When deriving the chiral anomaly in the non perturbative approach for a theory of massless Dirac fermions, you start by showing that the path-integral measure is not invariant unter the chiral transformation:

$\psi \rightarrow U(x) \psi = e^{i\alpha(x)\gamma^5}\psi\\\overline{\psi} \rightarrow \overline{\psi} \,\overline{U}(x) = \overline{\psi} \, \gamma^0 e^{-i\alpha(x)\gamma^5}\gamma^0 = \overline{\psi} \, U(x)$

and the measure transforms as:

$d\overline{\psi}d\psi \rightarrow \det[U \overline{U}]^{-1}\,d\overline{\psi}d\psi = \det[U]^{-2}\, d\overline{\psi}d\psi $

The you proceed with evaluating the determinant to find the anomaly term. (It is done like this for example in Weinberg chap. 22.2)

But now imagine we write the massles Dirac Lagrangian in its chiral components $\psi_L$ and $\psi_R$:

$\mathcal{L} = \overline{\psi} i \gamma^\mu D_\mu \psi= i \psi_L^\dagger \bar{\sigma}^\mu D_\mu\psi_L + i \psi_R^\dagger {\sigma}^\mu D_\mu\psi_R $

This is now a Lagrangian of two independent Weyl spinors, one left handed one right handed. The chiral transformation from a above now reads

$\psi_L \rightarrow e^{-\alpha(x)} \psi_L \,\,, \hspace{1cm} \psi_L^\dagger \rightarrow e^{\alpha(x)} \psi_L^\dagger\\ \psi_R \rightarrow e^{\alpha(x)} \psi_R \,\,, \hspace{1cm} \psi_R^\dagger \rightarrow e^{-\alpha(x)} \psi_R^\dagger$

We write the measure for the the path integral now as $d\psi_L^\dagger d\psi_L d\psi_R^\dagger d\psi_R $ which is lorentz and gauge invariant.

But it is invariant under the chiral transformation! So what happend, where is our Anomaly gone?

In rewriting the measure $d\overline{\psi}d\psi$ as two measures of Weyl fileds the $\gamma^0$ got dropped, which before was responsible for changing the sign in the exponent of the transformation once more and made the measure non-invariant. But i don't see anything wrong in rewriting the measure in this way. Even for the massive case this should be a legible procedure, or am i wrong?

Probably somebody can give me a lesson in constructing measures for Weyl spinors ot tell me else what is going on here. Cheers!

  • 3
    $\begingroup$ Sorry but your argument is a naive classic one and you could make the very same argument in the Dirac-spinor formalism, too. It would be equally wrong. You may rewrite Weinberg's discussion of anomalies in the Weyl-spinor 2-component formalism, too. See e.g. section 6.25 on page 111 by Howard Haber here: scipp.ucsc.edu/~haber/susybook/feyn115.pdf No $\gamma_5$ appears explicitly but all the triangle anomalies etc. are still there. $\endgroup$ Commented Jun 14, 2012 at 5:55
  • $\begingroup$ Thanks for the answer and the reference. In the section you are giving the anomaly is derived for Wely fields but in a diagrammatic way. I am not questioning these results i am wondering that in Fujikawas method in deriving the anomalies vi a PI the reasoning doesn't work if i just choose another measure which appears equally valid to me. So what would the "non-naive quantum mechanical" way to argue which measure to choose for the PI? $\endgroup$
    – user9784
    Commented Jun 16, 2012 at 0:34
  • $\begingroup$ The answer to your comment above is gauge invariance, but you are misunderstanding the source of the chiral variance. Both your thing and Weinberg's are trivially chirally invariant, the issue is that neither of them is properly regulated. Weinberg's discussion of the regulator can be confusing to a newcomer--- he makes many formal steps, and he starts in Minkowski space. The way to understand it is to think about defining a finite gauge invariant cutoff fermion measure, and then looking at the gauge variation of this. This is what Fujikawa is doing. $\endgroup$
    – Ron Maimon
    Commented Jun 16, 2012 at 8:46
  • $\begingroup$ I am sorry for the delay in the answer, but I got confused on something stupid, which is actually only tangential--- what happens if you do a forward derivative discretization of the 2-component formulation? Then you don't superficially get doubling. But then you get nonsense because the thing is not unitary. You need to stagger the Fermions somehow. $\endgroup$
    – Ron Maimon
    Commented Jun 20, 2012 at 17:11

2 Answers 2


Weinberg's presentation is not pedagogically ideal, because the steps are too formal, and he puts emphasis on ones that can be misleading to a student. The presentation might lead you think that the determinant of the U factors are somehow not equal to 1 naively because of some phase business, and this is categorically not true. This is what is confusing you--- the two component description makes it manifest that there is nothing with a non-unit determinant, and the four-component formulation hides this.

