2D harmonic oscillator: polar versus Cartesian eigenstates

Question

Consider the usual two-dimensional harmonic oscillator (2D HO) with the Hamiltonian $$ H = -\frac{1}{2}\nabla_x^2 + \frac{1}{2}x^2 -\frac{1}{2}\nabla_y^2 + \frac{1}{2}y^2. $$ In Cartesian coordinates, the solution is a family of states $$ |n_x\rangle |n_y\rangle, $$ where the labels $n_x$ and $n_y$ correspond to the number of excitations of the one-dimensional HOs. On the other hand, solving in polar coordinates, one obtains instead a family of states $$ |n,m\rangle, $$ where now these states are also eigenstates of the angular momentum operator with the eigenvalue $m$.

How to switch between the two families of solutions? In other words, what is the decomposition $$ |n,m\rangle = \sum_{n_x,n_y} c_{n,m,n_x,n_y} |n_x\rangle |n_y\rangle? $$

I know that for example $$ c_{0,0,0,0}=1 $$ and $$ c_{0,1,1,0}=1/\sqrt{2}, \quad c_{0,1,0,1}=1/\sqrt{2}, $$ but is there perhaps a table somewhere, or a systematic procedure to obtain this decomposition for arbitrary states?

Answer 1

You might want to investigate the operators $$ \hat a^\dagger_\pm=\frac{1}{\sqrt{2}}\left(a^\dagger_x\pm ia^\dagger_y\right)\, ,\qquad \hat a_\pm=\frac{1}{\sqrt{2}}\left(a_x\mp ia_y\right)\, , $$ and in particular $[\hat L_z,\hat a^\dagger_\pm]$. It should then be an easy matter to construct the $\vert n,m\rangle$ basis in the form $$ \left(a_-^\dagger\right)^\alpha\left(a_+^\dagger\right)^\beta\vert 0\rangle $$ and compute its overlap with the $\vert n_xn_y\rangle$ basis.

Answer 2

The answer which can be derived using the rotation operators provided by ZeroTheHero's answer is \begin{align} |n,m\rangle = & \frac{1}{\sqrt{n!(n+|m|)!}}\frac{1}{2^{\left(2n+|m|\right)/2}}\sum_{j=0}^{n}\sum_{k=0}^{n+|m|}\binom{n}{j}\binom{n+|m|}{k} \\ &\times i^{\mathrm{sng}(m)\cdot(k-j)}\sqrt{\left(2n+|m|-j-k\right)!\left(j+k\right)!} \,\, |2n+|m|-j-k\rangle_{x}|j+k\rangle_{y}, \end{align} which works for any $m$ (both positive and negative), and where I use standard notation for binomial coefficients, while the sign function is nonstandard. In particular, it is $1$ for zero or positive $m$, and $-1$ for negative $m$.