Here is a way to derive the increase in energy of a free-electron gas under a uniform magnetic field without calculating the explicit expression for free energy, as presented in Chapter 15 of Grosso & Parravicini. enter image description here

One first considers a slice of the Fermi sphere with constant $k_z$. As the uniform magnetic field turns on, the allowed energy levels within each slice collapse into discrete Landau levels. Within $\pm\hbar\omega_c/2$ of the Fermi energy $E_F$, where $\omega_c$ is the cyclotron frequency, there is one Landau level at $E_n$, $\varepsilon$ away from $E_F$ and it is the only one responsible for the energy change in each slice. So far, so good.

Now begins the part I have trouble with. The authors write that if we consider the case $\varepsilon>0$, the electrons belonging to the energy interval $[E_n-\hbar\omega_c/2,E_F]$ are shifted to the Fermi energy and redistributed to other slices. The case of $\varepsilon <0$ gives a similar increase in net energy. Furthermore, the excess of states transferred to the Fermi energy in the slices with $\varepsilon >0$ are accommodated by an equal number of extra states available in the slices with $\varepsilon <0$, so for $\hbar\omega_c << E_F$, the Fermi energy is independent of the magnetic field.

Now my questions in specific forms

  1. In the above figure, why are the electrons in the colored interval "redistributed to other slices"? Because of translational symmetry along z, which is not broken by the magnetic field, every state should retain its value of $k_z$ right?
  2. Isn't the figure wrong in what it labels as "$E_F$"? Shouldn't it be really "$E_F - \hbar^2 k_z^2/2m$"? The Fermi energy is defined by the radius of the Fermi sphere so it should be the same whatever $k_z$ slice you consider.
  • $\begingroup$ Does your first point (1.) imply that the uniform magnetic field is perpendicular to the z-direction (or that the field is parallel to the xy-plane)? $\endgroup$ – Deschele Schilder Feb 3 '20 at 7:38

Answer to question 1

Suppose that the system is periodic in the $z$ direction so that the multi-electron system has translation symmetry along $z$, despite being confined to a finite volume. (The finite volume condition is necessary in order to have a non-zero Fermi energy with a finite number of electrons.)

A magnetic field parallel to $z$ does not break this translation symmetry, but this does not imply that every occupied mode should retain its value of $k_z$ as the strength of the magnetic field is varied. The single-electron modes vary smoothly with the field strength, but which of those modes are occupied may change (necessarily discontinuously, because it's a discrete set). This happens because as the magnetic field is changed continuously, two total-energy eigenvalues may cross over each other, and then shifting an electron from one value of $k_z$ to another may be required in order to remain on the lowest-energy branch.

To be explicit, consider this expression for the energy of a single mode (equation (3.9) in Harrison's Solid State Theory, 1979): $$ E_1 = \hbar\omega_c(n+1/2)+\frac{\hbar^2 k_z^2}{2m} \tag{1} $$ where the cyclotron frequency $\omega_c$ is proportional to the strength of the magnetic field. Each pair $(n,k_z)$ specifies a mode, and only two electrons (one for each polarization) can occupy each mode. Now consider a three-electron system. Two electrons can go in the $(0,0)$ mode, but the third electron must go in a different mode. If the magnetic field is weak enough, then putting the third electorn in the mode $(1,0)$ gives the lowest energy. If the magnetic field is strong enough, then putting the third electron in the mode $(0,k_z^1)$ gives the lowest energy (where $k_z^1$ is the lowest non-zero value of $k_z$). Somewhere between those extremes, the levels cross, and the electron must switch from a zero value of $k_z$ to a non-zero value of $k_z$ in order to keep the energy minimized.

By the way, equation (1) ignores the finite-volume requirement in the $x$ and $y$ dimensions, which is why (1) doesn't become $\propto k_x^2+k_y^2+k_z^2$ as $\omega_c\to 0$ for fixed $n$. (The transformation between the $k_x,k_y$ basis and the Landau-mode basis depends on the magnetic field, so properly taking the zero-field limit requires consideration of this transformation as well as the energy levels.) However, in a sufficiently large volume, the idea behind the preceding argument still applies.

For additional confirmation, see section 5.2 in Dresselhaus, Solid State Physics Part III: Magnetic Properties of Solids (http://web.mit.edu/6.732/www/6.732-pt3.pdf).

Answer to question 2

I think part (b) of the figure shown in the OP is labeled correctly, if the "Landau level" $E_n$ is taken to include the $\hbar^2 k_z^2/2m$ term. The horizontal axis in that figure starts at $\hbar^2 k_z^2/2m$, so the left-hand edge of part (b) of the figure corresponds to the $k_z$-axis of the Fermi sphere shown in part (a), and the point labeled $E_F$ in part (b) corresponds to the surface of the sphere (at the given value of $k_z$) in part (a). In other words, the horizontal axis in part (b) represents the total energy of a mode with a given value of $k_z$, not just the additional energy beyond $\hbar^2 k_z^2/2m$.

  • $\begingroup$ So when an electron must switch from a zero value of $k_z$ to a non-zero value of $k_z$ in order to keep the energy minimized, this is actually carried by collision between electrons correct? So for an ideal non-interacting Fermi gas, we will not see the re-distribution. $\endgroup$ – wcc Feb 2 '20 at 4:40
  • $\begingroup$ @wcc You're right that some kind of interaction must be present in order for the multi-electron system to shed excess energy and remain in its ground state as the magnetic field is varied. Mutual interaction (collision) between the electrons themselves might not be required, but at least some interaction with the surrounding medium. The key assumption is that we can neglect those interactions when studying the ground-state energy as a function of the magnetic field. $\endgroup$ – Chiral Anomaly Feb 2 '20 at 20:42

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