When can I use Wick's theorem?

Wick's theorem means that for fermions, a four point correlation function (for example) can be written in terms of two point correlation functions:

\begin{equation} \langle b_l^\dagger b_l b_m^\dagger b_m \rangle = \langle b_l^\dagger b_l \rangle \langle b_m^\dagger b_m \rangle - \langle b_l^\dagger b_m^\dagger \rangle \langle b_l b_m \rangle+ \langle b_l^\dagger b_m \rangle \langle b_l b_m^\dagger \rangle \end{equation}

My question is when can I use this? In particular, I'm interested in finite temperature many-body perturbation theory and calculating the correlation functions from something like

\begin{equation} Z[\overline{f},f] = \int \mathcal{D} (\overline{\phi},\phi) e^{-S_0 +S_{int}+\int_0^\beta d\tau \sum_l (\overline{f}_l b_l + b_l^\dagger f_l)}, \end{equation}

where $S_0$ is the unperturbed part of the system, $S_{int}$ is the perturbation, and the final part of the exponential allows us to calculate the correlation functions via functional derivatives.

Are there any circumstances under which I would need to calculate the four point correlation function itself? Or can I always use Wick's theorem?

You can always use Wick's theorem when you're describing expectation values with respect to free field states (that is, non-interacting, like a single Slater determinant state). For example, in Hubbard-Stratonovich or (generally) Variational Mean-Field Theory (like Bogliubov-deGennes) you take the interaction terms and decouple them into a model that is quadratic in fermion operators, but in a background of classical fields.

In a path-integral sense then, what I'm saying is that Wick's theorem is a statement about expectation values with respect to Gaussian distributions. Since we generally can't calculate in anything other than Gaussian distributions this is handy. Consider the action

$S=ax^2+bx^4$

and we want to integrate $e^{-S}$ over all $x$. Well, this is clearly the expectation value of $e^{-bx^4}=1-bxxxx+\mathcal{O}(b^2)$ with respect to a Gaussian distribution, and Wick's trick tells me that (since $x$ is just a scalar there's no antisymmetry...): $$\langle xxxx\rangle=\langle xx\rangle+\langle xx\rangle + \langle xx\rangle$$ so to any order in $b$ we can calculate the integral and we only have to know $\langle xx\rangle$ and do a little bit of combinatorics.

The generalization of this to operators follows basically the same logic and is surprisingly straightforward (... for path integrals! Standard books always present Wick's theorem in an operator formulation that presumably doesn't even make sense at finite temperatures and I hate it.)

• Thanks @wsc, so as an example, an XX spin chain written in terms of spinless fermions (i.e. after a Jordan-Wigner transformation) $H_1 = J \sum_l (b_l^\dagger b_{l+1} +b_{l+1}^\dagger b_l)$ is an example of a non-interacting Hamiltonian, while an XXZ spin chain is interacting because the extra term, $H_1+ J_z \sum_l [(1-2 b_l^\dagger b_l) (1-2 b_{l+1}^\dagger b_{l+1})]$, means it's not a free fermion model and so Wick's theorem is not applicable. But if the $J_z$ term were treated as a perturbation (i.e. it is $S_{int}$), Wick's theorem could still be used. Is that correct? – Calvin Oct 11 '11 at 15:56
• Also, if that's right, is it correct that there would be no need to calculate something like $\langle b_l^\dagger b_l b_m^\dagger b_m \rangle$ for the XX spin chain with the $J_z$ term as a perturbation, since I can just calculate it from the two point correlation functions? (While for an XXZ spin chain, assuming I can solve it in path integral form, I couldn't do this and so calculating the full $\langle b_l^\dagger b_l b_m^\dagger b_m \rangle$ is necessary). – Calvin Oct 11 '11 at 16:04
• All of those statements are correct. By the way, I don't know if your actual problem is with spin chains and J-W transformations, but the correlation functions in fermion language will be much more exotic N-point functions, and the technology for calculating them is very well described (actually lots of things are well described!) in Tsvelik's book. – wsc Oct 11 '11 at 16:44
• Thanks @wsc What do you mean by "the correlation functions in fermion language will be much more exotic N-point functions"? I have Tsvelik's book if you have section titles. – Calvin Oct 11 '11 at 21:19
• I actually don't have it with me :( It should be easy to find -- my point is just that spins get mapped to "string operators." Tthat is, if you want to calculate $\langle \sigma^z_0 \sigma^z_{N}\rangle$ you have a product of $\sim N$ fermions to take the expectation value of, and you also want the $N\rightarrow \infty$ limit. Wick's theorem still works, of course, and you still only need 2-point functions, but the combinatorics requires the evaluation of Toeplitz determinants. – wsc Oct 11 '11 at 21:44

In the operator$^1$ formulation, the vital assumption, which makes the standard Wick's theorem hold, is the assumption that the contractions are in the center of the pertinent operator algebra. This is often stated casually as the contractions should be $c$-numbers, meaning that the contractions should (super)commute with all pertinent operators.

$^1$ A standard lore in physics states that the operator formalism is equivalent to the path integral formalism, although the actual map between the two formalisms may be quite subtle.

• Thanks for your answer @Qmechanic, but I don't really understand. By operator formulation, do you mean the Heisenberg picture? I thought that the path integral formulation was separate to that? In the case of my question above, $Z[\overline{f},f]$ can be rewritten in terms of Grassmann numbers rather than c-numbers. My question of when Wick's theorem can be used is from wondering if there are any restrictions on, for example, what $S_0$ or $S_{int}$ can be, and whether there are case where I must calculate the four point correlation function itself instead of using Wick's theorem to find it. – Calvin Oct 11 '11 at 15:25
• This is not true. The Wick theorem works when the contractions are pure bosons, because you fix the bosons to one configuration and do the path integral over the fermions. The bosonic fields are not c-numbers once you integrate over them. – Ron Maimon Oct 11 '11 at 16:05
• @Ron Maimon: Yes, the contractions may not (super)commute with all sectors of the theory. – Qmechanic Oct 11 '11 at 16:31
• @Qmechanic: of course the formalisms are equivalent! But the Wick theorem is best stated in path integrals, because it is absurdly complicated in the operator form. This is one of the reasons. Fermions always go by Wick theorem, because you don't have 4-fermi interactions. In 2d, they don't, unless you introduce new bosonic fields to mediate a local interaction. – Ron Maimon Oct 11 '11 at 16:42
• The advantage of the operator formulation is that the Wick's Theorem can be recast in a precise self-contained mathematical language, whose scope is not limited to a particular model. – Qmechanic Oct 11 '11 at 20:06