I was wondering about what the wavefunction of a particle in a magnetic field would look like, so after some quick work and a little research, I found the the Hamiltonian is(adsbygoogle = window.adsbygoogle || []).push({});

[itex]\hat{H}[/itex] = [itex]\frac{(\hat{p}+qA)^{2}}{2m}[/itex]

whereAis the vector potential such thatB=∇×A. I thought that the vector potential was a useful fiction, like the scalar potential for the electric field, as it is not unique (adding the gradient of a scalar field to the vector potential does not affect the magnetic field, for instance). However, does its presence as an operator on ψ not mean that it is in fact an observable?

I am further confused by the specific case of a constant magnetic field. Taking the field to be B along the z axis, it can be easily verified that the corresponding vector potential would be

A= cyi+dxjsuch that d-c = B

The choice of c and d would appear to be arbitrary (at least classically), however they seem to affect the wavefunction. Furthermore, taking c=0, d=B, the Schrodinger equation turns into the Hamiltonian of a free particle in the x direction, and the a simple harmonic oscillator in the y direction. This is not at all what I would have expected, since classically, the particle would be in simple harmonic motion in both direction (circular motion). Why does this apparently arbitrary choice ofAseem to make the particle unbound in the x direction?

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# Problems with a Charged Particle in a Magnetic Field

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