Ward Identity and Chiral Anomalies
This article is one of Quantum Field Thoery.
There is an old Eastern proverb:
水滴穿石
This means that the falling water drop penetrates the stone.
Chiral Anomaly is classical symmetry inconsist in quantum verson with fermion.
19. Perturbation Theory Anomalies
Massless Dirac Lagrangian has an enhanced symmetry associated with the separate number conservation of left- and right-handed fermions. This symmetry is generated by the axial vector current $j^{\mu5}=\bar{\psi}\gamma^\mu \gamma^5\psi$. Classically, $$\begin{align}\partial_\mu j^{\mu5}=0\end{align}$$ for zero-mass fermions. This equation of motion is true note only in free fermion theory but also, as a classical field equation, in massless QED and QCD.
19.1 The Axial Current in Teo Dimensions
We want to analyze the current conservation equation for the axial current in massless QCD. First, study the physics that violates axial current conservation in massless QED.
The Lagrantion of two-dimensional QED is $$\begin{align}\mathcal{L}=\bar{\psi}(iD\!\!\!/)\psi-\frac{1}{4}(F_{\mu\nu})^2,\end{align}$$ with $\mu,\nu=0,1$ and $D_\mu=\partial_\mu+ieA_\mu$. The Dirac matrices must be chosen to satisfy the Dirac algebra $$\begin{align}\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}.\end{align}$$ In two dimensions, this set of relations can be represented by $2\times 2$ matrices; we choose $$\begin{align}\gamma^0=\begin{pmatrix}0&-i\\ i&0\end{pmatrix},\ \gamma^1=\begin{pmatrix}0&i\\ i&0\end{pmatrix}.\end{align}$$ The Dirac spinors will be two-component fields.
The product of the Dirac matrices, which anticommutes with each of the $\gamma^\mu$, is $$\begin{align}\gamma^5=\gamma^0\gamma^1=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.\end{align}$$ Then, just as in four dimensions, there are two possible currents, $$\begin{align}j^\mu=\bar{\psi}\gamma^\mu\psi,\quad j^{\mu5}=\bar{\psi}\gamma^\mu\gamma^5\psi,\end{align}$$ and both are conserved if there is no mass term in the Lagrangian.
To make the conservation laws quite explicit, we label the components of the fermion field $\psi$ in this spinor basis as $$\begin{align}\psi=\begin{pmatrix}\psi_+\\ \psi_-\end{pmatrix}.\end{align}$$ The subscript indicates the $\gamma^5$ eigenvalues. Then, using the explicit representations (4) and (7), we can rewrite the fermionic part of (2) as $$\begin{align}\mathcal{L}=\psi^\dagger_+i(D_0+D_1)\psi_++\psi^\dagger_-i(D_0-D_1)\psi_-.\end{align}$$ In the free theory, the field equation of $\psi_+$ would be $$\begin{align}i(\partial_0+\partial_1)\psi_+=0;\end{align}$$ the solutions to this equation are waves that move to the right in the one-dimensional space at the speed of light. We will thus refer to the particles associated with $\psi_+$ as right-moving fermions. This distinction is analogous to the distinction between left- and right-handed particles which gives the physical interpretation of 7.5 in four dimensions. Since the Lagrangian (8) contains no terms that mix left- and right-moving fields, it seems obvious that the number currentes for these fields are separately conserved. Thus, $$\begin{align}\partial_\mu\left( \bar{\psi}\gamma^\mu\left(\frac{1-\gamma^5}{2}\right)\psi\right)=0,\quad \partial_\mu\left( \bar{\psi}\gamma^\mu\left(\frac{1+\gamma^5}{2}\right)\psi\right)=0.\end{align}$$
It is a curious property of two-dimensional spacetime that ther vector and axial vector fermionic currents are not independent of each other. Let $\epsilon^{\mu\nu}$ be the totally antisymmetric symbol in two dimensions, with $\epsilon^{01}=+1$. Then the two-dimensional Dirac matrices obey the identity $$\begin{align}\gamma^\mu\gamma^5=-\epsilon^{\mu\nu}\gamma_\nu.\end{align}$$ The currents $j^{\mu5}$ and $j^\mu$ have the same relation. Thus we can study the properties of the axial vector current by using results that we have already derived for the vector current.
Vacuum Polarization Diagrams
In Section 7.5, we computed the lowest-order vacuum polarization of QED in dimensional regularization. In the limit of zero mass, we found, in Eq. (7.90), $$\begin{align}i\Pi^{\mu\nu}(q)=-i(q^2g^{\mu\nu}-q^\mu q^\nu)\frac{2e^2}{(4\pi)^{d/2}}\mbox{tr}[1]\int^1_0dx\, x(1-x)\frac{\Gamma(2-\frac{d}{2})}{(-x(1-x)q^2)^{2-d/2}},\end{align}$$ where $\mbox{tr}[1]=4$ gives the convention for tracing over Dirac matrices given in Eq. (7.88). If we set $\mbox{tr}[1]=2$ to be consistent with (4) and set $d=2$ in (12), we find the finite and well-defined result $$\begin{align}i\Pi^{\mu\nu}(q)=-i(q^2g^{\mu\nu}-q^\mu q^\nu)\frac{2e^2}{4\pi}\cdot 2\cdot \frac{1}{q^2}=9\left( g^{\mu\nu}-\frac{q^\mu q^\nu}{q^2}\right)\frac{e^2}{\pi}.\end{align}$$ Notice that this expression has the structure of a photon mass $$\begin{align}m_\gamma^2=\frac{e^2}{\pi}.\end{align}$$ In the discussion below Eq. (7.72), we see $q^2=0$ pole can generate mass. (13) shows it arise from the infrared behavior.
We an find the expectation value of the current induced by a background electromagnetic field. This quantity is generated by the diagram of Fig. 19.1, which gives $$\begin{align}\int d^2xe^{iq\cdot x}\langle j^\mu(x)\rangle=\frac{i}{e}(i\Pi^{\mu\nu}(q))A_\nu(q)=-\left( g^{\mu\nu}-\frac{q^\mu q^\nu}{q^2}\right)\cdot \frac{e}{\pi}A_\nu(q),\end{align}$$ where $A_\nu(q)$ is the Fourier transform of the background field. This quantity manifestly satisfies the current conservation relation $q_\mu\langle j^\mu(q)\rangle=0$.
The identity (11) between the vector and axial vector currents allows us to derive from (15) the corresponding expectation value of $j^{\mu5}$. We find $$\begin{align}\langle j^{\mu5}(q)\rangle=-\epsilon^{\mu\nu}\langle j_\nu(q)\rangle=\epsilon^{\mu\nu}\frac{e}{\pi}\left( A_\nu(q)-\frac{q_\nu q^\lambda}{q^2}A_\lambda(q)\right).\end{align}$$ If the axial vector current were conserved, this object would satisfy the Ward identity. Instead, we find $$\begin{align}q_\mu \langle j^{\mu5}(q)\rangle=\frac{e}{\pi}\epsilon^{\mu\nu}q_\mu A_\nu(q).\end{align}$$ This is the Fourier transform of the field equation $$\begin{align}\partial_\mu j^{\mu5}=\frac{e}{2\pi}\epsilon^{\mu\nu}F_{\mu\nu}.\end{align}$$ Apparently, the axial vector current is not conserved in the presence of electromagnetic fields, as the resulf of an anomalous behavior of its vacuum polarization diagram.
How could this happen? The Feynman diagrams formally satisfy the Ward identity both for the vector and for the axial vector current. The problem must come in the regularization of the vacuum polarization diagram. By dimensional analysis, we knwo that this diagram has the form $$\begin{align}\sim\! \bigcirc\! \sim =ie^2\left( Ag^{\mu\nu}-B\frac{q^\mu q^\nu}{q^2}\right).\end{align}$$ The coefficient $B$ is a finite integral, and is, in any event, unambiguously determined by the low-energy structure of the theory since it is the residue of the pole in $q^2$. However, the integral $A$ is logarithmically divergent, so its value depends on the regularization. Dimensional regularization automatically subtracts this integral to set $A=B$; then the vector current Ward identity is satisfied. But then we are led directly to (17). We could, alternatively, regularize the integral $A$ so that $A=0$. (Ignore Ward identity) Working through the steps of the previous paragraph with this modification, we now find $q_\mu\langle j^{\mu5}(q)\rangle=0$, but $$\begin{align}q_\mu \langle j^\mu (q)\rangle=\frac{e}{\pi}q^\nu A_\nu(q).\end{align}$$ Though the result (17) is unpleasant, the result (20) would be a complete disaster, since it depends on the unphysical gauge degrees of freedom of the vector potential. We conclude that it is not possible to regularize two-dimensional QED so that, simultaneously, the theory is gauge invariant and the axial vector current is conserved. The price of requiring gauge invariance is the anomalous nonconservation of the axial current shown in (18).
