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...


1 Answer 1


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

  • $\begingroup$ 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? $\endgroup$
    – RedGiant
    Commented Aug 30, 2021 at 17:54
  • $\begingroup$ @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. $\endgroup$ Commented Aug 31, 2021 at 14:14
  • $\begingroup$ In the case of superconductors the interacting Hamiltonian is number-conserving, but the mean-field theory is not, due to spontaneous symmetry breaking. It conserves only parity, as you stated. Is it possible that spontaneous symmetry breaking leads to a ground state where not even parity is conserved in some special case? Or is there some fundamental law forbidding this? $\endgroup$ Commented Mar 16 at 18:36

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