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I know this is connected to the fact that fermions are represented by anticommuting operators, but I still cannot find the way to get this minus in Feynman rules.

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This is explained on Wikipedia, under "Feynman diagram". – Ron Maimon Dec 16 '11 at 19:06
up vote 8 down vote accepted

It's because a fermion loop with $N$ vertices along the loop corresponds to, up to bosonic factors everywhere (which never change the sign and mostly commute with everything else), $$ D = \langle \psi_1^\dagger \psi_1\cdot \psi_2^\dagger \psi_2 \cdots \psi_N^\dagger \psi_N \rangle $$ where each of the products $\psi^\dagger \psi$ comes from one vertex. Note that vertices are what produces the $\psi$ factors because they appear in the interaction Hamiltonian. However, in the Feynman rules, we need to revisualize the diagram so that it's composed from propagators which are factors of the type $\psi_1\psi_2^\dagger$ and so on, geometrically corresponding to links between adjacent vertices. We can make this form manifest if we move $\psi_1^\dagger$ to the very end: $$ D = - \langle \psi_1\psi_2^\dagger\cdot \psi_2 \psi_3^\dagger \cdots \psi_N \psi_1^\dagger\rangle $$ In this form, we have a product of $N$ nice factors $\psi_i \psi_{i+1}^\dagger$ that may be attributed to propagators. However, I had to add a minus sign because we needed to permute Grassmann-valued $\psi_1$ through $2N-1$ other fermionic operators and $2N-1$ is odd. Therefore I had to correct the sign by $(-1)^{2N-1}=-1$ for both expressions to be equal.

I was a bit schematic so I didn't indicate whether I used an operator formalism or the Feynman path integral approach. The logic for the sign is the same in both versions. In the Feynman path integral language, the $\psi$ objects in the demonstration above are Grassmann numbers, not operators, so they strictly anticommute with each other.

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