I will be happy if someone could clarify the mystery here.

Consider the following derivation of the anomalous Slavnov-Identity. It's based on lecture notes by Adel Bilal.

Suppose we have an action with a field (or fields) $\phi$, and a symmetry transformation which takes $$\phi \rightarrow \phi ' = \phi + \epsilon F(x, \phi(x)) ,$$ under which the action is invariant, $S[\phi '] = S[\phi]$.

However, let's suppose that the functional measure is anomalous: $${ \mathcal{D} \phi } \rightarrow { \mathcal{D} \phi } \space \space e^{i\epsilon \int \mathcal{A}(x)}.$$

Let's look at the following path integral (which is the generating functional of the connected diagrams):

$$ \int {\mathcal{D}\phi \space \text{exp}(iS[\phi] + i\int J(x)\phi(x))} = \int {\mathcal{D}\phi ' \space \text{exp}(iS[\phi '] + i\int J(x)\phi '(x))} = \int {\mathcal{D}\phi \space \text{exp} ({i\epsilon \int \mathcal{A}(x)} ) \space \text{exp}(iS[\phi] + i\int J(x)\phi(x) + i\epsilon \int J(x)F(x,\phi))} = \int {\mathcal{D}\phi \space \text{exp}(iS[\phi] + i\int J(x)\phi(x)) \left( 1 + i\epsilon \int A(x) + i\epsilon \int J(x)F(x,\phi)\right)}. $$

The first equality is just a change of integration variables, the second is the substitution of $\phi '$ in terms of $\phi$, and the third is just reorganization and first order expansion in $\epsilon$.

Now you get the Slavnov-Taylor identity (eq. 4.5 in the original paper):

$$ \int { (\mathcal{A}(x) + J(x) {\left< {F(x,\Phi)} \right>} _J) } = 0 \tag{4.5}$$

Can someone explain how this identity could be possible, since $\mathcal{A}$ doesn't depend on $J$, which is an auxiliary field, so substituting $J=0$ would give $ \int \mathcal{A} = 0$ which doesn't seem plausible.

The origin of the problem (if it is a problem as it seems) should be in the change of variables in the first equality above, but it seems valid. To simplify matters, I think the above equations could all be derived with $J=0$, to give:

$$ \int {\mathcal{D}\phi \space \text{exp}(iS[\phi])} = \int {\mathcal{D}\phi ' \space \text{exp}(iS[\phi '])} = \int {\mathcal{D}\phi \space \text{exp} ({i\epsilon \int \mathcal{A}(x)} ) \space \text{exp}(iS[\phi])} $$

Which doesn't seem plausible at all.

Maybe it's related to some regularization schemes that should usually be implemented in QFT, but it isn't immediate from these equations that you need any regularization here, so it's unclear to me what is wrong with these equations.


2 Answers 2


The derivation in the given reference indeed seems confused and inconsistent. The crucial error seems to me that $$ S[\phi + \epsilon\delta\phi] = S[\phi]$$ is just not true for an infinitesimal symmetry. The definition of a symmetry is that $S[\phi']=S[\phi]$ (modulo boundary terms) for the finite transformation $\phi\mapsto \phi'$. Writing this infintesimally, one gets1 $$ S[\phi'] = S[\phi] + \epsilon\frac{\delta S}{\delta \phi}\delta\phi + \mathcal{O}(\epsilon^2)$$ meaning $\frac{\delta S}{\delta \phi}\delta\phi = 0$ modulo boundary terms, but crucially this does not mean that $S[\phi+\epsilon\delta\phi] = S[\phi]$. We have $\frac{\delta S}{\delta \phi}\delta\phi = \int\partial_\mu j^\mu$ for the Noether current $j$.

A proper derivation of the S-T identities expands $S[\phi']$ as I did above to obtain $$ \langle \mathcal{A}(x) + \left(\frac{\delta S}{\delta \phi(x)} + J(x)\right)\delta\phi(x)\rangle_J = 0$$ which gives $\langle \mathcal{A}\rangle = \partial_\mu\langle j^\mu\rangle$ at $J=0$, which is the correct statement of anomalous non-conservation.

1The notation $\frac{\delta S}{\delta \phi}\delta\phi$ is short for $\int\frac{\delta S}{\delta \phi(x)}\delta\phi(x)$.

