Gauge choice in showing Landau level degeneracy via the algebraic method

Question

I'm trying to understand the algebraic method of formulating the Landau level problem better. I'm referring to David Tong's notes on the Quantum Hall effect for this (but not exactly following his notation). The relevant section to this discussion is Chapter 1, pages 14-17 and 22-24.

Working in natural units (I set the magnetic length $l_B=1$ and the cyclotron frequency $w_c=2$) the Landau level Hamiltonian $ \hat{H} = (\mathbf{\hat{p}}-\mathbf{\hat{A}})^2$ is rewritten as a quantum harmonic oscillator (QHO) Hamiltonian: $ \hat{H} = 2\left(\hat{b}^\dagger \hat{b}+\frac{1}{2}\right) $ where the bosonic operators are given by $\hat{b}=\frac{1}{\sqrt{2}}(\hat{\pi}_x-i\hat{\pi}_y)$ and $\hat{b}^\dagger=\frac{1}{\sqrt{2}}(\hat{\pi}_x+i\hat{\pi}_y)$ and $\mathbf{\hat{\pi}}=\mathbf{\hat{p}}-\mathbf{\hat{A}}$ is the canonical momentum.

The usual Fock space ladder of states and energies of the QHO however does not account for the degeneracy of the Landau levels. To account for this, Tong constructs another set of non gauge-invariant momenta $\mathbf{\hat{\Pi}}=\mathbf{\hat{p}}+\mathbf{\hat{A}}$ (see Eq. 1.28 in the notes) and shows that another set of bosonic operators (let me call these $\hat{d}, \hat{d}^\dagger$) constructed using these momenta commutes with the Hamiltonian (see Section 1.4.3). Specifically, the Landau level wavefunctions are then given by: $$ |n,m> = \frac{(\hat{b}^\dagger)^n(\hat{d}^\dagger)^m}{\sqrt{n!m!}}|0,0> $$ where $n,m$ are the eigenvalues of the number operators $\hat{b}^\dagger\hat{b}$ and $\hat{d}^\dagger\hat{d}$ respectively. The levels are degenerate because the energies of the Landau levels are still labeled only by $n$ (see Eq. 1.16 in the notes): $$ E_{n,m} = 2\left(n+\frac{1}{2}\right) $$

Crucially, however, (and as he remarks) this analysis only holds for the symmetric gauge. The $\hat{\Pi}$ operator only commutes with the Hamiltonian in this gauge. This is somewhat troubling to me.

How do we show the degeneracy of the Landau levels in other gauges---let's say the Landau gauge within the algebraic approach? What is the operator there that commutes with the bosonic operators of the QHO Hamiltonian?

Answer 1

There is a different, gauge-invariant algebraic approach that you will find Tong adopts a little later in the notes. The idea is to introduce so-called guiding center coordinate operators \begin{align} X &= x - \frac{\ell_B^2}{\hbar}\pi_y\\ Y &= y + \frac{\ell_B^2}{\hbar}\pi_x \end{align} that commute with $\pi_x$ and $\pi_y$, but not with each other: $\left[X,Y\right]=i\ell_B^2$. Since $H$ can be written exclusively in terms of $\pi_x$ and $\pi_y$, the guiding center coordinates commute with the hamiltonian. As the name suggests, these operators can be interpreted as position operators for the center of the cyclotron orbit (or, alternatively, as the position operators projected to a given Landau level). Since the guiding center operators commute with the hamiltonian, the position of the center of the orbit is conserved, as in the classical cyclotron motion. The degeneracy within a Landau level can be interpreted as a choice of where to place the center of the orbit. The failure of $X$ and $Y$ to commute means there is a quantum uncertainty in the guiding center position—we can localize the orbits at best to an area of $O(\ell_B^2)$. In other words, within a Landau level, real-space geometry is non-commutative.

We can define a separate set of guiding center ladder operators $b = (X + iY)/(\sqrt{2}\ell_B)$, $b^{\dagger} = (X - iY)/(\sqrt{2}\ell_B)$ obeying $\left[b^{\dagger},b\right] = 1$. Since the guiding center coordinates commute with the momenta $\pi_x$, $\pi_y$, the guiding center ladder operators commute with the Landau level ladder operators $a$, $a^{\dagger}$, and we have two independent Heisenberg algebras: \begin{align} [a^{\dagger},a] = 1,\,[b^{\dagger},b] = 1,\, [a,b] = [a,b^{\dagger}] = 0 \end{align} A standard way to construct the degenerate states within a Landau level is using coherent states, which are eigenstates of the the $b$ operators.