# Spherical Tensor Operators and the isotropic harmonic oscillator

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.

Answer 1: $V_q V^\dagger_{q'}$ is just the product of two operators. For instance, if $q=1$ and $q'=0$, then $$V_1 V^\dagger_{0}=\left(-\frac{a_x+ia_y}{\sqrt{2}}\right)a^\dagger_z\, .$$ Answer 2: Bad choice of subscripts, or incorrect. Assuming you really mean $k_x=j_1$ and $k_y=j_2$, then correct. If you mean the $x$ and $y$ operators, as in $a_x$ or $a_y^\dagger$, then incorrect. A much less confusing way of writing $T^k_q$ would be $$T^k_q=\sum_{m_1m_2}C^{j_1j_2;k}_{m_1m_2;q} X^{j_1}_{m_1}X^{j_2}_{m_2}$$ where $q$ is in the range $-k \le q\le k$, the $m_1,m_2$ are $\pm 1$ or $0$, and $j_1,j_2$ are both $1$ if you use the $V_m=T^1_m$ as you defined the $V$'s. The construction of your $T^0_0$ tensor as a product of $V$'s is thus correct.

Now... unfortunately your approach may not be the most productive in its current form. The reason for this is that you need to create a net number of excitation if you want to reach states above the vacuum. Thus, you will need tensors like $V_k$, $V_kV_m$ and $V_k V_m V_s$, can you will need to couple them properly. The simplest way to avoid the mess would be (in the case of the $N=3$ shell) to note that $(V_1)^3$ is proportional to $T^{\ell=3}_{m=3}$ so $(V_1)^3\vert 0$ *must * be proportional to the $\ell=3,m=3$ state of this shell. You can get the other $\ell=3, m$ states using the lowering operator $l_-$. The $\ell=1,m=1$ state must contain three excitation, so you will need to multiply $V_1$ by a scalar $$W^0 = \sum_{m_1m_2} C^{110}_{m_1m_20}V_{m_1}V_{m_2}$$ to guarantee you have three excitations of net angular momentum $1$. You can get the remaining $\ell=1, m\ne 1$ states by lowering.

It is of course also possible to construct directly an $\ell=3,m$ tensor from the triple coupling of $V_kV_mV_s$, or an $\ell=1,m$ tensor from this type of triple coupling, but in general it is simpler to construct one state (usually the $m=\ell$ state) and crank down from it.

A similar approach will work for the $N=2$ shell, except your tensor will be of degree $2$ in the $V$'s. I trust you already know the branching rules: $N=3$ contains $\ell=3,1$, $N=2$ contains $\ell=2,0$, $N=1$ contains $\ell=1$ only, and of course $N=0$ contains $\ell=0$.

In answer to a comment: multiple products of $V^\dagger$ will span the symmetric subspace of the decomposition of multiple products of $\ell=1$ spaces. This is why the $V^\dagger\otimes V^\dagger$ will not contain an $L=1$ part as this is in the antisymmetric subspace of $(\ell=1)\otimes(\ell=1)$.

• 2.Yes, I meant $k_1$ and $k_2$, and I agree this is a better notation.
– soap
Jan 26, 2017 at 17:21
• Upgrade of last comment: 1. My question is what is meant by "product representation", not "product", but I now think that the "product representation" is the vector space spanned by the spherical tensors constructed using my second theorem (and not simply spanned by the operators $V_qV^\dagger_{q'}$). 2. Yes, I meant $k_1$ and $k_2$, and I agree this is a better notation.
– soap
Jan 26, 2017 at 17:26
• Another thought about point 1: the product representation is the vector space spanned by the operators of the type $V_qV_{q'}$ (which is still a vector operator and thus a spherical tensor of rank 1), but can be spanned by three bases (which span three irreducible spaces-each of which with a well defined angular momentum): the bases of the spherical tensors constructed using my second theorem - one of the spaces will correspond to $J=0$, another to $J=1$ and another to $J=2$. Is this correct?
– soap
Jan 26, 2017 at 17:58
• @Simoes : see edit to answer. Jan 26, 2017 at 19:20