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Diagonalizing Spin Hamiltonian |
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| Jul11-10, 03:29 AM | #1 |
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Diagonalizing Spin Hamiltonian
How would one find the eigenstates/values for the following Hamiltonian?
[tex] H=A S_z + B S_x [/tex] where [itex]A,B[/itex] are just constants. Any help is appreciated. Thanks. |
| Jul11-10, 07:56 AM | #2 |
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I don't know how to use latex that well, so I'll try to give you the general idea as to how I got the solution.
The matrix for which we need to find the eigenvalues is [A B B -A] the eigenvalues come out to be sqrt(A^2 + B^2), with both negative and positive values of the square-root. Next, use the substitution cos(theta) = A/[sqrt(A^2 + B^2) and sin(theta) = B/sqrt(A^2 + B^2), while solving the equations for the components of the eigenvector for either value of the eigenvalue. You will have the second component of the eigenvector related to the first component multiplied by the tan of half the angle theta, with plus/minus sign for the corresponding sign of the eigenvalue. |
| Jul12-10, 12:51 PM | #3 |
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Thanks, I think I've figured it out now.
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