# Time-Dependent Degenerate Pertubation Theory for Spin System

1. Apr 7, 2013

### jcharles513

1. The problem statement, all variables and given/known data
Consider the so-called Spin Hamiltonian
H=AS2Z+B(S2X-S2Y)

for a system of spin 1. Show that the Hamiltonian in the SZ basis is the 3x3 matrix:

\hbar2*[(A,0,B; 0,0,0; B,0,A)].

Find the eigenvalues using degenerate pertubation theory.
2. Relevant equations
Spin Pauli Matrices

3. The attempt at a solution

I don't know where to start with this since the Pauli Matrices I know are 2x2 matrices. How did the Hamiltonian become a 3x3. What am I missing? Can someone get me started?

2. Apr 7, 2013

### TSny

See here where $\hbar$ has been set equal to 1.

3. Apr 7, 2013

### jcharles513

Thank you. That makes sense. I guess I wasn't searching for the right thing in google. I may have more questions after that but this gets me started.

4. Apr 7, 2013

### jcharles513

Degen. Perturb.

I'm confused about how to use degenerate perturbation on this system. My understanding of degenerate perturbation is shaky. I know that I need to take a subspace of this matrix that has the degeneracy and then transform the perturbation of that subspace to this new diagonal basis. But I'm not sure what my subspace of this matrix is?

5. Apr 7, 2013

### jcharles513

I used the 3x3 subspace and got eigenvalues 0,A+B, and A-B. Is this correct? And the correct procedure?

6. Apr 7, 2013

### TSny

I believe those are the correct eigenvalues for the Hamiltonian. Did you get them by finding the exact eigenvalues of the complete Hamiltonian?

I think the exercise is to find eigenvalues using degenerate perturbation theory where you treat B as a small parameter so that the unperturbed Hamiltonian is Ho=ASz2 and the perturbation is V = B(Sx2 - Sy2).

7. Apr 7, 2013

### jcharles513

I diagonalized the pertubation (B-part) and then added it the A-part. This made it diagonal. I'm not sure if that's the right way to do it or not? If not, can you get me started on the correct method?

8. Apr 7, 2013

### TSny

In degenerate perturbation theory, you would not diagonalize the B-part in the full 3-dimensional vector space of the hamiltonian. Rather, if you want the perturbation theory correction to the degenerate eigenvalue, A, of the unperturbed Hamiltonian (A-part hamiltonian) you diagonalize the B-part relative to the 2-dimensional space spanned by the degenerate eigenfunctions corresponding to the eigenvalue A of the unperturbed hamiltonian, as explained in standard texts.

In this problem, it turns out that degenerate perturbation theory yields the exact result, and diagonalizing the B-part with respect to the degenerate subspace is equivalent to diagonalizing the entire hamiltonian. But in general that wouldn't be true.

Unless you still have specific questions, I would recommend reviewing degenerate perturbation theory and then coming back with specific questions if you still have any. This is not the place to try to explain the general formalism of degenerate perturbation theory.

Last edited: Apr 7, 2013