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

We wish to verify that the states we have deduced are really eigenstates of the $S^2$ operator. We will really compute this in the most brute force.

$$\begin{eqnarray*} S^2&=&\left(\vec{S}_1 + \vec{S}_2 \right)^2 = S^2_1+S^2_2+2{\v... ...{(2)}_y + S^{(1)}_z S^{(2)}_z \right) \chi^{(1)}_+ \chi^{(2)}_+ \end{eqnarray*}$$

$$\begin{eqnarray*} S_x \chi_+ &=& {\hbar\over 2}\left(\matrix{0&1\cr 1&0}\right)\... ...\over 2}\left(\matrix{-i\cr 0}\right)=-i{\hbar\over 2}\chi_+ \\ \end{eqnarray*}$$

$$\begin{eqnarray*} S^2 \chi^{(1)}_+ \chi^{(2)}_+ &=& {3\over 2}\hbar^2\chi^{(1)}_... ...^{(1)}_+ \chi^{(2)}_+ = 2 \hbar^2 \chi^{(1)}_+ \chi^{(2)}_+ \\ \end{eqnarray*}$$

Note that $s(s+1)=2$, so that the $2\hbar$ is what we expected to get. This confirms that we have an s=1 state.

Now lets do the $\chi_{00}$ state.

$$\begin{eqnarray*} S^2 \chi_{00} &=& \left(S^2_1+S^2_2+2{\vec{S}_1\cdot \vec{S}_2... ...) \\ &=& \hbar^2\left(1 -1\right) \chi_{00}= 0\hbar^2\chi_{00} \end{eqnarray*}$$