Now, $[H,L_{i}]=0$ since $H$ is rotationally invariant. One can prove that {H,L^{2},L_{z}} is a C.S.C.O.

Now, $[H,L_{i}]=0$ since $H$ is rotationally invariant. One can prove that $\{H,L^{2},L_{z}\}$ is a C.S.C.O.

After reading what Emilio and ZeroTheHero and some other things, I tried another method which seems more general. Here it goes.

Let $$ \begin{cases} V_{1}=-\frac{a_{x}+ia_{y}}{\sqrt{2}}\\ V_{0}=a_{z}\\ V_{-1}=\frac{a_{x}-ia_{y}}{\sqrt{2}} \end{cases}\mbox{ and }\begin{cases} V_{1}^{\dagger}=-\frac{a_{x}^{\dagger}-ia_{y}^{\dagger}}{\sqrt{2}}\\ V_{0}^{\dagger}=a_{z}^{\dagger}\\ V_{-1}^{\dagger}=\frac{a_{x}^{\dagger}+ia_{y}^{\dagger}}{\sqrt{2}} \end{cases} $$

We have $$ [J_{z},V_{q}^{\dagger}]=q\hbar V_{q}^{\dagger} $$ $$ [J_{z},V_{q}]=q\hbar V_{q} $$ $$ [N,V_{q}^{\dagger}]=V_{q}^{\dagger} $$ $$ [N,V_{q}]=-V_{q} $$

So the “daggered” operators increase the eigenvalue $n$ of $N$ by one (and thus the energy by $\hbar\omega$ ), while the “undaggered” operators decrease it by one. Also, the $V_{q}^{(\dagger)}$ change the eigenvalue of J_{z} by $q\hbar$ .

With this in mind, we can face the problem.

Assumption: All states of the system can be obtained by applying these operators to the ground state $\vert0\rangle$ . I don't know why!-Please explain it if you do.

For $n=1$ , the biggest value for $m$ one can get for a state of $\mathcal{E}(n=1)$ is $1$, since $(V_{1}^{\dagger})^{1}\vert0\rangle$ is the only way to get the maximum $m$ while (increasing $n$ to $1$) by application of the operators defined above, and indeed $L_{z}V_{1}^{\dagger}\vert0\rangle=\hbar V_{1}^{\dagger}\vert0\rangle$ . Now, this state, having $m=1$ , must have $l\geq1$ . It cannot be $l>1$ since this would imply that I had states with $m>1$ (obtained by applying $L_{+}$ to $V_{1}^{\dagger}\vert0\rangle$ ) in $\mathcal{E}(n=1)$ (the Hamiltonian commutes with $L_{+}$ because of rotational invariance and so does N ), which is absurd. For $m=0$ , we easily see that there is only one state: $V_{0}^{\dagger}\vert0\rangle$ . Since there is only one, it must be the one of $\mathcal{E}(l=1)$ . There is no need to analyze other values for m (negative values will bring nothing new). In conclusion, the only value for $l$ for states with $n=1$ is $1$ .

Let us try something harder.

For $n=3$ , the biggest value for $m$ for a state of $\mathcal{E}(n=3)$ is three, corresponding to the state $(V_{1}^{\dagger})^{3}\vert0\rangle$ . Clearly, there are no other states with $m=3$ in $\mathcal{E}(n=3)$ . Just like before, this means that the space $\mathcal{E}(l=3)$ is in the direct sum decomposition of $\mathcal{E}(n=3)$ . Now, for $m=2$ we only have one state $V_{0}^{\dagger}(V_{1}^{\dagger})^{2}\vert0\rangle$ , so there is no $l=2$ representation in the direct product decomposition. For $m=1$ , we have $V_{-1}^{\dagger}(V_{1}^{\dagger})^{2}\vert0\rangle$ but also $(V_{0}^{\dagger})^{2}V_{1}^{\dagger}\vert0\rangle$ , so there is one more representation (besides $l=2$ ) with states with $m=1$ . It can't be $l=2$ because there are no $m=2$ states. It is therefore an $l=1$ representation. Finally, for $m=0$ I only have two states: $V_{-1}^{\dagger}V_{0}^{\dagger}(V_{1}^{\dagger})^{1}\vert0\rangle$ and $(V_{0}^{\dagger})^{3} \vert0\rangle$ , which must belong to the two representations already found. Therefore, $l\in\{1,3\}$ .