I know that if we start with a massive theory, the chiral states $L$ and $R$ remain coupled to each other in the massless limit. Because a charged Dirac particle of a given helicity can make a transition into a virtual state of the opposite helicity by emitting a real photon (this is the physical origin of the anomaly). Also, the masslessness of a Dirac field theory is expressed by the invariance under a chiral transformation


From Noether theorem, the chiral invarince gives a conserved axial-vector current


and from the E.o.M. for Heisenberg fields we find


where $j_5$ is the chiral density and $m$ is the mass. Here is what I don't understand: I expect that in the limit $m\rightarrow 0$ this should be true $\partial_{\mu}j_{5}^{\mu}\rightarrow 0$. But its not. All the text books give


I can't figure out this result. I don't know how to derive it or what its physical interpretation is.

  • 2
    $\begingroup$ Those same textbooks out to have derivations of the anomaly! See e.g. Srednicki chapters 75-77. Note that there are many different derivations of the anomaly, each of which highlights different aspects of the underlying physics. Zee's QFT in a Nutshell has the standard take on it in Ch IV.7, with a catalog of various derivations. Shifman has a great physical discussion focusing on the less often heard of IR side of the anomaly. $\endgroup$
    – Michael
    Commented Jun 14, 2013 at 22:58
  • $\begingroup$ The basic physical idea is that you need a regulator, but none exists which preserves the chiral invariance. The field strength term survives in the limit that the regulator goes away. This can be thought of as the gauge field causing a restructuring of the fermion vacuum (c.f. Shifman). $\endgroup$
    – Michael
    Commented Jun 14, 2013 at 23:01
  • $\begingroup$ These notes by McLerran and these notes by Shaposhnikov describe what I mean by restructuring of the vacuum. $\endgroup$
    – Michael
    Commented Jun 15, 2013 at 0:04
  • $\begingroup$ A more formal way is to use the Fujikawa point of view about the non-invariance of the path integral measure, see this lecture. $\endgroup$
    – Trimok
    Commented Jun 15, 2013 at 7:29

2 Answers 2


Let me add a few comments to Michael Brown's answer/comment. As he mentioned, a QFT is well defined with an action $and$ a regulator. We always wish to use regulators that preserve gauge invariance, since that is a redundancy of our description and should not be removed in our quantum theory. However, any regulator that preserves gauge invariance, necessarily violates chiral invariance. P&S mentions the possibility of having gauge non-invariant regulators that preserve chiral invariance but this is not a desirable definition of our theory.

Another way to see this, is that the usual regulators used to define a theory is dimensional regularization and Pauli-Villars. PV requires introducing a (large) fermion mass and it explicitly breaks chiral symmetry. The problem with Dimreg is more subtle. The chiral symmetry involves the $\gamma^5$ matrix which is only well defined in $d=4$. When one extends dimensions to $d=4-\epsilon$, one has to be careful with the treatment of $\gamma^5$. It turns out that while chiral symmetry is restored in the 4 dimensions it is not in the $-\epsilon$ dimensions (mathematically). This is what gives us the axial anomaly. P&S has a discussion on how to treat the $\gamma^5$ matrix in $d=4-\epsilon$.

All this is discussed in Chapter 19 of P&S


You asked the right question, I was thinking of it too when reading about axial anomaly.

Below is how I explained it to myself.

As you mentioned in your question, "... a charged Dirac particle of a given helicity can make a transition into a virtual state of the opposite helicity by emitting a real photon (this is the physical origin of the anomaly)".

Please note that

$\bar{F}^{\mu\nu}F_{\mu\nu}= - 2 \bf{E}\bf{B}$

where $\bf{E}$ and $\bf{B}$ are electric and magnetic field strengths.

In the case of photon $\bf{E} \perp \bf{B}$, and hence $ \bf{E}\bf{B} = 0$.

Consequently, in zero mass limit the chiral symmetry is conserved:



In fact, the spinorial equation for massless fermions (such as Dirac equation where $m=0$) is always equivalent to source-free Maxwell equations. See, e.g., this link.

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    $\begingroup$ I don't think this is correct. The extra term is a purely quantum mechanical effect and as explained in the other answers, is due to a choice of regularization. $\endgroup$
    – Prahar
    Commented Jun 19, 2013 at 18:07
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    $\begingroup$ I don't see how your comment is related to my answer. I did not mean that extra term is due to some reason other than inconsistency of regularization with simultaneous conservation of axial & vector currents. I only mean that in the case of photons (transverse EM waves) $\bar{F}^{\mu\nu}F_{\mu\nu}=0$. $\endgroup$ Commented Jun 20, 2013 at 9:22
  • $\begingroup$ Photon does not propagate only EM waves. Photons also propagate just magnetic fields or just electric fields. So, the statement that in the case of photons $E \cdot B = 0$ is wrong. You could however say, in the case of EM waves, $E \cdot B = 0$. $\endgroup$
    – Prahar
    Commented Jun 20, 2013 at 17:57
  • $\begingroup$ Don't you think that even in the case of just electric (and just magnetic) fields $\bf{E}\bf{B} = 0$? $\endgroup$ Commented Jun 21, 2013 at 16:13
  • 1
    $\begingroup$ Well, that was just a random example. There could be an arbitrary electric and magnetic field with $E \cdot B \neq 0$. $\endgroup$
    – Prahar
    Commented Jun 21, 2013 at 16:59

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