Expectation value of time-evolved number operator for ground state Coulomb system

Question

I'm going through "Advanced Quantum Mechanics" by Franz Schwabl, and he calculates the electron energy levels from the Coulomb interaction in a perturbative way (section 2.2.3). In the course of this, he calculates that: $$ \left< a^\dagger_{k + q, \sigma} (t) \: a_{k + q, \sigma} (t) \right> = \left< \phi_0 \right| a^\dagger_{k + q, \sigma} (t) \: a_{k + q, \sigma} (t) \left| \phi_0 \right> = n_{k + q, \sigma} $$ (eq. 2.2.18 to 2.2.19), where $\left| \phi_0 \right>$ is the filled Fermi sphere state, and $n_{k + q, \sigma}$ is 1 if $k + q < k_F$ (the Fermi momentum) and zero otherwise.

Why does this last equality hold? This is clearly true for $t = 0$ by how the creation and annihilation operators are defined, but it seems like this should be a time-dependent quantity if we are considering time-dependent creation and annihilation operators. That is, we should have to explicitly time-evolve these operators with the exponential of the Hamiltonian (given below) to calculate the expectation value. He seems to just do it without thinking. What am I missing?

The Hamiltonian (in case it is relevant) is just the Coulomb interaction Hamiltonian: $$ H = \sum_{\mathbf{k}, \sigma} \frac{\hbar^2 k^2}{2m} a^\dagger_{\mathbf{k} \,\sigma} a_{\mathbf{k} \, \sigma} + \frac{1}{2V} \sum_{\mathbf{q} \neq 0, \mathbf{p}, \mathbf{k}' \\ \sigma, \sigma'} \frac{4 \pi e^2}{q^2} a^\dagger_{\mathbf{p} + \mathbf{q} \, \sigma} a^\dagger_{\mathbf{k'} - \mathbf{q} \,\sigma} a_{\mathbf{k'} \, \sigma'} a_{\mathbf{p} \, \sigma} $$

Answer 1

I think he is implicitly assuming that the filled Fermi sea $\vert{\phi_0}\rangle$ is a good approximation for the ground state of the weakly interacting Hamiltonian. Therefore he is assuming that under time evolution, this state only picks up a phase.

In terms of Green's functions in terms of Feynman diagrams, we can say that he is approximating the interacting Green's function as the free Green's function $+$ the Fock Green's function. The Hartree term is zero because there is no interaction when $q=0$. (Ignore this paragraph if you do not know diagrammatic perturbation theory.)

In terms of usual perturbation theory, he is just finding the first-order correction to the ground state energy, which is given by $$ \Delta E = \langle \phi_0 \vert H_{\text{Coulomb}}\vert \phi_0\rangle $$ You can verify that this matches the correction to the energy he obtained. So we can say he is essentially doing first-order perturbation theory.