# Fermi level with Landau levels

So my question is regarding where the Fermi energy is when you have 2D electron gas in an applied magnetic field. My book explains that, using the Landau gauge, you find that the 2D density of states (normally the constant $g_{2D} = m/\pi \hbar^2$) collapses into a series of Dirac delta functions $g_{2D}\delta(E - E_n)$ for the harmonic oscillator levels $E_n = \hbar \omega_c (n + 1/2)$ ($\omega_c$ being the cyclotron frequency) because in this gauge, all the states with the same $n$ are degenerate. This is illustrated in the top half of this picture.

They also say that, due to scattering, the peaks can't be defined with such high precision, so you actually get shapes more like the bottom half of that picture. This all makes sense to me.

But I get confused when I try to find the Fermi energy for a given density and nonzero magnetic field (at $T = 0$, always...) without the scattering that widens the peaks. With no magnetic field, I would simply integrate the 2D DoS from $0$ to $E_F$ and set that equal to my number density, and solving for $E_F$, I'd find it somewhere. But with a magnetic field I run into a problem: My understanding of the Fermi energy is that it's the energy such that all states at or below it are occupied and all those above are unoccupied.

But this seems to only offer certain values of the number density for this case! When you integrate over a range of these Dirac deltas, you essentially pick up $g_{2D}\hbar \omega_c$ states for each delta you integrate over, so it seems like in the case of an applied B field, $T = 0$, and no scattering that widens the deltas to something more realistic, you can only have number densities of the form $n_{2D} = m g_{2D}\hbar \omega_c$, where $m$ is an integer.

I'm sure I'm not explaining this well and that I'm probably wrong, but what am I missing? Thanks!

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Could please add a reference to the book and the specific section you have in mind? – Philosophiæ Naturalis Dec 15 '15 at 16:14

I absolutely understand what you mean. Just assume that the last $\delta(E-E_n)$ is only partly occupied. It's actually a common strategy to widen such Delta functions a little bit when integrating numerically, and then the procedure is obvious.

There is nothing wrong with that, you just have to know how to handle the math - and the easiest way is to use broaden the peak just a tiny bit (which is not a physical, but a mathematical trick!).

You can also use this trick when calculating

$$D(E) = \int\delta(E-E(k))\,dk,$$

which you often have to do numerically, and you have to express the Delta term in a way that a computer can handle - which is a peak of finite height.

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Thanks for the response. It's definitely obvious to me how you do it when they have finite width, as they always will in real life, but does that mean the theory just breaks down without that mechanism? – declan Apr 7 '13 at 16:14
No, it just means that the Delta function is sometimes hard to handle in real physics --> see edit. – Rafael Reiter Apr 7 '13 at 16:36
Okay, I think I know what you're talking about and I think I've done it before -- taking the DD to be a Gaussian or something, and then taking the limit as the width goes to zero to get back the DD? But, I'm still confused about where the actual Fermi energy will end up -- with the DD's, it's either exactly on one, or in between them, which is equivalent to being exactly on the previous one. – declan Apr 7 '13 at 17:19
You're correct. The Fermi energy will be at one of the Delta functions. – Rafael Reiter Apr 7 '13 at 17:48