Commutator of the Hamiltonian and crystal momentum

Question

Equation (2.4) of "Berry Phase Effects on Electronic Properties" by Xiao, Chang and Niu uses the identity

$$\langle\frac{\partial n}{\partial q}| k\rangle (E_n-E_k)= \langle n| \frac{\partial H_q}{\partial q} |k\rangle \quad ,$$ where $E_i$ is the energy of the eigenstate $|i\rangle$ of the crystal Hamiltonian $H_{q}=e^{-ikr}He^{ikr}$ corresponding the crystal momentum vector $\mathbf{k}$. I have tried to reverse engineer this equation starting from the left hand side and have gotten as far as proving that

$$\frac{\partial H}{\partial q} = H^* \frac{\partial}{\partial q^*} -\frac{\partial}{\partial q*} H\quad ,$$

where I have used $*$ to denote the Hermitian conjugate. Then I made the assumption that $H$ and $q$ should be self-adjoint such that their eigenvalues are real and found that this required the commutator of $H$ and the derivative of $q$ to be

$$[H, \frac{\partial}{\partial q}] = \frac{1}{2} \quad ,$$

but I'm struggling to justify this claim. Is there another way to prove this identity / a reason the commutator takes this value?

Answer 1

The equation follows from the a general result called Hellman-Feynman theorem: suppose one has a Hamiltonian $H(\lambda)$ depending on a parameter $\lambda$ (e.g. your $q$), with eigenstates $|n(\lambda)\rangle$ and energies $E_n(\lambda)$. Then for $m\neq n$ (i.e. two orthogonal eigenstates),

$$ \langle m|\partial_\lambda H|n\rangle=(E_m-E_n)\langle \partial_\lambda m|n\rangle. $$

To prove this, just differentiate the eigenvalue equation $H(\lambda)|n(\lambda)\rangle=E_n(\lambda)|n(\lambda)\rangle$ with respect to $\lambda$ and then take the overlap with $|m(\lambda)\rangle$.

Answer 2

Additionally to the answer by @Meng Cheng, you can obtain the desired result by a straightforward calculation:

Start by defining

$$H(q) \equiv U(q) \, H\, U^\dagger(q) \quad , $$

with $U(q)\equiv e^{-iqx}$. To proceed, note that if $|n\rangle$ is an eigenvector of $H$ with eigenvalue $E_n$, then $$|n(q)\rangle \equiv U(q) |n\rangle $$ is an eigenvector of $H(q)$ with the same eigenvalue. We further compute $$\partial_q H(q) = -ix\, H(q) + H(q) \, ix $$ and hence $$\langle n(q)|\partial_q H(q)|k(q)\rangle = -i E_k \langle n(q)|x|k(q)\rangle + i E_n \langle n(q)|x|k(q)\rangle \quad .$$

Finally, it is not hard to realize that

$$\langle n(q)|ix = \partial_q \langle n(q)|\equiv \langle \partial_q n(q) |\quad ,$$ which yields

$$\langle n(q)|\partial_q H(q)|k(q)\rangle = \left(E_n-E_k\right)\, \langle \partial_q n(q)|k(q)\rangle \quad .$$