# Spin 3/2

1. May 19, 2005

### assyrian_77

I've been trying to work on this for a while:

Let us say I have an $$S=1/2$$ dimer with $$H=JS_{1}\cdot S_{2}$$. With a $$\hat{z}$$-diagonal basis, $$|\uparrow\uparrow\rangle$$, $$|\uparrow\downarrow\rangle$$, $$|\downarrow\uparrow\rangle$$, $$|\downarrow\downarrow\rangle$$, I can easily construct the H-matrix by either using the Pauli matrices or the S-operators. Diagonalizing the matrix gives me the energy eigenvalues and the eigenvectors. Although I can get the energies in an easier way.

My problem/dilemma/question is this: What if I have an $$S=3/2$$ dimer (same form on H)? What $$\hat{z}$$-diagonal basis (if any) can I use? And am I right in assuming that the matrices to use are the $$4\times4$$-matrices listed in e.g. Schiff: Quantum Mechanics (1968), page 203? (Don't feel like typing them right now)

And a bonus-question: Assuming now $$S=1$$. What happens?

*I feel a bit silly for not knowing this*

2. May 20, 2005

### werty

About the basis, you can use the $$|00\rangle, |01\rangle, |02\rangle,|03\rangle, |10\rangle,|11\rangle$$... basis and matrices are 4*4 dont know what though, also you shouldnt feel silly for having problems with this problem. I think most people settle for understanding the spin 1/2 problem and then refering a book when the spin gets higher.

Last edited: May 20, 2005
3. May 20, 2005

### dextercioby

U can use the 4*4 matrices without any problem for 3/2 spin.I don't have Schiff's 1968 book (i got the incomplete 1949 one),but angular momentum is described in zilllion of books,even special books on angular momentum in QM.

And for spin 1,there are 3 generators which are 3*3 matrices.

Daniel.

The basis is the standard basis:$|j,m\rangle$ which spans the irreducible space $\mathcal{E}_{j}$.

4. May 20, 2005

### assyrian_77

Let me get this straight...you are suggesting that I use the the total spin S (and m) as basis, i.e.

$|3\pm3\rangle, |3\pm2\rangle$ and so on all the way to $|00\rangle$

Right?

But what if would want to start from scratch so to say, i.e. just apply the Hamiltonian to the $|m_{1}m_{2}\rangle$ basis. For the $S=1/2$, I could just apply the $S$-operators to the kets or the $S$-matrices to the columns. How would I do something similar in the $S=3/2$ case? I guess my question is how to write the $|m_{1}m_{2}\rangle$ in this case as columns (i.e. in matrix form).

I guess it is not really necessary to do it this way. I mean, I can always look in a table of Clebsch-Gordan coefficients, but I would like to know.

Btw, thanks for the help earlier.

5. May 20, 2005

### dextercioby

So you're talking about composing 2 3/2 spins.Use the C-G theorem to get the irreducible spaces and then the C-G formula to find the vectors in the basis.

What $|m_{1},m_{2}\rangle$ basis...?There's no such thing. :uhh:

Daniel.

6. May 20, 2005

### assyrian_77

For the $$S=1/2$$ case I can let the $$H=J(S_1^xS_2^x+S_1^yS_2^y+S_1^zS_2^z)$$ operate on $$|\uparrow\uparrow\rangle=|\frac{1}{2}\frac{1}{2}\rangle$$ etc. by:

1) either using the operations $$S^x|\uparrow\rangle=\frac{1}{2}|\downarrow\rangle$$ etc.

2) or by writing $$S^x, S^y, S^z$$ as Pauli matrices and multiply with $$|\uparrow\rangle=\left(\begin{array}{cc}1\\0\end{array}\right)$$ and $$|\downarrow\rangle=\left(\begin{array}{cc}0\\1\end{array}\right)$$.

So I am wondering: for the $$S=3/2$$ case, is there an analogous way of doing this?

7. May 20, 2005

### dextercioby

Hold on.

$$\hat{S}_{x}|\uparrow\rangle \neq \frac{1}{2}|\downarrow\rangle$$

Do you see why?

Daniel.

