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A question regarding Landau tubes:
I cannot understand why pictures of the Landau tubes are in k space as the Fermi sphere, especially considering the fact that $k_y$ ceases to be a good quantum number to describe an electron in presence of a magnetic field.

EDIT- just to be clearer: the Fermi sphere is a depiction of the occupied k states; this makes sense when the electrons are free so that the k vector quantum numbers ($k_x$ $k_y$ $k_z$) are good quantum numbers to describe an eigenstate of the Hamiltonian. When we turn on a magnetostatic field clearly only 2 out of three of the k quantum numbers remain useful to describe an eigenstate of the Hamiltonian; so why are pictures of the landau tubes in the same k space as the Fermi sphere?

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    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented May 9, 2022 at 15:35

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My understanding is that the question is about figure like this one (taken from thread What's the coordinate of the Landau tube?):
enter image description here

Magnetic field does not change the number of electronic states or the energy of electrons, that is, the electron states are still filled up to the Fermi level (Fermi level is not the same as Fermi energy, but in case of a metal they coincide.) This means that at zero temperature all the states with energies below the Fermi level are filled, whereas the states above the Fermi level are empty.

Without magnetic field this means that $$ \frac{\hbar^2 k_x^2}{2m}+\frac{\hbar^2 k_y^2}{2m}+\frac{\hbar^2 k_z^2}{2m}\leq \epsilon_F,$$ as the states are characterized by quantum numbers $k_x,k_y,k_z$. One usually assumes periodic boundary conditions, so this can be thought of as a sphere of radius $k_F=\frac{1}{\hbar}\sqrt{2m\epsilon_F}$ filled with small cubes.

With magnetic field present, the filled states are defined by condition $$ \hbar\omega_c\left(n+\frac{1}{2}\right)+ \frac{\hbar^2 k_z^2}{2m}\leq \epsilon_F,$$ where the quantum numbers are now $n, k_z$ (the energy is degenerate in the third quantum number, the choice of which is gauge-dependent.)

Landau tubes is a visualization of this last relation, via superimposing it with the Fermi sphere that existed without the magnetic field applied. The electrons are confined to the sheets, depicted as the tubes, their quantum numbers being the number of the sheet (counted from the inner one, having index $n=0$, and the position along the z axis. The crossing between a tube and the Fermi sphere shows the range of the momenta $k_z$ of the states filled within each Fermi level.

This depiction is particularly convenient, e.g., when discussing De Haas-Van Alphen effect, where increasing the magnetic field corresponds to increasing the radius of the tubes, the reduction of the range of the momenta allowed for the outer tube, until the tube completely crosses the Fermi sphere. The tubes crossing one-by-one the Fermi sphere is then a convincing visual representation for the rise of the oscillatory behavior of susceptibilities with magnetic field (image source):
enter image description here

Thus, the visual appeal of this presentation is not related to how we solve the Schrödinger equation in magnetic field - which is probably what motivated the question. Indeed, discussing Landau gauge and Landau tubes alongside, and the fact that the name of Landau appears in both terms may create false impression that one should follow from the other. The argument behind Landau tubes is gauge independent, and perhaps the discussion would be less confusing, if the Schrödinger equation was solved in a symmetric gauge, $\mathbf{A}=\frac{1}{2}\mathbf{B}\times\mathbf{r}$, where it would give rise to Darwin-Fock states - but the discussion of these is somewhat more involved.

Remark: (in response to comments)
Landau tubes should not be taken literally, as a representation of the density-of-states in k-space. That is, it is incorrect to think that every point on a Landau tube represents a state with momenta $k_x,k_y,k_z$ - the interpretation that is correct without magnetic field. The correct representation of the Landau levels would be a diagram like this (image source):
enter image description here
Each parabola represents a Landau level, filled with electrons up to the point where the parabola intersects with the Fermi energy. The base of parabola is $\hbar\omega_c\left(n+\frac{1}{2}\right)$. De Haas-Van ALphen effect the corresponds to increasing spaces between the parabolas and the upper parabolas crossing the Fermi level. If necessary, this figure could be extended in the third dimension, to represent the degenerate quantum number (the choice of which depends on the choice of the gauge.)

Landau tubes are intended as an intuitive representation, presumably helpful to people with extensive experience in the theory of metals, who are used to think about metal in terms of the Fermi sphere, and those who need to make a link between the phenomena traditionally interpreted in terms of Fermi sphere and the phenomena in magnetic field.

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  • $\begingroup$ Hi Roger, thanks for the answer. I'm still trying to fully understand it, and I think your last paragraph is pointing towards my confusion. $\endgroup$
    – user196574
    Commented Apr 25, 2023 at 0:04
  • $\begingroup$ Why does filling $\hbar\omega_c\left(n+\frac{1}{2}\right)+ \frac{\hbar^2 k_z^2}{2m}\leq \epsilon_F$ correspond to filling up crisp tubes in a space marked by quantum numbers $k_x, k_y, k_z$? For example, if I were to describe the plotted tubes, it seems I'd use the equation $\hbar^2 (k_x^2 + k_y^2)/2m = \hbar \omega_c (n + 1/2)$ to draw the $n$th tube. Why is that? It seems like we can't have both $k_x$ and $k_y$ be good quantum numbers, so I don't see why the tube has a well-defined boundary in $k$-space. $\endgroup$
    – user196574
    Commented Apr 25, 2023 at 0:04
  • $\begingroup$ @user196574 I added a Remark in my answer in response to your comment. I think you are trying to attribute more meaning to Landau tubes than they actually carry. This is vintage Landau - where the "simple" explanation is often harder to understand than the math. I think Landau tubes are important to know in order to read articles, particularly those written by experimentalist or Landau students. But they are not a theoretical tool that one would ever need to use. $\endgroup$
    – Roger V.
    Commented Apr 25, 2023 at 8:06
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    $\begingroup$ Hi, thanks for the answer. I especially found the "Remark" paragraph really useful since it was the main point of my question. Thanks a lot $\endgroup$ Commented Apr 25, 2023 at 9:23
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    $\begingroup$ Thanks, I appreciate the figure with $k_z$, $E$, and $E_F$ - I think it makes the physics of quantum oscillations very clear. $\endgroup$
    – user196574
    Commented Apr 26, 2023 at 0:02

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