Finding the Error in Computing Spherical Tensor of Rank 0 Using General Formula

[Silicon-Based]

TL;DR

Coeffcient of spherical tensor of rank 0 from Clebsch_Gordan

This should be a trivial question. I am trying to compute the spherical tensor $T_0^{(0)} = \frac{(U_1 V_{-1} + U_{-1} V_1 - U_0 V_0)}{3}$ using the general formula (Sakurai 3.11.27), but what I get is:
$$
T_0^{(0)} = \sum_{q_1=-1}^1 \sum_{q_2=-1}^1 \langle 1,1;q_1,q_2|1,1;0,q\rangle U_{q_1}^{(1)} V_{q_2}^{(1)} \;\;\;\; \text{for} \; q = q_1 + q_2
$$
$$
T_0^{(0)} = \langle 1,1;1,-1|1,1;0,0\rangle U_{1}^{(1)} V_{-1}^{(1)} + \langle 1,1;0,0|1,1;0,0\rangle U_{0}^{(1)} V_{0}^{(1)} + \langle 1,1;-1,1|1,1;0,0\rangle U_{-1}^{(1)} V_{1}^{(1)}
$$
But this is $T_0^{(0)} = \frac{(U_1 V_{-1} + U_{-1} V_1 - U_0 V_0)}{\sqrt{3}}$, if I read the Clebsch-Gordan coefficients correctly. Where did I make a mistake?

 

[azntoon]

You made a mistake in the Clebsch-Gordan coefficient. The correct coefficient is $\langle 1,1;1,-1|1,1;0,0\rangle = \frac{1}{\sqrt{3}}$, not $1$. Therefore, you should have $$T_0^{(0)} = \frac{1}{\sqrt{3}}(U_{1}^{(1)} V_{-1}^{(1)} + U_{0}^{(1)} V_{0}^{(1)} + U_{-1}^{(1)} V_{1}^{(1)})$$