- #1

phi1123

- 9

- 0

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

where

**A**is the vector potential such that

**B**=

**∇**×

**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**= cy

**i**+dx

**j**such 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 of

**A**seem to make the particle unbound in the x direction?