Current Correlators and Polarization Tensors

The previous page derived Ward identities from local symmetry transformations. We now apply the same logic to the two-point function of a current. In QED this is the vacuum-polarization tensor. Here we calculate the time-ordered kernel generated by a vacuum in-out functional; a causal response to an applied field uses its retarded counterpart. For a massless Dirac fermion in two dimensions, the calculation also shows why vector gauge invariance and axial-current conservation cannot both survive quantization.

The central lesson is that a current-current correlator is not just a separated-point loop diagram. The product $Tj^\mu(x)j^\nu(y)$ is singular at $x=y$, so the full polarization tensor is a distribution. It may contain contact terms supported at coincident points. Those terms are local in position space and polynomial in momentum space; their allowed combinations are constrained by the Ward identities we preserve. A calculation that keeps only separated points can therefore get the nonlocal part right while still violating the exact Ward identity.

Thus the useful slogan is

$$\text{conserved vector current} \quad\Longrightarrow\quad \text{transverse polarization tensor},$$

but “polarization tensor” means the regulated loop plus its local counterterms.

Required background. Ward identities and chiral symmetries supplies the source-functional Ward identity and the two-dimensional vector/axial-current relation. Helpful background. Vacuum polarization and gauge-invariant counterterms gives the higher-dimensional one-loop prototype; current-source derivatives distinguishes the vacuum Hessian, its contact terms and retarded response.

Current response from a background source

Source and light-cone conventions. Define the connected source functional by

$$Z[A]=e^{iW[A]} =\left\langle\exp\left(+i\int d^dx\,A_\mu j^\mu\right)\right\rangle_0.$$

This follows from $D_\mu=\partial_\mu-iA_\mu$, with $\psi\mapsto e^{+i\alpha}\psi$ and $A_\mu\mapsto A_\mu+\partial_\mu\alpha$.

The quadratic part of $W[A]$ is

$$W[A]=W[0]+{1\over2}\int_q A_\mu(-q)\Pi^{\mu\nu}(q)A_\nu(q)+O(A^3), \qquad \int_q\equiv\int {d^dq\over(2\pi)^d}.$$

In the two-dimensional sections,

$$x^\pm=x^0\pm x^1, \qquad \partial_\pm={1\over2}(\partial_0\pm\partial_1), \qquad \widehat A_\pm=A_0\pm A_1, \qquad q_\pm=q_0\pm q_1.$$

Thus $\widehat A_\pm$ and $q_\pm$ are full sums of Cartesian covector components, with $q_+q_-=q^2$. The coordinate one-form components used in lesson 23 are $A_\pm^{\rm coord}=\widehat A_\pm/2$. The inherited inverse Fourier transform gives

$$\delta\widehat A_\pm(q)=-iq_\pm\alpha(q), \qquad \delta\widehat A_\pm(-q)=+iq_\pm\alpha(-q).$$

Retain both phases when varying a quadratic functional.

As in lesson 19, $d^2x$ in the Lorentzian formulas on this page means the Cartesian measure $dx^0dx^1$, whose absolute Jacobian is $dx^0dx^1=\frac12dx^+dx^-$.

For Dirac slash notation, write

$$p\!\!\!/\equiv\gamma^\mu p_\mu, \qquad q\!\!\!/\equiv\gamma^\mu q_\mu, \qquad D\!\!\!/\equiv\gamma^\mu D_\mu.$$

The ratios $q_+/q_-$ below use these full-sum momenta. As in lesson 19, the propagation labels are $\psi_+=P_-\psi$ and $\psi_- =P_+\psi$; the subscripts on $\psi_\pm$ are not chirality eigenvalues. For one free massless Dirac fermion with a positive unit coupling to the external source, the covariant result is

$$\Pi^{\mu\nu}(q)=\frac1\pi \left(\eta^{\mu\nu}-\frac{q^\mu q^\nu}{q^2}\right).$$

The positive magnitude $\kappa=1/(4\pi)$ corresponds to the negative same-chirality coefficients derived below. Local counterterms can change polynomial contact terms, but cannot reverse the nonlocal part. The normalization agrees with the regulated Dirac-loop calculation in Morais and Mota 2009, §§ II–III, pp. 2–4, Eqs. (1)–(2) and (12)–(15), PDF; their opposite source-charge sign cancels between the two current insertions.

The two-dimensional formulas concern the vacuum on the noncompact plane and the quadratic functional of smooth, compactly supported sources. First read the algebra at $q^2\ne0$. On the null cone, all ratios inherit the same Feynman boundary value: for example, $q_+/q_-$ means $q_+^2/(q^2+i0)$. This does not prescribe independent ordinary division by a vanishing momentum component. Compact flux sectors, harmonic modes and fermion zero modes require additional data.

Couple a conserved current to a classical background field,

$$Z[A]=e^{iW[A]} =\int\mathcal D\Phi\, \exp\left(iS_0+i\int d^dx\,A_\mu j^\mu\right).$$

Then

$$\langle j^\mu(x)\rangle_A=+{\delta W[A]\over\delta A_\mu(x)}$$

and

$$\Pi^{\mu\nu}(x-y) ={\delta^2 W[A]\over\delta A_\mu(x)\delta A_\nu(y)}\bigg|_{A=0}.$$

At separated points this is the connected time-ordered correlator,

$$\Pi^{\mu\nu}(x-y)\big|_{x\neq y} =i\langle Tj^\mu(x)j^\nu(y)\rangle_{\rm conn}.$$

As a distribution, however, it can also contain local terms such as derivatives of $\delta^{(d)}(x-y)$. In momentum space these are polynomials in $q$. They are the contact terms.

