Consider the operator $$\mathcal{P}=L_y L_z$$ and the hydrogen atom states $\left|n,l,m\right\rangle$. Evaluate the ratio $$\mathcal{R}=\frac{\left\langle3,1,-1|\mathcal{P}|3,1,0\right\rangle}{\left\langle3,1,0|\mathcal{P}|3,1,1\right\rangle}.$$
First, I noted that $\mathcal{P}\neq\mathcal{P}^\dagger$, so I put $$\mathcal{P}_\mathrm{sym}\equiv\frac{1}{2}\left\{L_y,L_z\right\},$$ since this is the observable quantity. Then I wrote $\mathcal{P}_\mathrm{sym}$ in terms of spherical tensors $L^{(k)}_q$ of rank $k$ (with $2k+1$ components), obtaining $$\mathcal{P}_\mathrm{sym}=\frac{i}{2}\left(L^{(2)}_1 + L^{(2)}_{-1}\right).$$ Now, using Wigner-Eckart theorem, I got $$\mathcal{R}_{\mathrm{sym}}=\frac{\frac{1}{\sqrt{10}}\left\langle3,1||L^{(2)}||3,1\right\rangle}{\sqrt{\frac{3}{10}}\left\langle3,1||L^{(2)}||3,1\right\rangle}=\frac{1}{\sqrt3}.$$
But, using a direct method, I also got (since $n=3$ is fixed) $$\mathcal{R}_{\mathrm{sym}}=\frac{\begin{pmatrix}0&0&1\end{pmatrix} \begin{pmatrix}&-1&\\1&&1\\&-1&\end{pmatrix} \begin{pmatrix}0\\1\\0\end{pmatrix}}{\begin{pmatrix}0&1&0\end{pmatrix} \begin{pmatrix}&-1&\\1&&1\\&-1&\end{pmatrix} \begin{pmatrix}1\\0\\0\end{pmatrix}}=-1.$$ Where do I go wrong?
