Total spin operator acting on two-particle state

Question

I have the following state on which I want to apply the total spin operator:

$\hat{S}^2|\alpha\rangle = \hat{S}^2\frac{1}{\sqrt{2}}\left(|\frac{3}{2},\frac{1}{2}\rangle-|\frac{1}{2},\frac{3}{2}\rangle\right)$

Now,the total spin operator can be rewritten in the following form,

$\hat{S}^2 = \hat{S}^2_1 + \hat{S}^2_2 + \hat{S}_{1+}\hat{S}_{2-} + \hat{S}_{1-}\hat{S}_{2+} + 2\hat{S}_{1z}\cdot\hat{S}_{2z}$

How does each operator stated above act on this state?

I am aware that, an operator $\hat{S}_{1}$ only acts on the first particle and so on, but how about the latter terms?

The answer should come out as:

$\hat{S}^2|\alpha\rangle = 12\hbar|\alpha\rangle$

Answer 1

Nondimensionalize $\hbar$ to 1 for simplicity, otherwise you are begging for more mistakes.

Your final answer is wrong in a maximally misleading way: the coupled basis state you are inspecting has total J=2 and m=2, (check it!), so the total Casimir should be 6, (and not 12 corresponding to the symmetric J=3 state). I warned you.

The rest is straightforward: the 1,2 operators act on the 1,2, sites of your uncoupled basis vectors, $|m_1,m_2\rangle$. And you know $S_+|1/2\rangle= \sqrt{3}|3/2\rangle$, and $S_-|3/2\rangle= \sqrt{3}|1/2\rangle$, $S_+|3/2\rangle= 0$.

Consequently, the non-diagonal terms $\small \hat{S}_{1+}\hat{S}_{2-}\! +\! \hat{S}_{1-}\hat{S}_{2+} $ involving raising and lowering operators interchange the terms in your antisymmetric sum, i.e., effectively result in a diagonal contribution to eigenvalues by -3 ! Neat, huh?

Consequently, $$ \hat{S}^2|\alpha\rangle = \left ( \hat{S}^2_1 + \hat{S}^2_2 + \hat{S}_{1+}\hat{S}_{2-} + \hat{S}_{1-}\hat{S}_{2+} + 2\hat{S}_{1z} \hat{S}_{2z}\right ) \frac{1}{\sqrt{2}}\left( |\frac{3}{2},\frac{1}{2}\rangle- |\frac{1}{2},\frac{3}{2}\rangle\right ) $$

$$ = ( 15/4+15/4 -3+ 2\cdot 3/4 ) \frac{1}{\sqrt{2}}\left(|\frac{3}{2},\frac{1}{2}\rangle-|\frac{1}{2},\frac{3}{2}\rangle\right) \\ = 6|\alpha\rangle , $$ as warned: 2$\times$3 = 6.