Expanding $W[A]$ gives

$$\begin{aligned} W[A]=W[0] &+\int_q A_\mu(-q)\langle j^\mu(q)\rangle \\ &+{1\over2}\int_q A_\mu(-q)\Pi^{\mu\nu}(q)A_\nu(q)+O(A^3). \end{aligned}$$

Therefore, if the one-point current vanishes,

$$\langle j^\mu(q)\rangle_A =+\Pi^{\mu\nu}(q)A_\nu(q)+O(A^2).$$

This is the linear change of the in-out expectation value. A causal expectation value instead involves a retarded current kernel, obtained with the appropriate continuation or an in-in source construction. In particular, the time-ordered Hessian is not automatically a retarded response function. Changing the source coupling from $+A_\mu j^\mu$ to $-A_\mu j^\mu$ changes the first-derivative relation, but its two source factors leave the quadratic Hessian’s sign unchanged.

For later use, keep the sign convention separate from the physics. With $H_{\rm ext}=\int U\rho$, a positive potential energy tends to lower the density, so static density response is negative. With a local chemical-potential source $\delta\mu$, the coupling is $-\int\delta\mu\rho$ and the same compressibility appears with the opposite sign.

The vacuum current-current correlator generates the quadratic in-out functional of a background source. Gauge invariance requires the full polarization tensor, including contact terms, to be transverse; a causal response additionally requires the retarded prescription.

If $A_\mu$ is a background gauge field, vector gauge invariance means

$$W[A+\partial\alpha]=W[A].$$

Infinitesimally,

$$0=\delta W[A] =\int d^dx\,{\delta W\over\delta A_\mu(x)}\partial_\mu\alpha(x) =-\int d^dx\,\alpha(x)\partial_\mu{\delta W\over\delta A_\mu(x)}.$$

Since $\alpha(x)$ is arbitrary,

$$\boxed{ \partial_\mu{\delta W[A]\over\delta A_\mu(x)}=0. }$$

Differentiating once more with respect to $A_\nu(y)$ and setting $A=0$ gives

$$\partial_\mu^x\Pi^{\mu\nu}(x-y)=0.$$

In momentum space,

$$\boxed{ q_\mu\Pi^{\mu\nu}(q)=0. }$$

This is the transverse Ward identity.

Tensor structure and local counterterms

Lorentz invariance constrains the parity-even two-point tensor to be

$$\Pi^{\mu\nu}(q)=A(q^2)\eta^{\mu\nu}+B(q^2)q^\mu q^\nu.$$

Transversality gives

$$q_\mu\Pi^{\mu\nu}(q)=\left[A(q^2)+q^2B(q^2)\right]q^\nu=0,$$

so $A(q^2)=-q^2B(q^2)$. Hence

$$\boxed{ \Pi^{\mu\nu}(q) =\left(q^\mu q^\nu-q^2\eta^{\mu\nu}\right)\Pi(q^2) +\Pi_{\rm loc}^{\mu\nu}(q). }$$

In Euclidean signature the same structure is

$$\Pi_{\mu\nu}^{E}(q) =\left(q^2\delta_{\mu\nu}-q_\mu q_\nu\right)\Pi_E(q^2) +\Pi^{E}_{{\rm loc},\mu\nu}(q).$$

The local part must also be transverse if the vector current is gauged. For example,

$$\Delta W[A]=-{c\over4}\int d^dx\,F_{\mu\nu}F^{\mu\nu}$$

shifts the polarization tensor by a local transverse polynomial,

$$\Delta\Pi^{\mu\nu}(q) =-c\left(q^2\eta^{\mu\nu}-q^\mu q^\nu\right),$$

where the sign follows from the displayed $\Delta W$ and the inherited inverse Fourier transform. By contrast,

$${1\over2}m_A^2\int d^dx\,A_\mu A^\mu$$

would contribute $m_A^2\eta^{\mu\nu}$ and is not transverse. A local photon mass term is forbidden by exact gauge invariance.

A useful checklist is: nonlocal terms can carry physical long-distance information, while local transverse terms encode scheme choices and counterterms. Local non-transverse terms are allowed only when the source is not gauged or when the symmetry is explicitly broken.

This distinction matters later in two dimensions. A massless fermion loop can generate

$$A_\mu(-q)\left(\eta^{\mu\nu}-{q^\mu q^\nu\over q^2}\right)A_\nu(q).$$

This behaves like a mass term after gauge fixing, but it is nonlocal and transverse. It is therefore compatible with gauge invariance.

