The <canonical commutation relations> are $[X_i,X_j]=[P_i,P_j]=0$ and $[X_i,P_j]=i\hbar\delta_{ij}$. Substitution into the definitions of the <creation and annihilation operators> gives
$$
\boxed{[A_i^\dagger,A_j^\dagger]=0,\qquad[A_i,A_j]=0,\qquad[A_i,A_j^\dagger]=\delta_{ij}.}
$$

The normalized <ground state> is defined by
$$
A_i|0\rangle=0\quad(i=1,2,3),
\qquad \langle0|0\rangle=1.
$$
Repeated use of $[A_i,A_i^\dagger]=1$ gives the <Cartesian number state of the three-dimensional isotropic harmonic oscillator>
$$
\boxed{|n_1,n_2,n_3\rangle
=\frac{(A_1^\dagger)^{n_1}(A_2^\dagger)^{n_2}(A_3^\dagger)^{n_3}}
{\sqrt{n_1!n_2!n_3!}}|0\rangle,}
$$
with <energy eigenvalue>
$$
E_{n_1n_2n_3}=\hbar\omega\left(n_1+n_2+n_3+\frac32\right).
$$

Inverting the ladder-operator definitions yields
$$
X_i=\sqrt{\frac{\hbar}{2\mu\omega}}(A_i+A_i^\dagger),
\qquad
P_i=i\sqrt{\frac{\mu\hbar\omega}{2}}(A_i^\dagger-A_i).
$$
The 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 operators>
$$
\boxed{L_i=-i\hbar\varepsilon_{ijk}A_j^\dagger A_k.}
$$
Write $|j\rangle=A_j^\dagger|0\rangle$. Then
$$
L_i|j\rangle=-i\hbar\varepsilon_{ikj}|k\rangle,
\qquad
L^2|j\rangle=2\hbar^2|j\rangle.
$$
Since the <orbital angular momentum> eigenvalue is $\ell(\ell+1)\hbar^2$, every first-excited state has
$$
\boxed{\ell=1.}
$$
Finally,
$$
L_z|x\rangle=i\hbar|y\rangle,
\qquad
L_z|y\rangle=-i\hbar|x\rangle.
$$
Therefore the normalized <magnetic quantum number> $m=+1$ state can be chosen as
$$
\boxed{|\psi\rangle=\frac{|x\rangle+i|y\rangle}{\sqrt2}
=\frac{A_x^\dagger+iA_y^\dagger}{\sqrt2}|0\rangle,}
$$
which obeys $L_z|\psi\rangle=\hbar|\psi\rangle$ and $\lVert|\psi\rangle\rVert=1$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 33B / Solution

{title2=$L_i=-i\hbar\varepsilon_{ijk}A_j^\dagger A_k$}

For the <three-dimensional isotropic harmonic oscillator>, substituting $X_i=\sqrt{\hbar/(2\mu\omega)}(A_i+A_i^\dagger)$ and $P_i=i\sqrt{\mu\hbar\omega/2}(A_i^\dagger-A_i)$ into $L_i=\varepsilon_{ijk}X_jP_k$ gives $L_i=-i\hbar\varepsilon_{ijk}A_j^\dagger A_k$. On the first-excited Cartesian states $|j\rangle=A_j^\dagger|0\rangle$, this is the three-dimensional vector representation and $L^2|j\rangle=2\hbar^2|j\rangle$, so the whole first-excited level has orbital quantum number $\ell=1$.

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 Incoming links: Orbital angular momentum in oscillator ladder operators