The Axial Vector Current Operator Equation
To understand what happend to the axial current from another viewpoint, we now study the operator equation for the divergence of $j^{\mu5}$. Varying the Lagrangian (2), we find the following equations of motion for the fermion fields: $$\begin{align} \partial\!\!\!/ \psi=-ieA\!\!\!/ \psi,\quad \partial_\mu\bar{\psi}\gamma^\mu=ie\bar{\psi}A\!\!\!/ .\end{align}$$ By using the equations of motion in the most straightforward way, it is easy to conclude that $\partial_\mu j^{\mu5}=0$. However, a closer look at these manipulations reveals some subtleties, which alter the final conclusion.
The axial vector current is a composite operator built out of fermion fields. In the previous chapter we saw that products of local operators are often singular, so we will define the current by placing the two fermion fields at distinct points separated by a distance $\epsilon$ and then carefully taking the limit as the two fields apporach each other. Explicitly, we define $$\begin{align}j^{\mu5}=\mbox{symm lim}_{\epsilon\rightarrow 0}\left\{ \bar{\psi}(x+\frac{\epsilon}{2})\gamma^\mu\gamma^5\exp \left[ -ie\int^{x+\epsilon/2}_{x-\epsilon/2}dz\cdot A(z)\right] \psi(x-\frac{\epsilon}{2})\right\}.\end{align}$$ Notice that, because we have placed $\psi$ and $\bar{\psi}$ at different points, we must introduce a Wilson line (15.53) in order that the operator be locally gauge invariant. To give $j^{\mu5}$ the correct transformation properties under Lorentz transformations, the limit $\epsilon\rightarrow 0$ should be taken symmetrically, $$\begin{align}\mbox{symm lim}_{\epsilon\rightarrow 0} \left\{ \frac{\epsilon^\mu}{\epsilon^2}\right\}=0,\quad \mbox{symm lim}_{\epsilon\rightarrow 0} \left\{ \frac{\epsilon^\mu\epsilon^\nu}{\epsilon^2}\right\}=\frac{1}{d}g^{\mu\nu},\end{align}$$ with $d=2$ in this case.
We now compute the divergence of the axial current defined as in (22): $$\begin{align}.\end{align}$$ $$\begin{align}\partial_\mu j^{\mu5}=\mbox{symm lim}_{\epsilon\rightarrow 0}\left\{ \left(\partial_\mu\bar{\psi}(x+\frac{\epsilon}{2})\right)\gamma^\mu\gamma^5\exp \left[ -ie\int^{x+\epsilon/2}_{x-\epsilon/2}dz\cdot A(z)\right] \psi(x-\frac{\epsilon}{2})\nonumber\\ +\bar{\psi}(x+\frac{\epsilon}{2})\gamma^\mu\gamma^5\exp \left[ -ie\int^{x+\epsilon/2}_{x-\epsilon/2}dz\cdot A(z)\right] \left(\partial_\mu\psi(x-\frac{\epsilon}{2})\right)\nonumber\\ +\bar{\psi}(x+\frac{\epsilon}{2})\gamma^\mu\gamma^5 \left[ -ie\epsilon^\nu\partial_\mu A(z)\right] \psi(x-\frac{\epsilon}{2}) \right\}.\end{align}$$ Using the equations of motion (21), and keeping terms up to order $\epsilon$, we can reduce this to $$\begin{align}\partial_\mu j^{\mu5}=\mbox{symm lim}_{\epsilon\rightarrow 0}\left\{ \bar{\psi}(x+\frac{\epsilon}{2})\left[ -ieA\!\!\! / (x+\frac{\epsilon}{2})-ieA\!\!\! / (x-\frac{\epsilon}{2}) \nonumber\\ -ie\epsilon^\nu \gamma^\mu\partial_\mu A_\nu(x)\right] \gamma^5\psi(x-\frac{\epsilon}{2})\right\}\nonumber\\ =\mbox{symm lim}_{\epsilon\rightarrow 0}\left\{ \bar{\psi}(x+\frac{\epsilon}{2})\left[ -ie\epsilon^\nu \gamma^\mu(\partial_\mu A_\nu-\partial_\nu A_\mu)\right] \gamma^5\psi(x-\frac{\epsilon}{2})\right\}.\end{align}$$
Expression (25) seems to vanish in the limit $\epsilon\rightarrow 0$. However, we must take account of the fact that the product of the fermion operators is singular. In two dimensions, the contraction of fermion fields is $$\begin{align}\langle T\psi(y)\bar{\psi}(z)\rangle=\int \frac{d^2k}{(2\pi)^2}e^{-ik\cdot (y-z)}\frac{ik\!\!\! /}{k^2}\nonumber\\ =-\partial\!\!\! / \left( \frac{i}{4\pi}\ln (y-z)^2\right) =\frac{-i}{2\pi}\frac{\gamma^\alpha(y-z)_\alpha}{(y-z)^2}.\end{align}$$ Thus, $$\begin{align}\langle T\bar{\psi}(x+\frac{\epsilon}{2})\Gamma \psi(x-\frac{\epsilon}{2})\rangle =\frac{-i}{2\pi}\mbox{tr}\left[ \frac{\gamma^\alpha \epsilon_\alpha}{\epsilon^2}\Gamma\right].\end{align}$$ Notice that the result (27) contains an extra minus sign from the interchange of fermion operators.
Because the contraction of fermion fields is singular as $\epsilon\rightarrow 0$, the terms of order $\epsilon$ in the last line of (25) can give a finite contribution. Thaking the contraction according to (27), we find $$\begin{align}\partial_\mu j^{\mu5}= \mbox{symm lim}_{\epsilon\rightarrow 0} \left\{ \frac{-i}{2\pi}\mbox{tr}\left[ \frac{\gamma^\alpha\epsilon_\alpha}{\epsilon^2}\gamma^\mu\gamma^5\right] \cdot (-ie\epsilon^\nu F_{\mu\nu})\right\}\end{align}$$. In two dimensions, $\mbox{tr}[\gamma^\alpha \gamma^\mu\ gamma^5]=2\epsilon^{\alpha\mu}$. Thus, $$\begin{align}\partial_\mu j^{\mu5}=\frac{e}{2\pi}\mbox{symm lim}_{\epsilon\rightarrow 0} \left\{ 2\frac{\epsilon_\mu\epsilon^\nu}{\epsilon^2}\right\} \epsilon^{\mu\alpha}F_{\nu\alpha}.\end{align}$$
An Example with Fermion Number Nonconservation
To complete our discussion of the two-dimensional axial vector current, we wil show that the nonconservation equation (18) also has a global aspect. In free fermion theory, the integral of the axial current conservation law gives $$\begin{align}\int d^2x\partial_\mu j^{\mu5}=N_R-N_L=0.\end{align}$$ This relation implies that the difference in the number of right-moving and left-moving fermions cannot be changed in any possible process. Combining this with the conservation law for the vector current, we conclude that the number of each type of fermion is separately conserved.
In two-dimensional QED, the conservation equation for the axial current is replaced by the anomalous nonconservation equation (18). If the right-hand side of this equation were the total derivative of a quantity falling off sufficiently rapidly at infinity, its integral would vanish and we would still retain the global conservation law. In fact, $\epsilon^{\mu\nu}F_{\mu\nu}$ is a total derivative: $$\begin{align}\epsilon^{\mu\nu}F_{\mu\nu}=2\partial_\mu(\epsilon^{\mu\nu}A_\nu).\end{align}$$ However, it is easy to imagine examples where the integral of this quantity does not vanish, for example, a world with a constant background electric field. In such a world, the conservation law (30) must be violated. But how can this happen?
Let us analyze this problem by thinking about fermions in one space dimension in a background $A^1$ field that is constant in space and has a very slow time dependence. We will assume that the system has a finite length $L$, with periodic boundary conditions. Notice that the constant $A^1$ field cannot be removed by a gauge transformation that satisfies the periodic boundary conditions. One way to see this is to note that the system gives a nonzero value to the Wilson line $$\begin{align}\exp\left[ -ie\int^L_0dx\, A_1(x)\right],\end{align}$$ which forms a gauge-invariant closed loop due to the periodic boundary conditions.