Ward contraction of the one-loop bubble

For a Dirac fermion with vector current

$$j^\mu=\bar\psi\gamma^\mu\psi,$$

write the raw quadratic Dirac kernel and its Feynman inverse as

$$K_0(p)=p\!\!\!/-m, \qquad G_F(p)=\frac{p\!\!\!/+m}{p^2-m^2+i0}, \qquad S(p)=iG_F(p).$$

Thus $K_0G_F=1$ as a boundary-value identity, whereas $K_0S=i$. The raw kernel $K_0$ is not the literal inverse of the contraction $S$. Expanding the fermion determinant, $W[A]=-i\operatorname{Tr}\log(K_0+\gamma^\mu A_\mu)$ up to an $A$-independent constant, gives the regulated loop

$$\begin{aligned} \Pi^{\mu\nu}_{\rm loop}(q) &=i\int_p \operatorname{tr}\left[ \gamma^\mu G_F(p+q)\gamma^\nu G_F(p) \right]\\ &=-i\int_p \operatorname{tr}\left[ \gamma^\mu S(p+q)\gamma^\nu S(p) \right], \end{aligned}$$

where the regulator and local completion are kept until the Ward identity is imposed. Contract with $q_\mu$. Since

$$q\!\!\!/=(p\!\!\!/+q\!\!\!/-m)-(p\!\!\!/-m) =K_0(p+q)-K_0(p),$$

the contracted bubble becomes

$$q_\mu\Pi^{\mu\nu}_{\rm loop}(q) =i\int_p\operatorname{tr}\left[ \left(K_0(p+q)-K_0(p)\right) G_F(p+q)\gamma^\nu G_F(p) \right].$$

Using $K_0G_F=1$ inside the trace, together with cyclicity, gives a difference of two one-propagator integrals. In terms of $S=iG_F$ it is

$$q_\mu\Pi^{\mu\nu}_{\rm loop}(q) = \int_p\operatorname{tr}\left[\gamma^\nu S(p)\right] - \int_p\operatorname{tr}\left[S(p+q)\gamma^\nu\right].$$

If these integrals were absolutely convergent, the second term could be shifted by $p\mapsto p-q$, and the difference would vanish. But the individual integrals are ultraviolet divergent. Shifting a divergent integral is not a harmless algebraic step; it is a statement about the regulator.

For a regulated divergent integral, shifting the integration variable removes one boundary strip and adds another. Their difference is the elementary origin of local surface terms in Ward-identity manipulations.

The one-dimensional analog is

$$I_\Lambda(a)=\int_{-\Lambda}^{\Lambda}dp\,\left[f(p+a)-f(p)\right].$$

For small $a$,

$$I_\Lambda(a) =a\int_{-\Lambda}^{\Lambda}dp\,f'(p)+O(a^2) =a\left[f(\Lambda)-f(-\Lambda)\right]+O(a^2).$$

If $f$ vanishes at both ends, the boundary term disappears. If $f$ approaches different limits, the boundary term survives. A divergent loop integral can behave the same way.

The lesson is not that gauge Ward identities are unreliable. It is that Ward identities are statements about regulated composite operators. A gauge-invariant regulator and a gauge-invariant counterterm prescription define the local terms so that

$$q_\mu\Pi^{\mu\nu}(q)=0.$$

If another classical symmetry is incompatible with this choice, that other symmetry becomes anomalous. This is why the anomaly is often visible as a finite “surface term” in a formally linearly divergent integral.

Vector and axial currents in two dimensions

In two dimensions the axial current is dual to the vector current. With the convention used on the previous page,

$$j_5^\mu=-\epsilon^{\mu\nu}j_\nu.$$

Suppose vector gauge invariance fixes the current-current correlator to be

$$\Pi^{\mu\nu}_{VV}(q) =C_V\left(\eta^{\mu\nu}-{q^\mu q^\nu\over q^2}\right).$$

For one massless Dirac fermion with the covariant unit-current normalization,

$$C_V={1\over\pi}.$$

The vector Ward identity is automatic:

$$q_\mu\Pi^{\mu\nu}_{VV}(q)=0.$$

Now dualize one index to form the axial-vector correlator,

$$\Pi^{\mu\nu}_{5V}(q) =-\epsilon^{\mu\rho}\Pi_{\rho}{}^\nu(q).$$

Its divergence is

$$\begin{aligned} q_\mu\Pi^{\mu\nu}_{5V}(q) &=-C_Vq_\mu\epsilon^{\mu\rho} \left(\delta_\rho{}^\nu-{q_\rho q^\nu\over q^2}\right) \\ &=-C_Vq_\mu\epsilon^{\mu\nu}, \end{aligned}$$

because $q_\mu\epsilon^{\mu\rho}q_\rho=0$. Equivalently,

$$\boxed{ q_\mu\Pi^{\mu\nu}_{5V}(q)=C_V\epsilon^{\nu\rho}q_\rho. }$$

Thus a nonzero transverse vector kernel implies a nonzero axial divergence. With the source term $+A_\nu j^\nu$, the in-out source expansion gives $\langle j_5^\mu\rangle_A=+\Pi_{5V}^{\mu\nu}A_\nu+O(A^2)$. In position space, for one positively charged unit Dirac fermion, this gives

$$\boxed{ \partial_\mu j_5^\mu =-{1\over2\pi}\epsilon^{\mu\nu}F_{\mu\nu} =-{1\over\pi}\epsilon^{\mu\nu}\partial_\mu A_\nu. }$$

The regulator has preserved vector gauge invariance. The axial current is anomalous.

