# Density of States

Discussion in 'Electronic Basics' started by Sunil, Feb 5, 2007.

1. ### SunilGuest

I was reading up on the density of states and i hit on a snag...

I dint understand how they take the "volume" of the density of
states...i mean density of states corresponds to the density of energy
level rite...how can u consider the "volume" of energy level??

Moreover....can you also give me the qualitative reason for the
density of states function??

thank you

2. ### redbellyGuest

Um, this really isn't a "basic electronics" question.

Regards,

Mark

3. ### Bob PownallGuest

I'd have to agree.

I'd suggest that the OP try posting to sci.engr.semiconductors instead.

Bob Pownall

4. ### Jon SlaughterGuest

Its exactly what it says. Given a specific energy level E we can find the
density of the states at that level per unit volume per unit energy(we have
to normalize so its more useful).

Now any time your talking about density you must count something and then
divide by some related factor that in some sense normalizes it. Mass
density is the the mass(which is effectively the "count" of the number of
atoms) divided by the volume that contains that mass. Ofcourse this is a
little easier to do as we don't have to know the mass of the atoms nor how
many because we have newtons laws to help us out. (we just weigh the thing
and measure its volume through whatever means we can).

So in this case we have to somehow count the states of the electrons(or
whatever were describing). Think of a material that has its electrons all
in different states. We sorta want to have a means of knowing the density of
those states in the material. Basically a proportion of the states that
contribute to the total energy in the volume.

That is, we need to find all the states that contribute to energy E and then
divide by the volume we looked at and divide by the total energy. Think of
an atom with its electrons in different states. These states correspond to
definite energy levels but what proportions correspond to what energy
levels?

For a free electron we have E = (hbar*k/2/m)^2. This gives us the energy in
terms of phase space.

The key here is that phase space is a space of states. Each unit in the
phase space corresponds to a specific state. That means now its really easy
to add up the states because we just have to integerate over a volume in it.
But what volume? Well we know that its spheres because constant energy
levels result in sphere in k space. Remember, if you solve shrodingers wave
equation you get that psi_k is dependent on k. Think of k as an independent
variable you then just have to add up the right number of psi_k's that
depend on E because thats what were after. (states in terms of energy)

So take our free particle,

E = (hbar*k)^2/2/m

and then integrate over a sphere with radius k = sqrt(2mE/hbar^2). Remember,
essentially we want to count all the states within our constant energy
surface. In k space we just have to integrate. Since we are dealing with
spheres in k space and volume tells us the number of k spaces(if say, we
were dealing with a 1x1x1 = 1 cube then we would have exactly one state, if
a 2x3x1 cube we would have 6 states). Therefor the number of states that
give us an energy less or equal to E is

2*4/3*pi* sqrt(2mE/hbar^2)^3

which is simply the volume of a sphere in k space with radius
sqrt(2mE/hbar^2). The extra factor of 2 comes in because each state is
actually 2 states(spin up and spin down for electron).

So now we are counting the number of states that have at most energy E. But
here we assumed that the unit length was a state when in fact its L/2/pi. (I
didn't scale my axis because I forgot). This means that my volume is off by
a factor of (L/2/pi)^3. (instead of counting unit cubes I have to count
different sized cubes and that means I just have to multiply by the scaling
factor).

So what I really have is

(L/2/pi)^3*2*4/3*pi* sqrt(2mE/hbar^2)^3
= L^3/3/pi^2/hbar^3*(2mE)^(3/2)

This counts the states who has energy <= E. Its not hard but you have to
just not think about it as being to hard. Just count the states. To do that
you just have to know 3 facts:

Each cube with edge L/2/pi is a state in k space.
A sphere of radius E represents a constant energy.
Each state needs to be counted twice for an valence electron.

So now we have counted the total number of states that give an energy <= E
in a volume of L^3(were talking about a free electron in a cube with edge
L).

Now we just have divide out the total volume and differentiate it w.r.t E to
get number of states that contribute to energy E(not <= E). Actually with
would be E*dE.

Dividing out the volume(effectively normalizing it w.r.t volume) gives

1/3/pi^2/hbar^3*(2mE)^(3/2)

and differentiating will give you the density of states of E. Now I'm not
going to differentiate it but approximate it. We are trying to find the
number of states that are between two energy levels that are close together,
say E + dE and E.

[1/3/pi^2/hbar^3*(2mE + dE)^(3/2) - 1/3/pi^2/hbar^3*(2mE)^(3/2)]/dE

but as you can see this really is just the derivative. We had to divide by
dE so the limit would not be zero but technically its the
derivative*dE(which you might not see the factor of dE there in your
forumlas).

The resulting equation for the density of states is

d/dE[1/3/pi^2/hbar^3*(2mE)^(3/2)] = [1/2/pi^2/hbar^3*(2m)^(3/2)*E^(1/2)]dE

Hopefully that made sense. Essentially we go into phase space where its much
easier to count states, add up all the states that give us an energy between
E and E + dE(which is easier if we first just count those that give us an
energy <= E), this evolves finding volumes over spheres, then substitute k =
K(E) into the equation to get it in terms of E and divide by the volume to
normalize it w.r.t to volume.

I'm not sure if I did a decent explination as it was off the top of my head
but I'm sure there are plenty of resources on the net. Its not necessarily
a hard thing to understand but you need not get confused on the details. Its
just a counting problem. (and if I didn't say it I was doing it for a free
particle in a box because thats easier)

Jon  