Applying the Operator to and .

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Applying the $\bgroup\color{black}$S^2$\egroup$ Operator to $\bgroup\color{black}$\chi_{1m}$\egroup$ and $\bgroup\color{black}$\chi_{00}$\egroup$ .

$\begin{displaymath}\bgroup\color{black} S^2=\left(\vec{S}_1 + \vec{S}_2 \right)^2 = S^2_1+S^2_2+2{\vec{S}_1\cdot \vec{S}_2}\egroup\end{displaymath}$

$\begin{displaymath}\bgroup\color{black} S^2 \chi^{(1)}_+ \chi^{(2)}_+ = s_1(s_1+... ... + 2{\vec{S}_1\chi^{(1)}_+\cdot \vec{S}_2} \chi^{(2)}_+ \egroup\end{displaymath}$

$\begin{displaymath}\bgroup\color{black} = {3\over 2}\hbar^2\chi^{(1)}_+ \chi^{(2... ...+ S^{(1)}_z S^{(2)}_z \right) \chi^{(1)}_+ \chi^{(2)}_+ \egroup\end{displaymath}$

$\begin{displaymath}\bgroup\color{black} S_x \chi_+ = {\hbar\over 2}\left(\matrix... ...0}\right) = {\hbar\over 2}\left(\matrix{0\cr 1}\right) \egroup\end{displaymath}$

$\begin{displaymath}\bgroup\color{black} S_y \chi_+ = {\hbar\over 2}\left(\matrix... ...0}\right) = {\hbar\over 2}\left(\matrix{0\cr i}\right) \egroup\end{displaymath}$

$\begin{displaymath}\bgroup\color{black} S^2 \chi^{(1)}_+ \chi^{(2)}_+ = {3\over ... ...1\cr 0}\right)_1 \left(\matrix{1\cr 0}\right)_2 \right] \egroup\end{displaymath}$

$\begin{displaymath}\bgroup\color{black} S^2 \chi^{(1)}_+ \chi^{(2)}_+ = {3\over ... ...}_+ \chi^{(2)}_+ = 2 \hbar^2 \chi^{(1)}_+ \chi^{(2)}_+ \egroup\end{displaymath}$

$\begin{displaymath}\bgroup\color{black}s(s+1)=2\egroup\end{displaymath}$

This confirms that we have an s=1 state.

James Branson
2001-09-17