I am studying the Fujikawa method of determining the chiral anomalies in a $U(1)$ theory. As we know the basis vectors selected are the eigenstates of the Dirac operator. One of the reasons given is that the eigenstates diagonalize the action which is needed for determining an exact quantity such as Ward-Takahashi identities. Anyone care to explain? I am referring to Path Integrals and Quantum Anomalies by Kazuo Fujikawa and Hiroshi Suzuki.

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    $\begingroup$ What exactly do you want us to explain? How the Ward-Takahashi identity follows? Why we need eigenstates of the Dirac operator to derive it? Please be more specific (and tell us what part exactly you don't understand). $\endgroup$ – ACuriousMind Jun 28 '14 at 13:08
  • $\begingroup$ yes. Why do we need eigenstates of the dirac operator to derive it? $\endgroup$ – SubhamDC Jun 29 '14 at 20:40

Alright, I will try to answer why we need Dirac eigenstates in this procedure, but I am not sure if it is anything more that the tautology that the Fujikawa method is precisely defined by using the Dirac eigenstates.

Let me briefly recap what the idea is: (All of this is for a Euclidean theory.)

We consider the infinitesimal local transformaton

$$\psi(x) \mapsto (1 + \mathrm{i}\alpha(x)\gamma_5)\psi(x) \equiv \psi'(x)$$

which, as it is a gauge transformation, should not change the value of the path integral:

$$ Z = \int \mathcal{D}\psi\mathcal{D}\bar\psi \mathrm{e}^{\int\bar\psi \mathrm{i}\not{D}\psi\mathrm{d}^4x} \overset{!}{=} \int\mathcal{D}\psi'\mathcal{D}\bar\psi'\mathrm{e}^{\int \bar\psi\not{D}\psi\mathrm{d}^4x + \int\alpha\partial_\mu j_5^\mu\mathrm{d}^4x}$$

To derive the anomaly term, we must examine what the change of the measure $\mathcal{D}\psi \mapsto \mathcal{D}\psi'$ is. In general, we can say that it is, by the usual transformation formulae for Grassmannian integrations, $\det(M)^{-1}$ for the operator acting as $\psi' = M\psi$.

Now, we must recall that the functional measure $\mathcal{D}\psi$ is only defined in the limit of some (UV) regularized theory, it does not exist on its own.. So to actually calculate the correct $\det(M)$, we must obtain it as the limit $\lim_{\Lambda \rightarrow \infty}\det(M_\Lambda)$ of some $M_\Lambda$ and some UV cutoff scale (not necessarily hard) $\Lambda$. UV regulators suppress high momenta. And what are the eigenstates of the Dirac operator? ... Right, they are the momentum modes! So, the most natural regularisation of our theory is to exponentially suppress states with high Dirac eigenvalues as per

$$ \widetilde{\psi}_n = \mathrm{e}^{-\frac{\not{D}^2}{2\Lambda}}\psi_n$$

where the $\psi_n$ are the unregularised Dirac eigenstates. And by choosing the regulator that way, the only consistent way to define the measure is to see it as the integral over the coefficients of these modes, i.e.

$$\mathcal{D}\psi = \lim_{\Lambda \rightarrow \infty}\prod_n \mathrm{d}\widetilde{a}_n$$

where $\psi = \sum_n \widetilde{a}_n\widetilde{\psi}_n$. (The sum over the $n$ is actually an integral if we use no IR cutoff, but it doesn't matter here, you could as well use one, but it clutters notation a bit and contributes nothing insightful.)

Now, the rest of the Fujikawa anomaly method follows through as hopefully described in your book. I hope this is at least an approximate answer to your question.

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  • $\begingroup$ My question was Why do we need the eigenstates for deriving the ward takahashi identities ,which usually follows from the path integral derivation without the need for any sort of basis eigenstates. Deriving the WT identities was one of the reasons given in the book for selecting the so called "standard basis" as fujikawa himself terms the basis set in his paper. $\endgroup$ – SubhamDC Jul 4 '14 at 8:55
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    $\begingroup$ The non-anomalous Ward-Takahashi identities follow without needing the Dirac eigenstates, they simply follow from invariance of the measure and my second equation (if your book says something else, I'm afraid it's probably wrong). Deriving the anomalous Ward-Takahashi identities amounts to determining $\det(M)$, and I think I have made plausible why we should choose Dirac eigenstates to calculate that. One could argue that one always needs to proceed as in the anomalous case to really prove that the measure is invariant, however. $\endgroup$ – ACuriousMind Jul 4 '14 at 9:34

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