3
$\begingroup$

I am reading about some canonical transformations of the Hamiltonian (of a system consisting of an electron interacting with an ionic lattice) due to Tomanaga and Lee, Low and Pines. One of the important considerations should be translation invariance (momentum conservation). In a paper by Lee, Low and Pines ("Motion of Slow Electrons in a Polar Crystal", 1953), there is a frequent mention of low-lying energy levels. Intuitively, I would like to think they are referring to energy states close to the ground-state, but I don't think that is quite correct.

Do low-lying energy levels have to do with, perhaps, energy of the electron with small momentum? The title of the paper by Lee, Low, and Pines is after all about "Slow Electrons".

I'm confused. Some clarification could be helpful.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Lee, Low, and Pines mention "low lying energy states of the system ($P^2/2m\ll\omega$, where $P$ is the total momentum of the system, $m$ is the mass of the electron, and $\omega$ is the frequency of lattice oscillations)." So yes, on the one hand, these states are "close to the ground state", on the other hand, they have electrons "with small momentum" in the conduction band. They discuss dielectrics, as far as I can understand, so the conduction band is empty in the ground state, so electrons can have low momentum in the low-lying energy states (the momentum of these electrons is low compared to the Debye momentum and, therefore, compared to the Fermi momentum as well.)

Improve this answer
$\endgroup$
2
  • $\begingroup$ Not sure what Debye and Fermi momentum are but I think your answer helps with some intution. thanks. -R $\endgroup$
    – r.g.
    Commented Feb 17, 2012 at 0:47
  • $\begingroup$ You may wish to google "Debye frequency" and "Fermi energy". The relevant momenta are defined by the standard formulas. For example, Fermi momentum is, roughly speaking, the largest momentum of electron for the ground state of the crystal. $\endgroup$
    – akhmeteli
    Commented Feb 17, 2012 at 3:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.