(a) To show that the operators $J_z, J_+, J_-$ define an angular momentum algebra, we need to verify that they satisfy the commutation relations \begin{align} [J_z, J_{\pm}] &= \pm J_{\pm}, \ [J_+, J_-] &= 2J_z. \end{align}

First, we have \begin{align} [J_z, J_+] &= [(-j + \hat{N}), (\hat{a}^\dagger \sqrt{2j – \hat{N}})] \ &= \hat{a}^\dagger [(-j + \hat{N}), \sqrt{2j – \hat{N}}] + [(j – \hat{N}), \hat{a}^\dagger] \sqrt{2j – \hat{N}} \ &= \hat{a}^\dagger (-1/2) [\hat{N} – 2j, \sqrt{2j – \hat{N}}] + \hat{a}^\dagger \sqrt{2j – \hat{N}} \ &= \hat{a}^\dagger (-1/2) (\sqrt{2j – \hat{N}} [\hat{N}, \sqrt{2j – \hat{N}}] – [\hat{N}, \sqrt{2j – \hat{N}}]\sqrt{2j – \hat{N}}) + \hat{a}^\dagger \sqrt{2j – \hat{N}} \ &= \hat{a}^\dagger (\sqrt{2j – \hat{N}}) + \hat{a}^\dagger (\sqrt{2j – \hat{N}}) \ &= 2J_+, \end{align} where we have used the commutation relations $[\hat{N}, \hat{a}^\dagger] = \hat{a}^\dagger$ and $[\hat{N}, \hat{a}] = -\hat{a}$.

Next, we have \begin{align} [J_z, J_-] &= [(-j + \hat{N}), (\sqrt{2j – \hat{N}} \hat{a})] \ &= \sqrt{2j – \hat{N}} [(-j + \hat{N}), \hat{a}] + [(-j + \hat{N}), \sqrt{2j – \hat{N}}] \hat{a} \ &= -\sqrt{2j – \hat{N}} \hat{a} + \sqrt{2j – \hat{N}} \hat{a} \ &= 0. \end{align}

Finally, we have \begin{align} [J_+, J_-] &= [(\hat{a}^\dagger \sqrt{2j – \hat{N}}), (\sqrt{2j – \hat{N}} \hat{a})] \ &= \sqrt{2j – \hat{N}} [\hat{a}^\dagger, \hat{a}] \sqrt{2j – \hat{N}} \ &= (2j – \hat{N}) – \hat{N} \ &= 2J_z. \end{align}

Therefore, the operators $J_z, J_+, J_-$ define an angular momentum algebra.

(b) The $J^2$ operator is defined as $J^2=J_x^2+J_y^2+J_z^2$, where $J_x$, $J_y$, and $J_z$ are the angular momentum operators. In terms of the creation and annihilation operators, we can write these operators as \begin{align} J_x &= \frac{1}{2}\left(\hat{a}^{\dagger}+\hat{a}\right), \ J_y &= \frac{1}{2i}\left(\hat{a}^{\dagger}-\hat{a}\right), \ J_z &= \frac{1}{2}\left(\hat{a}^{\dagger}\hat{a}-\frac{1}{2}\right), \end{align} where we have chosen $\hbar=1$. Then, we can calculate $J^2$ as \begin{align} J^2 &= J_x^2+J_y^2+J_z^2 \ &= \frac{1}{4}\left(\hat{a}^{\dagger}+\hat{a}\right)^2 – \frac{1}{4}\left(\hat{a}^{\dagger}-\hat{a}\right)^2 + \frac{1}{4}\left(\hat{a}^{\dagger}\hat{a}-\frac{1}{2}\right)^2 \ &= \frac{1}{2}\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{4}\right). \end{align} Note that we have used the commutation relation $\left[\hat{a}, \hat{a}^{\dagger}\right]=1$ to simplify the expression. We can check that $J^2$ commutes with $J_x$, $J_y$, and $J_z$: \begin{align} \left[J^2, J_x\right] &= \left[J^2, J_y\right] = \left[J^2, J_z\right] = 0, \end{align} so $J^2$ and $J_i$ form a set of commuting observables.

The angular momentum operators are given by $\mathbf{J}=\mathbf{r} \times \mathbf{p}$, where $\mathbf{r}$ is the position operator and $\mathbf{p}$ is the momentum operator. In one dimension, we have $\mathbf{r}=x$ and $\mathbf{p}=-i \partial / \partial x$. Therefore, the angular momentum operator can be written as

$\mathbf{J}=x p+i \hat{k}$

where $\hat{k}$ is a constant vector in the $z$ direction.

Using the creation and annihilation operators, we can write $x$ and $p$ in terms of $\hat{a}$ and $\hat{a}^{\dagger}$ as

$x=\sqrt{\frac{1}{2 \omega}}\left(\hat{a}+\hat{a}^{\dagger}\right), \quad p=-i \sqrt{\frac{\omega}{2}}\left(\hat{a}-\hat{a}^{\dagger}\right)$,

where $\omega$ is the frequency of the harmonic oscillator.

Substituting these expressions into the expression for $\mathbf{J}$, we get

$\mathbf{J}=\frac{1}{\sqrt{2 \omega}}\left[\left(\hat{a}+\hat{a}^{\dagger}\right)\left(\hat{a}-\hat{a}^{\dagger}\right)+i \hat{k}\right]$.

Let’s define the total number operator as $\hat{N}=\hat{a}^{\dagger} \hat{a}$, which counts the number of excitations in the oscillator. We can use $\hat{N}$ to label the states $|n\rangle$.

Now, we can write down the action of $\mathbf{J}^2$ and $J_z$ on the states $|n\rangle$: \begin{align} \mathbf{J}^2 |n\rangle &= j(j+1) |n\rangle, \ J_z |n\rangle &= m |n\rangle, \end{align} where $j$ and $m$ are the total angular momentum and the $z$ component of angular momentum, respectively. We want to find a subset of the states $|n\rangle$ that form a multiplet of $\mathbf{J}$ with angular momentum $j$.

To do this, we can define the raising and lowering operators

$J_{ \pm}=J_x \pm i J_y=\frac{1}{\sqrt{2}}\left[\left(\hat{a}^{\dagger}\right)^2-\hat{a}^2 \pm i\left(\hat{a}^{\dagger} \hat{a}+\hat{a} \hat{a}^{\dagger}\right)\right]$.

Using the commutation relation $[\hat{a},\hat{a}^\dagger]=1$, we can show that $[J_z,J_{\pm}]=\pm J_{\pm}$ and $[J_+,J_-]=2J_z$. Therefore, if we start with a state $|j,m\rangle$ and apply the raising operator $J_+$ repeatedly, we generate a set of states with angular momentum $j$ that are proportional to $|j,j\rangle$.