Completeness of Landau basis We know that the Landau Hamiltonian (uniform magnetic field) is diagonalized by wavefunctions $|n,m\rangle,n,m\in \mathbb{N}$ in the symmetric gauge. However, does this set of functions form a "complete" basis for $L^2(\mathbb{R}^2)$? I would think the answer is yes, but I can't quite think of a rigorous proof.
 A: I think I have found a solution, so I will provide a sketch of the proof. Notice that the ladder operators can be written as
$$
a^* = -i\left(D-\frac{\bar{z}}{2}\right),\quad b^* = -i\left(\bar{D} -\frac{z}{2}\right)
$$
where $z,\bar{z} = x\pm iy$ and $D,\bar{D} = D_z,D_{\bar{z}}$.
Let $N,M\in \mathbb{N}$ and
$$
V_{N,M}= \text{span}\left\{ z^\alpha  \bar{z}^\beta e^{-|z|^2/2}:\alpha,\beta \in \mathbb{N},\alpha-\beta =M,\beta \le N\right\}
$$
Using induction, it's not too hard to show that for all $M\ge 0$, the family of Landau states $|0,M\rangle,|1,M+1\rangle,...,|N,M+N\rangle$ form a basis for $V_{N,M}$. Hence, $|0,M\rangle,|1,M+1\rangle,...$ forms a basis for $V_M = \bigcup V_{N,M}$. A similar statement can be made for $M<0$. Hence, the Landau states form a  basis for
$$
V = \text{span}\left\{ z^\alpha  \bar{z}^\beta e^{-|z|^2/2}:\alpha,\beta \in \mathbb{N}\right\}
$$
Notice that $V$ is also equal to
$$
V = \text{span}\left\{x^\alpha y^\beta e^{-(x^2+y^2)/2}:\alpha,\beta \in \mathbb{N}\right\}
$$
It should be noted that (see Brian Hall Quantum theory for Mathematicians section 11.4 for proof)
$$
\sum_{\alpha \in \mathbb{N}} \frac{(ipx)^\alpha}{\alpha!}\to_{L^2(\mathbb{R})} e^{ipx-x^2/2}
$$
Therefore, we see that $e^{i(px+qy)} e^{-(x^2+y^2)/2}$ is in the $L^2$ closure of $V$.
Now suppose that $\psi \in L^2(\mathbb{R}^2)$ is orthogonal to all the Landau states. Then it must be orthogonal to $e^{i(px+qy)} e^{-(x^2+y^2)/2}$. Therefore,
$$
\iint dx dy e^{-i(px+qy)} e^{-(x^2+y^2)}\psi(x,y) =0
$$
Hence, the Fourier transform of $e^{-(x^2+y^2)}\psi(x,y)$ is zero everywhere, and thus $e^{-(x^2+y^2)}\psi(x,y)=0$ and thus $\psi(x,y)=0$.
