Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Ashcroft, Mermin Solid State Physics, Eq. 17.43 is

$$ \epsilon(\textbf{k}) = \frac{\hbar^2 k^2}{2m} - e\phi(\textbf{r}) $$

where $\textbf{k}$ is the wavevector and all other symbols have their usual meaning.

What does the wavevector tell me and why do we use it instead of position $\textbf{r}$?

share|cite|improve this question
up vote 3 down vote accepted

Well $\vec{r}$ is actually a position vector it points out a particles particular position with reference to some origin, the wave vector $\vec{k}$ actually tells us more as it is related to momentum $\vec{p}$ which gives us a sense of direction of travel of the wave.

share|cite|improve this answer
Imagine a plane wave, $\vec{r}$ will point from the origin to a spot on the plane wave not necessarily in the direction of travel, whereas the wave vector points from the wave front outwards. – Triatticus May 23 '13 at 8:53
Why is it useful to know that? @Dan: What can you use the direction of the travel of the wave for? – TMOTTM May 23 '13 at 8:59
Well it is useful to know in case of reflection and transmission problems when faced with differing interfaces. Knowing the incoming direction lets you decompose the wave into components and solve for the resulting transmission and reflection coefficients. – Triatticus May 23 '13 at 9:01
We can also then solve various problems using this wave solution like the transfer of energy of a wave via the poynting vector which requires knowledge of the direction of the electric field waves (the magnetic field is related to the direction of the electric field by $\vec{B} = \frac{1}{c} \hat{k} \times \vec{E} $) – Triatticus May 23 '13 at 9:06
I'm coming from a chemistry background, so I think in terms of discrete eigenfunctions. But I know in solids, the eigenfunctions start form continuous bands. Is it fair to say that $\textbf{k}$ is actually what points out an eigenfunction in a band? – TMOTTM May 23 '13 at 9:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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