Matrix element of the derivative of an operator between its eigenstates

Question

I want to calculate a matrix element of the derivative of the Hamiltonian between two eigenstates $\alpha$ and $\beta$ given by $u^\alpha(x,y)$ and $u^\beta(x,y)$ (called the Bloch functions): $$\langle \alpha | \frac{\partial \hat{H}}{\partial k_j}|\beta\rangle$$This is taken from Eq. (3.6) in the paper by Komoto, Topological Invariant and the Quantization of the Hall conductance. In the same paper in Eq. (3.4), they defined the matrix element of v between states $\alpha$ and $\beta$ as:

$$(v)_{\alpha \beta}=\delta_{k_1 k'_1} \delta_{k_2 k'_2} \int_0^{qa} dx \int_0^b dy u_{k_1 k_2}^{\alpha^*} v u_{k'_1 k'_2}^{\beta}$$

The result from the paper is: $(E^\beta -E^\alpha)\langle \alpha| \frac{\partial u^\beta}{\partial k_j} \rangle=-(E^\beta - E^\alpha)\langle \frac{\partial u^\alpha}{\partial k_j}| \beta\rangle$.

I tried using the expression for matrix element between states given in the paper but cannot obtain their result. I think there has to be an integration by parts involved in order to get $(E^\beta - E^\alpha)$ but integration by parts requires the presence of an integral over $k_j$ which is not the case here.

Answer 1

Consider the general matrix element $E^{\beta}\langle \alpha |\nabla \beta \rangle$. This can be simplified as: $$E^{\beta}\langle \alpha |\nabla \beta \rangle=\langle \alpha |\nabla (\hat{H} \beta) \rangle=\langle \alpha |\nabla \hat{H}| \beta \rangle + \langle \alpha |\hat{H} \nabla \beta \rangle=\langle \alpha |\nabla \hat{H}| \beta \rangle + E^{\alpha}\langle \alpha |\nabla \beta \rangle$$

Hence, $\langle \alpha |\nabla \hat{H}| \beta \rangle=(E^{\beta} - E^{\alpha})\langle \alpha |\nabla \beta \rangle$. Similarly if one repeats the calculation starting from $E^{\alpha}\langle \beta |\nabla \alpha \rangle$, one gets $\langle \alpha |\nabla \hat{H}| \beta \rangle=-(E^{\beta} - E^{\alpha})\langle \nabla \alpha | \beta \rangle$. This is exactly what is required to be proved.