Homework Help: Eigevalues and eigevectors of an Hamiltonian

1. Jul 1, 2011

19matthew89

1. The problem statement, all variables and given/known data
Hi,
I must find eigenvalues and eigenvector of this Hamiltonian, which describes a system of two 1/2-spin particles.

H = A($S_{1z}$ - $S_{2z}$) + B($S_{1}$ · $S_{2}$)

where $S_{1}$ and $S_{2}$ are the two spins, $S_{1z}$ and $S_{2z}$ are their z-components, and A and B are constants.

2. Relevant equations

Since it's easy finding eigenvalues and eigevectors of ($S_{1}$· $S_{2}$) in total spin base (I can write it as (0.5)($S^{2}$-${S_{1}}^{2}$-${S_{2}}^{2}$) I've thought I should use this base, but $S_{1z}$ and $S_{2z}$ do not commute.

3. The attempt at a solution

I've tried to solve this problem in this way. I know you can write this Hamiltonian using the basis {|++>; |+->; |-+>; |-->} and, if I know how H acts on this basis I can write a matrix which represents the action of H. So I computed how H acts using the basis {|++>; |+->; |-+>; |-->} to estimate the action of A($S_{1z}$ - $S_{2z}$) and the basis {|11>; |10>; |1-1>; |00>} to estimate the action of B($S_{1}$ · $S_{2}$) and the I've correlated the two bases using Clebcsh-Gordan coefficients. Is this way of proceeding correct?
At the end I've found a matrix in {|++>; |+->; |-+>; |-->} which, obviously, is not diagonal. How can now find eigenvalues and eigenvector? Diagonalizing? Is there a faster way to get a diagonal matrix?

Thanks

Last edited: Jul 1, 2011
2. Jul 1, 2011

vela

Staff Emeritus
Why do you say S1z and S2z don't commute?

What matrix did you end up with? It seems to me it should reduce down to a problem of diagonalizing a 2x2 matrix.

3. Jul 1, 2011

19matthew89

Because in the basis of {$S^{2}$, $S_{z}$, ${S_{1}}^{2}$, $S_{2}^{2}$} the operator $S_{1z}$ and $S_{2z}$ do not commute with the other elements of the basis. At the end I've found a 4x4 matrix (each basis is made of 4 elemnts). In the basis {|++>; |+->; |-+>, |-->} the matrix is the one shown in pdf file (I don't know how to insert a matrix eheheh).

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4. Jul 1, 2011

vela

Staff Emeritus
Your terminology is all messed up.
1. The operators aren't elements of the basis. The simultaneous eigenstates of the operators are the basis, i.e. $\{\vert 1 1 \rangle, \vert 1 0 \rangle, \vert 1 -1 \rangle, \vert 0 0 \rangle \}$ is one of the bases you're referring to.
2. It doesn't make sense to say the operators don't commute with elements of a basis. Operators commute or don't commute with other operators.
3. "S1z and S2z don't commute" means [S1z,S2z]≠0, which isn't true. S1z and S2z do commute.
4. What you meant was that S1z and S2z don't commute with S2. As a consequence, the first term of the Hamiltonian isn't represented by a diagonal matrix in the basis above.
Now on to your matrix. (You can right-click on the matrix and choose "Show Source" to see the LaTeX markup.)
\begin{pmatrix}
B/4 & 0 & 0 & 0 \\
0 & -B/4+A & B/2 & 0 \\
0 & B/2 & -B/4-A & 0 \\
0 & 0 & 0 & B/4
\end{pmatrix}
[strike]Can you show how you calculated the middle block? The B terms I get differ from yours by a factor of $\sqrt{2}$.[/strike]

EDIT: I think your matrix is fine. I missed a factor of $\sqrt{2}$ somewhere.

You should be able to see what two of the eigenvalues and corresponding eigenvectors are by inspection. You just need to diagonalize the submatrix to find the remaining two.

Last edited: Jul 1, 2011
5. Jul 2, 2011

19matthew89

Thank you so much!
Yes, you're right, I'm sorry! I should have written the eigenstates as basis while the list I've put down is a Complete Set of Commuting Observabels to which the basis

$\{\vert 1 1 \rangle, \vert 1 0 \rangle, \vert 1 -1 \rangle, \vert 0 0 \rangle \}$

corresponds.

Thanks for clarifying my terminology, too! I'm very grateful to you for letting me know that my line of reasoning, despite my lexicon, is correct. (I'll go and compute again the matrix to find the missing factor of $\sqrt{2}$).

Bye

P.S. I thank you also for the hint how to post a matrix

6. Jul 2, 2011

vela

Staff Emeritus
I think your matrix is correct, though I guess it can't hurt to recheck your work.