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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.

Qmechanic
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Bababeluma
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1 Answers1

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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.

Qmechanic
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