Chiral bubbles in light-cone variables

The same conclusion can be seen without gamma-matrix traces. A massless Dirac fermion in two dimensions splits into right- and left-moving components,

$$S_0=\int d^2x\, \left(2i\psi_+^\dagger\partial_-\psi_+ +2i\psi_-^\dagger\partial_+\psi_-\right).$$

The chiral currents are

$$J_+=\psi_+^\dagger\psi_+, \qquad J_-=\psi_-^\dagger\psi_-.$$

The $\psi_+$ propagator has the distributional form

$$G_+(p)={i\over p_-+i0\operatorname{sgn}(p_+)},$$

and the $\psi_-$ propagator is

$$G_-(p)={i\over p_++i0\operatorname{sgn}(p_-)}.$$

The sign of the infinitesimal imaginary part is crucial. It is the remnant of the relativistic Feynman prescription $1/(p^2+i0)$ after the massless propagator has been factorized into chiral pieces.

The sign requires both the contour integral and the determinant prefactor. Let $K_+=2i\partial_-$ be the raw chiral kernel; its source is $\widehat A_-$ because $j^0=J_++J_-$ and $j^1=-J_++J_-$. Expanding its determinant gives

$$\begin{aligned} W_+[\widehat A_-] &=-i\operatorname{Tr}\log(K_++\widehat A_-),\\ W_+^{(2)} &=\frac{i}{2}\operatorname{Tr} \left(K_+^{-1}\widehat A_-K_+^{-1}\widehat A_-\right). \end{aligned}$$

The $J_+J_+$ Hessian therefore has an overall $i$. The other prefactor is the positive Cartesian momentum Jacobian $|dp^0dp^1|=\frac12|dp_+dp_-|$. The factor $2$ in $K_+$ cancels the $1/2$ in $\partial_-$, leaving the raw inverse denominator $p_-$. For a positive temporary $\epsilon$, the separated loop is

$$\Pi_{++}(q) =\lim_{\epsilon\downarrow0} {i\over2} \int\frac{dp_+}{2\pi}\int\frac{dp_-}{2\pi} {1\over p_-+i\epsilon\operatorname{sgn}(p_+)} {1\over p_-+q_-+i\epsilon\operatorname{sgn}(p_++q_+)}.$$

Perform the $p_-$ integral first and then take the common boundary value. A finite sign-$\epsilon$ deformation is useful for checking residues; it is not itself a covariant regulator defining all local contact terms.

For fixed $p_+$, the two poles in the complex $p_-$ plane lie on the same side unless $p_+$ and $p_++q_+$ have opposite signs. For $q_+>0$, this happens only in the strip

$$-q_+<p_+<0.$$

In this strip the upper pole is $p_-=+i\epsilon$, while the other pole is $p_-=-q_--i\epsilon$. Closing above, with positive orientation, gives

$$\int_{\mathbb R}\frac{dp_-}{2\pi} \frac{1}{(p_--i\epsilon)(p_-+q_-+i\epsilon)} =\frac{i}{q_-+2i\epsilon}.$$

The remaining strip has width $q_+$, so the two factors of $i$ fix the sign:

$$\frac{i}{2}\frac{q_+}{2\pi} \frac{i}{q_-+2i\epsilon} =-\frac{q_+}{4\pi(q_-+2i\epsilon)}.$$

For $q_+<0$, the strip is $0<p_+<-q_+$ and the upper pole comes from the second denominator. The contour integral is $-i/(q_--2i\epsilon)$; multiplication by the positive strip width $-q_+$ gives the same boundary value. Exchanging the two chiralities then yields, away from the null cone,

$$\boxed{ \Pi_{++}(q)=-\kappa{q_+\over q_-}, \qquad \Pi_{--}(q)=-\kappa{q_-\over q_+}, \qquad \kappa=\frac{1}{4\pi}. }$$

Distributionally these are $-\kappa q_+^2/(q^2+i0)$ and $-\kappa q_-^2/(q^2+i0)$ with the common prescription specified above. Both chiralities have the same sign and magnitude. The full-sum source components and the Cartesian measure together fix this normalization; converting only one of them gives a wrong factor.

At separated points, the mixed right-left correlator vanishes,

$$\Pi_{+-}^{\rm sep}(q)=0.$$

The figure separates the support argument from the determinant phase. Follow the strip first, then the two factors of $i$ that determine the signed coefficient.

For $q_+>0$ and non-null external momentum, opposite-side poles restrict the $J_+J_+$ loop to $-q_+<p_+<0$. The contour contributes $i/q_-$ in the boundary limit; the determinant contributes another $i$ and the Cartesian Jacobian is $1/2$, giving $\Pi_{++}=-q_+/(4\pi q_-)$. The strip is schematic, not to scale; finite $\epsilon$ checks the contour before its boundary limit.

The nonlocal ratios $q_+/q_-$ and $q_-/q_+$ are robust. A local counterterm cannot change them. What a local counterterm can change is the mixed component $\Pi_{+-}$.

