# Band density of states

1. Aug 16, 2013

### roam

For an electron gas generated in the inversion layer of a semiconductor interface, my book gives the conduction band density of states for the two dimensional electron gas as:

$g(E)=\frac{L^2m^*}{\hbar^2 \pi}$​

Where m* is the effective mass of the electron. I can't follow how this was exactly derived.

So the density of state is given by

$g(E)=2g(k) \frac{dk}{dE}$​

Where

$E=\frac{(\hbar k)^2}{2m} \implies \frac{dE}{dk}= \frac{\hbar^2 k}{m}$

And also the density of states per spin: $g(k) k dk = 2 \frac{L_x L_y}{\pi} k dk$

Hence substituting I've got:

$g(E) = \frac{2\times 2 L^2 k m}{\pi \hbar^2 k} =\frac{4L^2 m}{\hbar^2 \pi}$​

But why do I end up with a "4" on the numerator? Did I make a mistake, or is that a typo in the book?

Any response is greatly appreciated.

2. Aug 16, 2013

### Jolb

I don't know why you have Lx and Ly--are you referring to spin operators? (They would typically have the symbol S rather than L which is orbital angular momentum). L here refers to the size of the 2DEG (L~"length" of a side) and not to anything related to angular momentum. The factor of 2 you have in your second formula is there to correct for spin degeneracy, so you've taken care of spin problems already.

The way I always do DOS calculations is by plotting the momentum eigenstates on the axes (kx, ky), then drawing an annular region of radius k and width dk and calculating how many eigenstates would on average lie inside the first quadrant of that annulus. This recipe immediately generalizes to higher dimension if you remember the surface areas of n-spheres. That was a quick description so let me spell it out.

So the calculation would go like this:
First solve the schrodinger equation for the 2-dimensional square with side length L using appropriate boundary conditions. You get
kx=nxπ/L
ky=nyπ/L.

Now go ahead and draw these on the (kx,ky) axes. Each pair (nx, ny) occupies a region of area (π/L)2. Thus there is, on average, one state per area (L/π)2 in k-space.

Now we want to know how many states lie in the region between k=(kx2+ky2)1/2 and k+dk. This region would have the shape of an annulus, but since kx and ky must be positive, we are only concerned with the first quadrant. The area of that region would be dk*(2πk)*(1/4)=πk/2 dk. The number of states in this area would thus be (L/π)2 * nk/2 dk = L2k/(2π) dk

Now we want to substitute back using E=([STRIKE]h[/STRIKE]k)2/(2m)
dE = [STRIKE]h[/STRIKE]2k/m dk = [strike]h[/strike] (2E/m)1/2 dk
Thus k = (2mE)1/2/[strike]h[/strike]
dk = dE/[strike]h[/strike] (m/2E)1/2

Thus we have that the number of states in the area is
L2k/(2π) dk = L2/(2π) (2mE)1/2/[strike]h[/strike] dE/[strike]h[/strike] (m/(2E))1/2 = L2m/(2π[strike]h[/strike]2) dE.

Now since we have done everything so far without regards to spin degeneracy, we multiply the last expression by 2 and we get the right answer.

PS: sorry if it is hard to follow when I write fractions--- if there is a space after a term in a denominator, then the space should be an implicit multiplication sign. As an example, rs/p yz = yzrs/p

Last edited: Aug 16, 2013
3. Aug 17, 2013

### roam

Thank you very much for your response. I see that the factor of 2 accounts for spin, and the calculation makes perfect sense now.

I have another relevant question: my book says that the conduction electron density is given by

$n = \frac{m^* k_B T}{\hbar^2 \pi} \ln (e^{(\mu-E_c)/k_B T}+1)$​

But I can't get to this using the equation we found for density of states. Electron density is given by density of states multiplied by the Fermi-Dirac distribution function (the most probable number of particles in a particular state with energy E):

$n(E)=g(E) f(E,T)$

So​

$n(E)=\frac{L^2m^*}{\hbar^2 \pi} \frac{1}{e^{(E-\mu)/k_B T}+1}$​

But this is not the right equation. Am I using the wrong formula? Or does it have to be calculated in a different way?

Much appreciated

4. Aug 17, 2013

### Cthugha

People are usually interested in the total electron density, not some density of electrons in some range of energies. So you might want to have a look at the integral:

$n=\int g(E) f(E,T) dE$​

5. Aug 18, 2013

### roam

Thank you very much for the helpful post, completely missed the integral.