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?
Best Answer
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.