# Three spin-1/2 particles

1. Jun 7, 2012

### Gunthi

1. The problem statement, all variables and given/known data
Find the energies for a 3 spin-1/2 particles with the Hamiltonean:
$$H=\frac{E_0}{\hbar^2}(\vec{S_1}.\vec{S_3}+\vec{S_2}.\vec{S_3})$$

3. The attempt at a solution

From the Clebsh-Gordon table one gets all the spin functions:
$$|\frac{3}{2},\frac{3}{2}>...|\frac{1}{2},\frac{1}{2}>...|\frac{3}{2},-\frac{3}{2}>$$ (6 states in total)

So, to get the matrix elements for the Hamiltonian I tried developing the dot product so I could work directly with the operators i.e.:

$$\vec{S_1}.\vec{S_3}=S_{1x}S_{3x}+S_{1y}S_{3y}+S_{1z}S_{3z}$$

Now the problem is that the states as they are defined represent only one particle and the spin operators act on each particle individually and there's no CG table for a 1/2x1/2x1/2 spin addition.

What's the best way to approach this?

2. Jun 8, 2012

### vela

Staff Emeritus
Hint: You can write the Hamiltonian as
$$\hat{H} = \frac{E_0}{\hbar^2}(\vec{S}_1+\vec{S}_2)\cdot\vec{S}_3.$$

3. Jun 8, 2012

### Gunthi

Yes but my question is how do these operators act on these states? The basis I'm using only shows the m numbers of two particles because I've already summed the spins of the first two...

4. Jun 8, 2012

### CFede

The usual trick here is to write

$S_{1}\cdot S_{2}=\frac{1}{2}((S_1+S_2)^2-S_1^2-S_2^2)$

Those operators you DO know how they act.

$S_i^2\mid s,m\rangle = s(s+1)\mid s,m\rangle=\frac{3}{4}\mid s,m\rangle$

The same goes for $(S_1+S_2)^2$, since you now its possible values from Clebsch-Gordan coefficients

Last edited: Jun 8, 2012
5. Dec 6, 2012

### finitepenguin

Sorry to revive this thread but I am also trying to work out 3 spin 1/2 particle addition and this comment confused me. How can you find $(S_1+S_2)^2$ in the |S,m> basis using C.G. coefficients when S represents the total spin $S_1+S_2+S_3$?

6. Dec 6, 2012

### finitepenguin

I think I partly answered my own question: I guess you can treat $(S_1+S_2) + S_3$ as an addition of two spin systems? If that is the case then $(S_1+S_2)^2|S,m>=h^2(S_1+S_2)(S_1+S_2+1)|S,m>$. But then how do we treat $S_1$ and $S_2$? And I don't seem to need any C.G. coefficients using this method?

Last edited: Dec 6, 2012
7. Dec 6, 2012

### TSny

You can plow through it by using the 8 "product states" as basis states:

$|+++\rangle\equiv |\uparrow\rangle_1|\uparrow\rangle_2|\uparrow \rangle_3$

$|-++\rangle\equiv |\downarrow\rangle_1|\uparrow\rangle_2|\uparrow \rangle_3$

etc. where $|\uparrow\rangle_1$ is spin up along z for particle 1.

Find the matrix representation of H in this basis and then find the eigenvalues of H. It's not too tedius. Most matrix elements are zero and H will be in block diagonal form. Not very elegant, but it works.

8. Dec 6, 2012

### finitepenguin

I agree that is probably the most straight forward way of going about this, I wanted to try to avoid explicit construction of the Hamiltonian in matrix form though with the hopes of possibly being able to generalize the process. Any hints on the route I was working on the previous post?

Thanks!