Following the derivation of the three-dimensional Hamiltonian, Eq. (3.84), we find that the Hamiltonian of this one-dimensional system is $$\begin{align}H=\int dx\, \psi^\dagger (-i\alpha^1D_1)\psi,\end{align}$$ where $\alpha=\gamma^0\gamma^1=\gamma^5$. In the components (7), $$\begin{align}H=\int dx \left\{ -i\psi^\dagger_+(\partial_1-ieA^1)\psi_++i\psi^\dagger_-(\partial_1-ieA^1)\psi_-\right\}.\end{align}$$ For a constant $A^1$ field, it is easy to diagonalize this Hamiltonian. The eigenstates of the covariant derivatives are wavefunctions $$\begin{align}e^{ik_nx},\quad \mbox{with}\ k_n=\frac{2\pi n}{L},\ n=-\infty,\cdots,\infty,\end{align}$$ to satisfy the periodic boundary conditions. Then the single-particle eigenstates of $H$ have energies $$\begin{align}\psi_+:\quad E_n=+(k_n-eA^1),\nonumber\\ \psi_-:\quad E_n=-(k_n-eA^1).\end{align}$$ Each type of fermion has an infinite tower of equally spaced levels. To find the ground state of $H$, we fill the negative energy levels and interpret holes created among these filled states as antiparticles.
Now adiabiatically change the value of $A^1$. The fermion energy levels slowly shift in accord with the relations (36). If $A^1$ changes by the finite amount $$\begin{align}\Delta A^1=\frac{2\pi}{eL},\end{align}$$ which brings the Wilson loop (32) back to ist original value, the spectrum of $H$ retruns to its original form. In this process, each level of $\psi_+$ moves down to the next position, and each level of $\psi_-$ moves up to the next position, as shown in Fig. 19.2. The remarkably, one right-moving fermion disappears from the vacuum and one extra left-moving fermion appears. At the same time, $$\begin{align}\int d^2x\left( \frac{e}{\pi}\epsilon^{\mu\nu}F_{\mu\nu}\right)=\int dt\, dx\, \frac{e}{\pi}\partial_0A_1=\frac{e}{\pi}L(-\Delta A^1)=-2,\end{align}$$ where we have inserted (37) in the last line. Thus the integrated form of the anomalous nonconservation equation (18) is indeed satisfied: $$\begin{align}N_R-N_L=\int d^2x\, \left( \frac{e}{2\pi}\epsilon^{\mu\nu}F_{\mu\nu}\right).\end{align}$$
Even in this simple example, we see that it is not possible to escape the question of ultraviolet regularization in analyzing the chiral conservation law. Right-moving fermions are lost and left-moving fermions appear from the depths of the fermionic spectrum, $E\rightarrow \infty$. In computing the changes in the separate fermion numbers, we have assumed that the vacuum cannot change the charge it contains at large negative energies. This prescription is gauge invariant, but it leads to the nonconservation of the axial vector current.
19.2 The Axial Current in Four Dimensions
The Axial Vector Current Operator Equation
Start with massless four-dimensional QED. We can still use (25). to gauge invariant. From this point, we must compute the singular terms in the operator product of the two fermion fields in the limit $\epsilon\rightarrow 0$. As in two dimensions, the leading term is given by contracting the two operators using a free-field propagator. This contribution gives $$\begin{align}\langle T\psi(y)\bar{\psi}(z)\rangle=\int \frac{d^4k}{(2\pi)^4}e^{-ik\cdot (y-z)}\frac{ik\!\!\! /}{k^2}\nonumber\\ =-\partial\!\!\! / \left( \frac{i}{4\pi^2}\frac{1}{(y-z)^2}\right) =\frac{-i}{2\pi^2}\frac{\gamma^\alpha(y-z)_\alpha}{(y-z)^4}.\end{align}$$ This is highly singular as $(y-z)\rightarrow 0$, but it gives zero when traced with $\gamma^\mu\gamma^5$. To find a nonzero result, we must consider terms of higher order in the expansion of the product of operators.
In a nonzero background gauge field, the contraction of fermion fields is given by the series of diagrams shown in Fig. 19.3. We have computed the leading term in this series in (40). The higher terms give singular terms as $(y-z)\rightarrow 0$. The second term in the series is given by $$\begin{align}=\int \frac{d^4k}{(2\pi)^4}\frac{d^4p}{(2\pi)^4}e^{-i(k+p)\cdot y}e^{ik\cdot z}\frac{i(k\!\!\! /+p\!\!\! /)}{(k+p)^2}(-ieA\!\!\! /(p))\frac{ik\!\!\! /}{k^2}.\end{align}$$ This contribution leads to $$\begin{align}\langle\bar{\psi}(x+\frac{\epsilon}{2})\gamma^\mu\gamma^5\psi(x-\frac{\epsilon}{2})\rangle\nonumber\\ =\int \frac{d^4k}{(2\pi)^4}\frac{d^4p}{(2\pi)^4} e^{ik\cdot \epsilon}e^{-ip\cdot x}\mbox{tr}\left[ (-\gamma^\mu\gamma^5)\frac{i(k\!\!\! /+p\!\!\! /)}{(k+p)^2}(-ieA\!\!\! / (p))\frac{ik\!\!\! /}{k^2}\right]\nonumber\\ ==\int \frac{d^4k}{(2\pi)^4}\frac{d^4p}{(2\pi)^4} e^{ik\cdot \epsilon}e^{-ip\cdot x} \frac{4e\epsilon^{\mu\alpha\beta\gamma}(k+p)_\alpha A_\beta(p)k_\gamma}{k^2(k+p)^2}.\end{align}$$ To evaluate the limit $\epsilon\rightarrow 0$, we can expand the integrand for large $k$. Then $$\begin{align}\langle\bar{\psi}(x+\frac{\epsilon}{2})\gamma^\mu\gamma^5\psi(x-\frac{\epsilon}{2})\rangle \sim 4e\epsilon^{\mu\alpha\beta\gamma}\int\frac{d^4p}{(2\pi)^4}e^{-ip\cdot x}p_\alpha A_\beta(p)\int \frac{d^4k}{(2\pi)^4}e^{ik\cdot \epsilon}\frac{k_\gamma}{k^4}\nonumber\\ = 4e\epsilon^{\mu\alpha\beta\gamma} (\partial_\alpha A_\beta(x))\frac{\partial}{\partial \epsilon^\gamma}\left( \frac{i}{16\pi^2}\ln \frac{1}{\epsilon^2}\right)\nonumber\\ =2e\epsilon^{\mu\alpha\beta\gamma} F_{\alpha\beta}(x)\left(\frac{-i}{8\pi^2}\frac{\epsilon_\gamma}{\epsilon^2}\right).\end{align}$$ Substituting this expression into (25), we find $$\begin{align}\partial_\mu j^{\mu5}=\mbox{symm lim}_{\epsilon\rightarrow 0} \left\{ \frac{e}{4\pi^2}\epsilon^{\alpha\beta\mu\gamma}F_{\alpha\beta}\left(\frac{-i\epsilon_\gamma}{\epsilon^2}\right)(-ie\epsilon^\nu F_{\mu\nu})\right\}.\end{align}$$ Now take the symmetric limit $\epsilon\rightarrow 0$ in four dimensions, We find $$\begin{align}\partial_\mu j^{\mu5}=-\frac{e^2}{16\pi^2}\epsilon^{\alpha\beta\mu\nu}F_{\alpha\beta}F_{\mu\nu}.\end{align}$$ This equation, which expresses the anomalous nonconservation of the four-dimensional axial current, is known as the Adler-Bell-Jackiw anomaly.
Triangle Diagrams
We can confirm the Adler-Bell-Jackiw relation by checking, in standard perturbation theory, that the divergence of the axial vector current has nonzero matrix element to create two photons. To do this, we must analyze the matrix element $$\begin{align}\int d^4x\, e^{-iq\cdot x}\langle p,k|j^{\mu5}(x)|0\rangle =(2\pi)^4\delta^{(4)}(p+k-q)\epsilon^*_\nu(p)\epsilon^*_\lambda(k)\mathcal{M}^{\mu\nu\lambda}(p,k).\end{align}$$ The leading-order diagrams contributing to $\mathcal{M}^{\mu\nu\lambda}$ are shown in Fig. 19.4. The first diagram gives the contribution $$\begin{align}\mathcal{M}^{\mu\nu\lambda}=(-1)(-ie)^2\int \frac{d^4l}{(2\pi)^4}\mbox{tr}\left[ \gamma^\mu\gamma^5 \frac{i(l\!\!\! /-k\!\!\! /)}{(l-k)^2}\gamma^\lambda\frac{il\!\!\! /}{l^2}\gamma^\nu \frac{i(l\!\!\! /+p\!\!\! /)}{(l+p)^2}\right],\end{align}$$ and the second diagram gives an identical contribution with $(p,\nu)$ and $(k,\lambda)$ interchanged.
