# Matrix elements of Tenser force $S_{12}$ in Spin-dependent Nucleon-Nucleon Interactions

I am aware of the fact, that the matrix elements of the tensor operator $$S_{12}= 3( \boldsymbol{\sigma}_1 \cdot \hat{\mathbf r})(\boldsymbol{\sigma}_2 \cdot \hat{\mathbf r}) - \boldsymbol{\sigma}_1 \cdot \boldsymbol{\sigma}_2, \quad \hat{\mathbf{r}} = \frac{\mathbf{r}}{r}$$ can be calculated via the Wigner-Eckart theorem. The exercise (see picture 1) implies, that there is another way to do that. The steps (a), (b), (c), (e) are clear for me. I struggle with (d) and (f). Before i ask my questiens to these parts, I summarize my solution for (a), (b), (c), (e) shortly:

The simultaneous eigenfunctions (or eigenstates) of $$\mathbf J^2$$, $$J_z$$, $$\mathbf S^2$$ and $$\mathbf L^2$$ are \begin{align} | \mathscr{Y}_{L,S,J}^M \rangle &= |LS;JM \rangle = \sum_{m_L=-L}^L \sum_{\sigma=-S}^S |LS;m_L\sigma \rangle \underbrace{C_{LS; m_L\sigma}^{JM }}_{\text{C. G. - coefficients}} \\ \Rightarrow \mathscr{Y}_{L,S,J}^M(\theta,\varphi) &= \langle \hat{\mathbf{r}}(\theta,\varphi)| \mathscr{Y}_{L,S,J}^M \rangle =\sum_{m_L=-L}^L \sum_{\sigma=-S}^S Y_L^{m_L}(\theta,\varphi)|S\sigma \rangle C_{LS; m_L\sigma}^{JM} . \end{align}

Solution (a): Write $$S_{12}$$ as $$2(3Q-\mathbf{S}^2)$$, where $$\mathbf{S} = 1/2(\boldsymbol{\sigma}_1 +\boldsymbol{\sigma}_2)$$ is the total spin, and $$Q=(\mathbf{S} \cdot \hat{\mathbf{r}})^2$$. Because the quadratic norm of the vector $$\mathbf{S}|00\rangle$$ vanishes $$\langle00|\mathbf{S}^2|00\rangle = 0,$$ $$\mathbf{S}|00\rangle$$ vanishes, too. $$\Rightarrow Q|00\rangle =0,S_{12}|00\rangle =0$$.

Solution (b): $$[\mathbf L^2,S_{12}]\ne 0$$, but $$[P,S_{12}]=0$$ ( $$P$$ is the parity operator). Hence, the eigenstates of $$S_12$$ can be found among the simultaneous eigenfunctions of $$\mathbf J^2$$, $$J_z$$, $$\mathbf S^2$$ and $$P$$ (we replaced $$\mathbf L^2$$ with $$P$$). For $$S=1$$ (Spin - Triplett) and $$P=(-1)^J$$, we have $$L=J$$ (because of the spherical harmonics). In this case $$\mathcal{Y}_{J,1,J}^M$$ is a possible eigenfunction of $$S_{12}$$. $$S_{12}\mathscr{Y}_{J,1,J}^M = a_{J,J} \mathscr{Y}_{J,1,J}^M$$ For $$S=1$$ and $$P=(-1)^{J+1}$$ the eigenfunction is a superposition of $$L=J+1$$- and $$L=J-1$$-states. For $$L=J\pm 1$$, $$S_{12}$$ "couples" these states: $$S_{12}\mathscr{Y}_{L,1,J}^M = a_{L,J+1} \mathscr{Y}_{J+1,1,J}^M + a_{L,J-1} \mathscr{Y}_{J-1,1,J}^M.$$

Solution (c): Showing, that $$[\mathbf{J},S_{12}]=0$$ is straight foreward. And than it is obvious I think.

Solution (e): Just use the "north pole formular" for spherical harmonics $$Y_{l}^m(\theta=0,\varphi)=Y_{l}^m(\hat{\mathbf{e}}_z) = \sqrt{\frac{2l+1}{4\pi}} \delta_{m0}$$ and $$Q \big \vert_{\theta=0} = S_z^2$$.

Question to (d): \begin{align} a_{J+1,J-1} &= \langle \mathscr{Y}_{J+1,1,J}^M|S_{12}| \mathscr{Y}_{J-1,1,J}^M \rangle \\ a_{J-1,J+1} &= \langle \mathscr{Y}_{J-1,1,J}^M|S_{12}| \mathscr{Y}_{J+1,1,J}^M \rangle\end{align} I can only see from this, that the constants are respectively complex conjugate: $$a_{J+1,J-1} = a_{J-1,J+1}^*.$$ Did i miss something?

