Consider an energy-wavenumber graph, typical in solid state physics, like the one below.

I can follow the mathematics in the derivations with a KP model. But I don't understand the physical interpretation of the graph.

What I might have understood is the following: we may associate a unique(?) real number, called the wavenumber, to every electron wavefunction, by the Bloch's theorem, assuming periodic potential. Then, to produce the graph, we may calculate the energies of the wavefunctions and somehow(?) realize that $k$ is equivalent to $k+\frac{2\pi}{a}$ for all $k$.


(1) Is the correspondence between the electron wavefunctions and the wavenumber bijective? I.e. (a) Can we associate an unique wavenumber to every wavefunction? (b) Can different wavefunctions be associated with the same wavenumber/the same equivalence class of wavenumbers?

(2) What does it physically mean to say that $k$ is equivalent to $k+\frac{2\pi}{a}$?

A mathematically rigorous explanation of the E-k graph would be highly appreciated.

An example of an energy-wavenumber graph from Grosso, page 12: An example of an energy-wavenumber graph from Grosso, page 12

  • 1
    $\begingroup$ Commenting with a partial answer: Due to lattice structure. $\psi(r,t)=\psi(r+na,t):n\in\mathbb{N}$. $k \equiv k + \frac{2\pi}{a}$ satisfies this equality. $\endgroup$ – acarturk May 21 '19 at 1:51
  • $\begingroup$ $k$ is a measure of momentum, and you are already familiar with energy vs momentum relationships. For example, a free electron has $E = p^{2}/2m$, so is a parabola in $E$ vs $k$. In a crystal it is no longer free, and then you have the symmetry-imposed conditions that @acarturk indicates above. $\endgroup$ – Jon Custer May 21 '19 at 13:28

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.