Conduction Band Density of States for 2D Electron Gas

In summary: It makes perfect sense now. In summary, the conduction band density of states for a two dimensional electron gas in the inversion layer of a semiconductor interface is given by ##g(E)=\frac{L^2m^*}{\hbar^2 \pi}##, where m* is the effective mass of the electron. The density of states per spin is expressed as ##g(k) k dk = 2 \frac{L_x L_y}{\pi} k dk## to account for spin degeneracy. This leads to the formula ##g(E) = \frac{4L^2 m}{\hbar^2 \pi}##. The total electron density is then calculated by integrating the density of
  • #1
roam
1,271
12
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? :confused:

Any response is greatly appreciated.
 
Physics news on Phys.org
  • #2
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)1/2Thus 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:
  • #3
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? :confused:

Much appreciated
 
  • #4
roam said:
##n(E)=\frac{L^2m^*}{\hbar^2 \pi} \frac{1}{e^{(E-\mu)/k_B T}+1}##[/CENTER]

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

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

What is the conduction band density of states for a 2D electron gas?

The conduction band density of states for a 2D electron gas is the number of available energy states per unit energy and per unit area in the conduction band of a 2D material. It represents the distribution of energy states for electrons in the conduction band in a 2D system.

Why is the conduction band density of states important?

The conduction band density of states is important because it determines the electronic properties of a 2D material. It affects the conductivity, mobility, and other transport properties of electrons in the material.

How is the conduction band density of states calculated?

The conduction band density of states can be calculated using the density of states function, which takes into account the effective mass and energy dispersion of electrons in the conduction band. It can also be obtained experimentally through techniques such as angle-resolved photoemission spectroscopy.

What factors affect the conduction band density of states in a 2D electron gas?

The conduction band density of states is affected by the material properties such as the band structure, lattice structure, and impurities. It can also be influenced by external factors such as temperature and electric or magnetic fields.

How does the conduction band density of states change in different 2D materials?

The conduction band density of states can vary significantly between different 2D materials due to their unique band structures and material properties. For example, graphene has a linear energy dispersion in its conduction band, resulting in a constant density of states, while other materials may have parabolic or other types of energy dispersions, leading to different density of states distributions.

Similar threads

Replies
2
Views
2K
  • Atomic and Condensed Matter
Replies
0
Views
393
  • Atomic and Condensed Matter
Replies
4
Views
2K
  • Atomic and Condensed Matter
Replies
4
Views
962
  • Calculus and Beyond Homework Help
Replies
7
Views
729
  • Advanced Physics Homework Help
Replies
6
Views
1K
  • Atomic and Condensed Matter
Replies
7
Views
5K
  • Atomic and Condensed Matter
Replies
7
Views
2K
  • Atomic and Condensed Matter
Replies
5
Views
2K
  • Atomic and Condensed Matter
Replies
3
Views
3K
Back
Top