I find anomalies an extremely unintuitive subject, because they're studied so indirectly. In the standard textbook presentation, one computes an abstract quantity that should be zero classically (say, some time-ordered correlation function) and finds that it isn't, and that's it.

I'm wondering how anomalies manifest at a basic level, say, at the level of the equations of motion. However, in this case I'm not sure how they can possibly appear.

Naively, upon canonical quantization, interacting quantum fields will obey equal-time canonical commutation relations, just as the corresponding classical fields have canonical Poisson brackets. Then the equations of motion for classical fields translate to operator equations for the fields, e.g. in $\phi^4$ theory we would have $$(\partial^2 + m^2) \hat{\phi} = \hat{\phi}^3$$ modulo formal issues involving multiplying operator-valued distributions.

Next, we may take the time-ordered vev of both sides. Pulling time derivatives past the time-ordering symbol produces the appropriate contact terms to arrive at the Schwinger-Dyson equations.

At the classical level, a conserved current obeys $\partial_\mu j^\mu = 0$. Hence by the same reasoning as for $\hat{\phi}$, this should hold as an operator equation, $\partial_\mu \hat{j}^\mu = 0$. Taking the time-ordered vev of both sides and pulling out the time derivative reproduces the Ward-Takahashi identity, with no anomaly terms.

I'm sure that almost everything I said above is extremely naive from the point of view of formal QFT, but morally speaking, where does the above logic go wrong? For the axial anomaly, the textbook arguments show that $$\partial_\mu \hat{j}^{\mu 5} \sim - \frac{g^2}{16 \pi^2} \epsilon^{\mu\nu\rho\sigma} \hat{F}_{\mu\nu} \hat{F}_{\rho \sigma}$$ where the $\sim$ indicates that only some matrix elements of the operators on both sides are shown to match. Is this in fact an operator equality? If so, where does the extra term slip in above?

  • $\begingroup$ Possible (though unanswered) duplicates: physics.stackexchange.com/q/107098/50583, physics.stackexchange.com/q/261956/50583 (If I get bored this weekend I might try to expand my comment to tparker's question into ananswer) Note that your "affect the equations of motion" is really the same as "in the operator formalism" since in the path-integral formalism we show validity of equations of motion as equations of quantum expectation values by the derivation of the Schwinger-Dyson equation under the assumption that the measure is invariant, i.e. the theory is non-anomalous. $\endgroup$
    – ACuriousMind
    Feb 14, 2019 at 20:55

2 Answers 2


Short answer: The operator $\hat j$ is meaningless as written, because it contains divergences. In order to manipulate this operator (and take its divergence) one must introduce a regulator which, when removed, happens to leave a finite piece: the anomaly.

The usual narrative is that $\hat j$, defined as $$ \hat j^\mu(x)\overset?=\bar\psi(x)\gamma^\mu\psi(x)\tag1 $$ is very much ill-defined, inasmuch as it contains operators at coinciding spacetime points. (As in the OP, $\psi,\bar\psi$ are operator-valued distributions with coinciding singular support, so their product is undefined, just like $\delta(x)^2$; see also this PSE post). Therefore, any manipulation is unjustified, and it may or may not lead to the "correct" answer.

One way to fix this is to require the equations of motion to be whatever the path integral predicts. This a consistent prescription, and it has the advantage of being manifestly covariant and non-perturbative. This is the point of view held by, for example, DeWitt in his QFT book (and it seems to me this is how physicists think nowadays).

A more down-to-earth approach is to introduce some regulator to your operator $\hat j$, perform all your manipulations using well-defined objects only, and take the physical limit at the very end. In the case above, one lets $$ \hat j^\mu(x):=\lim_{y\to x}\bar\psi(x) W_\gamma(x,y)\gamma^\mu\psi(y)\tag2 $$ where $W$ is a Wilson line with $\partial\gamma=x-y$ (to make $\hat j$ gauge invariant). Proceeding as in the OP, one derives (cf. this PSE post) $$ \mathrm d\hat j=e(F)\tag3 $$ the Euler class of the gauge field, as is well-known. This is a non-perturbative effect, although (being topological, cf. this PSE post) can be detected by a one-loop computation.

More generally, the "equations of motion" of a quantum field are ill-defined, and any formal manipulation is unjustified from a rigorous point of view. Luckily for us, some manipulations turn out to give the "correct" answer anyway (otherwise people would have ditched QFT in the 30s, before understanding how it really works!). A more conservative philosophy is to work with well-defined objects at all times or, even better, to work with distributions as they are supposed to be used (in the so-called causal approach).

