# Expectation value of fermionic creation/annihilation operator?

Let us consider a many-body system of interacting fermions, described by a Hamiltonian $$H$$. This system is in thermal equilibrium with a bath of temperature $$T$$. What is the expectation value of creation/annihilation operator? $$\langle c^\dagger_k \rangle = ?$$

Somewhere, I saw that this expectation value should be zero, but I don't know how to prove it?

By the way, this problem arose when I was working on DMFT. The idea was to show that starting from the atomic multiplets, we cannot treat the hopping term using mean-field theory if the system is fermionic...

The expectation value of an observable in thermal equilibrium is given by

$$\langle O\rangle =\text{Tr}(Oe^{-\beta H})$$ If the eigenstates of the Hamiltonian are $$|\psi_n\rangle$$, we can rewrite this as $$\langle O\rangle =\sum_n \langle\psi_n|O|\psi_n\rangle e^{-\beta E_n}$$

Now, let's assume our Hamiltonian is particle conserving. This is the same as saying that the Hamiltonian commutes with the number operator $$N:=\sum_k c_k^\dagger c_k$$. You can check that this is the same as saying every term in $$H$$ is the product of an equal number of creation and annihilation operators. All electronic systems are of this form.

In this case, we can simultaneously diagonalize both $$H$$ and $$N$$. So, without loss of generality, we assume $$N|\psi_n\rangle=N_n|\psi_n\rangle$$ for all eigenstates $$|\psi_n\rangle$$ of $$H$$. But in that case, for each $$n$$ we have

$$\langle\psi_n|c_k^\dagger|\psi_n\rangle = \frac{1}{N_n}\langle\psi_n|c_k^\dagger N|\psi_n\rangle =\frac{1}{N_n}\langle\psi_n| (N-1) c_k^\dagger|\psi_n\rangle=\frac{N_n-1}{N_n}\langle\psi_n|c_k^\dagger|\psi_n\rangle$$ from which it follows that $$\langle\psi_n|c_k^\dagger|\psi_n\rangle=0$$ for each $$n$$, hence $$\langle c_k^\dagger\rangle=0$$.

(What if $$N_n=0$$? I'll leave you to check that case for yourself!)

What if you don't have particle number conservation? In general, you won't have $$\langle c^\dagger_k\rangle=0$$! However, in many systems without particle number conservation, you still conserve particle number parity. This is true in effective theories for superconductors, for example, where you have terms like $$c_k^\dagger c_{k'}^\dagger$$ in your Hamiltonian, but no terms like $$c_k^\dagger c_{k'}^\dagger c_{k''}^\dagger$$.

In this case, you can prove that the parity operator $$P=(-1)^N$$ commutes with $$H$$, and so without loss of generality you can assume the eigenstates of the Hamiltonian satisfy $$P|\psi_n\rangle = P_n|\psi_n\rangle$$ with $$P_n=\pm 1$$. Now we can do a similar trick as before:

$$\langle\psi_n|c_k^\dagger|\psi_n\rangle=P_n\langle\psi_n|c_k^\dagger P|\psi_n\rangle=-P_n\langle\psi_n|Pc_k^\dagger|\psi_n\rangle=-\langle\psi_n|c_k^\dagger|\psi_n\rangle$$

Hence $$\langle\psi_n|c_k^\dagger|\psi_n\rangle=0$$ hence $$\langle c_k^\dagger\rangle=0$$.

• Thank you. This is very clear. However, I am particularly interested in the case when the particle number is non-conserving. Is it possible to see what happens in that case? Aug 30, 2021 at 17:54
• @RedGiant See edit. In general I don't think you can't say this is true, but for Hamiltonians made of an even number of creation/annihilation operators it is true. Aug 31, 2021 at 14:14