Fujikawa method

In physics, Fujikawa's method is a way of deriving the chiral anomaly in quantum field theory. It uses the correspondence between functional determinants and the partition function, effectively making use of the Atiyah–Singer index theorem.

Derivation
Suppose given a Dirac field $$\psi$$ which transforms according to a representation $$\rho$$ of the compact Lie group G; and we have a background connection form of taking values in the Lie algebra $$\mathfrak{g}\,.$$ The Dirac operator (in Feynman slash notation) is
 * $$D\!\!\!\!/\ \stackrel{\mathrm{def}}{=}\ \partial\!\!\!/ + i A\!\!\!/$$

and the fermionic action is given by
 * $$\int d^dx\, \overline{\psi}iD\!\!\!\!/ \psi$$

The partition function is
 * $$Z[A]=\int \mathcal{D}\overline{\psi}\mathcal{D}\psi\,e^{-\int d^dx\,\overline{\psi}iD\!\!\!/\,\psi}.$$

The axial symmetry transformation goes as
 * $$\psi\to e^{i\gamma_{d+1}\alpha(x)}\psi\,$$
 * $$\overline{\psi}\to \overline{\psi}e^{i\gamma_{d+1}\alpha(x)}$$
 * $$S\to S + \int d^dx \,\alpha(x)\partial_\mu\left(\overline{\psi}\gamma^\mu\gamma_{d+1}\psi\right)$$

Classically, this implies that the chiral current, $$j_{d+1}^\mu \equiv \overline{\psi}\gamma^\mu\gamma_{d+1}\psi$$ is conserved, $$0 = \partial_\mu j_{d+1}^\mu$$.

Quantum mechanically, the chiral current is not conserved: Jackiw discovered this due to the non-vanishing of a triangle diagram. Fujikawa reinterpreted this as a change in the partition function measure under a chiral transformation. To calculate a change in the measure under a chiral transformation, first consider the Dirac fermions in a basis of eigenvectors of the Dirac operator:
 * $$\psi = \sum\limits_{i}\psi_ia^i,$$
 * $$\overline\psi = \sum\limits_{i}\psi_ib^i,$$

where $$\{a^i,b^i\}$$ are Grassmann valued coefficients, and $$\{\psi_i\}$$ are eigenvectors of the Dirac operator:
 * $$D\!\!\!\!/ \psi_i = -\lambda_i\psi_i.$$

The eigenfunctions are taken to be orthonormal with respect to integration in d-dimensional space,
 * $$\delta_i^j = \int\frac{d^dx}{(2\pi)^d}\psi^{\dagger j}(x)\psi_i(x).$$

The measure of the path integral is then defined to be:
 * $$\mathcal{D}\psi\mathcal{D}\overline{\psi} = \prod\limits_i da^idb^i$$

Under an infinitesimal chiral transformation, write
 * $$\psi \to \psi^\prime = (1+i\alpha\gamma_{d+1})\psi = \sum\limits_i \psi_ia^{\prime i},$$
 * $$\overline\psi \to \overline{\psi}^\prime = \overline{\psi}(1+i\alpha\gamma_{d+1}) = \sum\limits_i \psi_ib^{\prime i}.$$

The Jacobian of the transformation can now be calculated, using the orthonormality of the eigenvectors
 * $$C^i_j \equiv \left(\frac{\delta a}{\delta a^\prime}\right)^i_j = \int d^dx \,\psi^{\dagger i}(x)[1-i\alpha(x)\gamma_{d+1}]\psi_j(x) = \delta^i_j\, - i\int d^dx \,\alpha(x)\psi^{\dagger i}(x)\gamma_{d+1}\psi_j(x).$$

The transformation of the coefficients $$\{b_i\}$$ are calculated in the same manner. Finally, the quantum measure changes as
 * $$\mathcal{D}\psi\mathcal{D}\overline{\psi} = \prod\limits_i da^i db^i = \prod\limits_i da^{\prime i}db^{\prime i}{\det}^{-2}(C^i_j),$$

where the Jacobian is the reciprocal of the determinant because the integration variables are Grassmannian, and the 2 appears because the a's and b's contribute equally. We can calculate the determinant by standard techniques:
 * $$\begin{align}{\det}^{-2}(C^i_j) &= \exp\left[-2{\rm tr}\ln(\delta^i_j-i\int d^dx\, \alpha(x)\psi^{\dagger i}(x)\gamma_{d+1}\psi_j(x))\right]\\

&= \exp\left[2i\int d^dx\, \alpha(x)\psi^{\dagger i}(x)\gamma_{d+1}\psi_i(x)\right]\end{align}$$ to first order in α(x).

Specialising to the case where α is a constant, the Jacobian must be regularised because the integral is ill-defined as written. Fujikawa employed heat-kernel regularization, such that
 * $$\begin{align}-2{\rm tr}\ln C^i_j &= 2i\lim\limits_{M\to\infty}\alpha\int d^dx \,\psi^{\dagger i}(x)\gamma_{d+1} e^{-\lambda_i^2/M^2}\psi_i(x)\\

&= 2i\lim\limits_{M\to\infty}\alpha\int d^dx\, \psi^{\dagger i}(x)\gamma_{d+1} e^{{D\!\!\!/\,}^2/M^2}\psi_i(x)\end{align}$$ ($${D\!\!\!\!/}^2$$ can be re-written as $$D^2+\tfrac{1}{4}[\gamma^\mu,\gamma^\nu]F_{\mu\nu}$$, and the eigenfunctions can be expanded in a plane-wave basis)
 * $$= 2i\lim\limits_{M\to\infty}\alpha\int d^dx\int\frac{d^dk}{(2\pi)^d}\int\frac{d^dk^\prime}{(2\pi)^d} \psi^{\dagger i}(k^\prime)e^{ik^\prime x}\gamma_{d+1} e^{-k^2/M^2+1/(4M^2)[\gamma^\mu,\gamma^\nu]F_{\mu\nu}}e^{-ikx}\psi_i(k)$$
 * $$= -\frac{-2\alpha}{(2\pi)^{d/2}(\frac{d}{2})!}(F)^{d/2},$$

after applying the completeness relation for the eigenvectors, performing the trace over γ-matrices, and taking the limit in M. The result is expressed in terms of the field strength 2-form, $$F \equiv \tfrac{1}{2}F_{\mu\nu}\,dx^\mu\wedge dx^\nu\,.$$

This result is equivalent to $$(\tfrac{d}{2})^{\rm th}$$ Chern class of the $$\mathfrak{g}$$-bundle over the d-dimensional base space, and gives the chiral anomaly, responsible for the non-conservation of the chiral current.