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In our lecture, we had the following scenario:

Suppose we have a 2D-free electron gas which is in a magnetic field $B$ perpendicular to the free electron system. Now the electron states have energies of $\hbar \omega_c\left( \lambda + \frac{1}{2} \right)$, where $\lambda \in \mathbb N_{0}$ is the Landau Level index and $\omega_c$ the cyclotron frequency.

Now the professor stated in his script (without proof) that for the internal energy $U$ (for $k_B T \ll \hbar\omega_c$, $k_BT\ll E_F$, $\mathbf{T=0}$, for high magnetic fields), the following formula holds:

$$ U = \hbar\omega_c \left[ n\left( s+\frac{1}{2}\right) - s n_L\left( s + 1 \right) \right], $$ where $$n_L = \frac{eB}{h}$$ is the Landau degeneracy, $n$ the electron density and $s$ the highest occupied state (thanks @Matteo!).

Any hints on how to prove this (I didn't find anything online or in an old lecture script of Statistical Physics) would be aprreciated!

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  • $\begingroup$ Sorry maybe it's a stupid comment but why is there the Landau level index $\lambda$ in the expression of $U$? A thermodynamic quantity should not depend on it. Is it maybe the index of the highest occupied Landau level? $\endgroup$
    – Matteo
    Jun 30, 2020 at 7:43
  • $\begingroup$ @Matteo No, the stupidity lies on my side, I updated the question! Thanks! :-) $\endgroup$
    – user248824
    Jun 30, 2020 at 11:02

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Let me change the notation with $g \equiv n_L = eB/h$. Now the internal energy is given by the sum of all the energies of all the particles in the system. Now since the temperature is much smaller than the other energy scales (magnetic field and Fermi energy), we can simply set $T=0$. Also ignore the spin of the electron (I'll comment on this later on).

So think about the electronic structure of such system: you have a bunch of levels labeled by a positive integer $\lambda$ with spacing given by $\hbar \omega_c$, each of them can accomodate $2g$ electrons. Now everything depends on how many electrons you have, for example if you have $N<2g$, all the electrons will occupy the ground state and the internal energy will be $U = N E_0 = N\hbar\omega_c/2$. If you have $2g<N<4g$ electrons, they will occupy all the available states with $\lambda = 0$ and they will occupy $N-2g$ levels with $\lambda=1$, their total energy will be $U=2g \hbar\omega_c/2 + (N-2g) 3\hbar\omega_c/2 $.

In general, if you have $2(\lambda_0-1)g < N < 2\lambda_0 g$ electrons, they will occupy all the available states with labels $\lambda=0,1,...,\lambda_0-1$ and will partially occupy the levels with label $\lambda_0$. Their internal energy will be: $$ U = 2g\sum_{\lambda=0}^{\lambda_0-1} \hbar\omega_c \left(\lambda + \frac{1}{2} \right) + (N-2g \lambda_0) \hbar\omega_c \left( \lambda_0 + \frac{1}{2} \right). $$ I'll leave the maths as an exercise for you, but it's very easy and you finally get your professor's result!

PS: a brief comment about spin: please notice that the comments above are correct if the Hamiltonian does not include the Pauli term $-\bf{m}\cdot \bf{S}$, where $\bf{m}$ is the magnetic moment of electrons and $\bf{S}$ the spin operator. When the Pauli term is included, there is a huge effect on Landau levels. The first effect is that the states are now labeled by $\hbar\omega_c \lambda$ (with vanishing zero-point energy), the second one is that the level with $\lambda=0$ has degeneracy $g$ while all the others have degeneracy $2g$.

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  • $\begingroup$ Thanks! I will take a look at it whenever I can and try to obtain the result! :-) $\endgroup$
    – user248824
    Jun 30, 2020 at 11:04
  • $\begingroup$ Ok! then let me know :) you might find useful the following result: $\sum_{n=0}^m n = m(m+1)/2$ $\endgroup$
    – Matteo
    Jun 30, 2020 at 11:50
  • $\begingroup$ Honestly, I don't know whether I agree with you on the spin-part. In the exercise sheet, it says that because of spin degeneracy, we have $2g$ electrons that fit into each Landau level ... Apart from that, I could verify the formula!! Thanks a lot for that. ;-) $\endgroup$
    – user248824
    Jun 30, 2020 at 14:02
  • $\begingroup$ Yes, that is how you account for spin if you don't include the Pauli term $-\bf{m}\cdot\bf{S}$ in the Hamiltonian, where $\bf{m}$ is the magnetic moment. If you introduce this term, all the levels are messed up $\endgroup$
    – Matteo
    Jun 30, 2020 at 15:57
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    $\begingroup$ If you can set $T=0$, then the electrons occupy the lowest energy levels. If otherwise $T \neq 0$, then there might be excitations, so not necessarily the electrons will occupy the levels with lowest energy! If for instance $N<2g$, then you may have $N-1$ particles in the $\lambda=0$ level, and $1$ particle in the $\lambda=1$ level and so on $\endgroup$
    – Matteo
    Jun 30, 2020 at 19:23

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