It is easy to give a formal argument that the matrix element of the divergence of the axial current vanishes at this orther. Taking the divergence of the axial current in (46) is equivalent to dotting this quantity with $iq_\mu$.($\partial_\mu=\Rightarrow q_\mu$) Now we operate on the right-hand side of (47) as we do to prove a Ward identity. Replace $$\begin{align}q_\mu \gamma^\mu\gamma^5=(l\!\!\! /+p\!\!\! /-l\!\!\! /+k\!\!\! /)\gamma^5=(l\!\!\! /+p\!\!\! /)\gamma^5+\gamma^5(l\!\!\! /-k\!\!\! /).\end{align}$$ Each momentum factor combines with the numerator adjacent to it to cancel the corresponding denominator. This brings (47) into the form $$\begin{align}iq_\mu\cdot\mathcal{M}^{\mu\nu\lambda}=e^2\int \frac{d^4l}{(2\pi)^4}\mbox{tr}\left[ \gamma^5\frac{(l\!\!\! /-k\!\!\! /)}{(l-k)^2}\gamma^\lambda\frac{l\!\!\! /}{l^2}\gamma^\nu +\gamma^5\gamma^\lambda\frac{l\!\!\! /}{l^2}\gamma^\nu\frac{(l\!\!\! /+p\!\!\! /)}{(l+p)^2}\right].\end{align}$$ Now pass $\gamma^\nu$ through $\gamma^5$ in the second term and shift the integral over the first term according to $l\rightarrow (l+k)$: $$\begin{align}iq_\mu\cdot\mathcal{M}^{\mu\nu\lambda}=e^2\int \frac{d^4l}{(2\pi)^4}\mbox{tr}\left[ \gamma^5\frac{l\!\!\! /}{l^2}\gamma^\lambda\frac{(l\!\!\! /+k\!\!\! /)}{(l+k)^2}\gamma^\nu -\gamma^5\frac{l\!\!\! /}{l^2}\gamma^\nu\frac{(l\!\!\! /+p\!\!\! /)}{(l+p)^2}\gamma^\lambda\right].\end{align}$$ This expression is manifestly antisymmetric under the interchange of $(p,\nu)$ and $(k,\lambda)$, so the contribution of the second diagram in Fig. 19.4 precisely cancels (47).
However, because this derivation involves a shift of the integration variable, we should look closely at whether this shift is allowed by the regularization. From (47), we see that the integral that must be shifted is divergent. If the diagram is regulated with a simple momentum cutoff, or even with Pauli-Villars regularization, it turns out that the shift leaves over a finite, nonzero term. In Chapter 7, we encountered a similar problem in our discussion of the QED vacuum polarization diagram. We evaded the problem there by using dimensional regularization. Dimensional regularization of the diagrams of Fig. 19.4 will automatically insure the validity of the QED Ward identities for the photon emission vertices, $$\begin{align}p_\nu\mathcal{M}^{\mu\nu\lambda}=k_\lambda\mathcal{M}^{\mu\nu\lambda}=0.\end{align}$$ But in the analysis of the axial vector current, even dimensional regularization has an extra subtlety, because $\gamma^5$ is an intrinsically four-dimensional object. In their original paper on dimensional regularization, 't Hooft and Veltman suggested using the definition $$\begin{align}\gamma^5=i\gamma^0\gamma^1\gamma^2\gamma^3\end{align}$$ in $d$ dimensions. This definition has the consequence that $\gamma^5$ anticommutes with for $\mu=0,1,2,3$ but commutes with $\gamma^\mu$ for other values of $\mu$.
In the evaluation of (47), the external indices and the momenta $p,k,q$ all live in the physical four dimensions, but the loop momentum $l$ has components in all dimensions. (Like dim. reg.) Write $$\begin{align}l=l_{\parallel}+l_{\perp},\end{align}$$ where the first term has nonzero components in dimensions $0,1, 2, 3$ and the second term has nonzero components in the other $d$-4 dimensions. Because $\gamma^5$ commutes with the $\gamma^\mu$ in these extra dimensions, identity (48) is modified to $$\begin{align}q_\mu\gamma^\mu\gamma^5=(l\!\!\! /+k\!\!\! /)\gamma^5+\gamma^5(l\!\!\! /-p\!\!\! /)-2\gamma^5l\!\!\! /_\perp.\end{align}$$ The first two terms cancel according to the argument given above; the shift in (50) is justified by the dimensional regularization. However, the third term of (54) gives an additional contribution: $$\begin{align}iq_\mu\cdot\mathcal{M}^{\mu\nu\lambda}=e^2\int \frac{d^4l}{(2\pi)^4}\mbox{tr}\left[ -2\gamma^5l\!\!\! /_\perp\frac{(l\!\!\! /+k\!\!\! /)}{(l+k)^2}\gamma^\lambda\frac{l\!\!\! /}{l^2}\gamma^\nu\frac{(l\!\!\! /+p\!\!\! /)}{(l+p)^2}\right].\end{align}$$ To evaluate this contribution, combine denominators in the standard way, and shift the integration variable $l\rightarrow l+P$, where $P=xk-yp$. In expanding the numerator, we must retain one factor each of $\gamma^\nu,\gamma^\lambda,p\!\!\! /$, and $k\!\!\! /$ to give a nonzero trace with $\gamma^5$. This leaves over one factor of and one factor of $l\!\!\! /_\perp$ which must also be evaluated with components in extra dimensions in order to give a nonzero integral. The factors $l\!\!\! /+\perp$ anticommute with the other Dirac matrices in the problem and thus can be moved to adjacent positions. Then we must evaluate the integral $$\begin{align} \frac{d^4l}{(2\pi)^4}\frac{l\!\!\! /_\perp l\!\!\! /_\perp}{(l^2-\Delta)^3},\end{align}$$ where $\Delta$ is a function of $k$, $p$, and the Feynman parameters. Using $$\begin{align}(l\!\!\! /_\perp)^2=l^2_\perp\rightarrow \frac{(d-4)}{d}l^2\end{align}$$ under the symmetrical integration, we can evaluate (56) as $$\begin{align}\frac{i}{(4\pi)^{d/2}}\frac{(d-4)}{2}\frac{\Gamma(2-\frac{d}{2})}{\Gamma(3)\Delta^{2-d/2}}\rightarrow_{d\rightarrow 4}\frac{-i}{2(4\pi)^2}.\end{align}$$ Notice the behavior in which a logarithmically divergent integral contributes a factor $(d-4)$ in the denominator and allows an anomalous term, formally proportional to $(d-4)$, to give a finite contribution. The remainder of the algebra in the evaluation of (55) is straightforward. The terms involving the momentum shift $P$ cancel, and we find $$\begin{align}iq_\mu\cdot\mathcal{M}^{\mu\nu\lambda}=e^2\left(\frac{-i}{2(4\pi)^2}\right)\mbox{tr}[2\gamma^5(-k\!\!\! /)\gamma^\lambda p\!\!\! /\gamma^\nu]=\frac{e^2}{4\pi^2}\epsilon^{\alpha\lambda\beta\nu}k_{\alpha}p_\beta.\end{align}$$ This term is symmetric under the interchange of $(p,\nu)$ with $(k,\lambda)$, so the second diagram of Fig. 19.4 gives an equal contribution. Thus, $$\begin{align}\langle p,k|\partial_\mu j^{\mu5}(0)|0\rangle =-\frac{e^2}{2\pi^2}\epsilon^{\alpha\nu\beta\lambda}(-ip_\alpha)\epsilon^*_\nu(p)(-ik_\beta)\epsilon^*_\lambda(k)\nonumber\\ =-\frac{e^2}{16\pi^2}\langle p,k|\epsilon^{\alpha\nu\beta\lambda}F_{\alpha\nu}F_{\beta\lambda}(0)|0\rangle,\end{align}$$ as we would expect from the Adler-Bell-Jackiw anomaly equation.
[Insert] 5.5 Compton Scattering - Photon Polarization Sum
$$=i\mathcal{M}(k)\equiv i\mathcal{M}^\mu(k)\epsilon^*_\mu(k).$$ Since the amplitude always contains $\epsilon^*_\mu(k)$, we have extracted this factor and defined $\mathcal{M}^\mu(k)$ to be the rest of the amplitude $\mathcal{M}$. The cross section will be proportional to $$\sum_\epsilon \left| \epsilon^*_\mu(k)\mathcal{M}^\mu(k)\right|^2=\sum_\epsilon \epsilon^*_\mu\epsilon_\nu \mathcal{M}^\mu(k)\mathcal{M}^{\nu*}(k).$$ For simplicity, we orient $k$ in the 3-direction: $k^\mu=(k,0,0,k)$. Then the two transvers polarization vectors, over which we are summing, can be chosen to be $$\epsilon^\mu_1=(0,1,0,0);\quad \epsilon^\mu_2=(0,0,1,0).$$ With these conventions, we have $$\sum_\epsilon \left| \epsilon^*_\mu(k)\mathcal{M}^\mu(k)\right|^2=\left| \mathcal{M}^1(k)\right|^2+\left| \mathcal{M}^2(k)\right|^2.$$
Now recall the chapter 4 that external photons are created by the interaction term $\int d^4x\, ej^\mu A_\mu$, where $j^\mu=\bar{\psi}\gamma^\mu \psi$ is the Dirac vector current. Therefore we expect $\mathcal{M}^\mu(k)$ to be given by a matrix element of the Heisenberg field $j^\mu$: $$\mathcal{M}^\mu(k)=\int d^4x\, e^{ik\cdot x}\langle f|j^\mu(x)|i\rangle,$$ where the initial and final states include all particles except the photon in equestion.