Finally, it bears mentioning that the modern approach to anomalies is through topological arguments. A nice review is given by e.g. the first few sections of 1808.00009. Cohomology, bordism groups, and all that.

  • $\begingroup$ Isn't W only a Wilson arc? I don't see anything looping.... $\endgroup$ Feb 19, 2019 at 9:39
  • $\begingroup$ @ArnoldNeumaier Yes, absolutely, I meant a line, not a loop. Thank you! $\endgroup$ Feb 19, 2019 at 14:26
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    $\begingroup$ Thanks as always! As usual, I’m cursed for only reading low rigor books (which make exactly the incorrect statements you point out) and medium rigor books (which mostly make correct statements but never explain or even mention what’s wrong with the wrong ones). $\endgroup$
    – knzhou
    Feb 19, 2019 at 22:22
  • $\begingroup$ For instance, I don’t think the issues you mention are explained in Srednicki or Schwartz. Is there any medium-rigor place I can go to get a handle on stuff like this, the OPE, what normal ordering means, etc.? $\endgroup$
    – knzhou
    Feb 19, 2019 at 22:23
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    $\begingroup$ @knzhou This is one of those rare cases where Peskin&Shröder does a decent job (§19). Ticciati doesn't do a bad job either and Weinberg, as usual, is a must-read. Although as I said above, the best formulation is through topological arguments (it's basically a cohomological exercise; quite fun if you're into diff geo and algebraic topology and all that), and classical books don't cover this approach. The reference in my answer is a fine review (which probably contains many more useful references). Witten's 1508.04715 is great too. $\endgroup$ Feb 20, 2019 at 2:36

The devil is in the details. When you say that

...modulo formal issues involving multiplying operator-valued distributions

right after you wrote the equations of motion you ignore a very important point. The right-hand side of the equation

$$ (\partial^2 + m^2) \hat{\phi} = \hat{\phi}^3 $$

is not well defined if you does not regularize and renormalize it. Turns out that in other to do it exactly you need in advance the complete (quantum) solution of the model. One way to circumvent this problem is by organize your computations by perturbation theory of the non-linearity. Then at each order in your computation, for a given regularization, you are going to need to introduce a counter-term to match some renormalization conditions (and avoid nonsensical divergences). The anomaly are going to appear by the fact that there is no regularization, and no finite conter-term, able to preserve the anomalous symmetry. This imply that any attempt to make sense of the equation of motion you wrote are going to break the classical symmetry.

This conclusion does not depends on the perturbation theory. You may try to make sense of this equation by putting it on the lattice and use computers to do non-perturbative computations. By putting the theory on the lattice you are regularizing the theory. The anomaly will appear as an impossibility to put this theory on the computer without breaking (explicitly) the symmetry.

A more abstract way to see what goes wrong is to work with renormalized operators. In free theory, operators like $\phi^3$ need to be normal ordered (renormalized). Once you introduce an interaction, the normal ordering are going to be modified, and you can calculate this modifications perturbativally. This calculation is obtained by looking at $\phi^3$ as a vertex of Feynman diagrams. The normal ordering of free theory is the subtraction of self-contraction of the legs that come out of the vertex. In the interacted case there will be also collision of this legs, with new vertices appearing on this collisions, and new subtractions are going to be required.

About the last equation, this is an identity between operators in the Heisenberg picture. Again, classically you expect that the left-hand side is identically zero as an operator equation. The fact that the current operator suffers from ordering problems in the case of chiral fermions, you are going to get a correction. What is interesting is that this correction is exact at one-loop computation. Usually equations like the first one you wrote will continue to get corrections indefinitely. You can think about this equations (already with the proper corrections) as equation of motions coming from varying the 1PI quantum effective action.

So, in your example of the chiral anomally, the 1PI quantum effective action at one-loop will have a term that explicitly breaks the chiral anomally and contribute with the equations of motion.

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    $\begingroup$ Thanks for the answer! Can you explain what normal ordering has to do with renormalization? In the usual QFT books, the only place normal ordering ever comes up is somewhere in the first chapter, where the free field Hamiltonian is normal ordered. Why does your new use of normal ordering come in? $\endgroup$
    – knzhou
    Feb 19, 2019 at 22:28
  • $\begingroup$ See Polchinski volume 1 charter 2 and 3. There he work it out the connection between renormalization and normal ordering for 2d QFTs. There is also a book by Tom Banks where he explain renormalization on that lines. $\endgroup$
    – Nogueira
    Feb 21, 2019 at 0:28

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