Let we have a crystalline solid whose Hamiltonian is given as (equ.1): $$H=\frac{p^2}{2m}+V$$ where $V$ is periodic which means eigenfunctions are Blochstates $\psi_{nk}$ with eigenvalues $E_{nk}$. One can write Schrodinger equation as $H|\psi_{nk}>=E_{nk}|\psi_{nk}>$ or equivalently $H|u_{nk}>=E_{nk}|u_{nk}>$ where $|u_{nk}>$ is cell periodic function and (equ.2) $$H=\frac{(p^2+\hbar k)^2}{2m}+V$$
In this article
Theory of polarization of crystalline solids, R.D. King-Smith and David Vanderbilt, Phys. Rev. B 47, 1651(R), 1993.
authors wrote a equations which I want to understand (equ. 3) $$<\psi_{nk}|p|\psi_{mk}>=\frac{m}{\hbar}<u_{nk}|[\partial_k,H]|u_{mk}>$$ in these equations $H$ is given by equ.2
My attempt:
write right hand side of equ.3 as:
$$<\psi_{nk}|p|\psi_{mk}>=\frac{m}{\hbar}<\psi_{nk}|[\partial_k,H]|\psi_{mk}>$$
now $H$ is given by equ.1. Right hand side of above equation is (using $p=\hbar k$)
$$=\frac{m}{\hbar}[<\psi_{nk}|\partial_kH|\psi_{mk}>-<\psi_{nk}|H\partial_k|\psi_{mk}>]$$
$$=\frac{m}{\hbar}[\frac{2\hbar}{2m}<\psi_{nk}|p|\psi_{mk}>-<\psi_{nk}|H\partial_k|\psi_{mk}>]$$
$$=<\psi_{nk}|p|\psi_{mk}>-\frac{m}{\hbar}<\psi_{nk}|H\partial_k|\psi_{mk}>$$
First term is exactly what I wanted but how to deal with second term? It should be zero. How?
