Symmetry in Landau's problem

Question

Suppose a particle is forced to move in the $x-y$ plane, under a constant magnetic field $\vec{B} = B\hat{z}$. The Hamiltonian can the be written as $$ H = \frac{\Pi^2}{2m} $$ where $\vec{\Pi} = \vec{P} - q\vec{A}$, where we choose the symmetric gauge $\vec{A} = \frac{B}{2}(-y,x,0)$. Using $\vec{\Pi}$, we can generate creation and annihilation operators $a = \frac{1}{\sqrt{2Bq\hbar}}\left(\Pi_x - i\Pi_y\right)$, such that $[a,a^{\dagger}] = 1$, and then $$ H = \hbar\omega_B\left(a^{\dagger}a + \frac{1}{2}\right) $$ Under this gauge, the operator $\widetilde{\Pi} = \vec{P} + q\vec{A}$ commutes with $\vec{\Pi}$, and we can also generate creation and annihilation operators $c = \frac{1}{\sqrt{2Bq\hbar}}\left(\widetilde{\Pi}_x - i\widetilde{\Pi}_y\right)$ such that $[c,c^{\dagger}] = 1$. Also, $c^{\dagger}c$ commutes with $[a^{\dagger},a]$, so we can find a common eigenbasis. This means that there are two quantum numbers, $n_a$ which are the eigenvalues of $a^{\dagger}a$ and $n_c$ which are the eigenvalues of $c^{\dagger}c$. Since the energy levels are dependent only on $n_a$, we conclude that there is infinite degeneracy in every level.

Now, the canonical angular momentum in the $z$ direction can be written as $L_z = -\hbar(a^{\dagger}a - c^{\dagger}c)$.

I am asked to explain the infinite degeneracy using symmetry.

The only symmetry I can see in this problem is rotations around the $z$ axis. I do notice that the energy levels are gauge-invariant, while the operators $c,c^{\dagger}$ are not, so there is no complete gauge symmetry. The fact that there is complete symmetry of rotations around $z$ cannot imply infinite degeneracy, as for example a $3D$ isotropic oscillator has complete symmetry of rotations around every axis, but does not have infinite degeneracy.

Which symmetry causes then the infinite degeneracy?

Answer 1

The existence of infinite degeneracy of an energy level implies the existence of a non-compact dynamical symmetry group commuting with the Hamiltonian. One may think about a non-compact generator as a translation operator (continuous or discrete), then all the translated versions of a state will have the same energy, therefore the degeneracy will become infinite.

The case of the planar Landau problem is very clearly explained by Joohan Lee.

The magnetic translation operators: $$T_i = \pi_i - \epsilon_{ij}Bx_j$$ ($x_i$ are the coordinates and $\pi_i$ are the canonical momenta), together with angular momentum operator $$J = \epsilon_{ij} x_i \pi_j + \frac{1}{2} B x_i^2$$ generate a central extension of the $E(2)$ Lie algebra. (It is a semidirect product of the Heisenberg-Weyl algebra and the rotation on the plane) $$ [T_i, T_j] = -i \epsilon_{ij}B $$ $$ [ J, T_i] = i \epsilon_{ij}T_j $$ The Hamiltonian can be expressed in terms of these operators: $$H = \frac{1}{2m}(T_i^2-2BJ)$$ The Hamiltonian commutes with all the generators, it is the Casimir of the algebra.

Since the operators do not commute we can diagonalize only one of them simultaneously with the Hamiltonian.

Suppose we choose to diagonalize $T_1$, then the eigenvectors of the Hamiltonian will be labeled by: $|E, t_1\rangle$, where $E$ is the Hamiltonian eigenvalue and $t_1$ is the translation operator $T_1$ eigenvalue. It is not hard to check that all the states: $$e^{i \alpha T_2}|E, t_1\rangle = |E, t_1+B\alpha\rangle$$ are degenerate for all $\alpha$.

Suppose we choose to diagonalize $J$ together with the Hamiltonian, then the eigenvectors of the Hamiltonian will be labeled by: $|E, j\rangle$, where $j$ is the angular momentum $J$ eigenvalue. It is not hard to check that all the states: $$(T_1+iT_2 )^n|E, j\rangle = |E, j+n\rangle$$ are degenerate for all $n$.

Thus in summary, An infinite degeneracy requires the system to have a non-compact symmetry group commuting with the Hamiltonian.

It is worthwhile to mention that the symmetry group itself (the central extension of $E(2)$) is finite dimensional, only its representation realized in the quantum Hilbert space is infinite dimensional. However, certain states of the second quantized version of the system possess infinite dimensional symmetry groups. In our case, quantum Hall systems, at least with certain fractional fillings, can be shown to have an infinite dimensional symmetry algebra: the $W_{\infty}$ algebra which is the Lie algebra of the area-preserving diffeomorphisms (please see Capelli, Trugenberger and Zemba). This symmetry expresses the fact that the second quantized state is incompressible.