Gauge-invariant completion in two dimensions

For the unrescaled projectors and Cartesian measure used above, $j^0=J_++J_-$ and $j^1=-J_++J_-$. The global source convention therefore gives the exact coupling

$$S_{\rm int}=+\int d^2x\, \left(\widehat A_-J_+ + \widehat A_+J_-\right).$$

The separated chiral bubbles give

$$\begin{aligned} W_{\rm sep}[A] =-{\kappa\over2}\int_q\bigg[& \widehat A_-(-q){q_+\over q_-}\widehat A_-(q) \\ &+\widehat A_+(-q){q_-\over q_+}\widehat A_+(q) \bigg]. \end{aligned}$$

Under the vector gauge transformation

$$\delta\widehat A_+(q)=-iq_+\alpha(q), \qquad \delta\widehat A_-(q)=-iq_-\alpha(q),$$

the negative-momentum leg varies with the opposite phase. Varying both legs and then relabeling $q\mapsto-q$ in one term gives the local variation

$$\delta W_{\rm sep} =-i\kappa\int_q \alpha(-q)\left[q_+\widehat A_-(q)+q_-\widehat A_+(q)\right].$$

This can be canceled by the local contact term

$$W_{\rm ct}[A] =+\kappa\int_q \widehat A_+(-q)\widehat A_-(q).$$

Indeed, its two-leg variation is $+i\kappa\int_q\alpha(-q)[q_+\widehat A_-(q)+q_-\widehat A_+(q)]$. Thus the gauge-invariant quadratic effective action is

$$\boxed{ \begin{aligned} W_{\rm gi}[A] ={\kappa\over2}\int_q\bigg[& -\widehat A_-(-q){q_+\over q_-}\widehat A_-(q) \\ &-\widehat A_+(-q){q_-\over q_+}\widehat A_+(q)\\ &+\widehat A_-(-q)\widehat A_+(q) +\widehat A_+(-q)\widehat A_-(q) \bigg]. \end{aligned} }$$

Equivalently,

$$\boxed{ W_{\rm gi}[A] ={\kappa\over2}\int_q \mathcal F_\Sigma(-q){1\over q_+q_-}\mathcal F_\Sigma(q), }$$

where the full-sum gauge-invariant momentum combination is

$$\mathcal F_\Sigma(q) :=q_+\widehat A_-(q)-q_-\widehat A_+(q).$$

The negative argument also reverses the momenta:

$$\mathcal F_\Sigma(-q) =-q_+\widehat A_-(-q)+q_-\widehat A_+(-q).$$

This minus sign produces the negative same-chirality terms when the bilinear is expanded. The two mixed terms become equal after integration and relabeling, not at an arbitrary individual momentum. Exercise 4 checks the expansion and a sign-sensitive timelike example.

With the inherited inverse Fourier transform and the coordinate components of lesson 23, $F_{+-}^{\rm coord}(q)=-i\mathcal F_\Sigma(q)/4$. This gives an explicit round trip from the full-sum notation to the curvature two-form.

The same round trip fixes the normalization without a convention-dependent remainder:

$${\mathcal F_\Sigma(-q)\mathcal F_\Sigma(q)\over q_+q_-} =4A_\mu(-q) \left(\eta^{\mu\nu}-{q^\mu q^\nu\over q^2}\right)A_\nu(q),$$

so $\kappa=1/(4\pi)$ reproduces the covariant kernel $\Pi^{\mu\nu}=(1/\pi)(\eta^{\mu\nu}-q^\mu q^\nu/q^2)$.

In the next figure, $X(q)$ denotes the sum of the two unsigned diagonal bilinears and $Y(q)$ the sum of both mixed bilinears:

$$\begin{aligned} X(q)&=\widehat A_-(-q)\frac{q_+}{q_-}\widehat A_-(q) +\widehat A_+(-q)\frac{q_-}{q_+}\widehat A_+(q),\\ Y(q)&=\widehat A_-(-q)\widehat A_+(q) +\widehat A_+(-q)\widehat A_-(q). \end{aligned}$$

Inspect the signs in $-X+Y$: the positive contact completes the negative chiral pieces into the curvature bilinear.

For the page’s full-sum sources and common boundary prescription, $W_{\rm sep}=-(\kappa/2)\int_qX$ and $W_{\rm ct}=(\kappa/2)\int_qY$. Their variations cancel and $-X+Y=\mathcal F_\Sigma(-q)\mathcal F_\Sigma(q)/q^2$. The negative-momentum argument is essential. This schematic sequence uses $\kappa=1/(4\pi)$ and the bilinears defined in the text; it does not represent an ordinary square at one momentum.

This is the cleanest way to see why the mixed polarization component is subtle. The separated diagram gives $\Pi_{+-}^{\rm sep}=0$. The full gauge-invariant distribution contains a local contact term. Without it, the vector Ward identity fails. This is the concrete two-dimensional example of the general warning at the start of the page: separated-point correlators and full response kernels are not the same object.

The expression $\mathcal F_\Sigma(1/q_+q_-)\mathcal F_\Sigma$ should be read as a nonlocal quadratic functional. After the explicit factor-four translation above, it is the two-dimensional version of $F(1/\Box)F$. It is gauge invariant because it is written in terms of the curvature combination, but it is not the same as a local Proca mass $A_\mu A^\mu$.

If $A_\mu$ is made dynamical, the induced action contains the transverse projector

$$A_\mu(-q) \left(\eta^{\mu\nu}-{q^\mu q^\nu\over q^2}\right) A_\nu(q).$$

The sign matters as well as transversality. With the rescaled Maxwell term $-F^2/(4e^2)$, the transverse inverse coefficient is

$$-\frac{q^2}{e^2}+\frac1\pi, \qquad q^2=m_\gamma^2=\frac{e^2}{\pi}>0.$$

For a timelike mode $q_\mu=(\omega,0)$ with $A_0=0$, $P_T^{11}=-1$: the same quadratic kernel is $K_{11}=\omega^2/e^2-1/\pi$. Reversing the induced coefficient would instead give a negative mass squared even though the tensor remains transverse. This is the massless Schwinger-model mechanism. The physical massive mode and the zero-flux, infinite-line assumptions are developed in the bosonization calculation.

