Conservation of crystal momentum with Peierl's substitution

Question

I am trying to understand Sec. 3 of Di Xiao's review paper (https://arxiv.org/abs/0907.2021). Specifically, I am interested in Sec. 3A on anomalous velocity (pages 13-14 in the arXiv pdf).

I can follow most of the derivation, but there is one point which exposes my lack of understanding of the conservation of crystal momentum. In the attached excerpt, it is mentioned that $q$ is a good quantum number and therefore $\dot{q}=0$. I fail to understand why this is the case.

I looked up Ashcroft and Mermin's classic book but I failed to find a derivation for $\dot{q}\propto [\hat{H},q]=0$ even for the case when there is no external field. My understanding is that $q$ is only conserved upto a reciprocal lattice vector $G$. So how can $\dot{q}$ be zero? Shouldn't $\dot{q}$ have occasional delta functions whenever a $G$ is added?

So my main question then is, how to prove $\dot{q}=0$:

a) in the absence of external fields

b) in the presence of external fields (like in Di Xiao's paper)

Answer 1

Perhaps the answers and comments to this related question, Is crystal momentum really momentum?, can help with the conceptual picture. The crystal momentum arises from a discrete translational symmetry of the total Hamiltonian and in that sense it is a good quantum number. It is in principle the eigenvalue of the generator of the discrete symmetry. In addition, when you have a macroscopic crystal and an atomic-scale lattice spacing, it often makes sense to assume the continuum limit as you switch to working in the Brillouin zone. Then the generator of the discrete translational symmetry becomes a quasi-momentum that commutes with the Hamiltonian and is conserved, wherefrom ${\dot q} = 0$.