configuration M (x) that dominates the full partition function. Under this assumption, which can be regarded as a mean-field approximation, the free energy F = −T ln Z takes the same form as Eq.(2.186). In a later chapter we will develop the theory of the effective action and we will return to this point. Thermodynamic stability requires that the stiffness K(T ) and the coefficient b(T ) of the quartic term must be positive. The second term has a

40

Classical Field Theory

coefficient a(T ) which can have either sign. A simple choice of these parameters is K(T ) ≃ K0 ,

b(T ) ≃ b0 ,

a(T ) ≃ a ¯ (T − Tc )

(2.187)

where Tc is an approximate value of the critical temperature. The free energy F (M ) defines a Classical, or Euclidean, Field Theory. In fact, by rescaling the field M (x) in the form √ Φ(x) = KM (x) (2.188) we can write the free energy as

1 d 2 F (Φ) = ∫ d x { (!Φ) + U (Φ)} 2

where the potential U (Φ) is 2

where m ¯ =

a(T ) K

U (Φ) =

and λ =

b . K2 2

(2.189)

2

m ¯ 2 λ 4 Φ + Φ + ... 2 4!

(2.190)

Except for the absence of the term involving

the canonical momentum Π (x), F (Φ) has a striking resemblance to the Hamiltonian of a scalar field in Minkowski space! We will see below that this is not an accident, and that the Ginzburg-Landau theory is closely 4 related to a quantum field theory of a scalar field with a Φ potential. U (Φ)

T > T0

T < T0 −Φ0

Φ0

Φ

Figure 2.4 The Landau free energy for the order parameter field Φ: for T > T0 the free energy has a unique minimum at Φ = 0 while for T < T0 there are two minima at Φ = ±Φ0

2.7 The Landau theory of phase transitions as a field theory

41

Let us now ask the following question: is there a configuration Φc (x) that gives the dominant contribution to the partition function Z? If so, we should be able to approximate Z = ∫ DΦ exp{−F (Φ)} ≈ exp{−F (Φc )}{1 + ⋯}

(2.191)

This statement is usually called the Mean Field Approximation. Since the integrand is an exponential, the dominant configuration Φc must be such that F has a (local) minimum at Φc . Thus, configurations Φc which leave F (Φ) stationary are good candidates (we actually need local minima!). The problem of finding extrema is simply the condition δF = 0. This is the same problem we solved for classical field theory in Minkowski space-time. Notice that in the derivation of F we have invoked essentially the same type of arguments as before: (a) invariance and (b) locality (differentiability). The Euler-Lagrange equations can be derived by using the same arguments that we employed in the context of a scalar field theory. In the case at hand they become −

δF δF )=0 + ∂j ( δΦ(x) δ∂j Φ(x)

(2.192)

For the case of the Landau theory the Euler-Lagrange Equation becomes the Ginzburg-Landau Equation 0 = − & Φc (x) + m ¯ Φc (x) + 2

2

λ 3 Φ (x) 3! c

(2.193)

The solution Φc (x) that minimizes the energy is uniform in space and thus has ∂j Φc = 0. Hence, Φc is the solution of the very simple equation 2

m ¯ Φc +

λ 3 Φ =0 3! c

(2.194)

2

Since λ is positive and m ¯ may have either sign, depending on whether T > Tc or T < Tc , we have to explore both cases. 2 For T > Tc , m ¯ is also positive and the only real solution is Φc = 0. This 2 is the paramagnetic phase. But, for T < Tc , m ¯ is negative and two new solutions are available, namely √ 6∣m ¯2 ∣ Φc = ± (2.195) λ

These are the solutions with lowest energy and they are degenerate. They both represent the magnetized (or ferromagnetic)) phase. We now must ask if this procedure is correct or, rather, when can we

42

Classical Field Theory

expect this approximation to work. The answer to this question is the central problem of the theory of phase transitions which describes the behavior of statistical systems in the vicinity of a continuous (or second order) phase transition. It turns out that this problem is also connected with a central problem of Quantum Field theory, namely when and how is it possible to remove the singular behavior of perturbation theory, and in the process remove all dependence on the short distance (or high energy) cutoff from physical observables. In Quantum Field Theory this procedure amounts to a definition of the continuum limit. The answer to these questions motivated the development of the Renormalization Group which solved both problems simultaneously.

2.8 Field theory and Statistical Mechanics We are now going to discuss a mathematical procedure that will allow us to connect quantum field theory to classical statistical mechanics. We will use a mapping called a Wick rotation. Let us go back to the action for a real scalar field Φ(x) in D = d + 1 space-time dimensions S = ∫ d x L(Φ, ∂µ Ψ) D

(2.196)

D

where d x is D

d

d x ≡ dx0 d x

(2.197)

Let us formally carry out the analytic continuation of the time component x0 of xµ from real to imaginary time xD (2.198)

x0 ↦ −ixD under which

Φ(x0 , x) ↦ Φ(x, xD ) ≡ Φ(x)

(2.199)

where x = (x, xD ). Under this transformation, the action (or rather i times the action) becomes iS ≡ i ∫ dx0 d x L(Φ, ∂0 Ψ, ∂j Φ) ↦ ∫ d x L(Φ, −i∂D Φ, ∂j Φ) d

D

(2.200)

If L has the form L=

1 1 1 2 2 2 (∂ Φ) − V (Φ) ≡ (∂0 Φ) − (!Φ) − V (Φ) 2 µ 2 2

(2.201)

2.8 Field theory and Statistical Mechanics

43

then the analytic continuation yields 1 1 2 2 L(Φ, −i∂D Ψ, &Φ) = − (∂D Φ) − (!Φ) − V (Φ) 2 2 Then we can write

(2.202)

1 1 2 2 D iS(Φ, ∂µ Φ) −−−−−−−−→ − ∫ d x [ (∂D Φ) + (!Φ) + V (Φ)] (2.203) 2 2 x0 → −ixD

This expression has the same form as (minus) the potential energy E(Φ) for a classical field Φ in D = d + 1 space dimensions. However it is also the same as the energy for a classical statistical mechanics problem in the same number of dimensions i.e. the Landau-Ginzburg free energy of the last section. In Classical Statistical Mechanics, the equilibrium properties of a system are determined by the partition function. For the case of the Landau theory of phase transitions the partition function is −E(Φ)/T Z = ∫ DΦ e

(2.204)

where the symbol “∫ DΦ” means sum over all configurations. We will discuss the meaning of this expression and the definition of the “measure” DΦ later on. If we choose for energy functional E(Φ) the expression

where

1 D 2 E(Φ) = ∫ d x [ (∂Φ) + V (Φ)] 2 (∂Φ) ≡ (∂D Φ) + (!Φ) 2

2

2

(2.205)

(2.206)

we see that the partition function Z is formally the analytic continuation of iS(Φ, ∂µ Φ)/h̵ Z = ∫ DΦ e

(2.207)

where we have used h̵ which has units of action (instead of the temperature). What is the physical meaning of Z? This expression suggests that Z should have the interpretation of a sum of all possible functions Φ(x, t) (i.e. the histories of the configurations of the field Φ) weighed by the phase factor exp{ hi̵ S(Φ, ∂µ Φ)}. We will discover later on that if T is formally identified ̵ then Z represents the path-integral quantization with the Planck constant h, of the field theory! Notice that the semiclassical limit h̵ → 0 is formally equivalent to the low temperature limit of the statistical mechanical system. The analytic continuation procedure that we just discussed is called a Wick rotation. It amounts to a passage from D = d+1-dimensional Minkowski

44

Classical Field Theory

space to a D-dimensional Euclidean space. We will find that this analytic continuation is a very powerful tool. As we will see below, a number of difficulties will arise when the theory is defined directly in Minkowski space. Primarily, the problem is the presence of ill-defined integrals which are given precise meaning by a deformation of the integration contours from the real time (or frequency) axis to the imaginary time (or frequency) axis. The deformation of the contour amounts to a definition of the theory in Euclidean rather than in Minkowski space. It is an underlying assumption that the analytic continuation can be carried out without problems. Namely, the assumption is that the result of this procedure is unique and that, whatever singularities may be present in the complex plane, they do not affect the result. It is important to stress that the success of this procedure is not guaranteed. However, in almost all the theories that we know of this assumption holds. The only case in which problems are known to exist is the theory of Quantum Gravity (which we will not discuss in this book).

3 Classical Symmetries and Conservation Laws

We have used the existence of symmetries in a physical system as a guiding principle for the construction of their Lagrangians and energy functionals. We will show now that these symmetries imply the existence of conservation laws. There are different types of symmetries which, roughly, can be classified into two classes: (a) spacetime symmetries and (b) internal symmetries. Some symmetries involve discrete operations, hence called discrete symmetries, while others are continuous symmetries. Furthermore, in some theories these are global symmetries, while in others they are local symmetries. The latter class of symmetries go under the name of gauge symmetries. We will see that, in the fully quantized theory, global and local symmetries play different roles. Spacetime symmetries are the most common examples of symmetries that are encountered in Physics. They include translation invariance and rotation invariance. If the system is isolated, then time-translation is also a symmetry. A non-relativistic system is in general invariant under Galilean transformations, while relativistic systems, are instead Lorentz invariant. Other spacetime symmetries include time-reversal (T ), parity (P ) and charge conjugation (C). These symmetries are discrete. In classical mechanics, the existence of symmetries has important consequences. Thus, translation invariance, which is a consequence of uniformity of space, implies the conservation of the total momentum P of the system. Similarly, isotropy implies the conservation of the total angular momentum L and time translation invariance implies the conservation of the total energy E. All of these concepts have analogs in field theory. However, in field theory new symmetries will also appear which do not have an analog in the classi-

52

Classical Symmetries and Conservation Laws

cal mechanics of particles. These are the internal symmetries, that will be discussed below in detail.

3.1 Continuous symmetries and Noether’s theorem We will show now that the existence of continuous symmetries has very profound implications, such as the existence of conservation laws. One important feature of these conservation laws is the existence of locally conserved currents. This is the content of the following theorem, due to Emmy Noether. Noether’s theorem: For every continuous global symmetry there exists a global conservation law. Before we prove this statement, let us discuss the connection that exists between locally conserved currents and constants of motion. In particular, let us show that for every locally conserved current there exist a globally µ conserved quantity, i.e. a constant of motion. To this effect, let j (x) be some locally conserved current, i.e. jµ (x) satisfies the local constraint ∂µ j (x) = 0 µ

V (T + ∆T )

(3.1)

T + ∆T

time Ω δΩ

V (T )

T space

Figure 3.1 A spacetime 4-volume.

Let Ω be a bounded 4-volume of spacetime, with boundary ∂Ω. Then, the

3.2 Internal symmetries

53

Divergence (Gauss) theorem tells us that 0 = ∫ d x ∂µ j (x) = ∮ 4

µ

Ω

dSµ j (x) µ

∂Ω

(3.2)

where the r.h.s. is a surface integral on the oriented closed surface ∂Ω (a 3volume). Let us suppose that the 4-volume Ω extends all the way to infinity in space and has a finite extent in time ∆T . µ If there are no currents at spacial infinity, i.e. lim∣x∣→∞ j (x, x0 ) = 0, then only the top (at time T + ∆T ) and the bottom (at time T ) of the boundary ∂Ω (shown in Fig. (3.1)) will contribute to the surface (boundary) integral. Hence, the r.h.s. of Eq.(3.2) becomes 0=∫

dS0 j (x, T + ∆T ) − ∫

dS0 j (x, T )

d x j (x, T + ∆T ) − ∫

d x j (x, T )

0

V (T +∆T )

3

V (T )

0

(3.3)

Since dS0 ≡ d x, the boundary contributions reduce to two oriented 3volume integrals 0=∫

V (T +∆T )

3

Thus, the quantity Q(T )

0

V (T )

Q(T ) ≡ ∫

is a constant of motion, i.e.

V (T )

3

0

(3.4)

d x j (x, T ) 3

Q(T + ∆T ) = Q(T )

0

(3.5)

∀ ∆T

(3.6) µ

Hence, the existence of a locally conserved current, satisfying ∂µ j = 0, implies the existence of a globally conserved charge (or Noether charge) 3 0 Q = ∫ d x j (x, T ), which is a constant of motion. Thus, the proof of the Noether theorem reduces to proving the existence of a locally conserved current. In the following sections we will prove Noether’s theorem for internal and spacetime symmetries.

3.2 Internal symmetries Let us begin, for simplicity, with the case of the complex scalar field φ(x) ≠ ∗ φ (x). The arguments that follow below are easily generalized to other cases. ∗ ∗ Let L(φ, ∂µ φ, φ , ∂µ φ ) be the Lagrangian density. We will assume that the Lagrangian is invariant under the continuous global internal symmetry

54

Classical Symmetries and Conservation Laws

transformation

φ(x) ↦φ (x) = e φ(x) ′

iα

φ (x) ↦φ (x) = e ∗

′∗

φ (x)

−iα ∗

(3.7)

where α is an arbitrary real number (not a function!). The system is invariant under the transformation of Eq.(3.7) if the Lagrangian L satisfies L(φ , ∂µ φ , φ , ∂µ φ ) ≡ L(φ, ∂µ φ, φ , ∂µ φ ) ′

′

′∗

′∗

∗

∗

(3.8)

Then we say that the transformation shown in Eq.(3.7) is a global symmetry of the system. In particular, for an infinitesimal transformation we have φ (x) = φ(x) + δφ(x) + . . . ,

φ (x) = φ (x) + δφ (x) + . . .

′

′∗

∗

∗

(3.9)

where δφ(x) = iαφ(x). Since L is invariant, its variation must be identically equal to zero. The variation δL is δL =

δL δL δL δL ∗ ∗ δφ + δ∂ φ + ∗ δφ + ∗ δ∂µ φ δφ δ∂µ φ µ δφ δ∂µ φ

(3.10)

Using the equation of motion δL δL )=0 − ∂µ ( δφ δ∂µ φ

(3.11)

and its complex conjugate, we can write the variation δL in the form of a total divergence δL δL ∗ δL = ∂µ [ δφ + δφ ] (3.12) ∗ δ∂µ φ δ∂µ φ ∗

∗

Thus, since δφ = iαφ and δφ = −iαφ , we get δL = ∂µ [ i (

δL ∗ δL φ− φ ) α] δ∂µ φ δ∂µ φ∗

(3.13)

Hence, since α is arbitrary, δL will vanish identically if and only if the 4µ vector j , defined by µ

j = i( is locally conserved, i.e.

δL δL ∗ φ ) φ− δ∂µ φ δ∂µ φ∗

δL = 0

iff

µ

∂µ j = 0

(3.14)

(3.15)

In particular, if L has the form

L = (∂µ φ) (∂ φ) − V (∣φ∣ ) ∗

µ

2

(3.16)

3.3 Global symmetries and group representations

55

which is manifestly invariant under the symmetry transformation of Eq.(3.7), µ we see that the current j is given by → µ µ ∗ ∗ µ ∗← j = i (∂ φ φ − φ ∂ φ) ≡ iφ ∂µ φ (3.17) Thus, the presence of a continuous internal symmetry implies the existence of a locally conserved current. Furthermore, the conserved charge Q is given by → 3 0 3 ∗← Q = ∫ d x j (x, x0 ) = ∫ d x iφ ∂0 φ

(3.18)

In terms of the canonical momentum Π(x), the globally conserved charge Q of the charged scalar field is Q = ∫ d x i(φ Π − φΠ ) 3

∗

∗

(3.19)

3.3 Global symmetries and group representations Let us generalize the result of the last subsection. Let us consider a scalar a field φ which transforms irreducibly under a certain representation of a Lie group G. In the case considered in the previous section the group G is the group of complex numbers of unit length, the group U (1). The elements of iα this group, g ∈ U (1), have the form g = e .

S1

R

|z| = 1 1

Figure 3.2 The U (1) group is isomorphic to the unit circle while the real numbers R are isomorphic to a tangent line.

This set of complex numbers forms a group in the sense that,

56

Classical Symmetries and Conservation Laws

1 . It is closed under complex multiplication, i.e. iα

g=e

∈ U (1)

and

′

iβ

g =e

∈ U (1) ⇒ g ∗ g = e ′

i(α+β)

∈ U (1). (3.20)

2 . There is an identity element, i.e. g = 1. iα 3 . For every element g = e ∈ U (1) there is a unique inverse element −1 −iα g =e ∈ U (1).

The elements of the group U (1) are in one-to-one correspondence with the points of the unit circle S1 . Consequently, the parameter α that labels the transformation (or element of this group) is defined modulo 2π, and it should be restricted to the interval (0, 2π]. On the other hand, transformations infinitesimally close to the identity element, 1, lie essentially on the line tangent to the circle at 1 and are isomorphic to the group of real numbers R. The group U (1) is compact, in the sense that the length of the natural parametrization of its elements is 2π, which is finite. In contrast, the group R of real numbers is not compact (see Fig. (3.2)). In the sequel, we will almost always work with internal symmetries with a compact Lie group. For infinitesimal transformations the groups U (1) and R are essentially identical. There are, however, field configurations for which they are not. A typical case is the vortex configuration in two dimensions. For a vortex the phase of the field on a large circle of radius R → ∞ winds by 2π. Such configurations would not exist if the symmetry group was R instead of U (1) (note that analyticity requires that φ → 0 as x → 0.) φ = φ0 eiθ φ=0

θ φ = φ0

Figure 3.3 A vortex

Another example is the N -component real scalar field φ (x), with a = 1, . . . , N . In this case the symmetry is the group of rotations in N -dimensional Euclidean space a

φ (x) = R φ (x) ′a

ab b

(3.21)

3.3 Global symmetries and group representations

57

a

The field φ is said to transform like the N -dimensional (vector) representation of the Orthogonal group O(N ). The elements of the orthogonal group, R ∈ O(N ), satisfy

1 . If R1 ∈ O(N ) and R2 ∈ O(N ), then R1 R2 ∈ O(N ), 2 . ∃I ∈ O(N ) such that ∀R ∈ O(N ) then RI = IR = R, −1 −1 t 3 . ∀R ∈ O(N ) ∃R ∈ O(N ) such that R = R , t

where R is the transpose of the matrix R. a Similarly, if the N -component vector φ (x) is a complex field, it transforms under the group of N × N Unitary transformations U φ (x) = U φ (x) ′a

ab b

(3.22)

The complex N × N matrices U are elements of the Unitary group U (N ) and satisfy

1 . U1 ∈ U (N ) and U2 ∈ U (N ), then U1 U2 ∈ U (N ), 2 .∃I ∈ U (N ) such that ∀U ∈ U (N ), U I = IU = U , −1 −1 † † t ∗ 3 . ∀U ∈ U (N ), ∃U ∈ U (N ) such that U = U , where U = (U ) . a

In the particular case discussed above φ transforms like the fundamental (spinor) representation of U (N ). If we impose the further restriction that ∣ det U ∣ = 1, the group becomes SU (N ). For instance, if N = 2, the group is SU (2) and φ=(

φ1 ) φ2

(3.23)

it transforms like the spin-1/2 representation of SU (2). In general, for an arbitrary continuous Lie group G, the field transforms like φa (x) = (exp [iλ θ ])ab φb (x) ′

k k

(3.24)

where the vector θ is arbitrary and constant (i.e. independent of x). The k matrices λ are a set of N × N linearly independent matrices which span the algebra of the Lie group G. For a given Lie group G, the number of such matrices is D(G), and it is independent of the dimension N of the representation that was chosen. D(G) is called the rank of the group. The k matrices λab are the generators of the group in this representation. In general, from a symmetry point of view, the field φ does not have to be a vector, as it can also be a tensor or for the matter transform under any representation of the group. For simplicity, we will only consider the case of the vector representation of O(N ), and the fundamental (spinor) and adjoint (vector) (see below) representations of SU (N ).

58

Classical Symmetries and Conservation Laws

For an arbitrary compact Lie group G, the generators {λ }, j = 1, . . . , D(G), † are a set of hermitean, λj = λj , traceless matrices, trλj = 0, which obey the commutation relations j

[λ , λ ] = if j

jkl

k

jkl l

λ

(3.25)

The numerical constants f are known as the structure constants of the Lie group and are the same in all its representations. In addition, the generators have to be normalized. It is standard to require the normalization condition 1 ab a b trλ λ = δ 2

(3.26)

In the case considered above, the complex scalar field φ(x), the symmeiα try group is the group of unit length complex numbers of the form e . This group is known as the group U (1). All its representations are onedimensional, and has only one generator. A commonly used group is SU (2). This group, which is familiar from nonrelativistic quantum mechanics, has three generators J1 , J2 and J3 , that obey the angular momentum algebra

with

[Ji , Jj ] = i(ijk Jk

1 tr(Ji Jj ) = δij 2

and

trJi = 0

(3.27)

(3.28)

The representations of SU (2) are labelled by the angular momentum quantum number J. Each representation J is a 2J + 1-fold degenerate multiplet, i.e. the dimension of the representation is 2J + 1. The lowest non-trivial representation of SU (2), i.e. J ≠ 0, is the spinor representation which has J = 21 and is two-dimensional. In this representation, the field φa (x) is a two-component complex spinor and the generators J1 , J2 and J3 are given by the set of 2 × 2 Pauli matrices Jj = 12 σj . The vector, or spin 1, representation is three-dimensional, and φa is a three-component vector. In this representation, the generators are very simple (Jj )kl = (jkl

(3.29)

Notice that the dimension of this representation (3) is the same as the rank (3) of the group SU (2). In this representation, known as the adjoint representation, the matrix elements of the generators are the structure constants. This is a general property of all Lie groups. In particular, for the group 2 2 SU (N ), whose rank is N − 1, it has N − 1 infinitesimal generators, and

3.3 Global symmetries and group representations

59

2

the dimension of its adjoint (vector) representation is N − 1. For instance, for SU (3) the number of generators is eight. Another important case is the group of rotations of N -dimensional Euclidean space, O(N ). In this case, the group has N (N − 1)/2 generators ij which can be labelled by the matrices L (i, j = 1, . . . , N ). The fundamental (vector) representation of O(N ) is N -dimensional and, in this representation, the generators are (L )kl = −i(δik δj$ − δi$ δjk ) ij

(3.30)

ij

It is easy to see that the L ’s generate infinitesimal rotations in N -dimensional space. Quite generally, in a given representation an element of a Lie group is labelled by a set of Euler angles denoted by θ. If the Euler angles θ are infinitesimal, then the representation matrix exp(iλ ⋅ θ) is close to the identity, and can be expanded in powers of θ. To leading order in θ the change a in φ is δφ (x) = i(λ ⋅ θ) φ (x) + . . . a

ab b

(3.31)

µ

If φa is real, the conserved current j is jµ (x) = k

δL k λ φb (x) δ∂µ φa (x) ab

(3.32)

k

where k = 1, . . . , D(G). Here, the generators λ are real hermitean matrices. In contrast, for a complex field φa , the conserved currents are jµ (x) = i ( k

δL δL k ∗ k λ φb (x) ) λab φb (x) − a δ∂µ φ (x) δ∂µ φa (x)∗ ab

(3.33)

k

Here the generators λ are hermitean matrices (but are not all real). Thus, we conclude that the number of conserved currents is equal to the number of generators of the group. For the particular choice L = (∂µ φa ) (∂ φa ) − V (φa φa ) ∗

µ

∗

(3.34)

the conserved current is

→ ∗← k k jµ = i λab φa ∂µ φb

(3.35)

and the conserved charges are → ∗← k 3 k Q = ∫ d x iλab φa ∂0 φb V

where V is the volume of space.

(3.36)

60

Classical Symmetries and Conservation Laws

3.4 Global and local symmetries: gauge invariance The existence of global symmetries assumes that, at least in principle, we a can measure and change all of the components of a field φ (x) at all points x in space at the same time. Relativistic invariance tells us that, although the theory may posses this global symmetry, in principle this experiment cannot be carried out. One is then led to consider theories which are invariant if the symmetry operations are performed locally. Namely, we should require that the Lagrangian be invariant under local transformations φa (x) → φa (x) = (exp [iλ θ (x) ])ab φb (x) ′

k k

(3.37)

For instance, we can demand that the theory of a complex scalar field φ(x) be invariant under local changes of phase φ(x) → φ (x) = e ′

iθ(x)

φ(x)

(3.38)

The standard local Lagrangian L

L = (∂µ φ) (∂ φ) − V (∣φ∣ ) ∗

2

µ

(3.39)

is invariant under global transformations with θ = constant, but it is not invariant under arbitrary smooth local transformations θ(x). The main problem is that since the derivative of the field does not transform like the field itself, the kinetic energy term is no longer invariant. Indeed, under a local transformation, we find ∂µ φ(x) → ∂µ φ (x) = ∂µ [e ′

iθ(x)

iθ(x)

φ(x)] = e

[∂µ φ + iφ∂µ θ]

(3.40)

In order to make L locally invariant we must find a new derivative operator Dµ , the covariant derivative, which transforms in the same way as the field φ(x) under local phase transformations, i.e. ′

′

iθ(x)

Dµ φ → Dµ φ = e

Dµ φ

(3.41)

From a “geometric” point of view we can picture the situation as follows. In order to define the phase of φ(x) locally, we have to define a local frame, or fiducial field, with respect to which the phase of the field is measured. Local gauge invariance is then the statement that the physical properties of the system must be independent of the particular choice of frame. From this point of view, local gauge invariance is an extension of the principle of relativity to the case of internal symmetries. Now, if we wish to make phase transformations that differ from point to point in spacetime, we have to specify how the phase changes as we go from one point x in spacetime to another one y. In other words, we have to define

3.4 Global and local symmetries: gauge invariance

61

a connection that will tell us how to parallel transport the phase of φ from x to y as we travel along some path Γ. Let us consider the situation in which x and y are arbitrarily close to each other, i.e. yµ = xµ + dxµ where dxµ is an infinitesimal 4-vector. The change in φ is φ(x + dx) − φ(x) = δφ(x)

(3.42)

If the transport of φ along some path going from x to x + dx is to correspond to a phase transformation, then δφ must be proportional to φ. So we are led to define δφ(x) = iAµ (x)dx φ(x) µ

(3.43)

where Aµ (x) is a suitably chosen vector field. Clearly, this implies that the covariant derivative Dµ must be defined to be Dµ φ ≡ ∂µ φ(x) − ieAµ (x)φ(x) ≡ (∂µ − ieAµ ) φ

(3.44)

where e is a parameter which we will give the physical interpretation of a coupling constant. How should Aµ (x) transform? We must choose its transformation law in iθ such a way that Dµ φ transforms like φ(x) itself. Thus, if φ → e φ we have D φ = (∂µ − ieAµ )(e φ) ≡ e Dµ φ ′ ′

′

iθ

iθ

(3.45)

This requirement can be met if

′

i∂µ θ − ieAµ = −ieAµ

(3.46)

Hence, Aµ should transform like 1 ′ Aµ → Aµ = Aµ + e ∂µ θ (3.47) But this is nothing but a gauge transformation! Indeed, if we define the gauge transformation Φ(x) 1 (3.48) Φ(x) ≡ e θ(x) we see that the vector field Aµ transforms like the vector potential of Maxwell’s electromagnetism. We conclude that we can promote a global symmetry to a local (i.e. gauge) symmetry by replacing the derivative operator by the covariant derivative. Thus, we can make a system invariant under local gauge transformations at the price of introducing a vector field Aµ , the gauge field, that plays the role of a connection. From a physical point of view, this result means that the impossibility of making a comparison at a distance of the phase of the

62

Classical Symmetries and Conservation Laws

field φ(x) requires that a physical gauge field Aµ (x) must be present. This procedure, that relates the matter and gauge fields through the covariant derivative, is known as minimal coupling. There is a set of configurations of φ(x) that changes only because of the presence of the gauge field. These are the geodesic configurations φc (x). They satisfy the equation Dµ φc = (∂µ − ieAµ )φc ≡ 0

(3.49)

which is equivalent to the linear equation (see Eq. (3.43)) (3.50)

∂µ φc = ieAµ φc

Let us consider, for example, two points x and y in spacetime at the ends of a path Γ(x, y). For a given path Γ(x, y), the solution of Eq. (3.49) is the path-ordered exponential of a line integral −ie ∫Γ(x,y) dzµ A (z)

φc (x) = e

µ

φc (y)

(3.51)

Indeed, under a gauge transformation, the line integral transforms like e∫

dzµ A ↦ e ∫ µ

Γ(x,y)

= e∫ −ie ∫Γ dzµ A

µ

1 µ dzµ e ∂ θ

µ

Γ(x,y)

Γ(x,y)

dzµ A (z) + θ(y) − θ(x) µ

Γ(x,y)

Hence, we get φc (y) e

dzµ A + e ∫

↦ φc (y) e

−ie ∫Γ dzµ A

iθ(x)

≡e

µ

φc (x)

(3.52)

−iθ(y) iθ(x)

e

e

(3.53)

as it should be. However, we may now want to ask how the change of phase of φc depends Γ Γ on the choice of the path Γ. Thus, let φc 1 (y) and φc 2 (y) be solutions of the geodesic equations for two different paths Γ1 and Γ2 with the same end points, x and y. Clearly, we have that the change of phase ∆γ given by ∆γ = −e ∫

dzµ A + e ∫ µ

Γ1

dzµ A ≡ −e ∮ µ

Γ2

µ

+

Γ

dzµ A

(3.54)

+

Here Γ is the closed oriented path +

+

−

Γ = Γ1 ∪ Γ2

(3.55)

and ∫

Γ− 2

dzµ Aµ = − ∫

Γ+ 2

dzµ Aµ

(3.56)

3.4 Global and local symmetries: gauge invariance

63

Σ

Γ Figure 3.4 A closed path Γ is the boundary of the open surface Σ. +

Using Stokes theorem we see that, if Σ is an oriented surface whose + + + boundary is the oriented closed path Γ , ∂Σ ≡ Γ (see Fig.(3.4), then ∆γ is given by the flux Φ(Σ) of the curl of the vector field Aµ through the + surface Σ , i.e. e µν ∆γ = − ∫ dSµν F = −e Φ(Σ) 2 Σ+ where F

µν

(3.57)

is the field tensor F

µν

µ

ν

ν

µ

=∂ A −∂ A

(3.58)

dSµν is the oriented area element, and Φ(Σ) is the flux through the surµν face Σ. Both F and dSµν are antisymmetric in their spacetime indices. µν In particular, F can also be written as a commutator of two covariant derivatives F µν

µν

i µ µ = e [D , D ]

(3.59)

Thus, F measures the (infinitesimal) incompatibility of displacements µν along two independent directions. In other words, F is a curvature tensor. µν These results show very clearly that if F is non-zero in some region of spacetime, then the phase of φ cannot be uniquely determined: the phase of φc depends on the path Γ along which it is measured.

64

Classical Symmetries and Conservation Laws

3.5 The Aharonov-Bohm effect The path dependence of the phase of φc is closely related to Aharonov-Bohm Effect. This is a subtle effect, which was first discovered in the context of elementary quantum mechanics, and plays a fundamental role in (quantum) field theory as well. Consider a quantum mechanical particle of charge e and mass m moving on a plane. The particle is coupled to an external electromagnetic field Aµ (here µ = 0, 1, 2 only, since there is no motion out of the plane). Let us consider the geometry shown in Fig.(3.5) in which an infinitesimally thin solenoid pierces the plane in the vicinity of some point r = 0. The

Σ+ r

Γ2

Φ

Γ1

r0 Γ+ Figure 3.5 Geometric setup of the Aharonov-Bohm effect: a magnetic flux + + Φ ≠ 0 is thread through the small hole in the punctured plane Σ . Here Γ + are the oriented outer and inner edges of Σ ; Γ1 and Γ2 are two inequivalent paths from r0 to r described in the text.

Schr¨ odinger Equation for this problem is HΨ = ih̵ where H=

∂Ψ ∂t

2 1 h̵ e ( * + c A) 2m i

(3.60)

(3.61)

is the Hamiltonian. The magnetic field B = B zˆ vanishes everywhere except at r = 0, B = Φ0 δ(r)

(3.62)

3.5 The Aharonov-Bohm effect

65

Using Stokes theorem we see that the flux of B through an arbitrary region + + Σ with boundary Γ is Φ=∫

Σ+

dS ⋅ B = ∮

Γ+

d$ ⋅ A

(3.63)

+

Hence, Φ = Φ0 for all surfaces Σ that enclose the point r = 0, and it is equal to zero otherwise. Hence, although the magnetic field is zero for r ≠ 0, the vector potential does not (and cannot) vanish. The wave function Ψ(r) can be calculated in a very simple way. Let us define iθ(r)

Ψ(r) = e

Ψ0 (r)

(3.64)

where Ψ0 (r) satisfies the Schr¨odinger Equation in the absence of the field, i.e. ∂Ψ0 H0 Ψ0 = ih̵ (3.65) ∂t with 2 h̵ 2 H0 = − * (3.66) 2m Since the wave function Ψ has to be differentiable and Ψ0 is single valued, we must also demand the boundary condition that lim Ψ0 (r, t) = 0

r→0

(3.67)

iθ

The wave function Ψ = Ψ0 e looks like a gauge transformation. But we will discover that there is a subtlety here. Indeed, θ can be determined as follows. By direct substitution we get e h̵ h̵ e iθ iθ ( * + c A) (e Ψ0 ) = e (h̵ *θ + c A + *) Ψ0 i i

(3.68)

Thus, in order to succeed in our task, we only have to require that A and θ must obey the relation e ̵ (3.69) h*θ + cA ≡ 0 Or, equivalently, e *θ(r) = − ̵ A(r) (3.70) hc However, if this relation holds, θ cannot be a smooth function of r. In fact, + the line integral of *θ on an arbitrary closed path Γ is given by ∫

Γ+

d$ ⋅ *θ = ∆θ

(3.71)

66

Classical Symmetries and Conservation Laws

where ∆θ is the total change of θ in one full counterclockwise turn around the path Γ. It is the immediate to see that ∆θ is given by e ∆θ = − ̵ ∮ d$ ⋅ A hc Γ+

(3.72)

We must conclude that, in general, θ(r) is a multivalued function of r which has a branch cut going from r = 0 out to some arbitrary point at infinity. The actual position and shape of the branch cut is irrelevant, but the discontinuity ∆θ of θ across the cut is not irrelevant. Hence, Ψ0 is chosen to be a smooth, single valued, solution of the Schr¨ odinger equation in the absence of the solenoid, satisfying the boundary condition of Eq. (3.67). Such wave functions are (almost) plane waves. Since the function θ(r) is multivalued and, hence, path-dependent, the wave function Ψ is also multivalued and path-dependent. In particular, let r0 be some arbitrary point on the plane and Γ(r0 , r) is a path that begins in r0 and ends at r. The phase θ(r) is, for that choice of path, given by e θ(r) = θ(r0 ) − ̵ ∫ dx ⋅ A(x) hc Γ(r0 ,r)

(3.73)

The overlap of two wave functions that are defined by two different paths Γ1 (r0 , r) and Γ2 (r0 , r) is (with r0 fixed) ⟨Γ1 ∣Γ2 ⟩ = ∫ d r ΨΓ1 (r) ΨΓ2 (r) 2

∗

ie 2 2 d$ ⋅ A − ∫ d$ ⋅ A)} ≡ ∫ d r ∣Ψ0 (r)∣ exp {+ ̵ (∫ hc Γ1 (r0 ,r) Γ2 (r0 ,r)

(3.74)

If Γ1 and Γ2 are chosen in such a way that the origin (where the solenoid is piercing the plane) is always to the left of Γ1 but it is also always to the right of Γ2 , the difference of the two line integrals is the circulation of A ∫

Γ1 (r0 ,r)

d$ ⋅ A − ∫

Γ2 (r0 ,r)

d$ ⋅ A ≡ ∮

Γ+ (r

0)

d$ ⋅ A

(3.75)

on the closed, positively oriented, contour Γ = Γ1 (r0 , r) ∪ Γ2 (r, r0 ). Since this circulation is constant, and equal to the flux Φ, we find that the overlap ⟨Γ1 ∣Γ2 ⟩ is ie ⟨Γ1 ∣Γ2 ⟩ = exp { ̵ Φ} (3.76) hc +

where we have taken Ψ0 to be normalized to unity. The result of Eq.(3.76) is known as the Aharonov-Bohm Effect.

3.6 Non-abelian gauge invariance

67

We find that the overlap is a pure phase factor which, in general, is different from one. Notice that, although the wave function is always defined up to a constant arbitrary phase factor, phase changes are physical effects. In addition, for some special choices of Φ the wave function becomes single valued. These values correspond to the choice e (3.77) ̵hc Φ = 2πn where n is an arbitrary integer. This requirement amounts to a quantization condition for the magnetic flux Φ, i.e. hc Φ = n ( e ) ≡ nΦ0

(3.78)

. where Φ0 is the flux quantum, Φ0 = hc e In 1931 Dirac considered the effects of a monopole configuration of magnetic fields on the quantum mechanical wave functions of charged particles (Dirac, 1931). In Dirac’s construction, a magnetic monopole is represented as a long thin solenoid in three-dimensional space. The magnetic field near the end of the solenoid is the same as that of a magnetic charge m equal to the magnetic flux going through the solenoid. Dirac argued that for the solenoid (the “Dirac string”) to be unobservable, the wave function must be single-valued. This requirement leads to the Dirac quantization condition for the smallest magnetic charge, ̵ me = 2π hc

(3.79)

which we recognize is the same as the flux quantization condition of Eq.(3.78). 3.6 Non-abelian gauge invariance Let us now consider systems with a non-abelian global symmetry. This means that the field φ transforms like some representation of a Lie group G, φa (x) = Uab φb (x) ′

(3.80)

where U is a matrix that represents the action of a group element. The local Lagrangian density L = ∂µ φa ∂ φ − V (∣φ∣ ) ∗ µ a

2

(3.81)

is invariant under global transformations. Suppose now that we want to promote this global symmetry to a local one. However, it is also correct for this general case as well that while the 2 potential term V (∣φ∣ ) is invariant even under local transformations U (x),

68

Classical Symmetries and Conservation Laws

the first term of the Lagrangian of Eq.(3.81) is not. Indeed, the gradient of φ does not transform properly (i.e. covariantly) under the action of the Lie group G, ∂µ φ (x) =∂µ [U (x)φ(x)] ′

= (∂µ U (x)) φ(x) + U (x)∂µ φ(x) =U (x)[∂µ φ(x) + U

−1

(x)∂µ U (x)φ(x)]

(3.82)

Hence ∂µ φ does not transform in the same way as the φ field. We can now follow the same approach that we used in the abelian case and define a covariant derivative operator Dµ which should have the property that Dµ φ should obey the same transformation law as the field φ, i.e. (Dµ φ(x)) = U (x) (Dµ φ(x)) ′

(3.83)

It is clear that Dµ is now both a differential operator as well as a matrix acting on the field φ. Thus, Dµ depends on the representation that was chosen for the field φ. We can now proceed in analogy with what we did in the case of electrodynamics, and guess that the covariant derivative Dµ should be of the form Dµ = I ∂µ − igAµ (x)

(3.84)

where g is a coupling constant, I is the N × N identity matrix, and Aµ is a matrix-valued vector field. If φ has N components, the vector field Aµ (x) is an N × N hermitian matrix which can be expanded in the basis of the group k generators λab (with k = 1, . . . , D(G), and a, b = 1, . . . , N ) which span the algebra of the Lie group G, (Aµ (x))ab = Aµ (x)λab k

k

(3.85)

Thus, the vector field Aµ (x) is parametrized by the D(G)-component 4k vectors Aµ (x). We will choose the transformation properties of Aµ (x) in such a way that Dµ φ transforms covariantly under gauge transformations, Dµ φ(x) ≡Dµ (U (x)φ(x)) = (∂µ − igAµ (x)) (U φ(x)) ′

′

′

′

=U (x)[∂µ φ(x) + U

≡U (x) Dµ φ(x)

−1

(x)∂µ U (x)φ(x) − igU

−1

(x)Aµ (x)U (x)φ(x)] ′

(3.86)

This condition is met if we require that U

−1

(x)igAµ (x)U (x) = igAµ (x) + U ′

−1

(x)∂µ U (x)

(3.87)

3.6 Non-abelian gauge invariance

69

or, equivalently, that Aµ obeys the transformation law Aµ (x) = U (x)Aµ (x)U ′

−1

i −1 (x) − g (∂µ U (x)) U (x)

Since the matrices U (x) are unitary and invertible, we have U

−1

(x)U (x) = I

(3.88)

(3.89)

we can equivalently write the transformed vector field Aµ (x) in the form Aµ (x) = U (x)Aµ (x)U ′

−1

′

i −1 (x) + g U (x) (∂µ U (x))

(3.90)

This is the general form of a gauge transformation for a non-abelian Lie group G. In the case of an abelian symmetry group, such as the group U (1), the iθ(x) matrix reduces a simple phase factor, U (x) = e , and the field Aµ (x) is a real number-valued vector field. It is easy to check that, in this case, Aµ transforms as follows i iθ(x) −iθ(x) ′ iθ(x) −iθ(x) ∂µ (e ) Aµ (x) =e Aµ (x)e + ge 1 ≡Aµ (x) + g ∂µ θ(x) (3.91) which recovers the correct form for an abelian gauge transformation. Returning now to the non-abelian case, we see that under an infinitesimal transformation U (x) (U (x))ab = [exp (iλ θ (x))]ab ≅ δab + iλab θ (x) + . . . k k

k

k

(3.92)

the scalar field φ(x) transforms as

δφa (x) ≅ iλab φb (x) θ (x) + . . . k

k

(3.93)

k

while the vector field Aµ now transforms as δAµ (x) ≅ if k

ksj

1 k j s Aµ (x) θ (x) + g ∂µ θ (x) + . . .

(3.94)

Thus, Aµ (x) transforms as a vector in the adjoint representation of the Lie group G since, in that representation, the matrix elements of the generators ksj k are the group structure constants f . Notice that Aµ is always in the adjoint representation of the group G, regardless of the representation in which φ(x) happens to be in. From the discussion given above, it is clear that the field Aµ (x) can be interpreted as a generalization of the vector potential of electromagnetism. Furthermore, Aµ provides for a natural connection which tell us how the k

70

Classical Symmetries and Conservation Laws

“internal coordinate system,” in reference to which the field φ(x) is defined, changes from one point xµ of spacetime to a neighboring point xµ + dxµ . a In particular, the configurations φ (x) which are solutions of the geodesic equation Dµ φb (x) = 0 ab

(3.95)

correspond to the parallel transport of φ from some point x to some point y. This equation can be written in the equivalent form ∂µ φa (x) = igAµ (x) λab φb (x) k

k

(3.96)

This linear partial differential equation can be solved as follows. Let xµ and yµ be two arbitrary points in spacetime and Γ(x, y) a fixed path with endpoints at x and y. This path is parametrized by a mapping zµ from the real interval [0, 1] to Minkowski space M (or any other space), zµ ∶ [0, 1] ↦ M, of the form zµ = zµ (t),

t ∈ [0, 1]

with the boundary conditions

zµ (0) = xµ

and zµ (1) = yµ

(3.97)

(3.98)

By integrating the geodesic equation Eq.(3.95) along the path Γ we obtain ∫

Γ(x,y)

dzµ

∂φa (z) µ = ig ∫ dzµ Aab (z) φb (z) ∂zµ Γ(x,y)

(3.99)

Hence, we find that φ(x) must be the solution of the integral equation φ(y) = φ(x) + ig ∫

dzµ A (z)φ(z) µ

Γ(x,y)

(3.100)

where we have omitted all the indices to simplify the notation. In terms of the parametrization zµ (t) of the path Γ(x, y) we can write 1

φ(y) = φ(x) + ig ∫ dt 0

dzµ µ A (z(t)) φ (z(t)) dt

(3.101)

We will solve this equation by means of an iteration procedure, similar to what it is used for the evolution operator in quantum theory. By substituting

3.6 Non-abelian gauge invariance

71

repeatedly the l.h.s. of this equation into its r. h. s. we get the series dzµ (t) µ A (z(t)) φ(x) dt 0 1 t1 dzµ1 (t1 ) dzµ2 (t2 ) µ1 2 µ + (ig) ∫ dt1 ∫ dt2 A (z(t1 )) A 2 (z(t2 )) φ(x) dt1 dt2 0 0 n 1 t1 tn−1 dzµj (tj ) µj n + . . . + (ig) ∫ dt1 ∫ dt2 . . . ∫ dtn ∏ ( A (z(tj ))) φ(x) dtj 0 0 0 1

φ(y) = φ(x) + ig ∫ dt

j=1

(3.102)

+ ...

µ

Here we need to keep in mind that the A ’s are matrix-valued fields which are ordered from left to right!. The nested integrals of Eq.(3.102) can be written in the form In = (ig) ∫ dt1 ∫ 1

n

0

t1

0

dt2 . . . ∫

1 (ig) ̂ [(∫ dt F (t)) ] P n! 0 n

n

≡

tn−1 0

dtn F (t1 )⋯F (tn ) (3.103)

̂ means the path-ordered prodwhere the F ’s are matrices and the operator P uct of the objects sitting to its right. If we formally define the exponential of an operator to be equal to its power series expansion, ∞

1 n A n! n=0

e ≡∑ A

(3.104)

where A is some arbitrary matrix, we see that the geodesic equation has the formal solution ̂ [exp (+ig ∫ dt φ(y) = P

dzµ µ A (z(t)))] φ(x) dt

1

0

or, what is the same

̂ [exp (ig ∫ φ(y) = P

Γ(x,y)

dzµ A (z))] φ(x) µ

(3.105)

(3.106)

Thus, φ(y) is given by an operator, the path-ordered exponential of the line µ integral of the vector potential A , acting on φ(x). By expanding the exponential in a power series, it is easy to check that, under an arbitrary local gauge transformation U (z), the path ordered ex-

72

Classical Symmetries and Conservation Laws

ponential transforms as follows ̂ [ exp(ig ∫ P

dz Aµ (z(t)))] ′

µ

Γ(x,y)

≡ U (y) Pˆ [exp (ig ∫

dz Aµ (z))] U µ

Γ(x,y)

−1

(x)

(3.107)

In particular we can consider the case of a closed path Γ(x, x), where x is ̂Γ(x,x) an arbitrary point on Γ. The path-ordered exponential W ̂Γ(x,x) = P ̂ [exp (ig ∫ W

dz Aµ (z))] µ

Γ(x,x)

(3.108)

is not gauge invariant since, under a gauge transformation it transforms as ′ ̂Γ(x,x) W = Pˆ [exp (ig ∫

dz Aµ (z))] ′

µ

Γ(x,x)

= U (x) Pˆ [exp (ig ∫

dz Aµ (t))] U µ

Γ(x,x)

̂Γ(x,x) transforms like a group element, Therefore W

−1

(x)

̂Γ(x,x) = U (x) W ̂Γ(x,x) U −1 (x) W

̂Γ (x, x), which we denote by However, the trace of W ̂Γ(x,x) ≡ tr P ̂ [exp (ig ∫ WΓ = tr W

dz Aµ (z))] µ

Γ(x,x)

(3.109)

(3.110)

(3.111)

not only is gauge-invariant but it is also independent of the choice of the point x. However, it is a functional of the path Γ. The quantity WΓ , which is known as the Wilson loop, plays a crucial role in gauge theories. In the quantum theory this object will become the Wilson loop operator. Let us now consider the case of a small closed path Γ. If Γ is small, then the minimal area a(Γ) enclosed by Γ and its length +(Γ) are both infinitesimal. In this case, we can expand the exponential in powers and retain only the leading terms. We get ̂Γ ≈ I + ig P ̂ ∮ dz µ Aµ (z) + W Γ

2 (ig) ̂ (∮ dz µ Aµ (z)) + ⋯ P 2! Γ 2

(3.112)

Stokes theorem says that the first integral, the circulation of the vector field Aµ on the closed path Γ, is given by ∮ dzµ A (z) = ∫ ∫ dx ∧ dx µ

Γ

µ

Σ

ν

1 (∂ A − ∂ν Aµ ) 2 µ ν

(3.113)

3.6 Non-abelian gauge invariance

73 µ

ν

where ∂Σ = Γ is the infinitesimal area element bounded by Γ, and dx ∧ dx is the oriented infinitesimal area element. Furthermore, the quadratic term in Eq.(3.112) can be expressed as follows 2 1 1 ̂ (∮ dz µ Aµ (z)) ≡ ∫ ∫ dxµ ∧ dxν (−[Aµ , Aν ]) + . . . P 2 2! Γ Σ

(3.114)

Therefore, for an infinitesimally small loop, we get ̂Γ(x,x′ ) ≈ I + W

ig µ ν 2 ∫ ∫ dx ∧ dx Fµν + O(a(Σ) ) 2 Σ

(3.115)

where Fµν is the field tensor, defined by

Fµν ≡ ∂µ Aν − ∂ν Aµ − ig[Aµ , Aν ] = i [Dµ , Dν ]

(3.116)

Keep in mind that since the fields Aµ are matrices, the field tensor Fµν is also a matrix. Notice also that now Fµν is not gauge invariant. Indeed, under a local gauge transformation U (x), Fµν transforms as a similarity transformation Fµν (x) = U (x)Fµν (x)U ′

−1

(x)

(3.117)

This property follows from the transformation properties of Aµ . However, alµν though Fµν itself is not gauge invariant, other quantities such as tr(Fµν F ) are gauge invariant. Let us finally note the form of Fµν in components. By expanding Fµν in k the basis of the group generators λ (hence, in the algebra of the gauge group), k

k

(3.118)

Fµν = Fµν λ k

we find that the components, Fµν are k

k

k

Fµν = ∂µ Aν − ∂ν Aµ + gf

k$m

$

m

Aµ Aν

(3.119)

The natural local, gauge-invariant, theory for a non-abelian gauge group is the Yang-Mills Lagrangian 1 µν L = − trFµν F (3.120) 4 for a general compact Lie group G, that we will call the gauge group. Notice that the apparent similarity with the Maxwell Lagrangian is only superficial, since in this theory it is not quadratic in the vector potentials. We will see in later chapters that other Lagrangians are possible in other dimensions if some symmetry, e.g. time-reversal, is violated.

74

Classical Symmetries and Conservation Laws

3.7 Gauge invariance and minimal coupling We are now in a position to give a general prescription for the coupling of matter and gauge fields. Since the issue here is local gauge invariance, this prescription is valid for both relativistic and non-relativistic theories. So far, we have considered two cases: (a) fields that describe the dynamics of matter and (b) gauge fields that describe electromagnetism and chromodynamics. In our description of Maxwell’s electrodynamics we saw that, if the Lagrangian is required to respect local gauge invariance, then only conserved currents can couple to the gauge field. However, we have also seen that the presence of a global symmetry is a sufficient condition for the existence of a locally conserved current. This is not only a necessary condition since a local symmetry also requires the existence of a conserved current. We will now consider more general Lagrangians that will include both matter and gauge fields. In the last sections we saw that if a system with Lagrangian L(φ, ∂µ φ) has a global symmetry φ → U φ, then by replacing all derivatives by covariant derivatives we promote a global symmetry into a local (or gauge) symmetry. We will proceed with our general philosophy and write down gauge-invariant Lagrangians for systems which contain both matter and gauge fields. I will give a few explicit examples

3.7.1 Quantum electrodynamics Quantum Electrodynamics (QED) is a theory of electrons and photons. The electrons are described by Dirac spinor fields ψα (x). The reason for this choice will become clear when we discuss the quantum theory and the SpinStatistics theorem. Photons are described by a U (1) gauge field Aµ . The Lagrangian for free electrons is just the free Dirac Lagrangian, LDirac ¯ = ψ(i ¯ ∂/ − m)ψ LDirac(ψ, ψ)

(3.121)

The Lagrangian for the gauge field is the free Maxwell Lagrangian Lgauge (A)

1 1 2 µν Lgauge (A) = − Fµν F ≡ − F (3.122) 4 4 The prescription that we will adopt, known as minimal coupling, consists in requiring that the total Lagrangian be invariant under local gauge transformations. The free Dirac Lagrangian is invariant under the global phase transformation (i.e. with the same phase factor for all the Dirac components) ψα (x) → ψα (x) = e ψα (x) ′

iθ

(3.123)

3.7 Gauge invariance and minimal coupling

75

if θ is a constant, arbitrary phase, but it is not invariant not under the local phase transformation ψα (x) → ψα (x) = e ′

iθ(x)

ψα (x)

(3.124)

As we saw before, the matter part of the Lagrangian can be made invariant under the local transformations ψα (x) → ψα (x) = e

ψα (x) 1 ′ Aµ (x) → Aµ (x) = Aµ (x) + e ∂µ θ(x) ′

iθ(x)

(3.125)

if the derivative ∂µ ψ is replaced by the covariant derivative Dµ Dµ = ∂µ − ieAµ (x)

(3.126)

The total Lagrangian is now given by the sum of two terms ¯ A) + Lgauge (A) L = Lmatter (ψ, ψ,

(3.127)

¯ A) is the gauge-invariant extension of the Dirac Lawhere Lmatter (ψ, ψ, grangian, i.e. ¯ A) = ψ(i ¯ D / − m)ψ Lmatter (ψ, ψ, ¯ µ ψAµ = ψ¯ (i∂/ − m) ψ + eψγ

(3.128)

/ is a shorthand for Dµ γ µ . Thus, Lgauge (A) is the usual Maxwell term and D the total Lagrangian for QED is 1 ¯ D / − m)ψ − F 2 LQED = ψ(i (3.129) 4 Notice that now both matter and gauge fields are dynamical degrees of freedom. The QED Lagrangian has a local gauge invariance. Hence, it also has a locally conserved current. In fact the argument that we used above to show that there are conserved (Noether) currents if there is a continuous global symmetry, is also applicable to gauge invariant Lagrangians. As a matter of fact, under an arbitrary infinitesimal gauge transformation δψ = iθψ δψ¯ = −iθ ψ¯ δAµ = 1e ∂µ θ

(3.130)

the QED Lagrangian remains invariant, i.e. δL = 0. An arbitrary variation of L is δL =

δL δL ¯ + δL δ∂µ Aν + δL δAµ δψ + δ∂µ ψ + (ψ ↔ ψ) δψ δ∂µ ψ δ∂µ Aν δAµ

(3.131)

76

Classical Symmetries and Conservation Laws

After using the equations of motion and the form of the gauge transformation, δL can be written in the form 1 µν δL 1 µ δL = ∂µ [j (x)θ(x)] − e F (x)∂µ ∂ν θ(x) + ∂ θ(x) δAµ e µ

where j (x) is the electron number current

(3.132)

µ

j = i( µ

∂L δL ψ − ψ¯ ) δ∂µ ψ δ∂µ ψ¯

(3.133) µν

For smooth gauge transformations θ(x), the term F ∂µ ∂ν θ vanishes because of the antisymmetry of the field tensor Fµν . Hence we can write 1 δL µ µ ] δL = θ(x)∂µ j (x) + ∂µ θ(x) [j (x) + e δAµ (x)

(3.134)

The first term tells us that since the infinitesimal gauge transformation θ(x) µ µ is arbitrary, the Dirac current j (x) locally is conserved, i.e. ∂µ j = 0. µ Let us define the charge (or gauge) current J (x) by the relation J (x) ≡ µ

δL δAµ (x)

(3.135)

which is the current that enters in the Equation of Motion for the gauge field Aµ , i.e. the Maxwell equations. The vanishing of the second term of Eq. (3.134), required since the changes of the infinitesimal gauge transformations are also arbitrary, tells us that the charge current and the number current are related by ¯ µψ Jµ (x) = −ejµ (x) = −e ψγ (3.136)

This relation tells us that since j (x) is locally conserved, then the global conservation of Q0 µ

Q0 = ∫ d xj0 (x) ≡ ∫ d x ψ (x)ψ(x) 3

3

†

(3.137)

implies the global conservation of the electric charge Q Q ≡ −eQ0 = −e ∫ d x ψ (x)ψ(x) 3

†

(3.138)

This property justifies the interpretation of the coupling constant e as the electric charge. In particular the gauge transformation of Eq.(3.125), tells us that the matter field ψ(x) represents excitations that carry the unit of charge, ±e . From this point of view, the electric charge can be regarded as a quantum number. This point of view becomes very useful in the quantum

3.7 Gauge invariance and minimal coupling

77

theory in the strong coupling limit. In this case, under special circumstances, the excitations may acquire unusual quantum numbers. This is not the case of Quantum Electrodynamics, but it is the case of a number of theories in one and two space dimensions, with applications in Condensed Matter systems such as polyacetylene, or the two-dimensional electron gas in high magnetic fields, i.e. the fractional quantum Hall effect, or in gauge theories with magnetic monopoles). 3.7.2 Quantum chromodynamics Quantum Chromodynamics (QCD) is the gauge field theory of strong interactions in hadron physics. In this theory the elementary constituents of i hadrons, the quarks, are represented by the Dirac spinor field ψα (x). The a theory also contains a set of gauge fields Aµ (x) that represent the gluons. The quark fields have both Dirac indices α = 1, . . . , 4 and color indices i = 1, . . . , Nc , where Nc is the number of colors. In the Standard Model of weak, strong interactions, and electromagnetic in particle physics, and in QCD, there are in addition Nf = 6 flavors of quarks, grouped into three generations, each labeled by a flavor index, and six flavors of leptons, also grouped into three generations. The flavor symmetry is a global symmetry of the theory. Quarks are assumed to transform under the fundamental representation of the gauge (or color) group G, say SU (Nc ). The theory is invariant under the group of gauge transformations. In QCD, the color group is SU (3) and so Nc = 3. The color symmetry is a non-abelian gauge symmetry. The gauge field Aµ is needed in order to enforce local gauge invariance. In coma a a ponents, we get Aµ = Aµ λ , where λ are the generators of SU (Nc ). Thus, 2 2 a = 1, . . . , D(SU (Nc )), and D(SU (Nc ) = Nc − 1. Thus, Aµ is an Nc − 1 2 dimensional vector in the adjoint representation of G. For SU (3), Nc −1 = 8 and there are eight generators. The gauge-invariant matter term of the Lagrangian, Lmatter is ¯ A) = ψ(i ¯ D / − m)ψ Lmatter (ψ, ψ,

(3.139)

/ = ∂/ − igA / ≡ ∂/ − igA / λ is the covariant derivative. The gauge field where D term of the Lagrangian Lgauge a a

1 1 a µν µν Lgauge (A) = − trFµν F ≡ − Fµν Fa (3.140) 4 4 is the Yang-Mills Lagrangian. The total Lagrangian for QCD is LQCD ¯ A) + Lgauge (A) LQCD = Lmatter (ψ, ψ,

(3.141)

78

Classical Symmetries and Conservation Laws

Can we define a color charge? Since the color group is non-abelian it has more than one generator. We showed before that there are as many conserved currents as generators are in the group. Now, in general, the group generators do not commute with each other. For instance, in SU (2) there is only one diagonal generator, J3 , while in SU (3) there are only two diagonal a generators, etc. Can all the global charges Q Q ≡ ∫ d x ψ (x)λ ψ(x) 3

a

†

a

(3.142)

be defined simultaneously? It is straightforward to show that the Poisson Brackets of any pair of charges are, in general, different from zero. We will a see below, when we quantize the theory, that the charges Q obey the same commutation relations as the group generators themselves do. So, in the quantum theory, the only charges that can be assigned to states are precisely the same as the quantum numbers that label the representations. Thus, if the group is SU (2), we can only assign to the states the values of the quadratic 2 Casimir operator J and of the projection J3 . Similar restrictions apply to the case of SU (3) and to other Lie groups. 3.8 Spacetime symmetries and the energy-momentum tensor

Until now we have considered only the role of internal symmetries. We now turn to spacetime symmetries, and consider the role of coordinate transformations. In this more general setting we will have to require the invariance of the action rather than only of the Lagrangian, as we did for internal symmetries. There are three continuous spacetime symmetries that will be important to us: a) translation invariance, b) rotation invariance and c) homogeneity of time. While rotations are a subgroup of Lorentz transformation, space and time translations are examples of inhomogeneous Lorentz transformations (in the relativistic case) and of Galilean transformations (in the nonrelativistic case). Inhomogeneous Lorentz transformations also form a group, known as the Poincar´e group. Note that the transformations discussed above are particular cases of more general coordinate transformations. However, it is important to keep in mind that, in most cases, general coordinate transformations are not symmetries of an arbitrary system. They are the symmetries of General Relativity. In what follows we are going to consider the response of a system to infinitesimal coordinate transformations of the form ′

xµ = xµ + δxµ

(3.143)

3.8 Spacetime symmetries and the energy-momentum tensor

79

where δxµ may be a function of the spacetime point xµ . Under a coordinate transformation the fields change as φ(x) → φ (x ) = φ(x) + δφ(x) + ∂µ φ δx ′

′

µ

(3.144)

where δφ is the variation of φ in the absence of a change of coordinates , i.e. a functional change. In this notation, a uniform infinitesimal translation by a constant vector aµ has δxµ = aµ and an infinitesimal rotation of the space axes is δx0 = 0 and δxi = (ijk θj xk . In general, the action of the system is not invariant under arbitrary changes in both coordinates and fields. Indeed, under an arbitrary change ′ 4 of coordinates xµ → xµ (xµ ), the volume element d x is not invariant and changes by a multiplicative factor of the form 4 ′

4

(3.145)

d x =d xJ

where J is the Jacobian of the coordinate transformation ′ 8 8 ′ ′ ∂xµ 888 ∂x1 ⋯x4 888 8 J= ≡ det ( )8 ∂x1 ⋯x4 8888 ∂xj 8888 8 8

For an infinitesimal transformation, xµ = xµ + δxµ (x), we get

(3.146)

′

′

∂xµ ν ν = gµ + ∂ δxµ ∂xν

(3.147)

Since δxµ is small, the Jacobian can be approximated by

′ 8 88 ∂xµ 888 88 ν ν ν 2 8 J = 88det ( )888 = ∣det (gµ + ∂ δxµ )∣ ≈ 1 + tr(∂ δxµ ) + O(δx ) ∂x 88 ν 88 8 8

Thus

J ≈ 1 + ∂ δxµ + O(δx ) µ

2

(3.148)

(3.149)

The Lagrangian itself is in general not invariant. For instance, even though we will always be interested in systems whose Lagrangians are not an explicit function of x, still they are not in general invariant under the given transformation of coordinates. Also, under a coordinate change the fields may also transform. Thus, in general, δL does not vanish. The most general variation of L is µ

δL = ∂µ L δx +

δL δL δφ + δ∂ φ δφ δ∂µ φ µ

(3.150)

80

Classical Symmetries and Conservation Laws

If φ obeys the equations of motion, δL δL = ∂µ δφ δ∂µ φ

(3.151)

then, the general change δL obeyed by the solutions of the equations of motion is δL µ δL = δx ∂µ L + ∂µ ( δφ) (3.152) δ∂µ φ The total change in the action is a sum of two terms

δS = δ ∫ d x L = ∫ δd x L + ∫ d x δL 4

4

4

(3.153)

where the change in the integration measure is due to the Jacobian factor, 4

4

δd x = d x ∂µ δx

µ

(3.154)

Hence, δS is given by δS = ∫ d x [(∂µ δx ) L + δx ∂µ L + ∂µ ( 4

µ

µ

δL δφ)] δ∂µ φ

(3.155)

Since the total variation of φ, δT φ, is the sum of the functional change of the fields plus the changes in the fields caused by the coordinate transformation, δT φ ≡ δφ + ∂µ φδx

µ

(3.156)

we can write δS as a sum of two contributions: one due to change of coordinates and another due to functional changes of the fields: δS = ∫ d x [(∂µ δx ) L + δx ∂µ L + ∂µ ( 4

µ

µ

δL ν (δT φ − ∂ν φδx ))] (3.157) δ∂µ φ

Therefore, for the change of the action we get δS = ∫ d x {∂µ [(gν L − 4

µ

δL δL ν ∂ν φ) δx ] + ∂µ [ δ φ]} δ∂µ φ δ∂µ φ T

(3.158)

We have already encountered the second term when we discussed the case of internal symmetries. The first term represents the change of the action S as a result of a change of coordinates. We will now consider a few explicit examples. To simplify matters we will consider the effects of coordinate transformations alone. For simplicity, here we will restrict our discussion to the case of the scalar field.

3.8 Spacetime symmetries and the energy-momentum tensor

81

3.8.1 Spacetime translations Under a uniform infinitesimal translation δxµ = aµ , the field φ does not change δT φ = 0

(3.159)

The change of the action now is δS = ∫ d x ∂µ (g L − 4

µν

δL ν ν ∂ φ) a δ∂µ φ

(3.160)

For a system which is isolated and translationally invariant the action must not change under a redefinition of the origin of the coordinate system. Thus, µν δS = 0. Since aµ is arbitrary, it follows that the tensor T T

µν

µν

≡ −g L +

δL ν ∂ φ δ∂µ φ

(3.161)

is conserved, ∂µ T

µν

=0

(3.162)

µν

The tensor T is known as the energy-momentum tensor. The reason for µν this name is the following. Given that T is locally conserved, by Noether’s ν theorem we can define the 4-vector P P =∫ d xT 3

ν

0ν

(x, x0 )

(3.163)

0

which is a constant of motion. In particular, P is given by P = ∫ d x T (x, x0 ) ≡ ∫ d x [−L + 0

But

δL δ∂0 φ

3

00

3

δL 0 ∂ φ] δ∂0 φ

(3.164)

is just the canonical momentum Π(x), Π(x) =

δL δ∂0 φ

(3.165)

Then we can easily recognize that the quantity in brackets in Eq.(3.164) in 0 the definition of P is just the Hamiltonian density H 0

H =Π∂ φ−L

(3.166)

ν

Therefore, the time component of P is just the total energy of the system P =∫ d xH 0

3

(3.167)

82

Classical Symmetries and Conservation Laws j

The space components P are P =∫ d xT j

Thus, since g

0j

3

= 0, we get

0j

= ∫ d x [−g L + 3

0j

δL j ∂ φ] δ∂j φ

P = ∫ d x Π(x) ∂φ(x) 3

(3.168)

(3.169)

The vector P is identified with the total linear momentum since (a) it is a constant of motion, and (b) it is the generator of infinitesimal space trans0j lations. For the same reasons we will denote the component T (x) with the j linear momentum density P (x). It is important to stress that the canonical j momentum Π(x) and the total linear momentum density P are obviously completely different physical quantities. While the canonical momentum is a field which is canonically conjugate to the field φ, the total momentum is the linear momentum stored in the field, i.e. the linear momentum of the center of mass.

3.8.2 Rotations If the action is invariant under global infinitesimal Lorentz transformations, of which spacial rotations are a particular case, ν

(3.170)

δxµ = ωµ xν µν

where ω is infinitesimal, and antisymmetric, then the variation of the action is zero, δS = 0. If φ is a scalar field, then δT φ is also zero. This is not the case for spinor or vector fields which transforms under Lorentz transformations. Because of their transformation properties of these fields, the angular momentum tensor that we define below will be missing the contribution representing the spin of the field (which vanishes for a scalar field). Here we will consider only the case of scalars. Then, since for a scalar field δT φ = 0, we find δS = 0 = ∫ d x ∂µ [(g L − 4

νρ

µν

δL ν νρ ∂ φ) ω xρ ] δ∂µ φ

(3.171)

Since ω is constant and arbitrary, the magnitude in brackets must also define a conserved current. µνρ Let M be the tensor defined by M

µνρ

≡T

µν ρ

x −T

µρ ν

x

(3.172)

3.8 Spacetime symmetries and the energy-momentum tensor

83 µνρ

in terms of which, the quantity in brackets in Eq.(3.171) becomes 21 ωνρ M . µνρ Thus, for arbitrary constant ω, we find that the tensor M is locally conserved, ∂µ M

µνρ

=0

(3.173)

In particular, the transformation δx0 = 0,

δxj = ωjk xk

(3.174)

represents an infinitesimal rotation of the spatial axes with (3.175)

ωjk = (jkl θl

µνρ

where θl are three infinitesimal Euler angles. Thus, we suspect that M must be related with the total angular momentum. Indeed, the local conserµνρ vation of the current M leads to the global conservation of the tensorial νρ quantity L L

νρ

≡∫ d xM 3

0νρ

(x, x0 )

(3.176)

In particular, the space components of Lνρ are

Ljk = ∫ d x (T0j (x) xk − T0k (x) xj ) 3

= ∫ d x (Pj (x) xj − Pk (x) xj ) 3

(3.177)

If we denote by Lj the (pseudo) vector Lj ≡ we get

1 ( L 2 jkl kl

(3.178)

Lj ≡ ∫ d x (jkl xk Pl (x) ≡ ∫ d x +j (x) 3

3

(3.179)

The vector Lj is the generator of infinitesimal rotations and is thus identified with the total angular momentum, whereas +j (x) is the corresponding (spacial) angular momentum density. Notice that, since we are dealing with a scalar field, there is no spin contribution to the angular momentum density. νρ The generalized angular momentum tensor L of Eq.(3.176) is not translationally invariant since under a displacement of the origin of the coordinate µ νρ ν ρ ρ ν system by a , L changes by an amount a P − a P . A truly intrinsic anµ gular momentum is given by the Pauli-Lubanski vector W , νλ

1 µνλρ L P µ W =− ( √ 2 P2

ρ

(3.180)

84

Classical Symmetries and Conservation Laws

which, in the rest frame P = 0, reduces to the angular momentum. µνλ Finally, we find that if the angular momentum tensor M has the form M

µνλ

=T

µν λ

x −T

µλ ν

(3.181)

x

µν

Then, the conservation of the energy-momentum tensor T and of the anguµνλ lar momentum tensor M together lead to the condition that the energymomentum tensor should be a symmetric second rank tensor, T

µν

=T

µν

(3.182)

Thus, we conclude that the conservation of angular momentum requires that µν the energy momentum tensor T for a scalar field be symmetric. µν The expression for T that we derived in Eq.(3.161) is not manifestly µν µν symmetric. However, if T is conserved, then the “improved” tensor T̃ µν µν µνλ T̃ = T + ∂λ K

(3.183)

is also conserved, provided the tensor K is antisymmetric in (µ, λ) and µνλ µν (ν, λ). It is always possible to find such a tensor K to make T̃ symmetric. The improved, symmetric, energy-momentum tensor is known as the Belinfante energy-momentum tensor. In particular, for the scalar field φ(x) whose Lagrangian density L is µνλ

1 2 L = (∂µ φ) − V (φ) 2

the locally conserved energy-momentum tensor T T

µν

µν

= −g L +

(3.184) µν

is

δL ν µν µ ν ∂ φ ≡ −g L + ∂ φ∂ φ δ∂µ φ

(3.185)

which is symmetric. The conserved energy-momentum 4-vector is P = ∫ d x (−g L + ∂ φ∂ φ) µ

Thus, we find that

3

0µ

0

µ

1 2 1 0 3 3 2 P = ∫ d x (Π ∂0 φ − L) = ∫ d x [ Π + (*φ) + V (φ)] 2 2

(3.186)

(3.187)

is the total energy of the field, and

P = ∫ d x Π(x) *φ(x) 3

(3.188)

is the linear momentum P of the field. Both are constants of motion.

3.9 The energy-momentum tensor for the electromagnetic field

85

3.9 The energy-momentum tensor for the electromagnetic field For the case of the Maxwell field Aµ , a straightforward application of these µν methods yields an energy-momentum tensor T of the form T

µν

µν

= −g L +

δL µ ∂ Aλ δ∂ν Aλ

1 µν αβ νλ µ = g F Fαβ − F ∂ Aλ 4

(3.189)

µν

It obeys ∂µ T = 0 and, hence, is locally conserved. However, this tensor is not gauge invariant. We can construct a gauge-invariant and conserved energy momentum tensor by exploiting the ambiguity in the definition of µν T . Thus, if we choose Kµνλ = Fνλ Aµ , which is anti-symmetric in the indices ν and λ, we can construct the required gauge-invariant and conserved energy momentum-tensor µν T̃ =

1 µν 2 ν µλ g F − Fλ F (3.190) 4 where we used the equation of motion of the free electromagnetic field, λ ∂ Fνλ = 0. Notice that this “improved” energy-momentum tensor is both gauge-invariant and symmetric. From here we find that the 4-vector P =∫ µ

x0 fixed

3 µ0 d x T̃

(3.191)

is a constant of motion. Thus, we identify P =∫ 0

d x T̃ 3

x0 fixed

00

=∫

3

d x x0 fixed

with the Hamiltonian, and P =∫

d x T̃ 3

i

x0 fixed

i0

=∫

x0 fixed

1 2 2 (E + B ) 2 d x (E × B)i 3

(3.192)

(3.193)

with the linear momentum (or Poynting vector) of the electromagnetic field. 3.10 The energy-momentum tensor and changes in the geometry µν

The energy momentum tensor T appears in classical field theory as a result of the translation invariance, in both space and time, of the physical system. µν We have seen in the previous section that for a scalar field T is a symmetric tensor as a consequence of the conservation of angular momentum. The µν definition that we found does not require that T should have any definite µν symmetry. However we found that it is always possible to modify T by

86

Classical Symmetries and Conservation Laws

adding a suitably chosen antisymmetric conserved tensor to find a symmetric µν version of T . Given this fact, it is natural to ask if there is a way to define the energy-momentum tensor in a such a way that it is always symmetric. This issue becomes important if we want to consider theories for systems which contain fields which are not scalars. µν It turns out that it is possible to regard T as the change in the action due to a change of the geometry in which the system lives. From classical physics, we are familiar with the fact that when a body is distorted in some manner, in general its energy increases since we have to perform some work against the body in order to deform it. A deformation of a body is a change of the geometry in which its component parts evolve. Examples of such changes of geometry are shear distortions, dilatations, bendings, and twists. On the other hand, there are changes that do not cost any energy since they are symmetry operations. Examples of symmetry operations are translations and rotations. These symmetry operations can be viewed as simple changes in the coordinates of the parts of the body which do not change its geometrical properties, i.e. the distances and angles of different points. Thus, coordinate transformations do not alter the energy of the system. The same type of arguments apply to any dynamical system. In the most general case, we have to consider transformations which leave the action invariant. This leads us to consider how changes in the geometry of the spacetime affect the action of a dynamical system. The information about the geometry in which a system evolves is encoded in the metric tensor of the space (and spacetime). The metric tensor is a symmetric tensor that specifies how to measure the distance ∣ds∣ between a pair of nearby points x and x + dx, namely ds = g (x) dxµ dxν 2

µν

(3.194)

Under an arbitrary local change of coordinates xµ → xµ + δxµ , the metric tensor changes as follows gµν (x ) = gλρ (x) ′

′

λ

ρ

∂x ∂x ∂x′µ ∂x′ν

(3.195)

For an infinitesimal change, this means that the functional change in the metric tensor δgµν is 1 ν µ λ δgµν = − (gµλ ∂ δxλ + gλν ∂ δxλ + ∂ gµν δxλ ) 2

(3.196)

4 √ The volume element, invariant under coordinate transformations, is d x g, where g is the determinant of the metric tensor.

