# Anomaly is due to the noninvariance of the path-integral under a symmetry. Is the noninvariance reflected on 1PI effective action?

When a symmetry is anomalous, the path integral $$Z=\int\mathcal{D}\phi e^{iS[\phi]}$$ is not invariant under that group of symmetry transformations $$G$$. This is because though the classical action $$S[\phi]$$ is invariant the measure may not be invariant. Since 1PI effective action $$\Gamma[\phi_{c}]$$ takes quantum corrections into account, I expect that it is not invariant under the symmetry. Is there a way to see/prove whether $$\Gamma[\phi_{c}]$$ is invariant under the anomalous symmetry? That requires one to know how $$\phi_c$$ changes under the symmetry.

The effective action $$\Gamma[\phi_c]$$ is the Legendre transform of the generator of connected correlation functions $$W[J]$$ which for most QFTs is defined as $$W[J]=-i\log Z[J]$$ $$Z[J]$$ is the generating functional/partition function of the theory. So you can write $$\Gamma[\phi_c]=W[J]-\int d^4x \ \phi_c(x) J(x)$$ where $$\phi_c=\langle \phi \rangle_{J}=\frac{\delta W}{\delta J}$$ is the vacuum expectation value in the presence of the external source $$J$$.

Ok so now, let's derive the Slavnov-Tylor identities in the case of an anomalous local gauge symmetry or a global symmetry: $$\phi'(x)\rightarrow \phi(x)+\delta \phi(x), \qquad \delta \phi(x) = \epsilon(x) F(x,\phi(x))$$ such that the classical action is left invariant $$S[\phi+\epsilon F]=S[\phi]$$

however if the symmetry is anomalous the functional measure will change: $$\mathcal{D}\phi\rightarrow \mathcal{D}(\phi+\epsilon F)=\mathcal{D}\phi\ e^{i\int d^4x\ \epsilon(x) \mathcal{A}(x)}$$ where $$\mathcal{A}$$ is the anomaly function in the parametrization we have given. Therefore the connected generating functional $$W$$ will transform as \begin{align} e^{i W[J]}&=\int \mathcal{D}\phi \exp\left[i S[\phi]+i\int d^4x\ \phi(x)J(x)\right]\\ &=\int \mathcal{D}\phi' \exp\left[i S[\phi']+i\int d^4x\ \phi'(x)J(x)\right]\\ &=\int \mathcal{D}\phi \exp\left[i S[\phi]+i\int d^4x \phi(x)J(x)+i\int d^4x \ \epsilon(x) F(x)J(x)+i\int d^4x\ \epsilon(x) \mathcal{A}(x)\right] \end{align} So, expanding to $$O(\epsilon)$$ we obtain the Ward identity $$\int d^4x \left(\mathcal{A}(x)+J(x)\langle F(x,\phi)\rangle _J\right)=0$$

now we can perform the Legendre tranformation and get the Slavnov-Tylor identity which is for the effective action: $$\int d^4x \left(\mathcal{A}(x)-\langle F(x,\phi)\rangle _{J_{\phi_c}}\frac{\delta \Gamma[\phi_c]}{\delta \phi_c(x)}\right)=0$$ the reason for this formula is that $$J=-\frac{\delta \Gamma[\phi_c]}{\delta \phi_c(x)}$$ from the Legendre tranform, while $${J_{\phi_c}}$$ means that $$J$$ is fixed by solving the equation $$\phi_c=\langle \phi \rangle_{J}=\frac{\delta W}{\delta J}$$ and so it correspond to a fixed value of $$\phi_c$$. So, the symmetry under which the effective action transforms is $$\phi_c\rightarrow \phi_c+\epsilon\langle F(x,\phi(x))\rangle_{J_{\phi_c}}$$

Up until now we have done all in general, however if we choose a linear transformation to be anomalous, i.e. $$F(x,\phi)=f(x)\phi(x)$$ depends linearly on the fields (which are most symmetry transformations), we obtain $$\int d^4x \left(\mathcal{A}(x)-F(x,\phi_c)\frac{\delta \Gamma[\phi_c]}{\delta \phi_c(x)}\right)=0$$ where the expectation value is replaced simply by $$F$$ calculated in $$\phi_c$$, since $$\langle f(x)\phi(x)\rangle=f(x)\phi_c$$.

Therefore, remembering that $$\epsilon F\equiv \delta \phi$$, for a linear simmetry we can write that the variation of the effctive action as $$\delta_\epsilon \Gamma[\phi_c]\equiv\int d^4x\ \delta\phi_c(x)\frac{\delta \Gamma[\phi_c]}{\delta \phi_c(x)}=\int d^4x\ \epsilon(x)\mathcal{A}(x)$$ so that if the symmetry was not anomalous the effective action would be invariant.

