# 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.

## 1 Answer

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.

Your ladder operators look ok.

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 '17 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 '17 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 '17 at 17:58
• @Simoes : see edit to answer. – ZeroTheHero Jan 26 '17 at 19:20