# Why is the density of states in $k$-space constant?

Why are the allowed states in $k$-space equidistant in every direction? As a consequence of this, the density of states for phonons in 3D is $$\frac{V}{(2\pi)^3}$$ while for electrons it is $$2 \frac{V}{(2\pi)^3}$$ because of the Pauli exclusion principle.

• This is only true if the system is isotropic (rotationally symmetric). The density of states in $\mathbf{k}$-space is not the same in every direction in general. – Mark Mitchison Feb 11 '15 at 17:35
• @mark mitchinson: That is true. I'm taking a look at the free electron gas, which is confined in a box, hence isotropic. I don't get intuitively what equidistant momentum states mean or why they are equidistant in 1D, 2D, and 3D. In energy space, the DOS sinks in 1D, remains constant in 2D, and increaases in 3D. – nomadStack Feb 11 '15 at 17:41
• I think you mean "why is the density of states in $\mathbf{k}$-space constant at every point"? Equidistant is a bit of a confusing term here. – Mark Mitchison Feb 11 '15 at 17:42
• @MarkMitchison I understand what you mean, but 'density of states into a direction' is ill-defined. Do you say population density is not the same in Manhattan towards north than east, if the blocks are not square but rectangular. Even if the distance between reciprocal lattice planes is different to different directions, the density of k-points in the asymptotic $L \rightarrow \infty$ limit is still the same everywhere. – Mikael Kuisma Jan 2 '16 at 8:23
• @MikaelKuisma If you have spin-orbit coupling (in a continuum model) the density of states really isn't isotropic in general (it depends on the direction of $\mathbf{k}$), hence the comment. – Mark Mitchison Jan 2 '16 at 14:16

## 1 Answer

Let me try to answer your question for the 1D case, which will help you understand the higher dimensional case. In 1D, the allowed $k$ vectors are evenly spaced. The reason is that only integral multiples of wavelengths are allowed. The corresponding $k$ vectors $(= \frac {2 \pi} {\lambda})$ are thus evenly spaced. Similar conditions apply for the higher dimensions, but visualization wouldn't be feasible.