= Solution
Use the source term $\int d^4x[J\phi+\bar\varsigma\psi+\bar\psi\varsigma]$ and left <Grassmann derivative>[Grassmann functional derivatives]. Differentiating its exponential inserts fields according to
$$
\boxed{\phi\ \longleftrightarrow\ \frac1i\frac\delta{\delta J},\qquad
\bar\psi\ \longleftrightarrow\ i\frac\delta{\delta\varsigma},\qquad
\psi\ \longleftrightarrow\ \frac1i\frac\delta{\delta\bar\varsigma}.}
$$
The sign for $\bar\psi$ follows because a <Left Grassmann derivative>[left derivative] with respect to $\varsigma$ must cross $\bar\psi$. Keep the order $\bar\psi\gamma_5\psi\phi$ when replacing these fields. Expand $e^{iS_{\mathrm{int}}}$ in the original <path integral>; each field monomial can now be generated by these ordered derivatives acting on the <Generating functional>[free functional]. Resumming gives
$$
\boxed{Z[J,\bar\varsigma,\varsigma]=\mathcal N\exp\left\{i\int d^4x\,\mathcal L_{\mathrm{int}}\left(\frac1i\frac\delta{\delta J},\ i\frac\delta{\delta\varsigma},\ \frac1i\frac\delta{\delta\bar\varsigma}\right)\right\}Z_0[J,\bar\varsigma,\varsigma].}
$$
The arguments refer respectively to $\phi,\bar\psi,\psi$ at $x$. The <Generating functional>[free functional] arises by performing the free <Gaussian integral>[Gaussian] field integrals, and $\mathcal N$ may be chosen so that $Z[0]=1$.
\b[No additional field-integration measure remains in this differential-operator formula.] Such a measure is printed outside the operator in the source even though $Z_0$ has already integrated out those fields. At $\lambda=0$, that literal expression would contain $\int\mathcal D\phi\mathcal D\bar\psi\mathcal D\psi\,1$ rather than simply $Z_0$, which is not the <Generating functional>[free generating functional]. Also, when $\mathcal L_{\mathrm{int}}$ denotes a density, the spacetime integral in the exponent is required.
Back to article page
