A particle of charge $Q$ is confined to move in the $x y$ plane, with electrostatic potential $\phi=0$ and vector potential A satisfying
$$
\boldsymbol{\nabla} \times \mathbf{A}=(0,0, B)
$$
Consider the operators $\rho_{x}, \rho_{y}, R_{x}$ and $R_{y}$, defined by
$$
\rho=\frac{1}{Q B} \hat{\mathbf{e}}_{z} \times(\mathbf{p}-Q \mathbf{A}) \quad \text { and } \quad \mathbf{R}=\mathbf{r}-\boldsymbol{\rho}
$$
where $\mathbf{r}$ and $\mathbf{p}$ are the usual position and momentum operators, and $\hat{\mathbf{e}}_{z}$ is the unit vector along $\mathbf{B}$. Show that the only non-zero commutators formed from the $x$ and $y$ components of these are
$$
\left[\rho_{x}, \rho_{y}\right]=\mathrm{i} r_{B}^{2} \quad \text { and } \quad\left[R_{x}, R_{y}\right]=-\mathrm{i} r_{B}^{2}
$$
where $r_{B}^{2}=\hbar / Q B$.
The operators $a, a^{\dagger}, b$ and $b^{\dagger}$ are defined via
$$
a=\frac{1}{\sqrt{2} r_{B}}\left(\rho_{x}+\mathrm{i} \rho_{y}\right) \quad \text { and } \quad b=\frac{1}{\sqrt{2} r_{B}}\left(R_{y}+\mathrm{i} R_{x}\right)
$$
Evaluate $[a, a \dagger]$ and $\left[b, b^{\dagger}\right] .$ Show that for suitably defined $\omega$, the Hamiltonian can be written
$$
H=\hbar \omega\left(a^{\dagger} a+\frac{1}{2}\right)
$$
Given that there exists a unique state $|\psi\rangle$ satisfying
$$
a|\psi\rangle=b|\psi\rangle=0
$$
what conclusions can be drawn about the allowed energies of the Hamiltonian and their degeneracies? What is the physical interpretation of these results?