3.10 The energy-momentum tensor and changes in the geometry

87

Coordinate transformations change the metric of spacetime but do not change the action. Physical or geometric changes, are changes in the metric tensor which are not due to coordinate transformations. For a system µν in a space with metric tensor g (x), not necessarily the Minkowski (or Euclidean) metric, the change of the action is a linear function of the inµν finitesimal change in the metric δg (x) (i.e. “Hooke’s Law”). Thus, we can write the change of the action due to an arbitrary infinitesimal change of the metric in the form 4 √ µν δS = ∫ d x g T (x) δgµν (x)

(3.197) µν

Below we will identify the proportionality constant, the tensor T , with the µν conserved energy-momentum tensor. This definition implies that T can be regarded as the derivative of the action with respect to the metric T

µν

(x) ≡

δS δgµν (x)

(3.198)

Since the metric tensor is symmetric, this definition always yields a symmetric energy momentum tensor. In order to prove that this definition of the energy-momentum tensor agrees with the one we obtained before (which was not unique!) we have to µν prove that this form of T that we have just defined is a conserved current for coordinate transformations. Under an arbitrary local change of coordinates, which leave the distance ds unchanged, the metric tensor changes by the δgµν given above. The change of the action δS must be zero for this case. We see that if we substitute the expression for δgµν in δS, then an integration by parts will yield a conservation law. Indeed, for the particular case of a flat metric, such as the Minkowski or Euclidean metrics, the change δS is 1 4 µν λ λ δS = − ∫ d x T (x) (gµλ ∂ν δx + gλν ∂µ δx ) (3.199) 2 √ since for a global coordinate transformation the Jacobian factor g and 4 the measure of spacetime d x are constant. Thus, if δS is to vanish for µν an arbitrary change δx, the tensor T (x) has to be a locally conserved µν current, i.e. ∂µ T (x) = 0. This definition can be extended to the case of more general spaces. We should note, however, that the energy-momentum tensor can be made symmetric only if the space does not have a property known as torsion. Finally, let us remark that this definition of the energy-momentum tensor

88

Classical Symmetries and Conservation Laws µν

also allows us to identify the spacial components of T with the stressenergy tensor of the system. Indeed, for a change in geometry which is does not vary with time, the change of the action reduces to a change of the total energy of the system. Hence, the space components of the energy momentum tensor tell us how much does the total energy change for a specific deformation of the geometry. But this is precisely what the stress energy tensor is!

4 Canonical Quantization

We will now begin the discussion of our main subject of interest: the role of quantum mechanical fluctuations in systems with infinitely many degrees of freedom. We will begin with a brief overview of quantum mechanics of a single particle.

4.1 Elementary Quantum Mechanics Elementary Quantum Mechanics describes the quantum dynamics of systems with a finite number of degrees of freedom. Two axioms are involved in the standard procedure for quantizing a classical system. Let L(q, q) ˙ be the Lagrangian of an abstract dynamical system described by the generalized coordinate q. In Chapter two, we recalled that the canonical formalism of Classical Mechanics is based on the concept of canonical pairs of dynamical variables. So, the canonical coordinate q has for partner the canonical momentum p: p=

∂L ∂ q˙

(4.1)

In the canonical formalism, the dynamics of the system is governed by the classical Hamiltonian H(q, p) = pq˙ − L(q, q) ˙

(4.2)

which is the Legendre transform of the Lagrangian. In the canonical (Hamiltonian) formalism the equations of motion are just Hamilton’s Equations, p˙ = −

∂H ∂q

q˙ =

∂H ∂p

(4.3)

The dynamical state of the system is defined by the values of the canonical

92

Canonical Quantization

coordinates and momenta at any given time t. As a result of these definitions, the coordinates and momenta satisfy a set of Poisson Bracket relations

where

{q, p}P B = 1

{q, q}P B = {p, p}P B = 0

{A, B}P B ≡

∂A ∂B ∂A ∂B − ∂q ∂p ∂p ∂q

(4.4)

(4.5)

In Quantum Mechanics, the primitive (or fundamental) notion is the concept of a physical state. A physical state of a system is a represented by a state vector in an abstract vector space, which is called the Hilbert space H of quantum states. The space H is a vector space in the sense that if two vectors ∣Ψ⟩ ∈ H and ∣Φ⟩ ∈ H represent physical states, then the linear superposition ∣aΨ + bΦ⟩ = a∣Ψ⟩ + b∣Φ⟩, where a and b are two arbitrary complex numbers, also represents a physical state and thus it is an element of the Hilbert space i.e. ∣aΨ + bΦ⟩ ∈ H. The Superposition Principle is an axiom of Quantum Mechanics. In Quantum Mechanics, the dynamical variables, i.e. the generalized coordinates, qˆ, and the associated canonical momenta pˆ, the Hamiltonian H, etc., are represented by operators that act linearly on the Hilbert space of states. Hence, Quantum Mechanics is a linear theory, even though the physical observables obey non-linear Heisenberg equations of motion. Let us denote by Aˆ an arbitrary operator acting on the Hilbert space H. The result of acting on the state ∣Ψ⟩ ∈ H with the operator Aˆ is another state ∣Φ⟩ ∈ H, ˆ A∣Ψ⟩ = ∣Φ⟩

(4.6)

The Hilbert space H is endowed with an inner product. An inner product is an operation that assigns a complex number ⟨Φ∣Ψ⟩ to a given pair of states ∣Φ⟩ ∈ H and ∣Ψ⟩ ∈ H. Since H is a vector space, there exists a set of linearly independent states {∣λ⟩}, called a basis, that spans the entire Hilbert space. Thus, an arbitrary state ∣Ψ⟩ can be expanded as a linear combination of a complete set of states that form a basis of H, ∣Ψ⟩ = ∑ Ψλ ∣λ⟩

(4.7)

λ

which is unique for a fixed set of basis states. The basis states can be chosen to be orthonormal with respect to the inner product, ⟨λ∣µ⟩ = δλµ

(4.8)

4.1 Elementary Quantum Mechanics

In general, if ∣Ψ⟩ and ∣Φ⟩ are normalized states ⟨Ψ∣Ψ⟩ = ⟨Φ∣Φ⟩ = 1

93

(4.9)

the action of an operator Aˆ on a state ∣Ψ⟩ is proportional to a (generally different) state ∣Φ⟩, ˆ A∣Ψ⟩ = α∣Φ⟩ (4.10)

The coefficient α is a complex number that depends on the pair of states ˆ This coefficient is the matrix element of Aˆ between and on the operator A. the state ∣Ψ⟩ and ∣Φ⟩, which we write with the notation ˆ α = ⟨Φ∣A∣Ψ⟩

(4.11)

Operators that act on a Hilbert space do not generally commute with each other. One of the axioms of Quantum Mechanics is the Correspondence Principle which states that in the classical limit, h̵ → 0, the operators should effectively become numbers, and commute with each other in the classical limit. The procedure of canonical quantization consists in demanding that to the classical canonical pair (q, p), that satisfies the Poisson Bracket {q, p}P B = 1, we associate a pair of operators qˆ and pˆ, acting on the Hilbert space of states H, which obey the canonical commutation relations [ˆ q , pˆ] = ih̵

[ˆ q, qˆ] = [ˆ p, pˆ] = 0

ˆ B] ˆ is the commutator of the operators Aˆ and B, ˆ Here [A, ˆ B] ˆ = AˆB ˆ−B ˆ Aˆ [A,

(4.12)

(4.13)

In particular, two operators that do not commute with each other cannot be diagonalized simultaneously. Hence, it is not possible to measure simultaneously with arbitrary precision in the same physical state two non-commuting observables. This is the the Uncertainty Principle. By following this prescription, we assign to the classical Hamiltonian H(q, p), which is a function of the dynamical variables q and p, an operator ˆ q , pˆ) obtained by replacing the dynamical variables with the correspondH(ˆ ing operators. Other classical dynamical quantities are similarly associated in Quantum Mechanics with quantum operators that act on the Hilbert space of states. Moreover, all operators associated with classical physical quantities, in Quantum Mechanics are hermitian operators relative to the inner product defined in the Hilbert space H. Namely, if Aˆ is an operator † and Aˆ is the adjoint of Aˆ ˆ ⟨Ψ∣Aˆ ∣Φ⟩ ≡ ⟨Φ∣AΨ⟩ , †

∗

(4.14)

94

Canonical Quantization

† then, Aˆ is hermitian iff Aˆ = Aˆ (with suitable boundary conditions). The quantum mechanical state of the system at time t, ∣Ψ(t)⟩, obeys the Schr¨ odinger Equation

ih̵

∂ ˆ q , pˆ) ∣Ψ(t)⟩ ∣Φ(t)⟩ = H(ˆ ∂t

(4.15)

The state ∣Ψ(t)⟩ is uniquely determined by the initial state ∣Ψ(0)⟩. Thus, in Quantum Mechanics, just as in Classical Mechanics, the Hamiltonian is the generator of the (infinitesimal) time evolution of the state of a physical system. It is always possible to choose a basis in which a particular operator is diagonal. For instance, if the operator is the canonical coordinate qˆ, a possible set of basis states are labelled by q and are its eigenstates, i.e. qˆ∣q⟩ = q ∣q⟩

The basis states {∣q⟩} are orthonormal and complete, ⟨q∣q ⟩ = δ(q − q ),

Iˆ = ∫ dq ∣q⟩⟨q∣

∣Ψ⟩ = ∑ Ψ(q) ∣q⟩ ≡ ∫

+∞

′

′

(4.16)

(4.17)

A state vector ∣Ψ⟩ can be expanded in an arbitrary basis. If the basis of states is {∣q⟩}, the expansion is −∞

q

dq Ψ(q) ∣q⟩

(4.18)

where we used that the eigenvalues of the coordinate q are the real numbers. The coefficients Ψ(q) of this expansion Ψ(q) = ⟨q∣Ψ⟩,

(4.19)

i.e. the amplitude to find the system at coordinate q in this state, are the values of the wave function of the state ∣Ψ⟩ in the coordinate representation. Since the canonical momentum pˆ does not commute with qˆ, it is not diagonal in this representation. Just as in Classical Mechanics, in Quantum Mechanics too the momentum operator pˆ is the generator of infinitesimal displacements. Consider the states ∣q⟩ and exp(− h̵i aˆ p) ∣q⟩. It is easy to prove that the latter is the state ∣q + a⟩ since ∞

i 1 −ia n n ( ) pˆ ∣q⟩ qˆ exp ( − ̵ aˆ p) ∣q⟩ ≡ qˆ ∑ n! h̵ h n=0

̵ it is easy to show that Using the commutation relation [ˆ q, pˆ] = ih, ̵ p [ˆ q , pˆ ] = ihnˆ n

n−1

(4.20)

(4.21)

4.2 Canonical quantization in Field Theory

95

Hence, we can write

Thus,

i i qˆ exp (− ̵ aˆ p) ∣q⟩ = (q + a) exp (− ̵ aˆ p) ∣q⟩ h h i exp (− ̵ aˆ p) ∣q⟩ = ∣q + a⟩ h

(4.22)

(4.23)

This is a unitary transformation, i.e.

−1 † i i (exp (− ̵ aˆ p)) = (exp (− ̵ aˆ p)) h h

(4.24)

We can now use this property to compute the matrix element i ⟨q∣ exp ( ̵ aˆ p) ∣Ψ⟩ ≡ Ψ(q + a) h

(4.25)

For a infinitesimally small, it can be approximated by i Ψ(q + a) ≈ Ψ(q) + ̵ a⟨q∣ˆ p∣Ψ⟩ + . . . h We find that the matrix element for pˆ has to satisfy ⟨q∣ˆ p∣Ψ⟩ =

Ψ(q + a) − Ψ(q) h̵ lim a i a→0

Thus, the operator pˆ is represented by a differential operator h̵ ∂ h̵ ∂ Ψ(q) = ⟨q∣Ψ⟩ ⟨q∣ˆ p∣Ψ⟩ ≡ i ∂q i ∂q

(4.26)

(4.27)

(4.28)

It is easy to check that the coordinate representation of the operator h̵ ∂ pˆ = (4.29) i ∂q

̵ and the coordinate operator qˆ satisfy the commutation relation [ˆ q , pˆ] = ih. 4.2 Canonical quantization in Field Theory

We will now apply the axioms of Quantum Mechanics to a Classical Field Theory. The result will be a Quantum Field Theory. For the sake of simplicity we will consider first the case of a scalar field φ(x). We have seen before that, given a Lagrangian density L(φ, ∂µ φ), the Hamiltonian can be found once the canonical momentum Π(x) is defined, Π(x) =

δL δ∂0 φ(x)

(4.30)

96

Canonical Quantization

On a given time surface x0 , the classical Hamiltonian is H = ∫ d x [Π(x, x0 )∂0 φ(x, x0 ) − L(φ, ∂µ φ)] 3

(4.31)

We quantize this theory by assigning to each dynamical variable of the classical theory a hermitian operator that acts on the Hilbert space of the ˆ quantum states of the system. Thus, the field φ(x) and the canonical mô mentum Π(x) are operators acting on a Hilbert space. These operators obey canonical commutation relations ˆ ̵ ̂ [φ(x), Π(y)] = ihδ(x − y)

(4.32)

In the field representation, the Hilbert space is the vector space of wave functions Ψ which are functionals of the configurations of the field (at a fixed time), which we will denote as {φ(x)}. In this notation, the wave functionals, i.e. the amplitude to find the state of the field in a given configuration, are Ψ[{φ(x)}] ≡ ⟨{φ(x)}∣Ψ⟩

(4.33)

In this representation, the field is a diagonal operator ˆ ⟨{φ}∣φ(x)∣Ψ⟩ ≡ φ(x) ⟨{φ(x)}∣Ψ⟩ = φ(x) Ψ[{φ}]

(4.34)

̂ The canonical momentum Π(x) is not diagonal in this representation but acts on the wave functionals as a functional differential operator, ̂ ⟨{φ}∣Π(x)∣Ψ⟩ ≡

h̵ δ Ψ[{φ}] i δφ(x)

(4.35)

What we just described is the Schr¨odinger Picture of quantum field theory. In this picture, as usual, the operators are time-independent but the states are time-dependent and satisfy the Schr¨odinger Equation ̵ 0 Ψ[{φ}, x0 ] = HΨ[{φ}, ̂ ih∂ x0 ]

(4.36)

For the particular case of a scalar field φ with the classical Lagrangian L 1 2 L = (∂µ φ) − V (φ) 2

(4.37)

ˆ is the quantum mechanical Hamiltonian operator H 1 1 2 2 ˆ ̂ (x) + (%φ(x)) ̂ = ∫ d3 x { Π + V (φ(x))} H 2 2

(4.38)

̂ The stationary states are the eigenstates of the Hamiltonian H. While it is possible to proceed further with the Schr¨odinger picture, the

4.3 Quantization of the free scalar field theory

97

manipulation of wave functionals becomes very cumbersome rather quickly. For this reason, the Heisenberg Picture is commonly used. In the Schr¨ odinger Picture the time evolution of the system is encoded in the time dependence of the states. In contrast, in the Heisenberg Picture the operators are time-dependent while the states are time-independent. The operators of the Heisenberg Picture obey quantum mechanical equations of motion. Let Aˆ be some operator that acts on the Hilbert space of states. Let us denote by AˆH (x0 ) the Heisenberg operator at time x0 , defined by ˆ ˆ Hx − Hx AˆH (x0 ) = e h̵ 0 Aˆ e h̵ 0 i

i

(4.39)

̂ It is straightforward for a system with a time-independent Hamiltonian H. ˆ to check that AH (x0 ) obeys the equation of motion ̵ 0 AˆH (x0 ) = [AˆH (x0 ), H] ̂ ih∂

(4.40)

Notice that in the classical limit, the dynamical variable A(x0 ) obeys the classical equation of motion ∂0 A(x0 ) = {A(x0 ), H}P B

(4.41)

where it is assumed that all the time dependence in A comes from the time dependence of the field (the “coordinates”) and of their canonical momenta. ˆ ̂ In the Heisenberg picture both φ(x, x0 ) and Π(x, x0 ) are time-dependent operators that obey the quantum mechanical equations of motion ˆ ˆ ̵ 0 φ(x, ̂ , ih∂ x0 ) = [φ(x, x0 ), H]

̵ 0 Π(x, ̂ ̂ ̂ ih∂ x0 ) = [Π(x, x0 ), H]

(4.42)

ˆ ̂ The Heisenberg field operators φ(x, x0 ) and Π(x, x0 ) (we will omit the subindex “H” from now on) obey equal-time commutation relations ˆ ̵ ̂ [φ(x, x0 ), Π(y, x0 )] = ihδ(x − y)

(4.43)

4.3 Quantization of the free scalar field theory We will now quantize the theory of a relativistic scalar field φ(x). In particular, we consider a free real scalar field φ whose Lagrangian density is 1 1 2 2 µ L = (∂µ φ)(∂ φ) − m φ 2 2

(4.44)

̂ for a free real scalar field is The quantum mechanical Hamiltonian H 2 1 2 1 1 2 2 ˆ ̂ = ∫ d3 x [ Π ̂ (x) + (%φ(x)) H + m φˆ (x)] 2 2 2

(4.45)

98

Canonical Quantization

̂ satisfy the equal-time commutation relations (in units with where φˆ and Π h̵ = c = 1) ˆ ̂ [φ(x, x0 ), Π(y, x0 )] = iδ(x − y)

(4.46)

̂ are time dependent operators In the Heisenberg representation, φˆ and Π while the states are time independent. The field operators obey the equations of motion ̂ ̂ ̂ x0 ) = [Π(x, x0 ), H] i∂0 Π(x,

ˆ ˆ ̂ i∂0 φ(x, x0 ) = [φ(x, x0 ), H],

(4.47)

These are operator equations. After some algebra, we find that the Heisenberg equations of motion for the field and for the canonical momentum are ˆ ̂ ∂0 φ(x, x0 ) =Π(x, x0 )

ˆ ̂ ∂0 Π(x, x0 ) = % φ(x, x0 ) − m φ(x, x0 ) 2

2ˆ

(4.48) (4.49)

Upon substitution we derive the field equation for the scalar field operator 2 2 ˆ (∂ + m ) φ(x) =0

(4.50)

ˆ Thus, the field operators φ(x) satisfy the Klein-Gordon equation as their Heisenberg equation of motion.

4.3.1 Field expansions Let us solve the field equation of motion by a Fourier transform 3

d k ˆ ik⋅x ˆ φ(x) =∫ φ(k, x0 ) e (2π)3

(4.51)

ˆ x0 ) are the Fourier amplitudes of φ(x). ˆ where φ(k, By demanding that the ˆ ˆ x0 ) should satisfy φ(x) satisfies the Klein-Gordon equation, we find that φ(k, 2ˆ 2 2 ˆ ∂0 φ(k, x0 ) + (k + m )φ(k, x0 ) = 0

(4.52)

ˆ ˆ x0 ) must satisfy Also, since φ(x, x0 ) is a real hermitian field operator, φ(k, † ˆ φˆ (k, x0 ) = φ(−k, x0 )

(4.53)

ˆ x0 ) is trivial. Let us write φ(k, ˆ x0 ) as the sum The time dependence of φ(k, of two terms ˆ x0 ) = φˆ+ (k)eiω(k)x0 + φˆ− (k)e−iω(k)x0 φ(k,

(4.54)

4.3 Quantization of the free scalar field theory

99

† The operators φˆ+ (k) and φˆ+ (k) are not independent since the reality conˆ dition of the field φ(x, x0 ) implies that † φˆ+ (k) = φˆ− (−k)

† φˆ+ (k) = φˆ− (−k)

(4.55)

This expansion is a solution of the equation of motion, the Klein-Gordon equation, provided ω(k) is given by √ (4.56) ω(k) = k2 + m2

Let us define the operator a ˆ(k) and its adjoint a ˆ (k) by †

a ˆ(k) = 2ω(k)φˆ− (k)

† † a ˆ (k) = 2ω(k)φˆ− (k)

(4.57)

The operators a ˆ (k) and a ˆ(k) obey the (generalized) creation-annihilation operator algebra †

[ˆ a(k), a ˆ (k )] = (2π) 2ω(k) δ (k − k ) ′

†

3

3

′

In terms of the operators a ˆ (k) and a ˆ(k) field operator becomes

(4.58)

†

ˆ φ(x) =∫

3

d k −iω(k)x0 +ik⋅x † iω(k)x0 −ik⋅x [ˆ a(k)e +a ˆ (k)e ] 3 (2π) 2ω(k)

(4.59)

where we have chosen to normalize the operators in such a way that the 3 d k . phase space factor takes the Lorentz-invariant form 2ω(k) The canonical momentum can also be expanded in a similar way 3

d k −iω(k)x0 +ik⋅x † iω(k) iω(k)x0 −ik⋅x ω(k)[ˆ a(k)e −a ˆ (k)e e ] 3 (2π) 2ω(k) (4.60) Notice that, in both expansions, there are terms with positive and negative frequency, and that the terms with positive frequency have creation operators † a ˆ (k) while the terms with negative frequency have annihilation operators a ˆ(k). This observation motivates the notation

̂ Π(x) = −i ∫

ˆ φ(x) = φˆ+ (x) + φˆ− (x)

(4.61)

where the expansion of φˆ+ has only positive frequency terms and the expansion of φˆ− has only negative frequency terms. This decomposition will turn out to be very useful.

100

Canonical Quantization

4.3.2 The Hamiltonian and its spectrum

We will now write the Hamiltonian in terms of the operators a ˆ(k) and a ˆ (k). The result is †

̂=∫ H

3

d k 1 † † ω(k) (ˆ a(k)ˆ a (k) + a ˆ (k)ˆ a(k)) (2π)3 2ω(k) 2

(4.62)

This Hamiltonian needs to be normal-ordered relative to a ground state which we will now define. A:Vacuum state

Let ∣0⟩ be the state that is annihilated by all the operators a ˆ(k), a ˆ(k)∣0⟩ = 0

(4.63)

Relative to this state, which we will call the vacuum state, the Hamiltonian can be written on the form ̂ =∶ H ̂ ∶ +E0 H

(4.64)

̂ ∶ is normal ordered relative to the state ∣0⟩. In other words, in ∶ H ̂∶ where ∶ H all the destruction operators appear the right of all the creation operators. ̂ ∶ annihilates the vacuum state Therefore the normal-ordered operator ∶ H ̂ ∶ ∣0⟩ = 0 ∶H

(4.65)

The real number E0 is the ground state energy. In this case it is equal to 3 1 E0 = ∫ d k ω(k) δ(0) 2

(4.66)

when δ(0), the delta function at zero momentum, is the infrared divergent quantity 3

d x ip⋅x V δ(0) = lim δ (p) = lim ∫ e = 3 p→0 p→0 (2π) (2π)3 3

(4.67)

where V is the (infinite) volume of space. Thus, E0 is extensive and can be written as E0 = ε0 V , where ε0 is the ground state energy density. We find ε0 = ∫

d k √ 2 d k ω(k) 1 k + m2 ≡ ∫ 2 (2π)3 (2π)3 2 3

3

(4.68)

Eq. (4.68) is the sum of the zero-point energies of all the oscillators. This quantity is formally divergent since the integral is dominated by the contributions with large momentum or, what is the same, short distances. This is an ultraviolet divergence. It is divergent because the system has an infinite

4.3 Quantization of the free scalar field theory

101

number of degrees of freedom even in a finite volume. We will encounter other examples of similar divergencies in field theory. It is important to keep in mind that they are not artifacts of our scheme but that they result from the fact that the system is in continuous space-time and has an infinite number of degrees of freedom. We can take two different points of view with respect to this problem. One possibility is simply to say that the ground state energy is not a physically observable quantity since any experiment will only yield information on excitation energies and, in this theory, they are finite. Thus, we may simply redefine the zero of the energy by dropping this term off. Normal ordering is then just the mathematical statement that all energies are measured relative to that of the ground state. As far as free field theory is concerned, this subtraction is sufficient since it makes the theory finite without affecting any physically observable quantity. However, once interactions are considered, divergencies will show up in the formal computation of physical quantities. This procedure then requires further subtractions. An alternative approach consists in introducing a regulator or cutoff. The theory is now finite but one is left with the task of proving that the physics is independent of the cutoff procedure. This is the program of the Renormalization group. Although it is not presently known if there should be a fundamental cutoff in these theories, i.e. if there is a more fundamental description of Nature at short distances and high energies, such as it is postulated by String Theory, it is clear that if quantum field theories are to be regarded as effective hydrodynamic theories valid below some high energy scale, then a cutoff is actually natural. B: Hilbert space We can construct the spectrum of states by inspecting the normal-ordered Hamiltonian 3 d k † ̂ ∶ H ∶= ∫ ω(k) a ˆ (k)ˆ a(k) (4.69) 3 (2π) 2ω(k)

̂ This Hamiltonian commutes with the linear momentum P ̂=∫ P

x0 fixed

3 ˆ ̂ d x Π(x, x0 )%φ(x, x0 )

(4.70)

which, up to operator ordering ambiguities, is the quantum mechanical version of the classical linear momentum P , P =∫

3

d xT x0

0j

≡∫

d x Π(x, x0 )%φ(x, x0 ) 3

x0

(4.71)

102

Canonical Quantization

̂ becomes In Fourier space P ̂=∫ P

3

d k † ka ˆ (k)ˆ a(k) 3 (2π) 2ω(k)

(4.72)

ˆ ∶ also commutes with the occupation The normal-ordered Hamiltonian ∶ H numbers of the oscillators, n ˆ (k), defined by n ˆ (k) ≡ a ˆ (k)ˆ a(k) †

(4.73)

ˆ commute with each other, we can Since {ˆ n(k)} and the Hamiltonian H use a complete set of eigenstates of {ˆ n(k)} to span the Hilbert space. We will regard the excitations counted by n ˆ (k) as particles that have energy and momentum (in more general theories they will also have other quantum numbers). Their Hilbert space has an indefinite number of particles and it is called Fock space. The states {∣{n(k)}⟩} of the Fock space, defined by ∣{n(k}⟩ = ∏ N (k)[ˆ a (k)] †

n(k)

k

∣0⟩,

(4.74)

with N (k) being normalization constants, are eigenstates of the operator n ˆ (k) n ˆ (k)∣{n(k)}⟩ = (2π) 2ω(k)n(k)∣{n(k)⟩ 3

(4.75)

These states span the occupation number basis of the Fock space. ̂ The total number operator N ̂≡∫ N

3

d k n ˆ (k) (2π)3 2ω(k)

(4.76)

̂ and it is diagonal in this basis i.e. commutes with the Hamiltonian H ̂∣{n(k)}⟩ = ∫ d3 k n(k) ∣{n(k)}⟩ N

(4.77)

The energy of these states is

3 ̂ H∣{n(k)}⟩ = [∫ d k n(k)ω(k) + E0 ] ∣{n(k)}⟩

(4.78)

Thus, the excitation energy ε({n(k)}) of this state is ε(k) = ∫ d k n(k)ω(k). 3

(4.79)

̂ has an operator ordering ambiguThe total linear momentum operator P ity. It will be fixed by requiring that the vacuum state ∣0⟩ be translationally

4.3 Quantization of the free scalar field theory

invariant, i.e.

̂∣0⟩ = 0. P

103

(4.80)

In terms of creation and annihilation operators the total momentum operator is ̂=∫ P

3

d k kn ˆ (k) (2π)3 2ω(k)

̂ is diagonal in the basis ∣{n(k)}⟩ since Thus, P

̂∣{n(k)}⟩ = [∫ d3 k k n(k)] ∣{n(k)}⟩ P

(4.81)

(4.82)

The state with lowest energy, the vacuum state ∣0⟩ has n(k) = 0, for all k. Thus the vacuum state has zero momentum and it is translationally invariant. The states ∣k⟩, defined by ∣k⟩ ≡ a ˆ (k)∣0⟩ †

(4.83)

have excitation energy ω(k) and total linear momentum k. Thus, the states {∣k⟩} are particle-like excitations which have an energy dispersion curve √ (4.84) E = k 2 + m2 ,

characteristic of a relativistic particle of momentum k and mass m. Thus, the excitations of the ground state of this field theory are particle-like, and have positive energy (relative to the vacuum state). From our discussion we can see that these particles are free since their energies and momenta are additive.

4.3.3 Causality The starting point of the quantization procedure was to impose equal-time ˆ commutation relations among the canonical fields φ(x) and their canonî cal momenta Π(x). In particular, two field operators on different spatial locations commute at equal times. But, do they commute at different times? To address this question let us calculate the commutator ∆(x − y) ˆ ˆ i∆(x − y) = [φ(x), φ(y)]

(4.85)

ˆ ˆ where φ(x) and φ(y) are Heisenberg field operators for space-time points x

104

Canonical Quantization

and y respectively. From the Fourier expansion of the fields, we know that the field operator can be split into a sum of two terms ˆ φ(x) = φˆ+ (x) + φˆ− (x)

(4.86)

where φˆ+ contains only creation operators and positive frequencies, and φˆ− contains only annihilation operators and negative frequencies. Thus the commutator is i∆(x − y) =[φˆ+ (x), φˆ+ (y)] + [φˆ− (x), φˆ− (y)] +[φˆ+ (x), φˆ− (y)] + [φˆ− (x), φˆ+ (y)]

(4.87)

The first two terms always vanish since the φˆ+ operators commute among themselves and so do the operators φˆ− . Thus, the only non-vanishing contributions are i∆(x − y) =[φˆ+ (x), φˆ− (y)] + [φˆ− (x), φˆ+ (y)]

′ † ′ ′ ′ = ∫ dk¯ ∫ dk¯ {[ˆ a (k), a ˆ(k )] exp (−iω(k)x0 + ik ⋅ x + iω(k )y0 − ik ⋅ y)

+[ˆ a(k), a ˆ (k )] exp (iω(k)x0 − ik ⋅ x − iω(k ), y0 + ik ⋅ y) } †

′

′

′

(4.88)

where ∫ dk¯ ≡ ∫

3

d k (2π)3 2ω(k)

(4.89)

Furthermore, using the commutation relations of the creation and annihilation operators we find that the operator ∆(x − y) is proportional to the identity operator, and hence, it is actually a function. It is given by iω(k)(x0 −y0 )−ik⋅(x−y) −iω(k)(x0 −y0 )+ik⋅(x−y) i∆(x − y) = ∫ dk¯ [e −e ] (4.90)

With the help of the Lorentz-invariant function )(k ), defined by 0

0

k 0 )(k ) = 0 ≡ sign(k ) ∣k ∣ 0

(4.91)

we can write ∆(x − y) in the manifestly Lorentz invariant form 4

d k 2 2 0 −ik⋅(x−y) δ(k − m ))(k )e i∆(x − y) = ∫ 3 (2π)

(4.92)

The integrand of Eq.(4.92) vanishes unless the mass shell condition 2

2

k −m =0

(4.93)

4.3 Quantization of the free scalar field theory

105

is satisfied. Notice that ∆(x − y) satisfies the initial condition ∂0 ∆∣x0 =y0 = −δ (x − y) 3

(4.94)

At equal times x0 = y0 the commutator vanishes, ∆(x − y, 0) = 0

(4.95)

Furthermore, by Lorentz invariance, ∆(x − y) also vanishes if the space-time 2 points x and y are separated by a space-like interval, (x − y) < 0. This must be the case since ∆(x−y) is manifestly Lorentz invariant. Thus if it vanishes 2 2 2 2 2 at equal times, where (x − y) = (x0 − y0 ) − (x − y) = −(x − y ) < 0, it 2 must vanish for all events with the negative values of (x − y) . This implies that, for events x and y which are not causally connected ∆(x − y) = 0, and that ∆(x − y) is non-zero only for causally connected events, i.e. in the forward light-cone shown in Fig.4.1. time forward light cone

x20 − x2 > 0 x20 − x2 < 0

x20 − x2 < 0

space

x20 − x2 > 0

backward light cone

Figure 4.1 The Minkowski space-time.

106

Canonical Quantization

4.4 Symmetries of the quantum theory In our discussion of Classical Field Theory we discovered that the presence of continuous global symmetries implied the existence of constants of motion. In addition, the constants of motion were the generators of infinitesimal symmetry transformations. We will now see what role symmetries play in the quantized theory. In the quantized theory all physical quantities are represented by operators that act on the Hilbert space of states. The classical statement that a quantity A is conserved if its Poisson Bracket with the Hamiltonian vanishes dA = {A, H}P B = 0 dt in the quantum theory becomes the operator identity

(4.96)

dÂH ̂ =0 = [Â H , H] (4.97) dt we are are using the Heisenberg representation. Then, the constants of motion of the quantum theory are operators that commute with the Hamiltô = 0. nian, [Â H , H] Therefore, the quantum theory has a symmetry if and only if the charge ̂ Q, which is a hermitian operator associated with a classically conserved µ current j (x) via the correspondence principle, i

̂=∫ Q

d x ˆj (x, x0 ) 3

x0 fixed

0

(4.98)

̂ is an operator that commutes with the Hamiltonian H ̂, H] ̂ =0 [Q

(4.99)

̂ constitute a representation of the generators of If this is so, the charges Q the algebra of the Lie group of the symmetry transformations in the Hilbert ̂(α) associated with the symmetry space of the theory. The transformations U ̂(α) = exp(iαQ ̂) U

(4.100)

are unitary transformations that act on the Hilbert space of the system. For instance, we saw that for a translationally invariant system the clasµ sical energy-momentum four-vector P P =∫

3

µ

d xT

0µ

(4.101)

x0 0

is conserved. In the quantum theory P becomes the Hamiltonian operator ̂. In the case of a free ̂ and P i becomes the total momentum operator P H,

4.4 Symmetries of the quantum theory

107

scalar field, we saw before that these operators commute with each other, ̂, H] ̂ = 0. Thus, the eigenstates of the system have well defined total [P energy and total momentum. Since P is the generator of infinitesimal translations of the classical theory, it is easy to check that its equal-time Poisson Bracket with the field φ(x) is {φ(x, x0 ), P }P B = ∂x φ j

j

(4.102)

In the quantum theory the equivalent statement is that the field operator ˆ ̂ satisfy the equal-time commutaφ(x) and the total momentum operator P tion relation ˆ ˆ ̂j ] = i∂xj φ(x, [φ(x, x0 ), P x0 ) (4.103) ˆ + a, x0 ) and φ(x, ˆ Consequently, the field operators φ(x x0 ) are related by ̂ −ia⋅P ˆ + a, x0 ) = eia⋅P̂ φ(x, ˆ φ(x x0 )e

(4.104)

Translation invariance of the ground state ∣0⟩ implies that it is a state with ̂∣0⟩ = 0. For a finite displacement a we get zero total momentum, P ̂ ia⋅P

e

∣0⟩ = ∣0⟩

(4.105)

which states that the state ∣0⟩ is invariant under translations, and belongs to a one-dimensional representation of the group of global translations. Let us discuss now what happens to global internal symmetries. The simplest case is the free complex scalar field φ(x) whose Lagrangian L is invariant under global phase transformations. If φ is a complex field, it can decomposed it into its real and imaginary parts 1 φ = √ (φ1 + iφ2 ) 2

(4.106)

The classical Lagrangian for a free complex scalar field φ is ∗ µ

2 ∗

L = ∂µ φ ∂ φ − m φ φ

(4.107)

now splits into a sum of two independent terms

L(φ) = L(φ1 ) + L(φ2 )

(4.108)

where L(φ1 ) and L(φ2 ) are the Lagrangians for the free scalar real fields φ1 ∗ and φ2 . Likewise, the canonical momenta Π(x) and Π (x) are decomposed into Π(x) =

1 δL = √ (φ˙ 1 − iφ˙ 2 ) δ∂0 φ 2

1 ∗ Π (x) = √ (φ˙ 1 + iφ˙ 2 ) 2

(4.109)

108

Canonical Quantization

† In the quantum theory the operators φˆ and φˆ are no longer equal to each ̂ and Π ̂ † . Still, the canonical quantization procedure other, and neither are Π ̂ (and φˆ† and Π ̂ † ) satisfy the equal-time canonical tells us that φˆ and Π commutation relations 3 ˆ x0 ), Π(y, ̂ [φ(x, x0 )] = iδ (x − y)

(4.110)

The theory of the free complex scalar field is solvable by the same methods that we used for a free real scalar field. Instead of a single creation annihilation algebra we must introduce now two algebras, with operators a ˆ1 (k) and † † ˆ a ˆ1 (k), and a ˆ2 (k) and a ˆ2 (k). Let a ˆ(k) and b(k) be defined by 1 1 † † † a ˆ(k) = √ (ˆ a1 (k) + iˆ a2 (k)) , a ˆ (k) = √ (ˆ a1 (k) − iˆ a2 (k)) 2 2 1 1 † † † ˆb(k) = √ (ˆ a1 (k) − iˆ a2 (k)) , ˆb (k) = √ (ˆ a1 (k) + iˆ a2 (k)) 2 2

(4.111)

which satisfy the algebra

† ′ † ′ 3 3 ′ [ˆ a(k), a ˆ (k )] = [ˆb(k), ˆb (k )] = (2π) 2ω(k) δ (k − k )

(4.112)

while all other commutators vanish. The Fourier expansion for the fields now is ˆ φ(x) =∫ † φˆ (x) = ∫

3

d k −ik⋅x † ik⋅x (ˆ a(k)e + ˆb (k)e ) 3 (2π) 2ω(k) 3

d k −ik⋅x † ik⋅x (ˆb(k)e +a ˆ (k)e ) (2π)3 2ω(k)

(4.113) √ where ω(k) = k2 + m2 and k0 = ω(k). In this notation, the normal-ordered Hamiltonian is ̂ ∶= ∫ ∶H

3

d k † † ω(k) (ˆ a (k)ˆ a(k) + ˆb (k)ˆb(k)) 3 (2π) 2ω(k)

(4.114)

̂ is given by the similar exThe normal-ordered total linear momentum P pression ̂=∫ P

3

d k † † k (ˆ a (k)ˆ a(k) + ˆb (k)ˆb(k)) 3 (2π) 2ω(k)

(4.115)

We see that there are two types of quanta, a and b. The field φˆ creates b-quanta and it destroys a-quanta. The vacuum state has no quanta and is annihilated by both operators, a ˆ(k)∣0⟩ = ˆb(k)∣0⟩ = 0. The one-particle

4.4 Symmetries of the quantum theory

109

† † states now have a two-fold degeneracy since the states a ˆ (k)∣0⟩ and ˆb (k)∣0⟩ have one particle of type a and one of type b respectively. These states have exactly the same energy, ω(k), and the same momentum k. Thus, for each value of the energy and of momentum, we have a two dimensional space of possible states. This degeneracy is a consequence of the symmetry: the states form multiplets. What is the quantum operator that generates this symmetry? The classically conserved current is ∗ ↔

jµ = iφ ∂µ φ

(4.116)

In the quantum theory jµ becomes the normal-ordered operator ∶ ˆjµ ∶. The ̂ is corresponding global charge Q ˆ ∶ ̂ = ∶ ∫ d3 x i (φˆ† ∂0 φˆ − (∂0 φˆ† )φ) Q 3

d k † † =∫ (ˆ a (k)ˆ a(k) − ˆb (k)ˆb(k)) 3 (2π) 2ω(k) ˆ ˆb =Na − N

(4.117)

ˆa and N ˆb are the number operators for quanta of type a and b where N ˆa − N ˆb is conserved. Since ̂, H] ̂ = 0, the difference N respectively. Since [Q this property is consequence of a symmetry, it is expected to hold in more general theories than the simple free-field case that we are discussing here, ˆa and N ˆb in general may not be ̂, H] ̂ = 0. Thus, although N provided that [Q ˆ ˆ conserved separately, the difference Na −Nb will be conserved if the symmetry is exact. Let us now briefly discuss how is this symmetry realized in the spectrum of states.

4.4.1 The vacuum state ̂ annihilates the The vacuum state has Na = Nb = 0. Thus, the generator Q vacuum ̂∣0⟩ = 0 Q

(4.118)

Therefore, the vacuum state is invariant (i.e. is a singlet) under the symmetry, ∣0⟩ = e ′

̂ iQα

∣0⟩ = ∣0⟩

(4.119)

110

Canonical Quantization

Because the state ∣0⟩ is always defined up to an overall phase factor, it spans a one-dimensional subspace of states which are invariant under the symmetry. This is the vacuum sector and, for this problem, is trivial. 4.4.2 One-particle states

There are two linearly-independent one-particle states, ∣+, k⟩ and ∣−, k⟩ defined by † † ∣+, k⟩ = a ˆ (k) ∣0⟩ ∣−, k⟩ = ˆb (k) ∣0⟩ (4.120)

̂-quantum Both states have the same momentum k and energy ω(k). The Q numbers of these states, which we will refer to as their charge, are

Hence

† † ˆa − N ˆb )ˆ ˆa a ̂∣+, k⟩ = (N Q a (k)∣0⟩ = N ˆ (k)∣0⟩ = +∣+, k⟩ ˆa − N ˆb ) ˆb† (k)∣0⟩ = −∣−, k⟩ ̂∣−, k⟩ = (N Q

̂ k⟩ = σ ∣σ, k⟩ Q∣σ,

(4.121)

(4.122)

† † where σ = ±1. Thus, the state a ˆ (k)∣0⟩ has positive charge while ˆb (k)∣0⟩ has negative charge. ̂(α) = exp(iαQ ̂) the states ∣±, k⟩ transUnder a finite transformation U form as follows

ˆ ∣+, k⟩ = e ∣+, k⟩ ̂(α) ∣+, k⟩ = exp(iαQ) ∣+, k⟩ = U ′ −iα ˆ ̂(α) ∣−, k⟩ = exp(iαQ)∣−, ∣−, k⟩ = U k⟩ = e ∣−, k⟩ ′

iα

(4.123)

ˆ The field φ(x) itself transforms as

since

′ ˆ ˆ ̂) φ(x) ̂) = eiα φ(x) φˆ (x) = exp(−iαQ exp(iαQ

ˆ ˆ ̂, φ(x)] [Q = −φ(x),

̂, φˆ† (x)] = φˆ† (x) [Q

(4.124)

(4.125)

Thus the one-particle states are doubly degenerate, and each state transforms non-trivially under the symmetry group. ˆ we see By inspection of the Fourier expansion for the complex field φ, that φˆ is a sum of two terms: a set of positive frequency terms, symbolized by φˆ+ , and a set of negative frequency terms, φˆ− . In this case all positive frequency terms create particles of type b (which carry negative charge) while

4.4 Symmetries of the quantum theory

111

the negative frequency terms annihilate particles of type a (which carry positive charge). The states ∣±, k⟩ are commonly referred to as particles and antiparticles: particles have rest mass m, momentum k and charge +1 while the antiparticles have the same mass and momentum but carry charge −1. This charge is measured in units of the electromagnetic charge −e (see the previous discussion on the gauge current). We finally note that this theory contains an additional operator, the ̂, which maps particles into antiparticles and charge conjugation operator C ̂, H] ̂ = 0. This vice versa. This operator commutes with the Hamiltonian, [C property insures that the spectrum is invariant under charge conjugation. In other words, for every state of charge Q there exists a state with charge −Q, all other quantum numbers being the same. Our analysis of the free complex scalar field can be easily extended to systems which are invariant under a more general symmetry group G. In all cases the classically conserved charges become operators of the quantum ̂a as generators are in theory. Thus, there are as many charge operators Q the group. The charge operators represent the generators of the group in the Hilbert (or Fock) space of the system. The charge operators obey the same commutation relations as the generators themselves do. A simple generalization of the arguments that we have used here tells us that the states of the spectrum of the theory must transform under the irreducible representations of the symmetry group. However, there is one important caveat that should be made. Our discussion of the free complex scalar field shows us that, in that case, the ground state is invariant under the symmetry. In general, the only possible invariant state is the singlet state. All other states are not invariant and transform non-trivially. But, should the ground state always be invariant? In elementary quantum mechanics there is a theorem, due to Wigner and Weyl, which states that for a finite system, the ground state is always a singlet under the action of the symmetry group. However, there are many systems in Nature, such as magnets, Higgs phases, superconductors, and many others, which have ground states which are not invariant under the symmetries of the Hamiltonian. This phenomenon, known as spontaneous symmetry breaking, does not occur in simple free field theories but it does happen in interacting field theories. We will return to this important question in chapter 12.

5 Path Integrals in Quantum Mechanics and Quantum Field Theory

In chapter 4 we discussed the Hilbert space picture of Quantum Mechanics and Quantum Field Theory for the case of a free relativistic scalar fields. Here we will present the Path Integral picture of Quantum Mechanics and of relativistic scalar field theories. The Path Integral picture is important for two reasons. First, it offers an alternative, complementary, picture of Quantum Mechanics in which the role of the classical limit is apparent. Secondly, it gives a direct route to the study regimes where perturbation theory is either inadequate or fails completely. In Quantum mechanics a standard approach to such problems is the WKB approximation, of Wentzel, Kramers and Brillouin. However, as it happens, it is extremely difficult (if not impossible) the generalize the WKB approximation to a Quantum Field Theory. Instead, the non-perturbative treatment of the Feynman path integral, which in Quantum Mechanics is equivalent to WKB, is generalizable to non-perturbative problems in Quantum Field Theory. In this chapter we will use path integrals only for bosonic systems, such as scalar fields. In subsequent chapters we will also give a full treatment of the path integral, including its applications to fermionic fields, abelian and non-abelian gauge fields, classical statistical mechanics, and non-relativistic many body systems. There is a huge literature on path integrals, going back to the original papers by Dirac (Dirac, 1933), and particularly Feynman’s 1942 PhD Thesis (Feynman, 2005), and his review paper (Feynman, 1948). Popular textbooks on path integrals include the classic by Feynman and Hibbs (Feynman and Hibbs, 1965), and Schulman’s book (Schulman, 1981), among many others.

5.1 Path Integrals and Quantum Mechanics

115

5.1 Path Integrals and Quantum Mechanics Consider a simple quantum mechanical system whose dynamics can be described by a generalized coordinate operator qˆ. We want to compute the amplitude F (qf , tf ∣qi , ti ) = ⟨qf , tf ∣qi , ti ⟩

(5.1)

known as the Wightman function. This function represents the amplitude to find the system at coordinate qf at the final time tf knowing that it was at coordinate qi at the initial time ti . The amplitude F (qf , tf ∣qi , ti ) is just a matrix element of the evolution operator ˆ i −tf )/h̵ iH(t

F (qf , tf ∣qi , ti ) = ⟨qf ∣e

∣qi ⟩

(5.2)

Let us set, for simplicity, ∣qi , ti ⟩ = ∣0, 0⟩ and ∣qf , tf ⟩ = ∣q, t⟩. Then, from the definition of this matrix element, we find out that it obeys lim F (q, t∣0, 0) = ⟨q∣0⟩ = δ(q) t→0

(5.3)

Furthermore, after some algebra we also find that ih̵

ˆ h̵ ∂F ∂ ∂ −iHt/ =ih̵ ⟨q, t∣0, 0⟩ = ih̵ ⟨q∣e ∣0⟩ ∂t ∂t ∂t ˆ h̵ ˆ −iHt/ =⟨q∣He ∣0⟩

ˆ ⟩⟨q ∣e = ∫ dq ⟨q∣H∣q ′

′

′

ˆ h̵ −iHt/

∣0⟩

(5.4)

where we have used that, since {∣q⟩} is a complete set of states, the identity operator I has the expansion, called the resolution of the identity, I = ∫ dq ∣q ⟩⟨q ∣ ′

′

′

(5.5)

Here we have assumed that the states are orthonormal, ⟨q∣q ⟩ = δ(q − q ) ′

Hence, ih̵

′

∂ ′ ˆ ′ ⟩F (q ′ , t∣0, 0) ≡ H ˆ q F (q, t∣0, 0) F (q, t∣0, 0) = ∫ dq ⟨q∣H∣q ∂t

(5.6)

(5.7)

In other words, F (q, t∣0, 0) is the solution of the Schr¨odinger Equation that satisfies the initial condition of Eq. (5.3). For this reason, the amplitude F (q, t∣0, 0) is called the Schr¨odinger Propagator.

116

Path Integrals in Quantum Mechanics and Quantum Field Theory

(qf , tf )

qf

q

(q , t ) ′

′

qi

(qi , ti )

ti

t

′

′

tf

t

Figure 5.1 The amplitude to go from ∣qi , ti ⟩ to ∣qf , tf ⟩ is a sum of products ′ ′ of amplitudes through the intermediate states ∣q , t ⟩.

The superposition principle tells us that the amplitude to find the system in the final state at the final time is the sum of amplitudes of the form F (qf , tf ∣qi , ti ) = ∫ dq ⟨qf , tf ∣q , t ⟩⟨q , t ∣qi , ti ⟩ ′

′

′

′

′

(5.8) ′

where the system is in an arbitrary set of states at an intermediate time t . Here we represented this situation by inserting the identity operator I at ′ the intermediate time t in the form of the resolution of the identity of Eq. (5.8). Let us next define a partition of the time interval [ti , tf ] into N subintervals each of length ∆t, tf − ti = N ∆t

(5.9)

ti = t0 ≤ t1 ≤ . . . ≤ tN ≤ tN +1 = tf

(5.10)

Let {tj }, with j = 0, . . . , N + 1, denote a set of points in the interval [ti , tf ], such that Clearly, tk = t0 + k∆t, for k = 1, . . . , N + 1. By repeating the procedure used in Eq.(5.8) of inserting the resolution of the identity at the intermediate times {tk }, we find F (qf , tf ∣qi , ti ) = ∫ dq1 . . . dqN ⟨qf , tf ∣qN , tN ⟩⟨qN , tN ∣qN −1 , tN −1 ⟩ × . . .

× . . . ⟨qj , tj ∣qj−1 , tj−1 ⟩ . . . ⟨q1 , t1 ∣qi , ti ⟩ (5.11)

5.1 Path Integrals and Quantum Mechanics

Each factor ⟨qj , tj ∣qj−1 , tj−1 ⟩ in Eq.(5.11) has the form ˆ j −tj−1 )/h̵ −iH(t

⟨qj , tj ∣qj−1 , tj−1 ⟩ = ⟨qj ∣e

∣qj−1 ⟩ ≡ ⟨qj ∣e

ˆ −iH∆t/ h̵

117

∣qj−1 ⟩

(5.12)

In the limit N → ∞, with ∣tf − ti ∣ fixed and finite, the interval ∆t becomes infinitesimally small and ∆t → 0. Hence, as N → ∞ we can approximate the expression for ⟨qj , tj ∣qj−1 , tj−1 ⟩ in Eq.(5.12) as follows ⟨qj , tj ∣qj−1 , tj−1 ⟩ = ⟨qj ∣e

∣qj−1 ⟩ ∆t ˆ 2 = ⟨qj ∣{I − i ̵ H + O((∆t) )}∣qj−1 ⟩ h ∆t ˆ j−1 ⟩ + O((∆t)2 ) = δ(qj − qj−1 ) − i ̵ ⟨qj ∣H∣q h −iH∆t/h̵

(5.13)

which becomes asymptotically exact as N → ∞. q(t) qf

t

qi ti

tf

∆t

Figure 5.2 A history q(t) of the system.

We can also introduce at each intermediate time tj a complete set of momentum eigenstates {∣p⟩} using their resolution of the identity I=∫

∞

−∞

dp ∣p⟩⟨p∣

Recall that the overlap between the states ∣q⟩ and ∣p⟩ is ⟨q∣p⟩ = √

1

2π h̵ For a typical Hamiltonian of the form

ipq/h̵ e

(5.14)

(5.15)

2

ˆ = pˆ + V (ˆ q) H 2m

(5.16)

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Path Integrals in Quantum Mechanics and Quantum Field Theory

its matrix elements are ˆ j−1⟩ = ∫ ⟨qj ∣H∣q

∞ −∞

2

pj dpj ipj (qj −qj−1 )/h̵ e [ + V (qj )] ̵ 2m 2π h

(5.17)

Within the same level of approximation we can also write

pj dpj qj + qj−1 [i ( ))] (p exp (q − q ) − ∆t H , j j−1 j 2 2π h̵ h̵ (5.18) where we have introduce the “mid-point rule” which amounts to the replacement qj → 12 (qj + qj−1 ) inside the Hamiltonian H(p, q). Putting everything together we find that the matrix element ⟨qf , tf ∣qi , ti ⟩ becomes ⟨qj , tj ∣qj−1 , tj−1 ⟩ ≈ ∫

⟨qf , tf ∣qi , ti ⟩ = lim ∫ ∏ dqj ∫ N

N →∞

∞ N +1

∏

−∞ j=1

dpj 2π h̵

⎧ ⎫ N +1 ⎪ qj + qj−1 ⎪ ⎪ ⎪ i ⎪ ⎪ )] [p (p exp ⎨ ⎬ ∑ j (qj − qj−1 ) − ∆t H j , ⎪ ⎪ ̵ 2 h ⎪ ⎪ ⎪ ⎪ j=1 ⎩ ⎭ (5.19) j=1

Therefore, in the limit N → ∞, holding ∣ti − tf ∣ fixed, the amplitude ⟨qf , tf ∣qi , ti ⟩ is given by the (formal) expression i tf ̵ ∫ dt [pq˙ − H(p, q)] ⟨qf , tf ∣qi , ti ⟩ = ∫ DpDq e h ti

(5.20)

where we have used the notation

N

DpDq ≡ lim ∏ N →∞

j=1

dpj dqj 2π h̵

(5.21)

which defines the integration measure. The functions, or configurations, (q(t), p(t)) must satisfy the initial and final conditions q(ti ) = qi ,

q(tf ) = qf

(5.22)

Thus the matrix element ⟨qf , tf ∣qi , ti ⟩ can be expressed as a sum over histories in phase space. The weight of each history is the exponential factor of Eq. (5.20). Notice that the quantity in brackets it is just the Lagrangian L = pq˙ − H(p, q)

(5.23)

5.1 Path Integrals and Quantum Mechanics

119

Thus the matrix element is just ⟨qf , tf ∣q, t⟩ = ∫ DpDq e h̵ i

S(q,p)

(5.24)

where S(q, p) is the action of each history (q(t), p(t)). Also notice that the sum (or integral) runs over independent functions q(t) and p(t) which are not required to satisfy any constraint (apart from the initial and final conditions) and, in particular they are not the solution of the equations of motion. Expressions of these type are known as path-integrals. They are also called functional integrals, since the integration measure is a sum over a space of functions, instead of a field of numbers as in a conventional integral. Using a Gaussian integral of the form (which involves an analytic continuation) p ∆t √ ∆t 2 i ̵ q˙ ) ̵ dp i(pq˙ − m 2m h e e 2h = ̵ 2π h̵ 2πih∆t 2

∫

∞

−∞

(5.25)

we can integrate out explicitly the momenta in the path-integral and find a formula that involves only the histories of the coordinate alone. Notice that there are no initial and final conditions on the momenta since the initial and final states have well defined positions. The result is ⟨qf , tf ∣qi , ti ⟩ = ∫ Dq e

i tf ∫ dt L(q, q) ˙ h̵ t i

(5.26)

which is known as the Feynman Path Integral (Feynman, 2005, 1948). Here L(q, q) ˙ is the Lagrangian, 1 2 L(q, q) ˙ = mq˙ − V (q) 2

(5.27)

and the sum over histories q(t) is restricted by the boundary conditions q(ti ) = qi and q(tf ) = qf . The Feynman path-integral tells us that in the correspondence limit, h̵ → 0, the only history (or possibly histories) that contribute significantly to the path integral must be those that leave the action S stationary since, otherwise, the contributions of the rapidly oscillating exponential would add up to zero. In other words, in the classical limit there is only one history qc (t) that contributes. For this history, qc (t), the action S is stationary, δS = 0, and qc (t) is the solution of the Classical Equation of Motion ∂L d ∂L − =0 ∂q dt ∂ q˙

(5.28)

120

Path Integrals in Quantum Mechanics and Quantum Field Theory q(tf )

q

ti

q(ti )

tf

t

Figure 5.3 Two histories with the same initial and final states.

In other terms, in the correspondence limit h̵ → 0, the evaluation of the Feynman path integral reduces to the requirement that the Least Action Principle should hold. This is the classical limit.

5.2 Evaluating path integrals in Quantum Mechanics Let us first discuss the following problem. We wish to know how to compute the amplitude ⟨qf , tf ∣qi , ti ⟩ for a dynamical system whose Lagrangian has the standard form of Eq. (5.27). For simplicity we will begin with a linear harmonic oscillator. The Hamiltonian for a linear harmonic oscillator is 2

2

p mω 2 H= + q 2m 2 and the associated Lagrangian is

(5.29)

2

L=

m 2 mω 2 q˙ − q 2 2

(5.30)

Let qc (t) be the classical trajectory. It is the solution of the classical equations of motion 2 d qc 2 + ω qc = 0 (5.31) 2 dt Let us denote by q(t) an arbitrary history of the system and by ξ(t) its deviation from the classical solution qc (t). Since all the histories, including the classical trajectory qc (t), obey the same initial and final conditions q(ti ) = qi

q(tf ) = qf

(5.32)

5.2 Evaluating path integrals in Quantum Mechanics

121

it follows that ξ(t) obeys instead vanishing initial and final conditions: ξ(ti ) = ξ(tf ) = 0

(5.33)

After some trivial algebra it is easy to show that the action S for an arbitrary history q(t) becomes tf d dqc d qc 2 [mξ ] + ∫ dt mξ ( 2 + ω qc ) dt dt dt ti ti (5.34) The third term vanishes due to the boundary conditions obeyed by the fluctuations ξ(t), Eq. (5.33). The last term also vanishes since qc is a solution of the classical equation of motion Eq. (5.31). These two features hold for all systems, even if they are not harmonic. However, the Lagrangian (and hence the action) for ξ, the second term in Eq. (5.34), in general is not the same as the action for the classical trajectory (the first term). Only for the ˙ has the same form as S(qc , q˙c ). harmonic oscillator S(ξ, ξ) Hence, for a harmonic oscillator, we get the path integral

˙ +∫ S(q, q) ˙ = S(qc , q˙c ) + S(ξ, ξ)

⟨qf , tf ∣qi , ti ⟩ = e h̵ i

2

tf

S(qc ,q˙c )

dt

∫

ξ(ti )=ξ(tf )=0

i

Dξ e h̵

˙ ∫ti dtL(ξ,ξ) tf

(5.35)

Notice that the information on the initial and final states enters only through the factor associated with the classical trajectory. For the linear harmonic oscillator, the quantum mechanical contribution is independent of the initial and final states. Thus, we need to do two things: 1) we need an explicit solution qc (t) of the equation of motion, for which we will compute S(qc , q˙c ), and 2) we need to compute the quantum mechanical correction, the last factor in Eq. (5.35), which measures the strength of the quantum fluctuations. For a general dynamical system, whose Lagrangian has the form of Eq. (5.27), the action of Eq. (5.34) takes the form ˙ qc ) S(q, q) ˙ = S(qc , q˙c ) + Seff (ξ, ξ; +∫

tf

dt

ti

2 tf d ∂V 999 dq d q [mξ c ] + ∫ dt (m 2c + 9 ) ξ(t) dt dt ∂q 999qc dt ti

(5.36)

where the Seff is the effective action for the fluctuations ξ(t) which has the form ˙ =∫ Seff (ξ, ξ)

tf ti

tf 99 ∂ V 1 2 1 tf ′ 99 ξ(t)ξ(t′ ) − O(ξ 3 ) dt mξ˙ − ∫ dt ∫ dt ′ 2 2 ti ∂q(t)∂q(t ) 99qc ti (5.37) 2

122

Path Integrals in Quantum Mechanics and Quantum Field Theory

Once again, the boundary conditions ξ(ti ) = ξ(tf ) = 0 and the fact the qc (t) is a solution of the equation of motion together imply that the last two terms of Eq. (5.36) vanish identically. 3 Thus, to the extent that we are allowed to neglect the O(ξ ) corrections (and higher), the effective action Seff can be approximated by an action that is quadratic in the fluctuation ξ. In general, this effective action will depend ′′ on the actual classical trajectory, since in general V (qc ) is not constant but is a function of time determined by qc (t). However, if one is interested in the quantum fluctuations about a minimum of the potential V (q), then qc (t) is constant (and equal to the minimum). We will discuss below this case in detail. Before we embark in an actual computation it is worthwhile to ask when 3 it should be a good approximation to neglect the terms O(ξ ) (and higher). Since we are expanding about the classical path qc , we expect that this approximation should be correct as we formally take the limit h̵ → 0. In the ̵ path integral the effective action always appears in the combination Seff /h. Hence, for an effective action that is quadratic in ξ, we can eliminate the dependence on h̵ by the rescaling √ ξ = h̵ ξ˜ (5.38) This rescaling leaves the classical contribution S(qc )/h̵ unaffected. However, n n/2 terms with powers higher than quadratic in ξ, say O(ξ˜ ), scale like h̵ . ̵ has an expansion of the form Thus the action (divided by h) S 1 (0) (2) ˜ n/2 (n) ˜ = ̵ S (qc ) + S (ξ; qc ) + ∑ h̵ S (ξ; qc ) h̵ h n=3 ∞

(5.39)

Thus, in the limit h̵ → 0, we can formally expand the weight of the path ̵ The matrix element we are calculating then takes integral in powers of h. the form ⟨qf , tf ∣qi , ti ⟩ = e

iS

(2)

(0)

(qc )/h̵

Z

(2)

̵ (qc ) (1 + O(h))

(5.40)

The quantity Z (qc ) is the result of keeping only the quadratic approximation. The higher order terms are a power series expansion in h̵ and are ̵ Here, I have used the fact that, by symmetry, in most analytic functions of h. cases of interest the odd powers in ξ in general do not contribute, although there are some cases where they do. Let us now calculate the effect of the quantum fluctuations to quadratic order. This is equivalent to the WKB approximation. Let us denote this

5.2 Evaluating path integrals in Quantum Mechanics

factor by Z, Z

(2)

(qc ) = ∫

˜ i )=ξ(t ˜ f )=0 ξ(t

(2) ˜ ˜ ˙ iSeff (ξ, ξ;qc )

Dξ e

123

(5.41)

It is elementary to show that, due to the boundary conditions, the action ˙ becomes Seff (ξ, ξ) 2

tf d ′′ ˜ ξ) ˜˙ = 1 ∫ dt ξ(t) ˜ ˜ Seff (ξ, [−m 2 − V (qc (t))] ξ(t) 2 ti dt

(5.42)

The differential operator

2

d ′′ Aˆ = −m 2 − V (qc (t)) dt

(5.43)

has the form of a Schr¨odinger operator for a particle on a “coordinate” t ′′ in a potential −V (qc (t)). Let ψn (t) be a complete set of eigenfunctions of Aˆ satisfying the boundary conditions ψ(ti ) = ψ(tf ) = 0. Completeness and orthonormality implies that the eigenfunctions {ψn (t)} satisfy ∑ ψn (t)ψn (t ) = δ(t − t ), ∗

′

′

n

∫

tf

ti

dt ψn (t)ψm (t) = δn,m ∗

(5.44)

˜ that satisfies the vanishing boundary conditions of An arbitrary function ξ(t) Eq.(5.33) can be expanded as a linear combination of the basis eigenfunctions {ψn (t)}, ˜ = ∑ cn ψn (t) ξ(t) (5.45) n

˜ i ) = ξ(t ˜ f ) = 0 as we should. Clearly, we have ξ(t For the special case of qi = qf = q0 , where q0 is a minimum of the potential ′′ V (q), V (q0 ) = ωeff > 0 is a constant, and the eigenvectors of the Schr¨odinger operator are just plane waves. For a linear harmonic oscillator ωeff = ω. Thus, in this case the eigenvectors are ψn (t) = bn sin(kn (t − ti ))

where

kn =

πn tf − ti

n = 1, 2, 3, . . .

√ and bn = 1/ tf − ti . The eigenvalues of Aˆ are An =

2 kn

2 − ωeff

(5.46)

(5.47)

2

π 2 2 n − ωeff = 2 (tf − ti )

(5.48)

124

Path Integrals in Quantum Mechanics and Quantum Field Theory

By using the expansion of Eq. (5.45), we find that the action S form 1 tf (2) ˜ Aˆ ξ(t) ˜ = 1 ∑ An c2n S = ∫ dt ξ(t) 2 ti 2

(2)

takes the (5.49)

n

where we have used the completeness and orthonormality of the basis functions {ψn (t)}. ˜ → cn . More The expansion of Eq.(5.45) is a canonical transformation ξ(t) to the point, the expansion is actually a parametrization of the possible histories in terms of a set of orthonormal functions, and it can be used to define the integration measure to be dcn D ξ˜ = N ∏ √ 2π n

(5.50)

with unit Jacobian. Here N is an irrelevant normalization constant that will be defined below. Finally, the (formal) Gaussian integral, which is defined by a suitable analytic continuation procedure, is ∫

∞ −∞

dcn i A c2 −1/2 √ e 2 n n = [−iAn ] 2π

(5.51)

can be used to write the amplitude as Z

(2)

= N ∏ An

−1/2

n

ˆ ≡ N (DetA)

−1/2

(5.52)

where we have used the definition that the determinant of an operator is equal to the product of its eigenvalues. Therefore, up to a normalization constant, we obtained the result Z

(2)

ˆ = (DetA)

−1/2

(5.53)

We have thus reduced the problem of the computation of the leading (Gaussian) fluctuations to the path-integral to the computation of a determinant of the fluctuation operator, a differential operator defined by the choice of classical trajectory. Below we will see how this is done. 5.2.1 Analytic continuation to imaginary time It is useful to consider the related problem obtained by an analytic continuation to imaginary time, t → −iτ . We saw before that there is a relation between this problem and Statistical Physics. We will now work out one example that will be very instructive.

5.2 Evaluating path integrals in Quantum Mechanics

125

Formally, upon the analytic continuation t → −iτ , the matrix element of the time evolution operator becomes i 1 − ̵ H(tf − ti ) − ̵ H(τf − τi ) h h ⟨qf ∣e ∣qi ⟩ → ⟨qf ∣e ∣qi ⟩

(5.54)

Let us choose

τf = β h̵

τi = 0

(5.55)

where β = 1/T , and T is the temperature (in units of kB = 1). Hence, we find that ̵ i , 0⟩ = ⟨qf ∣e ⟨qf , −iβ/h∣q

−βH

The operator ρˆ

∣qi ⟩

−βH

ρˆ = e

(5.56)

(5.57)

is the Density Matrix in the Canonical Ensemble of Statistical Mechanics for a system with Hamiltonian H in thermal equilibrium at temperature T . It is customary to define the Partition Function Z, ≡ ∫ dq ⟨q∣e

−βH

−βH

Z = tre

∣q⟩

(5.58)

where I inserted a complete set of eigenstates of qˆ. Using the results that were derived above, we see that the partition function Z can be written as a (Euclidean) Feynman path integral in imaginary time, of the form 1 β h̵ ∂q 2 1 Z = ∫ Dq[τ ] exp {− ̵ ∫ dτ [ m ( ) + V (q)]} 2 ∂τ h 0 ≡ ∫ Dq[τ ] exp {− ∫ dτ [ β

0

m ∂q 2 ( ) + V (q)]} 2h̵ 2 ∂τ

(5.59)

̵ where, in the last equality we have rescaled τ → τ /h. Since the Partition Function is a trace over states, we must use boundary conditions such that the initial and final states are the same state, and to sum over all such states. In other words, we must have periodic boundary conditions in imaginary time (PBC’s), q(τ ) = q(τ + β)

(5.60)

Therefore, a quantum mechanical system at finite temperature T can be described in terms of an equivalent system in classical statistical mechanics

126

Path Integrals in Quantum Mechanics and Quantum Field Theory

with Hamiltonian (or energy) H=

m ∂q 2 ( ) + V (q) 2h̵ 2 ∂τ

(5.61)

on a segment of length 1/T and obeying PBC’s. This effectively means that the segment is actually a ring of length β = 1/T . Alternatively, upon inserting a complete set of eigenstates of the Hamiltonian, it is easy to see that an arbitrary matrix element of the density matrix has the form ⟨q ∣e ′

−βH

∞

∣q⟩ = ∑ ⟨q ∣n⟩⟨n∣q⟩e ′

−βEn

n=0 ∞

=∑e

−βEn

n=0

ψn (q )ψn (q) −−−−−→ e β→∞ ∗

′

−βE0

ψ0 (q )ψ0 (q) ∗

′

(5.62)

where {En } are the eigenvalues of the Hamiltonian, E0 is the ground state energy and ψ0 (q) is the ground state wave function. Therefore, we can calculate both the ground state energy E0 and the ground state wave function from the density matrix and consequently from the (imaginary time) path integral. For example, from the identity 1 −βH ln tre β β→∞

E0 = − lim

(5.63)

we see that the ground state energy is given by β 1 m ∂q 2 E0 = − lim ln ∫ Dq exp {− ∫ dτ [ 2 ( ) + V (q)]} ∂τ β→∞ β 2h̵ q(0)=q(β) 0 (5.64) Mathematically, the imaginary time path integral is a better behaved object than its real time counterpart, since it is a sum of positive quantities, the statistical weights. In contrast, the Feynman path integral (in real time) is a sum of phases and, as such, is an ill-defined object. It is actually conditionally convergent, and to make sense of it convergence factors (or regulators) will have to be introduced. The effect of these convergence factors is actually an analytic continuation to imaginary time. We will encounter the same problem in the calculation of propagators. Thus, the imaginary time path integral, often referred to as the Euclidean path integral (as opposed to Minkowski), can be used to describe both a quantum system and a statistical mechanics system. Finally, we notice that at low temperatures T → 0, the Euclidean path integral can be approximated using methods similar to the ones we discussed

5.2 Evaluating path integrals in Quantum Mechanics

127

for the (real time) Feynman Path Integral. The main difference is that we must sum over trajectories which are periodic in imaginary time with period β = 1/T . In practice this sum can only be done exactly for simple systems such as the harmonic oscillator, and for more general systems one has to resort to some form of perturbation theory. Here we will consider a physical system described by a dynamical variable q and a potential energy V (q) which has a minimum at q0 = 0. For simplicity we will take V (0) = 0 2 ′′ and we will denote by mω = V (0) (in other words, an effective harmonic oscillator). The partition function is given by the Euclidean path integral 1 β Z = ∫ Dq[τ ] exp (− ∫ ξ(τ )AˆE ξ(τ )dτ ) 2 0

(5.65)

ˆ where AˆE is the imaginary time, or Euclidean, version of the operator A, and it is given by 2

m d ′′ (5.66) AˆE = − 2 2 + V (qc (τ )) ̵h dτ The functions this operator acts on obey periodic boundary conditions with period β. Notice the important change in the sign of the term of the potential. Hence, once again we will need to compute a functional determinant, although the operator now acts on functions obeying periodic boundary conditions. In a later chapter we will see that in the case of fermionic theories, the boundary conditions become antiperiodic.

5.2.2 The functional determinant

(2)

We will now do the computation of the determinant in Z . We will do the calculation in imaginary time and then we will carry out the analytic continuation to real time. We will follow closely the method is explained in detail in Sidney Coleman’s book, (Coleman, 1985). We want to compute 2

m d ′′ D = Det [− 2 2 + V (qc (τ ))] ̵h dτ

(5.67)

subject to the requirement that the space of functions that the operator acts on obeys specific boundary conditions in (imaginary) time. We will be interested in two cases: (a) Vanishing Boundary Conditions (VBC’s), which are useful to study quantum mechanics at T = 0, and (b) Periodic Boundary Conditions (PBC’s) with period β = 1/T . The approach is somewhat different in the two situations.

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Path Integrals in Quantum Mechanics and Quantum Field Theory

A: vanishing boundary conditions

h We define the (real) variable x = m τ . The range of x is the interval [0, L], √ ̵ with L = hβ/ m. Let us consider the following eigenvalue problem for the 2 Schr¨ odinger operator −∂ + W (x), ̵

(−∂ + W (x)) ψ(x) = λψ(x) 2

(5.68)

subject to the boundary conditions ψ(0) = ψ(L) = 0. Formally, the determinant is given by D = ∏ λn

(5.69)

n

where {λn } is the spectrum of eigenvalues of the operator −∂ + W (x) for a space of functions satisfying a given boundary condition. Let us define an auxiliary function ψλ (x), with λ a real number not necessarily in the spectrum of the operator, such that the following requirements are met: 2

a. ψλ (x) is a solution of Eq. (5.68), and b. ψλ obeys the initial conditions, ψλ (0) = 0 and ∂x ψλ (0) = 1.

It is easy to see that −∂ + W (x) has an eigenvalue at λn if and only if ψλn (L) = 0. (Because of this property this procedure is known as the shooting method.) Hence, the determinant D of Eq. (5.69) is equal to the product of the zeros of ψλ (x) at x = L. (1) (2) Consider now two potentials W and W , and the associated functions, (1) (2) ψλ (x) and ψλ (x). Let us show that 2

Det [−∂ + W 2

(1)

(x) − λ]

Det [−∂ 2 + W (2) (x) − λ]

=

(1)

ψλ (L) (2)

ψλ (L)

(5.70)

The l. h. s. of Eq. (5.70) is a meromorphic function of λ in the complex 2 (1) plane, which has simple zeros at the eigenvalues of −∂ +W (x) and simple 2 (2) poles at the eigenvalues of −∂ + W (x). Also, the l. h. s. of Eq. (5.70) approaches 1 as ∣λ∣ → ∞, except along the positive real axis which is where the spectrum of eigenvalues of both operators is. Here we have assumed that the eigenvalues of the operators are non-degenerate, which is the general case. Similarly, the right hand side of Eq. (5.70) is also a meromorphic function of λ, which has exactly the same zeros and the same poles as the left hand side. It also goes to 1 as ∣λ∣ → ∞ (again, except along the positive real axis), since the wave-functions ψλ are asymptotically plane waves in this limit . Therefore, the function formed by taking the ratio r. h. s. / l. h.