From the classical equations of motion, we know that the current $j^\mu$ is conserved: $\partial_\mu j^\mu(x)=0$. Provided that this property still holds in the quantum theory, we can dot $k_\mu$ to obtain $$k_\mu \mathcal{M}^\mu(k)=0.$$ The amplitude $\mathcal{M}$ vanishes when the polarization vector $\epsilon_\mu(k)$ is replaced by $k_\mu$. This famous relation is known as the Ward identity. If is essentially a statement of current conservation, which is a consequence of the gauge symemtry (4.6) of QED. We will give a formal proof of the Ward identity in Section 7.4.
[Insert] 9.6 Symmetries in the Functional Formalism
We will see that the functional integral gives, in a mose direct way, a quantum generalization of Noether's theorem. This result will lead to the analogue of the Ward-Takahashi identity for any symmetry of a general quantum field theory.
Equations of Motion
The quantum equations of motion follow from the functional integral formalism. As a first problem to study, let us examine the Green's functions of the free scalar field. To be specific, consider the three-point function: $$\langle \Omega|T\phi(x_1)\phi(x_2)\phi(x_3)|\Omega\rangle=Z^{-1}\int \mathcal{D}\phi e^{i\int d^4x\, \mathcal{L}[\phi]}\phi(x_1)\phi(x_2)\phi(x_3),$$ where $\mathcal{L}=\frac{1}{2}(\partial_\mu \phi)^2-\frac{1}{2}m^2\phi^2$ ans $Z$ is a shorthand for $Z[J=0]$, the functional integral over the exponential. In classical mechanics, we would derive the equations of motion by insisting that the action be stationary under an infinitesimal variation $$\phi(x)\rightarrow \phi'(x)=\phi(x)+\epsilon(x).$$ A change of variables does not alter the value of the integral. Nor does a shift of the integration variable alter the mesure: $\mathcal{D}\phi'=\mathcal{D}\phi$. Thus we can write $$\int \mathcal{D}\phi e^{i\int d^4x\, \mathcal{L}[\phi]}\phi(x_1)\phi(x_2)\phi(x_3)=\int \mathcal{D}\phi e^{i\int d^4x\, \mathcal{L}[\phi']}\phi'(x_1)\phi'(x_2)\phi'(x_3),$$ where $\phi'=\phi+\epsilon$. Expanding this equation to first order in $\epsilon$, we find $$0=\int \mathcal{D}\phi e^{i\int d^4x\, \mathcal{L}}\left\{ \left( i\int d^4x\, \epsilon(x)\left[ (-\partial^2-m^2)\phi(x)\right] \phi(x_1)\phi(x_2)\phi(x_3)\right) \\ +\epsilon(x_1)\phi(x_2)\phi(x_3)+\phi(x_1)\epsilon(x_2)\phi(x_3)+\phi(x_1)\phi(x_2)\epsilon(x_3)\right\}.$$ The last three terms can be combined with the first by writing, for instance, $\epsilon(x_1)=\int d^4x\, \epsilon(x)\delta(x-x_1)$. Noting that the right-hand side must vanish for any possible variation $\epsilon(x)$, we then obtain $$0=\int \mathcal{D}\phi e^{i\int d^4x\, \mathcal{L}}\left[ (\partial^2+m^2)\phi(x) \phi(x_1)\phi(x_2)\phi(x_3)\\ +i\delta(x-x_1)\phi(x_2)\phi(x_3)+i\phi(x_1)\delta(x-x_2)\phi(x_3)+i\phi(x_1)\phi(x_2)\delta(x-x_3)\right].$$ A similar equation holds for any number of fields $\phi(x_i)$.
Let see only $\phi(x_1)$. Notice that the derivatives acting on $\phi(x)$ can be pulled outside the functional integral. Then dividing by $Z$ yields the identity $$(\partial^2+m^2)\langle \Omega|T\phi(x)\phi(x_1)|\Omega\rangle=-i\delta(x-x_1).$$ The left-handed side of this relation is the Klein-Gordon operator acting on a correltation function of $\phi(x)$. The right-hand side is zero unless $x=x_1$; that is, the correlation function satisfies the Klein-Gordon equation except at the point where the arguments of the two $\phi$ fields coincide. The modification of the Klein-Gordon equation oat this point is called a contact term. In quantum field theory, the classical equations of motion for fields are satisfied by all quantum correlation functions of thoes fields, up to contact terms.
As an example, consider the identity that for an $(n+1)$-point correlation function of scalar fields: $$(\partial^2+m^2)\langle \Omega|T\phi(x)\phi(x_1)\cdots\phi(x_n)|\Omega\rangle=\sum^n_{i=1}\langle\Omega T\phi(x_1)\cdots\left(-i\delta(x-x_i)\right)\cdots\phi(x_n)|\Omega\rangle.$$ This identity says that the Klein-Gordon equation is obeyed by $\phi(x)$ inside any expectation value, up to contact terms associated with the time ordering. The result can also be derived from the Hamiltonian formalism.
For a general field theory of a field $\phi(x)$, governed by the Lagrangian $\mathcal{L}[\varphi]$, we get identity $$0=\int \mathcal{D}\varphi\, e^{i\int d^4x\, \mathcal{L}}\left\{ i\int d^4x\, \epsilon(x)\frac{\delta}{\delta \varphi(x)}\left( \int d^4x'\, \mathcal{L}\right)\cdot \varphi(x_1)\varphi(x_2)\\ +\epsilon(x_1)\varphi(x_2)+\varphi(x_1)\epsilon(x_2)\right\},$$ and similar identities for correlation functions of $n$ fields. By the rule for functional differentiation, the derivative of the action is $$\frac{\delta}{\delta\varphi(x)}\left( \int d^4x'\, \mathcal{L}\right)=\frac{\partial\mathcal{L}}{\partial\varphi}-\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\varphi)}\right);$$ this is the quantity that equals zero by the Euler-Lagrange equation of motion for $\varphi$. This lead to the set of identities $$\left\langle \left( \frac{\delta}{\delta\varphi(x)}\int d^4x'\, \mathcal{L}\right)\varphi(x_1)\cdots\varphi(x_n)\right\rangle=\sum^n_{i=1}\left\langle \varphi(x_1)\cdots \left( i\delta(x-x_i)\right)\cdots \varphi(x_n)\right\rangle.$$ In this equation, the angle-brackets denote a time-ordered correlation function in which derivatives on $\varphi(x)$ are placed outside the time-ordering symbol. This relation states that the classical Euler-Lagrange equations of the field ip are obeyed for all Green’s functions of $\varphi$, up to contact terms arising from the nontrivial commutation relations of field operators. These quantum equations of motion for Green’s functions, including the proper contact terms, are called Schwinger-Dyson equations.
Conservation Laws
Let us consider the theory of a free, complex-valued scalar field, with the Lagrangian $$\mathcal{L}=|\partial_\mu\phi|^2-m^2|\phi|^2.$$ This Lagrangian is invariant under the transformation $\phi\rightarrow e^{i\alpha}\phi$. To find the quantum formulae, consider the infinitesimal change of variables $$\phi(x)\rightarrow \phi'(x)=\phi(x)+i\alpha(x)\phi(x).$$
The measure of functional integration is invariant under the transformation, since this is a unitary tranmsformation of the variables $\phi(x)$. Thus, for the case of two fields, $$\int\mathcal{D}\phi\, e^{i\int d^4x\, \mathcal{L}[\phi]}\phi(x_1)\phi^*(x_2)=\left| \int\mathcal{D}\phi\, e^{i\int d^4x\, \mathcal{L}[\phi']}\phi'(x_1)\phi'^*(x_2)\right|_{\phi'=(1+i\alpha)\phi}.$$ Expanding this equation to first order in $\alpha$, we find $$0=\int \mathcal{D}\phi\, e^{i\int d^4x\, \mathcal{L}}\left\{ i\int d^4x\, \left[ (\partial_\mu \alpha)\cdot i(\phi\partial^\mu \phi^*-\phi^*\partial^\mu \phi)\right] \phi(x_1)\phi^*(x_2)\\ +[i\alpha(x_1)\phi(x_1)]\phi^*(x_2)+\phi(x_1)[-i\alpha(x_2)\phi^*(x_2)]\right\}.$$ Integrate the term involving $\partial_\mu\alpha$ by parts. (Total derivative 0 by boundary setting) Then taking the coefficient of $\alpha(x)$ and dividing by $Z$ gives $$\langle_\mu j^\mu(x)\phi(x_1)\phi^*(x_2)\rangle=(-i)\left\langle (i\phi(x_1)\delta(x-x_1))\phi^*(x_2)\\ +\phi(x_1)(-i\phi^*(x_2)\delta(x-x_2))\right\rangle,$$ where $$j^\mu=i(\phi\partial^\mu\phi^*-\phi^*\partial^\mu\phi)$$ is the Noether current. This relation is the classical conservation law plus contact terms, that is , the Schwinger-Dyson equation associated wit hcurrent conservation.
