# Density of states, volume

1. Apr 17, 2013

### Dr_Pill

In my text:

The number of states per unit volume of the real space & the reciprocal space is given by

1 / (4∏³)

No further explanation is given.

How do you get to this 4∏³

And how come the density of states is the same in real space & reciprocal space?

I think this is incorrect, they should be the reversed version of each other.

2. Apr 17, 2013

### Traur

Consider a cubic box of size L. Then periodic boundary conditions imply that kx=nx*2pi/L where nx=0,1,2,3,... The same follows for ky and kz.
Now you can see that in k-space a state with a given k occupies a k-space volume of (2pi/L)^3. Then the number of states per unit volume is (1/V) L^3 * 2 /(2pi)^3 and you get that result. The factor of 2 comes from the fact that the electron has 2 distinct values for the z component of the spin. Note that if you consider a rectangular box you would obtain an identical conclusion.

3. Apr 24, 2013

### mcodesmart

I have attached a figure from Ashcroft and Mermin.

Do this. In the case of 2D, take the reciprocal lattice and you draw a circle or radius k from the center.

How many electrons states you would have in that area encircled in k-space? If you can figure this out, you can easily find the answer you are looking for. Also, in the case of electrons, there are 2 electrons allowed for each state (spin).

If all else fails, you can use this formula. $g(E)dE = 2g(k)dk/\nabla_k E(k)$

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Last edited: Apr 24, 2013