5.2 Evaluating path integrals in Quantum Mechanics

129

s. is an analytic function on the entire complex plane and it approaches 1 as ∣λ∣ → ∞. Then, general theorems of the theory of functions of a complex variable tell us that this function is equal to 1 everywhere. From these considerations we conclude that the following ratio is independent of W (x), Det (−∂ + W (x) − λ) 2

ψλ (L)

(5.71)

We now define a constant N such that

Det (−∂ + W (x)) 2

Then, we can write

ψ0 (L)

N [Det (−∂ + W )] 2

−1/2

̵ 2 = π hN

(5.72)

̵ 0 (L)] = [π hψ

−1/2

(5.73)

Thus we reduced the computation of the determinant, including the normalization constant, to finding the function ψ0 (L). For the case of the linear harmonic oscillator, this function is the solution of [−

2

∂ 2 + mω ] ψ0 (x) = 0 ∂x2

with the initial conditions, ψ0 (0) = 0 and ψ0 (0) = 1. The solution is Hence,

′

ψ0 (x) = √ 2

and we find

(5.74)

√ 1 sinh( mωx) mω −1/2

∂ 2 Z = N [Det (− 2 + mω )] ∂x

̵ 0 (L)]−1/2 = [π hψ

(5.75)

(5.76)

−1/2 π h̵ (5.77) sinh(βω)] Z = [√ mω √ ̵ where we have used L == hβ/ m. From this result we find that the ground state energy is ̵ −1 hω E0 = lim ln Z = (5.78) 2 β→∞ β

as it should be. Finally, by means of an analytic continuation back to real time, we can

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use these results to find, for instance, the amplitude to return to the origin after some time T . Thus, for tf − ti = T and qf = qi = 0, we get −1/2 iπ h̵ sin(ωT )] ⟨0, T ∣0, 0⟩ = [ √ mω

(5.79)

B: periodic boundary conditions

Periodic boundary conditions imply that the histories satisfy q(τ ) = q(τ +β). Hence, these functions can be expanded in a Fourier series of the form ∞

q(τ ) = ∑ e

iωn τ

(5.80)

qn

n=−∞

where ωn = 2πn/β. Since q(τ ) is real, we have the constraint q−n = qn . For such configurations (or histories) the action becomes S = ∫ dτ [ β

0

∗

m ∂q 2 1 ′′ 2 ( ) + V (0)q ] 2 2 ∂τ 2h̵

m 2 β ′′ 2 ′′ 2 = V (0)q0 + β ∑ [ 2 ωn + V (0)] ∣qn ∣ 2 ̵ h n≥1

(5.81)

The integration measure now is

dReqn dImqn dq0 Dq[τ ] = N √ ∏ 2π 2π n≥1

(5.82)

where N is a normalization constant that will be discussed below. After doing the Gaussian integrals, the partition function becomes, ⎡ ∞ ⎢ ⎢ ⎢ Z =N√ ∏ ∏ =N⎢ ⎢ ⎢ ⎢ βV ′′ (0) n≥1 βm ω 2 + βV ′′ (0) ⎣n=−∞ h̵ 2 n

1/2 ⎤ ⎥ ⎥ ⎥ ⎥ βm 2 ′′ (0) ⎥ ⎥ ⎥ ω + βV ⎦ h̵ 2 n (5.83) Formally, the infinite product that enters in this equation is divergent. The normalization constant N eliminates this divergence. This is an example of what is called a regularization. The regularized partition function is

1

1

1

Z= Using the identity

√

2 ′′ m 1 h̵ V (0) ] ∏ [1 + √ mωn2 h̵ 2 β βV ′′ (0) n≥1

∏ (1 + n≥1

−1

(5.84)

2

a sinh a )= a 2 2 n π

(5.85)

5.3 Path integrals for a scalar field theory

131

we find Z=

1

⎛ β h̵ V ′′ (0) 2 sinh ⎜ ( m ) ⎝ 2

1/2

(5.86)

⎞ ⎟ ⎠

which is the partition function for a linear harmonic oscillator, see L. D. Landau and E. M. Lifshitz, Statistical Physics, (Landau and Lifshitz, 1959b).

5.3 Path integrals for a scalar field theory We will now develop the path-integral quantization picture for a scalar field theory. Our starting point will be the canonically quantized scalar field. As ˆ we saw before, in canonical quantization the scalar field φ(x) is an operator that acts on a Hilbert space of states. We will use the field representation, which is the analog of the conventional coordinate representation in Quantum Mechanics. Thus, the basis states are labelled by the field configuration at some fixed time x0 , a set of states of the form { ∣{φ(x, x0 )}⟩ }. The field operator ˆ φ(x, x0 ) acts trivially on these states, ˆ φ(x, x0 )∣{φ(x, x0 )}⟩ = φ(x, x0 )∣{φ(x, x0 )}⟩

(5.87)

The set of states { ∣{φ(x, x0 )}⟩ } is both complete and orthonormal. Completeness here means that these states span the entire Hilbert space. Consequently the identity operator Iˆ in the full Hilbert space can be expanded in a complete basis in the usual manner, which for this basis it means Iˆ = ∫ Dφ(x, x0 ) ∣{φ(x, x0 )}⟩⟨{φ(x, x0 )}∣

(5.88)

Since the completeness condition is a sum over all the states in the basis and since this basis is the set of field configurations at a given time x0 , we will need to give a definition for integration measure which represents the sums over the field configurations. In this case, the definition of the integration measure is trivial, Dφ(x, x0 ) = ∏ dφ(x, x0 )

(5.89)

x

Likewise, orthonormality of the basis states is the condition

⟨φ(x, x0 )∣φ (x, x0 )⟩ = ∏ δ (φ(x, x0 ) − φ (x, x0 )) ′

′

(5.90)

x

Thus, we have a working definition of the Hilbert space for a real scalar field.

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Path Integrals in Quantum Mechanics and Quantum Field Theory

In canonical quantization, the classical canonical momentum Π(x, x0 ), defined as δL Π(x, x0 ) = = ∂0 φ(x, x0 ) (5.91) δ∂0 φ(x, x0 )

becomes an operator that acts on the same Hilbert space as the field itself φ ˆ ˆ does. The field operator φ(x) and the canonical momentum operator Π(x) satisfy equal-time canonical commutation relations ˆ ˆ ̵ (x − y) [φ(x, x0 ), Π(y, x0 )] = ihδ 3

(5.92)

Here we will consider a real scalar field whose Lagrangian density is 1 2 (∂ φ) − V (φ) (5.93) 2 µ It is a simple matter to generalize what follows below to more general cases, such as complex fields and/or several components. Let us also recall that the Hamiltonian for a scalar field is given by L=

2 ˆ ˆ ˆ 2 (x) + 1 (&φ(x)) ˆ = ∫ d3 x [ 1 Π H + V (φ(x))] 2 2

(5.94)

For reasons that will become clear soon, it is convenient to add an extra term to the Lagrangian density of the scalar field, Eq. (5.93), of the form Lsource = J(x) φ(x)

(5.95)

The field J(x) is called an external source. The field J(x) is the analog of external forces acting on a system of classical particles. Here we will always assume that the sources J(x) vanish both at spacial infinity (at all times) and everywhere in both the remote past and in the remote future, lim J(x, x0 ) = 0

∣x∣→∞

The total Lagrangian density is

lim J(x, x0 ) = 0

x0 →±∞

L(φ, J) = L + Lsource

(5.96)

(5.97)

Since the source J(x) is in general a function of space and time, the Hamiltonian that follows from this Lagrangian is formally time-dependent. We will derive the path integral for this quantum field theory by following the same procedure we used for the case of a finite quantum mechanical system. Hence we begin by considering the Wightman function defined as the amplitude ′ J ⟨{φ(x, x0 )}∣{φ (y, y0 )}⟩J

(5.98)

In other words, we want the transition amplitude in the background of the

5.3 Path integrals for a scalar field theory

133

sources J(x). We will be interested in situations in which x0 is in the remote future and y0 is in the remote past. It turns out that this amplitude is intimately related to the computation of ground state (or vacuum) expectation values of time ordered products of field operators in the Heisenberg representation (N )

G

ˆ 1 ) . . . φ(x ˆ N )]∣0⟩ (x1 , . . . , xN ) ≡ ⟨0∣T [φ(x

(5.99)

which are known as the N -point functions (or correlators). In particular the 2-point function (2)

G

ˆ 1 )φ(x ˆ 2 )]∣0⟩ (x1 − x2 ) ≡ −i⟨0∣T [φ(x

(5.100)

is called the Feynman Propagator for this theory. We will see later on that all quantities of physical interest can be obtained from a suitable correlation function of the type of Eq. (5.99). ˆ 1 ) . . . φ(x ˆ N )] for the timeIn Eq. (5.99) we have use the notation T [φ(x ordered product of Heisenberg field operators. For any pair Heisenberg of ˆ ˆ operators A(x) and B(y), that commute for space-like separations, their time-ordered product is defined to be ˆ B(y)] ˆ ˆ B(y) ˆ ˆ A(x) ˆ T [A(x) = θ(x0 − y0 )A(x) + θ(y0 − x0 )B(y)

(5.101)

where θ(x) is the step (or Heaviside) function 1 θ(x) = { 0

if x ≥ 0, otherwise

(5.102)

This definition is generalized by induction to to the product of any number of operators. Notice that inside a time-ordered product the Heisenberg operators behave as if they were c-numbers. Let us now recall the structure of the derivation that we gave of the path integral in Quantum Mechanics. We will paraphrase that derivation for this field theory. We considered an amplitude equivalent to Eq. (5.98), and realized that this amplitude is actually a matrix element of the evolution operator, i x0 ′ ̂ ′0 ) − ̵ ∫ dx0 H(x ′ ′ h y0 ∣{φ (y)}⟩ J ⟨{φ(x, x0 )}∣{φ (y, y0 )}⟩J = ⟨{φ(x)}∣T e (5.103) ̂ ′0 ) where T stands for the time ordering symbol (not temperature!), and H(x

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Path Integrals in Quantum Mechanics and Quantum Field Theory

is the time-dependent Hamiltonian whose Hamiltonian density is ̂ 0) = H(x

2 1 1 2 ˆ ˆ ˆ ̂ (x, x0 ) + (&φ(x, Π x0 )) + V (φ(x, x0 )) − J(x, x0 )φ(x, x0 ) 2 2 (5.104) Paraphrasing the construction used in the case of Quantum Mechanics of a particle, we first partition the time interval in a large number of steps N , each of width ∆t, and then insert a complete set of eigenstates of the field ˆ since the field plays the role of the coordinate. As it turned out, operator φ, we also had to insert complete sets of eigenstates of the canonical momentum ˆ operator, which here means the canonical field operator Π(x). Upon formally taking the time-continuum limit, N → ∞ and ∆t → 0 while keeping N ∆t fixed, we obtain the result that the phase-space path integral of the field theory is J ⟨{φ(x, x0 )}∣{φ (y, y0 )}⟩J ′

=∫

b. c.

i 4 ˙ − H(φ, Π) + Jφ] ∫ d x [φΠ ̵ h DφDΠ e

(5.105) where b.c. indicates the boundary conditions specified by the requirement ′ that the initial and final states be ∣{φ(x, x0 )}⟩ and ∣{φ (y, y0 )}⟩, respectively. Exactly as in the case of the path integral for a particle, the Hamiltonian of this theory is quadratic in the canonical momenta Π(x). Hence, we can further integrate out the field Π(x), and obtain the Feynman path integral for the scalar field theory in the form of a sum over histories of field configurations: J ⟨{φ(x, x0 )}∣{φ (y, y0 )}⟩J ′

=N∫

b. c.

i S(φ, ∂µ φ, J) ̵ Dφ e h

(5.106)

where N is an (unimportant) normalization constant, and S(φ, ∂µ φ, J) is the action for a real scalar field φ(x) coupled to a source J(x), 1 2 4 S(φ, ∂µ φ, J) = ∫ d x [ (∂µ φ) − V (φ) + Jφ] 2

(5.107)

5.4 Path integrals and propagators In Quantum Field Theory we will be interested in calculating vacuum (ground state) expectation values of field operators at various space-time locations. ′ Thus, instead of the amplitude J ⟨{φ(x, x0 )}∣{φ (y, y0 )}⟩J we may be interested in a transition between an initial state, at y0 → −∞ which is the vacuum state ∣0⟩, i.e. the ground state of the scalar field in the absence of

5.4 Path integrals and propagators

135

the source J(x), and a final state at x0 → ∞ which is also the vacuum state of the theory in the absence of sources. We will denote this matrix element by Z[J] = J ⟨0∣0⟩J

(5.108)

This matrix element is called the vacuum persistence amplitude. Let us see now how the vacuum persistence amplitude is related to the Feynman path integral for a scalar field of Eq. (5.106). In order to do that ′ we will assume that the source J(x) is “on” between times t < t and that ′ ′ we watch the system on a much longer time interval T < t < t < T . For this interval, we can now use the Superposition Principle to insert complete ′ sets of states at intermediate times t and t , and write the amplitude in the form ′ ′ J ⟨{Φ (x, T )}∣{Φ(x, T )}⟩J

(5.109)

=

∫ Dφ(x, t) Dφ (x, t )⟨{Φ (x, T )}∣{φ (x, t )}⟩ ′

′

′

′

′

′

×J ⟨{φ (x, t )}∣{φ(x, t)}⟩J ⟨{φ(x, t)}∣{Φ(x, T )}⟩ ′

′

The matrix elements ⟨{Φ (x, T )}∣{φ (x, t )}⟩ and ⟨{φ(x, t)}∣{Φ(x, T )}⟩ are given by ′

′

′

′

∗ −iEn (t − T )/h̵ ⟨{φ(x, t)}∣{Φ(x, T )}⟩ = ∑ Ψn [{φ(x)}]Ψn [{Φ(x)}] e n

′ ′ ′ ′ ′ ′ ′ ∗ ′ −iEm (T − t )/h̵ ⟨{Φ (x, T )}∣{φ (x, t )}⟩ = ∑ Ψm [{Φ (x)}]Ψm[{φ (x)}] e m

(5.110)

where we have introduced complete sets of eigenstates ∣{Ψn }⟩ of the Hamiltonian of the scalar field (without sources) and the corresponding wave functions, {Ψn [Φ(x)]}. ′ At long times T and T these series expansions oscillate very rapidly and a definition must be provided to make sense on these expressions. To this end, we will now analytically continue T along the positive imaginary time axis, ′ and T along the negative imaginary time axis, as shown in figure 5.4. After carrying out the analytic continuation, we find that the following identities

136

Path Integrals in Quantum Mechanics and Quantum Field Theory Im t T

t

t

′

Re t

T

′

Figure 5.4 Analytic continuation.

hold,

−iE0 T /h̵ ∗ −iE0 t/h̵ lim e ⟨{φ(x, t)}∣{Φ(x, T )}⟩ = Ψ0 [{φ}] Ψ0 [{Φ}] e

T →+i∞

′ ′ ′ iE0 t /h̵ iE0 T /h̵ ′ ′ ′ ′ ∗ [{φ }] e lim e ⟨{Φ (x, T )}∣{φ(x, t )}⟩ = Ψ [{Φ }] Ψ 0 0 ′

T →−i∞

(5.111)

This result is known as the Gell-Mann-Low Theorem.(Gell-Mann and Low, 1951) In this limit, the contributions from excited states drop out provided the vacuum state ∣0⟩ is non-degenerate. This procedure is equivalent to the standard adiabatic turning on and off of the external sources. The restriction to a non-degenerate vacuum state can be done by lifting a possible degeneracy by means of an infinitesimally weak external perturbation, which is switched off after the infinite time limit is taken. We will encounter similar issues in our discussion of spontaneous symmetry breaking in later chapters.

5.4 Path integrals and propagators

137

Hence, in the same limit, we also find that the following relation holds ⟨{Φ (x, T )}∣Φ(x, T )}⟩ ̵ Ψ∗0 [{Φ}] Ψ0 [{Φ′ }] T →+i∞ T →−i∞ exp [−iE0 (T ′ − T )/h] ′

lim

′

lim ′

= ∫ DΦDΦ Ψ0 [{φ (x, t )}] Ψ0 [{φ(x, t)}] J ⟨{φ (x, t )}∣{φ(x, t)}⟩J ′

∗

′

′

′

′

≡J ⟨0∣0⟩J

(5.112)

Eq.(5.112) gives us a direct relation between the Feynman Path Integral and the vacuum persistence amplitude of the form ′

Z[J] = J ⟨0∣0⟩J = N

lim

lim ∫ Dφ e

T →+i∞ T ′ →−i∞

i T 4 ∫ d x [L(φ, ∂µ φ) + Jφ] h̵ T

(5.113) In other words, in this asymptotically long-time limit, the amplitude of Eq. (5.98) becomes identical to the vacuum persistence amplitude J ⟨0∣0⟩J , regardless of the choice of the initial and final states. Hence we find a direct relation between the vacuum persistence function Z[J] and the Feynman path integral, given by Eq. (5.113). Notice that, in this limit, we can ignore the “hard” boundary condition and work instead with free boundary conditions. Or equivalently, physical properties become independent of the initial and final conditions placed. For these reasons, from now on we will work with the simpler expression i 4 ∫ d x [L(φ, ∂µ φ) + Jφ] ̵ h Z[J] = J ⟨0∣0⟩J = N ∫ Dφ e

(5.114)

This is a very useful relation. We will see now that Z[J] is the generating function(al) of all the vacuum expectation values of time ordered products of fields, i.e. the correlators of the theory. In particular, let us compute the expression

2 2 1 i 2 1 δ Z[J] 9999 δ J ⟨0∣0⟩J 9999 ′ ( ) ⟨0∣T [φ(x)φ(x )]∣0⟩ = = 9 9 h̵ Z[0] δJ(x)δJ(x′ ) 9999J=0 ⟨0∣0⟩ δJ(x)δJ(x′ ) 9999J=0 (5.115) Thus, the 2-point function, i.e. the Feynman propagator or propagator of the scalar field φ(x), becomes

⟨0∣T [φ(x)φ(x )]∣0⟩ = −i ′

i 1 ′ ∫ Dφ φ(x) φ(x ) exp ( ̵ S[φ, ∂µ φ]) h ⟨0∣0⟩ (5.116)

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Path Integrals in Quantum Mechanics and Quantum Field Theory

Similarly, the N -point function ⟨0∣T [φ(x1 ) . . . φ(xN )]∣0⟩ becomes ̵ ⟨0∣T [φ(x1 ) . . . φ(xN )]∣0⟩ = (−ih) =

where

N

N 99 1 δ J ⟨0∣0⟩J 99 9 ⟨0∣0⟩ δJ(x1 ) . . . δJ(xN ) 9999J=0

1 i ∫ Dφ φ(x1 ) . . . φ(xN ) exp ( ̵ S[φ, ∂µ φ]) h ⟨0∣0⟩ (5.117)

i Z[0] = ⟨0∣0⟩ = ∫ Dφ exp ( ̵ S[φ, ∂µ φ]) h

(5.118)

Therefore, we find that the path integral always yields vacuum expectation values of time-ordered products of operators. The quantity Z[J] can thus be viewed as the generating functional of the correlation functions of this theory. These are actually general results that hold for the path integrals of all theories.

5.5 Path integrals in Euclidean space-time and Statistical Physics In the last section we saw how to relate the computation of transition amplitudes to path integrals in Minkowski space-time with specific boundary conditions dictated by the nature of the initial and final states. In particular we derived explicit expressions for the case of fixed boundary conditions. However we could have chosen other boundary conditions. For instance, for the amplitude to begin in any state at the initial time and to go back to the same state at the final time, but summing over all states. This is the same as to ask for the trace Z [J] = ∫ DΦJ ⟨{Φ(x, t )}∣{Φ(x, t)}⟩J ′

′

i 4 − ̵ ∫ d x (H − Jφ) h ≡Tr T e i 4 ∫ d x (L + Jφ) ̵ ≡∫ Dφ e h

(5.119)

PBC

where PBC stands for periodic boundary conditions on some generally finite ′ time interval t − t, and T is the time-ordering symbol. Let us now carry the analytic continuation to imaginary time t → −iτ , i.e. a Wick rotation. Upon a Wick rotation the theory has Euclidean invariance,

5.5 Path integrals in Euclidean space-time and Statistical Physics

139

imaginary time

i.e. rotations and translations in D = d + 1-dimensional space. Imaginary time plays the same role as the other d spacial dimensions. Hereafter we will denote imaginary time by xD , and all vectors will have indices µ that run from 1 to D. We will consider two cases: infinite imaginary time interval, and finite imaginary time interval.

0

β

space Figure 5.5 Periodic boundary conditions wraps space-time into a cylinder.

5.5.1 Infinite imaginary time interval In this case the path integral becomes − ∫ d x (LE − Jφ) Z [J] = ∫ Dφ e D

′

(5.120)

where D is the total number of space-time dimensions. For the sake of definiteness here we discuss the four-dimensional case but the results are obviously valid more generally. Here LE is the Euclidean Lagrangian 1 1 2 2 LE = (∂0 φ) + (&φ) + V (φ) 2 2

(5.121)

The path integral of Eq. (5.120) has two interpretations. One is simply the infinite time limit (in imaginary time) and therefore it

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Path Integrals in Quantum Mechanics and Quantum Field Theory

must be identical to the vacuum persistence amplitude J ⟨0∣0⟩J . The only difference is that from here we get all the N -point functions in Euclidean space-time (imaginary time). Therefore, the relativistic interval is 2

2

2

2

x0 − x → −τ − x < 0

(5.122)

which is always space-like. Hence, with this procedure we will get the correlation functions for space-like separations of its arguments. To get to time-like separations we will need to do an analytic continuation back to real time. This we will do later on. The second interpretation is that the path integral of Eq. (5.120) is the partition function of a system in Classical Statistical Mechanics in D dimensions with energy density (divided by T ) equal to LE − Jφ. This will turn out to be a very useful connection (both ways!).

5.5.2 Finite imaginary time interval In this case we have 0 ≤ x0 = τ ≤ β = 1/T

(5.123)

where T will be interpreted as the temperature. Indeed, in this case the path integral is ′ −βH Z [0] = Tr e (5.124)

and we are effectively looking at a problem of the same Quantum Field Theory but at finite temperature T = 1/β. The path integral is once again the partition function but of a system in Quantum Statistical Physics! The partition function thus is (after setting h̵ = 1) Z [J] = ∫ Dφ ′

β −∫ e 0

dτ (LE − Jφ)

(5.125)

where the field φ(x, τ ) obeys periodic boundary conditions in imaginary time, φ(x, τ ) = φ(x, τ + β)

(5.126)

This boundary condition will hold for all bosonic theories. We will see later on that theories with fermions obey instead anti-periodic boundary conditions in imaginary time. Hence, Quantum Field Theory at finite temperature T is just Quantum Field Theory on an Euclidean space-time which is periodic (and finite) in one direction, imaginary time. In other words, we have wrapped (or compactified)

5.6 Path integrals for the free scalar field

141

Euclidean space-time into a cylinder with perimeter (circumference) β = 1/T (in units of h̵ = kB = 1). The correlation functions in imaginary time (which we will call the Euclidean correlation functions) are given by N ′ 99 1 δ Z [J] 99 99 = ⟨φ(x1 ) . . . φ(xN )⟩ ′ δJ(x ) . . . J(x ) Z [J] 1 N 999J=0

(5.127)

which are just the correlation functions in the equivalent problem in Statistical Mechanics. Upon analytic continuation the Euclidean correlation functions ⟨φ(x1 ) . . . φ(xN )⟩ and the N -point functions of the QFT are related by ̵ ⟨0∣T φ(x1 ) . . . φ(xN )∣0⟩ ⟨φ(x1 ) . . . φ(xN )⟩ ↔ (ih) N

(5.128)

For the case of a quantum field theory at finite temperature T , the path integral yields the correlation functions of the Heisenberg field operators in imaginary time. These correlation functions are often called the thermal correlation functions (or propagators). They are functions of the spatial positions of the fields, x1 , . . . , xN and of their imaginary time coordinates, xD1 , . . . , xDN (here xD ≡ τ ). To obtain the correlation functions as a function of the real time coordinates x01 , . . . , x0N at finite temperature T it is necessary to do an analytic continuation. We will discuss how this is done later on.

5.6 Path integrals for the free scalar field We will consider now the case of a free scalar field. We will carry our discussion in Euclidean space-time (i.e. in imaginary time), and we will do the relevant analytic continuation back to real time at the end of the calculation. The Euclidean Lagrangian LE for a free field φ coupled to a source J is 1 2 2 1 2 (∂ φ) + m φ − Jφ 2 µ 2 where we are using the notation LE =

(∂µ φ) = ∂µ φ∂µ φ 2

(5.129)

(5.130)

Here the index is µ = 1, . . . , D for an Euclidean space-time of D = d + 1 dimensions. For the most part (but not always) we will be interested in the case of d = 3 and Euclidean space has four dimensions. Notice the way the Euclidean space-time indices are placed in Eq. (5.130). This is not a misprint!

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Path Integrals in Quantum Mechanics and Quantum Field Theory

We will compute the Euclidean Path Integral (or Partition Function) ZE [J] exactly. The Euclidean Path Integral for a free field has the form −∫

ZE [J] = N ∫ Dφ e

1 1 2 2 2 D d x [ (∂µ φ) + m φ − Jφ] 2 2

(5.131)

In Classical Statistical Mechanics this theory is known as the Gaussian model. In what follows I will assume that the boundary conditions of the field φ (and the source J) at infinity are either vanishing or periodic, and that the source J also either vanishes at spatial infinity or is periodic. With these assumptions all terms which are total derivatives drop out identically. Therefore, upon an integration by parts and after dropping boundary terms, the Euclidean Lagrangian becomes 1 2 2 LE = φ [−∂ + m ] φ − Jφ 2

(5.132)

Since this action is a quadratic form of the field φ this path integral can be calculated exactly. It has terms which are quadratic (or, rather bilinear) in φ and a term linear in φ, the source term. By means of the following shift of the field φ ¯ φ(x) = φ(x) + ξ(x)

(5.133)

the Lagrangian becomes 1 2 2 LE = φ [−∂ + m ] φ − Jφ 2 1 1 2 2 2 2 2 2 = φ¯ [−∂ + m ] φ¯ − J φ¯ + ξ [−∂ + m ] ξ + ξ [−∂ + m ] φ¯ − Jξ 2 2 (5.134)

Hence, we can decouple the source J(x) by requiring that the shift φ¯ be such that the terms linear in ξ cancel each other exactly. This requirement leads to the condition that the classical field φ¯ be the solution of the following inhomogeneous partial differential equation [−∂ + m ] φ¯ = J(x) 2

2

(5.135)

Equivalently, we can write the classical field φ¯ is terms of the source J(x) 2 2 through the action of the inverse of the operator −∂ + m , φ¯ =

1 J −∂ + m2 2

(5.136)

5.6 Path integrals for the free scalar field

143

The solution of Eq. (5.135) is D ′ E ′ ′ ¯ φ(x) = ∫ d x G0 (x − x ) J(x )

where

G0 (x − x ) = ⟨x∣ ′

E

(5.137)

1 ′ ∣x ⟩ 2 −∂ + m

(5.138)

2

2

2

is the correlation function of the linear partial differential operator −∂ +m . E ′ Thus, G0 (x − x ) is the solution of [−∂x + m ] G0 (x − x ) = δ (x − x ) 2

2

E

′

D

′

(5.139)

In terms of G0 (x − x ), the terms of the shifted action become, E

′

1¯ D 2 2 ¯ ¯ ∫ d x( φ(x) [−∂ + m ]φ(x) − J φ(x)) 2 1 D ¯ = − ∫ d xφ(x)J(x) 2 1 D D ′ E ′ ′ = − ∫ d x d x J(x) G0 (x − x ) J(x ) 2 (5.140)

Therefore the path integral for the generating function of the free Euclidean scalar field ZE [J], defined in Eq. (5.131), is given by 1 D D ′ E ′ ′ ∫ d x ∫ d x J(x) G0 (x − x ) J(x ) 2 ZE [J] = ZE [0] e

where ZE [0] is

ZE [0] = ∫ Dξ e

−

1 D 2 2 ∫ d x ξ(x) [−∂ + m ] ξ(x) 2

(5.141)

(5.142)

Eq. (5.141) shows that, after the decoupling, ZE [J] is a product of two factors: (a) a factor that is function of a bilinear form in the source J, and (b) a path integral, ZE [0], that is independent of the sources. 5.6.1 Calculation of ZE [0]

The path integral ZE [0] is analogous to the fluctuation factor that we found in the path integral for a harmonic oscillator in elementary quantum mechanics. There we saw that the analogous factor could be written as a determinant of a differential operator, the kernel of the bilinear form that entered in the action. The same result holds here as well. The only difference is that

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Path Integrals in Quantum Mechanics and Quantum Field Theory

2 2 the kernel is now the partial differential operator Aˆ = −∂ + m whereas in Quantum Mechanics is an ordinary differential operator. Here too, the operator Aˆ has a set of eigenstates {Ψn (x)} which, once the boundary conditions in space-time are specified, are both complete and orthonormal, and the associated spectrum of eigenvalues An is

[−∂ + m ] Ψn (x) =An Ψn (x) 2

2

∫ d x Ψn (x) Ψm (x) =δn,m D

∑ Ψn (x) Ψn (x ) =δ(x − x ) ′

′

(5.143)

n

Hence, once again we can expand the field φ(x) in the complete set of states {Ψn (x)} φ(x) = ∑ cn Ψn (x)

(5.144)

n

Hence, the field configurations are thus parametrized by the coefficients {cn }. The action now becomes, S = ∫ d x LE (φ, ∂φ) = D

1 2 ∑ An cn 2

(5.145)

n

Thus, up to a normalization factor, we find that ZE [0] is given by ZE [0] = ∏ An

−1/2

n

≡ (Det [−∂ + m ]) 2

2

−1/2

,

(5.146)

and we have reduced the calculation of ZE [0] to the computation of the 2 2 determinant of a differential operator, Det [−∂ + m ]. In chapter 8 we will discuss efficient methods to compute such determinants. For the moment, it will be sufficient to notice that there is a simple, but formal, way to compute this determinant. First, we notice that if we are interested in the behavior of an infinite system at T = 0, the eigenstates of 2 2 the operator −∂ + m are simply suitably normalized plane waves. Let L be the linear size of the system, with L → ∞. Then, the eigenfunctions are labeled by a D-dimensional momentum pµ (with µ = 0, 1, . . . , d)

with eigenvalues,

Ψp (x) =

1

ip x e µ µ

2

2

(2πL)D/2

Ap = p + m

(5.147)

(5.148)

5.6 Path integrals for the free scalar field

145

Hence the logarithm of determinant is ln Det [−∂ + m ] =Tr ln [−∂ + m ] 2

2

2

2

= ∑ ln(p + m ) 2

2

p

=V ∫ where V = L

D

d p 2 2 ln(p + m ) (2π)D D

(5.149)

is the volume of Euclidean space-time. Hence, V d p 2 2 ln ZE [0] = − ∫ ln(p + m ) D 2 (2π) D

(5.150)

This expression is has two singularities: an infrared divergence and an ultraviolet divergence. ln Z[0], Eq(5.150), diverges as V → ∞. This infrared (IR) singularity actually is not a problem since ln ZE [0] should be an extensive quantity which must scale with the volume of space-time. In other words, this is how it should behave. However, the integral in Eq.(5.150) diverges at large momenta unless there is an upper bound (or cutoff) for the allowed momenta. This is an ultraviolet (UV) singularity. It has the same origin of the UV divergence of the ground state energy. In fact ZE [0] is closely related to the ground state (vacuum) energy since ZE [0] = lim ∑ e

−βEn

β→∞

Thus,

−βE0

∼e

+ ...

(5.151)

n

1 1 d d p 2 2 E0 = − lim ln ZE [0] = L ∫ ln(p + m ) D 2 β β→∞ (2π) D

d

(5.152)

d

where L is the volume of space, and V = L β. Notice that Eq. (5.152) is UV divergent. Later in this chapter we will discuss how to compute expressions of the form of Eq. (5.152).

5.6.2 Propagators and correlators A number of interesting results are found immediately by direct inspection of Eq. (5.141). We can easily see that, once we set J = 0, the correlation (0) ′ function GE (x − x ) (0)

GE (x − x ) = ⟨x∣ ′

1 ′ ∣x ⟩ −∂ + m2 2

(5.153)

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Path Integrals in Quantum Mechanics and Quantum Field Theory

is equal to the 2-point correlation function for this theory (at J = 0), 2 δ ZE [J] 9999 1 (0) ′ 9 = GE (x − x ) ⟨φ(x)φ(x )⟩ = ZE [0] δJ(x)δJ(x′ ) 9999J=0 ′

(5.154)

Likewise we find that, for a free field theory, the N -point correlation function ⟨φ(x1 ) . . . φ(xN )⟩ is equal to 99 1 δ ZE [J] 99 ⟨φ(x1 ) . . . φ(xN )⟩ = ZE [0] δJ(x1 ) . . . δJ(xN ) 99J=0 =⟨φ(x1 )φ(x2 )⟩ . . . ⟨φ(xN −1 )φ(xN )⟩ + permutations N

(5.155)

Therefore, for a free field, up to permutations of the coordinates x1 , . . . , xN , the N -point functions reduces to a sum of products of 2-point functions. Hence, N must be a positive even integer. This result, Eq. (5.155), which we derived in the context of a theory for a free scalar field, is actually much more general. It is known as Wick’s Theorem. It applies to all free theories, theories whose Lagrangians are bilinear in the fields, it is independent of the statistics and on whether there is relativistic invariance or not. The only caveat is that, as we will see later on, for the case of fermionic theories there is a sign associated with each term of this sum. It is easy to see that, for N = 2k, the total number of terms in the sum is (2k − 1)(2k − 3) . . . ... =

(2k)!

(5.156)

k

2 k!

Each factor of a 2-point function ⟨φ(x1 )φ(x2 )⟩, a free propagator, and it is also called a contraction. It also common to use the notation ⟨φ(x1 )φ(x2 )⟩ = φ(x1 )φ(x2 )

(5.157)

to denote a contraction or propagator.

5.6.3 Calculation of the propagator

(0)

We will now calculate the 2-point function, or propagator, GE (x − x ) for infinite Euclidean space. This is the case of interest in QFT at T = 0. Later on we will do the calculation of the propagator at finite temperature. (0) ′ Eq. (5.139) tells us that GE (x − x ) is the Green function of the operator (0)

′

−∂ + m . We will use Fourier transform methods and write GE (x − x ) in 2

2

′

5.6 Path integrals for the free scalar field

the form (0) GE (x

147

′ d p E i pµ (xµ − xµ ) −x)=∫ G (p) e 0 (2π)D D

′

(5.158)

which is a solution of Eq. (5.139) if (0)

GE (p) =

1 p 2 + m2

(5.159)

Therefore the correlation function in real (Euclidean!) space is the integral (0) GE (x

i p (x − xµ ) D d p e µ µ −x)=∫ p 2 + m2 (2π)D ′

′

(5.160)

We will often encounter integrals of this type and for that reason we will do this one in some detail. We begin by using the identity 1 ∞ 1 −Aα = ∫ dα e 2 A 2 0

(5.161)

where A > 0 is a positive real number. The variable α is called a FeynmanSchwinger parameter. 2 2 We now choose A = p + m , and substitute this expression back in Eq. (5.160), which takes the form (0) GE (x

α 2 2 ′ D 1 ∞ d p − (p + m ) + ipµ (xµ − xµ ) 2 − x ) = ∫ dα ∫ e (5.162) 2 0 (2π)D ′

The integrand is a Gaussian, and the integral can be calculated by a shift of the integration variables pµ , i.e. by completing squares xµ − xµ 1 √ 1 xµ − xµ α 2 2 ′ (p +m )−ipµ (xµ −xµ ) = ( αpµ − i √ ) − ( √ ) (5.163) 2 2 2 α α ′

2

′

2

and by using the Gaussian integral

xµ − xµ 1 √ − ( αpµ − i √ ) D 2 d p α −D/2 e = (2πα) ∫ (2π)D ′

2

(5.164)

After all of this is done, we find the formula (0)

GE (x − x ) = ′

∣x − x ∣ 1 2 − − m α 2α 2 e ′ 2

1 2(2π)D/2

∫

∞

0

−D/2

dα α

(5.165)

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Path Integrals in Quantum Mechanics and Quantum Field Theory

Let us now define a rescaling of the variable α, (5.166)

α = λt by which

∣x − x ∣ 1 2 1 2 ∣x − x ∣ + m λt + m α= 2α 2 2 2λt ′ 2

If we choose

′ 2

(5.167)

∣x − x ∣ m , ′

λ= the exponent becomes

(5.168)

∣x − x ∣ 1 2 m∣x − x ∣ 1 (t + ) + m α= t 2α 2 2 ′ 2

′

(5.169)

After this final change of variables, we find that the correlation function is (0) GE (x

−x)= ′

1

(2π)D/2

D −1 m 2 ′ ( ) K D −1 (m∣x − x ∣) ′ 2 ∣x − x ∣

(5.170)

where Kν (z) is the Modified Bessel function, which has the integral representation z 1 − (t + ) 1 ∞ ν−1 2 t e Kν (z) = ∫ dt t (5.171) 2 0 where ν =

D 2

− 1, and z = m∣x − x ∣. ′

power-law decay

exponential decay

ξ = 1/m

a

distance

Figure 5.6 Behaviors of the Euclidean propagator.

There are two interesting regimes: (a) long distances, m∣x − x ∣ ≫ 1 , and ′ (b) short distances, m∣x − x ∣ ≪ 1. ′

A: long distance behavior

In this regime, z = m∣x − x ∣ ≫ 1, a saddle-point calculation shows that the Bessel Function Kν (z) has the asymptotic behavior, √ π −z e [1 + O(1/z)] (5.172) Kν (z) = 2z ′

5.6 Path integrals for the free scalar field

149

Thus, in this regime the Euclidean propagator (or correlation function) behaves like ′ √ D − 2 −m∣x − x ∣ π/2 m e 1 (0) ′ [1 + O ( )] GE (x − x ) = D−1 m∣x − x′ ∣ (2π)D/2 (m∣x − x′ ∣) 2 (5.173)

Therefore, at long distances, the Euclidean (or imaginary time) propagator has an exponential decay with distance (and imaginary time). The length scale for this decay is 1/m, which is natural since it is the only quantity with units of length in the theory. In real time, and in conventional units, this ̵ length scale is just the Compton wavelength, h/mc. In Statistical Physics this length scale is known as the correlation length ξ. B: short distance behavior In this regime we must use the behavior of the Bessel function for small values of the argument, Kν (z) =

Γ(ν)

ν 2 ( 2z )

+ O(1/z

ν−2

The correlation function now behaves instead like, (0) GE (x

−x)= ′

− 1) Γ( D 2

4π D/2 ∣x − x′ ∣D−2

) + ...

(5.174)

(5.175)

where . . . are terms that vanish as m∣x − x ∣ → 0. Notice that the leading term is independent of the mass m. This is the behavior of the free massless theory. ′

5.6.4 Behavior of the propagator in Minkowski space-time We will now find the behavior of the propagator in real time. This means that now we must do the analytic continuation back to real time. Let us recall that in going from Minkowski to Euclidean space we continued x0 → −ix4 . In addition, there is also factor of i difference in the definition of the propagator. Thus, the propagator in Minkowski space-time (0) ′ G (x − x ) is the expression that results from the analytic continaution (0)

G

(0)

(x − x ) = iGE (x − x )∣x4 →ix0 ′

′

(5.176)

We can also obtain this result from the path integral formulation in

150

Path Integrals in Quantum Mechanics and Quantum Field Theory

Minkowski space-time. Indeed, the generating functional for a free real massive scalar field Z[J] in D = d + 1 dimensional Minkowski space-time is 2

1 m 2 2 i ∫ d x [ (∂φ) − φ + Jφ] 2 2 Z[J] = ∫ Dφe D

(5.177)

Hence, the expectation value to the time-ordered product of two field is ⟨0∣T φ(x)φ(y)∣0⟩ = −

1 δZ[J] 99 99 Z[J] δJ(x)δJ(y) 99J=0

(5.178)

On the other hand, for a free field the generating function is given by (up to a normalization constant N ) i D D (0) ∫ d x ∫ d yJ(x)G (x − y)J(y) 2 Z[J] = N [Det (∂ + m )] e (5.179) where G0 (x − y) is the Green function of the Klein-Gordon operator and satisfies 2

−1/2

2

(0)

(∂ + m ) G 2

2

(x − y) = δ (x − y) D

(5.180)

Hence, we obtain the expected result

(0)

⟨0∣T φ(x)φ(y)∣0⟩ = −iG

(x − y)

(5.181)

Let us compute the propagator in D = 4 Minkowski space-time by analytic continuation from the D = 4 Euclidean propagator. The relativistic interval s is given by s = (x0 − x0 ) − (x − x ) 2

′ 2

′ 2

(5.182)

The Euclidean interval (length) ∣x − x ∣, and the relativistic interval s are related by √ √ ′ (5.183) ∣x − x ∣ = (x − x′ )2 → −s2 ′

Therefore, in D = 4 space-time dimensions, the Minkowski space propagator is √ i m (0) ′ G (x − x ) = 2 √ (5.184) K1 (m −s2 ) 4π −s2

5.6 Path integrals for the free scalar field

151

We will need the asymptotic behavior of the Bessel function K1 (z), √ 3 π −z [1 + K1 (z) = e + . . .] , for z ≫ 1 2z 8z 1 z 1 K1 (z) = z + (ln z + C − ) + . . . , for z ≪ 1 (5.185) 2 2

where C = 0.577215 . . . is the Euler-Mascheroni constant. Let us examine

Oscillatory Behavior time-like separations

light cone

Power law

Decay

Exponential

Exponential

Decay

Decay space-like separations

Oscillatory Behavior

Figure 5.7 Behaviors of the propagator in Minkowski space-time. 2

the behavior of Eq. (5.184) in the regimes: (a) space-like, s < 0, and (b) 2 time-like, s > 0, intervals. A: space-like intervals: (x − x ) = s < 0 ′ 2

2

This is the space-like domain. By√inspecting Eq. (5.184) we see that for space-like separations, the factor −s2 is a positive real number. Consequently the argument of the Bessel function is real (and positive), and the 2 propagator is pure imaginary. In particular we see that, for s < 0 the Minkowski propagator is essentially the Euclidean correlation function, (0)

G

(0)

(x − x ) = iGE (x − x ), ′

′

2

for s < 0

(5.186)

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Path Integrals in Quantum Mechanics and Quantum Field Theory 2

Hence, for s < 0 we have the asymptotic behaviors, √ √ 2 √ π/2 m (0) ′ −m −s2 G (x − x ) =i e , for m −s2 ≫ 1 4π 2 (m√−s2 )3/2 (0)

G

(x − x ) = ′

i , 2 4π (−s2 )

√ for m −s2 ≪ 1 (5.187)

B: time-like intervals: (x − x ) = s > 0 ′ 2

2

This is the time-like domain. The analytic continuation yields √ m (0) ′ (5.188) G (x − x ) = √ K1 (im s2 ) 4π 2 s2 For pure imaginary arguments, the Bessel function K1 (iz) is the analytic (1) continuation of the Hankel function, K1 (iz) = − π2 H1 (−z) . This function is oscillatory for large values of its argument. Indeed, we now get the behaviors √ √ 2 √ π/2 m (0) ′ im s2 G (x − x ) = e , for m s2 ≫ 1 4π 2 (m√s2 )3/2

√ 1 , for m s2 ≪ 1 (5.189) 4π 2 s2 Notice that, up to a factor of i, the short distance behavior is the same for both time-like and space like separations. The main difference is that at large time-like separations we get an oscillatory behavior instead of an exponential decay. The length scale of the oscillations is, once again, set by the only scale in the theory, the Compton wavelength. (0)

G

(x − x ) = ′

5.7 Exponential decays and mass gaps The exponential decay at long space-like separations (and the oscillatory behavior at long time-like separations) is not a peculiarity of the free field theory. It is a general consequence of the existence of a mass gap in the spectrum. We can see that by considering the 2-point function of a generic theory, for simplicity in imaginary time. The 2-point function is (2)

G

′ ′ ˆ ˆ ′ , τ ′ )∣0⟩ (x − x , τ − τ ) = ⟨0∣T φ(x, τ )φ(x

(5.190)

where T is the imaginary time-ordering operator. The Heisenberg representation of the operator φˆ in imaginary time is (h̵ = 1) Hτ ˆ −Hτ ˆ φ(x, τ) = e φ(x, 0) e (5.191)

5.7 Exponential decays and mass gaps

153

Hence, we can write the 2-point function as (2)

G

(x − x ,τ − τ ) = ′

′

′ Hτ ˆ −H(τ − τ ) ˆ ′ −Hτ = θ(τ − τ )⟨0∣e φ(x, 0) e φ(x , 0) e ∣0⟩ ′

′

′ Hτ ˆ ′ −H(τ − τ ) ˆ −Hτ + θ(τ − τ )⟨0∣e φ(x , 0) e φ(x, 0) e ∣0⟩ ′

′

′ E (τ − τ ) −H(τ − τ ) ˆ ′ ˆ = θ(τ − τ ) e 0 ⟨0∣φ(x, 0) e φ(x , 0)∣0⟩ ′

′

′ E (τ − τ ) ˆ ′ , 0) e−H(τ − τ ) φ(x, ˆ + θ(τ − τ ) e 0 ⟨0∣φ(x 0)∣0⟩ (5.192) ′

′

ˆ We now insert a complete set of eigenstates {∣n⟩} of the Hamiltonian H, with eigenvalues {En }. The 2-point function now reads, (2)

G

(x − x ,τ − τ ) = ′

′

′ ˆ ˆ ′ , 0)∣0⟩ e−(En − E0 )(τ − τ ) = θ(τ − τ ) ∑ ⟨0∣φ(x, 0)∣n⟩ ⟨n∣φ(x ′

n

′ −(En − E0 )(τ − τ ) ˆ ′ , 0)∣n⟩ ⟨n∣φ(x, ˆ + θ(τ − τ ) ∑ ⟨0∣φ(x 0)∣0⟩ e ′

n

(5.193)

Since iPˆ ⋅ x ˆ −iPˆ ⋅ x ˆ φ(x, 0) = e φ(0, 0)e

(5.194)

and, in a translation invariant system, the eigenstates of the Hamiltonian are also eigenstates of the total momentum P Pˆ ∣0⟩ = 0,

Pˆ ∣n⟩ = Pn ∣n⟩,

where Pn is the linear momentum of state ∣n⟩, we can write

(5.195)

ˆ ˆ ′ , 0)∣0⟩ = ∣⟨0∣φ(0, ˆ 0)∣n⟩∣2 e−iPn ⋅ (x − x ) ⟨0∣φ(x, 0)∣n⟩ ⟨n∣φ(x ′

(5.196)

Using the above expressions we can rewrite Eq. (5.193) in the form (2)

G

(x − x , τ − τ ) ′

′

ˆ 0)∣n⟩∣2 [θ(τ − τ ′ )e−iPn ⋅ (x − x ) e−(En − E0 )(τ − τ ) = ∑ ∣⟨0∣φ(0, n

′

′

′ −iPn ⋅ (x − x) −(En − E0 )(τ − τ ) + θ(τ − τ )e e ] ′

′

(5.197)

154

Path Integrals in Quantum Mechanics and Quantum Field Theory ′

Thus, at equal positions, x = x , we obtain the following simpler expression ′ in the imaginary time interval τ − τ (2)

G

−(En

ˆ 0)∣n⟩∣ × e (0, τ − τ ) = ∑ ∣⟨0∣φ(0, 2

′

n

− E0 )∣τ − τ ∣ ′

(5.198)

In the limit of large imaginary time separation, ∣τ − τ ∣ → ∞, there is always a largest non-vanishing term in the sums. This is the term for the state ∣n0 ⟩ that mixes with the vacuum state ∣0⟩ through the field operator ˆ and with the lowest excitation energy, the mass gap En − E0 . Hence, for φ, 0 ′ large imaginary time separations, ∣τ − τ ∣ → ∞, the 2-point function decays exponentially, (2)

G

′

′ ˆ 0)∣n0 ⟩∣2 × e−(En0 − E0 )∣τ − τ ∣ (0, τ − τ ) ≃ ∣⟨0∣φ(0, ′

(5.199)

a result that we already derived for a free field in Eq.(5.173). Therefore, if the spectrum has a gap, the correlation functions (or propagators) decay exponentially in imaginary time. In real time we will get, instead, an oscillatory behavior. This is a very general result. Finally, notice that Lorentz invariance in Minkowski space-time (real time) implies rotational (Euclidean) invariance in imaginary time. Hence, exponential decay in imaginary times, at equal positions, must imply (in general) exponential decay in real space at equal imaginary times. Thus, in a Lorentz invariant system the propagator at space-like separations is always equal to the propagator in imaginary time.

5.8 Scalar fields at finite temperature We will now discuss briefly the behavior of free scalar fields in thermal equilibrium at finite temperature T . We will give a more detailed discussion in Chapter 10 where we will discuss more extensively the relation between observables and propagators. We saw in Section 5.5 that the field theory is now defined on an Euclidean cylindrical space-time which is finite and periodic along the imaginary time direction with circumference β = 1/T , where T is the temperature (where we set the Boltzmann constant kB = 1). Hence the imaginary time dimension has been compactified.

5.8 Scalar fields at finite temperature

155

5.8.1 The free energy Let us begin by computing the free energy. We will work in D = d + 1 Euclidean space-time dimensions. The partition function Z(T ) is computed by the result of Eq.(5.146) except that the differential operator now is 2 2 2 Aˆ = −∂τ − ∂ + m ,

(5.200)

2

with the caveat that now ∂ denotes the the Laplacian operator that acts only on the spacial coordinates, x, and that the imaginary time is periodic. The mode expansion for the field in this Euclidean (cylinder) space is ∞

φ(x, τ ) = ∑ ∫ n=−∞

d

d p iω τ +ip⋅x φ(ωn , p)e n d (2π)

(5.201)

where ωn = 2πT n are the Matsubara frequencies and n ∈ Z. The field operator is periodic in the imaginary time τ with period β = 1/T . The Euclidean action now is β 1 1 2 2 1 2 2 d S = ∫ dτ ∫ d x [ (∂τ φ) + (∂φ) + m φ ] 2 2 2 0

β d p 2 2 2 ∫ (p + m )∣φ0 (p)∣ 2 (2π)d d

=

d p 2 2 2 2 ∑ (ωn + p + m ) ∣φ(ωn , p)∣ +β ∫ d (2π) n≥1 d

(5.202)

where we split the action into the sum of the contribution from the zerofrequency Matsubara mode, denoted by φ0 (p) = φ(0, p), and the contributions of the modes for the rest of the frequencies. Since the free energy is given by F (T ) = −T ln Z(T ), we need to compute (again, up to the usual UV divergent normalization constant) F (T ) =

T 2 2 2 ln Det[ − ∂τ − ∂ + m ] (5.203) 2 We can now expand the determinant in the eigenvalues of the operator 2 2 2 −∂τ − ∂ + m , and obtain the formally divergent expression ∞

1 d p 2 2 2 ∑ ln (β[ωn + p + m ]) F (T ) = V T ∫ d 2 (2π) n=−∞ d

(5.204)

where V is the spatial volume. This expression is formally divergent both in the momentum integrals and in the frequency sum and needs to be regularized. We already encountered this problem in our discussion of path integrals in Quantum Mechanics. As in that case we will recall that we have a formally

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Path Integrals in Quantum Mechanics and Quantum Field Theory

divergent normalization constant, N , which we have not made explicit here and that can be defined as to cancel the divergence of the frequency sum (as we did in Eq.(5.84)). The regularized frequency sum can now be computed F (T ) = V T ∫

d p p +m 2 2 1/2 ln [ (β (p + m ) ) ∏ (1 + )] d ωn2 (2π) n=1 2

∞

d

2

Using the identity of Eq.(5.85) the free energy F (T ) becomes √ d d p p 2 + m2 ln [2 sinh ( F (T ) = V T ∫ )] 2T (2π)d √ p 2 + m2 ⎞ ⎛ − d p ⎟ ⎜ ⎟ T F (T ) = V ε0 + V T ∫ ln ⎜ 1−e ⎜ ⎟ d ⎟ ⎜ (2π) ⎠ ⎝

(5.205)

(5.206)

which can be written in the form

d

where

d p √ 2 1 p + m2 ∫ 2 (2π)d

(5.207)

d

ε0 =

(5.208)

is the (ultraviolet divergent) vacuum (ground state) energy density. Notice that the ultraviolet divergence is absent in the finite temperature contribution. 5.8.2 The thermal propagator The thermal propagator is the time-ordered propagator in imaginary time. It is equivalent to the Euclidean correlation function on the cylindrical geometry. We will denote the thermal propagator by (0)

GT (x, τ ) = ⟨φ(x, τ )φ(0, 0)⟩T

(5.209)

It has the Fourier expansion

⟨φ(x, τ )φ(0, 0)⟩T =

∞

d

iω τ +ip⋅x

1 e n d p ∑ ∫ β n=−∞ (2π)d ωn2 + p2 + m2

(5.210)

where, once again, ωn = 2πT n are the Matsubara frequencies. We will now obtain two useful expressions for the thermal propagator. The expression follows from doing the momentum integrals first. In fact, by observing that the Matsubara frequencies act as mass terms of a field in one dimension lower, which allows us to identify the integrals in Eq.(5.210) with

5.8 Scalar fields at finite temperature

157

the Euclidean propagators of an infinite number of fields, each labeled by an integer n, in d Euclidean dimensions with mass squared equal to 2

2

2

(5.211)

mn = m + ω n

We can now use the result of Eq.(5.170) for the Euclidean correlator (now in d Euclidean dimensions) and write the thermal propagator as the following series d ∞ iωn τ −1 1 e m 2 (0) n ( ) GT (x, τ ) = ∑ K d −1 (mn ∣x∣) 2 β n=−∞ (2π)d/2 ∣x − x′ ∣

(5.212)

where mn is given in Eq.(5.211). Since the thermal propagator is expressed as an infinite series of massive propagators, each with increasing masses, it −1 implies that at distances large compared with the length scale λT = (2πT ) , known as the thermal wavelength, all the terms of the series become negligible compared with the term with vanishing Matsubara frequency. In this limit, the thermal propagator reduces to the correlator of the classical theory in d (spatial) Euclidean dimensions, (0)

GT (x, τ ) ≃ ⟨φ(x)φ(0)⟩,

for ∣x∣ ≫ λT

(5.213)

In other terms, at distances large compared with the circumference β of the the cylindrical Euclidean space-time, the theory becomes asymptotically equivalent to the Euclidean theory in one space-time dimension less.

C+ ! +i m2 + p2

C+

C− ! −i m2 + p2

C−

Figure 5.8

158

Path Integrals in Quantum Mechanics and Quantum Field Theory

We will now find an alternative expression for the thermal propagator by doing the sum over Matsubara frequencies shown in Eq.(5.210). We will use the residue theorem to represent the sum as a contour integral on the complex plane, as shown in Fig.5.8, ∞

1 1 z e n dz e = ∮ cot ( ) ∑ 2 2 2 2 2 2 2 2πi 2T β n=−∞ ωn + p + m z +m +p C+ ∪C− iω τ

izτ

(5.214)

where the (positively oriented) contour C = C− ∪ C+ in the complex z plane is shown in Fig.5.8. The black dots on the real axis of represent the integers √ z = n, while the black dots on the imaginary axis represent the poles at ±i m2 + p2 . Upon distorting the contour C+ to the negatively oriented + contour C of the upper half-plane, and the contour C− to the negatively − oriented contour C of the lower half-plane, we can evaluate the integrals by using the residue theorem once again, but now at the poles on the imaginary axis. This computation yields the following result for the thermal propagator √ p 2 + m2 √ coth ( ) d 2T d p −∣τ ∣ p2 + m2 ip ⋅ x GT (x, τ ) = ∫ e e √ (2π)d 2 p 2 + m2

(5.215)

This expression applies to the regime τ ≪ β = 1/T in which quantum fluctuations play a dominant role. After a little algebra, we can now write the thermal propagator as √ −∣τ ∣ p2 + m2 ip ⋅ x d p e e GT (x, τ ) = ∫ √ d 2 2 (2π) 2 p +m d

+∫

d

d p (2π)d

1 √ p 2 + m2 exp ( )−1 T

√ −∣τ ∣ p2 + m2 ip ⋅ x e e √ p 2 + m2

(5.216)

By inspection of Eq.(5.216), we see that the first term on the r.h.s is the T → 0 limit, and that it is just (as it should be!) the propagator in D = d + 1 (0) Euclidean space-time dimensions GE (x, τ ), after an integration over frequencies. The second term in the r.h.s. of Eq.(5.216) describes the contributions to the thermal propagator from thermal fluctuations, shown in the

5.8 Scalar fields at finite temperature

159

form of the Bose occupation numbers, the Bose-Einstein distribution, n(p, T ) =

1 √ 2 p + m2 exp ( )−1 T

(5.217)

The appearance of the Bose-Einstein distribution was to be expected (and, in fact, required) since the excitations of the scalar field are bosons. Finally, we can find the time-ordered propagator, in real time x0 , at finite (0) temperature T , that we will denote by G (x, x0 ; T ). By means of the analytic continuation τ → ix0 of the of the thermal propagator of Eq.(5.216), we find, (0)

G

(0)

(x, x0 ; T ) = GM (x) +∫ (0)

d

d p (2π)d

1 √ p 2 + m2 exp ( )−1 T

√ −i∣x0 ∣ p2 + m2 ip ⋅ x e e √ p 2 + m2

(5.218)

where GM (x) is the (Lorentz-invariant) Minkowski space-time propagator in D dimensions (i.e. at zero temperature), given in Section 5.6.4. Notice that the finite temperature contribution is not Lorentz-invariant. This result was expected since at finite temperature space and time do not play equivalent roles.

160

Path Integrals in Quantum Mechanics and Quantum Field Theory

Exercises 5.1 Path Integral for a particle in a double well potential Consider a particle with coordinate q, mass m moving in the onedimensional double well potential V (q) V (q) = λ (q − q0 ) 2

2 2

(5.219)

In this problem you will use the path integral methods, in imaginary time, that were discussed in class to calculate the matrix element, 1 − HT T −T h ⟨q0 , ∣ − q0 , ∣ − q0 ⟩ ⟩ = ⟨q0 ∣e ̵ 2 2

(5.220)

to leading order in the semiclassical expansion, in the limit T → ∞. 1 Write down the expression of the imaginary time path integral that is appropriate for this problem. Write an explicit expression for the Euclidean Lagrangian (the Lagrangian in imaginary time). How does it differ from the Lagrangian in real time? Make sure that you specify the initial and final conditions. Do not calculate anything yet! 2 Derive the Euler-Lagrange equation for this problem (always in imaginary time). Compare it with the equation of motion in real time. Find the explicit solution for the trajectory (in imaginary time) that satisfies the initial and final conditions. Is the solution unique? Explain. What is the physical interpretation of this trajectory and of the amplitude? Hint: Your equation of motion in imaginary time looks like a funny. A simple way to solve for the trajectory that you need in this problem is to think of this equation of motion as if imaginary time was real time, then to find the analog of the classical energy and to use the conservation of energy to find the trajectory. 3 Compute the imaginary time action for the trajectory you found above. 4 Expand around the solution you found above. Write a formal expression of the amplitude to leading order. Find an explicit expression for the operator that enters in the fluctuation determinant. 5.2 Path Integral for a charged particle moving on a plane in the presence of a perpendicular magnetic field. Consider a particle of mass m and charge −e moving on a plane in the presence of an external uniform magnetic field perpendicular to the plane and with strength B. Let r = (x1 , x2 ) and p = (p1 , p2 ) represent

6 Non-Relativistic Field Theory

In this Chapter we will discuss the field theory description of non-relativistic systems. The material that we will be presented here is, for the most part, introductory as this topic is covered in depth in many classic textbooks, such as Methods of Quantum Field Theory in Statistical Physics by A. A. Abrikosov, L. P. Gorkov and I. E. Dzyaloshinskii (Abrikosov et al., 1963), Statistical Mechanics, A Set of lectures by R. P. Feynman (Feynman, 1972) (among many others). The discussion of “second quantization” is very standard, and similar to Feynman’s presentation. It is presented here for pedagogical reasons.

6.1 Second quantization and the many-body problem Let us consider now the problem of a system of N identical non-relativistic particles. For the sake of simplicity I will assume that the physical state of each particle j is described by its position xj relative to some reference frame. This case is easy to generalize. The wave function for this system is Ψ(x1 , . . . , xN ). If the particles are 2 identical then the probability density, ∣Ψ(x1 , . . . , xN )∣ , must be invariant under arbitrary exchanges of the labels that we use to identify (or designate) the particles. In quantum mechanics, the particles do not have well defined trajectories. Only the states of a physical system are well defined. Thus, even though at some initial time t0 the N particles may be localized at a set of well defined positions x1 , . . . , xN , they will become delocalized as the system evolves. Furthermore the Hamiltonian itself is invariant under a permutation of the particle labels. Hence, permutations constitute a symmetry of a manyparticle quantum mechanical system. In other words, in quantum mechanics identical particles are indistinguishable. In particular, the probability density

164

Non-Relativistic Field Theory

of any eigenstate must remain invariant if the labels of any pair of particles are exchanged. If we denote by Pjk the operator which exchanges the labels of particles j and k, the wave functions must satisfy

Pjk Ψ(x1 , . . . , xj , . . . , xk , . . . , xN ) = e Ψ(x1 , . . . , xj , . . . , xk , . . . , xN ) (6.1) Under a further exchange operation, the particles return to their initial labels and we recover the original state. This sample argument then requires that φ = 0, π since 2φ must not be an observable phase. We then conclude that there are two possibilities: either Ψ is even under permutation and P Ψ = Ψ, or Ψ is odd under permutation and P Ψ = −Ψ. Systems of identical particles which have wave functions which are even under a pairwise permutation of the particle labels are called bosons. In the other case, Ψ odd under pairwise permutation, they are Fermions. It must be stressed that these arguments only show that the requirement that the state prioriΨ be either even or odd is only a sufficient condition. It turns out that under special circumstances the phase factor φ may take values different from 0 or π. These particles are called anyons, and will be discussed in Chapter 22. It turns out that the only cases in which they may exist is if the particles are restricted to move on a line or on a plane. In the case of relativistic quantum field theories, the requirement that the states have well defined statistics (or symmetry) is demanded by a very deep and fundamental theorem that links the statistics of the states of the spin of the field. This is the spin-statistics theorem and will be discussed later on. iφ

6.1.1 Fock space We will now discuss a procedure, known as Second Quantization, which will enable us to keep track of the symmetry of the states in a simple way. Let us consider sgsin a system of N particles. The wave functions in the coordinate representation are Ψ(x1 , . . . , xN ) where the labels x1 , . . . , xN denote both the coordinates and the spin states of the particles in the state ∣Ψ⟩. For the sake of definiteness we will discuss physical systems describable ˆ of the form by Hamiltonians H N ̵2 N ˆ = − h ∑ "2j + ∑ V (xj ) + g ∑ U (xj − xk ) + . . . H 2m j=1

j=1

j,k

(6.2)

Let {φn (x)} be the wave functions for a complete set of one-particle states. Then an arbitrary N -particle state can be expanded in a basis of the tensor

6.1 Second quantization and the many-body problem

165

product of the one-particle states, namely Ψ(x1 , . . . , xN ) = ∑ C(n1 , . . . nN )φn1 (x1 ) . . . φnN (xN ) {nj}

(6.3)

Thus, if Ψ is symmetric (antisymmetric) under an arbitrary exchange xj → xk , the coefficients C(n1 , . . . , nN ) must be symmetric (antisymmetric) under the exchange nj ↔ nk . A set of N -particle basis states with well defined permutation symmetry is the properly symmetrized or antisymmetrized tensor product 1 P ∣Ψ1 , . . . ΨN ⟩ ≡ ∣Ψ1 ⟩ × ∣Ψ2 ⟩ × ⋯ × ∣ΨN ⟩ = √ ∑ ξ ∣ΨP (1) ⟩ × ⋯ × ∣ΨP (N ) ⟩ N! P (6.4) where the sum runs over the set of all possible permutation P . The weight factor ξ is +1 for bosons and −1 for fermions. For fermions, the N -particle state vanishes if two particles are in the same one-particle state. This is the exclusion principle. The inner product of two N -particle states is ⟨χ1 , . . . χN ∣ψ1 , . . . , ψN ⟩ =

1 P +Q ∑ξ ⟨χQ(1) ∣ψP (1) ⟩⋯⟨χQ(N ) ∣ψP (1) ⟩ = N! P,Q

= ∑ ξ ⟨χ1 ∣ψP (1) ⟩⋯⟨χN ∣ψP (N ) ⟩ P

′

(6.5)

P′

which is nothing but the permanent (determinant) of the matrix ⟨χj ∣ψk ⟩ for symmetric (antisymmetric) states, %% ⟨χ1 ∣ψ1 ⟩ . . . ⟨χ1 ∣ψN ⟩ %% ⋮ ⋮ ⟨χ1 , . . . χN ∣ψ1 , . . . ψN ⟩ = %%%% %% %% ⟨χN ∣ψ1 ⟩ . . . ⟨χN ∣ψN ⟩

%% %% %% %% %% %%ξ

(6.6)

In the case of antisymmetric states, the inner product is the familiar Slater determinant. Let us denote by {∣α⟩} the complete set of one-particle states which satisfy ⟨α∣β⟩ = δαβ ,

∑ ∣α⟩⟨α∣ = 1 α

(6.7)

The N -particle states are by {∣α1 , . . . αN ⟩}. Because of the symmetry requirements, the labels αj can be arranged in the form of a monotonic sequence α1 ≤ α2 ≤ ⋯ ≤ αN for bosons, or in the form of a strict monotonic sequence α1 < α2 < ⋯ < αN for fermions. Let nj be an integer which counts how many particles are in the j-th one-particle state. The boson

166

Non-Relativistic Field Theory

states ∣α1 , . . . αN ⟩ must be normalized by a factor of the form 1 ∣α1 , . . . , αN ⟩, √ n1 ! . . . nN !

with

α1 ≤ α2 ≤ ⋯ ≤ αN

(6.8)

where nj are non-negative integers. For fermions the states are ∣α1 , . . . , αN ⟩,

with

α1 < α2 < ⋯ < αN

(6.9)

but now nj = 0, 1. These N -particle states are complete and orthonormal 1 ∑ ∣α1 , . . . , αN ⟩⟨α1 , . . . , αN ∣ = Iˆ N ! α ,...,α 1

(6.10)

N

where the sum over the α’s is unrestricted and the operator Iˆ is the identity operator in the space of N -particle states. We will now consider the more general problem in which the number of particles N is not fixed a-priori. Rather, we will consider an enlarged space of states in which the number of particles is allowed to fluctuate. In the language of Statistical Physics what we are doing is to go from the Canonical Ensemble to the Grand Canonical Ensemble. Thus, let us denote by H0 the Hilbert space with no particles, H1 the Hilbert space with only one particle and, in general, HN the Hilbert space for N -particles. The direct sum of these spaces H H = H0 ⊕ H1 ⊕ ⋯ ⊕ HN ⊕ ⋯

(6.11)

is called the Fock space. An arbitrary state ∣ψ⟩ in Fock space is the sum over the subspaces HN , ∣ψ⟩ = ∣ψ

(0)

⟩ + ∣ψ

(1)

⟩ + ⋯ + ∣ψ

(N )

⟩+⋯

(6.12)

The subspace with no particles is a one-dimensional space spanned by the vector ∣0⟩ which we will call the vacuum. The subspaces with well defined number of particles are defined to be orthogonal to each other in the sense that the inner product in the Fock space ∞

⟨χ∣ψ⟩ ≡ ∑⟨χ j=0

(j)

∣ψ

(j)

⟩

vanishes if ∣χ⟩ and ∣ψ⟩ belong to different subspaces.

(6.13)

6.1 Second quantization and the many-body problem

167

6.1.2 Creation and annihilation operators

Let ∣φ⟩ be an arbitrary one-particle state. Let us define the creation operator † a ˆ (φ) by its action on an arbitrary state in Fock space a ˆ (φ)∣ψ1 , . . . , ψN ⟩ = ∣φ, ψ1 , . . . , ψN ⟩ †

(6.14)

Clearly, a ˆ (φ) maps the N -particle state with proper symmetry ∣ψ1 , . . . , ψN ⟩ onto the N + 1-particle state ∣φ, ψ, . . . , ψN ⟩, also with proper symmetry . The destruction or annihilation operator a ˆ(φ) is defined to be the adjoint of † a ˆ (φ) , †

⟨χ1 , . . . , χN −1 ∣ˆ a(φ)∣ψ1 , . . . , ψN ⟩ = ⟨ψ1 , . . . , ψN ∣ˆ a (φ)∣χ1 , . . . , χN −1 ⟩ (6.15) Hence ∗

†

⟨χ1 , . . . , χN −1 ∣ˆ a(φ)∣ψ1 , . . . , ψN ⟩ =⟨ψ1 , . . . , ψN ∣φ, χ1 , . . . , χN −1 ⟩ = %% ⟨ψ1 ∣φ⟩ ⟨ψ1 ∣χ1 ⟩ ⋯ ⟨ψ1 ∣χN −1 ⟩ %% ⋮ ⋮ ⋮ = %%%% %% %% ⟨ψN ∣φ⟩ ⟨ψN ∣χ1 ⟩ ⋯ ⟨ψN ∣χN −1 ⟩ ∗

%%∗ %% %% %% %% %%ξ

(6.16)

We can now expand the permanent (or determinant) to get ⟨χ1 , . . . , χN −1 ∣ˆ a(φ)∣ψ1 , . . . , ψN ⟩ =

%% ⟨ψ1 ∣χ1 ⟩ ... ⟨ψ1 ∣χN −1 ⟩ %% %% ⋮ ⋮ k−1 = ∑ ξ ⟨ψk ∣φ⟩ %%%% ... (noψk ) ... %% k=1 %% ... ⟨ψN ∣χN −1 ⟩ %% ⟨ψN ∣χ1 ⟩ N

N

= ∑ξ

k−1

k=1

⟨ψk ∣φ⟩⟨χ1 , . . . , χN −1 ∣ψ1 , . . . ψˆk . . . , ψN ⟩

%%∗ %% %% %% %% %% %% %%ξ

(6.17)

where ψˆk denotes that the one-particle state ψk is absent. Thus, the destruction operator is given by N

a ˆ(φ)∣ψ1 , . . . , ψN ⟩ = ∑ ξ k=1

k−1

⟨φ∣ψk ⟩∣ψ1 , . . . , ψˆk , . . . , ψN ⟩

(6.18)

With these definitions, we can easily see that the operators a ˆ (φ) and a ˆ(φ) †

168

Non-Relativistic Field Theory

obey the commutation relations a ˆ (φ1 )ˆ a (φ2 ) = ξ a ˆ (φ2 )ˆ a (φ1 ) †

†

†

†

(6.19)

Let us introduce the notation

ˆ B] ˆ ≡ AˆB ˆ −ξ B ˆ Aˆ [A,

(6.20)

ξ

ˆ are two arbitrary operators. For ξ = +1 (bosons) we have where Aˆ and B the commutator [ˆ a (φ1 ), a ˆ (φ2 )]+1 ≡ [ˆ a (φ1 ), a ˆ (φ2 )] = 0 †

†

†

†

(6.21)

while for ξ = −1 it is the anticommutator

[ˆ a (φ1 ), a ˆ (φ2 )]−1 ≡ {ˆ a (φ1 ), a ˆ (φ2 )} = 0 †

†

†

†

(6.22)

Similarly for any pair of arbitrary one-particle states ∣φ1 ⟩ and ∣φ2 ⟩ we get [ˆ a(φ1 ), a ˆ(φ2 )]−ξ = 0

(6.23)

It is also easy to check that the following identity holds [ˆ a(φ1 ), a ˆ (φ2 )]−ξ = ⟨φ1 ∣φ2 ⟩ †

(6.24)

So far we have not picked any particular representation. Let us consider the occupation number representation in which the states are labelled by the number of particles nk in the single-particle state k. In this case, we have 1 ∣n1 , . . . , nk , . . .⟩ ≡ √ ∣1, . . . , 1, 2, . . . , 2, . . .⟩ n1 !n2 ! . . .

