A $d=3$ isotropic harmonic oscillator of mass $\mu$ and frequency $\omega$ has lowering operators

$$\mathbf{A}=\frac{1}{\sqrt{2 \mu \hbar \omega}}(\mu \omega \mathbf{X}+\mathrm{i} \mathbf{P})$$

where $\mathbf{X}$ and $\mathbf{P}$ are the position and momentum operators. Assuming the standard commutation relations for $\mathbf{X}$ and $\mathbf{P}$, evaluate the commutators $\left[A_{i}^{\dagger}, A_{j}^{\dagger}\right],\left[A_{i}, A_{j}\right]$ and $\left[A_{i}, A_{j}^{\dagger}\right]$, for $i, j=1,2,3$, among the components of the raising and lowering operators.

How is the ground state $|\mathbf{0}\rangle$ of the oscillator defined? How are normalised higher excited states obtained from $|\mathbf{0}\rangle$ ? [You should determine the appropriate normalisation constant for each energy eigenstate.]

By expressing the orbital angular momentum operator $\mathbf{L}$ in terms of the raising and lowering operators, show that each first excited state of the isotropic oscillator has total orbital angular momentum quantum number $\ell=1$, and find a linear combination $|\psi\rangle$ of these first excited states obeying $L_{z}|\psi\rangle=+\hbar|\psi\rangle$ and $\||\psi\rangle \|=1$.