# Construction of tensor from vector product

In one problem I'm working on, I came across the computation of the reduced matrix element for the component of the rank 2 tensor

$$\langle j'\|[J^{(1)}\times J^{(1)}]^{(2)}_{(1)}\|j\rangle$$

I'm having a bit of trouble computing the respective tensor:

$$[J^{(1)}\times J^{(1)}]^{(2)}$$

I tried the following. Given two vectors, I tried computing the tensor using the well-known formula:

$$T_Q^{(k)}=\sum_{q,m}\langle\ k,q;l,m|KQ\rangle A_q^{(k)}B_m^{(l)}$$

Replacing with the angular momentum operator, we get:

$$T_Q^{(2)}=\sum_{q,m}\langle\ 1,q;1,m|2Q\rangle A_q^{(1)}B_m^{(1)}$$

With $$Q=q+m$$, so we really have one sum. However, this is the case for the direct product of two vectors, and I'm looking for the vector product, so I introduced an epsilon factor $$\epsilon$$ according to the definition of vector product:

$$T_Q^{(2)}=\sum_{q,m}\langle\ 1,q;1,m|2Q\rangle \epsilon_{Qqm} A_q^{(1)}B_m^{(1)}$$

But my problem is when I compute the components, I get zero, so I'm not sure if I'm defining correctly the operator. For example, for the component $$T_1^{(2)}$$:

$$T_1^{(2)}=\langle 1,1;1,0|2 \rangle\epsilon_{1,0,0}J_1^{(1)}J_0^{(1)}+\langle 1,0;1,1|2 \rangle\epsilon_{1,0,1}J_0^{(1)}J_1^{(1)}+\langle 1,-1;1,2|2 \rangle\epsilon_{1,-1,2}J_{-1}^{(1)}J_2^{(1)}=0$$

Since $$\epsilon_{1,0,0}=\epsilon_{1,0,1}=0$$ and $$J_2^{(1)}=0$$ since it doesn't exist.

• this $\epsilon$ is non-sensical. your definition of $T^(2)_Q$ with the CG and without the $\epsilon$ is the correct one to use. Hint: the $Q=2$ component of the tensor is certainly $A^{(1)}_1B^{(1)}_1$ else you cannot construct such a component if you put in the $\epsilon$. – ZeroTheHero Nov 12 '18 at 20:22
• I see, so there's no need to introduce the epsilon in the first place and I can just operate with the previous expression. If that's the case, I find that component to be $T^{(2)}_1=\frac{1}{\sqrt{2}}(J_1J_0+J_0J_1)$. – Charlie Nov 12 '18 at 20:28

The definition of (composite) tensor operator I have is $$C^J_m=\sum_{m_a} A^{j_a}_{m_a}B^{j_b}_{m_b} \langle j_a m_a; j_b m_b\vert J m\rangle\, .$$ In particular, for $$(Jm)=(2,1)$$, we find $$C^2_1= \frac{1}{\sqrt{2}}A^1_1 B^1_0+\frac{1}{\sqrt{2}}A^1_0 B^1_1$$ Acting on states of the form $$\vert (j_{a_1} ; j_{b_1} )j_1m_1\rangle$$ the reduced matrix element \begin{align} \langle j_2 \Vert C^J\Vert j_1\rangle &= \langle j_{a_2} \Vert A^{j_a}\Vert j_{a_1}\rangle \langle j_{a_2} \Vert A^{j_a}\Vert j_{a_1}\rangle \sqrt{(2j_1+1)(2j_2+1)(2j+1)}\\ &\qquad \times \left\{\begin{array}{ccc} j_{a_1}&j_a&j_{a_2}\\ j_{b_1}&j_b&j_{b_2}\\ j_1&J&j_2 \end{array}\right\} \tag{1} \end{align} with a $$9j$$-coefficient on the last list of (1) in there.