Assume we are given are fermionic many-particle system with creation operators $c_1^\dagger,c_2^\dagger,...$ acting on some Hilbert space $\mathcal{H}$. That is, the $c_i^\dagger$ and their corresponding annihilation operators $c_i=(c_i^\dagger)^\dagger$ act as bounded operators on $\mathcal{H}$ and satisfy the CAR relations $$\{c_i,c_j\}=\{c^\dagger_i,c_j^\dagger\}=0\quad\text{and}\quad\{c_i^\dagger,c_j\}=\delta_{ij}\quad\forall i,j$$

In the bosonic case, the Wick theorem can be elegantly written in the form $$\prod_{j=1}^nc_{i_j}^{\sigma_j}=\sum_{Q\subset\mathbb{N}_n}\langle\prod_{j\in Q}^nc_{i_j}^{\sigma_j}\rangle:\prod_{j\not\in Q}^nc_{i_j}^{\sigma_j}:$$ for any $i_1,...,i_n\in\mathbb{N}$ and $\sigma_1,...,\sigma_n\in\{-1,1\}$, where $c^-_i:=c_i$, $c^+_i:=c^\dagger_i$ for any $i\in\mathbb{N}$, $\mathbb{N}_n:=\{1,\dots,n\}$, $\langle A\rangle:=\langle\Omega\mid A\Omega\rangle$ tenotes the vacuum expectation value and $:\cdots:$ denotes the Normal ordering. However, for fermions this formula has to be modified by attaching an appropriate sign to every of the summands. What is this sign?


Here's a trick to recover the fermionic signs, knowing the bosonic formulas. First, introduce a sufficient number of algebraically independent "c-numbers" $\epsilon_k$, that are only required to anti-commute among themselves and with the $c_i$, $c_i^\dagger$ generators. Then, according to the usual sign rules, the products $\epsilon c_i$ and $\epsilon c_j^\dagger$, where $\epsilon$ is any one of the new generators, will obey the bosonic canonical commutation relations with extra anti-commuting factors: $$ [\epsilon c_i, \eta c_j^\dagger] = -\epsilon \eta \{c_i, c_j^\dagger\} = -\epsilon \eta \delta_{ij} = -\epsilon \eta \langle c_i c_j^\dagger \rangle = \langle (\epsilon c_i) (\eta c_j^\dagger) \rangle, $$ for example. The extra intermediate $-$ sign appeared because the $c_i$ and $\eta$ exchanged positions while moving $\eta$ out of the commutator or moving it inside the expectation value. Note that $\epsilon$ and $\eta$ need to be chosen independently, to avoid writing down tautologies like $0=0$ due to the nilpotency $\epsilon^2 = 0$ of the new generators.

So, the bosonic Wick formula still works in the following way $$ \prod_{j=1}^n \epsilon_{k_j} c_{i_j}^{\sigma_j}=\sum_{Q\subset\mathbb{N}_n}\langle\prod_{j\in Q}^n \epsilon_{k_j} c_{i_j}^{\sigma_j}\rangle:\prod_{j\not\in Q}^n \epsilon_{k_j} c_{i_j}^{\sigma_j}: , $$ where the $\epsilon_{k_j}$ multipliers are all distinct for distinct $j$. I'll leave it as an exercise to obtain the exact sign factor that you were looking for, which you can get by factoring out $\prod_{j=1}^n \epsilon_{k_j}$ from both sides of the above equation.

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    $\begingroup$ This is not just a trick but in fact a universal recipe. One can prove in various ways (e.g. it is a basic exercise in categorical reasoning) that any fermionic anything --- Berezinian integrals, statements about supergeometry, etc. --- can be recovered from including c-numbers and then only working with bosonic combinations. $\endgroup$ – Theo Johnson-Freyd Jul 27 '16 at 12:24
  • $\begingroup$ From what I have understood, I coumpute as $[\epsilon c_i, \epsilon c_j^\dagger]=\epsilon c_i\epsilon c_j^\dagger - \epsilon c_j^\dagger\epsilon c_i=\epsilon^2[c_i,c_j^\dagger]=0$. And what is $\eta$? $\endgroup$ – Robert Rauch Jul 28 '16 at 7:05
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    $\begingroup$ Ok, I see now. $\eta$ is "just another $\epsilon$", so it makes sense. I was a bit confused because you wrote $[\epsilon c_i, \epsilon c_j^\dagger]$. $\endgroup$ – Robert Rauch Jul 28 '16 at 7:10
  • $\begingroup$ @RobertRauch, yes and sorry about the typo! Thanks for fixing it. $\endgroup$ – Igor Khavkine Jul 28 '16 at 7:22
  • $\begingroup$ I'm now figuring out how to "factor out $\prod_{j=1}^n \epsilon_{k_j}$" from the right hand side, but as far as I see we have not yet explained how the $\epsilon_k$ act on $\mathcal{H}$, so expressions like $\langle \epsilon_kc_j\rangle$ are kind of meaningless so far... $\endgroup$ – Robert Rauch Jul 28 '16 at 7:54

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