(6.25)

where the state has n1 particles in state 1, n2 particles in state 2, and so on. In the case of bosons, the nj ’s can be any non-negative integer, while for fermions they can only be equal to zero or one. In general we have that if ∣α⟩ is the αth single-particle state, then √ † a ˆα ∣n1 , . . . , nα , . . .⟩ = nα + 1∣n1 , . . . , nα + 1, . . .⟩ √ a ˆα ∣n1 , . . . , nα , . . .⟩ = nα ∣n1 , . . . , nα − 1, . . .⟩ (6.26)

Thus for both fermions and bosons, a ˆα annihilates all states with nα = 0, † while for fermions a ˆα annihilates all states with nα = 1. The commutation relations are [ˆ aα , a ˆβ ] = [ˆ aα , a ˆβ ] = 0 †

†

[ˆ aα , a ˆβ ] = δαβ †

(6.27)

6.1 Second quantization and the many-body problem

for bosons, and

{ˆ aα a ˆβ } = {ˆ aβ , a ˆβ } = 0 †

†

{ˆ aα a ˆβ } = δαβ †

169

(6.28)

ˆ B} ˆ is the anticommutator of the operators Aˆ and B ˆ for fermions. Here, {A, ˆ B} ˆ ≡ AˆB ˆ +B ˆ Aˆ {A,

(6.29)

If a unitary transformation is performed in the space of one-particle state vectors, then a unitary transformation is induced in the space of the operators themselves, i.e. if ∣χ⟩ = α∣ψ⟩ + β∣φ⟩, then a ˆ(χ) = α a ˆ(ψ) + β a ˆ(φ) ∗

∗

a ˆ (χ) = αˆ a (ψ) + βˆ a (φ) †

†

†

(6.30)

and we say that a ˆ (χ) transforms like the ket ∣χ⟩ while a ˆ(χ) transforms like the bra ⟨χ∣. For example, we can pick as the complete set of one-particle states the momentum states {∣p⟩}. This is “momentum space”. With this choice the commutation relations are †

[ˆ a (p), a ˆ (q)]ξ = [ˆ a(p), a ˆ(q)]ξ = 0 †

†

[ˆ a(p), a ˆ (q)]ξ = (2π) δ (p − q) d d

†

(6.31)

where d is the dimensionality of space. In this representation, an N -particle state is ∣p1 , . . . , pN ⟩ = a ˆ (p1 ) . . . a ˆ (pN )∣0⟩ †

†

(6.32)

On the other hand, we can also pick the one-particle states to be eigenstates of the position operators, i.e. ∣x1 , . . . xN ⟩ = a ˆ (x1 ) . . . a ˆ (xN )∣0⟩ †

†

(6.33)

In position space, the operators satisfy

[ˆ a (x1 ), a ˆ (x2 )]ξ =[ˆ a(x1 ), a ˆ(x2 )]ξ = 0 †

†

[ˆ a(x1 ), a ˆ (x2 )]ξ =δ (x1 − x2 ) †

d

(6.34)

This is the position space or coordinate representation. A transformation from position space to momentum space is the Fourier transform ∣p⟩ = ∫ d x ∣x⟩⟨x∣p⟩ = ∫ d x ∣x⟩e d

d

ip⋅x

(6.35)

170

Non-Relativistic Field Theory

and, conversely ∣x⟩ = ∫

d p −ip⋅x ∣p⟩e d (2π) d

(6.36)

Then, the operators themselves obey

a ˆ (p) = ∫ d x a ˆ (x)e d

†

a ˆ (x) = ∫ †

(1)

†

ip⋅x

d p † −ip⋅x a ˆ (p)e (2π)d d

(6.37)

6.1.3 General operators in Fock space

Let A be an operator acting on one-particle states. We can always define an extension of A acting on any arbitrary state ∣ψ⟩ of the N -particle Hilbert space as follows: (1) ̂ A∣ψ⟩ ≡ ∑ ∣ψ1 ⟩ × . . . × A ∣ψj ⟩ × . . . × ∣ψN ⟩ N

(6.38)

j=1

For instance, if the one-particle basis states {∣ψj ⟩} are eigenstates of A with eigenvalues {aj } we get ⎛N ⎞ ̂ A∣ψ⟩ = ⎜∑ aj ⎟ ∣ψ⟩ ⎝j=1 ⎠

(6.39)

We wish to find an expression for an arbitrary operator A in terms of creation (1) and annihilation operators. Let us first consider the operator Aαβ = ∣α⟩⟨β∣ (1)

which acts on one-particle states. The operators Aαβ form a basis of the space of operators acting on one-particle states. Then, the N -particle extension of Aαβ is N

Âαβ ∣ψ⟩ = ∑ ∣ψ1 ⟩ × ⋯ × ∣α⟩ × ⋯ × ∣ψN ⟩⟨β∣ψj ⟩

(6.40)

j=1

Thus

1234 Âαβ ∣ψ⟩ = ∑ ∣ψ1 , . . . , α , . . . , ψN ⟩ ⟨β∣ψj ⟩ N

j=1

j

(6.41)

6.1 Second quantization and the many-body problem

171

In other words, we can replace the one-particle state ∣ψj ⟩ from the basis with the state ∣α⟩ at the price of a weight factor, the overlap ⟨β∣ψj ⟩. This operator has a very simple expression in terms of creation and annihilation operators. Indeed, a ˆ (α)ˆ a(β)∣ψ⟩ = ∑ ξ N

†

k−1

k=1

⟨β∣ψk ⟩ ∣α, ψ1 , . . . , ψk−1 , ψk+1 , . . . , ψN ⟩

(6.42)

We can now use the symmetry of the state to write

1234 ∣α, ψ1 , . . . , ψk−1 , ψk+1 , . . . , ψN ⟩ = ∣ψ1 , . . . , α , . . . , ψN ⟩ k

ξ

k−1

(6.43)

Thus the operator Aαβ , the extension of ∣α⟩⟨β∣ to the N -particle space, † coincides with a ˆ (α)ˆ a(β) † Â αβ ≡ a ˆ (α)ˆ a(β)

(6.44) (1)

We can use this result to find the extension for an arbitrary operator A of the form (1)

A we find

(1)

= ∑ ∣α⟩⟨α∣A α,β

∣β⟩ ⟨β∣

† (1) Â = ∑ a ˆ (α)ˆ a(β)⟨α∣A ∣β⟩

(6.45)

(6.46)

α,β

(1)

Hence the coefficients of the expansion are the matrix elements of A between arbitrary one-particle states. We now discuss a few operators of interest. A: the identity operator ˆ The identity operator 1 of the one-particle Hilbert space ˆ1 = ∑ ∣α⟩⟨α∣

(6.47)

α

̂ becomes the number operator N

† ̂ = ∑a N ˆ (α)ˆ a(α)

(6.48)

α

In position and in momentum space we find ̂=∫ N

d p † d † d a ˆ (p)ˆ a(p) = ∫ d x a ˆ (x)ˆ a(x) = ∫ d x ρˆ(x) d (2π) d

(6.49)

172

Non-Relativistic Field Theory

where ρˆ(x) = a ˆ (x)ˆ a(x) is the particle density operator. †

B: the linear momentum operator

In the space H1 , the linear momentum operator is (1)

pˆj

=∫

d d p h̵ d p ∣p⟩⟨p∣ = ∫ d x ∣x⟩ ∂ ⟨x∣ j i j (2π)d

(6.50)

̂j is Thus, we get that the total momentum operator P ̂j = ∫ P

d d p h̵ † d † pa ˆ (p)ˆ a(p) = ∫ d x a ˆ (x) ∂j a ˆ(x) d j i (2π)

(6.51)

C: Hamiltonian

The one-particle Hamiltonian H H has the matrix elements ⟨x∣H

(1)

∣y⟩ = −

Thus, in Fock space we get

(1)

(1)

2

p + V (x) = 2m

2 h̵ 2 d d " δ (x − y) + V (x)δ (x − y) 2m

† ̂ = ∫ dd x a H ˆ (x) (−

2 h̵ 2 " +V (x)) a ˆ(x) 2m

(6.52)

(6.53)

(6.54)

in position space. In momentum space we can define ̃(q) = ∫ dd x V (x)e−iq⋅x V

(6.55)

the Fourier transform of the potential V (x), and get ̂=∫ H

2

d p d p p † d q † ̃(q)ˆ a ˆ (p)ˆ a(p)+∫ ∫ V a (p+q)ˆ a(p) (6.56) d d (2π) 2m (2π) (2π)d d

d

d

The last term has a very simple physical interpretation. When acting on a one-particle state with well-defined momentum, say ∣p⟩, the potential term yields another one-particle state with momentum p + q, where q is the momentum transfer, with amplitude V˜ (q). This process is usually depicted by the Feynman diagram of Fig.6.1.

6.1 Second quantization and the many-body problem a ˆ (p + q)

173

†

p+q q

a ˆ(p)

̃(q) V

p

Figure 6.1 One-body scattering.

D: two-Body interactions (2) A two-particle interaction is a operator Vˆ which acts on the space of two-particle states H2 , has the form V

(2)

=

1 (2) ∑ ∣α, β⟩V (α, β)⟨α, β∣ 2

(6.57)

α,β

(2)

The methods developed above yield an extension of V to Fock space of the form 1 † † (2) ̂= ∑ a ˆ (α)ˆ a (β)ˆ a(β)ˆ a(α) V (α, β) (6.58) V 2 α,β

In position space, ignoring spin, we get

1 † † (2) ̂ = ∫ dd x ∫ dd y a V ˆ (x) a ˆ (y) a ˆ(y) a ˆ(x) V (x, y) 2 1 1 d d (2) d (2) ≡ ∫ d x ∫ d y ρˆ(x)V (x, y)ˆ ρ(y) + ∫ d x V (x, x) ρˆ(x) (6.59) 2 2

while in momentum space we find ̂= V

d p 1 d q d k † † ̃(k) a ∫ ∫ ∫ V ˆ (p + k)ˆ a (q − k)ˆ a(q)ˆ a(p) (6.60) d d 2 (2π)d (2π) (2π) d

d

d

̃(k) is only a function of the momentum transfer k, represented where V in the Feynman diagram of Fig.6.2. This is a consequence of translation invariance. In particular for a Coulomb interaction, V

(2)

(x, y) =

2

e ∣x − y∣

(6.61)

174

Non-Relativistic Field Theory

for which we have ̃(k) = V a ˆ (p + k) †

2

4πe k2

(6.62) a ˆ (q − k) †

p+k

q−k k ̃(k) V

a ˆ(p)

p

q

a ˆ(q)

Figure 6.2 Two-body interaction.

6.2 Non-relativistic field theory and second quantization We can now reformulate the problem of an N -particle system as a nonrelativistic field theory. The procedure described in the previous section is commonly known as Second Quantization. If the (identical) particles are bosons, the operators a ˆ(φ) obey canonical commutation relations. If the (identical) particles are Fermions, the operators a ˆ(φ) obey canonical anticommutation relations. In position space, it is customary to represent a ˆ(φ) ˆ by the operator ψ(x) which obeys the equal-time algebra † d ˆ [ψ(x), ψˆ (y)]ξ =δ (x − y)

† † ˆ ˆ [ψ(x), ψ(y)] = [ψˆ (x), ψˆ (y)]ξ = 0 ξ

(6.63)

where we have used the notation

ˆ B] ˆ ≡ AˆB ˆ − ξB ˆ Aˆ [A, ξ

(6.64)

with ξ = 1 for bosons and ξ = −1 for fermions. In this framework, the one-particle Schr¨odinger equation becomes the classical field equation [ih̵

2 ∂ h̵ 2 + " −V (x)] ψˆ = 0 ∂t 2m

(6.65)

6.3 Non-relativistic fermions at zero temperature

175

Can we find a Lagrangian density L from which the one-particle Schr¨odinger equation follows as its classical equation of motion? The answer is yes and L is given by 2 h̵ † † !ψ ⋅ "ψ − V (x)ψ ψ 2m Its Euler-Lagrange equations are

̵ ∂t ψ − L = ihψ †

∂t

δL δL δL = −! ⋅ + † † † δ∂t ψ δ!ψ δψ

(6.66)

(6.67)

which are equivalent to the field Equation Eq. 6.65. The canonical momenta † Πψ and Πψ are Πψ =

δL ̵ † = ihψ δ∂t ψ

(6.68)

δL ̵ = −ihψ δ∂t ψ †

(6.69)

and Πψ† =

Thus, the (equal-time) canonical commutation relations are ˆ ˆ ψ (y)] = ihδ(x ̵ [ψ(x), Π − y) ξ

which require that

ˆ [ψ(x), ψˆ (y)] = δ (x − y) d

†

ξ

(6.70)

(6.71)

Notice that the canonical momenta are proportional to the field. This is a consequence of the fact that the Lagrangian is first order in time derivatives. It is also necessary for the equal-time commutation and anticommutation relations to be those of creation and annihilation operators.

6.3 Non-relativistic fermions at zero temperature The results of the previous sections tell us that the action for non-relativistic fermions with two-body interactions is, in D = d + 1 space-time dimensions, 2 h̵ † † † ˆ ˆ ˆ ̵ S = ∫ d x [ψ ih∂t ψ − !ψˆ ⋅ !ψˆ − V (x)ψˆ (x)ψ(x)] 2m D

1 D D ′ † † ′ ′ ˆ ′ ˆ − ∫ d x ∫ d x ψˆ (x)ψˆ (x )U (x − x )ψ(x )ψ(x) 2

(6.72)

176

Non-Relativistic Field Theory

where U (x − x ) represents instantaneous pair-interactions, ′

U (x − x ) ≡ U (x − x )δ(x0 − x0 ) ′

′

′

(6.73)

ˆ for this system is The Hamiltonian H

2 h̵ d † † ˆ H =∫ d x [ !ψˆ ⋅ !ψˆ + V (x)ψˆ (x)ψ(x)] 2m 1 d d ′ † † ′ ′ ˆ ′ ˆ + ∫ d x ∫ d x ψˆ (x)ψˆ (x )U (x − x )ψ(x )ψ(x) 2

(6.74)

† For fermions the fields ψˆ and ψˆ satisfy equal-time canonical anticommutation relations † ′ ˆ {ψ(x), ψˆ (x)} = δ(x − x ) (6.75)

while for bosons they satisfy

ˆ [ψ(x), ψˆ (x )] = δ(x − x ) †

′

′

(6.76)

ˆ commutes with the total number operator In both cases, the Hamiltonian H d † ˆ ˆ = ∫ d xψˆ (x)ψ(x) ˆ conserves the total number of particles. N since H

The Fock space picture of the many-body problem is equivalent to the Grand Canonical Ensemble of Statistical Mechanics. Thus, instead of fixing the number of particles we can introduce a Lagrange multiplier µ, the chemical potential, to weigh contributions from different parts of the Fock space, ˜ and we define the operator H ̃≡H ̂ − µN ̂ H

(6.77)

ˆ this amounts to a shift of the energy by µN . In a Hilbert space with fixed N We will now allow the system to choose the sector of the Fock space but ˆ is fixed to with the requirement that the average number of particles ⟨N⟩ ¯ be some number N . In the thermodynamic limit (N → ∞), the chemical potential µ represents the difference of the ground state energies between two sectors with N + 1 and N particles respectively. The modified Hamiltonian ˜ is H ̃ = ∫ dd x ∑ ψˆσ† (x) (− H σ

2 h̵ 2 " +V (x) − µ) ψˆσ (x) 2m

1 † d d † + ∫ d x ∫ d y ∑ ψˆσ (x)ψˆσ′ (y)U (x − y)ψˆσ′ (y)ψˆσ (x) 2 ′ σ,σ

(6.78)

6.3 Non-relativistic fermions at zero temperature

177

6.3.1 The ground state of free fermions E Single Particle Energy

Fermi Energy

EF

Fermi Sea

Single Particle States

Occupied States

Empty States

Figure 6.3 The Fermi Sea.

Let us discuss now the very simple problem of finding the ground state for a system of N spinless free fermions. In this case, the interaction potential U (x−y) vanishes, and, if the system is isolated and translationally invariant, the external potential V (x) also vanishes. In general there is a complete set of one-particle states of eigenstates of the single-particle Hamiltonian {∣α⟩}. ˆ is In this basis, the second quantized Hamiltonian H † ˆ = ∑ Eα a H ˆα aα

(6.79)

α

where the index α labels the one-particle states by increasing order of their single-particle energies E1 ≤ E2 ≤ ⋯ ≤ En ≤ ⋯

(6.80)

Since were are dealing with fermions, we cannot put more than one particle in each state. Thus the state the lowest energy is obtained by filling up all the first N single particle states. Let ∣gnd⟩ denote this ground state ∣gnd⟩ = ∏ a ˆα ∣0⟩ ≡ a ˆ1 ⋯ˆ aN ∣0⟩ = ∣1 . . . 1, 00 . . .⟩ N

†

†

†

(6.81)

α=1

where the ket on the extreme right hand side has the lowest N states occu-

178

Non-Relativistic Field Theory

pied and all other states are empty. The energy of this state is Egnd , where Egnd = E1 + ⋯ + EN

(6.82)

The energy of the top-most occupied single particle state, EN , is called the Fermi energy. The set of occupied states is called the filled Fermi sea.

6.3.2 Excited states

A state such as ∣ψ⟩

∣ψ⟩ = ∣1 . . . 1010 . . .⟩

(6.83)

is an excited state. Here the ket denotes a state with the lowest N − 1 states and the N + 1 state are occupied, and all other states are empty. It is obtained by removing one particle from the single particle state N (thus leaving a hole behind), and putting the particle in the unoccupied single particle state N + 1. This is a state with one particle-hole pair, and it has ∣gnd⟩

N N +1 Single Particle States

†

a ˆN +1 a ˆN ∣Ψ⟩

Single Particle States

Figure 6.4 An excited (particle-hole) state.

the form

∣1 . . . 1010 . . .⟩ = a ˆN +1 a ˆN ∣gnd⟩ †

(6.84)

The energy of this state is

Eψ = E1 + ⋯ + EN −1 + EN +1

(6.85)

Eψ = Egnd + EN +1 − EN

(6.86)

Hence

Since EN +1 ≥ EN , Eψ ≥ Egnd , the excitation energy εψ = Eψ − Egnd is positive, εψ = EN +1 − EN ≥ 0

(6.87)

6.3 Non-relativistic fermions at zero temperature

179

6.3.3 Normal-ordering and particle-hole transformation: construction of the physical Hilbert space

It is apparent that, instead of using the empty state ∣0⟩ for reference state, it is physically more reasonable to use instead the filled Fermi sea ∣gnd⟩ as the physical reference state or vacuum state. Thus this state is a vacuum in the sense of absence of excitations. These arguments motivate the introduction of the particle-hole transformation. Let us introduce the fermion operators bα such that † ˆbα = a ˆα

for α ≤ N

(6.88)

† Since a ˆα ∣gnd⟩ = 0 (for α ≤ N ) the operators ˆbα annihilate the ground state ∣gnd⟩, i.e. ˆbα ∣gnd⟩ = 0 (6.89)

The following canonical anticommutation relations hold

† ′ ′ {ˆ aα , a ˆα } = {ˆ aα , ˆbβ } = {ˆbβ , ˆbβ } = {ˆ aα , ˆbβ } = 0

{ˆ aα , a ˆα′ } = δαα′ , †

† {ˆbβ , ˆbβ ′ } = δββ ′

(6.90)

where α, α > N and β, β ≤ N . Thus, relative to the state ∣gnd⟩, a ˆα and ˆbβ behave like creation operators. An arbitrary excited state has the form ′

′

†

† † † † ∣α1 . . . αm , β1 . . . βn ; gnd⟩ ≡ a ˆα1 . . . a ˆαm ˆbβ1 . . . ˆbβn ∣gnd⟩

†

(6.91)

This state has m particles, in the single-particle states α1 , . . . , αm , and n holes, in the single-particle states β1 , . . . , βn . The ground state is annihilated by the operators a ˆα and ˆbβ a ˆα ∣gnd⟩ = ˆbβ ∣gnd⟩ = 0

with α > N and β ≤ N

(6.92)

† † † ̂ = ∑ Eα a H ˆα a ˆα = ∑ Eα + ∑ Eα a ˆα a ˆα − ∑ Eβ ˆbβ ˆbβ

(6.93)

ˆ is normal ordered relative to the empty state ∣0⟩, i. The Hamiltonian H ˆ e., H∣0⟩ = 0, but it is not normal ordered relative to the actual ground ˆ state ∣gnd⟩. The particle-hole transformation enables us to normal order H relative to ∣gnd⟩, α

α≤N

α>N

β≤N

where the minus sign in the last term reflects the Femi statistics. Thus ̂ = Egnd + ∶ H ̂∶ H

(6.94)

180

Non-Relativistic Field Theory

where N

Egnd = ∑ Eα

(6.95)

α=1

is the ground state energy, and the normal ordered Hamiltonian is † † ̂ ∶= ∑ Eα a ∶ H ˆa a ˆα − ∑ ˆbβ ˆbβ Eβ α>N

(6.96)

β≤N

̂ is also not normal-ordered relative to ∣gnd⟩ either. The number operator N Thus, we write † † † ̂ = ∑a N ˆα a ˆα = N + ∑ a ˆα aα − ∑ ˆbβ ˆbβ α

α>N

(6.97)

β≤N

We see that particles raise the energy while holes reduce it. However, if we ˆ = ̂ (i.e. [N ̂, H] deal with Hamiltonians that conserve the particle number N 0) for every particle that is removed a hole must be created. Hence particles and holes can only be created in pairs. A particle-hole state ∣α, β gnd⟩ is ∣α, β gnd⟩ ≡ a ˆαˆbβ ∣gnd⟩ † †

(6.98)

It is an eigenstate with an energy

† † ̂ β gnd⟩ = (Egnd + ∶ H ̂ ∶) a H∣α, ˆαˆbβ ∣gnd⟩

= (Egnd + Eα − Eβ ) ∣α, β gnd⟩

(6.99)

and the excitation energy is Eα − Eβ ≥ 0. Hence the ground state is stable to the creation of particle-hole pairs. This state has exactly N particles since ̂∣α, β gnd⟩ = (N + 1 − 1)∣α, β gnd⟩ = N ∣α, β gnd⟩ N

(6.100)

† Let us finally notice that the field operator ψˆ (x) in position space is † † † ψˆ (x) = ∑⟨x∣α⟩ˆ aα = ∑ φα (x)ˆ aα + ∑ φβ (x)ˆbβ α

α>N

β≤N

(6.101)

where {φα (x)} are the single particle wave functions. The procedure of normal ordering allows us to define the physical Hilbert space. The physical meaning of this approach becomes more transparent in the thermodynamic limit N → ∞ and V → ∞ at constant density ρ. In this limit, the Hilbert space is the set of states which is obtained by acting finitely with creation and annihilation operators on the ground state (the vacuum). The spectrum of states that results from this approach consists on the set of

6.3 Non-relativistic fermions at zero temperature

181

states with finite excitation energy. Hilbert spaces which are built on reference states with macroscopically different number of particles are effectively disconnected from each other. Thus, the normal ordering of a Hamiltonian of a system with an infinite number of degrees of freedom amounts to a choice of the Hilbert space. This restriction becomes of fundamental importance when interactions are taken into account. 6.3.4 The free fermi gas Let us consider the case of free spin one-half electrons moving in free space. The Hamiltonian for this system is ̵2 ˜ = ∫ dd x ∑ ψˆσ† (x)[− h "2 −µ]ψˆσ (x) H 2m

(6.102)

σ=↑,↓

where the label σ =↑, ↓ indicates the z-projection of the spin of the electron. The value of the chemical potential µ will be determined once we satisfy that the electron density is equal to some fixed value ρ¯. In momentum space, we get ψˆσ (x) = ∫

d p ˆ −ip⋅x/h̵ ψσ (p) e d (2π) d

(6.103)

† where the operators ψˆσ (p) and ψˆσ (p) satisfy the canonical equal-time anticommutation relations † d d ′ {ψˆσ (p), ψˆσ′ (p)} = (2π) δσσ′ δ (p − p )

† † ′ ˆ {ψˆσ (p), ψˆσ′ (p)} = {ψ(p), ψˆσ′ (p )} = 0

(6.104)

The Hamiltonian has the very simple form ˜ =∫ H where ε(p) is given by

d p † ∑ (ε(p) − µ) ψˆσ (p)ψˆσ (p) d (2π) ˆ d

(6.105)

σ=i,↓

2

p (6.106) 2m For this simple case, ε(p) is independent of the spin orientation. It is convenient to measure the energy relative to the chemical potential (i.e. the Fermi energy) µ = EF . The relative energy E(p) is ε(p) =

E(p) = ε(p) − µ

(6.107)

182

Non-Relativistic Field Theory

Hence, E(p) is the excitation energy measured from the Fermi energy EF = µ. The energy E(p) does not have a definite sign since there are states with ε(p) > µ as well as states with ε(p) < µ. Let us define by pF the value of ∣p∣ for which E(pF ) = ε(pF ) − µ = 0

(6.108)

This is the Fermi momentum. Thus, for ∣p∣ < pF , E(p) is negative while for ∣p∣ > pF , E(p) is positive. We can construct the ground state of the system by finding the state with lowest energy at fixed µ. Since E(p) is negative for ∣p∣ ≤ pF , we see that by filling up all of those states we get the lowest possible energy. It is then natural to normal order the system relative to a state in which all one-particle states with ∣p∣ ≤ pF are occupied. Hence we make the particlehole transformation ˆbσ (p) = ψˆσ† (p) for ∣p∣ ≤ pF a ˆσ (p) = ψˆσ (p) for ∣p∣ > pF

(6.109)

In terms of the operators a ˆσ and ˆbσ , the Hamiltonian is d p † † [E(p)θ(∣p∣−pF )ˆ aσ (p)ˆ aσ (p)+θ(pF −∣p∣)E(p)ˆbσ (p)ˆbσ (p)] d (2π) σ=↑,↓ (6.110) where θ(x) is the step function ˜ = ∑ ∫ H

d

θ(x) = {

1 x>0 0 x≤0

(6.111)

Using the anticommutation relations in the last term we get d p † † ˜gnd E(p)[θ(∣p∣−pF )ˆ aσ (p)ˆ aσ (p)−θ(pF −∣(p)∣)ˆbσ (p)ˆbσ (p)]+E d (2π) σ=↑,↓ (6.112) ˜ where Egnd , the ground state energy measured from the chemical potential µ, is given by ˜ = ∑ ∫ H

d

˜gnd = ∑ ∫ E σ=↑,↓

d p d d θ(pF − ∣p∣)E(p)(2π) δ (0) = Egnd − µN d (2π) d

Recall that (2π) δ(0) is equal to

(6.113)

d

(2π) δ (0) = lim (2π) δ (p) = lim ∫ d xe d d

d d

p→0

d

p→0

ip⋅x

=V

(6.114)

6.3 Non-relativistic fermions at zero temperature

183

˜gnd is extensive where V is the volume of the system. Thus, E ˜gnd = V ε˜gnd E

(6.115)

and the ground state energy density ε˜gnd is ε˜gnd = 2 ∫

d

∣p∣≤pF

d p E(p) = εgnd − µ¯ ρ (2π)d

(6.116)

where the factor of 2 comes from the two spin orientations. Putting everything together we get ε˜gnd = 2 ∫

d

∣p∣≤pF

2

2

pF p p d p d−1 Sd ( ( − µ) = 2 ∫ dp p − µ) (6.117) d 2m d 2m (2π) (2π) 0

where Sd is the area of the d-dimensional hypersphere. Our definitions tell us p

2

F ≡ EF where EF , is the Fermi energy. that the chemical potential is µ = 2m Thus the ground state energy density εgnd (measured from the empty state) is equal to

d+2

d

pF pF Sd pF Sd 1 Sd d+1 ∫ = 2EF Egnd = m dp p = d d m(d + 2) (2π) (2π) 0 (d + 2)(2π)d (6.118) How many particles does this state have? To find that out we need to look at the number operator. The number operator can also be normal-ordered with respect to this state

ˆ =∫ N =∫

d p † ∑ ψˆσ (p)ψˆσ (p) = (2π)d σ=↑,↓ d

d p † † ∑ {θ(∣p∣ − pF )ˆ aσ (p)ˆ aσ (p) + θ(pF − ∣p∣)ˆbσ (p)ˆbσ (p)} d (2π) σ=↑,↓ d

(6.119)

ˆ can also be written in the form Hence, N

ˆ =∶ N ˆ ∶ +N N

(6.120)

ˆ ∶ is where the normal-ordered number operator ∶ N ˆ ∶= ∫ ∶ N

d p † † ∑ [θ(∣p∣ − pF )ˆ aσ (p)ˆ aσ (p) − θ(pF − ∣p∣)ˆbσ (p)ˆbσ (p)] d (2π) σ=↑,↓ (6.121) d

184

Non-Relativistic Field Theory

and N , the number of particles in the reference state ∣gnd⟩, is d p 2 d Sd d d ∑ θ(pF − ∣p∣)(2π) δ (0) = pF V N =∫ d d (2π) σ=↑,↓ (2π)d d

Therefore, the particle density ρ¯ = ρ¯ =

N V

(6.122)

is

2 Sd d p d (2π)d F

(6.123)

This equation determines the Fermi momentum pF in terms of the density ρ¯. Egnd 2d = d+2 EF . Similarly we find that the ground state energy per particle is N The excited states can be constructed in a similar fashion. The state ∣+, σ, p⟩ ∣+, σ, p⟩ = a ˆα (p)∣gnd⟩ †

(6.124)

is a state which represents an electron with spin σ and momentum p while the state ∣−, σ, p⟩ † (6.125) ∣−, σ, p⟩ = ˆbσ (p)∣gnd⟩

represents a hole with spin σ and momentum p. From our previous discussion we see that electrons have momentum p with ∣p∣ > pF while holes have momentum p with ∣p∣ < pF . The excitation energy of a one-electron state is E(p) ≥ 0(for ∣p∣ > pF ), while the excitation energy of a one-hole state is −E(p) ≥ 0 (for ∣p∣ < pF ). Similarly, an electron-hole pair is a state of the form † ′ ′ † ′ ∣σp, σ p ⟩ = a ˆσ (p)ˆbσ′ (p )∣gnd⟩

(6.126)

with ∣p∣ > pF and ∣p ∣ < pF . This state has excitation energy E(p) − E(p ), which is positive. Hence, states which are obtained from the ground state without changing the density, can only increase the energy. This proves that ∣gnd⟩ is indeed the ground state. However, if the density is allowed to change, we can always construct states with energy less than Egnd by creating a number of holes without creating an equal number of particles. ′

′

7 Quantization of the Free Dirac Field

7.1 The Dirac equation and Quantum Field Theory The Dirac equation is a relativistic wave equation that describes the quantum dynamics of spinors. We will see in this section that a consistent description of this theory cannot be done outside the framework of (local) relativistic Quantum Field Theory. The Dirac Equation (i∂/ − m)ψ = 0,

(7.1)

can be regarded as the equation of motion of a complex field ψ. Much as in the case of the scalar field, and also in close analogy to the theory of non-relativistic many particle systems discussed in the last chapter, we will regard the Dirac field as an operator that acts on a Fock space of states. We have already discussed that the Dirac equation also follows from a least-action-principle. Indeed the Lagrangian i ¯ ¯ µ ψ] − mψψ ¯ ≡ ψ(i ¯ ∂/ − m)ψ [ψ ∂/ψ − (∂µ ψ)γ (7.2) 2 has the Dirac equation for its equation of motion. Also, the momentum Πα (x) canonically conjugate to ψα (x) is L=

Πα (x) = ψ

δL † = iψα δ∂0 ψα (x)

(7.3)

Thus, they obey the equal-time Poisson Brackets

{ψα (x), Πβ (y)}P B = δαβ δ (x − y) ψ

Thus

3

{ψα (x), ψβ (y)}P B = iδαβ δ (x − y) †

3

†

(7.4)

(7.5)

In other words the field ψα and its adjoint ψα are a canonical pair. This

190

Quantization of the Free Dirac Field

result follows from the fact that the Dirac Lagrangian is first order in time derivatives. Notice that the field theory of non-relativistic many-particle systems (for both fermions and bosons) also has a Lagrangian which is first order in time derivatives. We will see that, because of this property, the quantum field theory of both types of systems follows rather similar lines, at least at a formal level. As in the case of the many-particle systems, two types of statistics are available to us: Fermi-Dirac and Bose-Einstein. We will see that only the choice of Fermi statistics leads to a physically meaningful theory of the Dirac equation. The Hamiltonian for the Dirac theory is 3 H = ∫ d x ψ¯α (x)[ − iγ ⋅ ! + m]

αβ

ψβ (x)

(7.6)

† where the fields ψ(x) and ψ¯ = ψ γ0 are operators which act on a Hilbert space to be specified below. Notice that the one-particle operator in Eq. (7.6) is just the one-particle Dirac Hamiltonian obtained if we regard the Dirac Equation as a Schr¨ odinger Equation for spinors. We will leave the issue of their commutation relations (Fermi or Bose) open for the time being. In any event, the equations of motion are independent of that choice (do not depend on the statistics). In the Heisenberg representation, we find

iγ0 ∂0 ψ = [γ0 ψ, H] = (−iγ ⋅ ! + m)ψ

(7.7)

which is just the Dirac equation. We will solve this equation by means of a Fourier expansion in modes of the form 3 d p m −ip⋅x ip⋅x ψ(x) = ∫ (ψ˜+ (p)e + ψ˜− (p)e ) (7.8) (2π)3 ω(p)

where ω(p) is a quantity with units of energy, and which will turn out to √ be equal to p0 = p2 + m2 , and p ⋅ x = p0 x0 − p ⋅ x. In terms of ψ˜± (p), the Dirac equation becomes (p0 γ0 − γ ⋅ p ± m) ψ˜± (p) = 0

(7.9)

In other words, ψ˜± (p) creates one-particle states with energy ±p0 . Let us make the substitution ψ˜± (p) = (±/ p + m)φ˜ (7.10) We get

(p / ∓ m)(±p / + m)φ˜ = ±(p − m )φ˜ = 0 2

2

(7.11)

7.1 The Dirac equation and Quantum Field Theory

191

This equation has non-trivial solutions only if the mass-shell condition is obeyed 2

2

p −m =0 (7.12) √ Thus, we can identify p0 = ω(p) = p⃗2 + m2 . At zero momentum these states become ψ˜± (p0 , p = 0) = (±p0 γ0 + m)φ˜ (7.13)

where φ˜ is an arbitrary 4-spinor. Let us choose φ˜ to be an eigenstate of γ0 . Recall that in the Dirac representation γ0 is diagonal

(1)

Thus the spinors u

γ0 = (

(m, 0) and u

(2)

⎛1 ⎞ ⎜0 ⎟ (1) ⎟ ⎜ u (m, 0) = ⎜ ⎟ ⎜0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝0 ⎠

have γ0 -eigenvalue +1

and the spinors v

(i)

γ0 u (m, 0) = +u

(σ)

(m, 0)(σ = 1, 2)

have γ0 -eigenvalue −1, (i)

(σ)

(m, 0) = −v

(σ)

⎛0⎞ ⎜1⎟ (2) ⎟ ⎜ u (m, 0) = ⎜ ⎟ ⎜0⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠

(m, 0)

(1)

(i)

In terms of ϕ

1 =( ) 0

the solutions are

σ = 1, 2

(m, 0)

(7.16)

(2)

ϕ

(7.17)

σ = 1, 2

0 =( ) 1

⎛ (σ) (σ) u (m, 0) = ⎜ ψ˜+ (p) = u (p) = √ ⎝ 2m(p0 + m) (p / + m)

(7.15)

⎛0⎞ ⎜0⎟ (2) ⎟ ⎜ ⎟ v (m, 0) = ⎜ ⎜ ⎟ ⎜ ⎜0⎟ ⎟ ⎝1⎠

Let ϕ (m, 0) be the 2-spinors (σ = 1, 2) ϕ

(7.14)

(m, 0)

(σ)

⎛0⎞ ⎜0⎟ (1) ⎟ ⎜ ⎟ v (m, 0) = ⎜ ⎜ ⎟ ⎜ ⎜1⎟ ⎟ ⎝0⎠

γ0 v

I 0 ) 0 −I

(7.18)

(7.19) √p

0 +m

2m σ⋅p √ 2m(p0 +m)

(σ)

ϕ

(σ)

ϕ

(m, 0) ⎞ ⎟ (m, 0) ⎠ (7.20)

192

Quantization of the Free Dirac Field

and ⎛ √ σ⋅p (−p / + m) (σ) (σ) ˜ 0 +m) √ p +m v (m, 0) = ⎜ 2m(p ψ− (p) = v (p) = √ 0 ⎝ 2m(p0 + m) 2m

(σ)

(m, 0)⎞ ⎟ (σ) ϕ (m, 0)⎠ (7.21) √ 2 + m2 while the two p where the two solutions ψ˜+ (p) have energy +p = + 0 √ solutions ψ˜− (p) have energy −p0 = − p2 + m2 . Therefore, the one-particle states of the Dirac theory can have both positive and negative energy and, as it stands, the spectrum of the one-particle Dirac Hamiltonian, shown schematically in Fig. 7.1, is not positive. In addition, each Dirac state has a two-fold degeneracy due to spin. ϕ

E

m

−m

∣p∣

Figure 7.1 Single particle spectrum of the Dirac theory.

The spinors u

(i)

and v

(i)

obey the orthogonality conditions

u ¯

(σ)

(σ)

v¯ u ¯ †

†

(ν)

(p)u

(σ)

(p)v (p)v

(ν) (ν)

(p) =δσν

(p) = − δσν (p) =0

(7.22)

where u ¯ = u γ0 and v¯ = v γ0 . It is straightforward to check that the operators Λ± (p) Λ± (p) =

1 (±p + m) 2m /

(7.23)

are projection operators of the spinors onto the subspaces with positive (Λ+ )

7.1 The Dirac equation and Quantum Field Theory

and negative (Λ− ) energy respectively. These operators satisfy 2

Λ± = Λ±

Tr Λ± = 2

(σ)

Λ+ + Λ− = 1

193

(7.24)

(σ)

Hence, the four 4-spinors u and v are orthonormal and complete natural bases of the Hilbert space of single-particle states. We can use these results to write the expansion of the field operator ψ(x) = ∫

d p m (σ) (σ) −ip⋅x ip⋅x ∑ [aσ,+ (p)u+ (p)e +aσ,− (p)v− (p)e ] (7.25) 3 p0 (2π) σ=1,2 3

where the coefficients aσ,± (p) are operators with as yet unspecified commutation relations. The (formal) Hamiltonian for this system is H=∫

3

d p m † † ∑ p0 [aσ,+ (p)aσ,+ (p) − aσ,− (p)aσ,− (p)] 3 p0 (2π) σ=1,2

(7.26)

Since the single-particle spectrum does not have a lower bound, any attempt to quantize the theory with canonical commutation relations will have the problem that the total energy of the system is not bounded from below. In other words “Dirac bosons” do not have a ground state and the system is unstable since we can put as many bosons as we wish in states with arbitrarily large but negative energy. Dirac realized that the simple and elegant way out of this problem was to require the electrons to obey the Pauli Exclusion Principle since, in that case, there is a natural and stable ground state. However, this assumption implies that the Dirac theory must be quantized as a theory of fermions. Hence we are led to quantize the theory with canonical anticommutation relations {as,σ (p), as′ ,σ′ (p )} = 0 ′

† ′ 3 p0 3 ′ {as,σ (p), as′ ,σ′ (p )} = (2π) m δ (p − p )δss′ δσσ′

(7.27)

where s = ±. Let us denote by ∣0⟩ the state anihilated by the operators as,σ (p), as,σ (p)∣0⟩ = 0

(7.28)

We will see now that this state is not the vacuum (or ground state) of the Dirac theory.

194

Quantization of the Free Dirac Field

7.1.1 Ground state and normal ordering

We will show now that the ground state or vacuum ∣vac⟩ is the state in which all the negative energy states are filled (as shown in Fig.7.2) ∣vac⟩ = ∏ a−,σ (p)∣0⟩ †

(7.29)

σ,p

E

∣p∣

Figure 7.2 Ground State of the Dirac theory.

We will normal-order all the operators relative to the vacuum state ∣vac⟩. This amounts to a particle-hole transformation for the negative energy states. † Thus, we define the fermion creation and annihilation operators bσ (p), bσ (p) † and dσ (p), dσ (p) to be bσ (p) = aσ,+ (p)

dσ (p) = aσ,− (p) †

which obey

bσ (p)∣vac⟩ = dσ (p)∣vac⟩ = 0

(7.30)

(7.31)

The Hamiltonian now reads H =∫

3

d p m † † p0 ∑ [bσ (p)bσ (p) − dσ (p)dσ (p)] (2π)3 p0 σ=1,2

(7.32)

ˆ relative to the vacuum state We now normal order H H =∶ H ∶ +E0

(7.33)

7.1 The Dirac equation and Quantum Field Theory

195

with a normal-ordered Hamiltonian ∶ H∶ = ∫

3

d p m † † ∑ p0 [bσ (p)bσ (p) + dσ (p)dσ (p)] (2π)3 p0 σ=1,2

(7.34)

The constant E0 is the (negative and divergent) ground state energy √ 3 E0 = −2V ∫ d p p2 + m2 (7.35)

similar to the expression we already countered in the Klein-Gordon theory, but with opposite sign. The factor of 2 is due to spin. In terms of the operators bσ and dσ the Dirac field has the mode expansion ψα (x) = ∫

d p m (σ) −ip⋅x † (σ) ip⋅x ∑ [bσ (p)uα (p)e + dσ (p)vα (p)e ] (7.36) (2π)3 p0 σ=1,2 3

which satisfy equal-time canonical anticommutation relations {ψα (x), ψβ (x )} = δαβ δ (x − x ) †

3

′

′

{ψα (x), ψβ (x )} = {ψα (x), ψβ (x )} = 0 ′

†

†

′

(7.37)

7.1.2 One-particle states

The excitations of this theory can be constructed by using the same methods employed for non-relativistic many-particle systems. Let us first construct µ the total four-momentum operator P P =∫ d xT µ

3

0µ

Hence

3

d p m µ † † =∫ p ∑ ∶ bσ (p)bσ (p) − dσ (p)dσ (p) ∶ (7.38) 3 p 0 (2π) σ=1,2 3

d p m µ † † p ∑ [bσ (p)bσ (p) + dσ (p)dσ (p)] (7.39) (2π)3 p0 σ=1,2 √ † † The states bσ (p)∣vac⟩ and dσ (p)∣vac⟩ have energy p0 = p2 + m2 and momentum p, ∶ P ∶= ∫ µ

∶ H ∶ bσ (p)∣vac⟩ =p0 bσ (p)∣vac⟩ †

†

∶ H ∶ dσ (p)∣vac⟩ =p0 dσ (p)∣vac⟩ †

†

∶ P ∶ bσ (p)∣vac⟩ =p bσ (p)∣vac⟩ i

†

i †

∶ P ∶ dσ (p)∣vac⟩ =p dσ (p)∣vac⟩ i

†

i †

(7.40)

196

Quantization of the Free Dirac Field

We see that there are four different states which have the same energy and momentum. Let us find quantum numbers to classify these states. 7.1.3 Spin The angular momentum tensor Mµνλ for the Dirac theory is 1 νλ 3 µ ν λ λ ν ¯ Mµνλ = ∫ d x ψ(x) γ [i (x ∂ − x ∂ ) + σ ] ψ(x) 2

where σ

νλ

is the matrix

σ

νλ

=

i ν λ [γ , γ ] 2

The conserved angular momentum J J

νλ

0νλ

=M

νλ

(7.41)

(7.42)

is

1 νλ 3 † ν λ λ ν = ∫ d x ψ (x) [i(x ∂ − x ∂ ) + σ ] ψ(x) 2

(7.43)

ij

In particular, out of its space components J , we can construct the total angular momentum three-vector J 1 ijk jk 1 ijk jk i 3 † ijk j k J = * J = ∫ d x ψ (i* x ∂ + * σ ) ψ 2 2

(7.44)

It is easy to check that, in the Dirac representation, the last term represents the spin. In the quantized theory, the angular momentum operator is J =L+S

(7.45)

where L is the orbital angular momentum L = ∫ d x ψ (x)x × i∂ψ(x) 3

while S is the spin

†

S = ∫ d x ψ (x) Σ ψ(x) 3

†

(7.46)

(7.47)

where Σ is the 4 × 4 matrix

Σ=

1 1 σ 0 ( )≡ σ 2 0 σ 2

(7.48)

In order to measure the spin polarization of a state we first go to the rest µ frame in which p = 0. In this frame we can consider the four-vector W W = (0, mΣ) µ

(7.49)

7.1 The Dirac equation and Quantum Field Theory

197

Let n be the space-like 4-vector n = (0, n), where n has unit length. Thus, µ µ n nµ = −1. We will use n to fix the direction of polarization in the rest frame. µ The scalar product Wµ n is a Lorentz invariant scalar and, hence, its value is independent of the choice of frame. In the rest frame we have µ

µ

µ

Wµ n = −mn ⋅ Σ ≡ −

m m n⋅σ 0 ) n ⃗ ⋅σ =− ( 2 2 0 n⋅σ

(7.50)

µ

In particular, if n = ez then Wµ n is µ

Wµ n = −

m σ3 0 ) ( 2 0 σ3

(7.51)

1 W ⋅n is a Lorentz scalar which measures which is diagonal. The operator − m the spin polarization:

1 1 (1) (1) − m W ⋅ nu (p) = + u (p) 2 1 1 (2) (2) − m W ⋅ nu (p) = − u (p) 2 1 (1) 1 (1) − m W ⋅ nv (p) = + v (p) 2 1 1 (2) (2) − m W ⋅ nv (p) = − v (p) 2

(7.52)

1 It is straightforward to check that − m W ⋅ n is the Lorentz scalar

1 1 1 µ ν λρ −mW ⋅ n = * n p σ = γ n /p 4m µνλρ 2m 5 /

which enables us to write the spin projection operator P (n) where we used that

1 P (n) = (I + γ5 n /) 2

1 1 (σ) (σ) σ 1 (σ) γ n γ n / p u (p) = / u (p) = (−1) u (p) 2m 5 / 2 5 2 1 1 (σ) (σ) σ 1 (σ) γ n / p v (p) = − γ5 n / v (p) = (−1) v (p) 2m 5 / 2 2

(7.53)

(7.54)

(7.55)

198

Quantization of the Free Dirac Field

7.1.4 Charge The Dirac Lagrangian is invariant under the global (phase) transformation ′

iα

ψ→ψ =e ψ ′ −iα ψ¯ → ψ¯ = e ψ¯ (7.56) Consequently, it has a locally conserved current j

µ

µ ¯ µψ j = ψγ

(7.57)

which is also locally gauge invariant. As a result it has a conserved total 3 0 charge Q = −e ∫ d xj (x). The corresponding operator in the quantized theory Q is Q = −e ∫ d x j (x) = −e ∫ d x ψ (x)ψ(x) 3

0

3

†

(7.58)

ˆ The total charge operator Q commutes with the Dirac Hamiltonian H [Q, H] = 0

(7.59)

Hence, the eigenstates of the Hamiltonian H have a well defined charge. In terms of the creation and annihilation operators, the total charge operator Q becomes Q = −e ∫

3

d p m † † ∑ (bσ (p)bσ (p) + dσ (p)dσ (p)) 3 p 0 (2π) σ=1,2

(7.60)

which is not normal-ordered relative to ∣vac⟩. The normal-ordered charge operator ∶ Q ∶ is 3

d p m † † ∶ Q ∶= −e ∫ ∑ [bσ (p)bσ (p) − dσ (p)dσ (p)] 3 p (2π) 0 σ=1,2

(7.61)

and we can write

Q = ∶ Q ∶ +Qvac

(7.62)

where Qvac is the unobservable (and divergent) vacuum charge Qvac = −eV ∫

3

d p (2π)3

(7.63)

V being the volume of space. From now on we will define the charge to be the subtracted charge operator ∶ Q ∶= Q − Qvac

(7.64)

7.1 The Dirac equation and Quantum Field Theory

199

which annihilates the vacuum state

∶ Q ∶ ∣vac⟩ = 0

(7.65)

the vacuum is neutral. In other words, we measure the charge of a state relative to the vacuum charge which we define to be zero. Equivalently, this amounts to a definition of the order of the operators in ∶ Q ∶ 3 1 † ∶ Q ∶ = −e ∫ d x [ψ (x), ψ(x)] 2

(7.66)

The one-particle states bσ ∣vac⟩ and dσ ∣vac⟩ have well defined charge: †

†

∶ Q ∶ bσ (p)∣vac⟩ = − ebσ (p)∣vac⟩ †

†

∶ Q ∶ dσ (p)∣vac⟩ = + edσ (p)∣vac⟩ †

†

(7.67)

Hence we identify the state bσ (p)∣vac⟩ with an electron of charge −e, spin √ † σ, momentum p and energy p0 = p2 + m2 . Similarly, the state dσ (p)∣vac⟩ is a positron with the same quantum numbers and energy of the electron but with positive charge +e. †

7.1.5 The Dirac energy-momentum tensor The energy-momentum tensor of the Dirac theory is obtained following the standard approach presented earlier. On general grounds, we expect that the energy-momentum tensor should be given by T

µν

¯ µ∂ν ψ − L = iψγ

(7.68)

Using the equation of motion of the free Dirac filed, i.e the Dirac equation, we find that the energy-momentum tensor for the Dirac theory reduces to T

µν

µ ν

¯ ∂ ψ = iψγ

(7.69)

From this expression it follows that the Hamiltonian is ¯ ∂ ψ = ∫ d xψ γ (−iγ ⋅ ∂ + m)ψ H = ∫ d x T iψγ 3

00

0 0

3

† 0

(7.70)

which is indeed the Hamiltonian of the Dirac field.

7.1.6 Causality and the spin-statistics connection Let us finally discuss the question of causality and the spin-statistics connection in the Dirac theory. To this end we will consider the anticommutator

200

Quantization of the Free Dirac Field

of two Dirac fields at different times i∆αβ (x − y) = {ψα (x), ψβ (y)}

(7.71)

By using the field expansion we obtain the expression d p m (σ) −ip⋅(x−y) (σ) σ ip⋅(x−y) (σ) ∑ [e uα (p)¯ uβ (p)+e vα (p)¯ vβ (p)] (2π)3 p0 σ=1,2 (7.72) By using the (completeness) identities 3

i∆αβ (x−y) = ∫

(σ)

(σ)

∑ uα (p)¯ uβ (p) = (

σ=1,2

(σ)

(σ)

∑ vα (p)¯ vβ (p) = (

σ=1,2

p+m / ) 2m αβ p−m / ) 2m αβ

(7.73)

we can write the anticommutator in the form

3 p+m p d p m / /−m −ip⋅(x−y) ip⋅(x−y) ( ) ) e [( e + ] 3 p 2m 2m 0 (2π) αβ αβ (7.74) After some straightforward algebra, we obtain the result

i∆αβ (x − y) = ∫

i∆αβ (x − y) = ∫

d p 1 −ip⋅(x−y) ip⋅(x−y) (i∂/x + m) [e −e ] (2π)3 2p0 3

= (i∂/x + m)αβ ∫

3

d p −ip⋅(x−y) ip⋅(x−y) [e −e ] 3 (2π) 2p0

(7.75)

We recognize the integral on the r.h.s. of Eq.(7.75) to be the commutator of two free scalar (Klein-Gordon) fields, ∆KG (x − y). Hence, the anticommutator two Dirac fields of the Dirac theory is i∆αβ (x − y) = (i∂/ + m)αβ i∆KG (x − y)

(7.76)

Since ∆KG (x − y) vanishes at space-like separations, so does ∆αβ (x − y). Hence, the Dirac theory quantized with anticommutators obeys causality. On the other hand, had we had quantized the Dirac theory with commutators (which, as we saw, leads to a theory without a ground state) we would have also found a violation of causality. Indeed, we would have obtained instead the result ˜ − y) ∆αβ (x − y) = (i∂/ + m)αβ ∆(x

(7.77)

7.2 The Propagator of the Dirac spinor field

201

˜ − y) is given by where ∆(x ˜ − y) = ∫ ∆(x

3

d p −ip⋅(x−y) ip⋅(x−y) (e +e ) 3 (2π) 2p0

(7.78)

which does not vanish at space-like separations. Instead, at equal times and ̃ (x − y) decays as at long distances ∆ −mR

e ˜ ∆(R, 0) ≃

, for mR ≫ 1 (7.79) R2 Thus, if the Dirac theory were to be quantized with commutators, the field operators would not commute at equal times at distances shorter than the Compton wavelength. This would be a violation of locality. The same result holds in the theory of the scalar field if it is quantized with anticommutators. These results can be summarized in the Spin-Statistics Theorem: fields with half-integer spin must be quantized as fermions, obey canonical anticommutation relations, whereas fields with integer spin must be quantized as bosons, obey canonical commutation relations. If a field theory is quantized with the wrong spin-statistics connection, either the theory becomes nonlocal, with violations of causality, and/or it does not have a ground state, or it contains states in its spectrum with negative norm. Notice that the arguments we have used were derived for free local theories. It is a highly non-trivial task to prove that the spin-statistics connection also remains valid for interacting theories. Although this can be done by making sufficiently strong assumptions of the behavior of perturbation theory, in reality it must the spin-statistics connection must be regarded as an axiom of local relativistic quantum field theories. 7.2 The Propagator of the Dirac spinor field

We will now compute the propagator for a spinor field ψα (x). We will find that it is essentially the Green function for the the Dirac operator. The propagator is defined by ′ ′ Sαβ (x − x ) = −i⟨vac∣T ψα (x)ψ¯β (x )∣vac⟩

(7.80)

where we have used the time ordered product of two fermionic field operators, which is defined by ′ ′ ′ ′ ′ T ψα (x)ψ¯β (x ) = θ(x0 − x0 )ψα (x)ψ¯β (x ) − θ(x0 − x0 )ψ¯β (x )ψα (x) (7.81)

Notice the change in sign with respect to the time ordered product of bosonic operators. The sign change reflects the anticommutation properties of the

202

Quantization of the Free Dirac Field

field. In other words, inside a time ordered product, Fermi fields behave as if they were anticommuting c-numbers. We will show now that this propagator is closely connected to the propagator of the free scalar field, the Green function for the Klein-Gordon operator 2 2 ∂ +m . ′ By acting with the Dirac operator on Sαβ (x − x ) we find ′ ′ (i∂/ − m)αβ Sβλ (x − x ) = −i (i∂/ − m)αβ ⟨vac∣T ψβ (x)ψ¯λ (x )∣vac⟩

We now use that

∂ ′ ′ θ(x0 − x0 ) = δ(x0 − x0 ) ∂x0

(7.82)

(7.83)

and the fact that the equation of motion of the Heisenberg field operators ψα is the Dirac equation, (i∂/ − m)αβ ψβ (x) = 0

to show that (i∂/ − m)

αβ

(7.84)

′ ′ Sβλ (x − x ) = −i⟨vac∣T (i∂/ − m)αβ ψβ (x)ψ¯λ (x )∣vac⟩ ′ 0 ′ + δ(x0 − x0 )(⟨vac∣γαβ ψβ (x)ψ¯λ (x )∣vac⟩ ′ 0 + ⟨vac∣ψ¯λ (x )γαβ ψβ (x)∣vac⟩)

= δ(x0 − x0 )γαβ ⟨vac∣ {ψβ (x), ψν (x )} ∣vac⟩γνλ 0

′

†

0

′

= δ(x0 − x0 )δ (x − x )δαλ ′

3

′

(7.85)

Therefore we find that Sβλ (x − x ) is the solution of the equation ′

(i∂/ − m)αβ Sβλ (x − x ) = δ (x − x )δαλ 4

′

′

(7.86)

Hence Sβλ (x − x ) = −i⟨vac∣T ψβ (x)ψ¯λ (x )∣vac⟩ is the Green function of the Dirac operator. We saw before that there is a close connection between the Dirac and the Klein-Gordon operators. We will now use this connection to relate their ′ propagators. Let us write the Green function Sαλ (x − x ) in the form ′

′

Sαλ (x − x ) = (i∂/ + m)αβ Gβλ (x − x ) ′

Since Sαλ (x − x ) satisfies Eq.(7.86), we find that ′

′

(7.87)

(i∂/ − m)αβ Sβλ (x − x ) = (i∂/ − m)αβ (i∂/ + m)βν Gνλ (x − x ) ′

′

(7.88)

7.3 Discrete symmetries of the Dirac theory

But

(i∂/ − m)αβ (i∂/ + m)βν = − (∂ + m ) δαν 2

Hence, Gαν (x − x ) must satisfy

2

203

(7.89)

′

− (∂ + m ) Gαν (x − x ) = δ (x − x )δαν 2

2

4

′

Therefore Gαν (x − x ) is given by

′

(7.90)

′

(0)

Gαν (x − x ) = −G ′

(0)

(x − x )δαν ′

(7.91)

where G (x−x ) is the propagator for a free massive scalar field, the Green function of the Klein-Gordon equation ′

(0)

(∂ + m ) G 2

2

(x − x ) = δ (x − x ) ′

4

′

(7.92)

We then conclude that the Dirac propagator Sαβ (x − x ), and the Klein(0) ′ Gordon propagator G (x − x ) are related by (0)

Sαβ (x − x ) = − (i∂/ + m)αβ G ′

′

(x − x ) ′

(7.93)

In particular, this relationship implies that the have essentially the same asymptotic behaviors that we discussed for the free scalar field, a power-law behavior at short distances (albeit with a different power), and exponential (or oscillatory) behavior at large distances. The spinor structure of the Dirac (0) ′ propagator is determined by the operator in front of G (x−x ) in Eq.(7.93). In momentum space, the Feynman propagator for the Dirac field, given by Eq. (7.93), becomes Sαβ (p) = (

p+m / ) p − m2 + i* αβ 2

(7.94)

Hence we get the same pole structure in the time-ordered propagator as we did for the Klein-Gordon field.

7.3 Discrete symmetries of the Dirac theory We will now discuss three important discrete symmetries in relativistic field theories: charge conjugation, parity and time reversal. These discrete symmetries have a different role, and a different standing, than the continuous symmetries discussed before. In a relativistic quantum field theory the ground state, the vacuum, must be invariant under continuous Lorentz transformations, but it may not be invariant under C, P or T . However, in a local

204

Quantization of the Free Dirac Field

relativistic quantum field theory the product CPT is always a good symmetry. This is in fact an axiom of relativistic local quantum field theory. Thus, although C, P or T may or may not to be good symmetries of the vacuum state, CPT must be a good symmetry. As in the case of the symmetries we discussed before, these symmetries must also be realized in the Fock space of the quantum field theory.

7.3.1 Charge conjugation Charge conjugation is a symmetry that exchanges particles and antiparticles (or holes). Consider a Dirac field minimally coupled to an external electromagnetic field Aµ . The equation of motion for the Dirac field ψ is / − m) ψ = 0 (i∂/ − eA

We will define the charge conjugate field ψ

(7.95)

c

ψ (x) = Cψ(x)C c

−1

(7.96)

where C is the (unitary) charge conjugation operator, C c ψ obeys

−1

/ − m) ψ = 0 (i∂/ + eA c

† 0 Since ψ¯ = ψ γ obeys

†

= C , such that (7.97)

← − µ ψ¯ [γ (−i ∂ µ − eAµ ) − m] = 0

(7.98)

which, when transposed, becomes [γ

µT

T (−i∂µ − eAµ ) − m] ψ¯ = 0

(7.99)

where T is the transpose, and

T 0T ∗ ψ¯ = γ ψ

Let C be an invertible 4 × 4 matrix, where C write

such that

Hence,

C [γ

µt

(7.100) −1

(−i∂µ − eAµ ) − m] C C (γ ) C µ t

−1

= −γ

/ − m] C ψ¯ [(i∂/ + eA)

is its inverse. Then, we can

−1

µ

T

t

C ψ¯ = 0

(7.101)

(7.102)

=0

(7.103)

7.3 Discrete symmetries of the Dirac theory

205

For Eq. (7.103) to hold, we must have † c T 0T ∗ CψC = ψ = C ψ¯ = Cγ ψ

(7.104)

c

Hence the field ψ thus defined has positive charge +e. We can find the charge conjugation matrix C explicitly: C = iγ γ = ( 2 0

c

0 iσ −1 )=C 2 −iσ 0 2

(7.105)

In particular this means that ψ is given by

c T 0t ∗ 2 ∗ ψ = C ψ¯ = Cγ ψ = iγ ψ

(7.106)

Eq. (7.106) provides us with a definition for a charge neutral Dirac fermion, c i.e. ψ represents a neutral fermion if ψ = ψ . Hence the condition is 2

ψ = iγ ψ

∗

(7.107)

A Dirac fermion that satisfies the neutrality condition is known as a Majorana fermion. To understand the action of C on physical states we can look for instance, c at the charge conjugate u of the positive energy, up spin, and charge −e, spinor in the rest frame ϕ −imt u = ( )e 0

which is

u =( c

(7.108)

0 imt 2 ∗) e −iγ ϕ

(7.109)

which has negative energy, down spin and charge +e. At the level of the full quantum field theory, we will require the vacuum state ∣vac⟩ to be invariant under charge conjugation: C∣vac⟩ = ∣vac⟩

(7.110)

How do one-particle states transform? To determine that we look at the action of charge conjugation C on the one-particle states, and demand that particle and anti-particle states to be exchanged under charge conjugation Cbσ (p)∣vac⟩ = Cbσ (p)C †

†

−1

Cdσ (p)∣vac⟩ = Cdσ (p)C †

†

C∣vac⟩ ≡ dσ (p)∣vac⟩

−1

†

C∣vac⟩ ≡ bσ (p)∣vac⟩ †

(7.111)

206

Quantization of the Free Dirac Field

Hence, for the one-particle states to satisfy these rules it is sufficient to require that the field operators bσ (p) and dσ (p) satisfy Cbσ (p)C = dσ (p);

Cdσ (p)C = bσ (p)

†

Using that

†

uσ (p) = −iγ (vσ (p)) ; 2

(7.112)

vσ (p) = −iγ (uσ (p)) 2

∗

∗

(7.113)

we find that the field operator ψ(x) transforms as ¯ γ ) Cψ(x)C = (−iψγ

0 2 T

†

and

¯ C ψ(x)C = (−iγ γ ψ) 0 2

†

(7.114)

T

(7.115)

In particular the fermionic bilinears we discussed before satisfy the transformation laws: 5

†

5

†

¯ ¯ C ψψC = +ψψ,

¯ ψC = iψγ ¯ ψ, Ciψγ

¯ µ ψC † = −ψγ ¯ µ ψ, C ψγ

¯ µ γ 5 ψC † = +ψγ ¯ µγ 5ψ C ψγ

(7.116)

7.3.2 Parity −1

We will define as parity the transformation P = P which reverses the momentum of a particle but not its spin. Once again, the vacuum state is invariant under parity. Thus, we must require Pbσ (p)∣vac⟩ = Pbσ (p)P †

†

−1

Pdσ (p)∣vac⟩ = Pdσ (p)P †

†

P∣vac⟩ ≡ bσ (−p)∣vac⟩

−1

†

P∣vac⟩ ≡ dσ (−p)∣vac⟩ †

(7.117)

In real space this transformation is equivalent to Pψ(x, x0 )P

−1

= γ ψ(−x, x0 ), 0

¯ P ψ(x, x0 )P

−1

7.3.3 Time reversal

¯ = ψ(−x, x0 )γ0

(7.118)

Finally, we discuss time reversal T . We will define T as the operator iHx0 −1 −iHx0 Te T =e ,

(7.119)

We will require the time reversal operator to be unitary in the sense that T

−1

=T

†

(7.120)

7.4 Chiral symmetry

207

However we will also require the operator to act on c-numbers as complex conjugation, i.e. T (c-number) = (c-number) T ∗

(7.121)

An operator with these properties is said to be anti-linear (anti-unitary). As a result, time reversal is the operator that which reverses the momentum and the spin of the particles: T bσ (p)T = b−σ (−p),

T dσ (p)T = d−σ (−p)

†

†

(7.122)

while leaving the vacuum state invariant:

T ∣vac⟩ = ∣vac⟩

In real space this implies:

T ψ(x, x0 )T = −γ γ ψ (x, −x0 ) †

1 3

∗

(7.123)

(7.124)

7.4 Chiral symmetry We will now discuss a global symmetry specific of theories of spinors known as chiral symmetry. Let us consider again the Dirac Lagrangian L = ψ¯ (i∂/ − m) ψ

(7.125)

Let us define the chiral transformation

′ iγ θ ψ =e 5 ψ

(7.126)

where θ is constant phase in the range 0 ≤ θ < 2π. We wish to find the Lagrangian for the new transformed field ψ. From iγ5 θ

¯ L = ψe and the fact that

µ iγ θ (iγ ∂µ − m) e 5 ψ

{γµ , γ5 } = 0

(7.127)

(7.128)

after some simple algebra, which uses the identity µ iγ θ −iγ5 θ µ γ e 5 =e γ ,

(7.129)

we find that the field ψ satisfies a modified Dirac Lagrangian of the form 2iγ θ L = ψ¯ (i∂/ − me 5 ) ψ

(7.130)

208

Quantization of the Free Dirac Field

Thus for m ≠ 0, the form of the Dirac Lagrangian changes under a chiral transformation. However, if the theory is massless, the Dirac theory has an exact global chiral symmetry. It is also instructive to determine how various fermion bilinears transform ¯ and the under the chiral transformation. We find that the Dirac mass , ψψ, 5 ¯ axial mass, iψγ ψ, under a chiral transformation transform as follows ′ ′ ¯ + sin(2θ)iψγ ¯ 5ψ ψ¯ ψ = cos(2θ)ψψ ′ 5

′

¯ + cos(2θ)iψγ ¯ 5ψ iψ¯ γ ψ = − sin(2θ)ψψ

(7.131)

and, hence, are not invariant under the chiral transformation. Instead, under ¯ and the γ5 mass, iψγ ¯ 5 ψ, the chiral transformation the Dirac mass, ψψ, transform (rotate) into each other. Finally we note that, although the Dirac Lagrangian is not chiral invariant if m ≠ 0, the Dirac current ′ µ ′ ¯ µψ ψ¯ γ ψ = ψγ

(7.132)

is invariant under the chiral transformation, and so is the coupling of the Dirac field to a gauge field.

7.5 Massless fermions Let us look at the massless limit of the Dirac equation in more detail. Historically this problem grew out of the study of neutrinos (which we now know are not necessarily massless, at least not all of them). For an eigenstate of 4-momentum pµ , the Dirac equation is (/ p − m) ψ(p) = 0

(7.133)

In the massless limit m = 0, the Dirac equation simply becomes

which is equivalent to

pψ(p) = 0 /

γ5 γ0 / pψ(p) = 0

(7.134)

(7.135)

Upon expanding in components we find

γ5 p0 ψ(p) = γ5 γ0 γ ⋅ pψ(p)

(7.136)

However, since σ 0 ]=Σ γ5 γ0 γ = [ 0 σ

(7.137)

7.5 Massless fermions

209

we can write γ5 p0 ψ(p) = Σ ⋅ pψ(p)

(7.138)

Thus, the chirality γ5 is equivalent to the helicity Σ ⋅ p of the state (only in the massless limit!). This suggests the introduction of the chiral basis in which γ5 is diagonal, γ0 = (

0 −I ) −I 0

I 0 ), γ5 = ( 0 −I

0 σ ) γ=( −σ 0

(7.139)

In this basis the massless Dirac equation becomes

I 0 σ 0 ( ) ⋅ pψ(p) = ( ) p ψ(p) 0 −I 0 0 σ

(7.140)

Let us write the 4-spinor ψ in terms of two 2-spinors of the form ψ ψ = ( R) ψL

(7.141)

in terms of which the Dirac equation decomposes into two separate equations for each chiral component, ψR and ψL . Thus the right handed (positive chirality) component ψR satisfies the Weyl equation (σ ⋅ p − p0 ) ψR = 0

(7.142)

while the left handed (negative chirality) component satisfies instead (σ ⋅ p + p0 ) ψL = 0

(7.143)

Hence, one massless Dirac spinor is equivalent to two Weyl spinors. We conclude that a theory of free massless Dirac fermions has a global chiral symmetry. It is natural to assume that it must also have a locally 5 conserved chiral (axial) current, that we will denote by jµ . An elementary calculation derives the result 5 ¯ 5 γµ ψ jµ = iψγ

(7.144)

However, we also know that the Dirac theory has a global U (1) phase symmetry which, when coupled to the vector field Aµ , becomes local U (1) gauge ¯ µ ψ. We will invariance, and has a locally conserved gauge current, jµ = ψγ see in chapter 20 that in an interacting theory its UV divergencies makes it impossible for both currents to be simultaneously conserved and that, if the theory is to be gauge-invariant, then the chiral current cannot not conserved. In other words, a classically conserved current may not be conserved at the quantum level. The phenomenon of non-conservation at the quantum

210

Quantization of the Free Dirac Field

level of a classically conserved current is known as a quantum anomaly. In this case, the non-conservation of the chiral current is known as the chiral (or axial) anomaly.

8 Coherent State Path Integral Quantization of Quantum Field Theory

8.1 Coherent states and path integral quantization. The path integral that we have used so far is a powerful tool but it can only be used in theories based on canonical quantization. For example, it cannot be used in theories of fermions (relativistic or not), among others. In this chapter we discuss a more general approach based on the concept of coherent states. Coherent states, and their application to path integrals, have been widely discussed. Excellent references include the 1975 lectures by Faddeev (Faddeev, 1976), and the books by Perelomov (Perelomov, 1986), Klauder (Klauder and Skagerstam, 1985), and Schulman (Schulman, 1981). The related concept of geometric quantization is insightfully presented in the work by Wiegmann (Wiegmann, 1989).

8.2 Coherent states Let us consider a Hilbert space spanned by a complete set of harmonic † oscillator states {∣n⟩}, with n = 0, . . . , ∞. Let a ˆ and a ˆ be a pair of creation and annihilation operators acting on this Hilbert space, and satisfying the commutation relations [ˆ a, a ˆ ]=1, †

[ˆ a ,a ˆ ]=0, †

[ˆ a, a ˆ] = 0

†

(8.1)

These operators generate the harmonic oscillators states {∣n⟩} in the usual way, 1 † n ∣n⟩ = √ (ˆ a ) ∣0⟩, n!

a ˆ∣0⟩ = 0

where ∣0⟩ is the vacuum state of the oscillator.

(8.2)

8.2 Coherent states

215

Let us denote by ∣z⟩ the coherent state ∣z⟩ = e

†

zˆ a

∣0⟩,

⟨z∣ = ⟨0∣ e

z¯a ˆ

(8.3)

where z is an arbitrary complex number and z¯ is the complex conjugate. The coherent state ∣z⟩ has the defining property of being a wave packet with optimal spread, i.e. the Heisenberg uncertainty inequality is an equality for these coherent states. How does a ˆ act on the coherent state ∣z⟩? ∞

a ˆ∣z⟩ = ∑

z † n a ˆ (ˆ a ) ∣0⟩ n! n=0

Since

n

[ˆ a, (ˆ a ) ] = n (ˆ a) † n

we find

∞

† n−1

(8.4)

(8.5)

,

z † n−1 a ˆ∣z⟩ = ∑ n (ˆ a) ∣0⟩ ≡ z ∣z⟩ n! n=0 n

(8.6)

Therefore ∣z⟩ is a right eigenvector of a ˆ and z is the (right) eigenvalue. Likewise we get ∞

†

Thus,

†

∞

z z † n (ˆ a ) ∣0⟩ = ∑ n n! n! n=1 n=0

a ˆ ∣z⟩ = a ˆ ∑

n

n−1

a ˆ ∣z⟩ = ∂z ∣z⟩

(ˆ a ) ∣0⟩ † n

†

(8.7)

(8.8)

Therefore the operators z and ∂z provide a representation of the algebra of creation and annihilation operators. ′ Another quantity of interest is the overlap of two coherent states, ⟨z∣z ⟩, ⟨z∣z ⟩ = ⟨0∣e

∣0⟩

′ †

z¯a ˆ

′

za ˆ

e

(8.9)

We will calculate this matrix element using the Baker-Hausdorff formulas ˆ A

ˆ B+ ˆ 1 [A, ˆ B] ˆ A+ 2

ˆ B

ˆ B] ˆ [A,

e e =e

=e

ˆ B

ˆ A

e e

(8.10)

ˆ B] ˆ is a c-number, i.e. it is prowhich holds provided the commutator [A,

portional to the identity operator. Since [ˆ a, a ˆ ] = 1, we find †

⟨z∣z ⟩ = e ′

z¯z

′

⟨0∣e

′ †

za ˆ

e ∣0⟩ z¯a ˆ

(8.11)

216

Coherent State Path Integral Quantization of Quantum Field Theory

But z¯a ˆ

e Hence we get

∣0⟩ = ∣0⟩,

⟨0∣ e

′ †

za ˆ

⟨z∣z ⟩ = e

z¯z

′

′

= ⟨0∣

(8.12)

(8.13)

An arbitrary state ∣ψ⟩ of this Hilbert space can be expanded in the harmonic oscillator basis states {∣n⟩}, ∞

∞

ψn ψn † n ∣ψ⟩ = ∑ √ ∣n⟩ = ∑ (ˆ a ) ∣0⟩ n! n! n=0 n=0

The projection of the state ∣ψ⟩ onto the coherent state ∣z⟩ is ∞

ψn † n ⟨z∣ (ˆ a ) ∣0⟩ n!

⟨z∣ψ⟩ = ∑

n=0

Since

⟨z∣ a ˆ = z¯ ⟨z∣ †

we find

∞

⟨z∣ψ⟩ = ∑

n=0

(8.14)

(8.15)

(8.16)

ψn n z¯ ≡ ψ(¯ z) n!

(8.17)

Therefore the projection of ∣ψ⟩ onto ∣z⟩ is the anti-holomorphic (i.e. antianalytic) function ψ(¯ z ). In other words, in this representation, the space of states {∣ψ⟩} are in one-to-one correspondence with the space of anti-analytic functions. In summary, the coherent states {∣z⟩} satisfy the following properties a ˆ∣z⟩ = z∣z⟩ ⟨z∣ˆ a = ∂z¯⟨z∣ † † a ˆ ∣z⟩ = ∂z ∣z⟩ ⟨z∣ˆ a = z¯⟨z∣ ¯ ⟨z∣ψ⟩ = ψ(¯ z ) ⟨ψ∣z⟩ = ψ(z)

(8.18)

Next we will prove the resolution of identity

dzd¯ z −z¯ z Iˆ = ∫ e ∣z⟩⟨z∣ 2πi

Let ∣ψ⟩ and ∣φ⟩ be two arbitrary states ∞

ψn ∣ψ⟩ = ∑ √ ∣n⟩, n! n=0

(8.19)

∞

φn ∣φ⟩ = ∑ √ ∣n⟩ n! n=0

(8.20)

8.2 Coherent states

217

such that their inner product is ∞

⟨φ∣ψ⟩ = ∑

n=0

1 ¯ φ ψ n! n n

(8.21)

Let us compute the matrix element of the operator Iˆ given in Eq.(8.19), ˆ ⟨φ∣I∣ψ⟩ =∑

φ¯n ψn ˆ ⟨n∣I∣m⟩ n! m.n

Thus we need to compute

ˆ ⟨n∣I∣m⟩ =∫

dzd¯ z −∣z∣2 e ⟨n∣z⟩⟨z∣m⟩ 2πi

(8.22)

(8.23)

Recall that the integration measure is defined to be given by dzd¯ z dRez dImz = π 2πi

(8.24)

The overlaps are given by 1 z n ⟨n∣z⟩ = √ ⟨0∣ (ˆ a) ∣z⟩ = √ ⟨0∣z⟩ n! n! n

and

1 z¯ † m ⟨z∣m⟩ = √ ⟨z∣ (ˆ a ) ∣0⟩ = √ ⟨z∣0⟩ m! m!

(8.25)

m

Now, since ∣⟨0∣z⟩∣ = 1, we get

(8.26)

2

Thus,

2

−∣z∣ dzd¯ z e n m ˆ z z¯ ⟨n∣I∣m⟩ =∫ √ 2πi n!m! 2 −ρ ∞ 2π dϕ e n+m i(n − m)ϕ = ∫ ρdρ ∫ ρ e √ 2π 0 0 n!m! ˆ ⟨n∣I∣m⟩ =

Hence, we have found that

δn,m ∞ n −x ∫ dx x e = ⟨n∣m⟩ n! 0 ˆ ⟨φ∣I∣ψ⟩ = ⟨φ∣ψ⟩

(8.27)

(8.28)

(8.29)

for any pair of states ∣ψ⟩ and ∣φ⟩. Therefore Iˆ is the identity operator in this Hilbert space. We conclude that the set of coherent states {∣z⟩} is an over-complete set of states.