Therefore, for general non-linear symmetry transformation the relation between the field tranformation and the effective action is complicated and has no simple classical analogue since one has to take the expectation value of the transformation, on the other hand if the symmetry is linear then the effective action tranforms as the classical action with $$\phi_c$$ instead of $$\phi$$. In both cases the presence of the anomaly makes the variation of the effective action nonzero as we have shown.

EDIT: The derivation is perfectly valid for global symmetries too, just pull $$\epsilon$$ out of the integrals and ignore its $$x$$ dependence.

In the case of gauge symmetries is not that anomalies have to be removed, it is that for a gauge theory to be consistent the anomalies have to cancel. Suppose you have a gauge theory with massless fermions (only massless fermions contribute to the anomaly) $$\mathcal{L}=-\frac{1}{4} F^{\mu\nu}F_{\mu\nu}+\bar{\psi}\left(i\gamma_\mu D^{\mu}\right)\psi$$ where $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu+i g \left[A_\mu,A_\nu\right]$$ and $$D_\mu=\partial_\mu+i g A_\mu$$. The Fermionic effective action can be calculated exactley $$S_{eff} = -\frac{1}{4} \int d^4x\ F^{\mu\nu}F_{\mu\nu}+tr \log\left(i\gamma_\mu D^\mu[A]\right) = -\frac{1}{4} \int d^4x\ F^{\mu\nu}F_{\mu\nu}+S_F[A]$$

Now, if $$S_F[A]$$ is gauge invariant you can quantize the theory as usual with ghosts and all that and you get an unitary renormalizable theory. However, if $$S_F[A]$$ is not gauge invariant, the theory becomes inconsistent. In the presence of a gauge anomaly that's precisely the case, let's write the effective action before the path integration over the matter fields is done. And let'd do a gauge tranformation of the fields, with $$A'= A+\delta A$$ being a gauge tranformation of the gauge field \begin{align} e^{i S_F[A']}=&\int \mathcal{D}\psi\mathcal{D}\bar{\psi}\ e^{i\int d^4x\ \bar{\psi}\left(i\gamma_\mu D^{\mu}[A']\right)\psi} = \int \mathcal{D}\psi'\mathcal{D}\bar{\psi}'\ e^{i\int d^4x\ \bar{\psi}'\left(i\gamma_\mu D^{\mu}[A']\right)\psi'} \\ &=\int \mathcal{D}\psi\mathcal{D}\bar{\psi}\ e^{i\int d^4x\ \bar{\psi}\left(i\gamma_\mu D^{\mu}[A]\right)\psi} e^{i\int d^4x \ \epsilon(x)\mathcal{A}(x)} = e^{i\int d^4x \ \epsilon(x)\mathcal{A}(x)} e^{i S_F[A]} \end{align} where we have followed the same steps as before where the gauge anomaly pops out of the measure tranformation. Therefore the variation of the effective action under a gauge tranformation is $$\delta_\epsilon S_F[A] \equiv S_F[A+\delta A]-S_F[A]=\int d^4 x\ \epsilon(x) \mathcal{A}(x)$$ and so you see that the anomaly breaks the gauge invariance of the matter fields effective action. This in turn breaks unitarity etc.

Ok so now that we have a more precise definition of what inconsistent means, we can see that the anomaly just can't be removed, it has to vanish. This is a property only a few gauge groups have, including the standard model's $$SU(3)\times SU(2)\times U(1)$$. The reason for this is in the explicit form of gauge anomalies (I will not repeat the derivation here since it can be easily found in textbooks and is quite long) : $$\mathcal{A}_a(x)=-\frac{1}{32 \pi^2}D_{abc}\epsilon^{\mu\nu\rho\sigma}F^b_{\mu\nu}F^c_{\rho\sigma}$$ where $$a,b,c$$ are indices belonging to the gauge group. The anomaly cancels if the group theory factors $$D_{abc} = 0$$.

• What about anomalous global symmetries? I think anomalies of local gauge symmetries have to be removed for consistency. @Fra – SRS Dec 22 '18 at 10:11
• I have edited the answer with some detail regarding anomalies and gauge groups. – Fra Dec 22 '18 at 12:40
• @Fra Hi, sorry for this comment but this is the only way to contact you. Some days ago, I have seen your answer into my question physics.stackexchange.com/questions/450811/… but now it looks you deleted it. I didn't have the time to read the answer but it seemed to me there was a lot of useful material. Why have you deleted the answer? Was the result incorrect? Cheers. – newUser Jan 8 at 23:29
• Hey, I reinstated it, I did delete it because I wanted to check something but never had the time. – Fra Jan 9 at 10:19