Undergrad How to diagonalize Hamiltonian with Zeeman field

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The discussion focuses on diagonalizing the Hamiltonian of an electron in a Zeeman field, which couples the electron's spin to the magnetic field. The Hamiltonian is set up by combining the kinetic and potential energy terms with the Zeeman term, but there is uncertainty about solving it due to its matrix structure and spatial components. The approach suggested involves using a unitary transformation to align the magnetic field along the z-axis, simplifying the Hamiltonian to a diagonal form with only the sigma_z component remaining. While this method diagonalizes the interaction Hamiltonian in spin space, it does not fully diagonalize the entire Hamiltonian unless the magnetic field is constant. The discussion highlights the complexity of finding exact analytical expressions for eigenenergies in such systems.
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Recently I have been asked to solve the problem of an electron in a Zeeman-field that couples the spin of the electron to the magnetic field.
I am not sure how to correctly set up the problem. I think, however, that what I have done on the picture is correct. The usual p^2/2m + V term in the Hamiltonian is tensored with the identity matrix and the Zeeman term is added in the usual way.
I am however unsure how to solve this Hamiltonian. How do you solve something like this with both an eigenvalue equation in the matrix structure and in the spatial part of H?
 

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This is usually solved perturbatively. Most QM textbooks have a chapter on time-independent perturbation theory.
 
Okay but is it not possible to find the exact analytical expression for the eigenenergies?
 
Easy. Just find a unitary transformation which rotates B in the direction of the z axis. Then the hamiltonian will be diagonal as only the sigma_z component remains.
 
DrDu said:
Easy. Just find a unitary transformation which rotates B in the direction of the z axis. Then the hamiltonian will be diagonal as only the sigma_z component remains.
This diagonalizes the interaction Hamiltonian, but not the full Hamiltonian.
 
If B is constant, it does, at least in spin space.
 
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Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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