Question to (f): What do they mean with $$\hat{\mathbf r} = \hat{\mathbf{k}}$$ ??

If someone has annotations or questions, feel free to comment :-). I sincerely thank those who can help me with my questions.

This is taken from: Donnelly, T., Formaggio, J., Holstein, B., Milner, R., & Surrow, B. (2017). Foundations of Nuclear and Particle Physics. Cambridge: Cambridge University Press. doi:10.1017/9781139028264 Url: https://www.cambridge.org/highereducation/books/foundations-of-nuclear-and-particle-physics/779E1C84B94FD01E4B43500AA84C8703

Today, I finished the whole problem by myself. In part (f) one has to consider $$\theta=0$$ again and then, because $$\mathbf J$$ commutes with $$S_{12}$$, one can consider the special cases $$M\in \{-1,0,1\}$$. For those who came accross to the same problem, I will give a complete solution of the problem in follwing:
To determine the radial equation for a Hamiltonian, that contains a tensor force term \begin{align} V_T(r)S_{12} \end{align} we have to evaluate how $$S_{12}$$ acts on $$\lvert LS;JM\rangle$$ directly. Depending on the value of $$S$$ we have to classify the values after the parity. To do this, we consider \begin{align} \mathscr Y_{L,S,J}^{M}(\theta,\varphi) &= \langle\hat{\mathbf n}(\theta,\varphi)\vert LS;JM \rangle \notag \\ &= \sum_{m_l=-L}^L \sum_{\sigma=-S}^S \langle\hat{\mathbf n}(\theta,\varphi)\vert LS;m_L \sigma \rangle C_{LS;m_L \sigma}^{JM} \notag \\ &= \sum_{m_l=-L}^L \sum_{\sigma=-S}^S Y_L^{m_L}(\theta,\varphi) \vert S \sigma\rangle C_{LS;m_L \sigma}^{JM} \notag \\ &= \sum_{\sigma=-S}^S Y_L^{m_L}(\theta,\varphi) \vert S \sigma\rangle C_{LS;m_L \sigma}^{JM} , \end{align} and use, that the parity of the spherical harmonics $$Y_L^{m_L}(\theta,\varphi)$$ is $$(-1)^L$$.
The functions $$\mathscr{Y}_{L,S,M}^{M}(\theta,\varphi)$$ are eigenfunctions of $$\mathbf S^2$$, thus, in order to determine the action of $$S_{12}$$, it is sufficient to determine the action of $$Q=(\mathbf S \cdot \hat{\mathbf{n}})^2$$ on these functions. If $$S=0$$, then necessarily $$L=J$$ (therefor the parity is $$(-1)^J$$) and $$\mathscr{Y}_{J0J}^{J}(\theta,\varphi) = Y_J^{M}(\theta,\varphi) \vert 00\rangle$$ is necessarily an eigenfunction of $$Q$$. The quadratic norm $$\langle 00|\mathbf S^2|00\rangle$$ of $$\mathbf S |S=0,M_S=0\rangle$$ vanishes. Therefore $$\mathbf S|00\rangle$$ vanishes, too. Consequently, it results, \begin{align} Q\mathscr{Y}_{J0J}^{M}(\theta,\varphi) =QY_J^{M}(\theta,\varphi) |00\rangle = 0. \end{align} If $$S=1$$ there are two possible values for the parity $$(-1)^J$$ and $$(-1)^{J+1}$$. In the first case, necessarily $$L=J$$ holds, and $$\mathscr{Y}_{J1J}^{M}(\theta,\varphi)$$ is eigenfunction of $$Q$$ \begin{align} Q\mathscr{Y}_{J1J}^{M}(\theta,\varphi) = C_{J}^{M} \mathscr{Y}_{J1J}^{M}(\theta,\varphi). \end{align} For the second case, the only possible values of $$L$$ are $$J\pm1$$ (unless $$J=0$$, in which case there is Just one value $$J=1$$) so that the eigenfunctions are of the form \begin{align} R_{J-1}^{M}(r) \mathscr Y_{J-}^{M}(\theta,\varphi) + R_{J+1}^{M}(r) \mathscr Y_{J+}^{M}(\theta,\varphi), \end{align} where $$\mathscr Y_{J\pm}^{M}(\theta,\varphi) = \mathscr{Y}_{J\pm1,1,J}^{M}(\theta,\varphi)$$. In other words, $$Q$$ couples $$L = J \pm1$$ states: \begin{align} Q \mathscr{Y}_{J+}^{M} (\theta,\varphi) &= D_{J+1\,J+1}^{M} \mathscr Y_{J+}^{M}(\theta,\varphi) + D_{J+1\,J-1}^{M} \mathscr Y_{J-}^{M}(\theta,\varphi)\\ Q \mathscr{Y}_{J-}^{M} (\theta,\varphi) &= D_{J-1\,J+1}^{M} \mathscr Y_{J+}^{M}(\theta,\varphi) + D_{J-1\,J-1}^{M} \mathscr Y_{J-}^{M}(\theta,\varphi) \end{align} Because $$J_z$$ commutes with $$Q$$, the constants $$\{C_{J}^{M}, D_{J\pm1\,J\pm1}^{M}, D_{J\mp1\,J\pm1}^{M}\}$$ and the radial functions are independent of $$M$$. I.e. \begin{align} D_{J\pm1\,J\pm1} &= \int \mathrm{d} \Omega \mathscr Y_{J\pm}^{M*}(\theta,\varphi) Q \mathscr Y_{J\pm}^{M}(\theta,\varphi), \label{constantD:a} \\ D_{J\pm1\,J\mp1} &= \int \mathrm{d} \Omega \mathscr Y_{J\mp}^{M*}(\theta,\varphi) Q \mathscr Y_{J\pm}^{M}(\theta,\varphi). \label{constantD:b} \end{align} But since $$Q$$ is hermitean ($$Q^\dagger = (\hat{\mathbf n} \cdot \mathbf S)^2 = Q$$, because $$[\hat{\mathbf n},\mathbf S] = 0$$), both integrals on the right side of the first equation above are real valued, and the integrals on the right sides of the second equation are respectively complex conJugates, i.e. $$D^*_{J+1J-1} = D_{J-1J+1}$$.
Now we calculate the constants $$\{C_{J}, D_{J\pm1J\pm1}, D_{J\mp1J\pm1}\}$$. With the "north pole formula" of the spherical harmonics, we can derive $$Q\mathscr{Y}_{J1J}^{M}(\theta=0,\varphi)$$: \begin{align} Q \mathscr{Y}_{L1J}^{M}(\theta,\varphi) \bigg\vert_{\theta = 0} &= (\mathbf S \cdot \hat{\mathbf e}_z)^2 \mathscr{Y}_{L1J}^{M}(\theta=0,\varphi) \notag \\ &= \sum_{\sigma=-1}^1 Y_J^{M-\sigma}(\theta=0,\varphi) \sigma^2 |1 \sigma\rangle C_{L1;M-\sigma\, \sigma}^{JM} \notag \\ &= \sum_{\sigma=-1}^1 \sigma^2 \sqrt{\dfrac{2L+1}{4 \pi}} \delta_{M-\sigma0}|1 \sigma\rangle C_{L1;M-\sigma\, \sigma}^{JM} \notag \\ &= \sqrt{\dfrac{2L+1}{4 \pi}}M^2|1 M\rangle C_{L1;0 1}^{JM} \end{align} Similary, we calculate $$\mathscr{Y}_{L1J}^{M}(\theta=0,\varphi)$$: \begin{align} \mathscr{Y}_{L1J}^{M}(\theta,\varphi) \bigg\vert_{\theta = 0} = \sqrt{\dfrac{2L+1}{4 \pi}}|1 M\rangle