8. May 20, 2005

### assyrian_77

No, I don't. Apart from the fact that I didn't include hbar.

9. May 20, 2005

### dextercioby

Oh,u used $\hbar=1$.Sorry.

Daniel.

10. May 20, 2005

### dextercioby

Yes of course.Find the spin matrices for S=3/2 (i guess you have them in Schiff) and the basis vectors (which will be columns of 4 entries).

Daniel.

11. May 20, 2005

### assyrian_77

Ok, thanks. I'll try to find the basis vectors.

Another question: Do you know of a table of 6j-symbols (I will later go over to work on a trimer)? I haven't been able to find one.

12. May 20, 2005

### dextercioby

Hmm,Rose or Edmonds books on Angular Momentum in QM should have them.

Daniel.

13. May 20, 2005

### assyrian_77

Aargh! I feel stupid.... the weekend is closing in, I guess......

I have the $S=3/2$ matrices, but I don't know the basis vectors (the column matrices), and I can't find them. I am assuming they are not as simple as

$|\frac{3}{2}\rangle=\left(\begin{array}{cc}1\\0\\0\\0\end{array}\right), |\frac{1}{2}\rangle=\left(\begin{array}{cc}0\\1\\0\\0\end{array}\right), |-\frac{1}{2}\rangle=\left(\begin{array}{cc}0\\0\\1\\0\end{array}\right), |-\frac{3}{2}\rangle=\left(\begin{array}{cc}0\\0\\0\\1\end{array}\right)$

...right?

14. May 20, 2005

### assyrian_77

Ok, ignore my last post...I used the basis vectors from above, and after some tedious work, I got the correct energy eigenvalues and eigenvectors. Now off to the trimer.

15. May 20, 2005

### dextercioby

I wish i could ignore it,but in fact the standard basis is

$\left|\frac{3}{2},\frac{3}{2}\right\rangle=\left(\begin{array}{cc}1\\0\\0\\0\end{array}\right), \left|\frac{3}{2},\frac{1}{2}\right\rangle=\left(\begin{array}{cc}0\\1\\0\\0\end{array}\right), \left|\frac{3}{2},-\frac{1}{2}\right\rangle=\left(\begin{array}{cc}0\\0\\1\\0\end{array}\right), \left|\frac{3}{2},-\frac{3}{2}\right\rangle=\left(\begin{array}{cc}0\\0\\0\\1\end{array}\right)$

Actually u don't need the exact form of the basis vectors and neither the operators,but that's another story.

Daniel.

16. May 20, 2005

### assyrian_77

I know, that is the basis I used. I meant ignore the fact that I thought this wasn't the basis. I was wrong.

There is of course a simpler way of getting the energies. Simply by using

$H=\frac{1}{2}J(S_{tot}(S_{tot}+1)-S_1(S_1+1)-S_2(S_2+1))$

where $S_1=S_2=\frac{3}{2}$ and $S_{tot}$ takes the values 3,2,1, and 0.

However, doing it this way, how can I find out the degeneracies?

17. May 20, 2005

### dextercioby

How did u get that Hamiltonian...?

Daniel.

18. May 21, 2005

### assyrian_77

Ok, we have $H=J\bar{S}_1\cdot\bar{S}_2$.

However, we know that $\bar{S}_{tot}=\bar{S}_1+\bar{S}_2$.

Squaring both sides: $\bar{S}_{tot}^2=(\bar{S}_1+\bar{S}_2)^2=\bar{S}_1^2+\bar{S}_2^2+2\bar{S}_1\cdot\bar{S}_2$.

Hence, the Hamiltonian can be written $H=\frac{1}{2}J(\bar{S}_{tot}^2-\bar{S}_1^2-\bar{S}_2^2)$.

Therefore, the energy eigenvalues (and not the Hamiltonian, sorry) are

$$E=\frac{1}{2}J(S_{tot}(S_{tot}+1)-S_1(S_1+1)-S_2(S_2+1))$$

where I used $S^2|\Psi\rangle=S(S+1)|\Psi\rangle$ (with $\hbar=1$).

Still, doing it like this, I don't know how I can find out the degeneracies.