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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:

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

Qmechanic
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Ponny Hans
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  • Does this help? – Tobias Fünke Dec 09 '22 at 10:51
  • @TobiasFünke It clarifies the steps of obtaining the form of the operator, thanks for that! Further, my question is more directed on what comes out when the operators have acted on the state which is where my confusion begins – Ponny Hans Dec 09 '22 at 11:55
  • There is something flakey here. You are probably adding two spin 3/2s in the uncoupled basis, by all appearances, and your answer specifies the spin 3 representation, which is symmetric, not antisymmetric in the uncoupled states. Are you sure about the minus sign? The coefficient of your expected answer? – Cosmas Zachos Dec 09 '22 at 14:54
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    Well, with the minus sign you are in the total spin 2 representation, and your Casimir should be 6, not 12. – Cosmas Zachos Dec 09 '22 at 14:56
  • @CosmasZachos The minus sign is correct. I chose to omit the symmetric spatial part of the problem. But for more context, We are considering two spin-3/2 particles in a magnetic field both being in the ground state. Where I have constructed the state by the means of Pauli principle & exclusion principle. It is also assumed that the two particles are in the same spatial state, which infers the antisymmetric spin-state. I believe I wasn't clear enough in the description, but it is two spin-3/2 particles. – Ponny Hans Dec 09 '22 at 15:00
  • @CosmasZachos To add further to this, I have been in contact with the supervisor for the course, and he gave me the answer seen in the last equation. My problem is not really knowing what comes out when, say, the middle two terms in the second equation acts on the state. Sorry for the inconvenience and lack of information. – Ponny Hans Dec 09 '22 at 15:07
  • @PonnyHans Please consider that the supervisor for your course may be wrong, or otherwise confused. As stated in the comments, the state with $S=3$ and $M_S = 2$ is proportional to $|3/2,1/2\rangle+|1/2,3/2\rangle$ (in your notation), not $|3/2,1/2\rangle-|1/2,3/2\rangle$. The latter state still has total $M_S=2$, but it also has total $S=2$. – hft Dec 09 '22 at 18:02
  • @PonnyHans If you act with $S_{1-} + S_{2-}$ on the stretch state $|3/2,3/2\rangle$, you will see what I mean regarding the $S=3$ and $M_S=2$ state. And you should see how to construct the orthogonal state with $S=2$ and $M_S=2$. Alternatively, one of the answers explicitly shows that the state you provide has total $S=2$... – hft Dec 09 '22 at 18:04

1 Answers1

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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.

Cosmas Zachos
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