Regulator choice and the anomaly

A gauge-invariant Pauli–Villars definition makes the symmetry choice explicit and checks the normalization without relying on the chiral split. Start with a small positive physical mass $m$ and a large regulator mass $M$, using the same vector coupling in both determinants:

$$\frac{\det(iD\!\!\!/-m)}{\det(iD\!\!\!/-M)}.$$

Normalize by the corresponding zero-source ratio. Combine the two loop integrands before removing the regulator or shifting the loop momentum. Their leading ultraviolet terms cancel. The massless physical limit is taken after the regulated calculation.

For a nonzero Euclidean momentum $Q$, the Dirac trace, a Feynman parameter $x$, and the convergent momentum integral give

$$\begin{aligned} \Pi_{E,ab}^{m,M}(Q) ={}&\frac{Q^2\delta_{ab}-Q_aQ_b}{\pi} \int_0^1 dx\,x(1-x)\\ &\times\left[ \frac{1}{m^2+x(1-x)Q^2} -\frac{1}{M^2+x(1-x)Q^2} \right]. \end{aligned}$$

This regulated, integrated tensor is already transverse. As $m\to0$, the first integral tends to $1/Q^2$; the second is bounded above by $1/(6M^2)$ and vanishes as $M\to\infty$. Thus

$$\Pi_{E,ab}(Q)=\frac1\pi \left(\delta_{ab}-\frac{Q_aQ_b}{Q^2}\right).$$

The Wick rotation carries the source one-form with it: $t=-i\tau$ gives $A_{E,\tau}=-iA_{L,0}$ and $A_{E,x}=A_{L,1}$. Together with $W_E=-iW_L$ on the rotated contour, this maps the positive Euclidean coefficient to the positive Lorentzian coefficient used above. This is the vector-gauge-preserving mass limit evaluated in Morais and Mota 2009, § V, p. 8, Eqs. (42)–(48), PDF. It also shows why a local contact cannot repair the wrong sign of the nonlocal term.

The heavy regulator preserves vector gauge invariance, but its mass couples left and right movers. It therefore breaks the axial rotation that assigns opposite phases to $\chi_+$ and $\chi_-$. When the physical mass has gone to zero, the surviving regulator contribution leaves the exact vector Ward identity

$$\partial_\mu j^\mu=0,$$

and the anomalous axial Ward identity

$$\partial_\mu j_5^\mu =-{1\over2\pi}\epsilon^{\mu\nu}F_{\mu\nu}.$$

The anomaly is the finite trace left by the ultraviolet regulator in the contact term of a current-current correlator.

This viewpoint is deliberately operational. To compute a current correlator, first regulate it; then add the local counterterms required by the symmetry you insist on preserving; only then ask whether another classical Ward identity still holds.

Bridge to many-body response

The contour argument in the chiral bubble will reappear in the next page. A nonrelativistic Fermi gas has particle and hole propagators whose poles can lie on opposite sides of the energy contour. The corresponding loop is nonzero only in restricted kinematic regions near the Fermi surface. The same analytic idea underlies density response, current response, and Fermi-surface instabilities.

In relativistic field theory, current correlators organize Ward identities and anomalies. In many-body theory, related kernels describe the response of a filled sea. The state and causal prescription must be chosen for that problem: a vacuum time-ordered loop cannot simply be relabeled as a retarded density response. In both settings, poles, contact terms and symmetry constrain the calculation.

Summary

The time-ordered current kernel is the quadratic Hessian of the connected in-out functional $W[A]$ with respect to a background source. For a non-anomalous vector current,

$$q_\mu\Pi^{\mu\nu}(q)=0.$$

Lorentz invariance then gives the transverse structure

$$\Pi^{\mu\nu}(q)= \left(q^\mu q^\nu-q^2\eta^{\mu\nu}\right)\Pi(q^2) +\text{local transverse terms}.$$

At one loop, the Ward contraction reduces to a difference of shifted integrals. If the integrals are divergent, the shift can leave a finite boundary term. A regulator and local counterterms are therefore part of the definition of the Ward identity.

In two dimensions, chiral current bubbles give

$$\Pi_{++}=-\kappa{q_+\over q_-}, \qquad \Pi_{--}=-\kappa{q_-\over q_+}.$$

The full-sum convention gives $\kappa=1/(4\pi)$ and the positive covariant unit-current coefficient $C_V=1/\pi$. The separated mixed correlator vanishes, but vector gauge invariance requires the positive mixed contact term in this normalization. The completed answer is

$$W_{\rm gi}[A] ={\kappa\over2}\int_q \mathcal F_\Sigma(-q){1\over q_+q_-}\mathcal F_\Sigma(q).$$

Preserving vector gauge invariance forces the axial Ward identity to contain the anomaly.

Common pitfalls

Ignoring contact terms. The separated-point diagram is not the full polarization tensor. The missing term can be local and still required by a Ward identity.

Shifting divergent integrals without saying how they are regulated. A formal loop-momentum shift is safe only for convergent integrals or for regulators that make the shift legitimate.

