= Solution
{c}
Couple a nondynamical <background gauge field> $B_\mu$ to the <Dirac current>, choosing a plus sign for the source term:
$$
\boxed{Z[B]=\mathcal N\int\mathcal D\bar\psi\,\mathcal D\psi\,
\exp\left\{i\int d^4x\,[\bar\psi(i\not\partial-m)\psi+B_\mu\bar\psi\gamma^\mu\psi]\right\}.}
$$
There is no integration over $B_\mu$. The <gauge covariant derivative> is $\partial_\mu-iB_\mu$, so the simultaneous transformations are
$$
\psi'=e^{i\alpha(x)}\psi,\qquad
\bar\psi'=\bar\psi e^{-i\alpha(x)},\qquad
B'_\mu=B_\mu+\partial_\mu\alpha.
$$
The kinetic term changes by $-j^\mu\partial_\mu\alpha$ and the source term changes by $+j^\mu\partial_\mu\alpha$. The <action> is therefore invariant, exactly for finite $\alpha$ as well as infinitesimally. Changing integration variables and using the invariant <functional measure> from part <1/c> proves the <background gauge invariance of a current generating functional>:
$$
\boxed{Z[B+\partial\alpha]=Z[B].}
$$
The allowed transformations preserve the vacuum <boundary conditions>. Its infinitesimal version is $\partial_\mu\delta Z/\delta B_\mu=0$. Because $B_\mu$ is only a source, this statement needs no <gauge fixing> or photon kinetic term. The absence of a vector <gauge anomaly> is essential to the change-of-variables argument.
Back to article page
