 Solution

Use the source term $\int d^{4}x[J\varphi +\bar{\varsigma }\psi +\bar{\psi }\varsigma ]$ and left Grassmann functional derivatives. Differentiating its exponential inserts fields according toThe sign for $\bar{\psi }$ follows because a left derivative with respect to $\varsigma$ must cross $\bar{\psi }$. Keep the order $\bar{\psi }{\gamma }_5\psi \varphi$ when replacing these fields. Expand $e^{i{S}_{\mathrm{int}}}$ in the original path integral; each field monomial can now be generated by these ordered derivatives acting on the free functional. Resumming givesThe arguments refer respectively to $\varphi ,\bar{\psi },\psi$ at $x$. The free functional arises by performing the free Gaussian field integrals, and $\mathcal{N}$ may be chosen so that $Z[0]=1$.

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 $\int \mathcal{D}\varphi \mathcal{D}\bar{\psi }\mathcal{D}\psi 1$ rather than simply $Z_0$, which is not the free generating functional. Also, when ${\mathcal{L}}_{\mathrm{int}}$ denotes a density, the spacetime integral in the exponent is required.