Wigner-Eckart Theorem

Question

The Wigner-Eckart Theorem states that in state $| j = 1 , m \rangle$ : $$ \langle 1 ,m_1| Y^{m_{2}}_1 |1,m_{2}\rangle = \int d\Omega Y^{{*}^{m_1}}_1 Y^{m_2}_{1} Y^{m_3}_{1} = C^{1 1 1}_{m_{1} m_{2} m_{3}} \langle j|| Y^{m_{2}}_1||j\rangle $$

In the general case, to find $\langle j|| Y^{m_{2}}_1||j\rangle$ we set $m_1 = m_2 = m_3 = 0$ and get:

$$\int d\Omega Y^{{*}^{0}}_1 Y^{0}_{1} Y^{0}_{1} = C^{1 1 1}_{0 0 0} \langle j|| Y^{m_{2}}_1||j\rangle $$

The result given by the integration is zero, so this means that $\langle j|| Y^{m_{2}}_1||j\rangle$ or $ C^{1 1 1}_{0 0 0} = 0 $ (the Clebcsh-Gordon coeffiecient is zero), according to Wolfram $C^{1 1 1}_{0 0 0} = 0$.

So in this the Wigner-Eckart isn't applicable because we can't find the matrix element $\langle j|| Y^{m_{2}}_1||j\rangle$ or we have to set the $m$ to something else?

Answer 1

No. It means $\langle 1\Vert Y^1 \Vert 1 \rangle = 0$. This implies nothing about $\langle j \Vert Y^1\Vert j\rangle$ although because the CG $C^{j0}_{10;j0}$ is $0$ unless $2j+1$ is even you can deduce that $\langle j \Vert Y^1\Vert j\rangle =0$ when this case applies.

I should add that the reduced matrix element is a actually proportional to $C^{j0}_{10;j0}$ from the composition law of spherical harmonics, and this is why my conclusion above holds, i.e. the integration of three spherical harmonics as you have is proportional to $C^{j_1m_1}_{j_2m_2;j_3m_3} C^{j_10}_{j_20;j_30}$ in general, so that taking out the $m$-dependent CG leaves you with the $ C^{j_10}_{j_20;j_30}$ factor with the selection rule that this is $0$ unless $j_1+j_2+j_3$ is even.