218

Coherent State Path Integral Quantization of Quantum Field Theory

Furthermore, since † n m ′ n ′m ′ n ′m z ¯z ⟨z∣ (ˆ a ) (ˆ a) ∣z ⟩ = z¯ z ⟨z∣z ⟩ = z¯ z e

′

(8.30)

we conclude that the matrix elements in generic coherent states ∣z⟩ and ∣z ⟩ of any arbitrary normal ordered operator of the form ′

† m Aˆ = ∑ An,m (ˆ a ) (ˆ a) n

(8.31)

n,m

are equal to

ˆ ′ ⟩ = (∑ An,m z¯n z ′m ) ez¯z ⟨z∣A∣z

′

(8.32)

n,m

ˆ a, a Therefore, if A(ˆ ˆ ) is an arbitrary normal ordered operator (relative to the state ∣0⟩), its matrix elements are given by †

† ′ ′ z¯z ˆ a, a ⟨z∣A(ˆ ˆ )∣z ⟩ = A(¯ z, z ) e

′

(8.33)

where A(¯ z , z ) is a function of two complex variables z¯ and z obtained from Aˆ by the formal replacement ′

′

a ˆ↔z ,

′

†

a ˆ ↔ z¯

(8.34)

The complex function A(¯ z , z ) is oftentimes called the symbol of the (normalˆ ordered) operator A. † ˆ =a For example, the matrix elements of the number operator N ˆ a ˆ, which measures the number of excitations, is ′

† ′ ′ z ¯z ˆ ∣z ′ ⟩ = ⟨z∣ˆ ⟨z∣N aa ˆ∣z ⟩ = z¯z e

′

(8.35)

8.3 Path integrals and coherent states The concept of coherent states have been applied to broad areas of Quantum Mechanics (Perelomov, 1986; Klauder and Skagerstam, 1985). Here we will focus on its application to path integrals (Faddeev, 1976). We want to compute the matrix elements of the evolution operator U , T ˆ † −i ̵ H(ˆ a ,a ˆ) U =e h

(8.36)

ˆ a ,a where H(ˆ ˆ) is a normal ordered operator and T = tf − ti is the total time †

8.3 Path integrals and coherent states

219

lapse. Thus, if ∣i⟩ and ∣f ⟩ denote two arbitrary initial and final states, we can write the matrix element of U as T ˆ † −i ̵ H(ˆ a ,a ˆ) ⟨f ∣e h ∣i⟩ =

N ( ˆ † a ,a ˆ)) ∣i⟩ ⟨f ∣ (1 − i ̵ H(ˆ h "→0,N →∞

lim

(8.37)

However, instead of inserting a complete set of states at each intermediate time tj (with j = 1, . . . , N ), we will now insert an over-complete set of coherent states {∣zj ⟩} at each time tj through the insertion of the resolution of the identity, N ( ˆ † ⟨f ∣ (1 − i ̵ H(ˆ a ,a ˆ)) ∣i⟩ = h

− ∑ ∣zj ∣ N −1 N dzj d¯ zj ( ˆ † = ∫ (∏ )e j=1 [ ∏ ⟨zk+1 ∣ (1 − i ̵ H(ˆ a ,a ˆ)) ∣zk ⟩] 2πi h j=1

N

2

k=1

( ˆ † ( ˆ † × ⟨f ∣ (1 − i ̵ H(ˆ a ,a ˆ)) ∣zN ⟩⟨z1 ∣ (1 − i ̵ H(ˆ a ,a ˆ)) ∣zi ⟩ h h (8.38)

In the limit ( → 0 these matrix elements become ( ( ˆ † ˆ a† , a a ,a ˆ)) ∣zk ⟩ =⟨zk+1 ∣zk ⟩ − i ̵ ⟨zk+1 ∣H(ˆ ˆ)∣zk ⟩ ⟨zk+1 ∣ (1 − i ̵ H(ˆ h h ( =⟨zk+1 ∣zk ⟩ [1 − i ̵ H(¯ zk+1 , zk )] (8.39) h

where H(¯ zk+1 , zk ) is a function obtained from the normal-ordered Hamilto† nian by performing the substitutions a ˆ → z¯k+1 and a ˆ → zk . Hence, we can write the following expression for the matrix element of the evolution operator of the form T ˆ † −i ̵ H(ˆ a ,a ˆ) ⟨f ∣e h ∣i⟩ =

N

N −1

− ∑ ∣zj ∣ ∑ z¯j+1 zj N −1 ⎛ N dzj d¯ zj ⎞ j=1 ( ∏ [1 − i ̵ H(¯ ⎟e e j=1 zk+1 , zk )] = lim ∫ ⎜∏ 2πi h "→0,N →∞ ⎠ ⎝ j=1 j=1 2

ˆ N⟩ ˆ ( ⟨z1 ∣H∣i⟩ ( ⟨f ∣H∣z ×⟨f ∣zN ⟩⟨z1 ∣i⟩ [1 − i ̵ ] [1 − i ̵ ] h ⟨f ∣zN ⟩ h ⟨z1 ∣i⟩

(8.40)

220

Coherent State Path Integral Quantization of Quantum Field Theory

By further expanding the initial and final states in coherent states dzf d¯ zf −∣zf ∣2 e ψ¯f (zf )⟨zf ∣ 2πi dzi d¯ zi −∣zi ∣2 ∣i⟩ = ∫ e ψi (¯ zi )∣zi ⟩ 2πi

⟨f ∣ = ∫

(8.41)

we find the (formal) result

T ˆ † a ,a ˆ) −i ̵ H(ˆ ∣i⟩ = ⟨f ∣e h

i tf h̵ ̵h ∫t dt [ 2i (z∂t z¯ − z¯∂t z) − H(z, z¯)] i = ∫ DzD¯ ze 1 2 2 (∣zi ∣ + ∣zf ∣ ) ¯ 2 ×e ψf (zf )ψi (¯ zi )

(8.42)

This is the coherent-state form of the path integral. We can identify in this expression the Lagrangian L as the quantity L=

h̵ (z∂t z¯ − z¯∂t z) − H(z, z¯) 2i

(8.43)

It is easy to check that this expression is equivalent to the phase-space path integral derived in Section 5.1. Notice that the Lagrangian for the coherent-state representation presented in Eq.(8.43) is first order in time derivatives. Because of this feature we are not guaranteed that the paths are necessarily differentiable. This property leads to all kinds of subtleties that for the most part we will ignore in what follows.

8.4 Path integral for a non-relativistic Bose gas The field theoretic description of a gas of (spinless) non-relativistic bosons † is given in terms of the creation and annihilation field operators φˆ (x) and ˆ φ(x), that satisfy the equal time commutation relations (in d space dimensions) ˆ [φ(x), φˆ (y)] = δ (x − y) †

Relative to the empty state ∣0⟩, such that

d

ˆ φ(x)∣0⟩ =0

(8.44)

(8.45)

8.4 Path integral for a non-relativistic Bose gas

221

the normal ordered Hamiltonian (in the Grand canonical Ensemble) is 2 h̵ 2 d ˆ† ˆ ˆ ! − µ + V (x)] φ(x) H = ∫ d x φ (x) [− 2m

+

1 d d † † ˆ φ(x) ˆ ∫ d x ∫ d y φˆ (x)φˆ (y)U (x − y)φ(y) 2

(8.46)

where m is the mass of the bosons, µ is the chemical potential, V (x) is an external potential and U (x − y) is the interaction potential between pairs of bosons. Following our discussion of the coherent state path integral we see that it is immediate to write down a path integral for a thermodynamically large system of bosons. The boson coherent states are now labelled by a complex ¯ field φ(x) and its complex conjugate φ(x). † ∫ dx φ(x)φˆ (x) ∣{φ(x)}⟩ = e ∣0⟩

(8.47)

ˆ φ(x)∣{φ}⟩ = φ(x)∣{φ}⟩

(8.48)

which has the coherent state property of being a right eigenstate of the field ˆ operator φ(x),

as well as obeying the resolution of the identity in this space of states − ∫ dx ∣φ(x)∣ I = ∫ DφDφ e ∣{φ}⟩⟨{φ}∣ 2

∗

(8.49)

The matrix element of the evolution operator of this system between an arbitrary initial state ∣i⟩ and an arbitrary final state ∣f ⟩, separated by a time lapse T = tf − ti (not to be confused with the temperature!), now takes the form i ˆ − ̵ HT ⟨f ∣e h ∣i⟩ =

̵ i tf d h ¯ ¯ ¯ t) − φ(x, t)∂t φ(x, t)] − H[φ, φ])} ∫ DφD φ¯ exp { ̵ ∫ dt (∫ d x [φ(x, t)∂t φ(x, i h ti 1 2 2 ∫ dx(∣φ(x, tf )∣ + ∣φ(x, ti )∣ ) 2 ¯ ¯ f (φ(x, tf ))Ψi (φ(x, (8.50) ×Ψ ti )) e

222

Coherent State Path Integral Quantization of Quantum Field Theory

¯ is where the functional H[φ, φ] 2 h̵ 2 ¯ = ∫ dd x φ(x) ¯ H[φ, φ] [− ! − µ + V (x)] φ(x) 2m

1 d d 2 2 + ∫ d x ∫ d y ∣φ(x)∣ ∣φ(y)∣ U (x − y) 2

(8.51)

It is also possible to write the action S in the less symmetric but simpler D d form (where d x ≡ dtd x) 2 2 h̵ D ¯ ⃗ + µ − V (x)) φ(x) ̵ t+ & S = ∫ d x φ(x) (ih∂ 2m

−

1 D D 2 2 ∫ d x ∫ d y ∣φ(x)∣ ∣φ(y)∣ U (x − y) 2

(8.52)

where U (x − y) = U (x − y)δ(tx − ty ). Therefore the path integral for a system of non-relativistic bosons, with chemical potential µ, has the same form os the path integral of the charged scalar field we discussed before except that the action is first order in time derivatives. The fact that the field is complex follows from the requirement that the number of bosons is a globally conserved quantity, which is why one is allowed to introduce a chemical potential. This formulation is useful to study superfluid Helium and similar problems. Suppose for instance that we want to compute the partition function Z for this system of bosons at finite temperature T , ˆ −β H Z = tre

(8.53)

where β = 1/T (in units where kB = 1). The coherent-state path integral representation of the partition function is obtained by 1) restricting the initial and final states to be the same ∣i⟩ = ∣f ⟩ and arbitrary, 2) summing over all possible states, and 3) a Wick rotation to imaginary time t → ̵ with periodic boundary conditions in −iτ , with the time-span T → −iβ h, imaginary time. The result is the (imaginary time) path integral ¯ −S (φ, φ) Z = ∫ DφD φ¯ e E

(8.54)

8.5 Fermion coherent states

223

where SE is the Euclidean action 2 1 β h̵ 2 ¯ ¯ ̵ SE (φ, φ) = ̵ ∫ dτ ∫ dx φ [h∂τ − µ − ! + V (x)] φ 2m h 0

1 β 2 2 + ̵ ∫ dτ ∫ dx ∫ dy U (x − y) ∣φ(x)∣ ∣φ(y)∣ 2h 0

(8.55)

The fields φ(x) = φ(x, τ ) satisfy periodic boundary conditions in imaginary time ̵ φ(x, τ ) = φ(x, τ + β h) (8.56)

This requirement suggests an expansion of the field φ(x) in Fourier modes of the form ∞

iω τ φ(x, τ ) = ∑ e n φ(⃗ x, ω n )

(8.57)

n=−∞

where the frequencies ωn (the Matsubara frequencies) must be chosen so that φ obeys the required PBCs. We find ωn =

2π 2πT n= ̵ n, β h̵ h

(8.58)

n∈Z

where n is an arbitrary integer. In one of the problems at the end of this chapter you will evaluate this path integral using a semiclassical approximation. 8.5 Fermion coherent states In this section we will develop a formalism for fermions which follows closely what we have done for bosons while taking into account the anti-commuting nature of fermionic operators. † Let {ci } be a set of fermion creation operators, with i = 1, . . . , N , and {ci } the set of their N adjoint operators, the associated annihilation operators. † The number operator for the i-th fermion is ni = ci ci . Let us define the basis states states of the i-th fermion by the kets ∣0i ⟩ and ∣1i ⟩, which obey the obvious definitions: ci ∣0i ⟩ = 0,

ci ∣0i ⟩ = ∣1i ⟩, †

ci ci ∣0i ⟩ = 0, †

ci ci ∣1i ⟩ = ∣1i ⟩ †

(8.59)

For N fermions the Hilbert space is spanned by the anti-symmetrized states ∣n1 , . . . , nN ⟩. Let ∣0⟩ ≡ ∣01 , . . . , 0N ⟩

(8.60)

224

Coherent State Path Integral Quantization of Quantum Field Theory

be the empty state. A general state in this Hilbert space is ∣n1 , . . . , nN ⟩ = (c1 )

† n1

. . . (cN )

nN

†

∣0⟩

(8.61)

As we saw before, the wave function ⟨n1 , . . . , nN ∣Ψ⟩ is fully antisymmetric. If the state ∣Ψ⟩ is a product state, the wave function is a Slater determinant.

8.5.1 Definition of fermion coherent states We now define fermion coherent states. Let {ξ¯i , ξi }, with i = 1, . . . , N , be a set of 2N Grassmann variables, a set of symbols also known as the generators of a Grassmann algebra. By definition, Grassmann variables satisfy the following properties {ξi , ξj } = {ξ¯i , ξ¯j } = {ξi , ξ¯j } = ξi = ξ¯i = 0 2

2

(8.62)

Therefore, Grassmann variables behave a a set of time-ordered fermion operators. We will also require that the Grassmann variables anti-commute with the fermion operators: † † {ξi , cj } = {ξ¯i , cj } = {ξ¯i , c,j } = {ξ¯i , cj } = 0

(8.63)

Let us define the fermion coherent states to be −ξc ∣ξ⟩ ≡ e ∣0⟩ ¯ ξc ⟨ξ∣ ≡ ⟨0∣ e †

(8.64) (8.65)

As a consequence of these definitions we have: †

−ξc † e = 1 − ξc

(8.66)

Similarly, if ψ is a Grassmann variable, then ¯ −ψc† ¯ ξc ¯ = eξψ ⟨ξ∣ψ⟩ = ⟨0∣e e ∣0⟩ = 1 + ξψ

(8.67)

For N fermions we have,

N

∣ξ⟩ ≡ ∣ξ1 , . . . , ξN ⟩ =

† N −ξi ci Πi=1 e ∣0⟩

since the following commutator vanishes,

[ξi ci , ξj cj ] = 0 †

†

− ∑ ξi ci ∣0⟩ ≡ e i=1 †

(8.68)

(8.69)

8.5 Fermion coherent states

225

8.5.2 Analytic functions of Grassmann variables We will define ψ(ξ) to be an analytic function of the Grassmann variable if it has a power series expansion in ξ, 2

(8.70)

∀n ≥ 2

(8.71)

ψ(ξ) = ψ0 + ψ1 ξ + ψ2 ξ + . . . where ψn ∈ C. Since n

ξ = 0,

then, all analytic functions of a Grassmann variable reduce to a first degree polynomial, ψ(ξ) ≡ ψ0 + ψ1 ξ

(8.72)

Similarly, we define complex conjugation by ψ(ξ) ≡ ψ¯0 + ψ¯1 ξ¯

(8.73)

where ψ¯0 and ψ¯1 are the complex conjugates of ψ0 and ψ1 respectively. ¯ We can also define functions of two Grassmann variables ξ and ξ, ¯ ξ) = a0 + a1 ξ + a ¯ A(ξ, ¯1 ξ¯ + a12 ξξ

(8.74)

where a1 , a ¯1 and a12 are complex numbers; a1 and a ¯1 are not necessarily complex conjugates of each other. 8.5.3 Differentiation over Grassmann variables Since analytic functions of Grassmann variables have such a simple structure, differentiation is just as simple. Indeed, we define the derivative as the coefficient of the linear term ∂ξ ψ(ξ) ≡ ψ1

(8.75)

∂ξ¯ψ(ξ) ≡ ψ¯1

(8.76)

Likewise we also have

Clearly, using this rule we can write

¯ = −∂ξ (ξ ξ) ¯ = −ξ¯ ∂ξ (ξξ)

(8.77)

A similar argument shows that

¯ ξ) = a1 − a12 ξ¯ ∂ξ A(ξ, ¯ ξ) = a ∂ξ¯A(ξ, ¯1 + a12 ξ ¯ ξ) = −a12 = −∂ξ ∂ξ¯A(ξ, ¯ ξ) ∂ξ¯∂ξ A(ξ,

(8.78) (8.79) (8.80)

226

Coherent State Path Integral Quantization of Quantum Field Theory

from where we conclude that ∂ξ and ∂ξ¯ anti-commute, {∂ξ¯, ∂ξ } = 0,

and

∂ξ ∂ξ = ∂ξ¯∂ξ¯ = 0

(8.81)

8.5.4 Integration over Grassmann variables The basic differentiation rule of Eq. (8.75) implies that 1 = ∂ξ ξ

(8.82)

which suggests the following definitions: ∫ dξ 1 = 0,

∫ dξ ∂ξ ξ = 0,

∫ dξ ξ = 1

(8.83)

¯ Analogous rules also apply for the conjugate variables ξ. It is instructive to compare the differentiation and integration rules: ∫ dξ 1 = 0 ↔ ∂ξ 1 = 0 ∫ dξ ξ = 1 ↔ ∂ξ ξ = 1

(8.84)

Thus, for Grassmann variables differentiation and integration are exactly equivalent ∂ξ ⟺ ∫ dξ

(8.85)

These rules imply that the integral of an analytic function f (ξ) is ∫ dξ f (ξ) = ∫ dξ (f0 + f1 ξ) = f1

and

(8.86)

¯ ξ) = ∫ dξ (a0 + a1 ξ + a ¯ = a1 − a12 ξ¯ ∫ dξA(ξ, ¯1 ξ¯ + a12 ξξ)

¯ ξ, ¯ ξ) = ∫ dξ (a0 + a1 ξ + a ¯ =a ∫ dξA( ¯1 ξ¯ + a12 ξξ) ¯1 + a12 ξ

¯ ¯ ξ) = − ∫ dξdξA( ¯ ξ, ¯ ξ) = −a12 ∫ dξdξA( ξ,

(8.87) It is straightforward to show that with these definitions, the following expression is a consistent definition of a delta-function: ′ −η(ξ − ξ ) δ(ξ , ξ) = ∫ dη e ′

′

where ξ, ξ and η are Grassmann variables.

(8.88)

8.5 Fermion coherent states

227

Finally, given that we have a vector space of analytic functions we can define an inner product as follows: ¯ ¯ e−ξξ f¯(ξ) g(ξ) ¯ = f¯0 g0 + f¯1 g1 ⟨f ∣g⟩ = ∫ dξdξ (8.89) as expected.

8.5.5 Properties of fermion coherent states We defined above the fermion bra and ket coherent states − ∑ ξj cj ∣{ξj }⟩ = e j ∣0⟩, †

∑ ξ¯j cj ⟨{ξj }∣ = ⟨0∣ e j

(8.90)

After a little algebra, using the rules defined above, it is easy to see that the following identities hold: ci ∣{ξj }⟩ =ξi ∣{ξj }⟩

⟨{ξj }∣ ci =∂ξ¯i ⟨{ξj }∣

ci ∣{ξj }⟩ = −∂ξi ∣{ξj }⟩ †

⟨{ξj }∣

† ci

= ξ¯i ⟨{ξj }∣

The inner product of two coherent states, ∣{ξj }⟩ and ∣{ξj }⟩, is

(8.91) (8.92)

′

∑ ξ¯j ξj ′ ⟨{ξj }∣{ξj }⟩ = e j

(8.93)

Similarly, we also have the Resolution of the Identity (which is easy to prove) N

− ∑ ξ¯i ξi N ∣{ξi }⟩⟨{ξi}∣ I = ∫ (Πi=1 dξ¯i dξi ) e i=1

(8.94)

Let ∣Ψ⟩ be some state. Then, we can use Eq. (8.94) to expand the state ∣Ψ⟩ in fermion coherent states ∣ξ⟩, N

where

∣Ψ⟩ = ∫

N (Πi=1 dξ¯i dξi )

− ∑ ξ¯i ξi Ψ(ξ)∣{ξi }⟩ e i=1

¯ ≡ Ψ(ξ¯1 , . . . , ξ¯N ) Ψ(ξ)

(8.95)

(8.96)

We can use the rules derived above to compute the following matrix elements ¯ ⟨ξ∣cj ∣Ψ⟩ = ∂ξ¯j Ψ(ξ),

¯ ⟨ξ∣cj ∣Ψ⟩ = ξ¯j Ψ(ξ) †

which is consistent with what we concluded above.

(8.97)

228

Coherent State Path Integral Quantization of Quantum Field Theory

Let ∣0⟩ be the “empty state”. We will not call it the “vacuum” since it is not in the sector of the ground state of the systems of interest. Let † A ({cj }, {cj }) be a normal ordered operator (with respect to the state ∣0⟩). By using the formalism worked out above one can show without difficulty ′ that its matrix elements in the coherent states ∣ξ⟩ and ∣ξ ⟩ are ∑ ξ¯i ξi ′ A ({ξ¯j }, {ξj }) =e i ′

† ′ ⟨ξ∣A ({cj }, {cj }) ∣ξ ⟩

(8.98)

ˆ, For example, the expectation value of the fermion number operator N ˆ = ∑ c† cj N j

in the coherent state ∣ξ⟩ is

(8.99)

j

ˆ ∣ξ⟩ ⟨ξ∣N = ∑ ξ¯j ξj ⟨ξ∣ξ⟩ j

(8.100)

8.5.6 Grassmann gaussian integrals Let us consider a Gaussian integral over Grassmann variables of the form − ∑ ξ¯i Mij ξj + ξ¯i ζi + ζ¯i ξi ¯ ζ] = ∫ (∏ dξ¯i dξi ) e i,j Z[ζ,

(8.101)

∑ ζ¯i (M ¯ ζ] = (det M ) e ij Z[ζ,

(8.102)

N

i=1

where {ζi } and {ζ¯i } are a set of 2N Grassmann variables, and the matrix Mij is a complex Hermitian matrix. We will now show that −1

)ij ζj

Before showing that Eq. (8.102) is correct, let us make a few observations: 1. Eq. (8.102) looks like the familiar expression for Gaussian integrals for −1/2 bosons except that, instead of a factor of (det M ) , we get a factor of det M in the numerator. Except for the absence of a square root, that the fluctuation determinant appears in the numerator is the main effect of the Fermi statistics!. 2. Moreover, if we had considered a system of N Grassmann variables (instead of 2N ) we would have obtained instead a factor of (det M )

1/2

= Pf(M )

(8.103)

8.5 Fermion coherent states

229

where M would now be an N × N real anti-symmetric matrix, and Pf(M ) denotes the Pfaffian of the matrix M .

To prove that Eq. (8.102) is correct it will be sufficient to consider the case ζi = ζ¯i = 0, since the contribution from these sources is identical to the bosonic case (except for the ordering of the Grassmann variables). Using the Grassmann identities we can write the exponential factor as − ∑ ξ¯i Mij ξj e i,j = ∏ (1 − ξ¯i Mij ξj )

(8.104)

ij

The integral that we need to do is

Z[0, 0] = ∫ (∏ dξ¯i ξi ) ∏ (1 − ξ¯i Mij ξj ) N

i=1

(8.105)

ij

From the integration rules, we can easily see that the only non-vanishing terms in this expression are those that have the just one ξi and one ξ¯i (for each i). Hence we can write N Z[0, 0] =(−1) ∫ (∏ dξ¯i dξi ) ξ¯1 M12 ξ2 ξ¯2 M23 ξ3 . . . + permutations N

i=1

N

N =(−1) M12 M23 M34 . . . ∫ (∏ dξ¯i dξi ) ξ¯1 ξ2 ξ¯2 ξ3 ξ¯3 . . . ξ¯N ξ1 i=1

+ permutations

=(−1)

2N

M12 M23 M34 . . . MN,1 + permutations

(8.106)

What is the contribution of the terms labeled “permutations”? It is easy to see that if we permute any pair of labels, say 2 and 3, we will get a contribution of the form (−1)

2N

(−1)M13 M32 M24 . . .

(8.107)

Hence we conclude that the Gaussian Grassmann integral is just the determinant of the matrix M , − ∑ ξ¯i Mij ξj Z[0, 0] = ∫ (∏ dξ¯i dξi ) e ij = det M N

(8.108)

i=1

Alternatively, we can diagonalize the quadratic form and notice that the Jacobian is “upside-down” compared with the result we found for bosons.

230

Coherent State Path Integral Quantization of Quantum Field Theory

8.6 Path integrals for fermions We are now ready to give a prescription for the construction of a fermion path integral in a general system. Let H the a normal-ordered Hamiltonian, with respect to some reference state ∣0⟩, of a system of fermions. Let ∣Ψi ⟩ be the ket at the initial time ti and ∣Ψf ⟩ be the final state at time tf . The matrix element of the evolution operator can be written as a Grassmann path integral i − H(tf − ti ) ⟨Ψf , tf ∣Ψi , ti ⟩ =⟨Ψf ∣e h̵ ∣Ψi ⟩ i ¯ ψ) S(ψ, ¯ × projection operators ≡ ∫ D ψDψ e h̵ (8.109) where we have not written down the explicit form of the projection operators ¯ ψ) is onto the initial and final states. The action S(ψ, ¯ ψ) = ∫ S(ψ,

tf

ti

¯ t ψ − H(ψ, ¯ ψ)] dt [ih̵ ψ∂

(8.110)

This expression of the fermion path integral holds for any theory of fermions, relativistic or not. Notice that it has the same form as the bosonic path integral. The only change is that for fermions the determinant appears in the numerator while for bosons it is in the denominator!

8.7 Path integral quantization of the Dirac field We will now apply the methods we just developed to the case of the Dirac theory. Let ψα (x), with α = 1, . . . , 4 be a free massive Dirac field in 3+1 space-time dimensions. This field satisfies the Dirac Equation as an equation of motion, (i∂/ − m) ψ = 0

(8.111)

where ψ is a 4-spinor and ∂/ = γ ∂µ . Recall that the Dirac γ-matrices satisfy the algebra µ

{γ , γ } = 2g µ

µν

ν

µν

(8.112)

where g is the Minkowski space metric tensor (in the Bjorken-Drell form). We saw before that in the quantum field theory description of the Dirac theory, ψ is an operator acting on the Fock space of (fermionic) states. We

8.7 Path integral quantization of the Dirac field

231

also saw that the Dirac equation can be regarded as the classical equation of motion of the Lagrangian density L = ψ¯ (i∂/ − m) ψ

(8.113)

† 0 where ψ¯ = ψ γ . We also noted that the momentum canonically conjugate † to the field ψ is iψ , from where the standard fermionic equal time anticommutation relations follow

{ψα (x, x0 ), ψβ (y, x0 )} = δαβ δ (x − y) †

3

(8.114)

The Lagrangian density L for a Dirac fermion coupled to sources ηα and η¯α is ¯ + η¯ψ L = ψ¯ (i∂/ − m) ψ + ψη (8.115)

where the sources are “classical” (Grassmann) anticommuting fields. We can follow the same steps described in the preceding sections to find the following expression for the path-integral of the Dirac field in terms of ¯ Grassmann fields ψ(x) and ψ(x) (which here are independent variables!) 4 ¯ + η¯ψ) i ∫ d x (ψη 1 Z[¯ η , η] = ⟨0∣T e ∣0⟩ ⟨0∣0⟩

¯ ≡ ∫ D ψDψe

¯ + η¯ψ) iS + i ∫ d x(ψη 4

(8.116)

where S = ∫ d xL. From this result it follows that the Dirac propagator is given by 4

iSαβ (x − y) =⟨0∣T ψα (x)ψ¯β (y)∣0⟩ =

2 2 (−i) δ Z[¯ η , η] EEEE E Z[0, 0] δη¯α (x)δηβ (y) EEEEη¯=η=0

=⟨x, α∣

1 ∣y, β⟩ i∂/ − m

(8.117)

Similarly, the partition function for a theory of free Dirac fermions is ZDirac = Det (i∂/ − m)

(8.118)

In contrast, in the case of a free real scalar field we found Zscalar = [Det (∂ + m ) ] 2

2

−1/2

(8.119)

Therefore for the case of the Dirac field the partition function is a determinant whereas for the scalar field is the inverse of a determinant (actually,

232

Coherent State Path Integral Quantization of Quantum Field Theory

of its square root). The fact that one result is the inverse of the other is a consequence that Dirac fields are quantized as fermions whereas scalar fields are quantized as bosons. As we saw, this is a general result for Fermi and Bose fields regardless of whether they are relativistic or not. Moreover, from this result it follows that the vacuum (ground state) energy of a Dirac field is negative whereas the vacuum energy for a scalar field is positive. We will see that in interacting field theories the result of Eq.(8.118) leads to the Feynman diagram rule that each fermion loop carries a minus sign which reflects the Fermi-Dirac statistics, and hence this rule holds even in the absence of relativistic invariance. We now note that, due to the charge-conjugation symmetry of the Dirac theory, the spectrum of the Dirac operator is symmetric, i.e for every positive eigenvalue of the Dirac operator there is a negative eigenvalue of equal 5 µ magnitude. More formally, since {γ , γ } = 0, 2

γ5 (i∂/ − m) γ5 = (−i∂/ − m)

(8.120)

and γ5 = I (the 4 × 4 identity matrix), it follows that

Hence

Det (i∂/ − m) = Det (i∂/ + m)

Det (i∂/ − m) =[Det (i∂/ − m) Det (i∂/ + m) ] =[Det (∂ + m ) ] 2

2

(8.121)

1/2

2

(8.122)

where we used that the “square” of the Dirac operator is the Klein-Gordon operator multiplied by the 4 × 4 identity matrix, which is why the exponent of the r.h.s of the equation above is 2 = 4 × 12 . It is easy to see that this Dirac

result implies that the vacuum energy for the Dirac fermion E0 and the scalar vacuum energy of a scalar field E0 , with the same mass m, are related by Dirac

E0

scalar

= −4E0

(8.123)

Here we have ignored the fact that both the l.h.s. and the r.h.s. of this equation are divergent, as we saw before. However since they have the same divergence, or what is the same is they are regularized in the same way, the comparison is meaningful. On the other hand, if instead of Dirac fermions, which are charged (and hence complex) fields, we consider Majorana fermions, which are charge

8.7 Path integral quantization of the Dirac field

233

neutral, and hence are real fields, we would have obtained instead the results ZMajorana = Pf (i∂/ − m) = [Det (i∂/ − m) ]

1/2

(8.124)

where Pf is the Pfaffian or, what is the same, the square root of the determinant. Thus the vacuum energy of a Majorana fermion is half the vacuum energy of a Dirac fermion. Finally, since a massless Dirac fermion is equivalent to two Weyl fermions, one for each chirality, it follows that the vacuum energy of a Majorana Weyl fermion is equal and opposite to the vacuum energy of a scalar field. Moreover this relation holds for all the states of their spectra which are identical. This observation is the origin of the concept of supersymmetry in which the fermionic states and the bosonic states are precisely matched. As a result, the vacuum energy of a supersymmetric theory is zero. We end by discussing briefly the theory of Dirac fermions in Euclidean space-time. We will focus on the theory in four dimensions, but this can be done in any dimension. There are two equivalent ways to do this analytic continuation. One option is to define a set of four anti-hermitian Dirac gamma matrices j

γ4 = −iγ

γj = γ ,

0

(8.125)

with j = 1, 2, 3, that satisfy the algebra

{γµ , γν } = −2δµν

(8.126)

with µ = 1, . . . , 4. Similarly, we define the hermitian γ5 matrix 5

γ5 = γ = γ1 γ2 γ3 γ4

(8.127)

In this notation the partition function is ¯ Z = ∫ DψD ψ¯ exp(−SE (ψ, ψ))

(8.128)

where SE is the Euclidean for a Dirac spinor which coupled to an abelian gauge field becomes 4 ¯ / + m)ψ SE = ∫ d xψ(i D

(8.129)

where Dµ = ∂µ + iAµ , again with µ = 1, 2, 3, 4, with A4 = −iA0 . The Euclidean Dirac propagator in momentum space is given by (with p0 = −ip4 ) S(p) =

p 1 /+m = 2 −p + m / p + m2

(8.130)

Alternatively, we can define the four gamma matrices to be hermitian and

234

Coherent State Path Integral Quantization of Quantum Field Theory

satisfy the Clifford algebra {γµ , γν } = 2δµnu . In this notation the Euclidean action is ¯ D / + m)ψ SE = ∫ d xψ( 4

(8.131)

where the operator ∂/ is anti-hermitian. In this notation the propagator is S(p) =

ip 1 /+m = 2 −i/ p + m p + m2

(8.132)

8.8 Functional determinants We will now face the problem of how to compute functional determinants. We have discussed before how to do that for path-integrals with a few degrees of freedom (i.e. in Quantum Mechanics). We will now generalize these ideas to Quantum Field Theory. We will begin by discussing some simple determinants that show up in systems of fermions and bosons at finite temperature and density. 8.8.1 Functional determinants for coherent states ˆ at Consider a system of fermions (or bosons) with one-body Hamiltonian h non-zero temperature T and chemical potential µ. The partition function

where β = 1/kB T ,

and

ˆ − µN ˆ) −β (H Z = tr e

ˆ ψ(x) ˆ ˆ = ∫ dx ψˆ† (x) h H ˆ ˆ = ∫ dx ψˆ† (x) ψ(x) N

(8.133)

(8.134)

(8.135)

is the number operator. Here x denotes both spacial and internal (spin) labels. The functional (or path) integral expression for the partition function is i ∗ ˆ + µ) ψ ̵ t−h ∫ dt ψ (ih∂ ̵ h Z = ∫ Dψ Dψ e ∗

(8.136)

̵ In imaginary time set t → −iτ , with 0 ≤ τ ≤ β h: −∫

Z = ∫ Dψ Dψ e ∗

β h̵

0

ˆ − µ) ψ dτ ψ (∂τ + h ∗

(8.137)

8.8 Functional determinants

235

The fields ψ(τ ) can represent either be bosons, in which case they are just complex functions of x and τ , or fermions, in which case they are complex Grassmann functions of x and τ . The only subtlety resides in the choice of boundary conditions 1. Bosons: Since the partition function is a trace, in this case the fields (be complex or real) must obey the usual periodic boundary conditions in imaginary time, ̵ ψ(τ ) = ψ(τ + β h) (8.138)

2. Fermions: In the case of fermions the fields are complex Grassmann variables. However, if we want to compute a trace it turns out that, due to the anticommutation rules, it is necessary to require the fields to obey instead anti-periodic boundary conditions, ̵ ψ(τ ) = −ψ(τ + β h)

(8.139)

ˆ Let {∣λ⟩} be a complete set of eigenstates of the one-body Hamiltonian h, {ελ } be its eigenvalue spectrum with λ a spectral parameter. Hence, we ˆ and let have a suitable set of quantum numbers spanning the spectrum of h, {φλ (τ )} be the associated complete set of eigenfunctions. We now expand ˆ the field configurations in the basis of eigenfunctions of h, ψ(τ ) = ∑ ψλ φλ (τ )

(8.140)

λ

ˆ are complete and orthonormal. The eigenfunctions of h Thus, if we expand the fields, the path-integral of Eq. (8.136) becomes (with h̵ = 1) − ∫ dτ ∑ ψλ (−∂τ − ελ + µ) ψλ ∗ λ Z = ∫ (∏ dψλ dψλ ) e ∗

(8.141)

λ

which becomes (after absorbing all uninteresting constant factors in the integration measure) Z = ∏ [Det (−∂τ − ελ + µ)]

σ

(8.142)

λ

where σ = +1 for fermions and σ = −1 for bosons. λ Let ψn (τ ) be the solution of the equation

(−∂τ − ελ + µ) ψn (τ ) = αn ψn (τ ) λ

λ

(8.143)

236

Coherent State Path Integral Quantization of Quantum Field Theory

where αn is the (generally complex) eigenvalue. The eigenfunctions ψn (τ ) will be required to satisfy either periodic or anti-periodic boundary conditions, λ

ψn (τ ) = −σψn (τ + β) λ

λ

(8.144)

where, once again, σ = ±1. The eigenvalue condition, Eq. (8.143) is solved by

provided αn satisfies

iα τ ψn (τ ) = ψn e n αn = −iωn + µ − ελ

(8.145)

(8.146)

where the Matsubara frequencies are given by (with kB = 1) 2πT (n + 12 ) , for fermions ωn = { 2πT n, for bosons

(8.147)

Let us consider now the function ϕα (τ ) which is an eigenfunction of −∂τ − ελ + µ, (−∂τ − ελ + µ) ϕα (τ ) = α ϕα (τ )

which satisfies only an initial condition for ϕα (0), such as

(8.148)

λ

ϕα (0) = 1 λ

(8.149)

Notice that since the operator is linear in ∂τ we cannot impose additional conditions on the derivative of ϕα . The solution of ∂τ ln ϕα (τ ) = µ − ε − α λ

is

λ λ (µ − ελ − α) τ ϕα (τ ) = ϕα (0) e

(8.150)

(8.151)

After imposing the initial condition of Eq. (8.149), we find λ − (α + ελ − µ) τ ϕα (τ ) = e

(8.152)

But, although this function ϕα (τ ) satisfies all the requirements, it does not have the same zeros as the determinant Det(−∂τ + µ − ελ − α). However, the function λ

λ σ Fα (τ )

= 1 + σ ϕα (τ ) λ

(8.153)

8.8 Functional determinants

237

does satisfy all the requirements. Indeed, λ σ Fα (β)

− (α + ελ − µ) β =1+σ e

(8.154)

which vanishes for α = αn . Then, a version of Coleman’s argument, discussed in Sec.5.2.2, tells us that Det (−∂τ + µ − ελ − α) = constant λ σ Fα (β)

(8.155)

where the right hand side is a constant in the sense that it does not depend on the choice of the eigenvalues {ελ }. Hence, Det (−∂τ + µ − ελ ) = const. σ F0 (β) λ

(8.156)

The partition function is

−βF σ Z=e = ∏ [Det (−∂τ + µ − ελ )]

(8.157)

λ

where F is the free energy, which we find it is given by F = −σT ∑ ln Det (−∂τ + µ − ελ − α) λ

β(µ − ελ ) ) + f (βµ) = −σT ∑ ln (1 + σe

(8.158)

λ

which is the correct result for non-interacting fermions and bosons. Here, we have set ⎧ ⎪ fermions ⎪0 (8.159) f (βµ) = ⎪ ⎨ βµ ⎪ ⎪ −2T N ln (1 − e ) bosons ⎪ ⎩

where N is the number of states in the spectrum {λ}. In some cases the spectrum has √ the symmetry ελ = −ε−λ , e. g. the Dirac theory whose spectrum is ε± = ± p2 + m2 , and these expressions can be simplified further, ∏ Det (−∂t + iελ ) = ∏ [Det (−∂t + iελ ) Det (−∂t − iελ )] λ

λ>0

= ∏ Det (∂t + ελ ) 2

2

λ>0

≡ ∏ Det (−∂τ + ελ ) 2

λ>0

2

(8.160)

238

Coherent State Path Integral Quantization of Quantum Field Theory

where, in the last step, we performed a Wick rotation. This last expression we have encountered before. The result is ∏ Det (−∂τ + ελ ) = const. ψ0 (β) 2

2

(8.161)

λ>0

where ψ0 (τ ) is the solution of the differential equation (−∂τ + ελ ) ψ0 (τ ) = 0 2

which satisfies the initial conditions

The solution is

Hence ψ0 (β) =

In particular, since

ψ0 (0) = 0

ψ0 (τ ) =

2

∂τ ψ0 (0) = 1

sinh(∣ελ ∣τ ) ∣ελ ∣

∣ε ∣β e λ sinh(∣ελ ∣β) ⟶ ∣ελ ∣ 2∣ελ ∣

as β → ∞,

(8.162)

(8.163)

(8.164)

(8.165)

β ∑ ∣ελ ∣ −β ∑ ελ ∣ε ∣β e λ λ>0 λ0

(8.166)

EG = ∑ ελ

(8.167)

we get that the ground state energy EG is the sum of the single particle energies of the occupied (negative energy) states: λ 1

(8.169)

is the well known Riemann ζ-function. We will now use the eigenvalue spectrum of the operator Aˆ to define the generalized ζ-function ζA (s) = ∑ n

1 asn

(8.170)

where the sum runs over the labels (here denoted by n) of the spectrum of ̂ We will assume that the sum (infinite series) is convergent the operator A. which, in practice, will require that we introduce some sort of regularization at the high end (high energies) of the spectrum. Upon differentiation we find dζA d −s ln an ln an =∑ e = −∑ s an ds ds n n

(8.171) +

where we have assumed convergence. Then, in the limit s → 0 we find that the following identity holds lim+

s→0

dζA = − ∑ ln an = − ln ∏ an ≡ − ln DetA ds n n

(8.172)

Hence, we can formally relate the generalized ζ-function, ζA , to the functional determinant of the operator A dζA EEE = − ln DetA (8.173) E ds EEEs→0+ We have thus reduced the computation of a determinant to the computation of a function with specific properties. Let us define now the generalized Heat Kernel ˆ −a τ ∗ −τ A GA (x, y; τ ) = ∑ e n fn (x)fn (y) ≡ ⟨x∣e ∣y⟩ n

(8.174)

240

Coherent State Path Integral Quantization of Quantum Field Theory

where τ > 0. The Heat Kernel GA (x, y; τ ) clearly obeys the differential equation ˆ A (x, y; τ ) −∂τ GA (x, y; τ ) = AG (8.175) which can be regarded as a generalized Heat Equation. Indeed, for 2 −D ! , this is the regular Heat Equation (where D is the diffusion stant) and, in this case τ represents time. In general we will refer to proper time. The Heat Kernel GA (x, y; τ ) satisfies the initial condition lim+ GA (x, y; τ ) = ∑ fn (x)fn (y) = δ(x − y) ∗

τ →0

Aˆ = conτ as

(8.176)

n

where we have used the completeness relation of the eigenfunctions {fn (x)}. Hence, GA (x, y; τ ) is the solution of a generalized Heat Equation with kernel ˆ It defines a generalized random walk or Markov process. A. We will now show that GA (x, y; τ ) is related to the generalized ζ-function ζA (s). Indeed, let us consider the Heat Kernel GA (x, y; τ ) at short distances, y → x, and compute the integral (below we denote by D is the dimensionality of space-time) D −a τ D ∗ ∫ d x lim GA (x, y; τ ) = ∑ e n ∫ d x fn (x)fn (x) y→x

n

−a τ −τ Aˆ = ∑ e n ≡ tr e

(8.177)

n

where we assumed that the eigenfunctions are normalized to unity ∫ d x ∣fn (x)∣ = 1 2

D

(8.178)

i.e. normalized inside a box. We will now use that, for s > 0 and an > 0 (or at least that it has a positive real part) we can write ∫

∞

dτ τ

s−1

0

−a τ Γ(s) e n = s an

(8.179)

where Γ(s) is the Euler Gamma function: Γ(s) = ∫

∞

dτ τ

s−1

−τ e

(8.180)

0

Then, we obtain the identity ∫

∞

0

dτ τ

s−1

∫ d x lim GA (x, y; τ ) = ∑ D

y→x

n

Γ(s) asn

(8.181)

8.8 Functional determinants

241

Therefore, we find that the generalized ζ-function, ζA (s), can be obtained from the generalized Heat Kernel GA (x, y; τ ): ζA (s) =

∞ 1 s−1 D ∫ dτ τ ∫ d x lim GA (x, y; τ ) y→x Γ(s) 0

(8.182)

This result suggests the following strategy for the computation of deterˆ we solve the generalized Heat minants. Given the hermitian operator A, Equation Aˆ GA = −∂τ GA

(8.183)

subject to the initial condition lim+ GA (x, y; τ ) = δ (x − y) D

(8.184)

τ →0

Next we find the associated ζ-function, ζA (s), using the expression ζA (s) =

∞ 1 s−1 D ∫ dτ τ ∫ d x lim GA (x, y; τ ) y→x Γ(s) 0

(8.185)

where we recognize the identity

−τ Aˆ D tr e = ∫ d x lim GA (x, y; τ )

(8.186)

y→x

+

We next take the limit s → 0 to relate the generalized ζ-function to the determinant: dζA (s) = − ln det Aˆ (8.187) lim+ ds s→0

In practice we will have to exercise some care in this step since we will find singularities as we take this limit. Most often we will keep the points x and y apart by a small but finite distance a, which we will eventually attempt to take to zero. Hence, we will need to understand in detail the short distance behavior of the Heat Kernel. Furthermore, the propagator of the theory SA (x, y) = ⟨x∣ Aˆ

−1

∣y⟩

(8.188)

can also be related to the Heat Kernel. Indeed, by expending Eq.(8.188) in ˆ we find the eigenstates of A, SA (x, y) = ∑ n

⟨x∣n⟩ ⟨n∣y⟩ fn (x)fn (y) =∑ an an ∗

n

(8.189)

242

Coherent State Path Integral Quantization of Quantum Field Theory

We can now write the following integral of the Heat Kernel as ∫

∞

0

dτ GA (x, y; τ ) = ∑ fn (x)fn (y) ∫ ∗

n

∞ 0

fn (x)fn (y) −a τ dτ e n = ∑ a ∗

n

n

(8.190) Hence, the propagator SA (x, y) can be expressed as an integral of the Heat Kernel SA (x, y) = ∫

∞

0

dτ GA (x, y; τ )

(8.191)

Equivalently, we can say that since GA (x, y; τ ) satisfies the Heat Equation ˆ A = −∂τ GA , AG

then

Thus, SA (x, y) = ∫

∞ 0

−τ Aˆ GA (x, y; τ ) = ⟨x∣ e ∣y⟩

−τ Aˆ −1 dτ ⟨x∣ e ∣y⟩ = ⟨x∣ Aˆ ∣y⟩

(8.192)

(8.193)

ˆ and that it is indeed the Green function of A,

Aˆx S(x, y) = δ(x − y)

(8.194)

It is worth to note that the Heat Kernel GA (x, y; τ ), as can be seen from Eq. (8.192), is also the Gibbs density matrix of the bounded Hermitian operˆ As such it has an imaginary time (τ !) path-integral representation. ator A. Here, to actually insure convergence, we must also require that the spectrum of Aˆ be positive. In that picture we view GA (x, y; τ ) as the amplitude for the imaginary-time (proper time) evolution from the initial state ∣y⟩ to the final state ∣x⟩. In other words, we picture SA (x, y) as the amplitude to go from y to x in an arbitrary time. 8.9 The determinant of the Euclidean Klein-Gordon operator As an example of the use of the Heat Kernel method we will use it to compute the determinant of the Euclidean Klein-Gordon operator. Thus, we will take the hermitian operator Aˆ to be 2 2 Aˆ = − & +m

(8.195)

in D Euclidean space-time dimensions. This operator has a bounded positive spectrum. Here we will be interested in a system with infinite size L → ∞, D and a large volume V = L . We will follow the steps outlined above.

8.9 The determinant of the Euclidean Klein-Gordon operator

243

We begin by constructing the Heat Kernel G(x, y; τ ). By definition it is the solution of the partial differential equation (− & +m ) G(x, y; τ ) = −∂τ G(x, y; τ ) 2

2

(8.196)

satisfying the initial condition

lim+ G(x, y; τ ) = δ (x − y) D

(8.197)

τ →0

We will find G(x, y; τ ) by Fourier transforms,

d p ip ⋅ (x − y) G(x, y; τ ) = ∫ G(p, τ ) e D (2π) D

(8.198)

We find that in order for G(x, y; τ ) to satisfy Eq.(8.196), its Fourier transform G(p; τ ) must satisfy the differential equation −∂τ G(p; τ ) = (p + m ) G(p; τ ) 2

2

(8.199)

The solution of this equation, consistent with the initial condition of Eq.(8.197) is − (p + m ) τ G(p; τ ) = e 2

2

(8.200)

We can now easily find G(x, y; τ ) by simply finding the anti-transform of G(p; τ ): d p − (p 2 + m2 ) τ + ip ⋅ (x − y) e G(x, y; τ ) = ∫ (2π)D D

− (m τ + e 2

=

1

(4πτ )D/2

∣x − y∣ ) 4τ 2

(8.201)

Notice that for m → 0, G(x, y; τ ) reduces to the usual diffusion kernel (with unit diffusion constant) lim G(x, y; τ ) =

m→0

Next we construct the ζ-function ζ−&2 +m2 (s) =

∣x − y∣ 4τ

2

1

(4πτ )D/2

− e

∞ 1 s−1 D ∫ dτ τ ∫ d x lim G(x, y; τ ) y→x Γ(s) 0

(8.202)

(8.203)

244

Coherent State Path Integral Quantization of Quantum Field Theory

We first do the integral ∫

∞

dτ τ

∫ d x G(x, y; τ )

s−1

D

0

=

V

(4π)

D/2

∫

2

∞

dτ τ

s−1−D/2

0

R 2 − (m τ + ) 4τ e

(8.204)

where R = ∣x − y∣. Upon scaling the variable τ = λt, with λ = R/2m, we find that ∫

∞

dτ τ

0

where Kν (z),

s−1

G(x, y; τ ) =

D

R s− 2 ( ) K D −s (mR) 2m 2

2

(4π)D/2

z 1 − (t + ) 1 ∞ ν−1 t Kν (z) = ∫ dt t e 2 2 0

(8.205)

(8.206)

is a modified Bessel function. Its short argument behavior is Γ(ν) 2 ν (z ) + . . . Kν (z) ∼ 2

(8.207)

As a check, we notice that for s = 1 the integral of Eq.(8.204) does reproduce the Euclidean Klein-Gordon propagator that we discussed earlier in these lectures. The next step is to take the short distance limit lim ∫

R→0

∞

dτ τ 0

s−1

G(x, y; τ ) = lim

R→0

=

1−s

2

m

D−2s

(2π)D/2 (mR) 2 −s D

D ) 2 D/2 2s−D

Γ (s −

(4π)

K D −s (mR)

m

2

(8.208)

Notice that we have exchanged the order of the limit and the integration. Also, after we took the short distance limit R → 0, the expression above acquired a factor of Γ (s − D/2), which is singular as s−D/2 approaches zero (or any negative integer). Thus, a small but finite R smears this singularity. Finally we find the ζ-function by doing the (trivial) integration over space ζ(s) =

∞ 1 D ∫ dτ ∫ d x lim G(x, y; τ ) y→x Γ(s) 0

=Vµ

−2s

m

D

(4π)D/2

Γ (s −

D ) 2

Γ(s)

m −2s (µ)

(8.209)

8.10 Path integral for spin

245

where µ = 1/R plays the role of a cutoff mass (or momentum) scale that we will need to make some quantities dimensionless. The appearance of this quantity is also a consequence of the singularities. We will now consider the specific case of D = 4 dimensions. For D = 4 the ζ-function is 4

ζ(s) = V

m µ m −2s (µ) 2 16π (s − 1)(s − 2) −2s

(8.210)

We can now compute the desired (logarithm of the) determinant for D = 4 dimensions: ln Det [− & +m ] = − lim+ 2

2

s→0

4

dζ m 3 m [ln µ − ] V = 4 ds 16π 2 4

(8.211)

where V = L . A similar calculation for D = 2 yields the result m m 1 [ln µ − ] V ln Det [− & +m ] = − 2π 2 2

2

where V = L .

2

2

(8.212)

8.10 Path integral for spin We will now discuss the use of path integral methods to describe a quantum mechanical spin. Consider a quantum mechanical system which consists of a spin in the spin-S representation of the group SU (2). The space of states of the spin-S representation is 2S + 1-dimensional, and it is spanned by the 2 basis {∣S, M ⟩} which are the eigenstates of the operators S and S3 , i.e. S ∣S, M ⟩ =S(S + 1) ∣S, M ⟩ 2

S3 ∣S, M ⟩ =M ∣S, M ⟩

(8.213)

with ∣M ∣ ≤ S (in integer-spaced intervals). This set of states is complete ad it forms a basis of this Hilbert space. The operators S1 , S2 and S3 obey the SU (2) algebra, [Sa , Sb ] = i(abc Sc

(8.214)

where a, b, c = 1, 2, 3. The simplest physical problem involving spin is the coupling to an external magnetic field B through the Zeeman interaction HZeeman = µ B ⋅ S

(8.215)

246

Coherent State Path Integral Quantization of Quantum Field Theory

where µ is the Zeeman coupling constant ( i.e. the product of the Bohr magneton and the gyromagnetic factor). Let us denote by ∣0⟩ the highest weight state ∣S, S⟩. Let us define the spin ± raising and lowering operators S , ±

(8.216)

S = S1 ± iS2

The highest weight state ∣0⟩ is annihilated by S , +

S ∣0⟩ = S ∣S, S⟩ = 0 +

Clearly, we also have

+

S ∣0⟩ =S(S + 1)∣0⟩

(8.217)

2

S3 ∣0⟩ =S∣0⟩

n0

(8.218)

θ n

n0 × n

Figure 8.1 Geometry for a spin coherent state ∣n⟩ (see text).

Let us consider now the spin coherent state ∣n⟩, (Perelomov, 1986) iθn0 × n ⋅ S ∣n⟩ = e ∣0⟩

(8.219)

2

where n is a three-dimensional unit vector (n = 1), n0 is a unit vector pointing along the direction of the quantization axis (i.e. the “North Pole” of the unit sphere) and θ is the colatitude, (see Fig. 8.1) n ⋅ n0 = cos θ

(8.220)

As we will see the state ∣n⟩ is a coherent spin state which represents a

8.10 Path integral for spin

247

spin polarized along the n axis. The state ∣n⟩ can be expanded in the basis ∣S, M ⟩, S

M =−S

(S)

(S)

∣n⟩ = ∑ DM S (n) ∣S, M ⟩

(8.221)

Here DM S (n) are the representation matrices in the spin-S representation. It is important to note that there are many rotations that lead to the same state ∣n⟩ from the highest weight ∣0⟩. For example any rotation along the direction n results only in a change in the phase of the state ∣n⟩. These rotations are equivalent to a multiplication on the right by a rotation about the z axis. However, in Quantum Mechanics this phase has no physically observable consequence. Hence we will regard all of these states as being physically equivalent. In other words, the states form equivalence classes (or rays) and we must pick one and only one state from each class. These rotations are generated by S3 , the (only) diagonal generator of SU (2). Hence, the physical states are not in one-to-one correspondence with the elements of SU (2) but instead with the elements of the right coset SU (2)/U (1), with the U (1) generated by S3 . In the case of of coherent state of a more general Lie group, the coset is obtained by dividing out the maximal torus generated by all the diagonal generators of the group. In mathematical language, if we consider all the rotations at once, the spin coherent states are said to form a hermitian line bundle.

n1 n2 n3

Figure 8.2 Spherical triangle with vertices at the unit vectors, n1 , n2 and n3 .

A consequence of these observations is that the D matrices do not form

248

Coherent State Path Integral Quantization of Quantum Field Theory

a group under matrix multiplication. Instead they satisfy D

(S)

(n1 )D

(S)

(n2 ) = D

(S)

iΦ(n1 , n2 , n3 )S3 (n3 ) e

(8.222)

where the phase factor is usually called a cocycle. Here Φ(n1 , n2 , n3 ) is the (oriented) area of the spherical triangle with vertices at n1 , n2 , n3 . However, since the sphere is a closed surface, which area do we actually mean? “Inside” or “outside”? Thus, the phase factor is ambiguous by an amount determined by 4π, the total area of the sphere, i4πM e

(8.223)

However, since M is either an integer or a half-integer this ambiguity in Φ has no consequence whatsoever, i4πM e =1

(8.224)

We can also regard this result as a requirement that M be quantized to be an integer or a half-integer, i.e. the representations of SU (2). The states ∣n⟩ are coherent states which satisfy the following properties (Perelomov, 1986). The overlap of two coherent states ∣⃗ n1 ⟩ and ∣n2 ⟩ is ⟨n1 ∣n2 ⟩ = ⟨0∣D

= ⟨0∣D =(

(S)

(S)

(n1 ) D †

(S)

(n2 )∣0⟩

iΦ(n1 , n2 , n ⃗ 0 )S3 (n0 )e ∣0⟩

1 + n1 ⋅ n2 S iΦ(n1 , n2 , n0 )S ) e 2

(8.225)

The (diagonal) matrix element of the spin operator is ⟨n∣S∣n⟩ = S n

(8.226)

Finally, the (over-complete) set of coherent states {∣n⟩} have a resolution of the identity of the form Iˆ = ∫ dµ(n) ∣n⟩⟨n∣

(8.227)

where the integration measure dµ(n) is dµ(n) = (

2S + 1 2 3 ) δ(n − 1)d n 4π

(8.228)

Let us now use the coherent states {∣n⟩} to find the path integral for a spin. In imaginary time τ (and with periodic boundary conditions) the path integral is simply the partition function −βH Z = tre

(8.229)

8.10 Path integral for spin

249

where β = 1/T (T is the temperature) and H is the Hamiltonian. As usual the path integral form of the partition function is found by splitting up the imaginary time interval 0 ≤ τ ≤ β in Nτ steps each of length δτ such that Nτ δτ = β. Hence we have Z=

lim

Nτ →∞,δτ →0

−δτ H Nτ tr (e )

(8.230)

and insert the resolution of the identity at every intermediate time step, Z=

≃

⎛ Nτ ⎞ ⎛ Nτ ⎞ −δτ H ⎜∏ ∫ dµ(nj )⎟ ⎜∏ ⟨n(τj )∣e ∣n(τj+1 )⟩⎟ Nτ →∞,δτ →0 ⎝ ⎠ ⎝ j=1 ⎠ j=1 lim

⎛ Nτ ⎞ ⎛ Nτ ⎞ ⎜∏ ∫ dµ(⃗ nj )⎟ ⎜∏ [⟨n(τj )∣⃗ n(τj+1 )⟩ − δτ ⟨n(τj )∣H∣n(τj+1)⟩]⎟ Nτ →∞,δτ →0 ⎝ ⎠ ⎝ j=1 ⎠ j=1 (8.231) lim

However, since

and

⟨n(τj )∣H∣n(τj+1 )⟩ ≃ ⟨n(τj )∣H∣n(τj )⟩ = µSB ⋅ n(τj ) ⟨n(τj )∣n(τj+1 )⟩

(8.232)

1 + n(τj ) ⋅ n(τj+1 ) S iΦ(n(τj ), n(τj+1 ), n0 )S ) e ⟨n(τj )∣n(τj+1 )⟩ = ( 2 (8.233) we can write the partition function in the form Z= where SE [n] is given by

lim

Nτ →∞,δτ →0

−S [n] ∫ Dn e E

(8.234)

Nτ

−SE [n] = iS ∑ Φ(n(τj ), n ⃗ (τj+1 ), n0 ) j=1

Nτ

+S ∑ ln ( j=1

Nτ 1 + n(τj ) ⋅ n ⃗ (τj+1 ) ) − ∑ (δτ )µS n(τj ) ⋅ B 2

(8.235)

j=1

The first term of the right hand side of Eq. (8.235) contains the expression Φ(n(τj ), n ⃗ (τj+1 ), n0 ) which has a simple geometric interpretation: it is the sum of the areas of the Nτ contiguous spherical triangles. These triangles have the pole n0 as a common vertex, and their other pairs of vertices trace a spherical polygon with vertices at {n(τj )}. In the time continuum limit this spherical polygon becomes the history of the spin, which traces a closed

250

Coherent State Path Integral Quantization of Quantum Field Theory

oriented curve Γ = {n(τ )} (with 0 ≤ τ ≤ β). Let us denote by Ω the region of the sphere whose boundary is Γ and which contains the pole n0 . − The complement of this region is Ω and it contains the opposite pole −n0 . Hence we find that +

Φ(n(τj ), n(τj+1 ), n0 ) = A[Ω ] = 4π − A[Ω ] +

lim

Nτ →∞,δτ →0

−

(8.236)

where A[Ω] is the area of the region Ω. Once again, the ambiguity of the area leads to the requirement that S should be an integer or a half-integer. Ω+

n0

n(τj )

n(τj+1

Ω− Figure 8.3 The function n(τ, s) as arbitrary extension of the of the history n(τ ) to the upper cap Ω+ of the sphere S2 .

There is a simple an elegant way to write the area enclosed by Γ. Let n ⃗ (τ ) be a history and Γ be the set of points o the 2-sphere traced by n ⃗ (τ ) for 0 ≤ τ ≤ β. Let us define n(τ, s) (with 0 ≤ s ≤ 1) to be an arbitrary extension + of n(τ ) from the curve Γ to the interior of the upper cap Ω , as shown in Fig.8.3, such that n(τ, 0) = n(τ ),

n(τ, 1) = n0 ,

n(τ, 0) = n(τ + β, 0)

(8.237)

Then the area can be written in the compact form A[Ω ] = ∫ ds ∫ dτ n(τ, s) ⋅ ∂τ n(τ, s) × ∂s n(τ, s) ≡ SWZ [n] +

1

0

β

(8.238)

0

In Mathematics this expression for the area is called the (simplectic) 2-form, and in the literature is usually called a Wess-Zumino action (Witten, 1984), SWZ , or a Berry phase. (Berry, 1984; Simon, 1983) The coherent state path integral for spin is a special case of the method of geometric quantization (Wiegmann, 1989).

8.10 Path integral for spin

251

Total Flux Φ = 4πS

Figure 8.4 A hairy ball or monopole

Thus, in the (formal) time continuum limit, the action SE becomes (Fradkin and Stone, 1988) SE = −iS SWZ [n] +

β Sδτ β 2 ∫ dτ (∂τ n(τ )) + ∫ dτ µS B ⋅ n(τ ) (8.239) 2 0 0

Notice that we have kept (temporarily) a term of order δτ , which we will drop shortly. How do we interpret Eq. (8.239)? Since n(τ ) is constrained to be a point 2 on the surface of the unit sphere, i.e. n = 1, the action SE [n] can be interpreted as the action of a particle of mass M = Sδτ → 0 and n ⃗ (τ ) is the position vector of the particle at (imaginary) time τ . Thus, the second term is a (vanishingly small) kinetic energy term, and the last term of Eq. (8.239) is a potential energy term. What is the meaning of the first term? In Eq. (8.238) we saw that SWZ [n], the the so-called Wess-Zumino or Berry phase term in the action, is the area of the (positively oriented) region A[Ω+ ] “enclosed” by the “path” n(τ ). In fact, SWZ [n] = ∫ ds ∫ dτ n ⋅ ∂τ n × ∂s n 1

β

0

(8.240)

0

+

is the area of the oriented surface Ω whose boundary is the oriented path + Γ = ∂Ω (see Fig. 8.3). Using Stokes theorem we can write the the expression SA[n] as the circulation of a vector field A[n], ∮

∂Ω

dn ⋅ A[n(τ )] = ∬

Ω+

dS ⋅ !n × A[n(τ )]

(8.241)

252

Coherent State Path Integral Quantization of Quantum Field Theory

provided the “magnetic field” !n × A is “constant”, namely B = !n × A[n(τ )] = S n(τ )

(8.242)

In other words, this is the magnetic field of a magnetic monopole located at the center of the sphere. What is the total flux Φ of this magnetic field? Φ=∫

sphere

dS ⋅ !n × A[n] = S ∫ dS ⋅ n ≡ 4πS

(8.243)

Thus, the total number of flux quanta Nφ piercing the unit sphere is Nφ =

Φ = 2S = magnetic charge 2π

(8.244)

We reach the condition that the magnetic charge is quantized, a result known as the Dirac quantization condition. Is this result consistent with what we know about charged particles in magnetic fields? In particular, how is this result related to the physics of spin? To answer these questions we will go back to real time and write the action S[n] = ∫

T 0

dt [

dn M dn 2 ( ) + A[n(t)] ⋅ − µSn(t) ⋅ B] 2 dt dt

(8.245)

2

with the constraint n = 1 and where the limit M → 0 is implied. The classical hamiltonian associated to the action of Eq. (8.245) is H=

2 1 [n × (p − A[n]) ] + µSn ⋅ B ≡ H0 + µSn ⋅ B 2M

(8.246)

It is easy to check that the vector Λ,

satisfies the algebra

Λ = n × (p − A)

̵ abc (Λc − hSn ̵ c) [Λa , Λb ] = ih(

(8.247)

(8.248)

where a, b, c = 1, 2, 3, (abc is the (third rank) Levi-Civita tensor, and with Λ⋅n=n⋅Λ =0

(8.249)

the generators of rotations for this system are ̵ L = Λ + hSn

(8.250)

8.10 Path integral for spin

253

The operators L and Λ satisfy the (joint) algebra ̵ abc Ls [La , Lb ] = −ih( ̵ abc nc [La , nb ] = ih(

Hence

[La , L ] = 0 2

̵ abc Λc [La , Λb ] = ih(

[La , Λ ] = 0 ⇒ [La , H] = 0 2

(8.251)

(8.252)

since the operators La satisfy the angular momentum algebra, we can diago2 nalize L and L3 simultaneously. Let ∣m, 6⟩ be the simultaneous eigenstates 2 of L and L3 , L ∣m, 6⟩ = h̵ 6(6 + 1)∣m, 6⟩ ̵ L3 ∣m, 6⟩ = hm∣m, 6⟩ 2

2

H0 ∣m, 6⟩ =

h̵ 6(6 + 1) − S ( ) ∣m, 6⟩ 2 2S 2M R

(8.253)

(8.254)

2

(8.255)

where R = 1 is the radius of the sphere. The eigenvalues 6 are of the form + + 6 = S + n, ∣m∣ ≤ 6, with n ∈ Z ∪ {0} and 2S ∈ Z ∪ {0}. Hence each level is 26 + 1-fold degenerate, or what is equivalent, 2n + 1 + 2S-fold degenerate. Then, we get 2 2 2 2 2 2 2 2 Λ = L − n h̵ S = L − h̵ S

(8.256)

Since M = Sδt → 0, the lowest energy in the spectrum of H0 are those with the smallest value of 6, i. e. states with n = 0 and 6 = S. The degeneracy of this “Landau” level is 2S + 1, and the gap to the next excited states diverges as M → 0. Thus, in the M → 0 limit, the lowest energy states have the same 2 degeneracy as the spin-S representation. Moreover, the operators L and L3 become the corresponding spin operators. Thus, the equivalency found is indeed correct. Thus, we have shown that the quantum states of a scalar (non-relativistic) particle bound to a magnetic monopole of magnetic charge 2S, obeying the Dirac quantization condition, are identical to those of those of a spinning particle! (Wu and Yang, 1976) We close this section with some observations on the semi-classical motion. From the (real time) action (already in the M → 0 limit) S = −∫

T 0

dt µS n ⋅ B + S ∫

T 0

1

dt ∫ ds n ⋅ ∂t n × ∂s n

(8.257)

0

we can derive a Classical Equation of Motion by looking at the stationary

254

Coherent State Path Integral Quantization of Quantum Field Theory

configurations. The variation of the second term in Eq. (8.257) is δS = S δ ∫

T 0

1

dt ∫ ds n ⋅ ∂t n × ∂s n = S ∫ 0

T 0

dtδn(t) ⋅ n(t) × ∂t n(t) (8.258)

the variation of the first term in Eq. (8.257) is δ∫

T

0

dt µSn(t) ⋅ B = ∫

T

dt δn(t) ⋅ µSB

(8.259)

0

Hence, δS = ∫

T 0

dt δn(t) ⋅ ( − µSB + Sn(t) × ∂t n(t))

(8.260)

which implies that the classical trajectories must satisfy the equation of motion (8.261)

µB = n × ∂t n If we now use the vector identity

n × n × ∂t n = (n ⋅ ∂t n) n − n ∂t n 2

and

n ⋅ ∂t n = 0,

and

2

n =1

(8.262)

(8.263)

we get the classical equation of motion ∂t n = µB × n

(8.264)

Therefore, the classical motion is precessional with an angular velocity Ωpr = µB.

9 Quantization of Gauge Fields

We will now turn to the problem of the quantization of gauge theories. We will begin with the simplest gauge theory, the free electromagnetic field. This is an abelian gauge theory. After that we will discuss at length the quantization of non-abelian gauge fields. Unlike abelian theories, such as the free electromagnetic field, even in the absence of matter fields non-abelian gauge theories are not free fields and have highly non-trivial dynamics.

9.1 Canonical quantization of the free electromagnetic field The Maxwell theory was the first field theory to be quantized. The quantization procedure of a gauge theory, even for a free field, involves a number of subtleties not shared by the other problems that we have considered so far. The issue is the fact that this theory has a local gauge invariance. Unlike systems which only have global symmetries, not all the classical configurations of vector potentials represent physically distinct states. It could be argued that one should abandon the picture based on the vector potential and go back to a picture based on electric and magnetic fields instead. However, there is no local Lagrangian that can describe the time evolution of the system in that representation. Furthermore, is not clear which fields, E or B (or some other field) plays the role of coordinates and which can play the role of momentum. For that reason, and others, one sticks to the Lagrangian formulation with the vector potential Aµ as its independent coordinate-like variable. The Lagrangian for the Maxwell theory 1 µν L = − Fµν F 4

(9.1)

262

Quantization of Gauge Fields

where Fµν = ∂µ Aν − ∂ν Aµ , can be written in the form L= where

1 2 2 (E − B ) 2

Ej = −∂0 Aj − ∂j A0 ,

(9.2)

Bj = −"jk" ∂k A"

(9.3)

The electric field Ej and the space components of the vector potential Aj form a canonical pair since, by definition, the momentum Πj conjugate to Aj is δL = ∂0 Aj + ∂j A0 = −Ej (9.4) Πj (x) = δ∂0 Aj (x)

Notice that since L does not contain any terms which include ∂0 A0 , the momentum Π0 , conjugate to A0 , vanishes Π0 =

δL =0 δ∂0 A0

(9.5)

A consequence of this result is that A0 is essentially arbitrary and it plays the role of a Lagrange multiplier. Indeed, it is always possible to find a gauge transformation φ ′

′

(9.6)

∂0 φ = −A0

(9.7)

A0 = A0 + ∂0 φ

Aj = Aj − ∂j φ

′

such that A0 = 0. The solution is

which is consistent provided that A0 vanishes both in the remote part and in the remote future, x0 → ±∞. The canonical formalism can be applied to Maxwell electrodynamics if ′ we notice that the fields Aj (x) and Πj ′ (x ) obey the equal-time Poisson Brackets {Aj (x), Πj ′ (x )}P B = δjj ′ δ (x − x ) ′

3

′

(9.8)

or, in terms of the electric field E,

{Aj (x), Ej ′ (x )}P B = −δjj ′ δ (x − x ) ′

3

′

(9.9)

Thus, the spatial components of the vector potential and the components of the electric field are canonical pairs. However, the time component of the vector field, A0 , does not have a canonical pair. Thus, the quantization procedure treats it separately, as a Lagrange multiplier field that is imposing a constraint, that we will see is the Gauss Law. However, at the operator level, the condition Π0 = 0 must then be imposed as a constraint. This fact

9.1 Canonical quantization of the free electromagnetic field

263

led Dirac to formulate the theory of quantization of systems with constraints (Dirac, 1966). There is, however, another approach, also initiated by Dirac, consisting in setting A0 = 0 and to impose the Gauss Law as a constraint on the space of quantum states. As we will see, this amounts to fixing the gauge first (at the price of manifest Lorentz invariance). The classical Hamiltonian density is defined in the usual manner H = Πj ∂0 Aj − L

(9.10)

We find

1 2 2 H(x) = (E + B ) − A0 (x)! ⋅ E(x) (9.11) 2 Except for the last term, this is the usual answer. It is easy to see that the last term is a constant of motion. Indeed the equal-time Poisson Bracket between the Hamiltonian density H(x) and ! ⋅ E(y) is zero. By explicit calculation, we get {H(x), ! ⋅ E(y)}P B = ∫ d z[ − 3

But

δH(x) δ! ⋅ E(y) δH(x) δ! ⋅ E(y) + ] δAj (z) δEj (z) δEj (z) δAj (z) (9.12)

δH(x) δH(x) δBk (w) 3 3 w =∫ d w = ∫ d wBk (w)δ(x − w)"k"j %" δ(w − z) δAj (z) δBk (w) δAj (z)

= −"k"j %" ∫ d wBk (w)δ(x − w)δ(w − z) 3

z

(9.13)

Hence δH(x) z x = "j"k %" (Bk (x)δ(x − z)) = "j"k Bk (x) %" δ(x − z) δAj (z)

(9.14)

Similarly, we get

δ! ⋅ E(y) =0 δAj (z)

δ! ⋅ E(y) y = %j δ(y − z), δEj (z)

Thus, the Poisson Bracket is

(9.15)

{H(x), ! ⋅ E(y)}P B = ∫ d z[−"j"k Bk (x) %" δ(x − z) %j δ(y − z)] 3

x

= −"j"k Bk (x) %" %j δ(x − y) y

x

= "j"k Bk (x) %" %j δ(x − y) = 0 x

y

x

(9.16)

264

Quantization of Gauge Fields

provided that B(x) is non-singular. Thus, !⋅E(x) is a constant of motion. It is easy to check that ! ⋅ E generates infinitesimal gauge transformations. We will prove this statement directly in the quantum theory. Since ! ⋅ E(x) is a constant of motion, if we pick a value for it at some initial time x0 = t0 , it will remain constant in time. Thus we can write (9.17)

! ⋅ E(x) = ρ(x)

which we recognize to be Gauss’s Law. Naturally, an external charge distribution may be explicitly time dependent and then d ∂ ∂ (! ⋅ E) = (! ⋅ E) = ρ (x, x0 ) dx0 ∂x0 ∂x0 ext

(9.18)

Before turning to the quantization of this theory, we must notice that A0 plays the role of a Lagrange multiplier field whose variation yields the Gauss Law, ! ⋅ E = 0. Hence, the Gauss Law should be regarded as a constraint rather than an equation of motion. This issue becomes very important in the quantum theory. Indeed, without the constraint ! ⋅ E = 0, the theory is absolutely trivial, and wrong. Constraints impose very severe restrictions on the allowed states of a quantum theory. Consider for instance a particle of mass m moving freely in three dimensional space. Its stationary states have plane wave wave functions Ψp (r, x0 ), with an energy E(p) = 2m . If we constrain the particle to move only on the surface of a sphere of radius R, it becomes equivalent to a rigid ̵2 2 rotor of moment of inertia I = mR and energy eigenvalues ""m = h2I &(& + 1) 2 2 where & = 0, 1, 2, . . ., and ∣m∣ ≤ &. Thus, even the simple constraint r = R , does have non-trivial effects. Unlike the case of a particle forced to move on the surface of a sphere, the constraints that we have to impose when quantizing Maxwell electrodynamics do not change the energy spectrum. This is so because we can reduce the number of degrees of freedom to be quantized by taking advantage of the gauge invariance of the classical theory. This procedure is called gauge fixing. For example, the classical equation of motion p

2

∂ A − ∂ (∂ν A ) = 0 2

µ

µ

ν

(9.19)

in the Coulomb gauge, A0 = 0 and ! ⋅ A = 0, becomes 2

∂ Aj = 0

(9.20)

However the Coulomb gauge is not compatible with the Poisson Bracket {Aj (x), Πj , (x )}P B = δjj ′ δ(x − x ) ′

′

(9.21)

9.1 Canonical quantization of the free electromagnetic field

265

since the spatial divergence of the delta function does not vanish. It will follow that the quantization of the theory in the Coulomb gauge is achieved at the price of a modification of the commutation relations. Since the classical theory is gauge-invariant, we can always fix the gauge without any loss of physical content. The procedure of gauge fixing has the attractive that the number of independent variables is greatly reduced. A standard approach to the quantization of a gauge theory is to fix the gauge first, at the classical level, and to quantize later. However, a number of problems arise immediately. For instance, in most gauges, such as the Coulomb gauge, Lorentz invariance is lost, or at least it is manifestly so. Thus, although the Coulomb gauge, also known as the radiation or transverse gauge, spoils Lorentz invariance, it has the attractive feature that the nature of the physical states (the photons) is quite transparent. We will see below that the quantization of the theory in this gauge has some peculiarities. Another standard choice is the Lorentz gauge µ

∂µ A = 0

(9.22)

whose main appeal is its manifest covariance. The quantization of the system is this gauge follows the method developed by and Gupta and Bleuer. While highly successful, it requires the introduction of states with negative norm (known as ghosts) which cancel all the gauge-dependent contributions to physical quantities. This approach is described in detail in the book by Itzykson and Zuber (Itzykson and Zuber, 1980). More general covariant gauges can also be defined. A general approach consists not on imposing a rigid restriction on the degrees of freedom, but to add new terms to the Lagrangian which eliminate the gauge freedom. For instance, the modified Lagrangian 1 2 1 µ 2 L = − Fµν + (∂ A (x)) 4 2α µ

(9.23)

is not gauge invariant because of the presence of the last term. We can easily see that this term weighs gauge equivalent configurations differently and the parameter 1/α plays the role of a Lagrange multiplier field. In fact, in the limit α → 0 we recover the Lorentz gauge condition. In the path integral quantization of the Maxwell theory it is proven that this approach is equivalent to an average over gauges of the physical quantities. If α = 1, 2 the equations of motion become very simple, i.e. ∂ Aµ = 0. This is the Feynman gauge. In this gauge the calculations are simplest although, here too, the quantization of the theory has subtleties (such as ghosts, etc.).

