How to calculate the angular momentum states of isotropic quantum harmonic oscillator? While trying to calculate the angular momentum states for the first non trivial even and odd states ($N=2$ and $N=3$). When $N=n_x + n_y + n_z$
By solving the radial problem one can see that there 6 states for $N=2$ and 10 states for $N=3$, it stems in the degeneracy of states: $(N+1)(N+2)/2$
I want to work out the angular momentum states using Schwinger's annihilation and creation operators:
$$a_+ = \frac{-1}{\sqrt 2} (a_x - ia_y) \qquad a_- = \frac{1}{\sqrt 2}(a_x + ia_y)$$
For $N=3$ we have 7 states: $\{|n=3;\ l=3;\ m \rangle\}_{m=-3}^3$
We can get the highest using: $$|n=3;\ l=3;\ m=3 \rangle = \frac{1}{\sqrt{3!}}(a^\dagger_+)^3|n_x=0;\ n_y=0;\ n_z=0 \rangle $$
And then to apply $L_-$ until we get to $|n=3;\ l=3;\ m=-3 \rangle$ which also equals to: $$ \frac{1}{\sqrt{3!}}(a^\dagger_-)^3|n_x=0;\ n_y=0;\ n_z=0 \rangle $$
I don't fully understand the posts about spherical tensors $T_q^{(k)}$ and how to actually translate it into practical actions. 

Is there a better way to get the smaller subspace then the following:
As I said, we know that the possible $l$ states are $l=3$ and $l=1$. They are orthogonal, therefore we can build: 
$$| l=3;m=3\rangle = (a^\dagger_+)^3|n_x=0;\ n_y=0;\ n_z=0 \rangle $$
Then we can apply the operator:
$$L_-= L_x - iL_y = \hbar \left((a_x^\dagger - ia_y)a_z - a_z^\dagger (a_x +i a_y)\right) = \hbar \sqrt{2}(a_-^\dagger a_z + a_z^\dagger a_+)$$ 
$2\cdot3+1 = 7$ times to find all the states in the irreducible representation of $l=3$. Then we use the fact that $| l=3;m=1\rangle$ is orthogonal to $| l=1;m=1\rangle$ and jump to the other subspace and apply the operator: $L_-$ $2\cdot 1 +1 = 3$ times on $| l=1;m=1\rangle$ 
My concerns:
There should be more general way to jump between subspaces. Note that I didn't use the spherical tensor properties nor Wigner–Eckart theorem and Spherical harmonics $Y_l^m$. 
I would expect to have an expression for each angular momentum state as a  sequence of operators $a_+, a_+, a_+^\dagger, a_-^\dagger, a_z$ that builds an $|nlm\rangle$ state namely finding $\{n_i\}_{i=1}^6$ such that:
$$| l;m\rangle \propto (a_+)^{n_1} (a_+)^{n_2} (a_+^\dagger)^{n_3} (a_-^\dagger)^{n_4} (a_z)^{n_5} (a_z^\dagger)^{n_6} | n_x=0;n_y=0;n_z=0 \rangle$$
Where $\{n_i\}_{i=1}^6$ are integers between $0$ and $l$.
I am mostly missing the approach to build the state of $l=1$ in $N=3$ or $l=0$ for $N=2$, which is more complicated than simply applying $(L_-)$ $2l$ times. I can use the orthogonality, but I prefer to understand a more systematic approach to jump between subspaces. 
Also after reading this wonderful document I couldn't get an answer of how to jump between subspaces of angular momentum. I picked $N=2$ and $N=3$ because those are the first states that involves 2 subspaces, but any generalization of how to jump between 2 subspaces is most welcome. 
One of my motivation of understanding this problem is to be familiar more spherical tensors, lie algebra, metamorphism between the SHO, Angular momentum and bosons. 
Moreover I am thinking to implement the algorithm in Mathematica, using this code to calculate all of angular momentum in the Cartesian basis.

The question boils up to the following:

Let be $N$ be the number states of a 3D isotropic SHO. Is it possible
  to find the expression that will give the $|n=N; l,m \rangle$ in the
  base of $|n_x;n_y;n_z \rangle$. Or how to write the spherical tensor operator $T_m^q$ as boson operators. e.g $T_0^1=a_z^\dagger$

And:

How to jump between 2 subspaces. For example from $|l=3;m=1\rangle$ to 
  $|l=1;m=1\rangle$

 A: As mentioned in previous posts, given $N=n_x+n_y+n_z$, it's straightforward to calculate the $|n=N;l=N;m=N\rangle$ state if we use:
 $$\frac{1}{\sqrt{N!}}(a_+^\dagger)^N |0\rangle$$
Then we can apply $L_-$ until we get to $|n=N;l=N;m=-N\rangle$
In order to jump to the second subspace in which $|n=N;l=N-2;m=0\rangle$, we can use the irreducible spherical tensors theorem from [Sakurai 3.10.27]:
$$T_{q}^{(k)} = \sum_{q_1 q_2} \langle k_1, k_2 ; q_1 , q_2 |k_1,k_2; k, q \rangle T_{q_1}^{(k_1)} T_{q_2}^{(k_2)}$$
Where the followings hold:


*

*$k=N-2$

*$k_1+k_2 = N$

*$q_1 = -k_1, \dots , k_1$ and $q_2 = -k_2, \dots , k_2$

*We are using the $a_+, a_-, a_+^\dagger a_-^\dagger$ which are spherical tensor.


For example for $N=2, l=0$:
We choose define 2 positive integers  $k_1=1$, $k_2=1$ s.t. $k_1 + k_2 =N=2$.
$$T_0^{(0)} = \sum_{q_1=-1 q_2=-1}^{1,1} \langle 1, 1 ; q_1 , q_2 |1,1; k, q \rangle T_{q_1}^{(1)} T_{q_2}^{(1)}$$
And $T_0^{(0)}|0,0,0\rangle_C$ will give us the $|n=2;l=0;m=0\rangle$ states, hence we jumped into the second subspace and now we can apply $L_{\pm}$ to list all the angular momentum states within that subspace.
