I was working with the following expression related to the Wick's theorem for four fermionic operators. $$ \langle c^\dagger_i c_j c^\dagger_p c_q \rangle = \langle c^\dagger_i c_q \rangle \langle c_j c^\dagger_p \rangle + \langle c^\dagger_i c_j \rangle \langle c^\dagger_p c_q \rangle. $$

I tried to show this for the simplest case where the Hamiltonian is $$ H = \sum_{j} \epsilon_j c_j^\dagger c_j $$

and using the definition of the thermal average as $\langle \ldots \rangle = \mathrm{tr}(\rho \ldots)$, where the density operator is defined to be $$ \rho = e^{-\beta (H - \mu N)}/Z \\ Z = \mathrm{tr}(e^{-\beta (H - \mu N)}). $$

The approach I used is to start with the canonical anticommutation relations $$ \langle \{ c_q, c_i^\dagger c_q c_p^\dagger \} \rangle = \delta_{iq} \langle c_j c_p^\dagger \rangle + \delta_{pq} \langle c_i^\dagger c_j \rangle. $$

In the Heisenberg picture, I allow $c_q$ to depend on time, $c_q(t) = c_q e^{\frac{1}{i\hbar} \epsilon_q t}$. Taking the Fourier transform $$\langle \{ c_q(E), c_i^\dagger c_q c_p^\dagger \} \rangle = 2\pi \delta(E - \epsilon_q) [\delta_{iq} \langle c_j c_p^\dagger \rangle + \delta_{pq} \langle c_i^\dagger c_j \rangle ]. $$

Now apply the fluctuation-dissipation theorem to get $$ \langle c_i^\dagger c_q c_p^\dagger c_q(E) \rangle = 2\pi f(E) \delta(E - \epsilon_q) [\delta_{iq} \langle c_j c_p^\dagger \rangle + \delta_{pq} \langle c_i^\dagger c_j \rangle ] $$

where $f(E) = [e^{\beta(E-\mu)} + 1]^{-1}$ is the Fermi-Dirac function. Lastly, I obtain the desired expression by doing the inverse Fourier transform (and setting $t=0$). $$ \langle c_i^\dagger c_q c_p^\dagger c_q \rangle = f(\epsilon_q) [\delta_{iq} \langle c_j c_p^\dagger \rangle + \delta_{pq} \langle c_i^\dagger c_j \rangle ] \\ = \langle c^\dagger_i c_q \rangle \langle c_j c_p^\dagger \rangle + \langle c_p^\dagger c_q \rangle \langle c_i^\dagger c_j \rangle. $$

I believe Wick's theorem is quite general, so it should be true for other more complicated Hamiltonians such as the Hubbard's Hamiltonian? How do I show it in a more general way?

  • 4
    $\begingroup$ Wick's theorem is not true for interacting theories (such as the Hubbard model). It is a theorem about Gaussian (free) theories. $\endgroup$
    – jacob1729
    Commented Jan 30 at 10:26

1 Answer 1


The general proof basically also uses the commutation relations.

The general statement is: Let $A_i,(i=1, \ldots n)$ be annihilation or creation operators, or linear combinations thereof. Then $$ A_1 \ldots A_n=\sum_{P:\left(i_1<k_1\right) \ldots\left(i_p<k_p\right)} \delta\left(\sigma_P\right)\left(\prod_{l=1}^p\left\langle 0\left|A_{i_l} A_{k_l}\right| 0\right\rangle\right): A_1 \ldots \hat{A}_{i_1} \ldots \hat{A}_{k_p} \ldots A_n:, $$ Here, the sum runs over all pairings $P$ of $p=0, \ldots,[n / 2]$ pairs of indices from $\{1, \ldots, n\}$ and $\sigma_P$ is a permutation that brings together the contracted pairs, i.e. $\sigma_P\left(k_l\right)=\sigma_P\left(i_l\right)+1$, but leaves the order of the remaining indices unchanged ( $\wedge$ means omission). The double dots : indicate that the expression is normal ordered. Because of the multilinearity of both sides, it suffices to assume that each $A_i$ is either an annihilation or a creation operator; then, it is sufficient to extend the sum over pairings where each $A_{i_l}$ is an annihilation operator and each $A_{k_l}$ is a creation operator, since the other contractions vanish. Now we use normal ordering and the commutation relations. Namely, we use the fact that for $n>0$ $$ \left\langle 0\left|: A_1 \ldots A_n:\right| 0\right\rangle=0 . $$ Thus with the help of $$ A B= \pm B A+[A, B]_{ \pm} $$ one can bring annihilation operators to the right. Thereby, we find $$ \left[a(f), a^*(g)\right]_{ \pm}=\left\langle 0\left|\left[a(f), a^*(g)\right]_{ \pm}\right| 0\right\rangle=\left\langle 0\left|a(f) a^*(g)\right| 0\right\rangle $$ (here, $a(f): \otimes^n \mathscr{H} \rightarrow \otimes^{n-1} \mathscr{H}$ is the annihilation operator for an element in the 1-particle Hilbert-space $f \in \mathscr{H}$, and $a^*$ the corresponding creation operator). If one shifts the annihilation operators to the right, then one gets terms which can be characterized by these pairings. It should not take long to convince yourself of this.

Edit: I should say that of course, the proof and validity relies on the fact that the contractions one does are c-numbers (the rigorous statement would be that these are in the center of the operator algebra that is considered).

  • 2
    $\begingroup$ Can I get some clarification on where this breaks down for an interacting theory (say $\phi^4$ theory). I think the problem is that the `original' creation/anniilation operators (of the gaussian point) no longer satisfy $\langle 0 | : A_1 \dots A_n : |0\rangle=0$ when $|0\rangle$ is the ground state of the interacting theory. $\endgroup$
    – jacob1729
    Commented Jan 30 at 11:28
  • $\begingroup$ As I said in the edit, as far as I understand for this to work we need to use the fact that the contractions $\left\langle 0\left|a(f) a^*(g)\right| 0\right\rangle$ are in the center of the operator algebra, meaning that they commute with everything and we can move them around. This is not generally the case for theories like $\phi^4$, of course unless one considers them in a perturbative setting. $\endgroup$
    – mika
    Commented Jan 30 at 11:34
  • $\begingroup$ I believe this approach, as outlined here, can only prove Wick theorem for the vacuum. The ingredient is that the normal order expectation value (over the vacuum) is zero. To prove Wick theorem using normal order for other Gaussian states you have to consider another splitting. The terms $\left\langle 0\left|a(f) a^*(g)\right| 0\right\rangle$ are always complex numbers and so always commute with everything irrespective of the theory being interacting or quadratic. $\endgroup$
    – lcv
    Commented Feb 2 at 8:52
  • $\begingroup$ Just to be clear, let me reiterate that Wick theorem is false for truly interacting (i.e. non-quadratic) theories. $\endgroup$
    – lcv
    Commented Feb 2 at 8:56
  • $\begingroup$ I dont quite know what you mean by "for the vacuum". The statement is about operators in general. If I would take the vacuum expectation value the normal ordered part will vanish and the formula simplifies. $\endgroup$
    – mika
    Commented Feb 2 at 10:23

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