266

Quantization of Gauge Fields

Still, within the Hamiltonian or canonical quantization procedure, a third approach has been developed. In this approach one fixes the gauge A0 = 0. This condition is not enough to eliminate completely the gauge freedom. In this gauge a residual set of time-independent gauge transformations are still allowed. In this approach quantization is achieved by replacing the Poisson Brackets by commutators and Gauss Law condition becomes now a constraint on the space of physical quantum states. So, we quantize first and constrain later. In general, it is a non-trivial task to prove that all the different quantizations yield a theory with the same physical properties. In practice what one has to prove is that these different gauge choices yield theories whose states differ from each other by, at most, a unitary transformation. Otherwise, the quantized theories would be physically inequivalent. In addition, the recovery of Lorentz invariance may be a bit tedious in some cases. There is however, an alternative, complementary, approach to the quantum theory in which most of these issues become very transparent. This is the path-integral approach. This method has the advantage that all the symmetries are taken care of from the outset. In addition, the canonical methods encounter very serious difficulties in the treatment of the non-abelian generalizations of Maxwell electrodynamics. We will consider here two canonical approaches: 1) quantization in the Coulomb gauge and 2) canonical quantization in the A0 = 0 gauge in the Schr¨ odinger picture.

9.2 Coulomb gauge Quantization in the Coulomb gauge follows the methods developed for the scalar field very closely. Indeed, the classical constraints A0 = 0 and !⋅A = 0 allow for a Fourier expansion of the vector potential A(x, x0 ). In Fourier space, we write A(x, x0 ) = ∫

3

d p A(p, x0 ) exp(ip ⋅ x) (2π)3 2p0

(9.24)

where A(p, x0 ) = A (−p, x0 ). The Maxwell equations yield the classical equation of motion, the wave equation ∗

∂ A(x, x0 ) = 0 2

(9.25)

The Fourier expansion is consistent only if the amplitude A(p, x0 ) satisfies ∂0 A(p, x0 ) + p A(p, x0 ) = 0 2

2

(9.26)

9.2 Coulomb gauge

267

The constraint ! ⋅ A = 0 in turn becomes the transversality condition p ⋅ A(p, x0 ) = 0

(9.27)

Hence, A(p, x0 ) has the time dependence A(p, x0 ) = A(p)e

ip0 x0

−ip0 x0

(9.28)

+ A(−p)e

where p0 = ∣p∣. Then, the mode expansion takes the form A(x, x0 ) = ∫ µ

3

d p ∗ ip⋅x −ip⋅x [A (p)e + A(p)e ] (2π)3 2p0

(9.29)

where p ⋅ x = pµ x . The transversality condition, Eq.(9.27), is satisfied by introducing two polarization unit vectors !1 (p) and !2 (p), such that !1 ⋅!2 = 2 2 !1 ⋅ p = !2 ⋅ p = 0, and !1 = !2 = 1. Hence, if the amplitude A has to be orthogonal to p, it must be a linear combination of !1 and !2 , i.e. A(p) = ∑ !α (p)aα (p)

(9.30)

α=1,2

where the factors aα (p) are complex amplitudes. In terms of aα (p) and ∗ aα (p) the Hamiltonian looks like a sum of oscillators. In the coulomb gauge, the passage to the quantum theory is achieved by assigning to each amplitude aα (p) a Heisenberg annihilation operator a ˆα (p). ∗ † Similarly aα (p) maps onto the adjoint operator, the creation operator a ˆα (p). The expansion of the vector potential in modes now is ˆ A(x) =∫

3

d p −ip⋅x † ip⋅x ∑ !α (p)[ˆ aα (p)e +a ˆα (p)e ] 3 (2π) 2p0 α=1,2

(9.31)

with p = 0 and p0 = ∣p∣. The operators a ˆα (p) and a ˆα (p) satisfy canonical commutation relations 2

†

[ˆ aα (p), a ˆα′ (p )] =2p0 (2π) δ(p − p ) †

3

′

′

[ˆ aα (p), a ˆα′ (p )] =[ˆ aα (p), a ˆα′ (p )] = 0 ′

†

†

′

(9.32)

It is straightforward to check that the vector potential A(x) and the electric field E(x) obey the (unconventional) equal-time commutation relation [Aj (x), Ej ′ (x )] = −i (δjj ′ − ′

2

%j %j ′ %2

) δ (x − x ) 3

′

(9.33)

where the symbol 1/% represents the inverse of the Laplacian, i.e. the

268

Quantization of Gauge Fields

Laplacian Green function. In the derivation of this relation, the following identity was used ′ pj pj ′ j j ∑ "α (p)"α (p) = δjj ′ − 2 (9.34) p α=1,2 These commutation relations are an extension of the canonical commutation relation, and are a consequence of the transversality condition, ! ⋅ A = 0 In this gauge, the (normal-ordered) Hamiltonian is ˆ =∫ H

3

d p † p0 ∑ a ˆα (p)ˆ aα (p) 3 (2π) 2p0 α=1,2

(9.35)

The ground state, the vacuum state ∣0⟩, is annihilated by both polarizations † a ˆα (p)∣0⟩ = 0. The single-particle states are a ˆα (p)∣0⟩ and represent transverse photons with momentum p, energy p0 = ∣p∣ and with the two possible linear polarizations labelled by α = 1, 2. Circularly polarized photons can be constructed in the usual manner. The Coulomb gauge has the advantage that, in this picture, the electromagnetic field can be regarded as a collection of linear harmonic oscillators which are then quantized. This, of course, is a simple reflection of the fact that Maxwell electrodynamics is a free field theory. It has, however, several problems. One is that Lorentz invariance is violated from the outset, and has to be recovered afterwards in the computation of observables. The other is that, as we will discuss below, in non-abelian theories the Coulomb gauge does not exist globally. For these reasons, its usefulness is essentially limited to the Maxwell theory. 9.3 The gauge A0 = 0 In this gauge we will apply directly the canonical formalism. In what follows we will fix A0 = 0 and associate to the three spatial components Aj of the vector potential an operator, Aˆj which acts on a Hilbert space of states. ̂ j. Similarly, to the canonical momentum Πj = −Ej , we assign an operator Π These operators obey the equal-time commutation relations ̂ j ′ (x )] = iδ(x − x )δjj ′ [Â j (x), Π ′

′

(9.36)

Hence, the vector potential A and the electric field E do not are canonically conjugate operators, and do not commute with each other, ˆj ′ (x )] = −iδjj ′ δ(x − x ) [Aˆj (x), E ′

′

(9.37)

Let us now specify the Hilbert space to be the space of states ∣Ψ⟩ with

9.3 The gauge A0 = 0

269

wave functions which, in the field representation, have the form Ψ({Aj (x)}). When acting on these states, the electric field is the functional differential operator ˆj (x) ≡ i δ E (9.38) δAj (x) In this Hilbert space, the inner product is

⟨{Aj (x)}∣{Aj (x)}⟩⟩ ≡ Πx,j δ (Aj (x) − Aj (x))

(9.39)

This Hilbert space is actually much too large. Indeed states with wave functions that differ by time-independent gauge transformations Ψφ ({Aj (x)}) ≡ Ψ({Aj (x) − %j φ(x)})

(9.40)

are physically equivalent since the matrix elements of the electric field opˆj (x) and magnetic field operator B ˆj (x) = "jk" %k Aˆ" (x) are the erator E same for all gauge-equivalent states, i.e. ′ ˆj (x)∣Ψφ ({Aj (x)})⟩ = ⟨Ψ′ ({Aj (x)})∣E ˆj (x)∣Ψ({Aj (x)})⟩ ⟨Ψφ′ ({Aj (x)})∣E

′ ˆj (x)∣Ψφ ({Aj (x)})⟩ = ⟨Ψ′ ({Aj (x)})∣B ˆ j (x)∣Ψ({Aj (x)})⟩ ⟨Ψφ′ ({Aj (x)})∣B

(9.41)

ˆ The (local) operator Q(x)

ˆ ˆ Q(x) = ! ⋅ E(x)

(9.42)

commutes locally with the Hamiltonian and with each other ˆ ˆ = 0, [Q(x), H]

ˆ ˆ [Q(x), Q(y)] =0

(9.43)

ˆ Hence, all the local operators Q(x) can be diagonalized simultaneously with ˆ H. ˆ Let us show now that Q(x) generates local infinitesimal time-independent gauge transformations. From the canonical commutation relation ˆj ′ (x ′ )] = −iδjj ′ δ(x − x ′ ) [Aˆj (x), E

(9.44)

we get (by differentiation)

ˆ ′ )] = [Aˆj (x), %j E ˆj ′ (x ′ )] = i %xj δ(x − x ′ ) [Aˆj (x), Q(x

(9.45)

Hence, we also find

ˆ [i ∫ dzφ(z)Q(z), Aˆj (x)] = − ∫ dzφ(z) %j δ(z − x) = %j φ(x) z

(9.46)

270

Quantization of Gauge Fields

and ˆ ˆ i ∫ dzφ(z)Q(z) −i ∫ dzφ(z)Q(z) ˆ e Aj (x)e =

ˆk (z) ˆk (z) −i ∫ dz %k φ(z)E i ∫ dz) %k φ(z)E =e Aˆj (x) e =Aˆj (x) + %j φ(x) (9.47)

The physical requirement that states that differ by time-independent gauge transformations be equivalent to each other leads to the demand that we should restrict the Hilbert space to the space of gauge-invariant states. These states, which we will denote by ∣Phys⟩, satisfy ˆ ˆ Q(x)∣Phys⟩ ≡ ! ⋅ E(x)∣Phys⟩ =0

(9.48)

Thus, the constraint means that only the states which obey the Gauss Law are in the physical Hilbert space. Unlike the quantization in the Coulomb gauge, in the A0 = 0 gauge the commutators are canonical and the states are constrained to obey the Gauss Law. In the Schr¨ odinger picture, the eigenstates of the system obey the Schr¨odinger equation 2

1 δ 2 ∫ dx [ − + Bj (x) ]Ψ[A] = EΨ[A] 2 δAj (x)2

(9.49)

where Ψ[A] is a shorthand for the wave functional Ψ({Aj (x)}). In this notation, the constraint of Gauss law is xˆ x %j E j (x)Ψ[A] ≡ i %j

δ Ψ[A] = 0 δAj (x)

(9.50)

This constraint can be satisfied by separating the real field Aj (x) into lonL T gitudinal Aj (x) and transverse Aj (x) parts Aj (x) =

L Aj (x)

+

T Aj (x)

3

d p L T ip⋅x (Aj (p) + Aj (p)) e =∫ 3 (2π)

where Aj (x) and Aj (x) satisfy L

T

%j Aj (x) = 0

Aj (x) = %j φ(x)

T

L

L

(9.51)

(9.52) T

and φ(x) is, for the moment, arbitrary. In terms of Aj and Aj the constraint of Gauss law simply becomes x

%j

δ

δAL j (x)

Ψ[A] = 0

(9.53)

9.3 The gauge A0 = 0

271

and the Hamiltonian now is 2

2

δ δ 2 T T ˆ = ∫ d3 p 1 [ − − + p Aj (p)Aj (−p)] H T T L L 2 δAj (p)δAj (−p) δAj (p)δAj (−p) (9.54) We satisfy the constraint by looking only at gauge-invariant states. Their wave functions do not depend on the longitudinal components of A(x). T Hence, Ψ[A] = Ψ[A ]. When acting on those states, the Hamiltonian is 2

δ 3 1 2 T T HΨ = ∫ d p [ − + p Aj (p)Aj (−p)]Ψ = EΨ T T 2 δAj (p)δAj (−p)

(9.55)

Let !1 (p) and !2 (p) be two vectors which together with the unit vector np = p/∣p∣ form an orthonormal basis. Let us define the operators (α = 1, 2; j = 1, 2, 3) δ 1 α T a ˆ(p, α) = √ "j (p)[ T + ∣p∣Aj (p)] δAj (−p) 2∣p∣ δ 1 † α T a ˆ (p, α) = √ "j (p)[− T + ∣p∣Aj (−p) δAj (p) 2∣p∣

(9.56)

These operators satisfy the commutation relations

[ˆ a(p, α), a ˆ (p , α )] = δαα′ δ (p − p ) †

′

′

3

′

(9.57)

ˆ and the expansion of the In terms of these operators, the Hamiltonian H transverse part of the vector potential are † † ˆ = ∫ d3 p ∣p∣ ∑ [ˆ H a (p, α)ˆ a(p, α) + a ˆ(p, α)ˆ a (p, α)] 2 α=1,2

3

d p T j ip⋅x † −ip⋅x Aj (x) = ∫ √ ∑ "α (p)[ˆ a(p, α)e +a ˆ (p, α)e ] 3 (2π) 2∣p∣ α=1,2 (9.58)

We recognize these expressions to be the same ones that we obtained before in the Coulomb gauge (except for the normalization factors). It is instructive to derive the wave functional for the ground state. The ground state ∣0⟩ is the state annihilated by all the oscillators a ˆ(p, α). Hence its wave function Ψ0 [A] satisfies ⟨{Aj (x)}∣ˆ a(p, α)∣0⟩ = 0

(9.59)

272

Quantization of Gauge Fields

This equation is the functional differential equation j

∑(p)[ α

δ

δATj (−p)

+ ∣p∣Aj (p)]Ψ0 ({Aj (p)}) = 0 T

T

(9.60)

It is easy to check that the unique solution of this equation is 1 3 T T Ψ0 [A] = N exp[− ∫ d p∣p∣Aj (p)Aj (−p)] 2

(9.61)

Since the transverse components of Aj (p) satisfy

(9.62)

we can write Ψ0 [A] in the form

(9.63)

Aj (p) = "jk" T

pk A" (p) p × A(p) ) =( ∣p∣ ∣p∣ j

3

1 d p Ψ0 [A] = N exp[− ∫ (p × A(p)) ⋅ (p × A(−p))] 2 ∣p∣

It is instructive to write this wave function in position space, i.e. as a functional of the configuration of magnetic fields {B(x)}. Clearly, we have p × A(p) = − i ∫

3

d x

(2π)3/2 3

d x

p × A(−p) =i ∫

(2π)

3/2

(!x × A(x)) e

−ip⋅x

(!x × A(x)) e

ip⋅x

(9.64)

By substitution of these identities back into the exponent of the wave function, we get Ψ0 [A] = N exp ( −

1 3 3 ′ ′ ′ ∫ d x ∫ d x B(x) ⋅ B(x )G(x − x )) 2

where G(x, x ) is given by ′

−ip⋅(x−x )

3

d p e G(x − x ) = ∫ (2π)3 ′

(9.65)

′

∣p∣

(9.66)

This function has a singular behavior at large values of ∣p∣. We will define ′ a smoothed version GΛ (x − x ) to be 3

−ip⋅(x−x )

d p e GΛ (x − x ) = ∫ (2π)3 ′

′

∣p∣

−∣p∣/Λ

(9.67)

e

which cuts off the contributions with ∣p∣ ≫ Λ. Also, GΛ (x, x ) formally ′

9.4 Path integral quantization of gauge theories

273

goes back to G(x − x ) as Λ → ∞. GΛ (x − x ) can be evaluated explicitly to give ′

′

∞ ′ 1 −t/Λ∣x−x ∣ ∫ dt sin t e 2π 2 ∣x − x′ ∣2 0 1 1 = 2 Im [ 1 ] ′ 2 2π ∣x − x ∣ −i Λ∣x−x′ ∣

GΛ (x − x ) = ′

Thus,

lim GΛ (x − x ) = ′

Λ→∞

1 2π 2 ∣x − x′ ∣2

Hence, the ground state wave functional Ψ0 [A] is

(9.68)

(9.69)

1 3 3 B(x) ⋅ B(x ) Ψ0 [A] = N exp ( − 2 ∫ d x ∫ d x ) 4π ∣x − x′ ∣2 ′

(9.70)

which is only a functional of the configuration of magnetic fields.

9.4 Path integral quantization of gauge theories We have discussed at length the quantization of the abelian gauge theory, i.e. Maxwell electromagnetism, within canonical quantization in the A0 = 0 gauge and a modified canonical formalism in the Coulomb gauge. Although conceptually what we have done is correct, it poses a number of questions. The canonical formalism is natural in the gauge A0 = 0, and can be generalized to other gauge theories. However, this gauge is highly non-covariant and it is necessary to prove covariance of physical observables at the end. In addition, the gauge field propagator in this gauge is very complicated. The particle spectrum is most transparent in the transverse (or Coulomb) gauge. However, in addition of being non-covariant, it is not possible to generalize this gauge to non-abelian gauge theories (or even to abelian gauge theories on a compact gauge group) due to subtle topological problems known as Gribov ambiguities (or Gribov “copies”). The propagator is equally awful in this gauge. The commutation relations in real space look quite different from those in scalar field theory. In addition, for non-abelian gauge groups, even in the absence of matter fields, the theory is already non-linear and needs to be regularized in a manner in which gauge invariance is preserved. Although it is possible to use covariant gauges, such as the Lorentz µ gauge ∂µ A = 0, the quantization of the theory is these gauges requires an approach, known as Gupta-Bleuer quantization, of difficult generalization. At the root of this problems is the issue of quantizing a theory which has

274

Quantization of Gauge Fields

a local (or gauge) invariance in a manner in which both Lorentz and gauge invariance are kept explicitly. It turns out that path-integral quantization is the most direct approach to deal with these problems. We will now construct the path integral for the free electromagnetic field. However, formally the procedure that we will present will hold, at least formally, for any gauge theory. We will begin with the theory quantized canonically in the gauge A0 = 0 (Dirac, 1966). We saw above that, in the gauge A0 = 0, the electric field E is (minus) the momentum canonically conjugate to the vector potential A, the spatial components of the gauge field, and both fields obey equal-time canonical commutation relations [Ej (x), Ak (x )] = iδ (x − x ) ′

3

′

(9.71)

In addition, in this gauge, the Gauss Law becomes a constraint on the space of states, i.e. ! ⋅ E(x)∣Phys⟩ = J0 (x)∣Phys⟩

(9.72)

which defines the physical Hilbert space. Here J0 (x) is a charge density distribution. In the presence of a set of conserved sources Jµ (x), that satisfy µ ∂µ J = 0, the Hamiltonian of the free field theory is ˆ = ∫ d3 x 1 (E 2 + B 2 ) + ∫ d3 x J ⋅ A H 2

(9.73)

We will construct the path-integral in the Hilbert space of gauge-invariant states defined by the condition of Eq.(9.72). Let su denote by Z[Jµ ] the partition function

′

ˆ ˆ ⎞ ⎛ −i ∫ dx0 H −i ∫ dx0 H ′ Z[J] = tr T e ≡ tr ⎜ ⎜ T e Pˆ ⎟ ⎟ ⎜ ⎟ ⎝ ⎠

(9.74)

where tr means a trace (or sum) over the space of states that satisfy the Gauss Law, Eq.(9.72). We implement this constraint by means of the operator Pˆ that projects onto the gauge-invariant states, ˆ Pˆ = ∏ δ (! ⋅ E(x) − J0 (x))

(9.75)

x

We will now follow the standard construction of the path integral, while making sure that we only sum over histories that are consistent with the constraint. In principle all we need to do is to insert complete sets of states ˆ which are eigenstates of the field operator A(x) at all intermediate times. These states, denoted by ∣{A(x, x0 )}⟩, are not gauge invariant, and do not

9.4 Path integral quantization of gauge theories

275

satisfy the constraint. However, the projection operator Pˆ projects-out the unphysical components of these states. Hence, if the projection operator is included in the evolution operator, the inserted states actually are gauge-invariant. Thus, to insert at every k intermediate time x0 (k = 1, . . . , N with N → ∞ and ∆x0 → 0) a complete set of gauge-invariant states, amounts to writing Z[J] as N

Z[J] = ∏ ∫ DAj (x, x0 ) k

k=1 k ⟨{Aj (x, x0 )}∣ (1

ˆ ∏ δ (! ⋅ E(x, x0 ) − J0 (x, x0 )) ∣{Aj (x, x0 )}⟩ − i∆x0 H) k

k

k+1

x

(9.76)

As an operator, the projection operator Pˆ is naturally spanned by the eigenstates of the electric field operator ∣{E(x, x0 )}⟩, i.e. ∏ δ (! ⋅ E(x, x0 ) − J0 (x, x0 )) ≡ x

∫ DE(x, x0 ) ∣{E(x, x0 )}⟩⟨{E(x, x0)}∣ ∏ δ (! ⋅ E(x, x0 ) − J0 (x, x0 )) x

(9.77)

The delta function has the integral representation ∏ δ (! ⋅ E(x, x0 ) − J0 (x, x0 )) = x

i∆x0 ∫ d x A0 (x, x0 ) (! ⋅ E(x, x0 ) − J0 (x, x0 )) = N ∫ DA0 (x, x0 )e 3

(9.78)

Hence, the matrix elements of interest become ˆ ∏ δ (%j E ˆj − J0 ) ∣{A(x, x0 + ∆x0 )}⟩ ∫ DA ∏ ⟨{A(x, x0 )}∣ (1 − i∆x0 H) x0

x

= ∫ DA0 DADE ∏⟨{A(x, x0 )}∣{E(x, x0 )}⟩⟨{E(x, x0 )}∣{A(x, x0 + ∆x0 )}⟩ x0

× exp [i∆x0 ∫ d x A0 (x, x0 ) (! ⋅ E(x, x0 ) − J0 (x, x0 )) ] 3

× exp [ −

ˆ ⟨{A(x, x0 )}∣H∣{E(x, x0 )}⟩ ] ⟨{A(x, x0 )}∣{E(x, x0 )}⟩

(9.79)

276

Quantization of Gauge Fields

The overlaps are equal to i ∫ d x A(x, x0 ) ⋅ E(x, x0 ) ⟨{A(x, x0 )}∣{E(x, x0 )}⟩ = e 3

(9.80)

Hence, we find that the product of the overlaps is given by

∏ ⟨{A(x, x0 )}∣{E(x, x0 )}⟩⟨{E(x, x0)}∣{A(x, x0 + ∆x0 )}⟩ = x0

−i ∫ dx0 ∫ d x E(x, x0 ) ⋅ ∂0 A(x, x0 ) =e 3

(9.81)

The matrix elements of the Hamiltonian are

ˆ ⟨{A(x, x0 )}∣H∣{E(x, x0 )}⟩ 1 2 2 3 = ∫ d x [ (E + B ) + J ⋅ A] 2 ⟨{A(x, x0 )}∣{E(x, x0 )}⟩

(9.82)

Putting everything together we find that the path integral expression for Z[J] has the form iS[Aµ , E] Z[J] = ∫ DAµ DE e

(9.83)

DAµ = DADA0

(9.84)

where

and the action S[Aµ , E] is given by S[Aµ , E] = ∫ d x [−E ⋅ ∂0 A −

1 2 2 (E + B ) − J ⋅ A + A0 (! ⋅ E − J0 )] 2 (9.85) Notice that the Lagrange multiplier field A0 , which appeared when we introduced the integral representation of the delta function, has become the time component of the vector potential. Since the action is quadratic in the electric fields, we can integrate them out explicitly to find 4

1 2 4 i ∫ d x (− E − E ⋅ (∂0 A + !A0 )) 2 ∫ DE e =

4 1 2 i ∫ d x (−∂0 A − !A0 ) 2 = const. e (9.86)

9.5 Path integrals and gauge fixing

277

We now collect everything and find that the path integral is Z[J] = ∫ DAµ

i ∫ d xL e 4

(9.87)

where the Lagrangian is 1 µν µ L = − Fµν F + Jµ A (9.88) 4 which is what we should have expected. We should note here that this formal argument is valid for all gauge theories, abelian or non-abelian. In other words, the path integral is always the sum over the histories of the field Aµ with a weight factor which is the exponential of i/h̵ times the action S of the gauge theory. Therefore, we found that, at least formally, we can write a functional integral which will play the role of the generating functional of the N -point functions of this theories, ⟨0∣T Aµ1 (x1 ) . . . AµN (xN )∣0⟩

(9.89)

9.5 Path integrals and gauge fixing The expression for the path integral in Eq.(9.87) is formal because we are summing over all histories of the field without restriction. In fact, since the action S and the integration measure DAµ are both gauge invariant, histories that differ by gauge transformations have the same weight in the path integral, and the partition function has an apparent divergence of the V form v(G) , where v(G) is the volume of the gauge group G and V is the (infinite) volume of space-time. In order to avoid this problem we must implement a procedure that restricts the sum over configurations in such a way that configurations which differ by local gauge transformations are counted only once. This procedure is known as gauge fixing. We will follow that approach introduced by L. Faddeev and V. Popov (Faddeev and Popov, 1967; Faddeev, 1976). Although the method works for all gauge theories, the non-abelian theories have subtleties and technical issues that we will discuss below. We will begin with a general discussion of the method, and then we will specialize it first for the case of Maxwell theory, the U (1) gauge theory without matter fields, and later to the case of a general compact gauge group. Let the vector potential Aµ be a field that takes values in the algebra of a gauge group G, i.e. Aµ is a linear combination of the group generators. Let U (x) be an unitary-matrix field that takes values on a representation of the

278

Quantization of Gauge Fields

group G (please recall our earlier discussion on this subject in section 3.6). For the abelian group U (1), we have iφ(x) U (x) = e

(9.90)

where φ(x) is a real (scalar) field. A gauge transformation is, for a group G U

†

Aµ = U Aµ U − iU ∂µ U

For the abelian group U (1) we have U

Aµ = Aµ + ∂µ φ

†

(9.91)

(9.92)

In order to avoid infinities in Z[J], we must impose restrictions on the sum over histories such that histories that are related via a gauge transformation are counted exactly once. In order to do that we must find a way to classify the configurations of the vector field Aµ into classes. We will do this by defining gauge fixing conditions. Each class is labelled by a representative configuration and other elements in the class are related to it by smooth gauge transformations. Hence, all configurations in a given class are characterized by a set of gauge invariant data, such as field strengths in the case of the abelian theory. The set of configurations that differ from each other by a local gauge transformation belong to the same class. We can think of the class as a set obtained by the action on some reference configuration by the gauge group, and the elements of a class constitute an orbit of the gauge group. Mathematically, the elements of the gauge class form a vector bundle. We must choose gauge conditions such that the theory remains local and, if possible, Lorentz covariant. It is essential that, whatever gauge condition we use, that each class is counted exactly once by the gauge condition. It turns out that for the Maxwell gauge theory this is always (and trivially) the case. However, in non-abelian theories, and in gauge theories with an abelian compact gauge group, there are many gauges in which a class may be counted more than once. The origin of this problem is a topological obstruction first shown by I. Singer. This question is known as the Gribov problem. The Coulomb gauge is well known to always have this problem, except for the trivial case of the Maxwell theory. Finally we must also keep in mind that we are only fixing the local gauge invariance, but we should not alter the boundary conditions since they represent physical degrees of freedom. In particular, if the theory is defined on a closed manifold, e.g. a sphere, tori, etc., large gauge transformations, which wrap around the manifold, represent global degrees of freedom (or

9.5 Path integrals and gauge fixing

279

states). Large gauge transformations play a key role on gauge theories at finite temperature, where the transformation wrap around the (finite and periodic) imaginary time direction. Also, there is a class of gauge theories, known as topological field theories, whose only physical degrees of freedom are represented by large gauge transformations on closed manifolds. We will discuss these theories in chapter 22.

classes

gauge transformations

representative of a class Figure 9.1 The gauge fixing condition selects a manifold of configurations.

How do we impose a gauge condition consistently? We will do it in the following way. Let us denote the gauge condition by that we wish to impose by g(Aµ ) = 0

(9.93)

where g(Aµ ) is a local differentiable function of the gauge fields and/or of µ their derivatives. Examples of such local conditions are g(Aµ ) = ∂µ A for µ the Lorentz gauge, or g(Aµ ) = nµ A for an axial gauge. The discussion that follows is valid for all compact Lie groups G of volume v(G). For the special case of the Maxwell gauge theory, the gauge group is U (1). Up to topological considerations, the group U (1) is isomorphic to the real numbers R, even though the volume of the compact U (1) group is finite, v(U (1)) = 2π, while for the non-compact case the group are the real numbers R whose “volume” is infinite, v(R) = ∞. Naively, to impose a gauge condition would mean to restrict the path integral by inserting Eq. (9.93) as a delta function in the integrand, iS[A, J] Z[J] ∼ ∫ DAµ δ (g(Aµ )) e

(9.94)

280

Quantization of Gauge Fields

We will now see that in general this is an inconsistent (and wrong) prescription. Following Faddeev and Popov, we begin by considering the expression defined by following integral ∆g [Aµ ] ≡ ∫ DU δ (g(Aµ )) −1

U

(9.95)

where Aµ (x) are the configurations of gauge fields related by the gauge transformation U (x) to the configuration Aµ (x), i.e. we move inside one class. In other words, the integral of Eq.(9.95) is a sum over the orbit of the gauge group. Thus, by construction, ∆g [Aµ ] depends only on the class defined by the gauge-fixing condition g or, what is the same, it is gaugeinvariant. −1 Let us show that ∆g [Aµ ] is gauge invariant. We first observe that the integration measure DU , called the Haar measure, is invariant under the ′ composition rule U → U U , U

DU = D(U U ) ′

(9.96)

where U is and arbitrary but fixed element of G. For the case of G = U (1), U = exp(iφ) and DU ≡ Dφ. Using the invariance of the measure, Eq. (9.96), we can write ′

∆g [Aµ ] = ∫ DU δ (g(Aµ )) = ∫ DU δ (g(Aµ )) = ∆g [Aµ ] −1

U

′

′

′′

UU

U

′′

−1

(9.97)

where we have set U U = U . Therefore ∆g [Aµ ] is gauge invariant, i.e. it is a function of the class and not of the configuration Aµ itself. Obviously we can also write Eq. (9.95) in the form ′

′′

−1

1 = ∆g [Aµ ] ∫ DU δ (g(Aµ )) U

(9.98)

We will now insert the number 1, as given by Eq. (9.98), in the path integral for a general gauge theory to find iS[A, J] Z[J] = ∫ DAµ × 1 × e

U iS[A, J] = ∫ DAµ ∆g [Aµ ] ∫ DU δ (g(Aµ )) e

(9.99)

We now make the change of variables

U

Aµ → Aµ

′

(9.100)

9.5 Path integrals and gauge fixing

where U = U (x) is an arbitrary gauge transformation, and find ′

281

′

Z[J] = ∫ DU ∫

U DAµ

′

′

′ ′ iS[A , J] U UU e ∆g [Aµ ] δ (g(Aµ ))

U

(9.101)

(Notice that we have changed the order of integration.) We now choose ′ −1 U = U , and use the gauge invariance of the action S[A, J], of the measure DAµ and of ∆g [A] to write the partition as iS[A, J] Z[J] = [∫ DU ] ∫ DAµ ∆g [Aµ ] δ (g(Aµ )) e

(9.102)

The factor in brackets in Eq. (9.102) is the infinite constant ∫ DU = v(G)

V

(9.103)

where v(G) is the volume of the gauge group and V is the (infinite) volume of space-time. This infinite constant is nothing but the result of summing over gauge-equivalent states inside each class. Thus, provided the quantity ∆g [Aµ ] is finite, and that it does not vanish identically, we find that the consistent rule for fixing the gauge consists in dividing-out the (infinite) factor of the volume of the gauge group but, more importantly, to insert together with the constraint δ (g(Aµ )) the factor ∆g [Aµ ] in the integrand of Z[J], iS[A, J] Z[J] ∼ ∫ DAµ ∆g [Aµ ] δ (g(Aµ )) e

(9.104)

Therefore the measure DAµ has to be understood as a sums over classes of configurations of the gauge fields and not over all possible configurations. We are only left to compute ∆g [Aµ ]. We will show now that ∆g [Aµ ] is a determinant of a certain operator, and is known as the Faddeev-Popov determinant. We will only compute first this determinant for the case of the abelian theory U (1). Below we will also discuss the non-abelian case, relevant for Yang-Mills gauge theories. U We will compute ∆g [Aµ ] by using the fact that g[Aµ ] can be regarded as a function of U (x) (for Aµ (x) fixed). We will now change variables from U to g. The price we pay is a Jacobian factor since 0 with the same magnitudes for all the parameters as in 4.2. We see it is indeed continuous, and the transition occurs at Tc = T0 . Potential Energy for λ4 > 0 Figure 3: The yellow curve again marks the transition point at Tc = T0 . Notice there is only one zero here indicating a continuous phase transition.

Magnetization around Tc for λ4 > 0 Figure 4: This time we see there is a continuous transition at Tc = T0 = 1.

11

4. [4/20] Let’s use our previous results to analyze the phase diagram for λ4 (T − T0 ). For the upper quadrants, λ4 > 0 we know that the phase boundary between the ordered and disordered phase is located at T − T0 = 0. This is just a vertical line! Now for λ4 < 0 we know that the phase boundary can be found with: T − T0 =

|λ4 |2 8λ6

Inverting this we can find our relation, which is shown in Figure 5 below.

Figure 5: Phase boundary with λ4 given as a function of T − T0

The black line is the phase boundary for our continuous (second order) phase transition while the blue curve is the boundary for the discontinuous (first order) phase transition. The region to the left of the curves is the ordered phase while the region to the right represents the disordered phase since the temperature will be less than or greater than Tc , respectively. Notice how the critical temperature for the first order phase transition increases with more negative λ4 as it should! The origin is known as a tricritical point: varying λ4 takes us from a first order phase transition to a second order one!

12

5

Scalar Electrodynamics

The dynamics of a charged complex scalar field coupled to the electromagnetic field Aµ (x) is governed by the Lagrangian density:  ∗   1 λ L = Dµ φ(x) Dµ φ(x) − m2o |φ(x)|2 − |φ(x)|4 − F µν Fµν 2 4  ∗   1 µν µ = Dµ φ(x) D φ(x) − V (φ(x)) − F Fµν 4 where the covariant derivative is defined by: Dµ = ∂µ + ieAµ 1. [2/20] Let’s begin by showing the above Lagrangian is invariant under the gauge transformations: φ0 (x) = e−ieΛ(x) φ(x) φ∗0 (x) = eieΛ(x) φ∗ (x) A0µ (x) = Aµ (x) + Λ(x) The potential is obviously invariant under these transforms, so we just check the covariant derivatives and the field tensors. Starting with the covariant derivatives: Dµ0 φ0 (x) = ∂µ + ieA0µ )e−ieΛ(x) φ(x) = e−ieΛ(x) ∂µ − ie∂µ λ(x) + ieA0µ )φ(x) = e−ieΛ(x) ∂µ + ieAµ )φ(x) Similarly:  It follows:



∗ Dµ0 φ0 (x) = eieΛ(x) ∂µ − ieAµ )φ∗ (x)

(21)

∗    ∗   Dµ0 φ0 (x) D0µ φ0 (x) = Dµ φ(x) Dµ φ(x)

The electromagnetic term is straight forward to check as well; indeed, we just check the field tensor itself 4 :

0 Fµν (x) = ∂µ Aν (x) + ∂µ Λ(x) − ∂ν Aµ (x) + ∂ν Λ(x) = ∂µ Aν (x) − ∂ν Aµ (x)

(22)

= Fµν (x) The invariance of the Lagrangian follows. 2. [8/20] Let’s now find our equations of motion. I may have done this in a slightly different manner than yourself, but this method gives some good practice in differentiation of co and contra-variant tensors. Furthermore, I keep the indices on the tensors as general as they can be, but we will neglect any spatial dependence from here on out since this clears up a lot of clutter. We just keep in mind: δAµ (x) = δµν δ(x − y) δAν (y) and so on. So, we’ll understand there is a Dirac delta function lurking in the background. Let’s begin with 4 We

are of course assuming Λ(x) is second order continuously differentiable.

13

derivatives in respect to our gauge field : ∗   δ  δL = Dσ φ Dσ φ µ µ δA δA !
∗  ∗ δDσ φ δ Dσ φ σ = D φ + Dσ φ δAµ δAµ !
∗  ∗ δDσ φ δ Dλ φ σ = gσλ D φ + Dσ φ δAµ δAµ !  ∗ µλ ∗ σ µσ = ie − gσλ δ φ D φ + Dσ φ φ δ = ie



δL 1 =− δ∂ ν Aµ 4 1 =− 4 =−

1 4

Dµ φ

∗

! ∗

φ − φ Dµ φ

δFαβ δF αβ Fαβ + F αβ ν µ δ∂ ν Aµ δ∂ A

!

σγ δF αβ αβ δF F + g g F αβ ασ βγ δ∂ ν Aµ δ∂ ν Aµ ! δF αβ δF σγ Fαβ + Fσγ ν µ ν µ δ∂ A δ∂ A

!

Let’s evaluate the derivative of the field tensor separately: δ∂ α Aβ δ∂ β Aα δF αβ = − ν µ δ∂ ν Aµ δ∂ ν Aµ  δ∂ A να µβ = δ δ − δ νβ δ µα Substituting in we find:    1  να µβ δL νβ µα νσ µγ νγ µσ = − δ δ − δ δ F + F δ δ − δ δ αβ σγ δ∂ ν Aµ 4 ! 1 =− Fνµ − Fµν + Fνµ − Fµν 4

!

= Fµν Going back to the Euler Lagrange equation we see 5 : ! !  ∗ δL δL ν ν ∗ ∂ = ⇒ ∂ Fνµ = ie φ Dµ φ − Dµ φ φ := jµ δ∂ ν Aµ δAµ

(23)

Now let’s work out the derivatives in respect to our complex scalar field: ∗ δ Dµ φ
δV (φ) δL  = Dµ φ − δφ δφ δφ  ∗ µ 2 ∗ = ieA Dµ φ − mo φ − λ|φ|2 φ∗ Similarly:  ∗ δ D µ φ
δL = Dµ φ δ∂ ν φ δ∂ ν φ  ∗ = Dµ φ δ µν 5 I used the anti-symmetry of the electromagnetic field tensor to cast the final equation in its above form. This should be reminiscent of Maxwell’s equations in tensor form.

14

Throwing everything together we find: ! δL δL ∂ = ν δ∂ φ δφ  ∗  ∗ ⇒ ∂ ν Dν φ = ieAν Dν φ − m2o φ∗ − λ|φ|2 φ∗   ⇒ Dν∗ D∗ν + m2o φ∗ = −λ|φ|2 φ∗ ν

(24)

This looks like a Klein-Gordon equation with a source term! The final EOM we get from variations in respect to φ∗ gives the complex conjugate of the above expression. 3. [2/20] We now find the Hamiltonian density for this system. We begin by listing the momentums that are conjugate to our fields:  ∗ δL Π = 0 = D0 φ δ∂ φ δL Π∗ = 0 ∗ = D0 φ (25) δ∂ φ   δL 0 Πµ = 0 µ = Fµ0 = −Ei δ∂ A Applying a Legendre transformation we see: H = Π∂ 0 φ + Π∗ ∂ 0 φ∗ + Πµ ∂ 0 Aµ − L


1 ~ 2 ~ 2 − |E| = Π Π∗ − ieA0 φ + Π∗ Π + ieA0 φ∗ + F0µ ∂ 0 Aµ − Π Π∗ + |Di φ|2 + V (φ) + |B| 2  


1 ~ 2 ∗ 0 ∗ 0 ∗ 0 i i 0 i 0 ~ 2 = Π Π − ieA φ + Π Π + ieA φ − Ei ∂ A + Ei ∂ A − Ei ∂ A − Π Π∗ + |Di φ|2 + V (φ) + |B| − |E| 2 

 1 ~ 2 ∗ 2 2 0 i ∗ ∗ ~ = Π Π + |Di φ| + |E| + |B| + V (φ) − A ∂ Ei − ie Π φ − Πφ 2 (26) In the last step I integrated by parts on the term Ei ∂ i A0 so that the scalar potential becomes a Lagrange multiplier which enforces our Gauss’ Law constraint. Indeed, when we learn about Noether’s theorem, we will see that there are conserved charges which accompany our conserved currents j µ . In particular, the relationship is: Z Q = e d3 xj 0 Hence, ej 0 is a charge density. Variations of our scalar potential, A0 , leads us to the condition:
~ ·E ~ = ej 0 ∂ i Ei = ie Π∗ φ∗ − Πφ ⇒ ∇ 4. [4/20] We can always rewrite a complex function in terms of a magnitude and a phase: φ(x) = ρ(x)eiθ(x) With this we can easily cast our Lagrangian into it’s polar form; we just note:   Dµ φ = eiθ ∂µ ρ + iρ(∂µ + eAµ ) Performing the requisite substitutions we get: 1 L = (∂µ ρ)2 + ρ2 (∂µ θ + eAµ )2 − V (ρ) − Fµν F µν 4

(27)

From this we derive our EOM for ρ and θ respectively: 0 = ∂µ ∂ µ ρ − ρ(∂µ θ + eAµ )2 + m2o ρ + λρ3 
 0 = ∂ µ ρ2 ∂µ θ + eAµ 15

(28)

The EOM for θ enforces the conservation of the gauge current, and so we can eliminate it with a gauge transform. We choose the London gauge: θ = 0. The gauge fixed Lagrangian is then 6 : 1 L = (∂µ ρ)2 + ρ2 e2 A0µ A0µ − V (ρ) − Fµν F µν 4

(29)

I’ll drop the primes on the vector potential from here on out. Furthermore, we can see the gauge field has acquired a mass! This is the famous Higgs mechanism. 5. [4/20] For this final question we take m2o < 0, so then we can see, from our EOM, that we have the following solutions for ρ constant: ρ=0 r |m2o | ρ = ρ¯ = ± λ Note that I neglected the gauge field, which we need to do if ρ is constant since Aµ is our only remaining degree of freedom. After freezing the ρ-field at ρ¯, our new effective Lagrangian is: |m2o |e2 1 Aµ Aµ − Fµν F µν λ 4 1 2 2 µ = ρ¯ e Aµ A − Fµν F µν 4

Leff =

(30)

I dropped all constant terms from the above Lagrangian. Now we can make it more explicit that the coefficient of the term quadratic in Aµ can be interpreted as the photon mass. We can easily find the EOM with this remaining degree of freedom (we already did the hard part in 5.2): (31) 2¯ ρ2 e2 Aµ + ∂ ν Fνµ = 0 Quickly notice that if we contract both sides of the above equation with ∂ µ , then: ∂ µ Aµ = 0 This implies: ∂ ν Fνµ = ∂ ν ∂ν Aµ Thus:



 ∂ 2 + 2¯ ρ2 e2 Aµ = 0

This is a Klein-Gordon equation for the vector potential, so we interpret the photon.

6 Choosing

(32) p

2¯ ρ2 e2 =

q

2|m2o | λ e

the gauge θ = 0 is equivalent to the gauge transformation φ0 (x) = e−iθ φ(x) ⇒ A0µ = Aµ (x) + ei ∂µ θ.

16

as the mass of

Physics 582, Fall Semester 2021 Professor Eduardo Fradkin Problem Set No. 2: Symmetries and Conservation Laws Due Date: Friday September 24, 2021, 9:00 pm US Central Time Here you will look again at problem 5 of Problem Set 1 in which you studied some of the properties of the dynamics of a charged (complex) scalar field φ(x) coupled to the electromagnetic field Aµ (x). Recall that the Lagrangian density L for this system is L = (Dµ φ(x)) (Dµ φ(x)) − m20 |φ(x)|2 − ∗


2 1 λ |φ(x)|2 − F µν Fµν 2 4

(1)

where Dµ is the covariant derivative Dµ ≡ ∂µ + ieAµ

(2)

e is the electric charge and ∗ denotes complex conjugation. In this problem set you will determine several important properties of this field theory at the classical level. 1. Derive an expression for the locally conserved current jµ (x), associated with the global symmetry φ(x) → φ′ (x) =eiθ φ(x) φ∗ (x) → φ′∗ (x) =e−iθ φ∗ (x) Aµ (x) → A′µ (x) =Aµ (x)

(3)

in terms of the fields of the theory. 2. Show that the conservation of the current jµ implies the existence of a constant of motion. Find an explicit form for this constant of motion. 3. Consider now the case of the local (gauge) transformation φ(x) → φ′ (x) =eiθ(x) φ(x) φ∗ (x) → φ′∗ (x) =e−iθ(x) φ∗ (x) Aµ (x) → A′µ (x) =Aµ (x) + ∂µ Λ(x)

(4)

where θ(x) and Λ(x) are two functions. What should be the relation between θ(x) and Λ(x) for this transformation to be a symmetry of the Lagrangian of the system? 1

4. Show that, if the system has the local symmetry of the previous section, there is a locally conserved gauge current Jµ (x). Find an explicit expression for Jµ and discuss in which way it is different from the current jµ of item 1). Find an explicit expression for the associated constant of motion and discuss its physical meaning. 5. Find the Energy-Momentum T µν tensor for this system. Show that it can be written as the sum of two terms T µν = T µν (A) + T µν (φ, A)

(5)

where T µν (A) is the energy-momentum tensor for the free electromagnetic field and T µν (φ, A) is the tensor which results by modifying the energymomentum tensor for the decoupled complex scalar field φ by the minimal coupling procedure. 6. Find explicit expressions for the Hamiltonian density H(x) and the linear momentum density P(x) for this system. Give a physical interpretation for all of the terms that you found for each quantity. 7. Consider now the case of an infinitesimal Lorentz transformation xµ → x′µ + ωµν xν

(6)

where ωµν infinitesimal and antisymmetric. Show that the invariance of the Lagrangian of this system under these Lorentz transformation leads to the existence of a conserved tensor Mµνλ . Find the explicit form of this tensor. Give a physical interpretation for its spacial components. Does the conservation of this tensor impose any restriction on the properties of the energy-momentum tensor T µν ? Explain. Warning: Be very careful in how you treat the fields. Recall that not all of the fields are scalars! 8. In this section yo will consider again the same system but in a polar representation for the scalar field φ, i.e. φ(x) = ρ(x) eiω(x)

(7)

In problem set 1, problem 2, you showed that for m20 < 0 the lowest energy states of the system can be well approximated by freezing the amplitude mode ρ to a constant value ρ0 which you obtained by an energy minimization argument. In this section you are asked to find the form of (a) the conserved gauge current Jµ , (b) the total energy H, (c) the total linear momentum P in this limit.

2

9. Consider now the analytic continuation to imaginary time of this theory. Find the energy functional of the equivalent system in classical statistical mechanics. Give a physical interpretation for each of the terms of this energy functional. If D is the dimensionality of space-time for the original system, what is the dimensionality of space for the equivalent classical problem?. Warning : Be very careful in the way you perform the analytic continuation of the components of the vector potential.

3

PHYS 582 Fall 2021

General Field Theory

Homework 2 Solutions Matthew O’Brien Total Points: 100

We consider the scalar electrodynamics field theory given by the Lagrangian
2 1 1 L = |Dµ φ|2 − m20 |φ|2 − λ |φ|2 − F µν Fµν , 2 4 with the covariant derivative defined as . Dµ = ∂µ + ieAµ .

(0.1)

(0.2)

Note that this definition differs from the one used in the lecture notes by the sign of the coupling constant. 1. [12 Points] An arbitrary variation of the Lagrangian (with no change in the coordinates) can be written as δL =

δL δL δL δL δL δL δφ + δ∂µ φ + ∗ δφ∗ + δ∂µ φ∗ + δAν + δ∂µ Aν . (1.1) δφ δ∂µ φ δφ δ∂µ φ∗ δAν δ∂µ Aν

We now assume that all fields obey the Euler-Lagrange equations δL δL − ∂µ = 0, δφ δ∂µ φ δL δL − ∂µ = 0, ∗ δφ δ∂µ φ∗ δL δL − ∂µ = 0, δAν δ∂µ Aν

(1.2a) (1.2b) (1.2c)

which imply that   δL δL δL δL δL δφ = ∂µ δφ + δ∂µ φ = + δ∂µ φ, ∂µ δ∂µ φ δ∂µ φ δ∂µ φ δφ δ∂µ φ   δL δL δL δL δL ∂µ δφ∗ = ∂µ δφ∗ + δ∂µ φ∗ = ∗ + δ∂µ φ∗ , ∗ ∗ ∗ δ∂µ φ δ∂µ φ δ∂µ φ δφ δ∂µ φ∗   δL δL δL δL δL ∂µ δAν = ∂µ δAν + δ∂µ Aν = + δ∂µ Aν , δ∂µ Aν δ∂µ Aν δ∂µ Aν δAν δ∂µ Aν

(1.3a) (1.3b) (1.3c)

where we have expanded using the product rule, and then simplified using Eqs. (1.2). Therefore, the variation of the Lagrangian simplifies to   δL δL δL ∗ δL = ∂µ δφ + δφ + δA (1.4) ν . δ∂µ φ δ∂µ φ∗ δ∂µ Aν In the case of the specific global transformation φ(x) −→ φ0 (x) = eiθ φ(x) ' (1 + iθ)φ(x) ∗

0∗

φ (x) −→ φ (x) = e Aµ (x) −→

A0µ (x)

−iθ ∗

∗

φ (x) ' (1 − iθ)φ (x)

= Aµ (x),

(1.5a) (1.5b) (1.5c) 1 of 11

PHYS 582 Fall 2021

General Field Theory

where θ ∈ R is a constant, we have the field variations δφ∗ = −iθφ∗ ,

δφ = iθφ,

δAµ = 0.

Then, we see that the variation (1.4) is   δL = ∂µ i [(Dµ φ)∗ φ − (Dµ φ)φ∗ ] θ.

(1.6)

(1.7)

Since θ is arbitrary, we see that the variation of the Lagrangian can only vanish δL = 0, if the quantity j µ = i [(Dµ φ)∗ φ − φ∗ Dµ φ] = i(φ ∂ µ φ∗ − φ∗ ∂ µ φ) + 2eAµ |φ|2 ,

(1.8)

is a locally conserved current. That is, it satisfies the continuity equation ∂µ j µ = 0. 2. [10 Points] Define the quantity Z Q=

d3 x j 0 (x, t),

(2.1)

where the integral is over all of space (not time), and j 0 is the temporal component of the conserved current j µ (1.8), which we assume vanishes sufficiently rapidly as |x| → ∞. Since j µ satisfies the continuity equation ∂µ j µ (x) = ∂t j 0 (x, t) + ∇ · j(x, t) = 0,

(2.2)

we find that

Z Z dQ = d3 x ∂t j 0 (x, t) = − d3 x ∇ · j(x, t). dt Stokes’ theorem then tells us that Z I 3 d x ∇ · j(x, t) = dS · j(x, t) = 0,

(2.3)

(2.4)

∂R3

where ∂R3 is the “boundary” of space at |x| → ∞, and the last equality comes from our assumption that j µ vanishes at ∞. Therefore, Q is a constant of motion, that is dQ/dt = 0. For the scalar ED theory, we have   Z  3 ∗ ∗ 0 2 Q = d x i φ(x, t)∂t φ (x, t) − φ (x, t)∂t φ(x, t) + 2eA (x, t)|φ(x, t)| . (2.5) 3. [10 Points] We now consider the local gauge transformation φ(x) −→ φ0 (x) = eiθ(x) φ(x),

(3.1a)

φ∗ (x) −→ φ0 ∗ (x) = e−iθ(x) φ∗ (x),

(3.1b)

Aµ (x) −→

A0µ (x)

= Aµ (x) + ∂µ Λ(x),

(3.1c)

where Λ(x) is a sufficiently smooth function. First, note that |φ|2 is manifestly gauge invariant. Then, observe that 0 Fµν = ∂µ A0ν ∂ν A0µ

= ∂µ (Aν + ∂ν Λ) − ∂ν (Aµ + ∂µ Λ) = Fµν + ∂µ ∂ν Λ − ∂ν ∂µ Λ = Fµν ,

(3.2) 2 of 11

PHYS 582 Fall 2021

General Field Theory

so the field strength tensor is also gauge invariant (assuming the function Λ is sufficiently differentiable). Finally, we consider the derivative term. For it to be gauge invariant, we need |Dµ0 φ0 (x)|2 = |Dµ φ(x)|2

=⇒

Dµ0 φ0 (x) = eiθ(x) Dµ φ(x).

(3.3)

Expanding this out, we find Dµ0 φ0 = (∂µ + ieA0µ )φ0 = (∂µ + ieAµ + ie∂µ Λ)eiθ φ = eiθ [∂µ + ieAµ + i(∂µ θ + e∂µ Λ)] φ = eiθ Dµ φ + ieiθ (∂µ θ + e∂µ Λ)φ,

(3.4)

which implies the relation 1 ∂µ Λ(x) = − ∂µ θ(x). e

(3.5)

4. [12 Points] For this local gauge transformation, the infinitesimal variations of the fields are δφ(x) = iθ(x)φ(x),

δφ∗ (x) = −iθ(x)φ∗ (x),

1 δAµ (x) = − ∂µ θ(x). e

(4.1)

We note that it is easy to check that the variation and partial derivatives commute. That is, for example, δ∂µ φ ' ∂µ φ0 − ∂µ φ = ∂µ (φ0 − φ) ' ∂µ δφ. We then re-write the general variation of the Lagrangian (1.4) as   δL δL δL δL ∗ δL = ∂µ δφ + δφ + δ∂µ Aν + δAν , ∗ δ∂µ φ δ∂µ φ δ∂µ Aν δAν

(4.2)

(4.3)

so that specifically for scalar ED,   1 1 δL δL = ∂µ i [(Dµ φ)∗ φ − (Dµ φ)φ∗ ] θ − F µν ∂µ ∂ν θ + ∂ν θ e e δAν 1 1 δL ∂ν θ, = ∂µ (j µ θ) + F µν ∂µ ∂ν θ − e e δAν

(4.4)

where j µ is the locally conserved “matter current” from the previous sections. For a sufficiently smooth function θ(x), ∂µ ∂ν θ(x) is a symmetric tensor, so its contraction with the antisymmetric tensor F µν must vanish. Also, ∂µ j µ = 0, so   1 δL µ δL = (∂µ θ) j − . (4.5) e δAµ Since the Lagrangian is invariant under the gauge transformation, this term must also vanish, and since θ(x) is arbitrary, this implies that δL = ej µ (x). δAµ (x)

(4.6)

3 of 11

PHYS 582 Fall 2021

General Field Theory

This relation implies the existence of a gauge current . J µ (x) = −

δL , δAµ (x)

(4.7)

which is necessarily locally conserved, since it is proportional to the locally conserved matter current current j µ : J µ = −ej µ = ie [φ∗ (Dµ φ) − (Dµ φ)∗ φ] = ie(φ∗ ∂ µ φ − φ ∂ µ φ∗ ) − 2e2 Aµ |φ|2 .

(4.8)

The choice of minus sign in Eq. (4.7) is a matter of convention, chosen here so that the equations of motion for Aµ exactly match the Maxwell equations. To understand the difference between j µ and J µ , we observe the following: • Noether’s theorem connects global symmetries and local conservation laws. In particular, the global U (1) symmetry of the Lagrangian (0.1) is a sufficient condition for the existence of the locally conserved matter current j µ ; conservation of j µ does not require gauge invariance, and there is still a conserved current in the absence of the gauge field Aµ . • The Lagrangian (0.1) can be written schematically as 1 L = − F µν Fµν − Aµ J µ + terms involving φ. 4

(4.9)

(Note that this is not strictly correct for the scalar field theory, since J µ depends explicitly on Aµ , unlike in the Dirac theory). As discussed in Sec. 2.6 of the lecture notes/textbook, conservation of the current J µ is a requirement of the local U (1) gauge invariance of electromagnetism. That is, attempting to couple a gauge field to matter in any other way would violate gauge invariance, which is a core desideratum of the theory. Therefore, the current J µ does not exist without a notion of local U (1) symmetry. By inspection of (4.8), we immediately identify that the constant of motion associated with J µ is simply Z Qgauge = −eQ = −e d3 x j 0 . (4.10) where Q is given by (2.5). Since J µ is the gauge/electrical current [from the definition Eq. (4.7)], and Qgauge is a constant of motion proportional to the coupling constant e, we can identify Qgauge as the total electrical charge of the system. Therefore, U (1) symmetry enforces conservation of electrical charge. 5. [12 Points] The energy-momentum tensor is defined by a suitable generalisation of Eq. (3.161) of the lecture notes/textbook δL ν δL ν ∗ δL ν T˜µν = ∂ φ+ ∂ φ + ∂ Aλ − g µν L δ∂µ φ δ∂µ φ∗ δ∂µ Aλ

(5.1a)

= (Dµ φ)∗ ∂ ν φ + (Dµ φ)∂ ν φ∗ − F µλ ∂ ν Aλ − g µν L.

(5.1b)

We can add any total divergence to this expression without altering its physical content: . T µν = T˜µν + ∂λ (F µλ Aν ) = T˜µν + Aν ∂λ F µλ + F µλ ∂λ Aν .

(5.2) 4 of 11

PHYS 582 Fall 2021

General Field Theory

Then, consider the Euler-Lagrange equations − Jµ =

δL δL = ∂ν F µν , = ∂ν δAµ (x) δ∂ν Aµ

(5.3)

where J µ is the gauge current from the previous part. We can then use this to re-write the energy-momentum tensor obtained from Eq. (5.2) in a manifestly gauge invariant and symmetric form: T µν = (Dµ φ)∗ ∂ ν φ + (Dµ φ)∂ ν φ∗ − F µλ (∂ ν Aλ − ∂λ Aν ) − Aν J µ − g µν L = (Dµ φ)∗ ∂ ν φ + (Dµ φ)∂ ν φ∗ − F µλ F νλ + ieAν [(Dµ φ)∗ φ − φ∗ Dµ φ] − g µν L = (Dµ φ)∗ (∂ ν φ + ieAν φ) + (Dµ φ)(∂ ν φ∗ − ieAν φ∗ ) − F µλ F νλ − g µν L  
1 2 2 2 µ ∗ ν µ ν ∗ µν 2 2 |Dσ φ| − m0 |φ| − λ |φ| = (D φ) (D φ) + (D φ)(D φ) − g 2 (5.4) 1 µν αβ µσ ν + g F Fαβ − F F σ . 4 On the last line, we identify the contribution to the energy-momentum from the electromagnetic field δL ν ∂ Aλ + F µλ ∂λ Aν − g µν LMaxwell (A) δ∂µ Aλ 1 = g µν F αβ Fαβ − F µσ F νσ . (5.5) 4 Then, observe that the energy-momentum tensor for the decoupled scalar field is T µν (A) =

δL ν δL ν ∗ ∂ φ+ ∂ φ − g µν L(φ, φ∗ ) δ∂µ φ δ∂µ φ∗  
1 2 2 µ ∗ ν µ ν ∗ µν 2 2 2 . = (∂ φ )(∂ φ) + (∂ φ)(∂ φ ) − g |∂σ φ| − m0 |φ| − λ |φ| 2

T µν (φ) =

(5.6)

This expression can be made gauge invariant by replacing all the derivatives by covariant derivatives. That is, modifying it by the minimal coupling procedure:  
1 µν µ ∗ ν µ ν ∗ µν 2 2 2 2 2 T (φ, A) = (D φ) (D φ) + (D φ)(D φ) − g |Dσ φ| − m0 |φ| − λ |φ| , 2 (5.7) which is simply the scalar field contribution on the first line of Eq. (5.4). Therefore, we have shown that the energy-momentum tensor of scalar electrodynamics has the form T µν = T µν (A) + T µν (φ, A).

(5.8)

6. [10 Points] The Hamiltonian density is simply the temporal component of the energymomentum tensor (5.4) H = T 00 , written as a function of the canonical momenta: Π(x) =

δL = (D0 φ(x))∗ , δ∂0 φ(x)

Π∗ (x) =

δL = D0 φ(x). δ∂0 φ∗ (x)

(6.1)

We also note that F αβ Fαβ = 2(B 2 − E 2 ), F0i = −F 0i = E i , Aµ = (A0 , −A), and ∂ µ = (∂0 , −∇). Therefore, 
2 1  2 1 H = |Π|2 + |∇φ|2 + m20 |φ|2 + λ |φ|2 + E + B2 2 2 (6.2)   + ieA · φ∗ ∇φ − φ∇φ∗ + (eA)2 |φ|2 .

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The different contributions are as follows: • |Π|2 is the canonical momentum squared, and so is related to the kinetic energy of the scalar field. • |∇φ|2 is the “tension” stored in the scalar field from “stretching” it, similarly to a spring. • U (φ) = m20 |φ|2 + (λ/2)(|φ|2 )2 is the potential energy of the scalar field. • (E 2 + B 2 )/2 is the usual energy density of the electromagnetic field. • The terms on the second line represent the energy associated with the interaction between charges and electromagnetic fields. Apart from a factor of 2 in front of the term quadratic in A, these have the form J · A. This is not surprising, since classically, we have that the energy of a current density jelec passing through a static magnetic field B = ∇ × A, is Z 1 U= dV jelec · A. (6.3) 2 • This expression for the Hamiltonian may differ from your answer in Homework 1 by the terms E · ∇A0 + ieA0 (φ∗ Π∗ − φΠ). (6.4) However, we should understand that the Hamiltonian is an energy density, and appears underneath an integral over all space. Therefore, integrating the first term by parts yields   − A0 ∇ · E − ie(φ∗ Π∗ − φΠ) , (6.5) which must vanish, since we recognise the term in brackets as the conservation law for the temporal component of the gauge current ∂µ F 0µ = −J 0 , that is, Gauss’ law. The boundary term in the integration by parts is exactly the contribution which we cancelled when we added a total divergence to the energy-momentum tensor in the previous part. Incidentally, this implies that the component A0 of the gauge field can be interpreted as a Lagrange multiplier which enforces Gauss’ law as a constraint. Similarly, the linear momentum density P i = T i0 can be read off from Eq. (5.4). Using the fact that F iσ F 0σ = [B × E]i , we find that P = Π(−∇ + ieA)φ + Π∗ [(−∇ + ieA)φ]∗ + E × B

(6.6)

where the contributions are as follows: • The first two terms are the linear momentum density of the complex scalar field (and its complex conjugate). One can see this by setting Aµ = 0 and φ ∝ exp[i(k · x − ωt)], in which case P ∝ k. • E × B is the usual Poynting vector; the momentum density of the EM field. 7. [12 Points] Accounting for a variation of the coordinate system δxµ leads to a modification of the variation of the Lagrangian from part 1:   δL δL δL µ ∗ δL = ∂µ Lδx + δφ + δφ + δAν . (7.1) δ∂µ φ δ∂µ φ∗ δ∂µ Aν

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If we consider a Lorentz transformation Λµν = δ µν + ω µν , where ωµν is infinitesimal and antisymmetric, then δxµ = ω µν xν and the Jacobian determinant is equal to 1. Therefore, all variation in the Action comes from the variation of the Lagrangian. Now, a scalar field is, by definition, invariant under Lorentz transformations. Therefore, the total variation is . 0 = δT φ = δφ + δxµ ∂µ φ = δφ + ω µν xν ∂µ φ, =⇒ δφ = −ω µν xν ∂µ φ,

(7.2)

where δφ is the variation of the field in the absence of coordinate transformations. On the other hand, the vector potential Aµ transforms like a vector under Lorentz transformations: A0 µ (x0 ) = Λµν Aν (x), . =⇒ ω µν Aν = δT Aµ = δAµ + δxλ ∂λ Aµ , =⇒ δAµ = ωµ ν Aν − ω λσ xσ ∂λ Aµ .

(7.3)

Therefore, the variation of the Lagrangian becomes  i  h  δL = ∂µ Lω µσ xσ − (Dµ φ)∗ ∂λ φ + (Dµ φ)∂λ φ∗ ω λσ xσ − F µν ωνσ Aσ − ω λσ xσ ∂λ Aν h  i = ∂µ Lg µλ − (Dµ φ)∗ ∂λ φ − (Dµ φ)∂λ φ∗ + F µν ∂λ Aν xσ − F µν gνλ Aσ ω λσ i h  = ∂µ Lg µλ − (Dµ φ)∗ ∂ λ φ − (Dµ φ)∂ λ φ∗ + F µν ∂ λ Aν xσ − F µν gν λ Aσ ωλσ   (7.4) = −∂µ T˜µλ xσ + F µλ Aσ ωλσ , where T˜µν is the non-symmetrised energy-momentum tensor. Since ωµν is arbitrary, for the variation to be consistent with the Lorentz invariance of the Lagrangian, we must have   (7.5) ∂µ T˜µν xσ + F µλ Aσ = 0, and hence, the terms in the parentheses define a conserved tensor. Then, using the fact that ωµν is an antisymmetric tensor, we can re-write this expression as   . 1 T˜µν xσ + F µλ Aσ ωλσ = M µλσ ωλσ , (7.6) 2 where     M µνλ = T˜µν xλ − T˜µλ xν + F µν Aλ − F µλ Aν , (7.7) is the locally conserved tensor (∂µ M µνλ = 0). We can define a pseudo-vector from the spatial components of the above tensor:   1 Ji = − εijk M 0jk = −εijk T˜0j xk + F 0j Ak . (7.8) 2 Since T˜µν is the non-symmetrised tensor, T˜0j is not the same linear momentum density defined above. Instead, T˜0j = (D0 φ)∗ ∂ j φ + (D0 φ)∂ j φ∗ − F 0λ ∂ j Aλ . = Pφj − E · (∂ j A),

(7.9) 7 of 11

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Substituting this into Eq. (7.8) then yields J = x × P φ + E · (x × ∇)A + E × A.

(7.10)

Since the first two terms originated from T˜0j , we can identify them as the orbital angular momentum of the scalar and electromagnetic fields, respectively. By examining Eq. (7.4), we see that the third term only arose due to the transformation property of the vector field Aµ under Lorentz transformations. Therefore, we can identify E × A with the intrinsic or spin angular momentum of the electromagnetic field. Finally, if we define

  . M µνλ = T˜µν xλ − T˜µλ xν + S µνλ ,

(7.11)

where S µνλ represents the intrinsic AM contributions, then the conservation laws ∂µ M µνλ = 0,

∂µ T˜µν = 0,

(7.12)

imply that   ∂µ S µνλ = ∂µ T˜µν xλ − T˜µλ xν = T˜λν − T˜νλ .

(7.13)

Therefore, the asymmetry of the canonical tensor is directly related to the intrinsic angular momentum. This is the only explicit restriction on the components of T˜µν from the conservation of M µνλ . However, if we re-define M µνλ by adding a total divergence   M µνλ −→ M µνλ + ∂α F µα Aν xλ − F µα Aλ xν = T µν xλ − T µλ xν ,

(7.14)

where T µν is the “improved” energy-momentum tensor we worked with in part 5, then the conservation laws instead imply that T λν − T νλ = 0,

(7.15)

as expected, since the expression for T µν found in part 5 was manifestly symmetric. Note that if we had used this modified tensor to define Ji , then the total angular momentum density would instead be J −→ x × P,

(7.16)

where P is the linear momentum density calculated in part 6. In this form, we cannot identify the orbital and intrinsic contributions intuitively. 8. [10 Points] We now define the complex scalar field in the polar representation φ(x) = ρ(x)eiω(x) . In this case, the Lagrangian in terms of ρ(x) and ω(x) is 1 1 L = (∂µ ρ)2 + ρ2 (∂µ ω + eAµ )2 − m20 ρ2 − λρ4 − F µν Fµν . 2 4

(8.1)

When m20 < 0, minimising the potential energy 1 U (ρ) = −|m20 |ρ2 + λρ4 , 2

(8.2) 8 of 11

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p gives ρ0 = |m20 |/λ. From hereon we work in the unitary gauge where ω(x) = 0. Then, freezing ρ(x) = ρ0 , the covariant derivative becomes Dµ φ = (∂µ + ieAµ )ρ0 = ieρ0 Aµ ,

(8.3)

since ∂µ φ = 0. Therefore, we find the following: (a) The gauge current (4.8) simplifies to J µ = −2e2 ρ20 Aµ .

(8.4)

(b) Noting that the canonical momentum Π(x) = (D0 φ(x))∗ , the total energy follows from (6.2)   Z Z 1 2 3 3 2 2 2 2 2 H = d x H(x) = d x e ρ0 (A0 + A ) + (E + B ) , (8.5) 2 and we have dropped the constant term −|m20 |2 /2λ in the Hamiltonian density, since it gives a formally infinite contribution to the total energy. (c) Similarly, the total linear momentum follows from (6.6) Z Z   j 3 j P = d x P (x) = d3 x 2e2 ρ20 A0 Aj + [E × B]j . (8.6) 9. [12 Points] We now consider the analytic continuation of the scalar EM theory in D = d + 1 spacetime dimensions to D Euclidean dimensions. First, we define the imaginary time coordinate τ : t = −iτ,

=⇒

∂t = i∂τ .

(9.1)

Then, we need to make sure that the Euclidean theory will still be gauge invariant after the analytic continuation. The gauge transformation  1  A0 −→ A0 0 = A0 − ∂t θ(x),  1 e Aµ −→ A0 µ = Aµ − ∂ µ θ(x) =⇒ (9.2)  e 1 A0 −→ A0 = A + ∇θ(x), e implies that the Wick rotated vector potential A˜ should have an imaginary time component which transforms like the spatial components: 1 A˜τ −→ A˜0τ = A˜τ + ∂τ θ(x), e

(9.3)

A0 = −iA˜τ .

(9.4)

and hence, From hereon, we drop the tilde on the Wick rotated vector potential. Thus, the covariant derivative is . D0 = ∂t + ieA0 = i∂τ + eAτ = i(∂τ − ieAτ ) = iDτ . (9.5) Therefore, (Dµ φ)∗ (Dµ φ) = −[(Dτ φ)∗ (Dτ φ) + (Dj φ)∗ (Dj φ)]  = − (∂τ φ∗ )(∂τ φ) + ieAτ (φ∗ ∂τ φ − φ ∂τ φ∗ ) + e2 A2τ |φ|2 + (∇φ∗ ) · (∇φ) + ieA · (φ∗ ∇φ − φ ∇φ∗ ) + (eA)2 |φ|2



(9.6)

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We then consider the Maxwell terms. First, F 0j F0j = gjk (∂ 0 Aj − ∂ j A0 )(∂ 0 Ak − ∂ k A0 ) = −δjk (∂t Aj + ∂j A0 )(∂t Ak + ∂k A0 ) = −δjk (i∂τ Aj − i∂j Aτ )(i∂τ Ak − i∂k Aτ ) = δjk (∂τ Aj − ∂j Aτ )(∂τ Ak − ∂k Aτ ) . = Fτ j Fτ j ,

(9.7)

with implied summation over all repeated indices regardless of position, and where we have used the fact that the Minkowski metric is diag(+, −, −, −) to make the replacement g jk = −δjk . Similarly, F ij Fij = g ik g j` Fij Fk` = (−δik )(−δj` )Fij Fk` = Fij Fij .

(9.8)

Therefore, since F µν is antisymmetric, F 2 = F µν Fµν = 2Fτ j Fτ j + Fij Fij ,

(9.9)

is positive definite, as required in a Euclidean theory. Finally, the Euclidean energy functional E is defined so that Z Z d iS(φ, A) = i dtd x L(φ, A) = − dD x E(φ, A) = −E(φ, A),

(9.10)

where dD x = dτ dd x and E is the energy density. Therefore, putting all of the above together, we find  Z
2 1 1 D E(φ, A) = d x |∂τ φ|2 + |∇φ|2 + m20 |φ|2 + λ |φ|2 + F 2 2 4  (9.11) ∗ ∗ ∗ ∗ 2 2 2 2 + ieAτ (φ ∂τ φ − φ ∂τ φ ) + ieA · (φ ∇φ − φ ∇φ ) + e (Aτ + A )|φ| . We can also write this in a simpler way by using the D dimensional Euclidean (positive) metric signature:   Z
1 1 2 D 2 2 2 2 2 E(φ, A) = d x |∂φ − ieAφ| + m0 |φ| + λ |φ| + F , (9.12) 2 4 where ∂ = (∂τ , ∇) and A = (Aτ , A), which represents a classical theory in D spatial dimensions. In three dimensions the 2-form F is (Hodge) dual to a pseudo-vector Fij = εijk Bk ,

(9.13)

which we can identify with the magnetic field. In this case, Eq. (9.12) is exactly the Ginzburg-Landau (free) energy of a superconductor. Therefore, we can understand the physical meaning of each term by analogy: Consider a superconductor in an applied (external) magnetic field, then 10 of 11

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• The first term has two contributions: – |∂φ|2 is the energy cost of gradients in the superconducting condensate. – The gauge field terms in the covariant derivative represent the gauge-covariant coupling of the superconducting condensate to the applied magnetic field. In particular, δE = ie(φ∗ ∂φ − φ ∂φ∗ ) + 2e2 A|φ|2 , (9.14) δA(x) is the density of supercurrent, which couples to the magnetic field.
2 • U (φ) = m20 |φ|2 + λ |φ|2 /2 is the potential energy, where m20 ∝ (T − Tc ) determines the critical temperature, and minimising the potential leads to a finite superconducting condensate density |φ|2 when T < Tc . • F 2 /4 = B 2 /2 is the energy density of the applied magnetic field.

