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On a common domain where all products are defined, expansion of the operator commutator givesUse the usual quantum position and momentum operators with $p_j=-i\hslash {\partial }_j$, the canonical commutation relation $[x_i,p_j]=i\hslash {\delta }_{ij}$, and wavefunctions with the decay/domain conditions needed for adjoints. The cross-coordinate operators commute, soFor an eigenstate $L\psi =\lambda \psi$, hermiticity gives $\lambda \Vert \psi {\Vert }^2=\overline{\lambda }\Vert \psi {\Vert }^2$, hence its eigenvalue is real.

The product identity gives $[L,x]=i\hslash y$, $[L,y]=-i\hslash x$, $[L,p_x]=i\hslash p_y$, $[L,p_y]=-i\hslash p_x$. Thus the commutators with $x^2,y^2$ cancel, as do those with $p_x^2,p_y^2$, proving $[L,H]=0$. Each oscillator energy eigenspace is finite-dimensional and preserved by $L$; the Hermitian restriction of $L$ can be diagonalized within it. Therefore an energy eigenbasis can be chosen to diagonalize angular momentum as well.

For ${\varphi }_0=e^{-(x^2+y^2)/(2\hslash )}$, differentiation gives $\Delta {\varphi }_0=[(x^2+y^2)/{\hslash }^2-2/\hslash ]{\varphi }_0$, so $H{\varphi }_0=\hslash {\varphi }_0$. Also $[H,x]=-i\hslash p_x$ and $p_x{\varphi }_0=ix{\varphi }_0$, giving $H(x{\varphi }_0)=2\hslash x{\varphi }_0$; the same calculation holds for $y{\varphi }_0$. Directly $L=-i\hslash (x{\partial }_y-y{\partial }_x)$, henceThe simultaneous normalized states areTheir squared unnormalized norm is $\int r^2{e}^{-r^2/\hslash }{d}^{2}x=\pi {\hslash }^2$. Their inner product vanishes because the integral of $(x-iy{)}^2{e}^{-r^2/\hslash }$ is zero: the $xy$ term is odd and the $x^2,y^2$ terms cancel by rotational symmetry.

Put $a=e\mathcal{E}$. The new Hamiltonian hasFor a nonzero applied coupling $a$, a common normalizable eigenstate would be killed by the commutator, so $y\psi =0$. It would have to be supported on the line $y=0$, a set of zero planar measure, and would be the zero vector in $L^2$. Thus there are no nonzero common eigenstates in this case. Merely having a nonzero commutator would not prove that conclusion for arbitrary operators; the multiplication operator here has trivial kernel. If $a=0$, the original simultaneous eigenstates remain.

For the shifted two-dimensional oscillator in a uniform electric field, complete the square:Therefore ${\psi }_1(x^{\prime },y^{\prime })$ and ${\psi }_2(x^{\prime },y^{\prime })$ have the same shifted energyThe modified angular momentum isIt is angular momentum about the displaced centre, commutes with $H^{\prime }$, and has eigenvalues $\pm \hslash$ on the two translated states.