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Problems with a Charged Particle in a Magnetic Field

  1. Jan 9, 2013 #1
    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

    [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 = cyi+dxj 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?
  2. jcsd
  3. Jan 9, 2013 #2
    The vector potential is not an observable, for the reason you mention: you could make it whatever you want just by a gauge transformation.

    consider [itex]\vec{A}=B_{0}y\hat{x}\implies \vec{B}=-B_{0}\hat{z}[/itex] this is a legit vector potential.
    \hat{H}=\frac{1}{2m}\left( p^2 + q\vec{A}\cdot\vec{p} + q\vec{p}\cdot\vec{A} + q^2 A^2 \right)=\frac{1}{2} \left( \hat{p}^2 + 2q B_{0}\hat{y} \, \hat{p}_{x} + q^2 B_{0}^{2}\hat{y}^2 \right)
    where the hats denote operators at the end, not unit vectors, and [itex][\hat{y},\hat{p}_{x}]=0[/itex].
    Last edited: Jan 9, 2013
  4. Jan 10, 2013 #3
    In classical E&M we tend to take the view that E and B are fundamental, and A is a useful derived quantity. In quantum mechanics, for the reason you've found, we tend to take the view that A is fundamental and E and B are useful derived quantities. For example, see the Aharanov-Bohm effect, where a nonzero A causes quantum mechanical effects even when E and B are zero in the entire region of the experiment: http://en.wikipedia.org/wiki/Aharanov-Bohm_effect

    That said, A itself is not an observable, because you can make it whatever you want by a gauge transformation. Only quantities that are unchanged by gauge transformations are observable. These include, for example, closed line integrals of the A field, which are the quantities measured in the Aharanov-Bohm effect.

    I think you have to be careful about interpreting the wave functions here. First, the energy eigenstates you find may be superpositions of many degenerate eigenstates. So if you find delocalized eigenstates, that doesn't necessarily mean that there isn't a different basis of localized eigenstates that look like the classical orbiting electrons. Second, the physical momentum operator is no longer the gradient operator, so a plane wave does not necessarily represent a particle travelling unimpeded in a certain direction.
  5. Jan 10, 2013 #4


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    This is one of the most fascinating topics. Gauge invariance rules a lot of elementary particle physics, and electromagnetism is of course just the most simple case of such a theory. A very nice paper about gauge invariance, the AB effect, etc. is

    Wu, T. T., and Yang, C. N. Concept of nonintegrable phase factors and global formulation of gauge fields. Phys. Rev. D 12 (1975), 3845.
  6. Jan 11, 2013 #5
    In all cases,gauge invariance is preserved i.e. it will be only line integral of A(over closed loop) which will be observable and not the A.
  7. Apr 3, 2013 #6
    Okay, so I guess i'm still confused by the whole choice of Gauge thing. You say:

    Does this mean that there is a degenerate energy level for each choice of Gauge? If not, how does the same state correspond to several different wavefunctions?
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