# How to derive the Fermion generating function formally from operator formalism?

The generating functionals for fermions is: $$Z[\bar{\eta},\eta]=\int\mathcal{D}[\bar{\psi}(x)]\mathcal{D}[\psi(x)]e^{i\int d^4x [\bar{\psi}(i\not \partial -m+i\varepsilon)\psi+\bar{\eta}\psi+\bar{\psi}\eta]}.$$

I don't understand how this can be derived formally from operator formalism.

For Scalar field: $$Z[0]=1=\langle 0 \vert 0 \rangle = \int\mathcal{D}\Phi(x)e^{iS},$$ which is rigorously done by interpolating complete basis $$\int d\Phi(x)\vert \Phi\rangle\langle \Phi\vert=1$$ and $$\int d\Pi(x)\vert \Pi\rangle\langle \Pi\vert=1$$ which makes sense since $$\hat{\phi}(x)$$ and $$\hat{\pi}(x)$$are hermitian operators and their eigenstates supposedly span a complete basis of the Hilbert space. We also have $$[\hat{\phi}(x),\hat{\pi}(y)]=i\delta^3(x-y).$$ These are sufficient to give a path integral expression for $$\langle 0 \vert 0 \rangle$$.

None of above applies to the Fermion case. For a start, the Dirac Spinor is not even Hermitian, making it questionable to insert any complete basis of field eigenstates.

Both full derivation and reference to relevant materials will be appreciated.

• Briefly speaking, to heuristically perform the derivation of the path integral from the operator formalism for fermions, one uses fermionic coherent states. The main idea is still the same: Insert infinitely many completeness relations. See also this related Phys.SE post. Commented Oct 10, 2023 at 9:57
• @Qmechanic thanks! My textbook completely skipped this part. Is it true that the path integral for complex scalar field is also coherent states path integral? Commented Oct 10, 2023 at 10:09
• $\uparrow$ Yes. Commented Oct 10, 2023 at 10:21