Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.3.8.
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As another example of the harmonic oscillator, we’ll look at a charged particle moving in a magnetic field. The field is given in terms of the magnetic vector potential
The field is
If the particle is confined to the plane and the magnetic field provides the only force, the force is given by the Lorentz force law
Since is always perpendicular to the direction of motion , the magnetic force does no work, so the kinetic energy and hence the speed of the particle is constant. Classically, the particle is thus confined to move in a circle with providing the centripetal force, so we have
where is the charge, is the mass and is the radius of the circle. The period of the orbit is
which gives an angular frequency of
This is the result in SI units; Shankar uses Gaussian units, in which the magnetic field picks up a factor of , so in Shankar’s notation, this is
As the rest of the problem relies on Gaussian units, we’ll stick to them from now on.
Classically, the Hamiltonian for the electromagnetic force is
where is the electric potential, which is zero here. Thus using 1, we have for the quantum version in which and the position vector are replaced by operators
We can perform a canonical transformation by defining
We can verify that these coordinates are canonical by checking their commutator:
Thus and have the correct commutator for a pair of position and momentum variables.
Rewriting 12 in terms of and , we have
Thus has the same form as that for a one-dimensional harmonic oscillator with frequency , so the energy levels of this system must be
We can expand 12 in terms of the original position and momentum variables to get
where is the hamiltonian for a 2-dim harmonic oscillator with frequency . As we saw when solving that system, the Hamiltonian for the isotropic oscillator commutes with since the potential is radially symmetric, thus the eigenfunctions of are also eigenfunctions of . In terms of the present problem, this means that the eigenfunctions of are also eigenfunctions of and thus also eigenfunctions of . In our solution of the 2-dim isotropic oscillator, we found that the energy levels are given by
where and is the angular momentum (in units of ). Thus for the oscillator with Hamiltonian , the energy levels are
The energy levels of the original are therefore, from 24
[Shankar says the can be ‘any integer’, but from our original derivation of 25, we found that is a non-negative integer.] Equation 29 gives the same energies as 21, since if , we get , while if we have . Both and give the same sequence of values as . [I’m not quite sure the two methods are equivalent, though, since 21, being the solution of a one-dimensional system is non-degenerate, while 29, being a two-dimensional system does have degenerate energy levels.]