Let's see current conservation in more general situations. Consider a local field theory of a set of fields $\varphi_a(x)$, governed by a Lagrangian $\mathcal{L}[\varphi]$. An infinitesimal symmetry transformation on the fields $\varphi_a$ will be of the general form $$\varphi_a(x)\rightarrow \varphi_a(x)+\epsilon\Delta\varphi_a(x).$$ We assume that the action is invariant under this transformation. Then, if ther parameter $\epsilon$ is taken to be constant, the Lagrangian must be invariant up to a total divergence: $$\mathcal{L}[\varphi]\rightarrow \mathcal{L}[\varphi]+\epsilon\partial_\mu\mathcal{J}^\mu.$$ If the symmetry parameter $\epsilon$ depends on $x$, as in the analysis of the previous paragraph, the variation of the Lagrangian will be slightly more complicated: $$\mathcal{L}[\varphi]\rightarrow \mathcal{L}[\varphi]+(\partial_\mu\epsilon)\Delta\varphi_a\frac{\partial\mathcal{L}}{\partial(\partial_\mu\varphi_a)}+\epsilon\partial_\mu\mathcal{J}^\mu.$$ Summation over the index $a$ is understood. Then $$\frac{\delta}{\delta \epsilon(x)}\int d^4x\, \mathcal{L}[\varphi+\epsilon\Delta\varphi]=-\partial_\mu j^\mu(x),$$ where $j^\mu$ is the Noether current, $$j^\mu=\frac{\partial\mathcal{L}}{\partial)\partial_\mu\varphi_a)}\Delta\varphi_a-\mathcal{J}^\mu.$$ Use result, we find the Schwinger-Dyson equation: $$\langle\partial_\mu j^\mu(x)\varphi_a(x_1)\varphi_b(x_2)\rangle=(-i)\left\langle (\Delta\varphi_a(x_1)\delta(x-x_1))\varphi_b(x_2)\\ +\varphi_a(x_1)(\Delta\varphi_b(x_2)\delta(x-x_2))\right\rangle.$$ A similar equation can be found for the correlator of $\partial_\mu j^\mu$ with $n$ fields $\varphi(x)$. These give the full set of Schwinger-Dyson equations associated with the classical Noether theorem.
As an example of the use of this variational procedure to obtain the Noether current, consider the symmetry of the Lagrangian with respect to spacetime translations. Under the transformation $$\varphi_a\rightarrow \varphi_a+a^\mu(x)\partial_\mu\varphi_a$$ the Lagrangian transforms as $$\mathcal{L}\rightarrow \mathcal{L}+\partial_\nu a^\mu\partial_\mu \varphi_a\frac{\partial\mathcal{L}}{\partial(\partial_\nu\varphi_a)}+a^\mu\partial_\mu\mathcal{L}.$$ The variation of $\int d^4x\, \mathcal{L}$ with respect to $a^\mu$ then gives rise to the conservation equation for the energy-momentum tensor $\partial_\nu T^{\mu\nu}=0$, with $$T^{\mu\nu}=\frac{\partial\mathcal{L}}{\partial(\partial_\nu\varphi_a)}\partial^\mu\varphi_a-g^{\mu\nu}\mathcal{L}.$$
The Ward-Takahashi Identity
Consider the QED functional integral, the change of variables $$\psi(x)\rightarrow (1+ie\alpha(x))\psi(x),$$ without the corresponding term in the transformation law for $A_\mu$. The QED Lagrangian transforms according to $$\mathcal{L}\rightarrow \mathcal{L}-e\partial_\mu\alpha\bar{\psi}\gamma^\mu\psi.$$ The transformation thus leads to the following identity for the functional integral over two fermion fields: $$0=\int \mathcal{D}\bar{\psi}\, \mathcal{D}\psi\, \mathcal{D}A\, e^{i\int d^4x\, \mathcal{L}}\left\{ -i\int d^4x\, \partial_\mu\alpha(x)\left[ j^\mu(x)\psi(x_1)\bar{\psi}(x_2)\right]\\ +(ie\alpha(x_1)\psi(x_1))\bar{\psi}(x_2)+\psi(x_1)(-ie\alpha(x_2)\bar{\psi}(x_2))\right\},$$ with $j^\mu=e\bar{\psi}\gamma^\mu\psi$. As in our other examples, an analogous equation holds for any number of fermion fields.
Dividing above relation by $Z$, we find $$i\partial_\mu\langle 0|Tj^\mu(x)\psi(x_1)\bar{\psi}(x_2)|0\rangle =-ie\delta(x-x_1)\langle 0|T\psi(x_1)\bar{\psi}(x_2)|0\rangle \\ +ie\delta(x-x_2)\langle 0|T\psi(x_1)\bar{\psi}(x_2)|0\rangle.$$ To put thus equation into a more familiar form, compute its Fourier transform by integrating: $$\int d^4x\, e^{-ik\cdot x}\int d^4x_1\, e^{+iq\cdot x_1}\int d^4x_2\, e^{-ip\cdot x_2}.$$ Then the amplitudes are converted to the amplitudes $\mathcal{M}(k;p;q)$ and $\mathcal{M}(p;q)$. Indeed, it became form $$-ik_\mu\mathcal{M}^\mu(k;p;q)=-ie\mathcal{M}_0(p;q-k)+ie\mathcal{M}_0(p+k;q).$$ This is exactly the Ward-Takahashi identity for two external fermions.
[Again] 19.2 The Axial Current in Four Dimensions
Chiral Transformation of the Functional Integral
A third way of understanding the Adler-Bell-Jackiw anomaly comes from analyzing the conservation law for the axial vector current from the functional integral for the fermion fields.
We first review the standard derivation of the axial vector Ward identities following the method of Section 9.6. Starting from the fermionic functional integral $$\begin{align}Z=\int \mathcal{D}\, \psi\mathcal{D}\bar{\psi}\exp \left[ i\int d^4x\, \bar{\psi}(i\mathcal{D})\psi\right],\end{align}$$ make the change of variables $$\begin{align}\psi(x) \rightarrow& \psi'(x) =(1+i\alpha(x)\gamma^5)\psi(x),\nonumber\\ \bar{\psi}(x) \rightarrow& \bar{\psi}'(x) = \bar{\psi}(1+i\alpha(x)\gamma^5).\end{align}$$ Since the global chiral rotation, with constant $\alpha$, is a symmetry of the Lagrangian, the only new terms in the Lagrangain that result from (62) contain derivatives of $\alpha$. Thus, $$\begin{align}\int d^4x\, \bar{\psi}'(iD\!\!\! /)\psi'=\int d^4x\left[ \bar{\psi}(iD\!\!\! /)\psi-\partial_\mu \alpha(x)\bar{\psi}\gamma^\mu\gamma^5\psi\right] \nonumber\\ =\int d^4x\, \left[ \bar{\psi}(iD\!\!\! /)\psi +\alpha(x)\partial_\mu (\bar{\psi}\gamma^\mu\gamma^5\psi)\right].\end{align}$$ Then, by varying the Lagrangian with respect to $\alpha(x)$, we derive the classical conservation equation for the axial current. By carrying out a similar manipulation on the functional expression for a correlation function, we would derive the associated Ward identities.