The part where Weinberg talks about U being pseudounitary is the misleading thing, pseudounitary or unitary, the U determinant is equal to 1, and the formal introduction of a delta function doesn't make the determinant any less 1. The quantity $e^{i\gamma_5 \alpha(x)}$ has a unit determinant because $\gamma_5$ is traceless (as Weinberg says in the middle of the calculation), so that whether you split it up into components or not, nothing changes, the measure is naively chirally invariant.

In order to define the path integral, you need to cut off the measure, and you just can't do it in a pure two-component spinor formalism consistently, not when the 2-component spinor is coupled to a gauge field. The issue is that a gauge invariant cut-off breaks the chiral symmetry. This means that you just can't make a pure two-component gauge invariant cut-off.

This only happens when you insist on a gauge invariant cutoff, otherwise you can keep the chiral current conserved (so that you keep the two spinor foralism sensible) but at the cost of making the ordinary (gauge) current not conserved. This issue is also the reason discretizing the Dirac equation is annoying--- discretization introduces a cutoff, and the cutoff must introduce other degrees of freedom in order to be consistent.

Gauge invariant cutoffs

Suppose you have a 2-spinor field which is not coupled to a gauge field, just coupled to some scalar fields. Then you can introduce a massive 2-spinor field with opposite sign kinetic term by using a Majorana mass term (in 2-spinor notation):

$$ M(\psi^\alpha\psi^\beta \epsilon_{\alpha\beta} + CC)$$

Where CC is the hermitian conjugate, with $\bar\psi$ replacing $\psi$. You can use the massive Majorana spinor to do a Pauli-villars regulator, by subtracting very massive loops from the physical light loops, leaving convergent integrals (you sometimes need several Pauli-Villars fields). Then the whole thing is fine, and the regulator is chirally invariant, it works in 2-spinor language.

What goes wrong when you couple the thing to a gauge field is that a massive Majorana spinor can't be charged. If you do the cutoff using a massive Pauli-Villars regulator, you break gauge invariance in the mass term. So you can't cut-off a single 2-component spinor with a 2 component charged spinor, not without ruining the gauge invariance and making the coupling to the vector inconsistent.

So you have some options. One superficially attractive option is to make a lattice regulator, and keep the two-component spinors. This doesn't work, because if you actually make a lattice regulator and keep the action real, you need to discretize the derivatives using a centered difference as follows:

$$ \bar\psi \sigma\cdot D \psi $$

Where D is a shift operator forward minus a shift operator backwards, using the gauge field on the link to do the shifting covariantly. The result produces a propagator with more degrees of freedom than one chiral spinor, since the lattice regulated D operator has a determinant which is zero at several places in the Brillouin zone. This is called the fermion doubling problem, and it makes a lattice regulator a nightmare.

So what you do instead is make the 2-component spinor into a four-compoent spinor, and cut off the 4-component spinor with a massive wrong-sign 4-component spinor which is charged, Pauli-Villars style. This introduces a fictitious patner to the original 2-component spinor, to make a full 4-component spinor, and coupling the whole thing to the gauge field, you only have a consistent truncation to two components only when the chiral current is conserved in any background gauge field configuration.

But now using this gauge invariant cutoff, you can compute the divergence of the axial current, and see that it isn't zero:

$$ \partial_\mu \bar\psi \gamma_5 \gamma_\mu \psi = {1\over 16\pi^2} F_{\mu\nu}F_{\lambda\sigma}\epsilon^{\mu\nu\lambda\sigma}$$

This calculation is done later in Weinberg

Fujikawa regulator

But you were specifically asking about the Fujikawa way,and when you are using Fujikawa's regulator, you takes advantage of the background gauge field to define a new set of Grassman integration variables as follows:

$$ \prod d\bar{C}_k dC_k$$

Where the $C_k$'s are a linear combination of the $\psi$'s using the k-th eigenfunction of the Euclidean Dirac operator. The point is that you can then make a gauge invariant regulator by introducing a factor of

$$e^{-k^2\over M^2}$$

into any integral over k, and this factor cuts off the short distance modes in a manifestly gauge invariant way, since it is cutting off high k modes, where k is the gauge invariant Dirac operator eigenvalue. Note that the cutoff is slightly different for every different choice of background A field, this is how the chiral current divergence ends up proportional to something involving the F tensor.

