# The density of states in 2d?

1. Dec 17, 2009

### jeebs

Hi,
I am trying to find an expression for the density of states of free two-dimensional electrons, as a function of energy, and I am really struggling.

I get that what I am looking for is the number of states per unit area of k-space per unit energy, and in general (3D), this is expressed as

density of states $$g(E) = \frac{1}{V}\frac{dN}{dE}$$

However, since this is in 2D, the V is actually an area. In k-space, I think a unit of area is $$A=\frac{\pi}{L}\frac{\pi}{L} = \frac{\pi^2}{L^2}$$ since$$k=\frac{\pi}{L}$$ for the smallest allowed length in k-space.

So, what I need is some expression for the number of states, N(E), but presumably have to find it in terms of N(k) first.

So, what I said was that the free electron has energy $$E = \frac{\hbar^2}{2m_e}(k_x^2 + k_y^2)$$ so that when I know N in terms of k, I can easily convert it to N in terms of E.

So I think I eventually need to get $$g(E) = \frac{L^2}{\pi^2}\frac{dN(E)}{dE}$$

However, this is where I have ran out of steam. I haven't been able to come up with an expression for the number of states as a function of wave number, N(k).

Can anyone give me a hand here? It would make a lovely Christmas present
Cheers.

Last edited: Dec 17, 2009
2. Dec 17, 2009

### diazona

Well, the actual wavenumbers $k_x$ and $k_y$ of any given state are going to be multiples of the quantum $\pi/L$, right? (Seriously, correct me if I'm wrong, I'm not 100% sure I'm remembering this correctly) So the energy expression is
$$E = \frac{\hbar^2\pi^2}{2 m_e L^2}\bigl(n_x^2 + n_y^2\bigr)$$
Now you have a grid, basically, in which there is a state at each point with integer coordinates. You can make a one-to-one correspondence between points/states and 1x1 unit squares on the grid. So the number of unit squares within any large enough area is about the same as the number of states within that area.

Try finding an expression for N(n), then convert it to N(k), then to N(E). If you're still stuck after a while, I'm sure someone will get back to you.

3. Dec 17, 2009

### phsopher

Have you tried writing to Santa? :tongue2:

How many states are inside a circle with radius k?

4. Dec 17, 2009

### jeebs

a circle of radius k has an area $$A=\pi k^2$$

and i believe there is one state per unit area on that grid, right?

so is the number of states $$N = \frac{A}{ \pi ^2/L^2} = \frac{L^2k^2}{\pi}$$ ??

5. Dec 17, 2009

### diazona

Seems reasonable so far...

6. Dec 17, 2009

### jeebs

Right so, $$N(k) = \frac{A}{\pi ^2 / L^2}$$ and $$E = \frac{\hbar ^2k^2}{2m} =\frac{\hbar ^2n^2\pi ^2}{2mL^2}$$

so $$N(E) = \frac{\pi k^2}{\pi ^2 / L^2} = \frac{L^2}{\pi}\frac{2mE}{\hbar ^2}$$

so my density of states is given by $$g(E) = \frac{1}{A}\frac{dN(E)}{dE} = \frac{1}{A}\frac{2mL^2}{\pi \hbar ^2} = \frac{2mL^4}{\pi^3 \hbar ^2} ???$$

how can this be right if I am looking for g as a function of E, but E does not appear in this expression?

7. Dec 17, 2009

### phsopher

The expression before last is correct, I think. However the V in $$g(E) = \frac{1}{V}\frac{dN}{dE}$$ refers to the volume in position space (volume of the "box") not volume in momentum space. As to E dependence, it seems that for a 2D system the energy density is constant.