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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?

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I am getting different Clebsch-Gordan coefficients than you, so I suspect the error is there. For the numerator, you need (in the $(j_1,j_2,m_1,m_2|j,m)$ notation) $$(1,2,0,-1|1,-1) = -\sqrt{\frac{3}{10}}$$ and for the denominator, $$(1,2,1,-1|1,0) = \sqrt{\frac{3}{10}}.$$ The ratio is indeed $-1$.

Alternatively, if you use 3-$j$ symbols, the ratio you need to calculate is $$\frac{(-1)^{1-1} \begin{pmatrix} 1 & 2 & 1 \\ 0 & -1 & 1 \end{pmatrix}}{(-1)^{1-0} \begin{pmatrix} 1 & 2 & 1 \\ 1 & -1 & 0 \end{pmatrix}} = \frac{\frac{-1}{\sqrt{10}}}{-1 \cdot \frac{-1}{\sqrt{10}}}$$ and this is $-1$ again.

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  • $\begingroup$ I wrote the right CGCs, but didn't read carefully the table. Thank you. $\endgroup$ – Vincenzo Ventriglia Mar 1 '18 at 9:38

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