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Let's have the spherical spinors $\psi_{j, m, l = j \pm \frac{1}{2}}$, $$ Y_{j, m, l = j \pm \frac{1}{2}} = \frac{1}{\sqrt{2l + 1}}\begin{pmatrix} \pm \sqrt{l \pm m +\frac{1}{2}}Y_{l, m - \frac{1}{2}} \\ \sqrt{l \mp m +\frac{1}{2}}Y_{l, m + \frac{1}{2}} \end{pmatrix}. $$ How to find an action of $(\hat {\sigma} \cdot \hat {\mathbf L}$ on it without using very long derivatives by using expressions for spin-orbital operator $(\hat {\sigma} \cdot \hat {\mathbf L} )$ (it is part of the square of total angular momentum $\hat {\mathbf J} = \hat {\mathbf L} + \hat {\mathbf S}$) in spherical coordinate system, if it is possible?

P.S. I don't know that $\hat {\mathbf J}^2\psi_{j, m, l = j \pm \frac{1}{2}} = j(j + 1)\psi_{j, m, l = j \pm \frac{1}{2}}$, so I can't express $(\hat {\sigma} \cdot \hat {\mathbf L})$ from this expression.

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If you don't know that spherical spinors are eigenfunctions of $\mathbf{J}^2$, how do you define them in the first place. – suresh Feb 28 '14 at 0:51
Try writing $\sigma\cdot \mathbf{L}= \sigma_3 L_3 + \sigma_+ L_- + \sigma_-L+$. That action of $L_3$, $L_\pm$ on the $Y_{l,m}$ is well-defined. – suresh Feb 28 '14 at 0:54
@Suresh : I doing it as an exercise: I need to check that spherical spinors are the eigenfunctions directly. So an exercise predicts that I can't use relation for acting of square of angular momentum operator. – Andrew McAddams Feb 28 '14 at 15:56
Thank you, I will try. Unfortunately, the first attempt was uncussesful. I can't get the equal factors near first and second spherical functions after acting of my operator on spinor. – Andrew McAddams Feb 28 '14 at 15:59
Make sure that you determine the constants $c_\pm(l,m)$ that are given by $L_\pm\ Y_{l,m} = c_\pm (l,m)\ Y_{l,m\pm1}$ correctly. – suresh Mar 1 '14 at 3:00

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