How Is the Volume of an Element in K-Space Determined?

[Master J]

I've been trying to figure out the volume of an element in k-space (free electron gas in a crystal lattice).

In deriving the density of states, dn/dE = dn/dk . dk/dE. My dn = 4pi(k^2).dk / V, where V is the volume of an element in k-space.
I came across the spurious, Dk.Dx ~ 2pi (an uncertainty principle relation), where V was said to be (2pi)^3, but that UP doesn't make sense to me.


Could someone shed some light on this?

The ever so grateful MJ...

 

[Physics Monkey]

A relatively simple derivation follows from state counting. Consider a three dimensional cube with periodic boundary conditions and sides of length L. The single particle energy eigenstates are of the form exp(i k x) where k is given by k = \frac{2\pi}{L}(n_x, n_y,n_z) with arbitrary integers n.

Summing over all these states can be written as \sum_n = \sum_{n_x,n_y,n_z} \Delta n_x \Delta n_y \Delta n_z = \sum_{n_x,n_y,n_z} \Delta k_x \Delta k_y \Delta k_z \left(\frac{L}{2 \pi}\right)^3 \rightarrow \frac{V}{(2\pi)^3} \int d^3 k.

Thus we may approximate sums over discrete single particles states in a big box via an integral over three dimensional k space with the measure \frac{V}{(2\pi)^3} d^3 k.

Is this what you were looking for?

 

[Master J]

Ah, indeed.

The k value slightly confuses me tho. Should it not be k = pi / L (sorry I'm rubbish at LaTeX)??

If we let the electron wavefunction say be sin(kx) in the cube, with the condition taht it is zero at the edges, that gives k = n pi / L. Where does the 2 come from?

 

[Physics Monkey]

I used periodic boundary conditions while you are using hard wall boundary conditions. It is true that in your case all the allowed values of k make are \pi n/L but with n positive. Thus compared to my analysis you have a density of states of ( L^3/\pi^3) d^3 k but you're also allowed to only integrate over the totally positive k octant. In my formula I integrate over all 8 octants thus making up the missing factor of 2^3 = 8.

Hope this helps.