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Significance of free electron gas density of states in different dimensions? 
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#1
Dec2912, 11:06 AM

P: 146

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
Dec2912, 12:17 PM

Sci Advisor
P: 1,659

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 vanHove singularities. For example if you do spectroscopy on 1Dnanowires, these can show up as really sharp peaks in absorption spectra. 


#3
Dec3012, 05:08 PM

P: 146

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
Dec3012, 07:52 PM

Sci Advisor
P: 1,659

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



#5
Dec3012, 08:24 PM

P: 159

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



#6
Dec3012, 08:25 PM

P: 146

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
Dec3012, 08:26 PM

P: 146




#8
Dec3012, 08:45 PM

P: 159




#9
Dec3112, 02:00 AM

P: 185

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 [tex]g(E) = \int \frac{dS}{\nabla_k E(k)}[/tex] 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
Dec3112, 08:59 AM

P: 146

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
Dec3112, 10:28 AM

Sci Advisor
P: 1,659

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
Dec3112, 02:36 PM

P: 185

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 kspace. 


#13
Jan2013, 08:10 PM

P: 146




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