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The canonical commutation relations are $[X_i,X_j]=[P_i,P_j]=0$ and $[X_i,P_j]=i\hslash {\delta }_{ij}$. Substitution into the definitions of the creation and annihilation operators gives

The normalized ground state is defined byRepeated use of $[A_i,A_i^{†}]=1$ gives the Cartesian number state of the three-dimensional isotropic harmonic oscillatorwith energy eigenvalue

Inverting the ladder-operator definitions yieldsThe antisymmetry of the Levi-Civita symbol then cancels the two-creation and two-annihilation terms in $\mathbf{L}=\mathbf{X}\times \mathbf{P}$, leaving the orbital angular momentum in oscillator ladder operatorsWrite $|j⟩=A_j^{†}|0⟩$. ThenSince the orbital angular momentum eigenvalue is $\ell (\ell +1){\hslash }^2$, every first-excited state hasFinally,Therefore the normalized magnetic quantum number $m=+1$ state can be chosen aswhich obeys $L_z|\psi ⟩=\hslash |\psi ⟩$ and $\Vert |\psi ⟩\Vert =1$.

Inverting the definitions of the creation and annihilation operators givesSubstituting in the orbital angular momentum $L$ givesOperators belonging to distinct modes commute. Therefore the terms with two annihilation operators and those with two creation operators cancel, and $a_1^{\ast }{a}_2=a_2{a}_1^{\ast }$, $a_2^{\ast }{a}_1=a_1{a}_2^{\ast }$. The surviving terms give the orbital angular momentum in oscillator ladder operators:

Set ${\omega }_1={\omega }_2=\omega$. Inverting the creation and annihilation operators givesSubstitute these expressions in $L_3=X_1{P}_2-X_2{P}_1$. Terms with two creation operators or two annihilation operators cancel because different modes commute. The remaining orbital angular momentum in oscillator ladder operators isEquivalently it is $-i\hslash (a_{1,+}{a}_{2,-}-a_{2,+}{a}_{1,-})$, consistent with the circular ladder operators of a two-dimensional oscillator.