# The addition of 3 spins

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1. Nov 14, 2017

### Marvelant

1. The problem statement, all variables and given/known data
Given 3 spins, #1 and #3 are spin-1/2 and #2 is spin-1. The particles have spin operators $\vec{S}_i, i=1,2,3$. The particles are fixed in space. Let $\vec{S} = \vec{S}_1 + \vec{S}_2 + \vec{S}_3$ be the total spin operator for the particles.
(ii) Find the eigenvalues of $\vec{S}^2$ and their multiplicities.

2. Relevant equations
$\vec{S}^2 \rvert s,m \rangle = s(s+1)\rvert s,m \rangle$, Spin matrices

3. The attempt at a solution
So, for each total spin $s$ there are a number of combination of the $s_i$ values associated with each spin, $\{ \pm 1/2 \}$ for #1 and #3, $\{ 0, \pm 1 \}$ for #2. Then for each combination there are $2s_i+1$ choices for $m$ which yields $(2s_1+1)(2s_2+1)(2s_3+1)$ possibilities
This comes down to eigenvalues
$6 \hbar^2$ with degeneracy 24
$2 \hbar^2$ with degeneracy 56
$0 \hbar^2$ with degeneracy 32
I am skeptical of these values and feel that there is probably a better way.

Last edited: Nov 14, 2017
2. Nov 14, 2017

### Orodruin

Staff Emeritus
The representation space is 2x2x3=12-dimensional. Clearly no eigenvalue can have a multiplicity higher than 12.

How would you do this if there were only two spins? One spin 1/2 and one spin 1.

3. Nov 15, 2017

### Marvelant

I suppose all the different combinations would be $\{ -1/2, 1/2 \} + \{ -1, 0, 1 \} = \{ -3/2, -1/2, 1/2, 3/2 \}$ which translates to $s = \{ 1/2, 3/2 \}$ with $\vec{S}^2$ eigenvalues of $(3/4) \hbar^2, (15/4) \hbar^2$

4. Nov 15, 2017

### Orodruin

Staff Emeritus
This notation is very confusing. Could you explain what you mean by it? The original basis is the direct product of a vector space of dimension 2 and one of dimension 3, the final vector space should therefore have dimension 6.

Also, how do you deduce the degeneracies of the eigenvalues?

5. Nov 15, 2017

### Marvelant

Sorry for the confusing notation, all I meant was that I added together every possible pair, one element from each set, and discarded any repeated values.
I suppose discarding the repeated values was the wrong move, since the degeneracy would depend on them. So I'm going to guess degeneracies of the eigenvalues would be 4 for $(3/4) \hbar^2$ and 2 for $(15/4) \hbar^2$ but then wouldn't there also be degeneracy from the different values of $m$ in $\rvert s, m \rangle$?

6. Nov 15, 2017

### Orodruin

Staff Emeritus
This will only give you the possible values of the $z$-components, not the eigenvalues of the $\vec S^2$ operator. You have to decompose the product space into irreps of $SU(2)$, i.e., multiplets with a fixed $s$. Every state in such a multiplet will have the same values for the eigenvalue of $\vec S^2$. What you have written down are the $S_z$ eigenvalues of all the states, not the spins of the irreps.

7. Nov 15, 2017

### Orodruin

Staff Emeritus
As an example, consider the product space of two spin-1/2 irreps. The product space is then $\newcommand{\ket}[1]{|#1\rangle} 2\otimes 2$ spanned by the states $\ket \pm \otimes \ket \pm$. Decomposing this into irreps leaves you with one singlet (spin-0) and one triplet (spin-1) representation. The triplet corresponds to $s = 1$ and is spanned by the symmetric states
$$\ket + \otimes \ket +, \ \frac{1}{\sqrt{2}} \left( \ket + \otimes \ket - + \ket - \otimes \ket +\right), \ \ket - \otimes \ket -$$
while the singlet with $s = 0$ is the anti-symmetric state $(\ket + \otimes \ket - - \ket - \otimes \ket +)/\sqrt 2$.

In terms of the decomposition into irreps, you would write this as
$$2\otimes 2 = 3 \oplus 1,$$
i.e., the product space of two spin-1/2 irreps is decomposed into one triplet and one singlet representation. You can do the corresponding exercise both for $2\otimes 3$ and $2\otimes 2 \otimes 3$.

8. Nov 15, 2017

### Marvelant

Can I generate these by applying the lowering operator to the highest ket like this?
$$S_- \rvert 3/2,3/2 \rangle = (S_{1-}+S_{2-}) \rvert 1/2,1/2 \rangle_1 \rvert 1,1 \rangle_2$$

9. Nov 15, 2017

### Orodruin

Staff Emeritus
That will generate the states in the $s = 3/2$ irrep. Those will not be the full set of states. How do you plan to construct the remaining states and what $s$ will they correspond to?

10. Nov 15, 2017

### Marvelant

I suppose I could do the same for $\rvert 1/2,1/2 \rangle$.

11. Nov 15, 2017

### Orodruin

Staff Emeritus
In order to do that you need to know which state is the $|1/2,1/2\rangle$ state. Which is it?

12. Nov 15, 2017

### Marvelant

Would it be a linear combination of $\rvert 1/2,-1/2 \rangle_1 \rvert 1,1 \rangle_1$ and $\rvert 1/2,1/2 \rangle_1 \rvert 1,0 \rangle_1$?

13. Nov 15, 2017

### Orodruin

Staff Emeritus
Yes, but you need to figure out which linear combination before you start applying $S_-$ to it. Otherwise you will get part of the $s = 3/2$ state in it as well.

To be honest, you do not really need to figure out exactly what the states are. What you need to figure out is which irreps are present in the products and what dimensions each irrep has. So, which are the irreps in the $2 \otimes 3$ product? What are their corresponding $\vec S^2$ eigenvalues and dimensions?

14. Nov 15, 2017

### Marvelant

I'm not confident enough in tensor products to work at that level, unfortunately.

15. Nov 15, 2017

### Orodruin

Staff Emeritus
So instead figure out the eigenstates of $\hat S^2$ and the degeneracy of the corresponding eigenvalues, that will also work but it is a lot more work.