I am studying Spherical Tensor Operators. In Sakurai's book ("Modern Quantum Mechanics") there is a theorem which can be used as the definition of spherical tensor operators. I will state it here for completeness, so that you know what I mean by a spherical tensor operator.
Theorem: $T^{(k)}$ is a spherical tensor operator if and only if $[J_{k},T_{q}^{(k)}]=\hbar qT_{q}^{(k)}$ and $[J_{\pm},T_{q}^{(k)}]=T_{q\pm1}^{(k)}\hbar\sqrt{k(k+1)-q(q\pm1)}$
I now want to solve a problem about the isotropic harmonic oscillator using spherical tensor operators.
Considering the Hamiltonian for the isotropic harmonic oscillator: $$H = \hbar\omega\,(a^{\dagger}_xa_x + a^{\dagger}_ya_y + a^{\dagger}_za_z + \frac{3}{2})$$
The problem: I want to identify the representations of the angular momentum present in the first three energy levels of the isotropic harmonic oscillator, and write the states $J=0$ on the $\vert n_x, n_y, n_z \rangle$ basis. This must be solved using the vector operators $\mathbf{V}=(a_{x},a_{y},a_{z}) $ and $\mathbf{V}^{\dagger}=(a_{x}^{\dagger},a_{y}^{\dagger},a_{z}^{\dagger}) $ and considering the product representation $V_{q}V_{q'}^{\dagger} $, with $V_{q}\,,V_{q'}^{\dagger}$ the spherical components of $\mathbf{V}$ and $\mathbf{V^{\dagger}}$. Explicitly:
$$\begin{cases} V_{1}=-\frac{a_{x}+ia_{y}}{\sqrt{2}}\\ V_{0}=a_{z}\\ V_{-1}=\frac{a_{x}-ia_{y}}{\sqrt{2}} \end{cases}\mbox{ and }\begin{cases} V_{1}^{\dagger}=-\frac{a_{x}^{\dagger}-ia_{y}^{\dagger}}{\sqrt{2}}\\ V_{0}^{\dagger}=a_{z}^{\dagger}\\ V_{-1}^{\dagger}=\frac{a_{x}^{\dagger}+ia_{y}^{\dagger}}{\sqrt{2}} \end{cases} $$
First Question: what exactly is meant by the product representation $V_{q}V_{q'}^{\dagger}$?
My attempt:
As far as I understand from what I read in Sakurai's and Baym's books, we can state the following as a theorem:
Theorem: Let $X^{(k_{x})}$ and $Y^{(k_{y})}$ be spherical tensors of ranks $k_{x}$ and $k_{y}$, respectively. Then, $$ T_{q}^{(k)}=\underset{q_{x},q_{y}}{\sum}\langle\overset{\overset{j_{1}}{\downarrow}}{k_{x}}\overset{\overset{j_{2}}{\downarrow}}{k_{y}};\overset{\overset{m_{1}}{\downarrow}}{q_{x}}\overset{\overset{m_{2}}{\downarrow}}{q_{y}}|\underbrace{\overset{\overset{j_{1}}{\downarrow}}{k_{x}}\overset{\overset{j_{2}}{\downarrow}}{k_{y}}}_{\text{not needed}};\overset{\overset{J}{\downarrow}}{k}\overset{\overset{M}{\downarrow}}{q}\rangle X_{q_{x}}^{(k_{x})}Y_{q_{y}}^{(k_{y})} =\underset{q_{1},q_{2}}{\sum}C_{q_{1}q_{2}q}^{k_{x}k_{y}k}\,X_{q_{1}}^{(k_{x})}Y_{q_{2}}^{(k_{y})} $$ is a spherical tensor of rank k
This theorem lets us construct tensor operators from two. Notice that the values for k will not be arbitrary, since the only Clebsch Gordan Coefficients which can be non-zero are the ones with $k\in\{|k_{x}-k_{y}|,|k_{x}-k_{y}|+1,...,k_{x}+k_{y}\}$. For the same reason, $q\in\{-k,-k+1,...,k\}.$
Second Question: Is this completely correct?
I can thus construct the spherical tensor operators which span the so called $J=0$ representation using the rank-1 spherical tensor operators given. in fact, it's just one and it's of the form $$T^{(0)}_{Q=M=0}=\underset{q_{1},q_{2}}{\sum}C_{q_{1}q_{2}0}^{1\,1\,0}\,V_{q_{1}}V^{\dagger}_{q_{2}}$$ This is easy to determine using a CG coefficients table.
I think I must use this in order to get my answer. Second question: How?
Another idea: I know how to construct the ladder operators $l_{+}$ and $l_{-}$ using the creation and destruction operators. A little bit of manipulation lets me write $$ l_{\pm}=\hbar\sqrt{2}(-V_{\pm1}V_{0}^{\dagger}-V_{0}V_{\mp1}^{\dagger}) $$ Now, we also know how to write $l_z$ in terms of the reaction and destruction operators and so also in terms of the spherical tensors. Since $l^2=\frac{l_+l_-+l_-l_+}{2}+l_z^2$, we can write $l^2$ in terms of the spherical tensors. Again, I think I must use this. Third question: How?
NOTE: If the question is still not clear, please tell me and I will try and edit it. Maybe my attempt is completely missing the point - please tell me if that is the case.
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