In the argument just given, we assumed that the functional measure does not change when we change variables from $\psi'(x)$ to $\psi$. This seems reasonable, because the relation of $\psi'$ and $\psi$ in (62) looks like a unitary transformation. However, we should examine this point more closely. First, we must carefully define the functional measure. To do this, expand the fermion field in a basis of eigenstates of $D\!\!\!\! /$. Define right and left eigenvectros of $D\!\!\!\! /$ by $$\begin{align}(iD\!\!\!\! /)\phi_m=\lambda_m\phi_m,\quad \hat{\phi}_m(iD\!\!\!\! /)=-iD_\mu\hat{\phi}_m\gamma^\mu=\lambda_m\hat{\phi}_m.\end{align}$$ For zero background $A_\mu$ field, these eigenstates are Dirac wavefunctions of definite momentum; the eigenvalues satisfy $$\begin{align}\lambda^2_m=k^2=(k^0)^2-(\mathbf{k})^2.\end{align}$$ For a fixed $A_\mu$ field, this is also the asymptotic form of the eigenvalues for large $k$. These eigenfunctions give us a basis we can use to expand $\psi$ and $\bar{\psi}$: $$\begin{align}\psi(x)=\sum_m a_m\phi_m(x),\quad \bar{\psi}(x)=\sum_m \hat{a}_m\hat{\phi}_m(x),\end{align}$$ where $a_m,\hat{a}_m$ are anticommuting coefficients multiplying the $c$-number eigenfunctions (64). The functional measure over $\psi,\bar{\psi}$ can then be defined as $$\begin{align}\mathcal{D}\psi\mathcal{D}\bar{\psi}=\prod_m da_m\, d\hat{a}_m,\end{align}$$ and the functional measure over $\psi',\bar{\psi}'$ can be defined in the same way.
If $\psi'(x)=(1+i\alpha(x)\gamma^5)\psi(x)$, the expansion coefficients of $\psi$ and $\psi'$ are related by a infinitesimal linear transformation $(1+C)$, computed as follows: $$\begin{align}a'_m=\sum_n \int d^4x\, \phi^\dagger_m(x)(1+i\alpha(x)\gamma^5)\phi_n(x)a_n=\sum_n(\delta_{mn}+C_{mn})a_n.\end{align}$$ Then $$\begin{align}\mathcal{D}\psi'\, \mathcal{D}\bar{\psi}'=\mathcal{J}^{-2}\cdot \mathcal{D}\psi\mathcal{D}\bar{\psi},\end{align}$$ where $\mathcal{J}$ is the Jacobian determinant of the transformation $(1+C)$. THe inverse of $\mathcal{J}$ appears in (69) as a result of the fermionic integration. To evaluate $\mathcal{J}$, we write $$\begin{align}\mathcal{J}=\det (1+C)=\exp [\mbox{tr}\ln (1+C)]=\exp \left[ \sum_n C_{nn}+\cdots\right],\end{align}$$ and we can ignore higher order terms in the last line because $C$ is infinitesimal. Thus, $$\begin{align}\ln \mathcal{J}=i\int d^4x\, \alpha(x)\sum_n \phi^\dagger_n(x)\gamma^5\phi_n(x).\end{align}$$ The coefficient of $\alpha(x)$ looks like $\mbox{tr}[\gamma^5]=0$. However, we must regularize the sum over eigenstates $n$ in a gauge-invariant way. The natural choice is $$\begin{align}\sum_n\phi^\dagger_n(x)\gamma^5\phi_n(x)=\lim_{M\rightarrow \infty}\sum_n \phi^\dagger_n(x)\gamma^5\phi_n(x)e^{\lambda^2_n/M^2}.\end{align}$$ As (65) indicates, the sign of $\lambda^2_n$ will be negative at large momentum after a Wick rotation; thus, the sign in the exponent of the convergence factor is given correctly. We can write (72) in an operator form $$\begin{align}\sum_n\phi^\dagger_n(x)\gamma^5\phi_n(x)=\lim_{M\rightarrow\infty}\sum_n\phi^\dagger_n(x)\gamma^5e^{(iD\!\!\! /)^2/M^2}\phi_n(x)\nonumber\\ =\lim_{M\rightarrow \infty}\langle x|\mbox{tr}\left[ \gamma^5e^{(iD\!\!\! /)^2/M^2}\right]|x\rangle,\end{align}$$ where, in the second line, we trace over Dirac indices.
To evaluate (73), we rewrite $(iD\!\!\!\! /)^2$ as $$\begin{align}(iD\!\!\!\! /)^2=-D^2+\frac{e}{2}\sigma^{\mu\nu}F_{\mu\nu},\end{align}$$ with $\sigma^{\mu\nu}=\frac{i}{2}[\gamma^\mu,\gamma^\nu]$. Since we are taking the limit $M\rightarrow \infty$, we can concentrate our attention on the asymptotic part of the spectrum, where the momentum $k$ is large and we can expand in powers of the gauge field. To obtain a nonzero trace with $\gamma^5$, we must bring down four Dirac matrices from the exponent. The leading term is given by expanding the exponent to order $(\sigma\cdot F)^2$, and then ignoring the background $A_\mu$ field in all other terms. This gives $$\begin{align}\lim_{M\rightarrow \infty}\langle x|\mbox{tr}\left[ \gamma^5e^{(-D^2+(e/2)\sigma\cdot F)/M^2}\right]|x\rangle\nonumber\\ = \lim_{M\rightarrow \infty} \mbox{tr}\left[ \gamma^5\frac{1}{2!}\left( \frac{e}{2M^2}\sigma^{\mu\nu}F_{\mu\nu}(x)\right)^2\right] \langle x|e^{-\partial^2/M^2}|x\rangle.\end{align}$$ The matrix element in (75) can be evaluated by a Wick rotation: $$\begin{align}\langle x|e^{-\partial^2/M^2}|x\rangle=\lim_{x\rightarrow y}\int \frac{d^4k}{(2\pi)^4}e^{-ik\cdot (x-y)}e^{k^2/M^2}\nonumber\\ =i\int \frac{d^4k_E}{(2\pi)^4}e^{-k_E^2/M^2}=i\frac{M^4}{16\pi^2}.\end{align}$$ Then (75) reduces to $$\begin{align}\lim_{M\rightarrow \infty}\frac{-ie^2}{8\cdot 16\pi^2}M^4\mbox{tr}\left[ \gamma^5\gamma^\mu\gamma^\nu\gamma^\lambda\gamma^\sigma \frac{1}{(M^2)^2}F_{\mu\nu}F_{\lambda\sigma}(x)\right]\nonumber\\ =-\frac{e^2}{32\pi^2}\epsilon^{\alpha\beta\mu\nu}F_{\alpha\beta}F_{\mu\nu}(x).\end{align}$$ Thus, $$\begin{align}\mathcal{J}=\exp \left[ -i\int d^4x\, \alpha(x)\left( \frac{e^2}{32\pi^2}\epsilon^{\mu\nu\lambda\sigma}F_{\mu\nu}F_{\lambda\sigma}(x)\right)\right].\end{align}$$ In all, we find that, after the change of variables (62), the functional integral (61) takes the form $$\begin{align}Z=\int \mathcal{D}\psi\, \mathcal{D}\bar{\psi}\, \exp \left[ i\int d^4x\, \left( \bar{\psi}(iD\!\!\!\! /)\psi +\alpha(x)\left\{ \partial_\mu j^{\mu5}+\frac{e^2}{16\pi^2}\epsilon^{\mu\nu\lambda\sigma}F_{\mu\nu}F_{\lambda\sigma}\right\}\right)\right].\end{align}$$ Varying the exponent with respect to $\alpha(x)$, we find precisely the Adler-Bell-Jackiw anomaly equation.
This derivation of the axial vector anomaly is especially interesting because it generalizes readily to any even dimensionality. The functional derivation always picks out for the right-hand side of the anomaly equation the pseudoscalar operator built from gauge fields that has the same dimension, d, as the divergence of the current. In two dimensions, this derivation leads immediately to (18). As long as d is even, we can always construct a matrix $\gamma^5$ that anticommutes with all of the Dirac matrices by taking their product. Then, the functional derivation leads straightforwardly to the result $$\begin{align}\partial_\mu j^{\mu5}=(-1)^{n+1}\frac{2e^n}{n!(4\pi)^n}\epsilon^{\mu_1\mu_2\cdots\mu_{2n}}F_{\mu_1\mu_2}\cdots F_{\mu_{2n-1}\mu_{2n}},\end{align}$$ where $n=d/2$.
At the end of the previous section, we argued that the axial vector anomaly leads to global nonconservation of fermionic charges in a two-dimensional system with a macroscopic electric field. In the same way, the four-dimensional anomaly equation leads to global nonconservation of the number of left- and right-handed fermions in background fields in which the right-hand side of (45) is nonzero. These are field configurations with parallel electric and magnetic fields. In Problem 19.1, we work out an example of four-dimensional massless fermions in a simple situation of this type and show that the fermion numbers are indeed violated, in a manner similar to what we saw at the end of Section 19.1, in accord with the Adler-Bell-Jackiw anomaly.
