Diagonalizing Spin Hamiltonian

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SUMMARY

The discussion focuses on diagonalizing the spin Hamiltonian represented by the equation H = A S_z + B S_x, where A and B are constants. The eigenvalues of the corresponding matrix [A B; B -A] are determined to be ±sqrt(A^2 + B^2). The discussion further details the substitution of cos(theta) = A/sqrt(A^2 + B^2) and sin(theta) = B/sqrt(A^2 + B^2) to derive the components of the eigenvector, establishing a relationship between the components through the tangent of half the angle theta.

PREREQUISITES
  • Understanding of quantum mechanics, specifically spin operators.
  • Familiarity with eigenvalue problems in linear algebra.
  • Knowledge of matrix representation of Hamiltonians.
  • Basic trigonometric identities and their application in quantum mechanics.
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  • Study the derivation of eigenvalues and eigenvectors for 2x2 matrices.
  • Explore the application of spin Hamiltonians in quantum mechanics.
  • Learn about the role of trigonometric functions in quantum state transformations.
  • Investigate numerical methods for solving Hamiltonian systems in quantum physics.
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Quantum physicists, students of quantum mechanics, and researchers working on spin systems or Hamiltonian dynamics will benefit from this discussion.

antibrane
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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.
 
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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.
 
Thanks, I think I've figured it out now.
 

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