# Significance of free electron gas density of states in different dimensions?

1. Dec 29, 2012

### VortexLattice

Hi all,

I was deriving the free electron gas for practice in 1, 2, and 3 dimensions, and I started wondering why they have different dependencies on energies and what that means. I got:

1D: $g(E) = \frac{1}{\pi\hbar} \sqrt{\frac{2m}{E}}$

2D: $g(E) = \frac{m}{\pi\hbar^2}$

3D: $g(E) = \frac{1}{2\pi^2} (\frac{2m}{\hbar^2})^{3/2} \sqrt{E}$

Why does this happen, though? I mean, I did all the math so I see where it comes from, but I don't have a good intuitive reason... It's very strange to me that in 3D, it has a 'reasonable' dependence on the energy, but then it becomes constant when confined to 2D, and then gets energy dependence again when confined to 1D.

Also, what does it mean that in the 1D case, the density of states diverges at $E = 0$? Does this mean that at low energies, there are (approaching) infinite states for electrons to fill? That starts to contradict what I know about the density of states.

Thank you!

2. Dec 29, 2012

### Cthugha

Everyone considers different things as intuitive. Did you try to make a sketch of the surfaces of equal energy in momentum space? These will be a sphere in 3D, a circle in 2D and, well, a point in 1D. As the dispersion is quadratic in k, these surface volumes grow "stronger", in the same manner or "weaker" than the dispersion does.

And, yes, the DOS can have singularities. These are called van-Hove singularities. For example if you do spectroscopy on 1D-nanowires, these can show up as really sharp peaks in absorption spectra.

3. Dec 30, 2012

### VortexLattice

Hi, I knew the shape of the equi-energy surfaces in k-space, but I still kind of don't see why that should change the energy dependence of how many states there are at some E, for a given dimension.

Like, in the dispersion relation, the $k$ is only the magnitude of the wave vector. So why does how many dimensions it's in matter?

So, for the singularities, what are peaks caused by? Is it that, at those energies, the crystal can have many states, so many electrons can be excited by the incoming photons to be in those states?

Thank you!

4. Dec 30, 2012

### Cthugha

Yes, indeed it is only the magnitude of the wave vector. That is the important point. It is equal to $$\sqrt{k_x^2 + k_y^2 + k_z^2}$$ in three dimensions. So you obviously have much more ways to 'combine' these three components to arrive at one certain wave vector magnitude than you have in one dimension, where you have only one component of the wave vector which also gives the magnitude. Each of these combinations is a state leading to the same energy, so you obviously get more states at equal energy with more dimensions.

5. Dec 30, 2012

### alemsalem

should these be multiplied by the volume( L, L^2, L^3)?

6. Dec 30, 2012

### VortexLattice

Hmmmm, so this makes sense to me. It makes sense to me that if there are more ways to get the same magnitude $k$, there should be more states at that energy. But why, physically, do we actually see decreasing number of states as energy increases, for the 1D case? Some sort of dependence like:

1D ~ $\sqrt{E^0}$, 2D ~ $\sqrt{E^1}$, 3D ~ $\sqrt{E^2}$

(or something) would make sense to me. But if you have electrons confined to an infinitely long wire, why do their states start getting limited as the energy gets higher?

7. Dec 30, 2012

### VortexLattice

It's pretty common to divide out by the volume, since you often use boundary conditions to impose a volume that doesn't actually exist.

8. Dec 30, 2012

### alemsalem

I guess its because the energy will go like n^2.. the levels will be farther apart and you will have less states for a fixed energy interval dE,, it doesn't happen in three dimensions because as Cthugha mentioned because the volume of the spherical shell in momentum space will grow fast with increasing energy, while in 1d the volume in momentum space will not grow as fast

9. Dec 31, 2012

### daveyrocket

It might help to think of the density of states as a distribution function, g(E) dE is the number of states between energy E and E + dE. Look at the dispersion in 1D, E = k^2. And remember that every distinct value of k is a different state. You're counting the number of states between energy E and E + dE, if you take E = 0 you're going to have a very large number of possible k values between 0 and dE because E(k) = k^2 is very flat there.

One way to write the density of states is $$g(E) = \int \frac{dS}{|\nabla_k E(k)|}$$
where dS is an integral over the constant energy surface E. So in regions where E(k) varies very slowly, you can get a large density of states. This is balanced by area of the surface integral. In 1D, the area of this 'surface' is the same for all energies, just two points. Then the denominator is equal to sqrt(E), so this gives the E^-(1/2) dependence.

In 2D, the area of this 'surface' is a circle, with circumference proportional to |k| = sqrt(E). The denominator is still sqrt(E) so they cancel and you get a constant value.

In 3D, the surface is a sphere, with surface area k^2 = E. You can do the math.

10. Dec 31, 2012

### VortexLattice

Thank you, but this is still just a mathematical explanation. It's still just saying, the slope of $E$ increases as $k$ increases, so in this mathematical definition of the density of states, you find these relations.

Sorry, I know it's nitpicky, but I'm trying to think of an intuitive explanation for why the 1D case actually has decreasing states for increasing k. I could buy that they're different, and 2D and 3D have more states, but I don't see why the baseline, 1D, actually has decreasing.

Thank you!

11. Dec 31, 2012

### Cthugha

Hmm, I somewhat do not see your problem as alemsalem has already explained the other part, but let me try to rephrase.

The dispersion of the free electron gas shows a quadratic dependence on the magnitude of the wave vector in any dimension. If we had discrete energy levels, the spacing between adjacent levels would grow larger. For continuous energy levels, this instead means that the density of states necessarily decreases with increasing energy. This decrease is the same independent of dimensionality.

Now this effect may be countered by the effect that your wave vector may have more than one component and you may find more ways to combine the wavevector components to a certain wave vector magnitude and therefore also energy.

In two dimensions, these effects exactly cancel, giving you a constant DOS. In three dimensions, this effect overcompensates the other one, giving you a DOS increasing with energy. In one dimension, you have just one wavevector component and this effect does not compensate the decrease of the DOS caused by the quadratic dispersion mentioned above. Therefore you are left with a decreasing DOS in one dimension.

12. Dec 31, 2012

### daveyrocket

I'm not sure what you're looking for. These are mathematical quantities, so the relationship between them is going to be mathematical. Reread my first paragraph above. Consider the density of states as a distribution function. g(E) dE is the number of states between E and E + dE.

Consider a discrete situation instead of the continuous situation. Say you have a 1D system with, say, N = 100, then your allowed k values are 1/N apart. Consider a finite dE, say 0.01. Count the number of allowed k values for 0 <= E < dE = 0.01. E = k^2 so there are 10 allowed k values (k = 0 to k = 0.1, so g(0) dE = 10. Then count the number of allowed k values for 0.01 <= E < 0.02 and there are 5 (from k = 0.1 to k = 0.14). Then count the number of allowed k values for 0.02 <= E < 0.03. There are 3.

Because E(k) increases faster at larger k values, at larger values of E there are fewer available states "nearby" in k-space.

13. Jan 20, 2013

### VortexLattice

Ok, this actually helped me visualize it a lot. Thanks a bunch!