The angular momentum operators $\mathbf{J}=\left(J_{1}, J_{2}, J_{3}\right)$ obey the commutation relations

$$\begin{aligned} &{\left[J_{3}, J_{\pm}\right]=\pm J_{\pm}} \\ &{\left[J_{+}, J_{-}\right]=2 J_{3}} \end{aligned}$$

where $J_{\pm}=J_{1} \pm i J_{2}$.

A quantum mechanical system involves the operators $a, a^{\dagger}, b$ and $b^{\dagger}$ such that

$$\begin{gathered} {\left[a, a^{\dagger}\right]=\left[b, b^{\dagger}\right]=1} \\ {[a, b]=\left[a^{\dagger}, b\right]=\left[a, b^{\dagger}\right]=\left[a^{\dagger}, b^{\dagger}\right]=0 .} \end{gathered}$$

Define $K_{+}=a^{\dagger} b, K_{-}=a b^{\dagger}$ and $K_{3}=\frac{1}{2}\left(a^{\dagger} a-b^{\dagger} b\right)$. Show that $K_{\pm}$and $K_{3}$ obey the same commutation relations as $J_{\pm}$and $J_{3}$.

Suppose that the system is in the state $|0\rangle$ such that $a|0\rangle=b|0\rangle=0$. Show that $\left(a^{\dagger}\right)^{2}|0\rangle$ is an eigenstate of $K_{3}$. Let $K^{2}=\frac{1}{2}\left(K_{+} K_{-}+K_{-} K_{+}\right)+K_{3}^{2}$. Show that $\left(a^{\dagger}\right)^{2}|0\rangle$ is an eigenstate of $K^{2}$ and find the eigenvalue. How many other states do you expect to find with same value of $K^{2}$ ? Find them.