Eigenstates, Charged Particle/Bead on a Loop in B-Field

[logic smogic]

Problem
A one dimensional particle of mass m and charge q moves on a circle of radius R. A magnetic field pierces the circle. The total magnetic flux through the circle is

\Phi = B \pi R^{2}.

Determine the ground state energy E_{0} (\Phi) and show that it is periodic with period \Phi_{0} = \frac{h c}{q}, called the magnetic flux quantum.

Attempt at Solution
One can use

B = \nabla \times A

and

\Phi = \int B \cdot da = \oint A \cdot dl

to evaluate the vector potential A on the circle. Using a gauge in which A is rotationally symmetric and points in the azimuthal direction,

A = \frac{B R}{2} \hat{\phi} = \frac{\Phi}{2 \pi R} \hat{\phi}

where R is the fixed radius of the loop. The time-independent Hamiltonian for a charged particle in a magnetic field is

H = \frac{(p - \frac{q}{c} A)^{2} }{2m}

Since the motion is 1-dimensional on the circle, the only degree of freedom is x = R \phi, with conjugate momentum,

p = -\imath \hbar \frac{\partial}{\partial x} = - \imath \hbar \frac{1}{R} \frac{\partial}{\partial \phi}

Inserting this and the Hamiltonian into the time-independent Schrodinger Equation yields:

\frac{1}{2m} ( -\frac{\imath \hbar}{R} \frac{\partial}{\partial \phi} - \frac{q}{c} \frac{\Phi}{2 \pi R} )^{2} \psi (\phi) = E \psi (\phi)

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Question: Noting that this resembles a quantum harmonic oscillator, and that \psi(2 \pi) = \psi(0), how do I find an expression for the eigenstates \psi_{n} (\phi)?

 

[Avodyne]

If you can find an eigenstate of p - \frac{q}{c} A, it will automatically be an eigenstate of H.

So, first, disregarding the condition that \psi(2 \pi) = \psi(0), what are the eigenstates of p - \frac{q}{c} A?

Then, what are they if you take into account this boundary condition?

 

[logic smogic]

I'm not sure how to find eigenstates of p - \frac{q}{c} A other than to make the canonical substitution for momentum, and then solve the second order, linear, homogenous differential equation.

Solving this equation is very messy (although there is some minor cancelling), and after attempting it very carefully, I arrived at this:

\psi( \phi )_{+} = \displaystyle{e^{ -\frac{\imath q \Phi + \sqrt{8 m E (\pi c R)^{2} - 2 (q \Phi)^{2} }}{2 \pi c \hbar} \phi }}

where I have not taken into consideration boundary conditions yet. Still, this seems a bit too muddled to be correct. I'll check the algebra again, but it would be more enlightening if there were way to 'see' the solution in the time-independent Schrodinger Equation listed above.---
Ah, but can we use the 'periodicity' condition of \psi (2 \pi) = \psi (0) to guess the form of the wavefunction (minus a normalization constant, of course).

From the differential equation solution, for some complex K, the form will be:

\psi (\phi) = e^{K \phi}

For the periodicity condition to be satisfied, K must be \imath n, for n an integer. So,

\psi (\phi) = e^{\imath n \phi}

Is that sensible?

 

[javierR]

Maybe it's better to use a gauge of the vector potential A in cartesian coordinates x and y (which is easiest to determine from \nabla\times A =B where B is in the z direction). You have been hinted to make the Hamiltonian look like a modified harmonic oscillator, which has the Hamiltonian like p^2/2m+ x^2 with constant sprinkled around...so clearly we should expand the Hamiltonian H~(p-eA)dot(p-eA)/2m as a start.
* The next key is to see that H can be split into parts
H~H(xy)+H(z)...what is the significance of the H(z) part seperating off like that?...and what does that tell you about the form of the wavefunction w.r.t. z? And the energy contribution from the z-motion of the particle?
* As for the H(xy) part, it should look like the harmonic oscillator H with extra terms that can be identified as r\times \nabla in the z direction; i.e. the angular momentum operator Lz.
* The wavefunctions will be the Hermite polynomials with additional factors from the Hz part and angular momentum (which we are ignorant of). The energy spectrum is therefore the harmonic oscillator part labeled by n values, an energy contribution from the particle's z-degree of freedom, and some contribution from the angular "l" quantum number.
*However, for the ground state, the ang momentum quantum number is...? Therefore, the energy of the ground state of the entire system is...?