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Physics
Quantum Physics
Quantum numbers of Landau levels
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[QUOTE="vanhees71, post: 5619934, member: 260864"] It's much easier to solve the energy-eigen problem in the gauge, where ##\vec{A}## depends only on one spatial component, e.g., $$\vec{A}=-By \vec{e}_x,$$ because of the higher symmetry of the corresponding Hamiltonian. In this gauge the Hamiltonian reads $$\hat{H}=\frac{1}{2m} [(-\mathrm{i} \partial_x+q By)^2-\partial_y^2-\partial_z^2],$$ i.e., you can seek for common eigenvectors of ##\hat{p}_x##, ##\hat{p}_z## and ##\hat{H}##. This leads to the ansatz $$u_{E,p_x,p_z}(\vec{x})=C \exp(\mathrm{i} p_x x+\mathrm{i} p_y y) Y(y),$$ leading to an effective one-dimennsional harmonic-oscillator energy-eigenvalue problem for ##Y(y)##. Of course, the energy eigenvalues are gauge invariant, and the wave function transforms with ##\psi'(t,\vec{x})=\exp[\pm \mathrm{i} q\chi(\vec{x})] \psi(t,\vec{x})## while ##\vec{A}'=\vec{A} + \vec{\nabla} \chi##. I'm a bit unsure about the sign in the phase factor. Just figure it out by direct calculation. :wink: [/QUOTE]
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Quantum numbers of Landau levels
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