19.3 Goldstone Bosons
Schwarz intro
Spontaneous Breaking of Chiral Symmetry
Anomalies of Chiral Currents
19.4 Chiral Anomalies and Chiral Gauge Theories
We will focus primarily on theories of massless fermions. If the Lagrangian contains no fermion mass terms, it has no terms that mix the two helicity states of a Dirac fermion. Thus, in a theory that contains massless Dirac fermions $\psi_i$, we can write the kinetic energy term in the helicity basis (3.36) as $$\begin{align}\mathcal{L}=\psi^\dagger_{Li}i\bar{\sigma}\cdot\partial\psi_{Li}+\psi^\dagger_{Ri}i\sigma\cdot\partial\psi_{Ri}.\end{align}$$ There is no difficulty in coupling this system to a gauge field by assigning the left-handed fields $\psi_{Li}$ to one representation of the gauge group $G$ and assigning the right-handed fields to a different representation. For example, we might assign the left-handed fields to a representation $r$ of $G$ and take the right-handed fields to be invariant under $G$. This gives $$\begin{align}\mathcal{L}=\psi^\dagger_{Li}i\bar{\sigma}\cdot D\psi_{Li}+\psi^\dagger_{Ri}i\sigma\cdot \partial\psi_{Ri},\end{align}$$ with $D_\mu=\partial_\mu-igA^a_\mu t^a_r$. In more conventional notation, (82) becomes $$\begin{align}\mathcal{L}=\bar{\psi}i\gamma^\mu\left( \partial_\mu-igA^a_\mu t^a_r\left( \frac{1-\gamma^5}{2}\right)\right)\psi.\end{align}$$ It is straightforward to verify that the classical Lagrangian (82) is invariant to the local gauge transformation $$\begin{align}\psi\rightarrow \left( 1+i\alpha^a t^a\left( \frac{1-\gamma^5}{2}\right)\right)\psi,\nonumber\\ A^a_\mu\rightarrow A^a_\mu+\frac{1}{g}\partial_\mu \alpha^a+f^{abc}A^b_\mu \alpha^c,\end{align}$$ which generalizes (15.46). Since the right-handed fields are free fields, we can even eliminate these fields and write a gauge-invariant Lagrangian for purely left-handed fermions.
To work out the general properties of chirally coupled fermions, it is useful to rewrite their Lagrangian with one further transformation. Below Eq. (3.38), we noted that the quantity $\sigma^2\psi^*_R$ transforms under Lorentz transformations as a left-handed field. Thus it is useful to rewrite the right-handed components in (81) as new left-hnaded fermions, by defining $$\begin{align}\psi'_{Li}=\sigma^2\psi^*_{Ri},\quad \psi'^\dagger_{Li}=\psi^T_{Ri}\sigma^2.\end{align}$$ This transformation relables the right-handed fermions as antifermions and calls their left-handed antiparticles a new species of left-handed fermions. By using (3.38), we can rewrite the Lagrangian for the right-handed fermions as $$\begin{align}\int d^4x\, \psi^\dagger_{Ri}i\sigma\cdot \partial\psi_{Ri}=\int d^4x\, \psi'^\dagger_{Li}i\bar{\sigma}\cdot \partial \psi'_{Li}.\end{align}$$ The minus sign from fermion interchange cancels the minus sign from integration by parts. Notice that, if the fermions are coupled to gauge fields in the representation $r$, this manipulation changes the covariant derivative as follows: $$\begin{align}\psi^\dagger_Ri\sigma\cdot (\partial-igA^at^a_r)\psi_R=\psi'^\dagger i\bar{\sigma}\cdot (\partial+igA^a(t^a_r)^T)\psi'_L\nonumber\\ \psi'^\dagger_Li\bar{\sigma}\cdot (\partial-igA^a t^a_\bar{r})\psi'_L.\end{align}$$ Thus the new fields $\psi'_L$ belong to the conjugate representation to $r$, for which the representation matrices are given by (15.82). In this notation, QCD with $n_f$ flavors of massless fermions is rewritten as an $SU(3)$ gauge theory coupled to $n_f$ amssless fermions in the $3$ and $n_f$ massless fermions in the $\bar{3}$ representation of $SU(3)$. The most general gauge theory of massless fermions would simply assign left-handed fermions to an arbitrary, reducible representation $R$ of the gauge group $G$. We have just seen that rewriting a system of Dirac fermions leads to $R=r\oplus \bar{r}$, a real representation in the sense described below be rewritten in terms of Dirac fermions and is intrinsically chiral.
The rewriting (85) transforms the mass term of QCD Lagrangian as follows: $$\begin{align}m\bar{\psi}_i\psi_i=m(\psi^\dagger_R\psi_L+\mbox{h.c.})=-m(\psi'^T_{Li}+\mbox{h.c.}).\end{align}$$ This has the form of the Majorana mass term. The most general mass term that can be built purely from left-handed fermion fields is $$\begin{align}\Delta \mathcal{L}_M=M_{ij}\psi_{Li}^T \sigma^2 \psi_{Lj} + \mbox{h.c.}\end{align}$$ The matrix $M_{ij}$ is symmetric under $i\leftrightarrow j$, since the minus sign from the antisymmetry of $\sigma^2$ is compensated by a minus sign from fermion interchange. This mass term is gauge invariant if $M_{ij}$ is invariant under $G$. For example, the mass term in (88) couples $3$ and $\bar{3}$ indices together in an $SU(3)$ singlet combination. In general, a gauge-invariant mass term exists if the representation containing the fermions is strictly real, in sense described below (15.82). In an intrinsically chiral theory, there is no possible gauge-invariant mass term. We will see in the next chapter that, in the gauge theory of the weak interactions, mass terms for the quarks and leptons are forbidden by gauge invariance.
At the classical level, there is no restriction on the representation $R$ of the left-handed fermions. However, at the level of one-loop corrections, many posible choices become inconsistent due to the axial vector anomaly. In a gauge theroy of left-handed massless fermions, consider computing the diagrams of Fig. 19.9, in which the external fields are non-Abelian gauge bosons and the marked vertex represents the gauge symmetry current $$\begin{align}j^{\mu a}=\bar{\psi}\gamma^\mu \left(\frac{1-\gamma^5}{2}\right) t^a\psi.\end{align}$$ The gauge boson vertices also contain factors of $(1-\gamma^5)/2$. The three pojectors can be moved together into a single factor. Then, if we regularize this diagram as in Section 19.2, the term containing a $\gamma^5$ has an axial vector anomaly that leads to the relation $$\begin{align}\langle p,\nu,b;k,\lambda,c|\partial_\mu j^{\mu a}|0\rangle =\frac{g^2}{8\pi^2}\epsilon^{\alpha\nu\beta\lambda}p_\alpha k_\beta\cdot \mathcal{A}^{abc},\end{align}$$ where $\mathcal{A}^{abc}$ is a trace over group matrices in the representation $R$: $$\begin{align}\mathcal{A}^{abc}=\mbox{tr}[t^a\{ t^b,t^c\}].\end{align}$$ This equation implies that, unless $\mathcal{A}^{abc}$ vanishes, the current $j^{\mu a}$ is not conserved. The factor (92) is totally symmteric in $(a,b,c)$, so this condition is independent of which current is treated as an external operator.
Since the whole construction of a theory with local gauge invariance is based on the existence of an exact global symmetry, the violation of the conservation of $j^{\mu a}$ does violence to the structure of the theory. For example, triangle diagrams of the form of Fig. 19.9 will now generate divergent gauge boson mass terms and will upset the delicate relations between three- and four-point vertices discussed in Chapter 16. These relations, following from the Ward identity, were necessary to insure the cancellation of unphysical states and the unitarity of the $S$-matrix. The only way to avoid this problem is to insist that $\mathcal{A}^{abc} = 0$ as a fundamental consistency condition for chiral gauge theories. Gauge theories satisfying this condition are said to be anomaly free.
If the fermion representation $R$ is real, $R$ is equivalent to its conjugate representation $\bar{R}$. Thu,s as we described below (15.82), $t^a_R$ is related by a unitary transformation to $t^a_\bar{R}=-(t^a_R)^T$. Since (92) is invariant to unitary transformations of the $t^a$, we can replace $t^a_R$ by $t^a_\bar{R}$. Then $$\begin{align}\mathcal{A}^{abc}=\mbox{tr}[(-t^a)^T\{(-t^b)^T,(-t^c)^T\}]=-\mbox{tr}[\{t^c,t^b\} t^a]=-\mathcal{A}^{abc}.\end{align}$$ Thus, is $R$ is real, the gauge theory is automatically anomaly free. As a special case, any gauge theory of Dirac fermions is anomaly free.
Reference
Peskin&Schroeder - An Introduction to Quantum Field Theory