 Solution

Use the path integral with source termThe sources $\eta ,\bar{\eta }$ are Grassmann variables; all derivatives below are left functional derivatives. Because $\eta$ stands to the right of $\bar{\psi }$, differentiating with respect to it introduces a minus sign. The insertion rules are thereforeFor example ${\delta }_L(\bar{\psi }\eta )/\delta \eta =-\bar{\psi }$. The order of the odd functional derivatives matters: the rightmost derivative acts first, and the displayed product inserts ${\bar{\psi }}_{\alpha }{\psi }_{\beta }$ in that order.

Expand $\exp [-ie\int \bar{\psi }{\gamma }^{\mu }\psi A_{\mu }]$ in its power series, replace each field by its source derivative, and pull the derivatives outside the free path integral. Re-exponentiating givesThis identity concerns the unnormalized generating functional. Normalized correlation functions divide by its zero-source value so that vacuum contributions cancel.