  • $\begingroup$ I think there is some confusion between symmetry of the action, and symmetry of the action subject to enforcing the classical equations of motion. $\endgroup$ Jan 25, 2016 at 20:37
  • $\begingroup$ @itamarhason: There is no such thing as a "symmetry subject to enforcing the e.o.m.". By definition of a solution of the e.o.m., $\delta S= 0$ for any transformation you do (the solution to the e.o.m. are critical points of the action functional). You recognize symmetries by $\delta S=0$ off-shell, not by what they do on-shell. Also, inside the path integral, you surely cannot enforce the e.o.m.. $\endgroup$
    – ACuriousMind
    Jan 25, 2016 at 20:42
  • $\begingroup$ you are absolutely right. $\endgroup$ Jan 25, 2016 at 20:45
  • $\begingroup$ I think I accept your answer, though I would put more emphasis on the fact that $\frac{\delta S}{\delta \phi (x)}$ isn't zero but $\partial_\mu j^\mu (x)$ $\endgroup$ Jan 25, 2016 at 20:50
  • $\begingroup$ Maybe you can write as follows: instead of $\frac{\delta S}{\delta \phi (x)} = 0$ (which is what is falsely assumed above) , we have $\frac{\delta S}{\delta \phi (x)} \delta \phi (x) = \epsilon(x) \partial_\mu j^\mu (x)$, the transformation infinitesimal parameter times the divergence of the noether current $j_\mu$, which is classically conserved, could be conserved in the quantum theory too if $\mathcal{A}$ would vanish, but isn't conserved if it doesn't. $\endgroup$ Jan 25, 2016 at 21:10

OP is basically asking (v4) the following.

How the equation $$ \int\! \mathrm{d}^4x~ \left(i{\cal A}(x) + i\sum_r J_r(x) \langle F^r\left[x,\Phi\right] \rangle_J\right) ~=~0, \tag{4.5} $$ or equivalently, $$ \int\! \mathrm{d}^4x~ \left(i{\cal A}(x) + i\sum_r J_r(x) F^r\left[x,\frac{\delta}{i \delta J}\right]\right)Z[J] ~=~0, \tag{4.5'} $$ can be fulfilled for arbitrary sources $J$? In particular, if the anomaly ${\cal A}$ is non-zero and source $J=0$ is zero?

In Eq. (4.5'), we introduce the partition function

$$ Z[J]~=~e^{iW[J]}~=~\int\! \left[ \prod_s {\cal D} \phi^s \right] \exp\left[i S[\phi] +i \int\! \mathrm{d}^4x\sum_r J_r(x) \phi^r(x) \right]. \tag{4.4} $$

Rather than essentially repeating the derivation of Ref. 1, perhaps a naive toy-model in a 0-dimensional spacetime is in order instead.

Example. Let the action $S$ be a constant independent of the fields $\phi^r$. Then we certainly have a symmetry of the action! Since the action is constant, let us for simplicity choose it to be zero $S\equiv 0$. The partition function becomes a multi-dimensional Dirac delta distribution

$$ Z[J]~=~ \delta(J)$$

(modulo factors of $2\pi$ that we ignore). Let us consider a linear symmetry

$$ F^r(\phi) ~=~\sum_s A^r{}_s ~\phi^s, $$

because Ref. 1 assumes that the anomaly not depend on the fields $\phi^s$. The "anomaly" becomes simply the trace $$ i{\cal A}~=~\sum_s \frac{\partial F^s(\phi)}{\partial \phi^s}~=~\sum_s A^s{}_s ,$$ which is not necessarily zero. Then eq. (4.5') reads $$\left( \sum_s A^s{}_s + \sum_{r,s} J_r A^r{}_s \frac{\partial}{ \partial J^s}\right)\delta(J)~=~ \sum_{r,s}\frac{\partial}{ \partial J^s} \left(J_r A^r{}_s~ \delta(J)\right)~=~0, $$ which indeed is a well-known identity for the Dirac delta distribution. In contrast, eq. (4.5) does not make sense in the example, because it is meaningless to divide with a Dirac delta distribution. We attribute this to the crudeness of the toy-model.


  1. A. Bilal, Lectures on Anomalies, arXiv:0802.0634.
  • $\begingroup$ This is nice, but doesn't really answer the question in my opinion. I think that you managed to give an example were it works only because the current $j_\mu$ vanishes. As explained above, when the current doesn't vanish (which is I think the interesting case anyway), (4.5) would be wrong. $\endgroup$ Jan 26, 2016 at 17:20
  • $\begingroup$ Note that ${\cal A}$ does not necessarily vanish in the example. $\endgroup$
    – Qmechanic
    Jan 27, 2016 at 11:21
  • $\begingroup$ I see, but does it have any significance? The theory is trivial... $\endgroup$ Jan 27, 2016 at 14:17

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