C_{L1;0 1}^{JM}. \end{align} The special cases $$M=0, 1$$ are useful. For $$L=J$$ and $$M=1$$ we obtain \begin{align} Q \mathscr{Y}_{J1J}^{M}(\theta,\varphi) \bigg\vert_{\theta = 0} =\sqrt{\dfrac{2J+1}{4 \pi}} |1 1\rangle C_{J1;0 1}^{JM}= \mathscr{Y}_{J1J}^{M}(\theta,\varphi) \bigg\vert_{\theta = 0}, \end{align} i.e. $$C_{J} = 1$$. Similary, we consider the special case $$\theta =0$$ for the four constants $$D_{J\pm1\,J\pm1}, D_{J\mp1\,J\pm1}$$: Inserting $$L=J\pm 1$$ and $$M=1$$ we obtain \begin{align} Q \mathscr Y_\pm^{1}(\theta,\varphi)\bigg\vert_{\theta = 0} &= \mathscr Y_\pm^{1}(\theta,\varphi)\bigg\vert_{\theta = 0}\notag \\ &= \sqrt{\dfrac{2(J\pm1)+1}{4\pi}} |11\rangle C_{J\pm1 \, 1;01}^{J1}. \end{align} $$M=0$$ yields \begin{align} Q\mathscr Y_\pm^{0}(\theta,\varphi)\bigg\vert_{\theta = 0} &= 0, \\ \mathscr Y_\pm^{0}(\theta,\varphi)\bigg\vert_{\theta = 0} &= \sqrt{\dfrac{2(J\pm1)+1}{4\pi}} |10\rangle C_{J\pm1,1;00}^{J0}. \end{align} The following system equation follows: \begin{align} 1&=D_{J+1,J+1} + D_{J+1,J-1}\sqrt{\dfrac{2J-1}{2J+3}} \dfrac{C_{J-1\, 1;01}^{J1}}{C_{J+1\, 1;01}^{J1}} \label{Dconsteqs:a}\\ 1&=D_{J-1,J+1}\sqrt{\dfrac{2J+3}{2J-1}}\dfrac{C_{J+1\, 1;01}^{J1}}{C_{J-1\, 1;01}^{J1}} + D_{J-1,J-1} \label{Dconsteqs:b}\\ 0&=D_{J+1,J+1} + D_{J+1,J-1}\sqrt{\dfrac{2J-1}{2J+3}} \dfrac{C_{J-1\, 1;00}^{J0}}{C_{J+1\, 1;00}^{J0}}\label{Dconsteqs:c} \\ 0&=D_{J-1,J+1}\sqrt{\dfrac{2J+3}{2J-1}}\dfrac{C_{J+1\, 1;00}^{J0}}{C_{J-1\, 1;00}^{J0}} + D_{J-1,J-1}\label{Dconsteqs:d} \end{align} Befor we write down, the solution of this system linear equations, we calculate the C.G. coefficients. There for we use the recursion formulas: \begin{align} \label{cgrecursiona} &c_+(J,M) C_{j_1j_2;m_1m_2}^{JM+1} =c_-(j_1,m_1) C_{j_1j_2;m_1- 1m_2}^{JM} c_-(j_2,m_2) C_{j_1j_2;m_1m_2- 1}^{JM} \end{align} and \begin{align} \label{cgrecursionb} &c_-(J,M) C_{j_1j_2;m_1m_2}^{JM-1} =c_+(j_1,m_1) C_{j_1j_2;m_1+ 1m_2}^{JM} + c_+(j_2,m_2) C_{j_1j_2;m_1m_2+ 1}^{JM}, \end{align} where \begin{align} c_+(j,m) = \sqrt{j(j+1)-m(m+1)}, \qquad c_-(j,m) = \sqrt{j(j+1)-m(m-1)}. \end{align} Using the first recoursion formular, we obtain \begin{align} \label{cgcoffJ+1} C_{J+1\,1;M+1,-1}^{JM} &= \dfrac{c_+(J,M)}{c_-(J+1,M+2)} C_{J+1\,1;M+2,-1}^{JM+1} \notag\\ &\hspace{0,2cm}\vdots \notag\\ &= \prod_{n=0}^{J-M} \dfrac{c_+(J,M+n)}{c_-(J+1,M+n+2)} \underbrace{C_{J+1\,1;J+1,-1}^{JJ}}_{=1} \notag \\ &= \prod_{n=0}^{J-M} \sqrt{\dfrac{J+M+n+1}{J+M+n+3}} \notag \\ &= \sqrt{\dfrac{(J+M+1)(J+M+2)}{2(J+1)(2J+3)}}. \end{align} Similar with the second one \begin{align} \label{cgcoffJ-1} C_{J-1\,1;M-1\,1}^{JM} &= \dfrac{c_+(J-1,M-1)}{c_-(J,M+1)} C_{J-1\,1;M\,1}^{JM+1} \notag\\ &\hspace{0,2cm}\vdots \notag\\ &= \prod_{n=0}^{J-M-1} \dfrac{c_+(J-1,M+n-1)}{c_-(J,M+n+1)} \underbrace{C_{J-1\,1;J-1\,1}^{JJ}}_{=1} \notag \\ &= \prod_{n=0}^{J-M-1} \sqrt{\dfrac{J+M+n-1}{J+M+n+1}} \notag \\ &= \sqrt{\dfrac{(J+M)(J+M-1)}{2J(2J-1)}}, \end{align} follows. Again we use the recursion formulas to obtain $$C_{J-1,1;01}^{J1},C_{J-1,1;00}^{J0},C_{J+1,1;01}^{J1}$$ and $$C_{J+1,1;00}^{J0}$$: \begin{align} C_{J-1,1;01}^{J1} &= \sqrt{\dfrac{(J+1)}{2(2J-1)}},\\ C_{J-1,1;0,0}^{J0} &= \dfrac{c_+(J,0)}{c_-(1,1)} C_{J-1,1;0,1}^{J,1} -\dfrac{c_-(J-1,0)}{c_-(1,1)}C_{J-1,1;-1,1}^{J0} \notag \\ &= \sqrt{\dfrac{J}{ (2 J-1) }}. \end{align} The remaining coefficient results from a symmetry property of the "Wigner 3-j symbols" (https://en.wikipedia.org/wiki/3-j_symbol): \begin{align} \begin{pmatrix} J_1 & J_2 & J_3 \\ m_1 & m_2 & m_3 \end{pmatrix} = (-1)^{J_1+J_2+J_3}\begin{pmatrix} J_1 & J_2 & J_3 \\ -m_1 & -m_2 & -m_3 \end{pmatrix}. \end{align} This implies \begin{align} C_{J+1,1;01}^{J1} &= (-1)^J \sqrt{2J+1}\begin{pmatrix} J+1 & 1 & J \\ 0 & 1 & -1 \end{pmatrix}\notag \\ &= (-1)^J (-1)^{2(J+1)} \sqrt{2J+1}\begin{pmatrix} J+1 & 1 & J \\ 0 & -1 & 1 \end{pmatrix}\notag \\ &=C_{J+1,1;0,-1}^{J-1} \notag \\ &= \sqrt{\dfrac{J}{2(2J+3)}}. \end{align} Finally we obtain \begin{align} C_{J+1,1;00}^{J0} &= \dfrac{c_-(J,0)}{c_+(1,-1)} C_{J+1,1;0,-1}^{J,-1} -\dfrac{c_+(J+1,0)}{c_+(1,-1)}C_{J+1,1;1,-1}^{J0} \notag \\ &=-\sqrt{\dfrac{(J+1) }{ (2 J+3)}} \end{align} If we subtract the third from the first equation we have: \begin{align} D_{J+1J-1} = \sqrt{\dfrac{2J+3}{2J-1}} \dfrac{C_{J+1,1;01}^{J1} C_{J+1,1;00}^{J0}}{C_{J-1,1;01}^{J1}C_{J+1,1;00}^{J0} -C_{J-1,1;00}^{J0} C_{J+1,1;01}^{J1}} =\dfrac{\sqrt{J(J+1)}}{2J+1}. \end{align} The last constatns are \begin{alignat}{2} &D_{J-1J+1} &&= D^*_{J+1J-1} =\dfrac{\sqrt{J(J+1)}}{2J+1} \\ &D_{J-1,J-1} &&= - D_{J+1J-1} \sqrt{\dfrac{2J+3}{2J-1}}\dfrac{ C_{J-1,1;01}^{J1}}{C_{J+1,1;01}^{J1}}= \dfrac{J+1}{2J+1}, \\ &D_{J+1,J+1}&&=1- D_{J+1J-1} \sqrt{\dfrac{2J+3}{2J-1}}\dfrac{ C_{J-1,1;01}^{J1}}{C_{J+1,1;01}^{J1}} = \dfrac{J}{2J+1}. \end{alignat} Conclusion: \begin{align} S_{12} \mathscr Y_{J,0,J}^M(\theta,\varphi) &= 0 \\ S_{12} \mathscr Y_{J,0,J}^M(\theta,\varphi) &= 2 \mathscr Y_{J,0,J}^M(\theta,\varphi) \\ S_{12} \mathscr Y_{J+1,1,J}^M(\theta,\varphi) &= - \dfrac{2(J+2)}{2J+1} \mathscr Y_{J+1,1,J}^M(\theta,\varphi) + \dfrac{6\sqrt{J(J+1)}}{2J+1} \mathscr Y_{J-1,1,J}^M(\theta,\varphi) \\ S_{12}\mathscr Y_{J-1,1,J}^M(\theta,\varphi) &= \dfrac{6\sqrt{J(J+1)}}{2J+1} \mathscr Y_{J+1,1,J}^M(\theta,\varphi)- \dfrac{2(J-1)}{2J+1} \mathscr Y_{J-1,1,J}^M(\theta,\varphi). \end{align}