# Dispersion Relation (e vs. k) clarification (crystal momentum or electron momentum)

If we get the dispersion relation from the Fourier transform of the lattice vectors then how do we get electrons information? Specifically, for the $k=0$ point of the graph, does this mean the electron has zero momentum(I am pretty sure that electrons don't have zero momentum in that case due to Heisenberg's uncertainty principle)?

I was thinking (since crystal momentum and electron momentum are different) that the $k$ is how the overall electrons wave packets move in regards to crystal momentum, or that crystal momentum dictates the electrons momentum.

As you can probably tell I am still very shaky on this, so anything is really appreciated!

## 2 Answers

$k$ is just a quantum number. $\hbar k$ gets its name "crystal impulse" from the fact, that the formula for a band structure without interaction (free electrons) coincides with the formula you get with the definition of classical impulse in terms of $k$, but it is NOT an actual impulse.

For a free electron we have the energy dispersion: $$\epsilon(k) = \frac{\hbar^2k^2}{2m}$$ The wave definition of classical impulse is $$p = \hbar k$$ which suggests the connection between energy and impulse $$p = \frac{m}{\hbar}\nabla_k \cdot\epsilon(k)$$

However this is $\neq \hbar k$ in most dispersion relations. Take for example the lowest conduction band in silicon: the minimum is not at $k = 0$. With the classical relation this would mean $p \neq 0$, but since its a minimum $\nabla_k\cdot\epsilon(k) = 0$ and therefore $p = 0$. This answers your question, whether the electrons impulse at $k = 0$ is also $0$ (it also has nothing to do with Heisenberg's principle).

This tells us that the $k$ in dispersion relations doesn't give anything about the impulse of an electron DIRECTLY and therefore not about the overal wave packet. To obtain the impulse you have to look at the relation with $\nabla$ given above.

• Thank you very much!! Very informative! I would thumbs up your question as well but I have no rep points. So if I understood your answer it is not the $k$ that tells you the electrons momentum, it is the really the gradient of the dispersion that gives the electrons momentum. Commented Feb 20, 2014 at 6:06
• and the $k$ is a just a quantum number that is found (one way) through the fourier transform of the crystal lattice vectors. Commented Feb 20, 2014 at 6:13

The dispersion relation gives you information regarding the relation between momentum of electrons, and energy of such electron. Heisenberg's uncertainty principle relates uncertainty in the position versus uncertainty in momentum, which is a very different issue.

If you consider a single massive free particle, it also possesses a dispersion relation in the form of $E=\dfrac{\hbar^2 k^2 }{2m}$, and this is in no way in contradiction with Heisenberg's principle.

The only difference in the case of electrons in a crystal is that this relation is modified, and is a bit more complex insofar as it acquires a band structure.