Mistaking a transverse nonlocal term for a forbidden local mass. The two-dimensional induced action behaves like a mass after gauge fixing, but it is written with a transverse projector and is gauge invariant.

Losing a Fourier or loop phase. The argument $-q$ reverses the momentum in $\mathcal F_\Sigma(-q)$. The residue supplies one $i$ and the fermion determinant supplies another. Transversality alone will not detect reversing the entire induced action; the covariant normalization and positive mass pole do.

Trying to conserve vector and axial currents simultaneously in the regulated theory. For the massless two-dimensional Dirac fermion, preserving vector gauge invariance forces the axial current to be anomalous.

Comparing absolute light-cone coefficients before translating conventions. The ratios $q_+/q_-$ and $q_-/q_+$ are invariant under harmless rescalings, but their prefactor is not. Compare the covariant tensor $C_V(\eta^{\mu\nu}-q^\mu q^\nu/q^2)$ or explicitly translate coordinates, currents, sources, and the measure.

Exercises

Exercise 1: Derive transversality from source gauge invariance

Starting from

$$W[A]=W[0]+{1\over2}\int_q A_\mu(-q)\Pi^{\mu\nu}(q)A_\nu(q)+O(A^3),$$

show that invariance under $A_\mu(q)\mapsto A_\mu(q)-iq_\mu\alpha(q)$ implies $q_\mu\Pi^{\mu\nu}(q)=0$. Keep the transformation of the negative-momentum leg and use $\Pi^{\mu\nu}(q)=\Pi^{\nu\mu}(-q)$.

Solution

The variation of the quadratic term is

$$\begin{aligned} \delta W^{(2)} =\frac12\int_q\big[ &+iq_\mu\alpha(-q)\Pi^{\mu\nu}(q)A_\nu(q)\\ &-iA_\mu(-q)\Pi^{\mu\nu}(q)q_\nu\alpha(q) \big]. \end{aligned}$$

In the second term, relabel $q\mapsto-q$, exchange $\mu$ and $\nu$, and use the stated symmetry. The reversed momentum supplies the extra minus sign, so the two contributions add:

$$\delta W^{(2)} =i\int_q \alpha(-q)q_\mu\Pi^{\mu\nu}(q)A_\nu(q).$$

Since $\alpha$ and $A_\nu$ are arbitrary,

$$q_\mu\Pi^{\mu\nu}(q)=0.$$

Exercise 2: Construct the transverse tensor structure

Assume Lorentz invariance and write

$$\Pi^{\mu\nu}(q)=A(q^2)\eta^{\mu\nu}+B(q^2)q^\mu q^\nu.$$

Use $q_\mu\Pi^{\mu\nu}=0$ to determine the transverse form.

Solution

Contracting with $q_\mu$ gives

$$q_\mu\Pi^{\mu\nu} =\left[A(q^2)+q^2B(q^2)\right]q^\nu.$$

For arbitrary $q^\nu$, transversality requires

$$A(q^2)=-q^2B(q^2).$$

Therefore

$$\Pi^{\mu\nu}(q) =B(q^2)\left(q^\mu q^\nu-q^2\eta^{\mu\nu}\right).$$

Renaming $B(q^2)=\Pi(q^2)$ gives the standard tensor structure.

Exercise 3: Locate the chiral bubble support strip

For $q_+>0$, consider the chiral bubble integral

$$I(q)=\int {dp_+dp_-\over(2\pi)^2} {1\over p_-+i0\operatorname{sgn}(p_+)} {1\over p_-+q_-+i0\operatorname{sgn}(p_++q_+)}.$$

Explain why only the strip $-q_+<p_+<0$ contributes. Replace $0$ temporarily by $\epsilon>0$ and evaluate the phase and coefficient of $I(q)$, integrating over $p_-$ first. Which additional factors convert it into $\Pi_{++}$?

Solution

For fixed $p_+$, the poles in the $p_-$ plane are

$$p_-=-i0\operatorname{sgn}(p_+), \qquad p_-=-q_- - i0\operatorname{sgn}(p_++q_+).$$

If $p_+$ and $p_++q_+$ have the same sign, the poles lie on the same side of the real $p_-$ axis, so the contour can be closed in the empty half-plane and gives zero.

For $q_+>0$, the signs are opposite only when

$$-q_+<p_+<0.$$

In that strip the upper pole is $p_-=+i\epsilon$. Its residue is $1/(q_-+2i\epsilon)$; closing above gives $i/(q_-+2i\epsilon)$ after the $1/(2\pi)$ integration factor. The remaining $p_+$ integral gives

$$\int_{-q_+}^{0}\frac{dp_+}{2\pi}=\frac{q_+}{2\pi}.$$

Hence

$$\begin{aligned} I_\epsilon(q)&=\frac{iq_+}{2\pi(q_-+2i\epsilon)},\\ \Pi_{++}(q)&=\lim_{\epsilon\downarrow0}\frac{i}{2}I_\epsilon(q) =-\frac{q_+}{4\pi q_-}. \end{aligned}$$

The extra $i$ is from the determinant Hessian and the extra $1/2$ is the Cartesian momentum Jacobian. The last ratio has the common boundary prescription; finite $\epsilon$ only defines this intermediate contour test.