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Physics 582, Fall Semester 2021 Professor Eduardo Fradkin Problem Set No. 3: Canonical Quantization Due Date: Friday October 8, 2021, 9:00 pm US Central Time 1

Spin waves in a quantum Heisenberg antiferromagnet

In this problem you will consider the Heisenberg model of a one-dimensional quantum antiferromagnet. I first give you a brief summary on the Heisenberg model. You do not need to have any previous knowledge on quantum magnets (or the Heisenberg model) to do this problem. You will be able to solve this problem with the methods discussed in class. The one-dimensional Heisenberg model is defined on a linear chain ( a onedimensional lattice) with N sites. The lattice spacing will be taken to be equal to one. The quantum mechanical Hamiltonian for this system is N/2

ˆ =J H

X

Sˆk (j) · Sˆk (j + 1)

(1)

j=−N/2+1

where the exchange constant J > 0 ( i.e. an antiferromagnet) and the operators Sˆk (k = 1, 2, 3) are the three angular momentum operators in the spin-S representation (S is integer or half-integer) which satisfy the commutation relations [Sˆj , Sˆk ] = iǫjkl Sˆl

(2)

For simplicity we will assume periodic boundary conditions, Sˆk (j) ≡ Sˆk (j + N ). In the semi-classical limit, S → ∞, the operators act like real numbers since the commutators vanish. In this limit, the state with lowest energy has nearby spins which point in opposite (but arbitrary) directions in spin space. This is the classical N´eel state. In this state the spins on the even sites sub-lattice point up along some direction in space while the spins on the odd sites sublattice point down. At finite values of S, the spins can only have a definite projection along one axis but not along all three at the same time. Thus we should expect to see some zero-point motion precessional effect that will depress the net projection of the spin along any axis but, if the state is stable, even sites will have predominantly up spins while odd sites will have predominantly down spins. This observations motivate the following definition of a set of basis states for the full Hilbert space of this system. 1

The states |Ψi of the Hilbert space of this chain are spanned by Q the tensor product of the Hilbert spaces of each individual j th spin |Ψj i, |Ψi = j ⊗|Ψj i. The latter are simply the 2S + 1 degenerate multiplet of states with angular momentum S of the form {|S, M (j)i} (|M (j)| ≤ S) which satisfy ~ 2 (j)|S, M (j)i = S(S + 1)|S, M (j)i, S

S3 (j)|S, M (j)i = M (j)|S, M (j)i (3)

The states in this multiplet can be obtained from the highest weight state |S, Si by using the lowering operator Sˆ− = Sˆ1 − iSˆ2 . Its adjoint is the raising operator Sˆ+ (j) = Sˆ1 (j)+iSˆ2 (j). For reasons that will become clear below, it is convenient to define for j even ( even site) the spin-deviation operator n ˆ (j) ≡ S − Sˆ3 (j). For an odd site ( j odd) the spin deviation operator is n ˆ (j) ≡ S + Sˆ3 (j). For j even, the highest weight state |S, Si is an eigenstate of n ˆ (j) with eigenvalue zero while the state |S, −Si has eigenvalue 2S n ˆ (j)|S, Si =(S − Sˆ3 (j))|S, Si = 0 n ˆ (j)|S, −Si =(S − Sˆ3 (j))|S, −Si = 2S |S, −Si

(4)

whereas for j odd the state |S, −Si has zero eigenvalue while the state |S, Si has eigenvalue 2S. In terms of the operators n ˆ (j), the basis states are {|S, M (j)i} ≡ {|n(j)i}, ˆ ± where M (j) = S ∓n(j). For even sites, the raising and lowering operators S(j) act on the states of this basis as follows     12 n−1 + ˆ S |ni = 2S 1 − n |n − 1i 2S h  n i 12 Sˆ− |ni = 2S(n + 1) 1 − |n + 1i (5) 2S For odd sites the action of the above two operators is interchanged. The action of the operators Sˆ± is somewhat similar to that of annihilation and creation operators in harmonic oscillator states. For this reason we define a set of creation and annihilation operators as follows. Since we have two sublattices and the operators Sˆ± are different on each sub-lattice, it is useful to introduce two types of creation and annihilation operators: the operators a ˆ† (j) † and a ˆ(j) which act on even sites, and ˆb (j) and ˆb(j) which act on odd sites. They obey the commutation relations i   h a ˆ(j), a ˆ† (k) = ˆb(j), ˆb† (k) = δjk h i h i [ˆ a(j), a ˆ(k)] = ˆb(j), ˆb(k) = a ˆ(j), ˆb(k) = 0 (6) and similar equations for their hermitian conjugates. It is easy to check that the action of raising and lowering operators on the states {|ni} is the same as the action of the following operators on the same states:

2

1. On even sites:  1 √ n ˆ (j) 2 a ˆ(j), Sˆ+ (j) = 2S 1 − 2S Sˆ3 (j) =S − n ˆ (j),

Sˆ− (j) =

 1 √ n ˆ (j) 2 2Sˆ a† (j) 1 − 2S

n ˆ (j) = a ˆ† (j)ˆ a(j)

(7)

2. On odd sites:  1 √ n ˆ (j) 2 ˆ − ˆ S (j) = 2S 1 − b(j), 2S Sˆ3 (j) = − S + n ˆ (j),

 1 √ n ˆ (j) 2 † ˆ 2S b (j) 1 − 2S † ˆ ˆ n ˆ (j) = b (j)b(j) (8) Sˆ+ (j) =

Notice that although the integers n can now range from 0 to infinity, the Hilbert space is still finite since (for even sites) Sˆ− |n = 2Si = 0. Similarly, for odd sites, the state |n = 2Si is anihilated by the operator Sˆ+ . 1. Derive the quantum mechanical equations of motion obeyed by the the spin operators Sˆ± (j), Sˆ3 (j) in the Heisenberg representation, for both j even and j odd. Are these equations linear? Explain your result. 2. Verify that the definition for the operators S ± and S3 of equations (7) and (8) are consistent with those of equation (5). 3. Use the definitions given above to show that the Heisenberg Hamiltonian can be written in terms of two sets of creation and annihilation operators a ˆ† (j) and a ˆ(j) (which act on even sites), and ˆb† (j) and ˆb(j) which act on odd sites. 4. Find an approximate form for the Hamiltonian which is valid in the semiclassical limit S → ∞ ( or S1 → 0). Include terms which are of order S1 (relative to the leading order term). Show that the approximate Hamiltonian is quadratic in the operators a and b. 5. Make the approximations of part 4 on the equations of motion of part 1. Show that the equations of motion are now linear. Of what order in S1 are the terms that have been neglected? 6. Show that the Fourier transforms r 2 X iqj e a ˆ(j), a ˆ(q) = N j even

ˆb(q) =

r

2 X −iqj ˆ e b(j) N

(9)

j odd

followed by the canonical (Bogoliubov) transformation cˆ(q) = cosh(θ(q)) a ˆ(q) + sinh(θ(q)) ˆb† (q) ˆ = cosh(θ(q)) ˆb(q) + sinh(θ(q)) a d(q) ˆ† (q)

3

(10)

yields a diagonal Hamiltonian HSW of the form Z + π2 dq HSW = E0 + ω(q)(ˆ nc (q) + n ˆ d (q)) 2π −π 2

(11)

ˆ where n ˆ c (q) = cˆ† (q)ˆ c(q), n ˆ d (q) = dˆ† (q)d(q), provided that the angle θ(q) is chosen properly. Here we haveprescaled the Fourier transformed operators c(q) and d(q) by a factor of N/2 and taken the thermodynamic limit ˆ and their hermiN → ∞. Check that the rescaled operators cˆ(q) and d(q) † ′ tian conjugates obey the algebra [c(q), c (q )] = [d(q), d† (q ′ )] = 2πδ(q − q ′ ) (here δ(q) is the Dirac delta function, not a Kronecker delta!). Derive an explicit expression for the angle θ(q) and for the frequency ω(q). 7. Find the ground state for this system in this approximation (usually called the spin-wave approximation). 8. Find the single particle eigenstates within this approximation. Determine the quantum numbers of the excitations. Find their dispersion (or energymomentum) relations. Find a set of values of the momentum q for which the energy of the excited states goes to zero. Show that the energy of these states vanish linearly as the momentum approaches the special points and determine the spin-wave velocity vs at these points. Note: The approach we used here is the semi-classical (spin-wave) approximation. Eq.(7) and Eq.(8) are known as the Holstein-Primakoff identities.

2

Two-Component Complex Scalar Field

In this problem, you will consider the theory of a two-component complex scalar field φa (x) (a = 1, 2) whose Lagrangian is L=

1 m2 (∂µ φa (x))∗ (∂ µ φa (x)) − 0 φ∗a (x) φa (x) 2 2

(12)

In class we showed that this theory is invariant under the classical global symmetry φa (x) → φ′a (x) =Uab φb (x) −1 ∗ φ∗a (x) → φ′∗ a (x) =Uab φb (x)

L(φ′ ) =L(φ)

(13)

where U is a 2 × 2 unitary matrix, i.e. U −1 = U † . This is the symmetry group SU (2). Thus φ transforms like the fundamental (spinor) representation of SU (2). The matrices U can be expanded in the basis of 2 × 2 Pauli matrices (σk )ab and are parametrized by three Euler angles θk (k = 1, 2, 3): Uab = [exp (iθ · σ)]ab = cos(|θ|)δab + i 4

θ · σab sin(|θ|) |θ|

(14)

1. Use the classical canonical formalism to find: (a) the canonical momenta Πa , conjugate to the fields φa , (b) the Hamiltonian H, and (c) the total momentum Pj . 2. Derive the classical constants of motion associated with the global symmetry SU (2). Relate these constants of motion with the generators of infinitesimal SU (2) transformations. How many constants of motion do you find? Explain your results. 3. Quantize this theory by imposing canonical commutation relations. Write an expression for the quantum mechanical Hamiltonian and total momentum operators in therms of the field and canonical momentum operators. 4. Derive an expression for the quantum mechanical generators of global infinitesimal SU (2) transformations in the Hilbert space of states of the system. Explain what relation, if any, do they have withe the conserved charges of the classical theory. 5. Derive the quantum mechanical equations of motion of the Heisenberg representation operators. 6. Find an expansion of the field and canonical momentum operators in terms of a suitable set of creation and annihilation operators. How many species of creation and annihilation operators do you need?. Justify your results. 7. Find an expression for the SU (2) generators in terms of creation and annihilation operators. 8. Find the ground state of the system and its quantum numbers. Find the normal ordered Hamiltonian, total momentum and the SU (2) generators relative to this state. 9. Find the spectrum of single particle states. Give an expression for their energies and assign quantum numbers to these states. Do you find any degeneracies?. What is the degree of this degeneracy and why?

5

582 Homework 3 Solutions October 2021

Contents 1 Spin waves in a quantum Heisenberg antiferromagnet

2

2 Two component complex scalar field

13

A Foiling out Bogoliubov Transform

24

B Local SU(2) Transformation

25

1

1

Spin waves in a quantum Heisenberg antiferromagnet ˆ 1. First let’s recall the Heisenberg equation of motion for an operator A: dAˆ i ˆ ˆ = − [A, H] + ∂t Aˆ dt ~ In our case we have operators which do not depend explicitly on time; thus, the above equation can be written as: i~

dAˆ ˆ H] ˆ = [A, dt

The Hamiltonian for the Heisenberg antiferromagnet is: ˆ =J H

N/2 X

Sˆk (j) · Sˆk (j + 1)

j=−N/2+1

We are interested in the case that Aˆ is a spin-operator. Since Sˆk (j) belongs to the spin-S representation irrespective of the parity of j, we do not have to consider each sublattice individually. Furthermore, we can simplify our calculation if we rewrite our Hamiltonian using ladder operators: i Xh 1  ˆ+ ˆ− ˆ =J S (j)S (j + 1) + Sˆ− (j)Sˆ+ (j + 1) H Sˆ3 (j)Sˆ3 (j + 1) + 2 j It is easy to show for any pair of sites, j and j 0 : [Sˆ3 (j), Sˆ± (j 0 )] = ±~Sˆ± (j)δjj 0 [Sˆ± (j), Sˆ∓ (j 0 )] = ±2~Sˆ3 (j)δjj 0 Using these we can find our EoM! (a) This is not hard but organization is important in what follows1 :  dSˆ3 (j) J X ˆ + 0 ˆ− 0 − 0 ˆ+ 0 ˆ ˆ ˆ = [S3 (j), S (j )S (j + 1)] + [S3 (j), S (j )S (j + 1)] dt 2i~ j 0 J X ˆ = [S3 (j), Sˆ+ (j 0 )]Sˆ− (j 0 + 1) + Sˆ+ (j 0 )[Sˆ3 (j), Sˆ− (j 0 + 1)] 2i~ j 0  − 0 ˆ+ 0 − 0 ˆ + 0 ˆ ˆ ˆ ˆ + [S3 (j), S (j )]S (j + 1) + S (j )[S3 (j), S (j + 1)]  

J X ˆ+ ˆ− 0 − + 0 − 0 ˆ+ + 0 ˆ− ˆ ˆ ˆ ˆ = S (j)S (j + 1) − S (j)S (j + 1) δjj 0 + S (j )S (j) − S (j )S (j) δj,j 0 +1 2i j 0 

J  ˆ− = S (j + 1) + Sˆ− (j − 1) Sˆ+ (j) − Sˆ+ (j + 1) + Sˆ+ (j − 1) Sˆ− (j) 2i (1) Note: I used the fact spin operators on distinct lattice sites commute to get the EoM in the above form. 1

I drop commutators involving the same k’s in Sk (j) since these obviously commute.

2

(b) Proceeding in an analogous manner to (a):    i dSˆ± (j) J h ˆ± =± S (j + 1) + Sˆ± (j − 1) Sˆ3 (j) − 2 Sˆ3 (j + 1) + Sˆ3 (j − 1) Sˆ± (j) dt 2i

(2)

So we have a system of first order differential equations for operators which depend on time. Since our equations contain products of operators which also depend on time, we can not express these differential equation as a differential operator which acts linearly; that is, they are not linear. 2. I probably did this problem different than the way you decided to do it. Letting the provided operators speak for themselves is all you needed to do. Here we motivate the alternative form handed to you since it is fairly easy. We recall how the raising and lowering operators act on our 2S + 1 degenerate multiplet of states: S ± |S, M (j)i =

p

(s ∓ m)(s ± m + 1)|S, M (j) ± 1i

Here I use lower case s’s and m’s without indices in the radical to save on clutter. We will do this when convenient. In this problem we are interested in how our system deviates from perfect antiferromagentism, so we consider the following operator used to label our eigenstates in terms of its eigenvalues: n ˆ (j) = S + (−1)j−1 Sˆ3 (j) The staggered nature of the n ˆ (j)’s reflect the fact we are working with an antiferromagnet. We also have the corresponding quantum numbers: M (j) = S + (−1)j−1 n(j) Suppressing the j dependence on our eigenvalues, m = s + (−1)j−1 n, we will substitute this into our above equation. Starting with even j: S + |s, mi =

p

(s − m)(s + m + 1)|s, m + 1i

r

n − 1
2sn 1 − |s, m + 1i 2s r n − 1
+ ⇒ S |ni ≡ 2sn 1 − |n − 1i 2s p S − |s, mi = (s + m)(s − m + 1)|s, m − 1i r n
= 2s n + 1)(1 − )n |s, m − 1i 2s r n
⇒ S − |ni ≡ 2s n + 1)(1 − )n |n + 1i 2s =

(3)

(4)

Notice the lowering ladder operator increases n here and vice-versa for the raising operator. This makes sense since even sites where elected to be spin up and so the spin deviation for these states are increased when we decrease M . This is what lead me to label the eigenstates with n ± 1 in the final lines of (3) and (4) after the ladder operators act on the states.

3

Now for odd j: p (s − m)(s + m + 1)|s, m + 1i r n
= 2s n + 1)(1 − )n |s, m + 1i 2s r n
⇒ S + |ni ≡ 2s n + 1)(1 − )n |n + 1i 2s S + |s, mi =

(5)

Now the raising ladder operator sends n → n + 1 here. Similarly (and again counter to what we saw before): p S − |s, mi = (s + m)(s − m + 1)|s, m − 1i r n − 1
= 2sn 1 − |s, m − 1i (6) 2s r n − 1
S − |ni ≡ 2sn 1 − |n − 1i 2s The action of the ladder operators on our quantum numbers and eigenstates are reminiscent of the Harmonic oscillator, which inspires us to introduce the following creation and annihilation operators. First, for even sites2 : aˆ† |ni =

√

n + 1|n + 1i √ a ˆ|ni = n|n − 1i

and for odd sites,

√ bˆ† |ni = n + 1|n + 1i √ ˆb|ni = n|n − 1i

Clearly, [ˆ a, ˆb† ] = 0. All the remaining commutation relations we suspect from the harmonic oscillator are immediately satisfied from the above equations and operators on different sites commute. Using the above definitions we express our ladder operators in the following form (even sites): r

n − 1
2sn 1 − |n − 1i 2s r n ˆ
√ n|n − 1i = 2s 1 − 2s r n ˆ
= 2s 1 − a ˆ|ni 2s r n ˆ
+ ⇒ S = 2s 1 − a ˆ 2s

S + |ni =

(7)

This is the relation we are after! We can easily justify the second line above. Indeed, it follows if we can expand a function, f , as a ˆ with it’s corresponding Taylor series. It is easy to prove with induction for a Hermitian operator, A, eigenstates, |αi: Aˆk |αi = αk |αi. We use this to send n − 1 → n ˆ since the eigenstate is currently |n − 1i, and the square-root is analytic. We then use the fact n is a real number to permute it with 2

Reminder: these do have indices and we will reinstate them momentarily.

4

n ˆ , then we can use the provided properties for the annihilation operator. We can find S − in a similar fashion: r n
S − |ni = 2s(n + 1) 1 − |n + 1i 2s r n
† = 2s 1 − a ˆ |ni 2s r n
† =a ˆ 2s 1 − |ni 2s r n ˆ
=a ˆ† 2s 1 − |ni 2s r n ˆ
− † ⇒S =a ˆ 2s 1 − 2s

(8)

We can rinse and repeat the above procedure, but instead, we recall the odd sites act counter to the even, so we just swap the roles of S + ↔ S − . For completeness, for odd sites: r n ˆ
+ † S = ˆb 2s 1 − 2s r (9)
n ˆ ˆb S − = 2s 1 − 2s This illustrates we can rewrite our ladder operators with the spin excess operator in place of the run-of-the-mill spin operators! 3. We now rewrite the Hamiltonian in terms of these operators. First recall: ˆ =J H

N/2 X

h

Sˆ3 (j)Sˆ3 (j + 1) + Sˆ1 (j)Sˆ1 (j + 1) + Sˆ2 (j)Sˆ2 (j + 1)

i

h

i 1  ˆ+ ˆ− S (j)S (j + 1) + Sˆ− (j)Sˆ+ (j + 1) Sˆ3 (j)Sˆ3 (j + 1) + 2

j=−N/2+1

=J

N/2 X j=−N/2+1

Caution, these operators are site dependent; i.e. even or odd, so we need to be careful here! First, let’s note: " #" # Sˆ3 (j)Sˆ3 (j + 1) = (−1)j−1 (ˆ n(j) − S) (−1)(j+1)−1 (ˆ n(j + 1) − S) = −(ˆ n(j) − S)(ˆ n(j + 1) − S)
= −S 2 + S n ˆ (j) + n ˆ (j + 1) − n ˆ (j)ˆ n(j + 1) Thus, this term is independent of the parity of j. We can easily see: J

N/2 X j=−N/2+1

N/2 X

Sˆ3 (j)Sˆ3 (j + 1) = −N JS 2 + J

j=−N/2+1

5

!


S n ˆ (j) + n ˆ (j + 1) − n ˆ (j)ˆ n(j + 1)

The ladder operators are where we should be careful, so let’s start by considering even sites alone. q (j) First, we define 1 − nˆ2S = fˆ(j) to reduce clutter. This results in: J

 X 1 Sˆ+ (j)Sˆ− (j + 1) + Sˆ− (j)Sˆ+ (j + 1) 2 j even  X  † † ˆ ˆ ˆ ˆ ˆ ˆ = SJ f (j)f (j + 1)ˆ a(j)b(j + 1) + a ˆ (j)b (j + 1)f (j)f (j + 1) j even

All that is different when summing the odd sites is the j + 1 terms correspond to a ˆ and the j terms 3 ˆ to b. Performing the substitutions :  X
ˆ = −JN S 2 + J H S n ˆ (j) + n ˆ (j + 1) − n ˆ (j)ˆ n(j + 1) j

+ SJ

X 

fˆ(j)fˆ(j + 1)ˆ a(j)ˆb(j + 1) + a ˆ† (j)ˆb† (j + 1)fˆ(j)fˆ(j + 1)

 (10)

j even

+ SJ

X

fˆ(j)fˆ(j + 1)ˆ a(j + 1)ˆb(j) + a ˆ† (j + 1)ˆb† (j)fˆ(j)fˆ(j + 1)



j odd

4. Now that we have an expression for our Hamiltonian in terms of our creation and annihilation operators we take the limit S → ∞. Looking at our above sum the leading order term is clearly S 2 , so let’s consider terms linear in S and neglect those of lower order (i.e. S 0 , S −1 , . . . ). Following this line of reasoning the summation resulting from the Sˆ3 ’s is trivial: drop n ˆ (j)ˆ n(j + 1). Notice the order of our creation and annihilation operators has been reduced for this part of the Hamiltonian! The portion containing the ladder operators takes a bit more work. S shows up as an overall multiplicative factor and also inside our function fˆ(j). Thus, we are interested in: r n ˆ (j) n ˆ (j + 1) S fˆ(j)fˆ(j + 1) = S 1 − 1− 2S 2S h ih
n ˆ (j) 1 n ˆ (j + 1) 1
i =S 1− +O 2 1− +O 2 4S S 4S S h i 1
1 ˆ (j) + n ˆ (j + 1) + O =S− n 4 S r

So, within our approximation we simply replace these products with S! Thus, within the so called spin-wave approximation we find: i i Xh X h 2 † † ˆ ˆ ˆ H = −JN S + SJ n ˆ (j) + n ˆ (j + 1) + SJ a ˆ(j)b(j + 1) + a ˆ (j)b (j + 1) j

j even

+ SJ

i   (11) Xh a ˆ(j + 1)ˆb(j) + a ˆ† (j + 1)ˆb† (j) + O S 0

j odd

This is clearly quadratic in our creation and annihilation operators. 3

Note: n ˆ (j) = a ˆ† (j)ˆ a(j) for j even and n ˆ (j) = ˆb† (j)ˆb(j) for j odd.

6

5. Let’s formally apply the S → ∞ limit to our equations of motion. Since we have rewritten our Hamiltonian in terms of our creation and annihilation operators, we will be looking at the EoM for said operators. Thus, we are about to see that the dynamics of these operators are also simplified in the spin-wave approximation since they are linear. We just calculate the following commutator: " i Xh † † † † ˆ ˆ [ˆ a (`), H] =SJ a ˆ (j)[ˆ a (`), a ˆ(j)] + [ˆ a (`), a ˆ(j)]b(j + 1) jeven

# i   Xh + a ˆ† (j + 1)[ˆ a† (`), a ˆ(j + 1)] + [ˆ a† (`), a ˆ(j + 1)]ˆb(j) + O S 0 jodd

"

#

  ˆ ˆ = −SJ 2ˆ a (`) + b(k + 1) + b(` − 1) + O S 0 †

where ` is an arbitrary lattice site. From this result we can see: " #   dˆ a† (`) = SJ 2ˆ a† (`) + ˆb(` + 1) + ˆb(` − 1) + O S 0 i~ dt

(12)

We
are retaining the same approximation as 1.3, so we are dropping terms of order 0 S −1 . To leading order in S we have a linear equation of motion. The time evolution of the annihilation operator can be found from the Hermitian conjugate of this above equation: " #   dˆ a(`) † † ˆ ˆ = SJ 2ˆ a(`) + b (` + 1) + b (k − 1) + O S 0 (13) −i~ dt We can rinse and repeat for ˆb(`), but this is the same as ˆb† (l) → a ˆ† (`) and a ˆ(`) → ˆb(`) in the above expressions. Thus, all the EoM for the relevant operators are linear (up to order S). 6. Let’s now rewrite our creation and annihilation operators using the following Fourier transforms: X X ˆb(q) = eiqj ˆb(j) (14) a ˆ(q) = e−iqj a ˆ(j), j even

j odd

Notice that I use a different normalization than the one used in the problem set. To be more specific, we write jeven = 2r and jodd = 2r + 1, where the integer r takes the values r = − N4 + 1, . . . , N4 , where we assume that there is an even number of even sites (and the same for the odd sites), and hence that N is a multiple of 4. Periodic boundary conditions require that a(2r) = a(2r + N ),

and

b(2r + 1) = b(2r + 1 + N )

which imply that eiqN = 1 Hence, the momentum q is restricted to the values qm =

2π m N

where m = − N4 + 1, . . . , N4 . Hence, the allowed momenta qm take the values (−

N 2π N 2π + 1) ≤ qm ≤ 4 N 4 N 7

Or, what is the same π π 2π + ≤ qm ≤ 2 N 2 2π The allowed values of the momenta have a spacing ∆q = N . We are interested in chains with periodic boundary conditions in the thermodynamic limit N → ∞. In this limit the spacing of the momenta ∆q → 0 and the allowed values of the momenta q fill densely the real interval − π2 < q ≤ π2 , known as the first Brillouin zone. −

The inverse of the Fourier transforms of Eq.(14) are 2 a(2r) = N

N/4 X

iqm 2r

a(qm )e

2 b(2r + 1) = N

,

m=−N/4+1

N/4 X

b(qm )e−iqm (2r+1)

(15)

m=−N/4+1

We now recognize that the sums shown in Eq.(15) have the form of Riemann sums which in the thermodynamic limit N → ∞ converge to the integrals Z

π/2

a(2r) = −π/2

Z

dq iq2r e a(q), π

π/2

b(2r + 1) = −π/2

dq −iq(2r+1) e b(q) π

(16)

where we used that the spacing is ∆q = 2π/N . We will now check the commutation relations of the operators a(q) with its Hermitian conjugate: [a(q), a† (q 0 )] =

N/4 X

0

[a(2r), a† (2r0 )]eiq2r−iq 2r

0

(17)

r,r0 =−N/4+1

=

N/4 X

0

e−i(q−q )2r

(18)

r=−N/4+1

≡πδP (q − q 0 )

(19)

and similarly we obtain [b(q), b† (q 0 ) = πδP (q − q 0 ) and all other commutators vanish. Here we introduced the periodic delta-function δP (q) ≡

∞ X

δ(q + πm)

m=−∞

In particular, we find the delta function at zero momentum is δP (0) =

N 2π

We will now proceed to express the operators that appear in the effective Hamiltonian of Eq.(11) in terms of the operators a(k) and b(k) (and of their Hermitian conjugates). We will work directly in

8

the thermodynamic limit. We begin with the sum of the local densities n(j) on the even sublattice: N/4 X

Z

π/2

n(2r) = −π/2

r=−N/4+1

Z

π/2

= −π/2 π/2

Z =

−π/2 π/2

Z =

−π/2

dq π

Z

dq π

Z

π/2

−π/2 π/2

−π/2

dk π

N/4 X

ei2r(q−k) a ˆ† (q)a(k)

r=−N/4+1

dk πδP (q − k)a† (q)a(k) π

dq † a (q)a(q) π dq na (q) π

A similar result works for j odd: N/4 X

Z

π/2

n(2r + 1) = −π/2

r=−N/4+1

dq † b (q)b(q) = π

Z

π/2

−π/2

dq nb (q) π

Similarly, we obtain N/4 X

2

Z h i n(2r) + n(2r + 1) = 2

π/2

−π/2

r=−N/4+1

i dq h † † 2 a (q)a(q) + b (q)b(q) π

These terms in Eq.(11) conserve the number of excitations, i.e. have as many creation ans annihilation operators. We will next perform the Fourier transform of the off-diagonal terms in Eq.(11). However, these terms do not conserve the number of excitations. They are N/4 X

+ SJ



 a(2r)b(2r + 1) + a† (2r)b† (2r + 1) + a(2r)b(2r − 1) + a† (2r)b† (2r − 1)

r+−N/4+1 N/4 X

= SJ

Z

r=−N/4+1

Z

π/2

= SJ −π/2

π/2

dq π

−π/2

Z

π/2

−π/2

  dk † † πδP (q − k)2 cos q a(q)b(q) + a (q)b (q) π

  dq † † 2 cos q a(q)b(q) + a (q)b (q) π

Combining these expressions we find 2

π/2

Z

H = −JN S + 2JS −π/2 π/2

= −JN S 2 − 2JS

Z

−π/2

Z

i dq h † a (q)a(q) + b† (q)b(q) + cos qa(q)b(q) + cos qa† (q)b† (q) π dq πδP (0) π

π/2

i dq h † † † † + 2SJ a (q)a(q) + b(q)b (q) + cos(q)a(q)b(q) + 2 cos(q)a (q)b (q) −π/2 π      Z π/2  1 1 dq  † cos q a(q) 2 a ˆ (q) b(q) = −JN S 1 + + 2JS cos q 1 b† (q) S −π/2 π 9

(20)

where in the second line we reordered the operators of a term in the first line and where we used that δP (0) = N/(2π). We can diagonalize the Hamiltonian by introducing the following Bogoliubov transformation4 :

      c(q) cosh θ(q)
sinh θ(q)
a(q) = d† (q) b† (q) sinh θ(q) cosh θ(q) and its inverse transformation

      a(q) cosh θ(q)
− sinh θ(q)
c(q) = b† (q) d† (q) − sinh θ(q) cosh θ(q) This transformation preserves our commutation relations 5 ! These new operators are interpreted as quasi-particles. We restate that the quasi particle operators, c(q) and d(q), satisfy the same algebra as a(q) and b(q). Now we are interested in the following multiplication:

 

   cosh θ(q)
−sinh θ(q)
1 cos(q) cosh θ(q)
−sinh θ(q)
cos(q) 1 −sinh θ(q) −sinh θ(q) cosh θ(q) cosh θ(q)



  cosh 2θ(q)
− sinh 2θ(q) cos(q)
cosh 2θ(q)
cos(q) − sinh
2θ(q) = cosh 2θ(q) cos(q) − sinh 2θ(q) cosh 2θ(q) − sinh 2θ(q) cos(q) To diagonalize the Hamiltonian we choose: 1 θ(q) = tanh−1 (cos(q)) 2 Notice this means:



sech2 2θ(q) = 1 − tanh2 2θ(q) = 1 − cos2 (q) = sin2 (q) and,


sinh 2θ(q) = cosh 2θ(q) tanh 2θ(q)
= cosh 2θ(q) cos(q) The diagonal terms are now a breeze to find:



cosh 2θ(q) − sinh 2θ(q) cos(q) =


 1
1 − cos2 q sech 2θ(q)

= |sin(q)| 4

See Appendix A if you prefer foiling out the expressions over matrix multiplication. The latter method is much easier, but the former is here in case you did/tried this problem that way and want to check it. 5 This is a direct consequence of the identity: cosh2 (x) − sinh2 (x) = 1.

10

In terms of our quasi-particles our Hamiltonian becomes:      Z π/2  dq  † 1 1 cos(q) a(q) 2 ˆ a (q) b(q) + 2SJ H = −JN S 1 + cos(q) 1 b† (q) S −π/2 π     Z π/2  †  c(q) 1 dq 2 = −JN S 1 + + 2SJ | sin(q)| c (q) d(q) d† (q) S −π/2 π   Z π/2   1 dq 2 = −JN S 1 + + 2SJ | sin(q)| c† (q)c(q) + d(q)d† (q) S −π/2 π   Z π/2 Z π/2   1 dq dq 2 = −JN S 1 + + 2SJ | sin(q)|πδP (0) + SJ | sin(q)| c† (q)c(q) + d† (q)d(q) S −π/2 π −π/2 π Thus, in this limit the Hamiltonian takes the simpler form π/2

Z H = E0 + 2SJ

−π/2

  dq ˆ | sin(q)| cˆ† (q)ˆ c(q) + dˆ† (q)d(q) π

(21)

Where: 2

Z

π/2

E0 = −JN S + JN S + 2SJ −π/2

   dq 1 2 2 |sin(q)|πδP (0) = −JN S 1 + 1− π S π

The second term is the leading quantum correction (to order 1/S) to the classical ground state energy. From this we can see that the Hamiltonian is quadratic and diagonal in the creation and annihilation operators for our quasi particles. 7. We can define the ground state, |0i, to be the state which satisfies: n ˆ c (q)|0i = n ˆ d (q)|0i = 0. Thus, the energy of the ground state is:    2 1 2 E0 = −JN S 1 + 1− S π 8. There are two types of a single particle states: |nc (k) = 1i = c† (k)|0i,

and

|nd (k) = 1i = d† (k)|0i

Let’s find the energy of these states using our above Hamiltonian. We just need to note:   nc (q)c† (k)|0i = c† (q)c(q)c† (k)|0i = c† (q) πδP (k − q) + c† (k)c(q) |0i = πc† (q)δP (k − q)|0i

so after subtracting off the ground state energy and integrating, the excitation energies of our single particle states are of the form: E(k) = 2JS|sin(k)| Note that: k → 0 ⇒ E(k) → 0 (restricting ourselves to the first Brillouin zone, i.e. neglect k = nπ with n 6= 0), so in the long wave length limit, k → 0, both types of single particle states have energies equal to the ground state: they are gapless!

11

Furthermore, we can expand our energy in the immediate neighborhood of k = 0 to obtain: E(k) ≈ 2JS|k| Thus, we see the energy vanishes linearly with the momentum k! Finally, the velocity of these excitations as k → 0 is given by: dE(k) vs = = 2JS d|k| k=0

This is the spin wave velocity.

12

2

Two component complex scalar field

Consider the Lagrangian:

 ∗ L = ∂µ φa ∂ µ φa − V (|φa |2 )

Where: V (|φa |2 ) = m2 φa φ∗a 1. Let’s begin by finding: (a) The canonical momentum Πa conjugate to φa . This is simple enough: Πa =

δL = ∂ 0 φ∗a δ∂0 φa

Since the field is complex, we can not forget about: Π∗a =

δL = ∂ 0 φa δ∂0 φ∗a

As a quick note, we see that we neglect any sort of index which indicates if we are talking about a co- or contra-variant canonical momentum. We just make the correct choice where it matters, and forgo any additional labels to avoid too much clutter. (b) With σ representing the extra degree of freedom we get from the complex fields we see the Hamiltonian of this system is: X X Π∗a Πa − L (∂0 φσa )Πσa − L = 2 H= a

a,σ

From here on out I drop all summation symbols with the understanding we are summing over the components of the field. With this in mind we see:   ∗ ∗ H = Πa Πa + ∇φa · ∇φa + V (|φ|2 ) So then:

Z H=

d3 xH

(c) Now for the total linear momentum the momentum density is given by: P k = Πa ∂ k φa + h.c. We could arrive here from the energy-momentum tensor as we did in the last homework. The Hermitian conjugate is needed to keep things real. From this we have: Z k P = d3 x P k 2. We begin by considering a two component complex scalar field:   φ Φ= 1 φ2 clearly: ∀ U ∈ SU (2), |U Φ|2 = |Φ|2 13

Furthermore, in the case of a global symmetry: L(U Φ) = L(Φ) (see Appendix B for the case of a local symmetry). We know from Noether’s theorem that a symmetry implies the existence of conserved currents. To find these currents we express the elements of our group in the following form6 :

 k k  [U (x)]ab = eiλ θ (x) ab ≈ δab + iλkab θk (x)   k k [U † (x)]ab = e−iλ θ (x) ab ≈ δab − iλkab θk (x) Of course, we’re working with small θk (x)’s. Since we are considering a 2-component complex scalar field, we need the following two transformation rules:   U (x)Φ a = [U (x)]ab φb   (U (x)Φ)† a = φ∗b [U † (x)]ba Taking our representations of the spin-1/2 doublet representation of SU (2) to be the Pauli matrices, k , the explicit transformation for our complex scalar field is given by: i.e. λkab = σab
k δφa = [U (x)]ab − δab φb ≈ iσab φb θk (x) Note, we are summing over b. Also, for the complex conjugate:
k k δφ∗a = φ∗b [U † (x)]ba − δab ≈ −iφ∗b σba θ (x) We are interested in a global symmetry; thus, our θk ’s are constant. Applying this variation to our Lagrangian, exploiting the Euler-Lagrange equation, and finally taking the necessary functional derivatives we find7 : δL δL δL = δφa + δ∂ν φa + φa ↔ φ∗a δφa δ∂ν φa h δL δL i δφa + δφ∗a = ∂ν δ∂ φ δ∂ν φ∗a (22) h ν a
∗
i k k ν ∗ k ν = i∂ν σab ∂ φa φb − φb σba ∂ φa θ h i

ν T ∗ k T ∗ k ν = i∂ν ∂ Φ σ Φ − Φ σ ∂ Φ θk So we have 3 conserved 4-currents (one for each Pauli Matrix) indexed with k. Because we want to (eventually) show our corresponding conserved charges are the generators of SU (2), we introduce an additional minus sign, and define/rewrite as: jνk

≡ −i




k † k ∂ν Φ σ Φ − Φ σ ∂ν Φ †

This is clearly conserved by the above relation. 6

The specific representation is not yet determined, but they are Hermitian. Notice the order of products in the second term of the second line. This can behseen in a few ways. For one, the h.c. of i δL φ lives in the dual space, so it must left multiply. You can also write this line as: ∂ν δ∂ν φa δφa + h.c. . Finally, just plugging in φ + δφ into our action achieves the same ordering of terms. 7

14

We can express our constants of motion in terms of these conserved currents in the usual way: Z Z  
k 3 k Q = d xj0 = −i d3 x ∂0 Φ† σ k Φ − Φ† σ k ∂0 Φ Z   3 T k † k ∗ = −i d x Π σ Φ − Φ σ Π Z   k = −iσab d3 x Πa φb − φ∗a Π∗b This means there are 3 classical constants of motion associated with global SU(2) transformations. 3. Now let’s go ahead and Quantize our theory. All we need to do is promote our fields to operators, and then impose equal time commutation relations. The result of which is: ˆ = H

Z

  ˆ 2) ˆ aΠ ˆ ∗ + ∇φˆa · ∇φˆa ∗ + V (|φ| d3 x Π a

ˆ b (~x0 , t)] = [φˆ∗a (~x, t), Π ˆ ∗b (~x0 , t)] = iδab δ 3 (~x − ~x0 ) [φˆa (~x, t), Π All other commutators are zero. We can also see that the total momentum operator becomes: Z k ˆ a ∂ k φˆa + h.c. Pˆ = d3 xΠ 4. The quantum mechanical generators of global infinitesimal SU (2) symmetry are related to the classical conserved quantities above, Qk , in a very intimate way; indeed, these are just the off-springs of our classical conserved charge. We promote the conserved charge to an operator by using the quantized field operators; thus: Z Z  

k 3 0k ˆ = d xˆj = −i d3 x ∂ 0 Φ ˆ T ∗σk Φ ˆ T ∗σk ∂ 0Φ ˆ− Φ ˆ Q Z  
ˆ T σk Φ ˆ− Φ ˆ T ∗σk Π ˆ∗ = −i d3 x Π (23) Z   k 3 ∗ ˆ∗ ˆ ˆ ˆ = −iσab d x Πa φb − φa Πb These guys are indeed the generators we are after which is evident from their comˆ i, Q ˆ j ] = 2iεijk Q ˆ k . These are clearly generators of our SU (2) mutation relations: [Q symmetry since we have the same structure constant as the Pauli matrices! To work out these commutation relations we note at equal times:     0 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ Πa φb (~x), Πc φd (~x ) = i Πa φd (~x)δbc − Πc φb (~x)δad δ(~x − ~x0 ) Which implies:     i ˆ ˆ a σab ˆ c σ j φˆd (~x0 ) = −i Π ˆ a φˆd (~x)σ i σ j − Π ˆ c φˆb (~x)σ i σ j δ(~x − ~x0 ) − Π φb (~x), Π ab bd ab ca cd   i j  j i  ˆ ˆ ˆ ˆ = −i Πa φd (~x) σ σ ad − Πc φb (~x) σ σ bc δ(~x − ~x0 ) h i ˆ a σi, σj = −iΠ φˆd (~x)δ(~x − ~x0 ) ad   ijk k ˆ ˆ = 2i − iΠa σad φd (~x) δ(~x − ~x0 ) 15

We can finally show our desired result: Z h i  i j k ˆ∗ k ˆ ˆ ,Q ˆ = d3 xd3 x0 2iijk i φˆ∗a σad ˆ a σad Q φd (~x) δ(~x − ~x0 ) Πa (~x) − Π ˆk = 2iijk Q Again, these conserved charges satisfy the same algebra as the Pauli matrices; therefore, they are generators of SU (2) transformations! 5. Let’s now use the Heisenberg equation of motion to find dynamics of our field operators: (a) 1 ˆ x0 , t)] ∂0 φˆa (~x, t) = [φˆa (~x, t), H(~ i Clearly φˆa only doesn’t commute with the momentum operator; thus, we just need to compute: ˆ b (~x0 , t)Π ˆ ∗ (~x0 , t)] = [φˆa (~x, t), Π ˆ b (~x0 , t)]Π ˆ ∗ (~x0 , t) = iδab δ 3 (~x − ~x0 )Π ˆ ∗ (~x0 , t) [φˆa (~x, t), Π b b b It follows: ˆ ∗ (~x, t) ∂0 φˆa (~x, t) = Π a As it should be! (b) The equation of motion for the complex conjugate field follows the exact same procedure, and is just the complex conjugate of the above equation: ˆ a (~x, t) ∂0 φˆ∗a (~x, t) = Π (c) ˆ a (~x, t), H(~ ˆ x0 , t)] ˆ a (~x, t) = 1 [Π ∂0 Π i The commutator with the potential is the exact same procedure as (a) and (b) except the canonical commutation relation is reversed, so there will be an overall minus sign. Because of this, I will not do it explicitly. The following commutator requires an integration by parts8 : Z ˆ a (~x, t), ∇0 φˆb (x~0 , t)∇0 φˆ∗ (x~0 , t)] d3 x0 [Π b Z ˆ a (~x, t), ∇0 φˆb (x~0 , t)]∇0 φˆ∗b (x~0 , t) = d 3 x0 [ Π (24) Z 0 ˆ a (~x, t), φˆb (x~0 , t)]∇ 2 φˆ∗ (x~0 , t) = − d 3 x0 [ Π b =i∇2 φˆ∗a (~x, t) The last line comes from commutation relations, and then using both of the delta functions. As a result, we find: ˆ a (~x, t) = ∇2 φˆ∗a (~x, t) − m2 φˆ∗a ∂0 Π 8

I throw out all boundary terms immediately because our field operators evaluate to zero there.

16

(d) Now we can rinse and repeat... Or take a complex conjugate of the above expression; I opt for the latter: ˆ ∗ (~x, t), H(~ ˆ x0 , t)] ˆ ∗ (~x, t) = 1 [Π ∂0 Π a i a = ∇2 φˆa (~x, t) − m2 φˆa (~x, t) Finally, we can use our two equations for each of the fields above (four in total) and write:
∂ 2 + m2 φˆa (x) = 0
∂ 2 + m2 φˆ∗ (x) = 0

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a

6. We’re now in a position to introduce a set of creation and annihilation operators, which will show our analogy between fields and harmonic oscillators still holds in the complex case. Since we have already quantized the real scalar field, we quickly note some similarities and some differences. We first consider the case of a single component complex scalar field, then we generalize our result to multiple components. Even with a single component however we will see a complex scalar field requires two sets of creation and annihilation operators opposed to one as was the case for a real scalar field ; this has to do with the fact we ”secretly” have a set of two quantum fields opposed to just one. Following Peskin and Schroeder we expand our field operator out in terms of its Fourier modes, treating each mode as an independent oscillator, which is the same idea we had for the real scalar field: ˆ x, x0 ) = φ(~

Z

i d3 k h ˆ ~ i~k·~ x ˆ− (~k, x0 )e−i~k·~x φ ( k, x )e + φ + 0 (2π)3

From our equations of motion in 2.5 (eq. (21)) we know that each part must satisfy the Klein-Gordon equation: ∂02 φˆ± (~k, x0 ) + (k 2 + m2 )φˆ± (~k, x0 ) = 0 We deduce the following time dependence: ~ φˆ± (~k, x0 ) = φˆ± (~k)e∓iω(k)x0

√ where ω(~k) = k 2 + m2 . There are 4 combinations total, but we only keep the solutions which are Lorentz invariant; that is, exponents which are multiples of: ωx0 − ~k · ~x = k · x. Thus: Z d3 k h ˆ ~ −ik·x ˆ ~ ik·x i ˆ φ(x) = φ+ (k)e + φ− (k)e (2π)3 ˆ+ (~ ˆ†− (−~ We now recall that for the case of a real scalar field we had the conditions: φ k) = φ k) † ˆ− (~ ˆ+ (−~ and φ k) = φ k). These conditions are no longer necessary because our complex scalar field is not necessarily Hermitian. This reaffirms the prediction that the complex field will have a set of two creation and annihilation operators! Furthermore, we can again see the necessity of a set of two operators after considering the mode expansion of the complex conjugate of the above field: Z i d3 k h ˆ∗ ~ ∗ i~k·~ x ∗ ~ −i~k·~ x ˆ ˆ φ (~x, x0 ) = φ ( k, x )e + φ ( k, x )e 0 0 − (2π)3 + 17

and so, ∂02 φˆ∗± (~k, x0 ) + (k 2 + m2 )φˆ∗± (~k, x0 ) = 0 which leads us to:

Z

ˆ∗

φ (x) =

d3 k h ˆ∗ ~ −ik·x ˆ∗ ~ ik·x i φ (k)e + φ− (k)e (2π)3 +

We are now in a position to define our creation and annihilation operators; our constraint this time is our field operators must be complex conjugates of one another . Exploiting orthogonality to equate the coefficients of our exponentials we see9 : φˆ† = φˆ∗ → φˆ∗+ = φˆ†− , φˆ∗− = φˆ†+ This motivates: a ˆ(~k) = 2ω(~k)φˆ+ and, ˆb† (~k) = 2ω(~k)φˆ− We can show that if these guys satisfy the following commutation relations: [ˆ a(~k), a ˆ† (~k 0 )] = (2π)3 ω(~k)δ 3 (~k − ~k 0 ) [ˆb(~k), ˆb† (~k 0 )] = (2π)3 ω(~k)δ 3 (~k − ~k 0 ) with all others zero, then our canonical commutation relations are preserved. With these the field operators become: d3 k

ˆ φ(x) =

Z

ˆ∗

Z



(2π)3 2ω(~k)

a ˆ(~k)e−ik·x + ˆb† (~k)eik·x



and, d3 k

φ (x) =

(2π)3 2ω(~k)

  ˆb(~k)e−ik·x + a ˆ† (~k)eik·x

The generalization to a 2-component complex field, let alone a n-component one, is trivial: just introduce subscripts and impose [ˆ am , a ˆ†n ] etc. are all proportional to δmn (where appropriate). Then we find: φˆm (x) =

Z

φˆ∗m (x)

Z =

d3 k

  −ik·x † ~ ik·x ~ ˆ a ˆm (k)e + bm (k)e

(2π)3 2ω(~k)   d3 k † ~ ik·x ˆbm (~k)e−ik·x + a ˆm (k)e (2π)3 2ω(~k)

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From these relations we can easily calculate our canonical momentum operators in terms of these creation and annihilation operators: d3 k  † ~ ik·x ˆ ~ −ik·x  a ˆ (k)e − bm (k)e 2(2π)3 m Z  d3 k  ∗ −ik·x † ~ ik·x ~ ˆ ˆ (x) = −i Π a ˆ ( k)e − b ( k)e m m m 2(2π)3 ˆ m (x) = i Π

Z

9

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There is no real difference between ∗ and † here (for single component complex scalar fields). They’re more like additional labels I use to relate the Fourier modes. If we didn’t use a †, the equations below would be something like: φˆ∗+ = φˆ∗− otherwise.

18

ˆ Pˆ and Q ˆ k in terms of our creation and annihilation opera7. Now let’s go ahead and express H, ˆk: Π ˆ m φˆn (x) and ˆ k . We calculate two types of terms from the definition of Q tors. We’ll start with Q ˆ ∗n (x). We’ll do this in steps: −φˆ∗m Π Z Z    d3 k d3 k 0 ∗ ˆ∗ −ik·x † ~ ik·x † ~ 0 ik0 ·x ˆ ~ ˆ ˆbm (~k 0 )e−ik0 ·x + a −φm Πn (x) = i ( k)e ( k )e a ˆ ( k)e − b ˆ n n m 2(2π)3 (2π)3 2ω(~k 0 ) Let’s now foil everything out:   0 0 0 0 i ˆbm a ˆn (~k)(~k 0 )e−i(k+k) ·x − ˆbm (~k 0 )ˆb†n (~k)ei(k−k) ·x + a ˆ†m (~k 0 )ˆ an (~k)e−i(k+k )·x − a ˆ†m (~k 0 )ˆb†n (~k)ei(k+k )·x   ~ ~ an (~k) − a ˆ†m (~k)ˆb†n (−~k)e2iω(k)xo ˆ†m (~k)ˆ → i ˆbm (~k)ˆ an (−~k)e−2iω(k)xo − ˆbm (~k)ˆb†n (~k) + a This last simplification results from the identity: Z ~ ~0 d3 xe±i(k±k )·~x = (2π)3 δ 3 (~k ± ~k 0 ) and the fact that we’re integrating over spatial coordinates (see Eq. 23). The arrow represents a spatial integration as well us the collapse of the ~k 0 integral using the resulting delta function. In a similar fashion: " # Z Z    3 3 0 d k d k 0 0 ˆ m φˆn (x) = i Π a ˆ†m (~k)eik·x − ˆbm (~k)e−ik·x a ˆn (~k 0 )e−ik ·x + ˆb†n (~k 0 )eik ·x 2(2π)3 (2π)3 2ω(~k 0 ) Expanding:   † ~ 0 i(k−k0 )·x 0 −i(k+k0 )·x † ~ ˆ† ~ 0 −i(k+k0 )·x † ~ 0 −i(k−k0 )·x ~ ˆ ~ ~ ˆ ~ ˆ i a ˆm (k)ˆ an (k )e − bm (k)ˆ an (k )e +a ˆm (k)bn (k )e − bm (k)bn (k )e   ~ ~ →i a ˆ†m (~k)ˆ an (~k) − ˆbm (~k)ˆ an (−~k)e−2iω(k)xo + a ˆ†m (~k)ˆb†n (−~k)e2iω(k)xo − ˆbm (~k)ˆb†n (~k) Since there is a factor of i multiplying these above two results, adding these two expressions together is a cynch: Z Z     d3 k k k 3 ∗ ˆ∗ k ˆ ˆ ˆ ˆ a ˆ†m (~k)ˆ an (~k) − ˆbm (~k)ˆb†n (~k) Q = −iσmn d x Πm φn − φm Πn = σmn (2π)3 2ω(~k) This is distinct, but reminiscent of, our conserved charge in the case of a U (1) symmetry, which described particle number conservation. We can actually see particle conservation if we notice that there is another linearly independent matrix past our Pauli matrices: the identity matrix! Notice the cancellation of the time dependent factors, so this conserved charge is indeed time independent! That is, it’s a constant of motion. 8. Let’s now express the Hamiltonian in terms of our creation and annihilation operators. For simplicity I’ll work out the case of a single component complex scalar field as we did in 2.6. We begin by noting we’ll need the following sum of terms:

ˆΠ ˆ ∗ (x) + ∇φˆ · ∇φˆ∗ (x) = Π

Z Z

  d3 k d3 k 0 0 0 ~ ~ ~ ~ ω(k)ω(k ) + k · k 2ω(~k)(2π)3 2ω(~k 0 )(2π)3    −ik·x † ~ ik·x † ~0 ik0 ·x −ik0 ·x 0 ~ ˆ ˆ ~ × a ˆ(k)e − b (k)e a ˆ (k )e − b(k )e 19

ˆ ˆ We find ∇φ(x) in the same way we found Π(x). The parentheses foil out to: ~ ~0 ~ ~0 ~ ~0 ~ ~0 a† (k~0 )e−i(k+k )·~x ei(ω(k)+ω(k ))x0 a ˆ(~k)ˆ a† (k~0 )ei(k−k )·~x e−i(ω(k)−ω(k ))x0 − ˆb† (~k)ˆ

! −a ˆ(~k)ˆb(k~0 )e

i(~k+~k0 )·~ x −i(ω(~k)+ω(~k0 ))x0

e

ˆ†

~0

+ b (~k)ˆb(k )e

−i(~k−~k0 )·~ x i(ω(~k)−ω(~k0 ))x0

e

Let’s just put a pin in this guy for now, and move on to the next expansion: Z Z    d3 k d3 k 0 0 0 ∗ ˆ ˆ φφ (x) = ˆ† (k~0 )eik ·x + ˆb(k~0 )e−ik ·x a ˆ(~k)e−ik·x + ˆb† (~k)eik·x a 2ω(~k)(2π)3 2ω(~k 0 )(2π)3 We foil out the terms to find: ~ ~0 ~ ~0 ~ ~0 ~ ~0 a† (k~0 )e−i(k+k )·~x ei(ω(k)+ω(k ))x0 a ˆ(~k)ˆ a† (k~0 )ei(k−k )·~x e−i(ω(k)−ω(k ))x0 + ˆb† (~k)ˆ

! ~ ~0

~

~0

~ ~0

~

~0

+a ˆ(~k)ˆb(k~0 )ei(k+k )·~x e−i(ω(k)+ω(k ))x0 + ˆb† (~k)ˆb(~k 0 )e−i(k−k )·~x ei(ω(k)−ω(k ))x0

Now we recall to obtain the Hamiltonian we integrate the Hamiltonian density over spatial coordinates. Using the identity, Z ~ ~0 d3 xei(k±k )·~x = (2π)3 δ 3 (~k − ~k 0 ) we can seriously clean up this mess! Indeed, making the necessary substitutions, and performing the integral over d3 x we find:  Z   ∗ ∗ 2 3 ∗ ∗ 2 ˆ2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ d x Π · Π + ∇φ · ∇φ + V (|φ| ) = d x Π · Π + ∇φ · ∇φ + m |φ| Z Z d3 k 0 d3 k = 2ω(~k)(2π)3 2ω(~k 0 ) "    ~ ~0 ~ ~0 δ 3 (~k − ~k 0 ) ω(~k)ω(~k 0 ) + ~k~k 0 + m2 a ˆ(~k)ˆ a† (k~0 )ei(ω(k)−ω(k ))xo + ˆb† (~k)ˆb(~k 0 )e−i(ω(k)−ω(k ))xo

ˆ = H

Z

3



   ~ ~0 ~ ~0 + δ 3 (~k + ~k 0 ) − ω(~k)ω(~k 0 ) − ~k~k 0 + m2 ˆb† (~k)ˆ a† (k~0 )e−i(ω(k)+ω(k ))xo + a ˆ(~k)ˆb(k~0 )ei(ω(k)+ω(k ))xo

#

After collapsing the delta functions, the second term goes away. This is due to the fact the delta function sends ~k 0 → −~k, and our frequency is just the energy, which we know from our equations of motion is given by: ω 2 (~k) = ~k 2 + m2 → −ω 2 (~k) + ~k 2 + m2 = 0. Similarly, for the first term we find that ~k 0 → ~k so then ω(~k)ω(~k 0 ) + ~k~k 0 + m2 → 2ω 2 (~k). Thus, we find: ˆ = H

Z

d3 k (2π)3 2ω(~k)

"

#

ω(~k) a ˆ(~k)ˆ a† (~k) + ˆb† (~k)ˆb(~k)

20

Finally, for a n-component complex scalar field we find: Z   ˆ ˆm · Π ˆ ∗ + ∇φˆm · ∇φˆ∗ + V (|φˆm |2 ) H = d3 x Π m m # " Z d3 k ω(~k) a ˆm (~k)ˆ a†m (~k) + ˆb†m (~k)ˆbm (~k) = 3 ~ (2π) 2ω(k) # " Z XZ d3 k ω(~k) 3 † ~ ˆ ~ † ~ ~ ˆ ~ = am (k) + bm (k)bm (k) + d3 k ω(k) a ˆm (k)ˆ δ (0) 2 (2π)3 2ω(~k) m The delta function is related to the infinite volume of space: δ 3 (0) = V /(2π)3 . Note, we are summing over m! Thus, the normal ordered Hamiltonian is: ˆ := :H

Z

#

"

d3 k (2π)3 2ω(~k)

am (~k) + ˆb†m (~k)ˆbm (~k) ω(~k) a ˆ†m (~k)ˆ

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The ground state is defined by: a ˆ†m (~k)ˆ am (~k)|0i = ˆb†m (~k)ˆbm (~k)|0i = 0 R ~ This tells us that the ground state energy of our system is Eo = 2 d3 k ω(2k) δ 3 (0), and has occupation numbers na~k = nb~k = 0. With normal ordering we shift the ground state energy to zero. What about the momentum? Well we go back through the riveting procedure of substituting in our creation and annihilation operators once more! Recall: Z ˆ ˆ φˆ + h.c. P = d3 xΠ∇ Z Z Z    d3 k 0 d3 k † ~0 −ik0 ·x 3 ˆb(k~0 )eik0 ·x + h.c. ~k 0 a ~k)eik·x − ˆb† (~k)e−ik·x a ˆ ( ˆ ( k )e − = dx 2(2π)3 (2π)3 2ω(k~0 ) So foiling again: 0

0

0

0

a ˆ(~k)ˆ a† (k~0 )ei(k−k )·x − ˆb† (~k)ˆ a† (k~0 )e−i(k+k )·x − a ˆ(~k)ˆb(k~0 )ei(k+k )·x + ˆb† (~k)ˆb(k~0 )e−i(k−k )·x Which becomes on substituting back into our integral: " Z Z 3 3 0 d k d k ~0 ~0 ~ ~ ~k 0 a Pˆ = ˆ(~k)ˆ a† (k~0 )δ 3 (~k − ~k 0 )ei(ω(k)−ω(k ))·xo − ˆb† (~k)ˆ a† (k~0 )δ 3 (~k + ~k 0 )e−i(ω(k)+ω(k ))·xo 3 2(2π) 2ω(k~0 ) # 0 0 ~ ~ ~ ~ −a ˆ(~k)ˆb(k~0 )δ 3 (~k + ~k 0 )ei(ω(k)+ω(k ))·xo + ˆb† (~k)ˆb(k~0 )δ 3 (~k − ~k 0 )e−i(ω(k)−ω(k ))·xo + h.c. Z =

d3 k 2(2π)3 2ω(~k)

" ~ ~ ~k a ˆ(~k)ˆ a(~k)† + ˆb† (~k)ˆ a(−~k)e−i(ω(k)−ω(−k))·xo

# +a ˆ(~k)ˆb† (−~k)e

i(ω(~k)−ω(−~k))·xo

21

+ ˆb(~k)ˆb† (~k) + h.c.

Since ω depends on the square of k, we can drop all the exponentials. Furthermore, when we do this, the remaining term:   ~k a ˆ(~k)ˆb† (−~k) + a ˆ(−~k)ˆb† (~k) is odd under ~k, so it integrates to zero. Thus, we finally find: " # Z 3 d k ~k a ˆ(~k)ˆ a† (~k) + ˆb(~k)ˆb† (~k) + h.c. Pˆ = 3 ~ 2(2π) 2ω(k) # " Z d3 k ~k a ˆ† (~k)ˆ a(~k) + ˆb† (~k)ˆb(~k) = 3 ~ (2π) 2ω(k) = : Pˆ :

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The last equality follows from the creation and annihilation operator commutation relations and: Z d3 k ~kδ 3 (0) = 0 3 ~ (2π) 2ω(k) as well as adding the Hermitian conjugate. Our momentum operator is already normal ordered! The expected result follows: Pˆ |0i =: Pˆ |0i := 0 Finally we normal order our SU (2) generators in respect to this very same ground state: k ˆ k = σmn Q

ˆk

→: Q : =

k σmn

Z Z

d3 k



a ˆ†m (~k)ˆ an (~k)

− ˆbm (~k)ˆb†n (~k)



(2π)3 2ω(~k)   d3 k a ˆ†m (~k)ˆ an (~k) − ˆb†n (~k)ˆbm (~k) (2π)3 2ω(~k)

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Clearly: ˆ k : |0i = 0 :Q 9. Using our creation operators we can build up our single particle states from the vacuum in the usual fashion: a ˆ†i (~q)|0i = |nai (~q) = 1i ≡ |ai (~q)i ˆb† (~q)|0i = |nbi (~q) = 1i ≡ |bi (~q)i i

Using the commutation relations of our creation and annihilation operators we find the excitation spectrum for these single particles states to be: p ˆ :i = ω(~q) = |~q|2 + m2 h: H Note this is independent of i (species) as well as if we are talking about particles or anti-particles; thus, there is a four-fold degeneracy . If we focus on just the particles or the anti-particles alone, we see that the degeneracy is 2. This means the single particle states of a single ”type” (same i but complex conjugates) form a j = 1/2 representation of SU (2). Indeed, one can show:

ˆ k : |ai (~q)i = σ k |ai (~q)i :Q ii k ˆ : |bi (~q)i = −σ k |bi (~q)i :Q ii

22

That is the particles are the spin-up states while the anti-particles are the spin down states 10 , which is indeed what you’d expect for the j = 1/2 states! We could also show: : Pˆ : |ai (~q)i = ~q|ai (~q)i : Pˆ : |bi (~q)i = ~q|bi (~q)i This completes our list of Quantum numbers as well as this assignment!

10 We should note that we have particles and antiparticles in another sense due to the underlying U (1) symmetry, which was previously pointed out.

23

Appendix

A

Foiling out Bogoliubov Transform

We begin by multiplying everything out:


†
cˆ(q) = cosh θ(q) a ˆ(q) + sinh θ(q) ˆb† (q) → cˆ† (q) = cosh θ(q) a ˆ (q) + sinh θ(q) ˆb(q)

†

ˆ = cosh θ(q) ˆb(q) + sinh θ(q) a d(q) ˆ (q) → dˆ† (q) = cosh θ(q) ˆb† (q) + sinh θ(q) a ˆ(q) ˆ Let’s invert these equations so that we have an explicit expression for a ˆ and ˆb in terms of cˆ and d. 2 2 This is easy enough if we recall: cosh (x) − sinh (x) = 1. Note this identity also preserves our desired commutation relations! We find:



ˆ a ˆ(q) = cosh θ(q) cˆ(q) − sinh θ(q) dˆ† (q) → a ˆ† (q) = cosh θ(q) cˆ† (q) − sinh θ(q) d(q)



ˆb(q) = cosh θ(q) d(q) ˆ − sinh θ(q) cˆ† (q) → ˆb† (q) = cosh θ(q) dˆ† (q) − sinh θ(q) cˆ(q) An astute choice of θ(q) will cancel all cross terms resulting in a diagonalized Hamiltonian. Let’s expand each term in our summation individually, then we can group like terms, and see where this brings us:  



ˆ cosh θ(q) cˆ(q) − sinh θ(q) dˆ† (q) 1. a ˆ† (q)ˆ a(q) = cosh θ(q) cˆ† (q) − sinh θ(q) d(q) 



 ˆ dˆ† (q) − cosh θ(q) sinh θ(q) cˆ† (q)dˆ† (q) + d(q)ˆ ˆ c(q) = cosh2 θ(q) cˆ† (q)ˆ c(q) + sinh2 θ(q) d(q)



ˆ − sinh θ(q) cˆ† (q) 2. ˆb† (q)ˆb(q) = cosh θ(q) dˆ† (q) − sinh θ(q) cˆ(q) cosh θ(q) d(q) 



 ˆ + sinh2 θ(q) cˆ(q)ˆ ˆ c(q) = cosh2 θ(q) dˆ† (q)d(q) c† (q) − cosh θ(q) sinh θ(q) cˆ† (q)dˆ† (q) + d(q)ˆ
 


ˆ − sinh θ(q) cˆ† (q) 3. a ˆ(q)ˆb(q) = cosh θ(q) cˆ(q) − sinh θ(q) dˆ† (q) cosh θ(q) d(q) 



 ˆ + sinh2 θ(q) cˆ† (q)dˆ† (q)) − cosh θ(q) sinh θ(q) cˆ(q)ˆ ˆ = cosh2 θ(q) cˆ(q)d(q) c† (q) + dˆ† (q)d(q) i 



ˆ 4. a ˆ† (q)ˆb† (q) = cosh θ(q) cˆ† (q) − sinh θ(q) d(q) cosh θ(q) dˆ† (q) − sinh θ(q) cˆ(q) 
†


 † 2 † † ˆ ˆ ˆ ˆ = cosh θ(q) cˆ (q)d (q) + sinh θ(q) cˆ(q)d(q)) − cosh θ(q) sinh θ(q) cˆ (q)ˆ c(q) + d(q)d (q) 2

Substituting these expressions into our Hamiltonian and normal ordering we find (this is why you should use a matrix!): h i a ˆ† (q)ˆ a(q) + ˆb† (q)ˆb(q) + cos(q)ˆ a(q)ˆb(q) + cos(q)ˆ a† (q)ˆb† (q) h i 

ˆ = 2 cosh 2θ(q) − sinh 2θ(q) cos(q) 1 + cˆ† (q)ˆ c(q) + dˆ† (q)d(q) h 



i 2 2 † † ˆ ˆ + cosh θ(q) + sinh θ(q) cos(q) − 2cosh θ(q) sinh θ(q) cˆ(q)d(q) + cˆ (q)d (q)

24

This shows us what choice we need to make on our angle θ(q); indeed, setting the coefficient of the cross term to zero gives:

2cosh θ(q) sinh θ(q)
cos(q) = cosh2 θ(q) + sinh2 θ(q)
2tanh θ(q) (31) = 2 1 + tanh θ(q)
= tanh 2θ(q) This implies: 1 θ(q) = tanh−1 (cos(q)) 2 This is what we got before with 10 times the work!

B

Local SU(2) Transformation

We are interested in the following symmetry: L(U Φ) = L(Φ). Our Lagrangian density above is only invariant for constant transforms; i.e. ∂µ Uab (x) = 0 ∀a, b. In order to promote our global symmetry to a local one we will need to define a covariant derivative. Inspired by our simpler case of a U (1) symmetry we define: Dµ = I∂µ − igAµ where I in the 2x2 identity matrix and Aµ is a 2x2 matrix valued vector field. We can figure out how Aµ transforms if we invoke:
0
Dµ Φ = U Dµ Φ This tells us that the field itself transforms as the covaraint derivative does provided we choose the transformation properties of our gauge field wisely. A simple calculation shows: h

i 0 −1 0 I∂µ − igAµ U Φ = U I∂µ Φ + iU I(∂µ U ) − igAµ U Φ
= U ∂µ Φ − igAµ Φ From this we deduce, for our transformation properties to be realized, that: i i A0µ = U Aµ U −1 − (∂µ U )U −1 = U Aµ U −1 + U (∂µ U −1 ) g g We can actually expand this vector field using the generators of SU(2); with the above generators of our group we find: (Aµ (x))ab = Akµ (x)λkab Here a, b and k all run over N. Following the work in problem 2.2 we see that under a local symmetry we’d have also need to consider how the gauge field varies. In the adjoint representation the gauge fields transforms like: 1 δAkµ (x) ≈ if ksj Ajµ (x)θs (x) + I ∂µ θk (x) g  i j So since σ , σ = 2iεijk σ k for SU(2), we see: k δφa ≈ iσab φb θk (x)

25

and, 1 δAkµ (x) ≈ 2iεksj Ajµ (x)θs (x) + I ∂µ θk (x) g We use these relations to derive conserved currents, and thus, charges. We know already: δL =

δL δL δL δL δφa + δ∂ν φa + φa ↔ φ∗a + δAkµ + δ∂ν Akµ k δφa δ∂ν φa δAµ δ∂ν Akµ

Using our equations of motion we can see: h δL i δL δL δL ∗ δφa + δ∂ν φa + φa ↔ φ∗a = ∂ν δφa + δφ δφa δ∂ν φa δ∂ν φa δ∂ν φ∗a a
∗
∗ k i i h k ν k ν = ∂ν σab D φa φb − σba D φa φb θ (x) 2 h i ≡ ∂ν j νk θk (x) We can see the Lagrangian does not depend on ∂ν Aµ ; thus: δL δL δL δAkµ + δ∂ν Akµ = δAk k k δAµ δ∂ν Aµ δAkµ µ we put everything together and find: h i h i δL 1 δL = ∂ν j νk θk (x) + 2iεksj Ajµ (x)θs (x) + I ∂µ θk (x) g δAkµ For θ constant this reduces to the answer in problem 2.2 (after choosing a gauge where Akµ = 0).

26

Physics 582, Fall Semester 2021 Professor Eduardo Fradkin Problem Set No. 4: Path Integrals in Quantum Mechanics and in Quantum Field Theory Due Date: Friday October 22, 2021, 9:00 pm US Central Time 1

Path Integral for a particle in a double well potential.

Consider a particle with coordinate q, mass m moving in the one-dimensional double well potential V (q) V (q) = λ q 2 − q02


2

(1)

In this problem you will use the path integral methods, in imaginary time, that were discussed in class to calculate the matrix element, τ τ hq0 , | − q0 , − i = hq0 |e−Hτ /~ | − q0 i 2 2

(2)

to leading order in the semiclassical expansion, in the limit τ → ∞. 1. Write down the expression of the imaginary time path integral that is appropriate for this problem. Write an explicit expression for the Euclidean Lagrangian (i.e. the Lagrangian in imaginary time). How does it differ from the Lagrangian in real time? Make sure that you specify the initial and final conditions. Do not calculate anything yet! 2. Derive the Euler-Lagrange equation for this problem (always in imaginary time). Compare it with the equation of motion in real time. Find the explicit solution for the trajectory (in imaginary time) that satisfies the initial and final conditions. Is the solution unique? Explain. What is the physical interpretation of this trajectory and of the amplitude? Hint: Your equation of motion in imaginary time looks funny. A simple way to solve for the trajectory that you need in this problem is to think of this equation of motion as if imaginary time was real time, then to find the analog of the classical energy and to use the conservation of energy to find the trajectory. 3. Compute the imaginary time action for the trajectory you found above.

1

4. Expand around the solution you found above. Write a formal expression of the amplitude to leading order. Find an explicit expression for the operator that enters in the fluctuation determinant. Do not compute the determinant.

2

Path Integral for a charged particle moving on a plane in a perpendicular magnetic field.

Consider a particle of mass m and charge −e moving on a plane in the presence of an external uniform magnetic field perpendicular to the plane and with strength B. Let r = (x1 , x2 ) and p = (p1 , p2 ) represent the coordinates and momentum of the particle. The Hamiltonian of the system is H(q, p) =

1 e (p + A(r))2 2m c

(3)

~ · A(r) = 0, the vector where A(r) is the vector potential. In the gauge ▽ potential is given by A1 (r) = −

B x2 , 2

A2 (r) =

B x1 2

(4)

1. Derive a path integral formula for the transition amplitude of the process in which the particle returns to its initial location r0 at time tf having left that point at ti i.e., hr0 , tf |r0 , ti i (5) where r0 is an arbitrary point of the plane and |tf − ti | → ∞. 2. Consider now the “ultra-quantum” limit m → 0. Find the form of the action S in this limit for a path which begins and ends at r0 . Find a geometric interpretation for this formula. (Hint: at some point you may have to use Stokes theorem). 3. Suppose that the particle actually is not on a two dimensional plane but is instead restricted to be on the surface of a sphere of large radius R → ∞, and that there us a uniform magnetic field normal to the surface of the sphere. Strictly speaking this requieres to have a magnetic monopole inside the sphere. Al though the associated vector potential reduces to the expression given in Eq.(4) only locally, i.e. on a plane locally tangent to the sphere at given point. Question: are there any ambiguities involved in the evaluation of the expression for the action of the particle?. Think of the regions enclosed by the path and left outside of the path. What i condition should satisfy the field strength B, so that the amplitude e ~ S is free from any ambiguities?. How is this condition related to the flux quantization?

2

3

Path Integrals for a Scalar Field Theory

Consider the problem of a one-component charged (i.e. complex) free scalar field φ(x) in D = 4 space-time dimensions. The action of this theory coupled to complex sources J(x) is Z i h (6) S = d4 x ∂µ φ(x)∗ ∂ µ φ(x) − m2 φ(x)∗ φ(x) − J(x)∗ φ(x) − J(x)φ(x)∗ 1. Find an explicit formula for the vacuum persistence amplitude J h0|0iJ for this theory in the form of a path integral in Minkowski space. Find the form of the path integral in Euclidean space (i.e., imaginary time). 2. Evaluate, using the methods discussed in class, the path integral you just wrote down, for a general set of complex sources J(x), both in Minkowski and in Euclidean space-time. 3. Use the formulas you just derived to compute the following two-point functions (in the limit J → 0) G2 (x − x′ ) =h0|T φ(x)φ∗ (x′ )|0i

(7)

G∗2 (x − x′ ) =h0|T φ∗ (x)φ(x′ )|0i

(8)

′

G2 (x − x′ ) =h0|T φ(x)φ(x′ )|0i ′

G2∗ (x

′

∗

∗

(9)

′

− x ) =h0|T φ (x)φ (x )|0i

(10)

Here x and x′ are two arbitrary point in space-time. Write your solutions in terms of a Green function. Do not calculate the Green function yet. 4. Find the equation that is satisfied by the Green function in both Euclidean and Minkowski space-times. Solve the equation for the Euclidean case and find the solution in Minkowski space-time by analytic continuation. Discuss th different asymptotic behaviors of the two-point functions in both Euclidean and Minkowski space-times. 5. Find explicit expressions for the following four-point functions Ga4 (x1 , x2 , x3 , x4 ) =h0|T φ(x1 )∗ φ(x2 )∗ φ(x3 )φ(x4 )|0i ′

G4 (x1 , x2 , x3 , x4 ) =h0|T φ(x1 )φ(x2 )φ(x3 )φ(x4 )|0i ′

(11) (12)

G4∗ (x1 , x2 , x3 , x4 ) =h0|T φ(x1 )∗ φ(x2 )∗ φ(x3 )∗ φ(x4 )∗ |0i

(13)

Gb4 (x1 , x2 , x3 , x4 ) Gc4 (x1 , x2 , x3 , x4 )

(14) (15)

∗

∗

=h0|T φ(x1 ) φ(x2 )φ(x3 ) φ(x4 )|0i =h0|T φ(x1 )∗ φ(x2 )φ(x3 )φ(x4 )∗ |0i

in terms of the two point functions. Find relations among these four point functions.

3

PHYS 582 Fall 2021

General Field Theory

Homework 4 Solutions Matthew O’Brien Total Points: 100

Question 1 (Particle in a Double Well Potential) Consider a particle with coordinate q and mass m moving in the potential V (q) = λ(q 2 − q02 )2 .

(1.0.1)

We are interested in the imaginary time transition amplitude hq0 , T /2|−q0 , −T /2i = hq0 | e−HT /~ |−q0 i ,

(1.0.2)

in the limit that T → ∞. 1. [10/35] The transition amplitude has the path integral form # " ˆ ˆ 1 T /2 −HT /~ dτ LE (q, q) ˙ , hq0 | e |−q0 i = Dq(τ ) exp − ~ −T /2

(1.1.1a)

q(−T /2)=−q0 q(T /2)=q0

m LE (q, q) ˙ = 2



dq dτ

2

m + V (q) = 2



dq dτ

2

+ λ(q 2 − q02 )2 ,

(1.1.1b)

where the subscripts under the functional integral are the boundary conditions for the sum over configurations q(−T /2) = −q0 and q(T /2) = +q0 , and LE is the Euclidean Lagrangian. For comparison, the Lagrangian in real time is     m dq 2 m dq 2 L= − V (q) = − λ(q 2 − q02 )2 . 2 dt 2 dt

(1.1.2)

and so the imaginary time Lagrangian looks like a conventional classical Lagrangian with an inverted potential. 2. [10/35] The imaginary time equation of motion follows from the Euler-Lagrange equation ∂LE d ∂LE − =0 ∂q dτ ∂ q˙ d2 q ∂V (q) =⇒ m 2 = = 4λq(q 2 − q02 ). dτ ∂q

(1.2.1)

This shows that the positions q = ±q0 are unstable equilibria, and that a particle which starts at q = −q0 will “roll” down the inverted potential before rolling up the hill towards q = +q0 . In contrast, the real time equation of motion is m

d2 q = −4λq(q 2 − q02 ). dt2

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In this case, q = ±q0 are stable equilibria, and the region in between the double well is classically forbidden for a particle with a small kinetic energy. The integral of motion which is analogous to the classical energy is the Hamiltonian function corresponding to the Euclidean Lagrangian ∂LE m HE = q˙ − LE = ∂ q˙ 2



dq dτ

2 − V (q).

