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

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: $$H = \sum_{\mathbf{k}, \sigma} \frac{\hbar^2 k^2}{2m} a^\dagger_{\mathbf{k} \,\sigma} a_{\mathbf{k} \, \sigma} + \frac{1}{2V} \sum_{\mathbf{q} \neq 0, \mathbf{p}, \mathbf{k}' \\ \sigma, \sigma'} \frac{4 \pi e^2}{q^2} a^\dagger_{\mathbf{p} + \mathbf{q} \, \sigma} a^\dagger_{\mathbf{k'} - \mathbf{q} \,\sigma} a_{\mathbf{k'} \, \sigma'} a_{\mathbf{p} \, \sigma}$$

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.