This cutoff is manifestly gauge invariant, and manifestly regulates all the Fermion integration, since it effectively cuts off the eigenvalues of the $D$-slash operator at some effective value M. So there are only a finite number of fermionic variables effectively contributing to the intermediate states of the path-integration.

Then Weinberg is evaluating the leading order correction at large M to the measure transformation using this invariant cutoff. This is why the D-slash operator appears in the cutoff that he uses, and he explains this obliquely. So when you transform the measure chirally, using the cutoff, you get the factor Weinberg gives:

$$ i\alpha(x)\gamma_5 e^{-(D\cdot\gamma)^2/M^2}\delta(x-y) $$

This Fujikawa regulator shifts the cutoff as you change the A-field, so that it is always at a given M-sized eigenvalue of $D\cdot\gamma$, and this is what allows you to calculate the anomaly when you take the cutoff to infinity. If you want to do this in 2-spinor formalism, you need to have both 2-spinors which together make the charged Fermion, so that the cutoff, either in Pauli-Villars language or in Fujikawa language, ends up being gauge invariant.

  • $\begingroup$ Hi Ron, this is wonderful prose. I have always wondered about Fujikawa's analysis of the anomaly using two-component spinors. Is there are place in literature that discusses this in more detail? $\endgroup$
    – QuantumDot
    Commented Nov 22, 2012 at 4:53
  • $\begingroup$ @QuantumDot: I wrote a longer answer (I did the calculation, although it was tedious and it overlapped Weinberg too much), I don't know anyplace, that's why I took time. The issue is the "it's the phase transformation in the measure" business, which doesn't emphasize the right thing. It's the point-splitting regulator which is the real cause. I think the best place (ironically enough) is Adler's original 1960s article, which emphasizes the detailed operator algebra. I worked this out some years ago and wrote it in an answer sheet while TAing QFT. But there is a minor research question here. $\endgroup$
    – Ron Maimon
    Commented Nov 22, 2012 at 21:32
  • $\begingroup$ ... The main minor research thing is fixing the operator differential algebra so that it is anomaly consistent. This is done case by case in books, using diagrams for specific operators, but the full algebra is not fixed up. Fixing up the algebra for this doesn't just need to fix the OPE, but also the differentiation product rule, so that you get the anomaly from the eqns of motion. This is not in the literature, and I never fully worked it out. The goal is to use a calculus for operators which has the anomaly terms built in, like Ito calculus does for the canonical commutation relations. $\endgroup$
    – Ron Maimon
    Commented Nov 22, 2012 at 21:34

Regularisation is not the problem. The OP's measure just isn't Lorentz-invariant.

The measure chosen by the OP is: $$ \mathcal D \psi^\dagger \mathcal D \psi \, .$$ Purely by formal manipulations, one can apply the chiral transformation $\psi \mapsto e^{i \gamma^5 \beta(x)} \psi$ and see that the measure is preserved: $$ \mathcal D \psi^\dagger \mathcal D \psi \mapsto \mathcal D (\psi^\dagger e^{-i \gamma^5 \beta(x)}) \mathcal D (e^{i \gamma^5 \beta(x)} \psi) = \mathcal D \psi^\dagger \mathcal D (e^{-i \gamma^5 \beta(x)} e^{i \gamma^5 \beta(x)} \psi) = \mathcal D \psi^\dagger \mathcal D \psi \, .$$ (I'm relatively confident that it's valid to do a formal manipulation like this, but the proof would be longer than I want to take for this answer.) Likewise for the gauge transformation: $$ \mathcal D \psi^\dagger \mathcal D \psi \mapsto \mathcal D (\psi^\dagger e^{-i \alpha(x)}) \mathcal D (e^{i \alpha(x)} \psi) = \mathcal D \psi^\dagger \mathcal D \psi \, .$$ It does not matter what regularisation is chosen. The measure is still preserved as shown. The regularisation quantifies the anomaly when the anomaly is nonzero.

The issue is that the OP's measure is not Lorentz-invariant. The quantity $\psi^\dagger \psi$ is not Lorentz-invariant, which is the reason we use $\overline \psi \psi$ in the first place, both in the measure and in the action functional. Therefore the OP's measure introduces a Lorentz anomaly. The precise value of the Lorentz anomaly would be dependent on the regularisation. The OP's suggestion is a fine one, and successfully preserves both vector and chiral gauge symmetries. It just gives up Lorentz-invariance.


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