(1.2.3)

Since we are interested in the limit T → ∞, we impose the usual requirement that q˙ = 0 as τ → ±∞ to ensure a well-behaved/finite action. In this case, since q(−T /2) = −q0 , and V (−q0 ) = 0, the initial value of the “Euclidean energy” is HE (−T /2) = 0, and since this is a constant of motion, we have, for all imaginary time,  HE = 0,

=⇒

dq dτ

2 =

2V (q) . m

(1.2.4)

We can integrate this to find the solution for the classical trajectory qc , in the limit T → ∞. Defining τ0 to be the value of imaginary time when the particle passes through the origin, qc (τ0 ) = 0, we obtain r ˆ ˆ qc (τ ) dq 2λ τ 0 = ± dτ , m τ0 q 2 − q02 0 r   2λ 1 qc (τ ) (τ − τ0 ), =⇒ − arctanh =± q0 q0 m "r # 2λq02 =⇒ qc (τ ) = q0 tanh (τ − τ0 ) , (1.2.5) m where in the last step, we chose the sign so that qc (τ ) → q0 as τ → ∞ and qc (τ ) → −q0 as τ → −∞, as required. The solution (1.2.5) is manifestly not unique, since the location τ0 is arbitrary. Note that there is no contradiction with Picard’s theorem on the uniqueness of solutions to ordinary differential equations: Eq. (1.2.4) is a first order nonlinear ODE, and hence, given some initial condition q(τ1 ) = q1 , one can only guarantee uniqueness of the corresponding solution on the interval τ ∈ [τ1 − ε, τ1 + ε] for some ε > 0. This is precisely the case once we have fixed the crossover point τ0 . Seen another way, it is not sufficient to only specify the behavior at τ → ±∞ in order to have a well-posed initial value problem. The imaginary time trajectory describes a tunneling event between the minima of the double well potential. This is evident from the real time equations of motion since, as we observed above, the region between ±q0 is classically forbidden. However, since the potential is inverted in imaginary time, the particle is able to “move” between the two minima. Consequently, hq0 , T /2|−q0 , −T /2i is the probability amplitude for the particle to tunnel quantum mechanically from the well with minimum at −q0 to the well at +q0 .

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3. [5/35] Using the identity (1.2.4), we can easily evaluate the classical action: ˆ ∞ dτ LE S[qc (τ )] = −∞ ∞

ˆ

 dqc 2 dτ m = dτ −∞ ˆ q0 dqc dqc m = dτ −q0 ˆ q0 p = dq 2mV (q) 

−q0

4 √ = q03 2mλ, 3

(1.3.1)

where we have chosen the same sign for the square root as above, which ensures the imaginary time action is strictly non-negative, as it must be. 4. [10/35] We now expand the path integral around the classical trajectory by defining q(τ ) = qc (τ ) + ξ(τ ),

(1.4.1)

where ξ(−∞) = ξ(∞) = 0, since the boundary conditions are carried by the classical solution. We can start by expanding the kinetic term: "     2 # ˆ ˆ m dq 2 m dqc dξ dξ dqc 2 dτ = dτ +2 + 2 dτ 2 dτ dτ dτ dτ " 2  2 # ˆ m dqc dξ = dτ + 2 dτ dτ   ˆ ˆ d dqc d2 qc + dτ mξ − dτ mξ 2 dτ dτ dτ " 2  2 # ˆ ˆ m dqc dξ d2 qc = dτ + − dτ mξ 2 , (1.4.2) 2 dτ dτ dτ where the boundary term evaluates to zero because ξ(−∞) = ξ(∞) = 0. Then, we expand the potential term: ˆ ˆ ˆ ∂V (q) dτ V (q) = dτ V (qc ) + dτ ξ(τ ) ∂q qc (1.4.3) ˆ ˆ 1 ∂ 2 V (q) 0 0 + dτ dτ ξ(τ )ξ(τ ) + . . . 2 ∂q(τ )∂q(τ 0 ) qc When added to the kinetic term above, the second term in this expansion gives the contribution " # ˆ d2 qc ∂V (q) − dτ ξ(τ ) m 2 − = 0, (1.4.4) dτ ∂q qc which vanishes since the term in the brackets is simply the Euler-Lagrange equation for the classical trajectory. Then, since the potential V (q) is a simple polynomial in q,

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the third term in the expansion of the potential yields ˆ ˆ ˆ ˆ 2 ∂ 2 V (q) 0 0 0 ∂ V (q) dτ dτ ξ(τ )ξ(τ ) = dτ dτ δ(τ − τ 0 )ξ(τ )ξ(τ 0 ) 0 2 ∂q(τ )∂q(τ ) qc ∂q(τ ) qc ˆ = dτ V 00 (qc )ξ 2 . (1.4.5) Therefore, putting everything together, the semi-classical expansion of the action is "  # ˆ "   #  ˆ m dqc 2 m dξ 2 1 00 S[q(τ )] ' dτ + V (qc ) + dτ + V (qc )ξ 2 2 dτ 2 dτ 2 # "   ˆ 4 √ m dξ 2 + 2λ(3qc2 (τ ) − q02 )ξ 2 . (1.4.6) = q03 2mλ + dτ 3 2 dτ Thus, the leading order tunneling amplitude is ˆ lim hq0 , T /2|−q0 , −T /2i = Dq(τ ) e−S[q(τ )]/~ T →∞

q(−∞)=−q0 q(∞)=q0

'

√ 4q03 2mλ e− 3~

ˆ

 ˜ ) exp − Dξ(τ

˜ ξ(−∞)=0 ˜ ξ(∞)=0

ˆ



∞

−∞

dτ 

m 2

dξ˜ dτ



!2

+ 2λ(3qc2 (τ ) − q02 )ξ˜2  ,

(1.4.7) √ where we have re-scaled ξ˜ = ξ/ ~. Integrating the kinetic term in the action by parts gives   ˆ ˆ ∞ 1 ∞ d2 . 1 2 2 ˜ ˜ ˜ ) AˆE ξ(τ ˜ ), (1.4.8) − dτ ξ(τ ) −m 2 + 4λ(3qc (τ ) − q0 ) ξ(τ ) = − dτ ξ(τ 2 −∞ dτ 2 −∞ ˜ ) yields which lets us identify that the functional integral over ξ(τ lim hq0 , T /2|−q0 , −T /2i = e

T →∞

−

√ 4q03 2mλ 3~

Det AˆE


−1/2

,

where the fluctuation determinant is   d2 2 2 ˆ Det AE = Det −m 2 + 4λ(3qc (τ ) − q0 ) dτ " "r # !# 2 d2 2λq 0 (τ − τ0 ) − 1 , = Det −m 2 + 4λq02 3 tanh2 dτ m

(1.4.9)

(1.4.10)

˜ ˜ over the space of functions obeying the boundary conditions ξ(−∞) = ξ(∞) = 0. It turns out that the eigenfunctions of this operator can be solved for exactly (see this Wikipedia page).

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Question 2 (Charged Particle in a Magnetic Field) Consider a particle with mass m and charge −e confined to a plane, such that it has position r = (x1 , x2 ) and momentum p = (p1 , p2 ). An external magnetic field B = B zˆ is applied, so that the Hamiltonian is 2 1  e (2.0.1a) H(r, p) = p + A(r) , 2m c B ˆ − x1 y) ˆ . A(r) = − (x2 x (2.0.1b) 2 1. [15/30] Consider the return amplitude hr0 , tf |r0 , ti i = hr0 | e−iH(tf −ti )/~ |r0 i .

(2.1.1)

In constructing the (one-dimensional) path integral, one considers intermediate matrix elements of the form ˆ dpj hqj | H |qj−1 i = hqj | H |pj i hpj |qj−1 i , (2.1.2) 2π~ and hqj | H |pj i is simple to evaluate for a Hamiltonian of the form H = p2 /2m + V (q), where p and q appear separately. The Hamiltonian for our present problem does not have this form. However, by expanding the Hamiltonian, we find   2 2 1 eB eB 1 H= p1 − x2 + p2 + x1 2m 2c 2m 2c  2 eB 1 eB 1 2 2 (p1 + p2 ) + (x1 p2 − x2 p1 ) + (x21 + x22 ), (2.1.3) = 2m 2mc 2m 2c since [x1 , p2 ] = [x2 , p1 ] = 0. Therefore, we can make all position coordinates appear to the left of the momenta, and hence, the usual path integral derivation is still valid. From this observation, we can immediately write down  ˆ tf  ˆ  i dt p · r˙ − H(r, p) . (2.1.4) hr0 , tf |r0 , ti i = Dr(t)Dp(t) exp ~ ti r(ti )=r0 r(tf )=r0

Since the exponent is quadratic in the momenta, we can perform the functional integral over p. We can re-write the integrand in the exponent as   eB p2 p · r˙ − H(r, p) = x˙ 1 + x2 p1 − 1 2mc 2m   eB p2 + x˙ 2 − x1 p2 − 2 (2.1.5) 2mc 2m  2 1 eB (x21 + x22 ), − 2m 2c from which we can infer the following useful change of coordinates:    1 eB P1 = √ p1 − m x˙ 1 + x2 , 2mc 2m    1 eB P2 = √ p2 − m x˙ 2 − x1 , 2mc 2m

(2.1.6a) (2.1.6b)

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such that the integrand becomes 1 −P12 −P22 + m 2

2 2  2   eB eB 1 eB 1 x˙ 1 + x2 + m x˙ 2 − x1 − (x21 +x22 ), (2.1.7) 2mc 2 2mc 2m 2c

in which case we have an elementary Gaussian integral over the two now-factorized shifted momenta P1 and P2 . This leaves only the terms which are independent of P1 and P2 , which, when expanded out, give 1 eB m(x˙ 21 + x˙ 22 ) − (x1 x˙ 2 − x2 x˙ 1 ). 2 2c Therefore, the path integral for the return amplitude is   ˆ tf  ˆ i 1 e 2 hr0 , tf |r0 , ti i = Dr(t) exp dt mr˙ − r˙ · A . ~ ti 2 c

(2.1.8)

(2.1.9)

r(ti )=r0 r(tf )=r0

2. [10/30] In the limit m → 0, the action [the integrand of (2.1.9)] becomes e S=− c

ˆ

tf

dt ti

dr · A, dt

(2.2.1)

where r(t) is any trajectory which satisfies r(ti ) = r(tf ) = r0 . This integral can be simplified using Stokes’ theorem. Denoting the region with boundary defined by the trajectory r(t) by Ω, ˛ e S=− dr · A c ∂Ω ˆ e =− dS · (∇ × A) c Ω eB = ∓ Area(Ω), c Φ = ∓2π~ (2.2.2) Φ0 where the ∓ sign corresponds to the orientation of the enclosed area—the direction of the path induces an orientation for the unit normal dS in either the zˆ (counterclockwise when viewed from above) or −zˆ (clockwise from above) direction—Φ = B Area(Ω) is the flux enclosed by the trajectory, and Φ0 = hc/e is the flux quantum; the superconducting flux quantum differs by a factor of 2 since Cooper pairs have charge 2e. Therefore, the action measures the (signed) area enclosed by the trajectory of the particle. 3. [5/30] Consider a trajectory which is negatively oriented with respect to the zˆ unit normal. On the one hand, as noted in the previous part, the amplitude associated with this trajectory must be     B Area(Ω) Φ iSin /~ = exp 2πi , (2.3.1) e = exp 2πi Φ0 Φ0 where Area(Ω) is area enclosed on the inside of the trajectory. 6 of 13

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On the other hand, if we suppose that the particle is actually on the surface of a sphere with large radius R—the radius must be large for the Hamiltonian (2.0.1), which is written in terms of Cartesian coordinates, to be a good approximation—then the trajectory also forms the boundary of another closed region which is positively ˆ with area [4πR2 − Area(Ω)]; this is the region left “outside” oriented with respect to z, of the trajectory. Therefore, we could equally associate to this same trajectory the amplitude   B[4πR2 − Area(Ω)] eiSout /~ = exp −2πi . (2.3.2) Φ0 To prevent any ambiguities, we should require that these two amplitudes be equal. That is, the phase associated with the “inside” and “outside” regions must be equal, modulo 2π, (Sin − Sout )/~ = 2πn, for n ∈ Z, (2.3.3) which implies that the field strength must be quantized according to B=n

Φ0 , 4πR2

for n ∈ Z.

(2.3.4)

In other words, the flux through the surface of the sphere is quantized in units of Φ0 . This is the same phenomenon as the quantization of magnetic flux through superconductors with a non-trivial topology (e.g., SQUIDs).

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Question 3 (Path Integrals for a Scalar Field Theory) Consider the action for a free complex scalar field in D = 4 spacetime dimensions ˆ h i S = d4 x |∂µ φ(x)|2 − m2 |φ(x)|2 − J(x)∗ φ(x) − J(x)φ(x)∗ .

(3.0.1)

Note: The sign of the source coupling terms is the opposite to the conventional definition in Minkowski space. In this question, we work in units of ~ = 1. 1. [4/35] Writing the field and sources as i 1 h φ(x) = √ φ1 (x) + iφ2 (x) , φ(x)∗ = 2 i 1 h J(x) = √ J1 (x) + iJ2 (x) , J(x)∗ = 2

i 1 h √ φ1 (x) − iφ2 (x) , 2 i 1 h √ J1 (x) − iJ2 (x) , 2

(3.1.1a) (3.1.1b)

where φ1,2 and J1,2 are real, the action decouples into the sum of the actions of two independent real scalar fields: S[φ, φ∗ , J, J ∗ ] = S[φ1 , J1 ] + S[φ2 , J2 ],   ˆ 1 2 1 2 2 4 S[φα , Jα ] = d x (∂µ φα (x)) − m φα (x) − Jα (x)φα (x) . 2 2

(3.1.2a) (3.1.2b)

Therefore, the vacuum persistence function in Minkowski space will take the simple form ˆ  ˆ  . ∗ iS[φ1 ,J1 ] iS[φ2 ,J2 ] h0|0i = Z [J, J ] = Dφ (x) e Dφ (x) e 1 2 M J J ˆ = Dφ1 (x)Dφ2 (x) ei(S[φ1 ,J1 ]+S[φ2 ,J2 ]) ˆ ∗ ∗ = Dφ(x)Dφ(x)∗ eiS[φ,φ ,J,J ] , (3.1.3) where S[φ, φ∗ , J, J ∗ ] is the action given at the beginning of this question. In the last step, we have changed the functional integral to be over φ and φ∗ instead of φ1 = Re(φ) and φ2 = Im(φ), but this is a trivial change of basis with Jacobian equal to 1. The vacuum persistence function in Euclidean space follows by analytic continuation of the action from real to imaginary time t = −iτ :  ˆ ˆ h ∗ ∗ ZE [J, J ] = Dφ(x)Dφ(x) exp − d4 x |∂τ φ(x)|2 + |∇φ(x)|2 (3.1.4) i 2 2 ∗ ∗ + m |φ(x)| + J(x) φ(x) + J(x)φ (x) . 2. [10/35] To evaluate these path integrals, define ¯ φ(x) = φ(x) + ξ(x),

¯ ∗ + ξ ∗ (x). φ(x)∗ = φ(x)

(3.2.1)

Then the Lagrangian (in Minkowski spacetime) becomes ¯ 2 + ∂µ φ∂ ¯ µ ξ ∗ + ∂µ φ¯∗ ∂ µ ξ + |∂µ ξ|2 − m2 |φ| ¯ 2 + φξ ¯ ∗ + φ¯∗ ξ + |ξ|2 L = |∂µ φ|

− J ∗ φ¯ + ξ − J φ¯∗ + ξ ∗ .


(3.2.2)

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We can integrate this by parts to obtain



1 1 L = − φ¯∗ ∂ 2 + m2 φ¯ − φ¯ ∂ 2 + m2 φ¯∗ − ξ ∂ 2 + m2 φ¯∗ − ξ ∗ ∂ 2 + m2 φ¯ 2 2



1 ∗ 2 1 2 − ξ ∂ + m ξ − ξ ∂ 2 + m2 ξ ∗ − J ∗ φ¯ + ξ − J φ¯∗ + ξ ∗ , 2 2

(3.2.3)

¯ 2 and |∂µ ξ|2 terms to make the final where ∂ 2 = ∂µ ∂ µ , and we split up the |∂µ φ| expression more symmetric between the fields and their complex conjugates. Now, we impose that φ¯ and φ¯∗ obey the classical equations of motion generated by the Lagrangian:

∂ 2 + m2 φ¯ = −J, ∂ 2 + m2 φ¯∗ = −J ∗ , (3.2.4) so that the Lagrangian greatly simplifies to

1 1 1 1 L = − ξ ∗ ∂ 2 + m2 ξ − ξ ∂ 2 + m2 ξ ∗ − J ∗ φ¯ − J φ¯∗ . 2 2 2 2

(3.2.5)

To go further, first note that, defining Aˆ = ∂ 2 + m2 , and ξ = ξ1 + iξ2 ,

1 ∗ˆ ˆ ∗ = 1 ξ1 Aξ ˆ 1 + ξ2 Aξ ˆ 2 . ξ Aξ + ξ Aξ 2 2 Then, observe that the solutions to the equations of motion above are     ˆ ¯ φ(x) J(y) 4 M ¯ ∗ = − d y G0 (x − y) J(y)∗ , φ(x) where GM 0 (x − y) = hx|

∂2

1 |yi . + m2

Therefore, we can write the path integral as ˆ  2 ˆ   i ZM [J, J ∗ ] = Dξ exp − d4 x ξ(x) ∂ 2 + m2 ξ(x) 2  ˆ  4 4 ∗ M × exp i d xd y J(x) G0 (x − y)J(y)  ˆ  
 2 2 −1 4 4 ∗ M = Det ∂ + m exp i d xd y J(x) G0 (x − y)J(y) ,

(3.2.6)

(3.2.7)

(3.2.8)

(3.2.9)

up to a formally divergent normalization constant. The corresponding expression in Euclidean spacetime is then obtained via analytic continuation: ˆ  2 ˆ   1 ∗ 4 2 2 ZE [J, J ] = Dξ exp − d x ξ(x) − ∂ + m ξ(x) 2 ˆ  × exp d4 xd4 y J(x)∗ GE (x − y)J(y) 0 ˆ  
 2 2 −1 4 4 ∗ E = Det − ∂ + m exp d xd y J(x) G0 (x − y)J(y) , (3.2.10) where ∂ 2 = ∂τ2 + ∇2 and the Euclidean Green’s function is GE 0 (x − y) = hx|

1 |yi . −∂ 2 + m2

(3.2.11) 9 of 13

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3. [5/35] In Minkowski spacetime, the two-point functions are G2 (x − x0 ) = h0| T φ(x)φ(x0 )∗ |0i 1 1 δ 2 Z[J, J ∗ ] = (i)2 Z[0] δJ(x)∗ δJ(x0 ) J=0 ˆ  1 δ 4 ∗ M 0 = d yJ(y) G0 (y − x ) i δJ(x)∗  ˆ  0 0 × exp i d4 yd4 y 0 J(y)∗ GM (y − y )J(y ) 0

J=0

= G∗2 (x

−iGM 0 (x

0

0

− x ), ∗

(3.3.1a)

0

− x ) = h0| T φ(x) φ(x ) |0i δ 2 Z[J, J ∗ ] 1 1 = (i)2 Z[0] δJ(x)δJ(x0 )∗ J=0 ˆ  1 δ 4 0 M 0 0 0 = d y G0 (x − y )J(y ) i δJ(x)  ˆ  4 4 0 ∗ M 0 0 × exp i d yd y J(y) G0 (y − y )J(y )

J=0

0 = −iGM 0 (x − x),

G02 (x

0

(3.3.1b)

0

− x ) = h0| T φ(x)φ(x ) |0i δ 2 Z[J, J ∗ ] 1 1 = (i)2 Z[0] δJ(x)∗ δJ(x0 )∗ J=0  ˆ 1 δ 4 0 M 0 0 0 = d y G0 (x − y )J(y ) i δJ(x)∗  ˆ  4 4 0 ∗ M 0 0 × exp i d yd y J(y) G0 (y − y )J(y ) ˆ  ˆ  J=0 0 0 0 0 0 = d4 y 0 GM d4 y 0 GM 0 (x − y )J(y ) 0 (x − y )J(y )  ˆ  4 4 0 ∗ M 0 0 × exp i d yd y J(y) G0 (y − y )J(y )

J=0

= 0, G02∗ (x

0

(3.3.1c) ∗

0 ∗

− x ) = h0| T φ(x) φ(x ) |0i δ 2 Z[J, J ∗ ] 1 1 = (i)2 Z[0] δJ(x)δJ(x0 ) J=0 = 0,

(3.3.1d)

0 with the Green’s function GM 0 (x − x ) defined in the previous part.

The Euclidean two-point functions follow by the same logic, without the factors of i: 0 E 0 GE 2 (x − x ) = G0 (x − x ),

(3.3.2a)

∗ 0 E 0 GE 2 (x − x ) = G0 (x − x),

(3.3.2b)

G02E (x

(3.3.2c)

0

−x)=

G02E ∗ (x

0

− x ) = 0,

0 with GE 0 (x − x ) defined in the previous part.

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4. [12/35] The equations satisfied by the Green’s functions follow from the matrix elements of the inverse differential operators shown above: 1 |yi , ∂ 2 + m2
(4) =⇒ ∂x2 + m2 GM 0 (x − y) = δ (x − y), 1 |yi , GE 0 (x − y) = hx| 2 −∂ + m2
(4) =⇒ − ∂x2 + m2 GE 0 (x − y) = δ (x − y),

GM 0 (x − y) = hx|

(3.4.1a)

(3.4.1b)

where in the Minkowski case ∂ 2 = ∂02 − ∇2 , and in the Euclidean case ∂ 2 = ∂τ2 + ∇2 . To solve for the Euclidean Green’s function, we expand both sides of the differential equation using a Fourier transform:
(4) − ∂x2 + m2 GE 0 (x − y) = δ (x − y), ˆ ˆ
d4 p ip·(x−y) d4 p E ip·(x−y) G e =⇒ − ∂x2 + m2 (p)e = , (2π)4 0 (2π)4 ˆ ˆ
E d4 p ip·(x−y) d4 p 2 2 ip·(x−y) p + m G (p)e = e , =⇒ 0 (2π)4 (2π)4 1 , =⇒ GE 0 (p) = 2 p + m2 ˆ d4 p eip·(x−y) E =⇒ G0 (x − y) = . (3.4.2) (2π)4 p2 + m2 We can then re-write this integral using a Feynman-Schwinger parameter to bring the denominator into the exponential: ˆ ˆ h α i d4 p 1 ∞ 2 2 E dα exp − (p + m ) + ip · (x − y) G0 (x − y) = 2 0 (2π)4 2 " #   ˆ ˆ ∞ 4 d p 1 √ 1 x − y 2 |x − y|2 1 2 dα = exp − αp − i √ − − m α 2 0 (2π)4 2 2α 2 α   ˆ ∞ 1 1 |x − y|2 1 2 = dα exp − − m α . (3.4.3) 2 0 (2πα)2 2α 2 Then, changing variables by letting |x − y| 1 , m t

α=

(3.4.4)

the integral can then be evaluated   ˆ ∞    1 m 1 m|x − y| 1 E G0 (x − y) = dt t exp − t+ (2π)2 |x − y| 2 0 2 t  
1 m = 2 K1 m|x − y| , 4π |x − y| where

1 Kν (z) = 2

ˆ

∞ ν−1

dt t 0



z exp − 2

  1 t+ , t

(3.4.5)

(3.4.6) 11 of 13

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is the modified Bessel function of the second kind. At large separations m|x − y|  1, we can use the asymptotic expansion r π −z Kν (z) ' e , 2z

(3.4.7)

which yields p GE 0 (x

− y) '

π/2 m2 e−m|x−y| , 4π 2 (m|x − y|)3/2

(3.4.8)

which is the usual exponential decay at large (Euclidean) distances of a theory with a mass gap, with correlation length ξ = 1/m. At short distances m|x − y|  1, we can use   Γ(ν) 2 ν Kν (z) ' , (3.4.9) 2 z which yields GE 0 (x − y) '

1 . − y|2

4π 2 |x

(3.4.10)

On these very short scales, the theory is insensitive to the mass gap, which corresponds to a much longer length scale than |x − y|. Therefore, the correlations have the same power law decay as a massless theory. We can then analytically continue back to Minkowski spacetime to find the Green’s function GM 0 (x − y). We write p p |x − y| = |x − y|2 = −s2 , (3.4.11) where s2 = (x0 − y0 )2 − |x − y|2 is the relativistic interval. From the previous part, we also note that the Green’s function in Minkowski space also picks up an additional factor of i relative to the time ordered vacuum expectation value. Therefore, we find that p i m √ (x − y) = K (m −s2 ). (3.4.12) GM 1 0 4π 2 −s2 For space-like separations, s2 < 0, so the Minkowski propagator has the exact same asymptotic behavior as the Euclidean propagator (up to a factor of i): p √ p π/2 m2 −m −s2 M √ G0 (x − y) ' i , for m −s2  1, (3.4.13a) e 4π 2 (m −s2 )3/2 p i GM , for m −s2  1. (3.4.13b) 0 (x − y) ' 4π 2 (−s2 ) Therefore, the two-point functions share the same asymptotic form in both Euclidean and Minkowski space in this regime. For time-like separations, s2 > 0, we can simplify the expression above to give GM 0 (x − y) =

√ 1 m 1 m (2) √ 2 2) = − √ √ H (m s ), K (im s 1 4π 2 s2 8π s2 1

(3.4.14)

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√ (2) where H1 (z) = J1 (z) − iY1 (z) is a Hankel function. For large separations m s2  1, we can use the asymptotic expansion (see, for example, https://dlmf.nist.gov/10.7) r h  2 νπ π i (2) Hν (z) ' exp −i z − − , (3.4.15) πz 2 4 so that, GM 0 (x

p √ e−iπ/4 π/2 m2 −im s2 √ e . − y) ' 4π 2 (m s2 )3/2

(3.4.16)

Therefore, we see that the two-point functions in Minkowski space are oscillatory, with a length scale of oscillations once again set by the mass: λ ∼ 1/m. For small separations, we use   Γ(ν) 2 ν (1) Hν (z) ' , (3.4.17) iπ z so that

i (3.4.18) 4π 2 s2 Therefore, the short distance behavior of the two-point function is the same for both space-like and time-like separations. GM 0 (x − y) '

5. [4/35] To find the four-point functions, it suffices to apply Wick’s theorem, since we are working with a free field theory. First, observe that the Green’s function (in Minkowski or Euclidean spacetime) is even in its argument: G0 (x − y) = G0 (y − x). Therefore, the two non-trivial two-point functions are equal: G2 (x − x0 ) = G∗2 (x − x0 ). And, recalling that the two other two-point functions vanish identically, we find that Ga4 (x1 , x2 , x3 , x4 ) = h0| T φ(x1 )∗ φ(x2 )∗ φ(x3 )φ(x4 ) |0i = G2 (x1 − x3 )G2 (x2 − x4 ) + G2 (x1 − x4 )G2 (x2 − x3 ),

(3.5.1a)

Gb4 (x1 , x2 , x3 , x4 ) = h0| T φ(x1 )∗ φ(x2 )φ(x3 )∗ φ(x4 ) |0i = G2 (x1 − x2 )G2 (x3 − x4 ) + G2 (x1 − x4 )G2 (x2 − x3 ) Gc4 (x1 , x2 , x3 , x4 )

∗

= h0| T φ(x1 ) φ(x2 )φ(x3 )φ(x4 ) |0i = G2 (x1 − x2 )G2 (x3 − x4 ) + G2 (x1 − x3 )G2 (x2 − x4 )

G04 (x1 , x2 , x3 , x4 )

(3.5.1b)

∗

=

G04∗ (x1 , x2 , x3 , x4 )

= 0.

(3.5.1c) (3.5.1d)

In this form, it is simple to deduce the relations Ga4 (x1 , x2 , x3 , x4 ) = Gb4 (x1 , x3 , x2 , x4 ) = Gc4 (x1 , x4 , x3 , x2 ).

(3.5.2)

That is, the three non-trivial four-point functions are equivalent up to a permutation of their arguments.

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Physics 582, Fall Semester 2021 Professor Eduardo Fradkin Problem Set No. 5: Due Date: Friday November 12, 2021, 9:00 pm US Central Time 1

Fermions in one dimension

In this problem we will consider an application of the Dirac theory to a problem in condensed matter physics: polyacetylene. Polyacetylene is a long polymer chain of the type (CH)n . The motion of the conduction electrons in polyacetylene can be described by the following model due to Su, Schrieffer and Heeger. In this model, one considers a linear chain of carbon atoms (C) with classical equilibrium positions at the regularly spaced sites {x(n)|x(n) = n a0 } (where a0 is the lattice constant). The carbon atoms share their π-orbital electrons, one per carbon atom. These electrons are allowed to hop from site to site. This hopping process is modulated by the lattice vibrations. Since the mass M of the atoms is much larger than the mass of the electrons (or, what is the same, the tunneling hopping (kinetic) energy t of the electrons is much larger than the kinetic energy of the atoms), we can give an approximate description by treating the atoms classically while treating the electrons as quantum mechanical objects. The Hamiltonian , for a lattice with N (even) sites, is N

H

=

2 X

−

X

σ=↑,↓ n=− N 2 +1 N

+

2 X

n=− N 2

+1



  [t − α (x(n) − x(n + 1))] c†σ (n)cσ (n + 1) + h.c.

Pn2 D 2 + (x(n) − x(n + 1)) 2M 2



(1)

where c†σ (n) and cσ (n) are fermion operators which create and destroy a πelectron with spin σ at the nth site of the chain, the x(n)’s are the coordinates of the carbon atoms (measured from their classical equilibrium positions), P (n) are their momenta, M is the carbon mass, D is the elastic constant, t is the electron hopping matrix element (for the undistorted lattice) and α is the electron-phonon coupling constant. Polyacetylene has one electron per carbon atom and, hence, it is a half-filled system and there are N electrons in a chain with N atoms. The study of this problem is greatly simplified by considering a continuum version of the model. If the coupling constant α is not too large, the only physical processes which are important are those which mix nearly degenerate 1

states, i.e. the only electronic states that will matter are those within a narrow band of width 2Ec centered at the Fermi energy EF = 0. In this limit, the single particle dispersion law becomes E(p) ≈ vF (p ± pF ). These states are right moving electrons ( with p ≈ pF ) and left moving electrons ( with ≈ −pF ). Here vF is the Fermi velocity. These considerations motivate the following way of writing the electron operators cσ (n) = eipF n Rσ (n) + e−ipF n Lσ (n)

(2)

Likewise, since the only processes in which phonons mix electrons near ±pF have momentum q ≈ 0 (forward scattering) or q ≈ 2pF (backward scattering), it is also natural to split the phonon fields into two terms x(n) = δ(n) + e2ipF n ∆+ (n) + e−2ipF n ∆− (n)

(3)

where pF = 2aπ0 . Within this set of approximations, it is natural to write an effective continuum Hamiltonian which only involves the left and right moving fermions and the phonons with p ≈ aπ0 . After Fourier transforming back to position space we get   X Z ∂ † σ3 ψσ (x) H= dx ψσ (x) −ivF ∂x σ=↑,↓  2  Z X Z p Π (x) 1 2 + dx (4) + ∆ (x) + dx 2g∆(x) ψ¯σ (x)ψσ (x) 2 8M a0 2 σ=↑,↓

where the two-component spinor ψσ (x)   Rσ (x) ψσ (x) ≡ Lσ (x)

(5)

is a Fermi field which represents the right and left moving electrons close to the Fermi energy. They obey the equal-time canonical anticommutation relations n o ψσα (x), ψσ† ′ α′ (x′ ) = δσσ′ δαα′ δ(x − x′ ) (6) while all other anticommutators are zero. Here the label α = 1 indicates upper (or right moving) component and α = 2 indicates the lower (or left moving) √ component. These fields differ from the ones we used above by a factor of 1/ 2a0 which gives these fields units of length−1/2 . The Bose field ∆(x) represents lattice vibrations with momentum close to 2pF . Thus, ∆(x) represents small fluctuations around a staggered distortion of the position of the atoms, Π(x) is the phonon canonical momentum and ∆ and Π obey canonical commutation relations [∆(x), Π(y)] = iδ(x − y)

2

  1 0 Notice that, in this approx0 −1 imation, the electrons are effectively a relativistic Dirac field with “speed of light” equal to the Fermi velocity. It is useful to define a set of 2 × 2 “Dirac” matrices,     0 −i 0 i γ0 = σ2 = , γ1 = iσ1 = , γ5 = γ0 γ1 = σ3 i 0 i 0

The matrix σ3 is the 2 × 2 Pauli matrix

We will also use the notation ψ¯ = ψ † γ0 = ψ † σ2 . The new effective electronphonon coupling constant is g = √αtD . 1. Consider the limit in which the mass of the carbon atoms is very heavy (M → ∞). In this adiabatic limit the phonon kinetic energy term of the effective Hamiltonian of Eq.(2) can be neglected and the phonon coordinates ∆(x) become classical variables. Find the ground state vector |gndi and energy of the system in this limit by calculating the ( constant) value of ∆ for which the ground state energy is minimized. You will have to cutoff some of the integrals at a relative momenta ±Ec /pF . Note: The spontaneous staggered distortion of the lattice is known as dimerization and this phenomenon is called the Peierls Instability. 2. Determine the energy spectrum and quantum numbers of the single-particle electronic states in this approximation. 3. Show that the continuum Hamiltonian (and the associated Lagrangian) is invariant under the global discrete symmetry transformation ψ(x) → γ5 ψ(x),

∆(x) → −∆(x)

(7)

with γ5 = σ3 . Show that, in terms of the lattice model, this transformation amounts to a shift of all the fields by one lattice site. In that language, the symmetry corresponds to an ambiguity in the way the dimerized structure is placed on the lattice Show that the operator ψ¯σ (x)ψσ (x) is odd under the discrete symmetry and, hence, it is an order parameter. 4. Compute the ground state expectation value for the order parameter of the previous section X hgnd|ψ¯σ (x)ψσ (x)|gndi (8) σ

in the M → ∞ approximation. Show that this order parameter has nonvanishing expectation value only is ∆ 6= 0 and establish a connection between both quantities.

3

2

Grassmann Stuff 1. Let a and a∗ be a pair of Grassmann variables. Let g(a∗ ) be an ”analytic function” of a single Grassmann variable a∗ , i.e. g(a∗ ) = g0 + g1 a∗

(9)

and let f (a) be another such function. Show that the inner product hf |gi defined by Z ∗ hf |gi = da∗ da e−a a f ∗ (a)g(a∗ ) (10) implies that

hf |gi = f¯0 g0 + f¯1 g1

(11)

where x¯ stands for the complex conjugate of x. 2. Show that (Af )(a∗ ) =

Z

∗

dα∗ dαA(a∗ , α)f (α∗ )e−α

α

= g(a∗ )

(12)

is equivalent to    A00 g0 = A01 g1

A10 A11

  f0 f1

(13)

and that (A B) (a∗ , a) =

Z

∗

dα∗ dα e−α

α

A(a∗ , α) B(α∗ , a) = C(a∗ , a)

(14)

is equivalent to the standard definition of the product of two 2×2 matrices. 3. Show that the operators a ˆ∗ and a ˆ, defined by a ˆ∗ f (α∗ ) = a∗ f (a∗ )

a ˆf (a∗ ) =

d f (a∗ ) da∗

(15)

satisfy canonical anticommutation relations,i.e. a ˆ∗ a ˆ∗ = a ˆ a ˆ = 0 and ∗ {ˆ a ,a ˆ} = 1. 4. Show that, if {ξj } is a set of N Grassmann variables (j = 1, . . . , N ), then Z=

Z Y N

dξj∗ dξj exp{−

j=1

3

N X

ξk∗ Mkl ξl } = detM

(16)

k,l=1

Dirac Fermions

The Lagrangian density L for the free massive Dirac field in 4-dimensional Minkowski space is
L = ψ¯ i∂/ − m ψ (17) 4

1. Consider the path integral for a free Dirac field in four space-time dimensions, coupled to a set of Grassmann sources η¯α (x) and ηα (x). Derive an expression for this generating function in terms of the sources and a Fermion determinant. Do not compute the determinant. 2. Use the results of previous question to show that the Feynman propagator of the Dirac theory is given by SFαβ (x − y) = hx, α|

1 |y, βi i∂/ − m

(18)

3. Use the results of the first part of this problem to derive an expression for the four point function (4) SF (x1 , x2 , x3 , x4 )α,β,γ,δ = h0|ψα (x1 )ψβ (x2 )ψ¯γ (x3 )ψ¯δ (x4 )|0i

(19)

in terms of products of propagators. Beware of the signs!!!!!!

4

Functional Determinants and the Casimir Effect

In this problem we are going to consider a free scalar field φ(x, t) in 1 + 1 space-time dimensions. The Lagrangian density L is L=

1 1 ∂µ φ(x)∂ µ φ(x) − m2 φ(x)2 2 2

(20)

where x ≡ (x, t). Consider the case in which the total length of the system along the space coordinate is equal to L and assume periodic boundary conditions, i.e. φ(x, t) = φ(x + L, t)

(21)

for all times t. 1. Calculate the classical value of the ground state energy of the system with the boundary conditions specified above. 2. Use path integral methods to derive a formal expression for the total ground state energy density (i.e. energy per unit length). This formula should contain a determinant which you should not compute for the moment. 3. Use the method of the ζ-function to compute the quantum correction to the ground state energy density. Consider the massless limit m → 0 only. Write your answer down in the form of an extensive piece and a finite-size term which vanishes as L → ∞ like A/Lη , with η > 0 . Find the value of this exponent η as well as the value of the coefficient A, sign included. Note: You may have to keep a dependence on the mass in one of the two 5

terms. Keep just the leading behavior in the small mass limit. Note: At some point of the calculation the following result may be useful: Poisson Summation Formula: Z ∞ ∞ ∞ X X f (n) = dx f (x) e2πimx (22) n=−∞

m=−∞

−∞

4. If you interpret the dependence of the ground state energy on the linear size of the system L as a potential energy for the“walls” that confine the system, what can you say about the force that the zero-point fluctuations exert on these “walls”. Note: In order to speak about walls we should have used vanishing, instead of periodic boundary conditions as we have done. The calculation is somewhat more complicated in that case. This effect, i.e. a force exerted on the walls of a system by the zero point motion of a field is known as the Casimir effect.

5

The Weakly-Interacting Bose Gas

Consider a gas of non-relativistic Bose particles at fixed density ρ inside a very large box of linear size L in three space dimensions. Let φ† (x) and φ(x) be a set of boson creation and annihilation operators. The second quantized Hamiltonian is   2 Z Z Z 1 pˆ 3 3 † − µ φ(x) + d x d3 x′ n ˆ (x)V (x − x′ )ˆ n(x′ ). (23) H = d x φ (x) 2m 2 where µ is the chemical potential, n ˆ = φ† φ and V (r) is a rotationally invariant short range interaction which we will take to be equal to V (x − x′ ) = λδ 3 (x − x′ )

(24)

The positive constant λ is the scattering amplitude and will play the role of a coupling constant for this system. 1. Use the method of Bose coherent states to find a path integral formula for the partition function of this system at temperature T . Do not compute the path integral at this stage. Assume constant boundary conditions at spacial infinity (i.e. that the field amplitude approaches a constant value at the boundaries). Carefully specify the boundary conditions in the imaginary time dimension. Write your answer down in the form Z ∗ Z = Dφ∗ Dφ e−SE (φ , φ) (25) and give an explicit expression for the Euclidean action SE .

6

2. Use the method of semiclassical quantization (i.e. the saddle point expansion) to determine the classical path at temperature T . What condition should be satisfied by φ(x) in order for it to be such a classical path?. Find the relationship between the ground state of the system at T = 0 and this classical path in the limit T → 0. Is the solution unique?. Justify your answer. Hint: Think of the symmetries of the Lagrangian. This will give you an idea about the uniqueness of the classical path. You may find it convenient to write the classical path φ(x) in the form of an amplitude times a phase. 3. Compute the time-ordered Green function ˆ φˆ† (y)i G(x − y) = −i hTˆφ(x)

(26)

at T = 0, in the semiclassical limit. What is the asymptotic value of G in the limit of equal times and large space separation?. Give a physical interpretation of this result. 4. Consider small quantum fluctuations around the classical path found in the previous sections. Write an arbitrary (but close) configuration φ(x) in the form p φ(x) = ρ0 + δρ(x) eiθ(x) (27)

Expand the action in powers of δρ and θ up to second order in both. Check the cancellation of the linear terms. Integrate out the density fluctuations and find an effective action for the phase variable Z (28) e−Seff (θ(x)) = Dδρ e−SE which is quadratic in θ.

5. Show that, for configurations {θ(x)} which are slowly varying, the effective action has the form Z i 1 h 2 2 Seff = d4 x K (∂τ θ) + v 2 (∇θ(x)) (29) 2 and calculate the coefficients K and v. Find the analytic continuation of this expression back in real time. Show that v is the velocity of propagation of the excitations. Find the time ordered propagator of the phase field θ(x). What equation of motion does it satisfy?. Draw an analogy between this equation and the equation of motion for a relativistic massless scalar field.

7

582 Homework 5 TA: Marcus Rosales November 2021

Contents 1 Fermions in one dimension

2

2 Grassmann Stuff

8

3 Dirac Fermions

11

4 Functional Determinants and the Casimir Effect

15

5 Weakly interacting Bose Gas

21

1

1

Fermions in one dimension

We will be using the effective continuum Hamiltonian: " # Z   2 XZ XZ p ∂ Π (x) 1 † 2 dx ψσ (x) − ivf σ3 ψσ (x) + dx + ∆ (x) + H= dx 2g∆(x)ψ¯σ (x)ψσ (x) 2 ∂x 8M a0 2 σ=± σ=±

(1.1)

1. 10/20: To simplify matters even further we consider the case of heavy carbon atoms: M → ∞. When we do this, we can drop the kinetic term, which amounts to disregarding the quantum fluctuations. Therefore, we can work with a constant classical field, so we replace ∆(x) with ∆0 . Our Hamiltonian reduces to: Z  XZ XZ p ∂  1 dx ψσ† (x) − ivf σ3 ψσ (x) + dx ∆20 + H= dx 2g∆0 ψ¯σ (x)ψσ (x) ∂x 2 σ=± σ=± Z Z (1.2)    p X XZ ∂ 1 dx ψσ† (x) − ivf = σ3 ψσ (x) + dx ∆20 + dx ψσ† (x) 2g∆0 σ2 ψσ (x) ∂x 2 σ=± σ=± Let’s group some terms together and write this out in two separate ways: !  √ 1 2 − 2g∆0 ψσ (x) + ∆0 H = dx ∂ vf ∂x 2 σ=± Z Z   X p 1 = dxψ¯σ − iγ1 ∂x + 2g∆0 ψσ + dx ∆20 2 σ=±   XZ p L = dxψ¯σ − iγ1 ∂x + 2g∆0 ψσ + ∆20 2 σ=± Z

X

iψσ† (x)

 −v ∂ √ f ∂x 2g∆0

(1.3)

√ Notice that the second term looks like a Dirac Hamiltonian with a mass term: m = 2g∆0 . We could solve this as was done in lecture, but we will opt for a straight forward Fourier expansion instead: Z dp m ψσ (x) = ψσ (p)eipx (1.4) 2π ω(p) We note that this Fourier modes are 2-component spinors. ω(p) will turn out to be the absolute value of our eigen energies. These factors are chosen to make the integration measure Lorentz invariant. In order for the anticommutation relations to be preserved, we need: o n ω(p) ψσα (p), ψσ† 0 α0 (p0 ) = 2π δσσ0 δαα0 δ(p − p0 ) m

(1.5)

With all other anti-commutation relations zero. Throwing in our Fourier transforms, and performing the necessary manipulations, we arrive at the following Hamiltonian:   √ X Z dp m L pv −i 2g∆0 H= iψσ† (p) √ f ψσ (p) + ∆20 (1.6) i 2g∆0 −pvf 2π ω(p) 2 σ=± We focus on the matrix equation for now:

Det

 pv √f − E i 2g∆0

 √ h i h i p p −i 2g∆0 = − (pvf + E)(pvf − E) + ( 2g∆0 )2 = − (pvf )2 − E 2 + ( 2g∆0 )2 = 0 −pvf − E

Which means: q E(p) = ± (pvf )2 + 2g∆20 ≡ ±ω(p)

2

This is the relativistic Pythagorean identity! Let’s push this analogy a bit further by defining: q p 1 = (pvf /ω(p))2 + ( 2g∆o /ω(p))2 q ≡ sin2 (θ) + cos2 (θ) There are many ways to proceed here! Do not worry too much if your method is different. Let’s express our field as a linear combination of eigenvectors for positive and negative eigenvalues (energies). We’ll call them u(p) and v(p), respectively: Z  dp m  a ˆσ (p)u(p)e−ipx + ˆbσ (p)v(p)eipx ψσ = 2π ω(p) I choose to set:

ω(p) δσσ0 δ(p − p0 ) m ω(p) δσσ0 δ(p − p0 ) {ˆbσ (p), ˆb†σ0 (p0 )} = 2π m {ˆ aσ (p), a ˆ†σ0 (p0 )} = 2π

(1.7)

etc. The factor of ω/m in the mode expansion as well as the above anti-commutation relations will set our normalization of our eigenvectors. Indeed: o n o n o Z dpdq  m2 n † † −i(px−qx0 ) ˆbσ (p), ˆb† 0 (q) v(p)v † (q)ei(px−qx0 ) a ˆ (p), a ˆ (q) u(p)u (q)e + ψσ (x), ψσ† 0 (x0 ) = 0 σ σ σ (2π)2 ω(p)ω(q) Z   0 0 dp m = δσσ0 u(p)u† (p)e−ip(x−x ) + v(p)v † (p)eip(x−x ) 2π ω(p) = δσσ0 δ(x − x0 ) ⇒ u(p)u† (p) = v(p)v † (p) =

ω(p) 2m

We will remember this for later. We solve the following matrix equations: 

√pvf i 2g∆0  pv −ω(p)v(p) = √ f i 2g∆0 ω(p)u(p) =

 √ −i 2g∆0 u(p) −pvf  √ −i 2g∆0 v(p) −pvf

Re-expressing in terms of sin and cos: 

 sin(θ) −icos(θ) u(p) icos(θ) −sin(θ)   sin(θ) −icos(θ) −v(p) = v(p) icos(θ) −sin(θ) u(p) =

These two systems only differ by an overall minus sign on the left. We cast the two systems into the following form and solve1 :      a sin(θ) −icos(θ) a± ± ± = b± icos(θ) −sin(θ) b± ⇒ ±a± = a± sin(θ) − ib± cos(θ) ±b± = ia± cos(θ) − b± sin(θ)

.˙. 1 do

icos(θ) ⇒ b± = a± sin(θ) ± 1     a± sin(θ) ± 1 = c± b± icos(θ)

not confuse a± , b± with our creation and annihilation operators!

3

Here c± is a normalization constant, which can be determined by recalling the normalization we want: h i ω(p) 1 = = |c± |2 (sin(θ) ± 1)2 + cos2 (θ) 2m 2cos(θ) = 2|c± |2 (1 ± sin(θ)) 1 1 ⇒ |c± | = q 2 cos(θ) 1 ± sin(θ)
=

1 ω(p) p 2 m(ω(p) ± pvf )

We arrive at our solutions in terms of our original parameters:   1 pvf + ω(p) u(p) = p im 2 m(ω(p) + pvf ) and

  1 pvf − ω(p) v(p) = p im 2 m(ω(p) − pvf )

Re-expressing our Hamiltonian in terms of the mode expansion we arrive at:

H=

h i L X Z dp m ω(p) a ˆ†σ a ˆσ − ˆb†σ ˆbσ + ∆20 2π ω(p) 2 σ=±

(1.8)

Notice I replaced the integral over x with the length of our polymer chain L = N a0 . We can now find the ground state with the usual prescription for a fermionic system: we define hole creation and annihilation operators: dˆσ = ˆb†σ , dˆ†σ = ˆbσ So we have: ˆb† ˆbσ = dˆσ dˆ† σ σ using ω(p) δσσ0 δ(p − p0 ) {dˆσ (p), dˆ†σ0 (p0 )} = 2π m We normal order our Hamiltonian:

H=

h i 1Z X Z dp m ω(p) ω(p) a ˆ†σ (p)ˆ aσ (p) + dˆ†σ (p)dˆσ (p) − 2π δσσ δ(p − p) + dx∆20 (x) 2π ω(p) m 2 σ=±

(1.9)

We now define our ground state in the usual manner: a ˆσ (p)|gndi = dˆσ (p)|gndi = 0 This can be used to determine an expression for our ground state energy; we have: XZ

Λ

q L dpδσσ δ(p − p) 2g∆20 + (pvf )2 + ∆20 2 σ=± −Λ Z vf Λ q L d¯ p L =− 2g∆20 + p¯ 2 + ∆20 π −vf Λ vf 2

Egnd = −

(1.10)

The last line is realized by noting the Dirac delta function in momentum space gives us the factor of L/2π, and we summed over sigma. Λ is our momentum cutoff. 4

We can finally find the distortion of the lattice in the ground state by minimizing the above equation in respect to the classical coordinate ∆0 : Z p 1 vf Λ d¯ 2g∆0 1 dEgnd p =− + ∆0 = 0 L d∆0 π −vf Λ vf 2g∆20 + p¯ 2 Z vf Λ 1 πvf d¯ pp ⇒ = 2g 2g∆20 + p¯ 2 −vf Λ Of course ∆0 = 0 satisfies this equation, but let’s find a non-trivial solution. The first integral is easy to recognize if we rewrite it as: Z

vf Λ

−vf Λ

d¯ pp

Z

1 2g∆20 + p¯ 2

vf Λ

d¯ p 1 q
2 p¯ 2g∆ 0 −vf Λ 1 − i √2g∆ 0 vf Λ p¯  1 −1  i√ = sin i 2g∆0 −vf Λ vf Λ  p¯  = sinh−1 √ 2g∆0 √

=

−vf Λ

−1

= 2sinh

 v Λ  √f 2g∆0

Plugging in we can solve for the ∆0 which minimizes our ground state energy:  v Λ  πvf f = 2sinh−1 √ 2g 2g∆0 "√ !#−1 2g πvf ⇒ ∆0 = sinh Λvf 4g r πvf 2 Λvf e− 4g ≈ g Since we will use this later, we make the simple observation that:  πv  √ ∆0 √ f g = 2exp − 0. The heat kernel has the following initial condition: X lim GA (x, y; τ ) = fn (x)fn∗ (y) = δ(x − y) τ →0+

n

We now relate our heat kernel to our generalized ζ-function in the following way: Z Z X X ˆ dD x lim GA (x, y; τ ) = e−an τ dD x fn (x)fn∗ (x) = e−an τ ≡ tr e−τ A y→x

n

We know that:

n

Γ(s) = asn

∞

Z

dτ τ s−1 e−an τ

0

So then we see by the above procedure: Z Z ∞ X Γ(s) Z ∞ X s−1 −an τ s−1 dD x lim GA (x, y; τ ) = dτ τ e = dτ τ y→x asn 0 0 n n We finally have: 1 ζA (s) = Γ(s)

∞

Z

dτ τ

s−1

Z

dD x lim GA (x, y; τ ) y→x

0

There are two ways we can proceed: (a) We find the eigenfunctions for Aˆ to build the generalized heat kernel. (b) We use the eigenvalues of Aˆ to build the generalized ζ-function. I will find the eigenfunctions, but we will construct the generalized ζ-function. Also, recall we are in 1 + 1-D, so we only have two components. To begin, we note that our eigenfunctions need to satisfy periodic boundary conditions in our spatial coordinate; i.e. (two labels for the two coordinates) fln (x + L, x0 ) = fln (x, x0 ). Here x0 is the label I choose to use for imaginary time since τ is already taken. This function also needs to be periodic in imaginary time with period β, which we will take to infinity in due time. Our eigenfunctions are very simple here! They are just complex exponentials. We could separate variables and show the imaginary time and the spatial components are both complex exponentials. If we did that, we’d get for fln (~x) ≡ fln (x, x0 ) = hl (x0 )gn (x): ˆ ln (x, x0 ) = an fln (x, x0 ) ⇒ Af

1 1 ∂ 2 hl (x0 ) + ∂ 2 gn (x) = aln − m2 ≡ k 2 hl (x0 ) 0 gn (x) x

We then define kx2 + ω 2 = k 2 , and do all the things we learned way back in intro to ODE’s. I’m just going to cut to the chase and say our normalized eigenfunctions are: eix·k fln (~x) = √ Lβ 2π with eigenvalues: aln = k 2 + m2 . Note, the eigenvalues are: k = (kn , ωl ) = ( 2π L n, β l). This gives us our periodic boundary conditions.

Since we know our eigenvalues, we can now write out our expression for the generalized ζ-function: ζAˆ (s) =

X 1 asln l,n

1 X = Γ(s)

Z

β X = Γ(s) n

Z

l,n

∞

dτ τ s−1 e−aln τ

0

0

∞

Z

∞

−∞

18

dτ dω s−1 −(ω2 +kn2 +m2 )τ τ e 2π

Notice that we replaced the sum with an integral in the last step by using the density of states; hence, ωl → ω. We can easily perform the integral over ω since it is just Gaussian: r Z β X ∞ dτ dω s−1 π −(kn2 +m2 )τ ζAˆ (s) = τ e Γ(s) n 0 2π τ Z β X ∞ dτ s−3/2 −(kn2 +m2 )τ √ τ e = Γ(s) n 0 2 π We are now in a position to use Poisson’s Summation formula: X XZ ∞ ¯ dye2πimy f (y) f (n) = n

−∞

m ¯

We identify: f (n) = exp(−kn2 τ ). Thus, substituting y in for n in kn explicitly we find: Z Z ∞
2 2π 2 2 dτ 2πimy β X ∞ √ e ¯ τ s−3/2 e− ( L ) y +m τ dy ζAˆ (s) = Γ(s) m 2 π −∞ 0 ¯ Z Z ∞ "
# 2 2 1 βL X ∞ dτ τ s−3/2 dk √ einkL e− k +m τ = 4πΓ(s) n 0 π −∞ Z (nL)2 2 βL X ∞ = dτ τ s−2 e−m τ − 4τ 4πΓ(s) n 0 The penultimate line was achieved simply from a change of variables, k = (2π/L)y, and I sent m ¯ → n to avoid confusion with our mass term (It is not an eigenvalue either!). The final step just relies on completing the square to find the Gaussian integral. We need to consider the massless limit, but there will be issues at n = 0; essentially, we have no regulator to tame the term τ s−2 . Thus, we imagine we have a very, very small mass which allows us to neglect it for all nonzero n. Specifically, we are taking the limit where we fix L, but send m to zero. Applying this line of reasoning, and recalling the Euler Γ-function: Z ∞ Γ(s) = dzz s−1 e−z 0

allows us to finish the calculation. The n = 0 term is (almost) automatic, but the n 6= 0 terms only require a 2 substitution of variables: z = (nL) 4τ . Furthermore, this is even in n, so we just sum over positive values. Thus: "Z # ∞ XZ ∞ (nL)2 βL s−2 −m2 τ s−2 − 4τ ζAˆ (s) = dτ τ e +2 dτ τ e (4.1) 4πΓ(s) 0 n=1 0 " # X  2 2(1−s) Z ∞ βL Γ(s − 1) = +2 dτ z −s e−z (4.2) 4πΓ(s) (m2 )s−1 nL 0 n=1 " #  2 2(1−s) βL Γ(s − 1) = +2 ζ(2 − 2s)Γ(1 − s) (4.3) 4πΓ(s) (m2 )s−1 L We can rewrite this expression in several ways; I chose the following. First, we’ll use (Euler’s reflection formula): Γ(1 − s)Γ(s) =

π sin(πs)

We can also use the relation which is used to analytically continue the Γ-function to the negative real numbers: Γ(s + 1) = sΓ(s) which implies: Γ(s − 1) 1 = Γ(s) s−1 19

" #  2 2(1−s) βL Γ2 (1 − s) sin(πs) 1 ζAˆ (s) = ζ(2 − 2s) +2 4π (m2 )s−1 (s − 1) L π

(4.4)

We need to take a derivative of this expression, then take the limit as s → 0. Mathematica is a good choice here. However, we can do this by hand if we keep linear order terms in a Taylor expansion. We start off by defining a mass scale µ (used momentarily), so that we can expand the first term as follows: (m2 )−s =(m2 )−s (1 + s + . . . ) 1−s =(e−s·log(m

2

/µ2 )

)(1 + s + . . . )

2

=(1 − s · log(m /µ2 ) + . . . )(1 + s) + . . . =1 + s(1 − log(m2 /µ2 )) + . . . That was fairly easy, but the next term looks like a nightmare! We’re going to cheat a little by permuting the limit and the derivative (or use Mathematica). This happens to work out fine here, but these do not commute in general! I repeat: these are dangerous actions I am taking here! The terms containing s are:  2 2(1−s) L

ζ(2 − 2s)

Γ2 (1 − s) sin(πs) s→0  2 2 −−−→ ζ(2)Γ2 (1)s π L

Using the famous result: ζ(2) =

π2 6

We find our final expression: # "   4π 2 βL dζ A =− − m2 1 − log m2 /µ2 ) + ln detAˆ = − lim+ ds 4π 3L2 s→0 1 ˆ + ... ln det(A) 2β  π m2 L  1 − log m2 /µ2 ) − + ... = 8π 6L

⇒ Efluc = EG − E0 =

(4.5)

We thus have a extensive piece and a term proportional to L−1 which goes to zero as m → 0. Note that E0 is singular, but I basically just redefined my reference of energy. 4. 2/20: Ignoring the fact we used periodic boundary conditions, opposed to vanishing boundary conditions, we analyze the pressure exerted by the zero point fluctuations on the walls enclosing the system. We expect the “outside” contributions to the pressure to overtake the “inside” contributions because there are more modes allowed in the infinite region “outside” the system: the momenta does not need to be multiples of 2π/L. The “outside” contribution should be divergent; hence, we’ll associate the divergent part of Efluc with the “outside” region. Similarly, the contribution from the “inside” region is due to the second term. Finally, in 1 + 1D pressure and force have the same units, so we have: FCasmir =

π Ein − Eout =− 2 L 6L

(4.6)

The minus sign tells us there is an attractive force between the walls, due to the above considerations. This is the famous Casmir effect!

20

5

Weakly interacting Bose Gas

Consider the Hamiltonian: Z H=

Z  i  hp2 1 ˆ d3 x0 n ˆ (x)V (x − x0 )ˆ n(x0 ) − µ φ(x) + d3 x φˆ† (¯ h2 x) 2m 2

We’ll take: V (x − x0 ) = λδ 3 (x − x0 ). 1. 5/20: In this problem we will be using the Bose coherent states: |{φ(x)}i = e

R

ˆ d3 xφ(x)φ(x)

|0i

As was the case with creation and annihilation operators, we replace the field operators with complex scalar fields (after doing the necessary work of course). Slicing up the time intervals in the usual way, one can show4 :

i ˆ −h ¯ H∆t

hf |e

Z |ii =

DφDφ¯ exp

( Z !) Z i i tf hh 3 ¯ ¯ ¯ ¯ d x φ(x, t)∂t φ(x, t) − φ(x, t)∂t φ(x, t) − H[φ, φ] dt h ti ¯ i  ¯ ti ))e × ψ¯f (φ(x, tf ))ψi (φ(x,

1 2

R

d3 x |φ(x,tf )|2 +|φ(x,ti )|2



I decided to carry around the ¯h for this problem (mainly because it was there when I took the class). We can write this in a simpler form, as was done in lecture: Z S=

Z Z   ¯2 ~ 2 h 1 ¯ ∇ + µ φ(x) − d4 x d4 x0 |φ(x)|2 |φ(x0 )|2 V (x − x0 ) d4 xφ(x) i¯ h∂t + 2m 2

Where, V (x − x0 ) = λδ 3 (x − x0 )δ(t − t0 ), and d4 x = dtd3 x. We can use the coherent-state path integrals to represent the partition function at finite temperature T: ˆ

Z = tre−β H We’d also like to work with the Euclidean version of the partition function. What we need to do is: (a) Set: |ii = |f i (b) Sum over all possible intermediate states. (c) Perform a Wick rotation t → −iτ . Since we are eventually going to derive a partition function, we are going to need a trace of our path integral. This is the same thing as saying our particle starts and ends in the same state; i.e. periodic boundary conditions in time. As we know from chapter 5, equation (5.130), Bosonic theories which have been Wick rotated obey periodic boundary conditions in imaginary time. That is: φ(x, τ ) = φ(x, τ + β) We compactified our imaginary time coordinate. The Wick rotated action is achieved in the usual way: t → −iτ ⇒∂t → i∂τ ⇒dtd3 x → −idτ d3 x Implementing this is easy enough. To indicate these substitutions I use small square braces around terms which change (in first line): 4 The

set up is in an Appendix.

21

Z Z Z   ¯2 ~ 2 h 1 β 3 ¯ [−idτ ] d x d3 y|φ(x)|2 |φ(y)|2 V (x − y) [−idτ ] d xφ(x) i¯ ∇ + µ φ(x) − i h[i∂τ ] + iS →i 2m 2 0 0 "Z # Z Z Z β   h2 ~ 2 ¯ 1 β 3 ¯ 3 3 2 2 =− dτ d xφ(x) ¯h∂τ − dτ d x d y|φ(x)| |φ(y)| V (x − y) ∇ − µ φ(x) + 2m 2 0 0 Z

β

Z

3

Note the use of the interaction potential (a delta function) to collapse the y-integral below. Our Euclidean action is given by:   ¯2 ~ 2 h ¯ d3 xφ(x) h∂τ − ¯ ∇ − µ φ(x) + 2m 0 Z β Z   ¯h2 ~ 2 ¯ ∇ − µ φ(x) + = dτ d3 xφ(x) h∂τ − ¯ 2m 0 Z β Z   ¯h2 ~ 2 ¯ dτ d3 xφ(x) h∂τ − ¯ = ∇ − µ φ(x) + 2m 0 Z β Z   ¯ ¯ = dτ d3 x ¯hφ(x)∂ τ φ(x) + H(φ, φ) Z

SE =

β

Z

dτ

β

1 2

Z

λ 2

Z

λ 2

Z

dτ d3 x

Z

d3 y|φ(x)|2 |φ(y)|2 V (x − y)

0 β

Z dτ

d3 x|φ(x)|4

0 β

Z dτ

d3 x|φ(x)|4

0

0

The trailing product in our above expression, which depends on the initial and final states, drops out because it is just a constant due to our boundary conditions. We arrive at our integral formula for our partition function: Z ¯ ¯ −SE (φ,φ) Z = DφDφe 2. 2/20: Now that we have a form for our generating functional let’s go ahead and find our saddle points of our Euclidean action above. We can hand pick the Lagrangian density from the above expression for SE : ! 2   h ¯ λ ¯ ~ 2 φ(x) + LE = φ(x) h∂τ − ¯ ∇ |φ(x)|2 − µ |φ(x)|2 2m 2 
¯h2 ~ 2  ¯ = φ(x) h∂τ − ¯ ∇ φ(x) + Vef f φ(x) 2m This guy is clearly invariant under the transformation: φ0 (x) = φ(x)eiθ So we have a global U (1) symmetry, and so our ground state is not unique (just change the phase). Furthermore, we can view the Lagrangian as the energy density for the Euclidean theory. The first term is a kinetic term, so it is positive definite. For the classical path we have φc (x) is a constant (which clearly satisfies our boundary conditions). To figure out what constant we minimize Vef f . This implies5 : √ φc (x) = ρ0 eiθ0 /2 r µ iθ0 /2 = e λ We also write the field in terms of the square-root of its density (at the classical level) so that we can say: µ = ρ0 λe−iθ0 This will be used below. Finally, we also note θ0 is an arbitrary constant phase, and we will eventually take θ0 = 0. For now we leave it undetermined. In the limit as T → 0, these configurations correspond to the degenerate ground states of the system since they minimize the energy. 5 The

phase angle is divided by 2 for later convenience. It’s purpose is shown immediately below.

22

3. 5/20: Let’s analyze the Green function in the semi-classical limit: φ(x) = φc (x) + δφ(x) ¯ ¯ φ(x) = φ¯c (x) + δ φ(x) Here δφ(x) is a small, but arbitrary, fluctuation about the classical path. Furthermore, since the fluctuations are small (semi-classical), we will Taylor expand the action to quadratic order, and then find the corresponding (real time) Green function. First, recall: Z Z Z β Z   λ β ¯h2 ~ 2 ¯ dτ d3 x|φ(x)|4 ∇ − µ φ(x) + S= dτ d3 xφ(x) ¯h∂τ − 2m 2 0 0 For simplicity, we write dx4E = dτ d3 x, but leave the function dependence as φ(x). To expand about the classical path we use the following Taylor expansion: # " Z Z Z 2 δ S δS δS 4 4 ¯ ¯ + 2 d x d y δ φ(x) S =S(φc ) + d4 xE δφ(y) δφ(x) + δ φ(x) E E ¯ ¯ δφ(x) δ φ(x) δ φ(x)δφ(y) φc φc φc Z Z Z Z 2 2 δ S δ S ¯ ¯ + d 4 xE d 4 yE dx4E d4 yE δ φ(x)δ φ(y) δφ(x)δφ(y) + ¯ ¯ + . . . δφ(x)δφ(y) δ φ(x)δ φ(y) φc φc " # Z 2     h ¯ 2 ¯ ¯ ¯ =S(φc ) + d4 xE 2δ φ(x) h∂τ − ¯ ∇2 − µ + 2λ|φc |2 δφ(x) + λ δφ(x)δφ(x)φ¯2c + δ φ(x)δ φ(x)φ + ... c 2m " Z     h2 2 ¯ ¯h2 2 3 ¯ ¯ =S(φc ) + d xE δ φ(x) ¯h∂τ − ∇ + ρ0 λ δφ(x) + δφ(x) − ¯h∂τ − ∇ + ρ0 λ δ φ(x) 2m 2m #   −iθ0 iθ0 ¯ ¯ + ρ0 λ e δφ(x)δφ(x) + e δ φ(x)δ φ(x) + ... ¯ ≡S(φc ) + δS(δφ, δ φ) Notice I integrated by parts in the last line; the purpose is to write the fluctuating part of the action as a matrix: # "  Z h ¯2 ¯   ∇2 + ρ0 λ δ φ(x) ρ0 λeiθ0 ¯h∂τ − 2m 4 ¯ ¯ (5.1) δS(δφ, δ φ) = d xE δ φ(x), δφ(x) h ¯2 δφ(x) −¯h∂τ − 2m ∇2 + ρ 0 λ ρ0 λe−iθ0 We now use this result to find our Green function6 :
G(x − y) = −ihT φ(x)φ† (y) i

¯ = −ihT φc (x)φ¯c (y) i − ihT δφ(x)δφ(y) i The first term is easy to find:

µ
hT φc (x)φ¯c (y) i = = ρ0 λ The other propagator is going to take some work! The focus will be on the fluctuating part of the action. We begin by introducing some sources: # "   ! Z h ¯2 2 iθ0 ¯   J   ∇ + ρ λ δ φ(x) ρ λe h ¯ ∂ − 0 τ 0 4 ¯ ¯ 2m ¯ = d xE δ φ(x), δφ(x) ¯ δφ(x) − δ φ(x), δS(δφ, δ φ) h ¯2 δφ(x) J −¯h∂τ − 2m ∇ 2 + ρ0 λ ρ0 λe−iθ0 Z   ˆ ≡ d4 xE δφT Hδφ − δφT J The last line is short hand for our full expression. The first thing we note is simply: 2 ¯
1 δ Z[J, J] ¯ hT δφ(x)δφ(y) i= ¯ Z[0, 0] δ J(x)δJ(y) J=0

6 You

should, in principle, also look at the mixed propagators. These terms will correspond to expectation values of our fluctuating field, but our generating functional will be shown to be quadratic. It follows these terms are zero.

23

We should be very familiar with the procedure at this point: we want the action to be in quadratic form, so we shift the fields: δφ(x) = δφ0 (x) + ξ(x) ¯ ¯ δ φ(x) = δ φ¯0 (x) + ξ(x) We can define column vectors in analogy to before   δ φ¯0 (x) δφ0 (x)   ¯ δ ξ(x) ξ= δξ(x)

δφ0 =

and we make the usual requirement that our shift satisfies (look at which currents are correlated with which fields by referencing our above definition): ˆ 0=J Hδφ Making the necessary substitutions we find: Z δS =

  ˆ − δφ0 T J d3 xdτ ξ T Hξ

ˆ In order to find the final term we need to explicitly solve for δφ0 in Hδφ 0 = J . This is a coupled set of equations which can be solved via the introduction of the following Fourier transforms7 : Z 4 Z 4 d pE d pE ip·x δφ (p)e , J(x) = J(p)eip·x δφ0 (x) = 0 4 (2π) (2π)4

δ φ¯0 (x) =

Z

d4 pE ¯ ¯ δ φ0 (−p)eip·x , J(x) = (2π)4

Z

d4 pE ¯ J(−p)eip·x (2π)4

Here: p = (ω, p). Substituting in we find the following matrix equation8 : " #    h ¯2 2 ρ0 λeiθ0 i¯hω + 2m p + ρ0 λ δ φ¯0 (−p) J(p) = ¯ h ¯2 2 δφ0 (p) J(−p) −i¯ hω + 2m p + ρ0 λ ρ0 λe−iθ0

⇒

    δ φ¯0 (−p) J(p) ˜ = G(p) ¯ δφ0 (p) J(−p)

Where9 : "

1

˜ G(p) =

" (ρ0 λ)2 −

hω ¯


2

+



h ¯ 2 p2 2m

ρ0 λe−iθ0 # h ¯2 2 2 i¯hω − 2m p − ρ0 λ + ρ0 λ

2

h ¯ −i¯hω − 2m p2 − ρ0 λ ρ0 λeiθ0

#

We can relate this to our fluctuations: Z δφ0 (x) =

    Z 4 d4 pE δ φ¯0 (−p) ipx d pE ˜ J(p) e = G(p) eipx ¯ J(−p) (2π)4 δφ0 (p) (2π)4 "Z #   Z 4 0 d pE ˜ 4 0 J(x ) −ipx0 = G(p) d x ¯ 0 e eipx J(x ) (2π)4   Z J(x0 ) 4 0 0 = d xE G(x − x ) ¯ 0 J(x )

7 Note the dot products in the exponents are in Euclidean 4-space! Also, our fields/currents are dressed up enough so we just denote the Fourier transforms by their arguments, and likewise for the currents sources. 8 Understanding orthogonality of distinct Fourier components are at play here. 9 Take the determinant to come up with the prefactor, then you can eye ball it. Of course, Gauss-Jordan elimination works too.

24

Where: G(x − x0 ) =

Z

0 d4 pE ˜ G(p)eip(x−x ) (2π)4

Making the above substitutions we can see: ¯ = Z[0, 0]e− Z[J, J]

R R

d4 xE d4 x0E J(x)G(x−x0 )J(x0 )

We are finally in a position to calculate our Green function. Going back to a previous part: " # 2 2 R R 4 ¯
4 0 0 0 δ 1 δ Z[J, J] − d z d z J(z)G(z−z )J(z ) E ¯ E = ¯ hT δφ(x)δφ(y) i= e ¯ Z[0, 0] δ J(x)δJ(y) δ J(x)δJ(y) J=0 J=0 " Z Z # δ2 4 4 0 0 0 = ¯ − d zE d zE J (z)G(z − z )J (z ) δ J(x)δJ(y) J=0 h i h i = G(x − y) + G(y − x) 21 12 h i = 2 G(x − y) 21

The last line can be seen by expanding the components of the Green functions in terms of their Fourier transforms and performing a change of variables. Looking at our above Green function we see we still need to evaluate an integral. We will do this by residues, but first let’s explicitly state the integral: d3 pdω (2π)4

Z

h i G(x − y)

=

21

h ¯2 2 2m p

i¯hω − " (ρ0 λ)2 −

¯hω


2

+



− ρ0 λ h ¯ 2 p2 2m

+ ρ0 λ

2

# eip(x−y)

2

h ¯ p2 + ρ0 λ ip(x−y) d3 pdω −i¯hω + 2m

2 e 2 (2π)4 ¯hω + ¯hωp

Z = where:

s ¯hωp =

 ¯ 2 p2  ¯h2 p2 h + 2ρ0 λ 2m 2m

We can evaluate the ω integral with contours using the residue theorem. The poles are at: ω = ±iωp We next need to determine our contour. The goal is to use a semicircular contour and Jordan’s lemma. The choice of contour boils down to what the sign of the relevant exponent is. This will depend on the difference: x − y ≡ x − y + ∆τ and in particular on ∆τ . If ∆τ > 0, Jordan’s lemma calls for a semi-circle in the positive half plane giving a pole at just ω = iωp . Similarly, a ∆τ < 0 calls for a semi-circle in the negative half plane, and a pole at ω = −iωp . Using the residue theorem for both of these cases gives: h

2

h ¯ hωp + 2m p2 + ρ0 λ −ωp |∆τ | ip(x−y) d3 p sgn(∆τ )¯ e e (2π)3 2ωp

Z

i G(x − y)

= 21

We are not out of the woods yet. We next convert to spherical polar coordinates10 : h

G(x − y)

i 21

=e

−ωp |∆τ |

Z 0

e−ωp |∆τ | = 4π 2 |x − y| 10 Do

∞

2

h ¯ p2 + ρ0 λ p2 dp sgn(∆τ )¯hωp + 2m (2π)2 2ωp

Z 0

∞

sgn(∆τ )¯ hωp + p dp ωp

h ¯2 2 2m p

+ ρ0 λ

not confuse p here with the 4-momenta! It is the magnitude of the 3-momenta.

25

Z

1

d(cos(θ))eip cos(θ)|x−y|

−1

sin(p|x − y|)

The integral, as it stands, is still a mess; however, we are interested in equal times! We then say: h

i G(x − y) 21

Z

1 = 2 4π |x − y|

∞

p dp

h ¯2 2 2m p

0

+ ρ0 λ sin(p|x − y|) ωp

For some characteristic length scale ξ of our system we consider the limit: ξ