Solving an S=3/2 Dimer: A Dilemma and Bonus Question

  • Thread starter assyrian_77
  • Start date
In summary: I see, they only cover spin 1/2. :(In summary, Daniel is explaining to the user how to diagonalize an S-matrix to get the energy eigenvalues and eigenvectors for a particular dimer form. He also explains that for a dimer with an S-value of 3/2, there are 3 generators of angular momentum. He also mentions that for spin 1, there are 3 generators as well.
  • #1
assyrian_77
115
0
I've been trying to work on this for a while:

Let us say I have an [tex]S=1/2[/tex] dimer with [tex]H=JS_{1}\cdot S_{2}[/tex]. With a [tex]\hat{z}[/tex]-diagonal basis, [tex]|\uparrow\uparrow\rangle[/tex], [tex]|\uparrow\downarrow\rangle[/tex], [tex]|\downarrow\uparrow\rangle[/tex], [tex]|\downarrow\downarrow\rangle[/tex], 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 [tex]S=3/2[/tex] dimer (same form on H)? What [tex]\hat{z}[/tex]-diagonal basis (if any) can I use? And am I right in assuming that the matrices to use are the [tex]4\times4[/tex]-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 [tex]S=1[/tex]. What happens?


*I feel a bit silly for not knowing this*
 
Physics news on Phys.org
  • #2
About the basis, you can use the [tex]|00\rangle, |01\rangle, |02\rangle,|03\rangle, |10\rangle,|11\rangle[/tex]... basis and matrices are 4*4 don't know what though, also you shouldn't feel silly for having problems with this problem. I think most people settle for understanding the spin 1/2 problem and then referring a book when the spin gets higher.
 
Last edited:
  • #3
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:[itex] |j,m\rangle [/itex] which spans the irreducible space [itex] \mathcal{E}_{j} [/itex].
 
  • #4
Let me get this straight...you are suggesting that I use the the total spin S (and m) as basis, i.e.

[itex]|3\pm3\rangle, |3\pm2\rangle[/itex] and so on all the way to [itex]|00\rangle[/itex]

Right?

But what if would want to start from scratch so to say, i.e. just apply the Hamiltonian to the [itex]|m_{1}m_{2}\rangle[/itex] basis. For the [itex]S=1/2[/itex], I could just apply the [itex]S[/itex]-operators to the kets or the [itex]S[/itex]-matrices to the columns. How would I do something similar in the [itex]S=3/2[/itex] case? I guess my question is how to write the [itex]|m_{1}m_{2}\rangle[/itex] 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
assyrian_77 said:
Let me get this straight...you are suggesting that I use the the total spin S (and m) as basis, i.e.

[itex]|3\pm3\rangle, |3\pm2\rangle[/itex] and so on all the way to [itex]|00\rangle[/itex]

Right?

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

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 [itex] |m_{1},m_{2}\rangle [/itex] basis...?There's no such thing. :uhh:


Daniel.
 
  • #6
For the [tex]S=1/2[/tex] case I can let the [tex]H=J(S_1^xS_2^x+S_1^yS_2^y+S_1^zS_2^z)[/tex] operate on [tex]|\uparrow\uparrow\rangle=|\frac{1}{2}\frac{1}{2}\rangle[/tex] etc. by:

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

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

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

*Thanks for your help*
 
  • #7
Hold on.

[tex]\hat{S}_{x}|\uparrow\rangle \neq \frac{1}{2}|\downarrow\rangle [/tex]

Do you see why?

Daniel.
 
  • #8
No, I don't. Apart from the fact that I didn't include hbar.
 
  • #9
Oh,u used [itex] \hbar=1 [/itex].Sorry.

Daniel.
 
  • #10
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
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
Hmm,Rose or Edmonds books on Angular Momentum in QM should have them.

Daniel.
 
  • #13
Aargh! I feel stupid... :frown: the weekend is closing in, I guess...

I have the [itex]S=3/2[/itex] 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


[itex]|\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)[/itex]

...right? :confused:
 
  • #14
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. :smile:
 
  • #15
I wish i could ignore it,but in fact the standard basis is

[itex]\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)[/itex]

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

Daniel.
 
  • #16
I know, that is the basis I used. I meant ignore the fact that I thought this wasn't the basis. I was wrong. :smile:

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

[itex]H=\frac{1}{2}J(S_{tot}(S_{tot}+1)-S_1(S_1+1)-S_2(S_2+1))[/itex]

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

However, doing it this way, how can I find out the degeneracies?
 
  • #17
How did u get that Hamiltonian...?

Daniel.
 
  • #18
Ok, we have [itex]H=J\bar{S}_1\cdot\bar{S}_2[/itex].

However, we know that [itex]\bar{S}_{tot}=\bar{S}_1+\bar{S}_2[/itex].

Squaring both sides: [itex]\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[/itex].

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

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

[tex]E=\frac{1}{2}J(S_{tot}(S_{tot}+1)-S_1(S_1+1)-S_2(S_2+1))[/tex]


where I used [itex]S^2|\Psi\rangle=S(S+1)|\Psi\rangle[/itex] (with [itex]\hbar=1[/itex]).

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

1. What is an S=3/2 dimer?

An S=3/2 dimer refers to a molecule or system composed of two atoms with a total spin of S=3/2. Spin is a quantum mechanical property of particles and can have values of 0, 1/2, 1, 3/2, etc.

2. What is the dilemma in solving an S=3/2 dimer?

The dilemma in solving an S=3/2 dimer arises from the fact that this system has four possible states (magnetic sublevels) instead of the usual two states found in a spin 1/2 system. This makes the calculations more complex and difficult to solve.

3. How is an S=3/2 dimer typically solved?

An S=3/2 dimer is typically solved using quantum mechanical methods such as perturbation theory or the variational method. These methods involve making approximations and solving equations to determine the energy levels and states of the dimer.

4. What is the significance of solving an S=3/2 dimer?

Solving an S=3/2 dimer has practical applications in fields such as chemistry, physics, and materials science. It can help us understand the behavior of molecules and materials with multiple atoms and how they interact with each other.

5. What is the bonus question related to solving an S=3/2 dimer?

The bonus question related to solving an S=3/2 dimer is often a more challenging problem that requires additional calculations or considerations. It may involve factors such as external magnetic fields, spin-orbit coupling, or other complexities in the system.

Similar threads

Replies
2
Views
987
  • Quantum Physics
Replies
1
Views
818
Replies
9
Views
1K
  • Quantum Physics
Replies
2
Views
681
  • Quantum Physics
Replies
3
Views
7K
  • Quantum Physics
Replies
21
Views
2K
  • Quantum Physics
Replies
6
Views
2K
  • Quantum Physics
Replies
17
Views
2K
Replies
1
Views
518
Replies
4
Views
3K
Back
Top