Using the Lowering Operator to Find Total Spin States

The total spin lowering operator is

$$\bgroup\color{black}S_-=S^{(1)}_-+S^{(2)}_-.\egroup$$

First lets remind ourselves of what the individual lowering operators do.

$$\bgroup\color{black}S^{(1)}_-\chi^{(1)}_+ = \hbar \sqrt{ {1... ...({-1\over 2}\right)} \chi^{(1)}_- = \hbar \chi^{(1)}_- \egroup$$

Now we want to identify $\chi_{11}=\chi^{(1)}_+\chi^{(2)}_+$. Lets operate on this equation with $S_-$. First the RHS gives

$$\bgroup\color{black}S_- \chi^{(1)}_+ \chi^{(2)}_+ = \left(S^{... ...1)}_- \chi^{(2)}_+ + \chi^{(1)}_+ \chi^{(2)}_- \right). \egroup$$

Operating on the LHS gives

$$\bgroup\color{black}S_-\chi_{11}=\hbar\sqrt{(1)(2)-(1)(0)}\chi_{10}=\sqrt{2}\hbar\chi_{10}. \egroup$$

So equating the two we have

$$\bgroup\color{black}\sqrt{2}\hbar\chi_{10}=\hbar\left( \chi^{(1)}_- \chi^{(2)}_+ + \chi^{(1)}_+ \chi^{(2)}_- \right).\egroup$$

$$\bgroup\color{black}\chi_{10}={1\over\sqrt{2}}\left( \chi^{(1)}_- \chi^{(2)}_+ + \chi^{(1)}_+ \chi^{(2)}_- \right).\egroup$$

Now we can lower this state. Lowering the LHS, we get

$$\bgroup\color{black}S_-\chi_{10}=\hbar\sqrt{(1)(2)-(0)(-1)}\chi_{1(-1)}=\sqrt{2}\hbar\chi_{1,-1}. \egroup$$

Lowering the RHS, gives

$$\bgroup\color{black}S_-{1\over\sqrt{2}}\left( \chi^{(1)}_+ \c... ...2)}_- \right) = \sqrt{2}\hbar\chi^{(1)}_- \chi^{(2)}_- \egroup$$

$$\bgroup\color{black}\Rightarrow\qquad \chi_{1,-1} = \chi^{(1)}_- \chi^{(2)}_- \egroup$$

Therefore we have found 3 s=1 states that work together. They are all symmetric under interchange of the two particles.

There is one state left over which is orthogonal to the three states we identified. Orthogonal state:

$$\bgroup\color{black}\chi_{00}={1\over\sqrt{2}}\left( \chi^{(1)}_+ \chi^{(2)}_- - \chi^{(1)}_- \chi^{(2)}_+ \right)\egroup$$

We have guessed that this is an $s=0$ state since there is only one state and it has m=0. We could verify this by using the $S^2$ operator.