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...
 A: 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 $\langle c_k^\dagger\rangle=0$.
