 Fujikawa's Method
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Fujikawa's Method

Fujikawa's method is a way of deriving the chiral anomaly in quantum field theory.

Suppose given a Dirac field ? which transforms according to a ? representation 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

$D\!\!\!\!/\ {\stackrel {\mathrm {def} }{=}}\ \partial \!\!\!/+iA\!\!\!/$ and the fermionic action is given by

$\int d^{d}x\,{\overline {\psi }}iD\!\!\!\!/\psi$ The partition function is

$Z[A]=\int {\mathcal {D}}{\overline {\psi }}{\mathcal {D}}\psi \,e^{-\int d^{d}x\,{\overline {\psi }}iD\!\!\!/\,\psi }.$ The axial symmetry transformation goes as

$\psi \to e^{i\gamma _{d+1}\alpha (x)}\psi \,$ ${\overline {\psi }}\to {\overline {\psi }}e^{i\gamma _{d+1}\alpha (x)}$ $S\to S+\int d^{d}x\,\alpha (x)\partial _{\mu }\left({\overline {\psi }}\gamma ^{\mu }\gamma _{d+1}\psi \right)$ 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:

$\psi =\sum \limits _{i}\psi _{i}a^{i},$ ${\overline {\psi }}=\sum \limits _{i}\psi _{i}b^{i},$ 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,

$\delta _{i}^{j}=\int {\frac {d^{d}x}{(2\pi )^{d}}}\psi ^{\dagger j}(x)\psi _{i}(x).$ The measure of the path integral is then defined to be:

${\mathcal {D}}\psi {\mathcal {D}}{\overline {\psi }}=\prod \limits _{i}da^{i}db^{i}$ Under an infinitesimal chiral transformation, write

$\psi \to \psi ^{\prime }=(1+i\alpha \gamma _{d+1})\psi =\sum \limits _{i}\psi _{i}a^{\prime i},$ ${\overline {\psi }}\to {\overline {\psi }}^{\prime }={\overline {\psi }}(1+i\alpha \gamma _{d+1})=\sum \limits _{i}\psi _{i}b^{\prime i}.$ The Jacobian of the transformation can now be calculated, using the orthonormality of the eigenvectors

$C_{j}^{i}\equiv \left({\frac {\delta a}{\delta a^{\prime }}}\right)_{j}^{i}=\int d^{d}x\,\psi ^{\dagger i}(x)[1-i\alpha (x)\gamma _{d+1}]\psi _{j}(x)=\delta _{j}^{i}\,-i\int d^{d}x\,\alpha (x)\psi ^{\dagger i}(x)\gamma _{d+1}\psi _{j}(x).$ The transformation of the coefficients $\{b_{i}\}$ are calculated in the same manner. Finally, the quantum measure changes as

${\mathcal {D}}\psi {\mathcal {D}}{\overline {\psi }}=\prod \limits _{i}da^{i}db^{i}=\prod \limits _{i}da^{\prime i}db^{\prime i}{\det }^{-2}(C_{j}^{i}),$ 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{aligned}{\det }^{-2}(C_{j}^{i})&=\exp \left[-2{\rm {tr}}\ln(\delta _{j}^{i}-i\int d^{d}x\,\alpha (x)\psi ^{\dagger i}(x)\gamma _{d+1}\psi _{j}(x))\right]\\&=\exp \left[2i\int d^{d}x\,\alpha (x)\psi ^{\dagger i}(x)\gamma _{d+1}\psi _{i}(x)\right]\end{aligned}} to first order in ?(x).

Specialising to the case where ? is a constant, the Jacobian must be regularised because the integral is ill-defined as written. Fujikawa employed heat-kernel regularization, such that

{\begin{aligned}-2{\rm {tr}}\ln C_{j}^{i}&=2i\lim \limits _{M\to \infty }\alpha \int d^{d}x\,\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^{d}x\,\psi ^{\dagger i}(x)\gamma _{d+1}e^{{D\!\!\!/\,}^{2}/M^{2}}\psi _{i}(x)\end{aligned}} (${D\!\!\!\!/}^{2}$ 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)

$=2i\lim \limits _{M\to \infty }\alpha \int d^{d}x\int {\frac {d^{d}k}{(2\pi )^{d}}}\int {\frac {d^{d}k^{\prime }}{(2\pi )^{d}}}\psi ^{\dagger i}(k^{\prime })e^{ik^{\prime }x}\gamma _{d+1}e^{-k^{2}/M^{2}+1/(4M^{2})[\gamma ^{\mu },\gamma ^{\nu }]F_{\mu \nu }}e^{-ikx}\psi _{i}(k)$ $=-{\frac {-2\alpha }{(2\pi )^{d/2}({\frac {d}{2}})!}}({\tfrac {1}{2}}F)^{d/2},$ after applying the completeness relation for the eigenvectors, performing the trace over ?-matrices, and taking the limit in M. The result is expressed in terms of the field strength 2-form, $F\equiv F_{\mu \nu }\,dx^{\mu }\wedge dx^{\nu }\,.$ 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.