1
$\begingroup$

I am trying to follow along a derivation (E. I. Blount, Solid State Phys. 13, 305 (1962)) in which he derives the matrix elements of the true momentum $p_{n,n'}(k,k')$ (not the crystal momentum). He arrives at the following expression:

$p_{n,n'}(k,k') = \delta(k-k')(\hbar k\delta_{n,n'}-i\hbar\int u_n^*\frac{\partial u_{n'}}{\partial x}\,d\tau)$,

where $n$ labels the band index, $k$ is the Bloch vector, $u_n$ is the Bloch function with the periodicity of the lattice, and the integral is over a unit cell. Can anyone help me to derive this result?

Crucially, the problem boils down to evaluating

$\int dx \, e^{-ikx}u_{nk}^*(x)e^{ik'x}u_{n'k'}(x)$,

which, according to Blount ought to equal $\delta_{n,n'}\delta(k,k')$. This makes sense if our wave functions

$\psi_{n,k}(x) = e^{ikx}u_{nk}(x)$

(i.e. Bloch waves) are to be normalized, but I just can't seem to figure out how to 'pluck' those two delta functions out from the integration.

$\endgroup$
  • $\begingroup$ The wave functions $\psi_{n, k}$ are, by definition, eigenfuntions of a (periodic) Hamiltonian, and so are orthogonal $\endgroup$ – By Symmetry May 12 at 10:48
  • $\begingroup$ Yes, its clear to me that the $\psi_{n,k}$ should be orthogonal, but it is unclear to me how to show orthogonality when you write the $\psi_{n,k}$ as Bloch waves. $\int dx e^{i(k-k’)x} = \delta(k-k’)$ but why explicitly is $\int dx e^{i(k-k’)x}u_{nk}^*u_{n’k’} = \delta(k-k’)\delta_{n,n’}$? $\endgroup$ – aRockStr May 12 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.