Exercise 4: Complete the chiral bubbles gauge-invariantly

Starting from $\mathcal F_\Sigma(q)=q_+\widehat A_-(q)-q_-\widehat A_+(q)$, derive its expression at $-q$ and show that

$$\begin{aligned} \frac{\mathcal F_\Sigma(-q)\mathcal F_\Sigma(q)}{q_+q_-} ={}&-\widehat A_-(-q)\frac{q_+}{q_-}\widehat A_-(q) -\widehat A_+(-q)\frac{q_-}{q_+}\widehat A_+(q)\\ &+\widehat A_-(-q)\widehat A_+(q) +\widehat A_+(-q)\widehat A_-(q). \end{aligned}$$

Check the sign for $q_\mu=(\omega,0)$ with $A_0(\pm q)=0$ and $A_1(\pm q)=a$, and compare with $4A_\mu(-q)P_T^{\mu\nu}A_\nu(q)$.

Solution

Reversing the argument reverses the full-sum momenta:

$$\mathcal F_\Sigma(-q) =-q_+\widehat A_-(-q)+q_-\widehat A_+(-q).$$

Multiplying by $\mathcal F_\Sigma(q)$ gives negative diagonal terms $-q_+^2\widehat A_-(-q)\widehat A_-(q)$ and $-q_-^2\widehat A_+(-q)\widehat A_+(q)$, and positive mixed terms. Dividing by $q_+q_-$ yields the stated expression. The mixed terms combine into twice either one only after integration and $q\mapsto-q$ relabeling.

For the timelike example, $\widehat A_-=-a$, $\widehat A_+=a$ and $q_+=q_-=\omega$. Therefore

$$\begin{aligned} \mathcal F_\Sigma(q)&=-2\omega a, &\mathcal F_\Sigma(-q)&=+2\omega a,\\ \frac{\mathcal F_\Sigma(-q)\mathcal F_\Sigma(q)}{\omega^2}&=-4a^2. \end{aligned}$$

The covariant expression agrees because $P_T^{11}=-1$. Replacing the opposite-momentum bilinear by an ordinary square would instead give $+4a^2$. With the Maxwell term, the correct sign yields $K_{11}=\omega^2/e^2-1/\pi$ and the positive pole $\omega^2=e^2/\pi$.

Exercise 5: Recover the axial divergence from vector response

Let

$$\Pi_{VV}^{\mu\nu}(q) =C_V\left(\eta^{\mu\nu}-{q^\mu q^\nu\over q^2}\right), \qquad \Pi_{5V}^{\mu\nu}(q)=-\epsilon^{\mu\rho}\Pi_{\rho}{}^\nu(q).$$

Show that

$$q_\mu\Pi_{5V}^{\mu\nu}(q)=C_V\epsilon^{\nu\rho}q_\rho.$$

Solution

Compute

$$q_\mu\Pi_{5V}^{\mu\nu} =-C_Vq_\mu\epsilon^{\mu\rho} \left(\delta_\rho{}^\nu-{q_\rho q^\nu\over q^2}\right).$$

The second term vanishes because

$$q_\mu\epsilon^{\mu\rho}q_\rho=0.$$

Thus

$$q_\mu\Pi_{5V}^{\mu\nu} =-C_Vq_\mu\epsilon^{\mu\nu}.$$

Using $q_\mu\epsilon^{\mu\nu}=-\epsilon^{\nu\rho}q_\rho$, we obtain

$$q_\mu\Pi_{5V}^{\mu\nu}=C_V\epsilon^{\nu\rho}q_\rho.$$

References

• C. W. Morais and A. L. Mota, “Momentum Space Regularizations and the Indeterminacy in the Schwinger Model,” International Journal of Modern Physics A 26 (2011) 1991–2006, DOI: 10.1142/S0217751X11053067. Preprint: arXiv:0910.4322v1 (2009), Open PDF; page and equation locators above refer to this version.
• W. Pauli and F. Villars, “On the Invariant Regularization in Relativistic Quantum Theory”, Reviews of Modern Physics 21 (1949) 434–444.
• J. Schwinger, “Gauge Invariance and Mass. II”, Physical Review 128 (1962) 2425–2429.
• Y. Takahashi, “On the Generalized Ward Identity”, Il Nuovo Cimento 6 (1957) 371–375.
• J. C. Ward, “An Identity in Quantum Electrodynamics”, Physical Review 78 (1950) 182.

Further reading

• S. Coleman, Lectures of Sidney Coleman on Quantum Field Theory, B. Gin-ge Chen, D. Derbes, D. Griffiths, B. Hill, R. Sohn, and Y.-S. Ting (eds.), World Scientific (2019), lectures on Ward identities, regularization, and two-dimensional fermions.
• A. M. Polyakov, Gauge Fields and Strings, Harwood Academic Publishers (1987), discussions of two-dimensional field theory, gauge invariance, and anomalies.
• M. D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press (2014), chapters on vacuum polarization, Ward–Takahashi identities, and anomalies.
• M. Srednicki, Quantum Field Theory, Cambridge University Press (2007), sections on vacuum polarization, QED Ward identities, and anomalies.
• S. Weinberg, The Quantum Theory of Fields, Vols. I–II, Cambridge University Press (1995–1996), chapters on current correlators, gauge-theory renormalization, and anomalies.
• A. Zee, Quantum Field Theory in a Nutshell, 2nd ed., Princeton University Press (2010), discussions of vacuum polarization, two-dimensional fermions, and anomalies.