Free particle on a ring

Main Question or Discussion Point

So the free particle wave functions are of the type:

ψ(x) = Aexp(ikx) + Bexp(-ikx) (1)

In a problem I am doing I am supposed to find the energy levels for a particle which is sliding on a frictionless ring and the exercise says that to do so I should use the fact that

ψ(x+L)=ψ(x) (2)

BUT! Since the phase of the wave-function in QM carries no physical significance, shouldn't the most general treatment add a phase to (1) such that:

ψ(x+L)=exp(iα)ψ(x) (3) , where α is an arbitrary real number.

Unfortunately when I do so and solve for the energy levels of the system I don't get the same result as when I use (1). α modifies the energy levels which of course shouldn't happen if the phase carries no physical significance. So is (3) a wrong assumption and if so, why? Here's what I did btw, maybe I made an error somewhere. Using (3) on (1):

Aexp(ikx)exp(ikL) + Bexp(-ikx)exp(-ikL) = exp(iα)(Aexp(ikx) + Bexp(-ikx))

This holds particularly for x=0 and x=k/2π yielding:

Aexp(ikL) + Bexp(-ikL) = exp(iα)(A+B)

Aexp(ikL) - Bexp(-ikL) = exp(iα)(A-B)

Which gives:

2Aexp(ikL) = 2Aexp(iα)

So either A=0 or kL-α = 2πn

But the appearance of α in the last equation alters the possible values of k and hence the possible energy values for the system.

Related Quantum Physics News on Phys.org
Simon Bridge
Homework Helper
If we suppose that $\psi$ is an energy eigenfunction, then: $$\hat H e^{i\alpha}\psi(x) = e^{i\alpha}\hat H \psi(x) = E e^{i\alpha}\psi(x)$$ ... so alpha does not affect the energy levels.

Notice:
$\psi(x+L)$ is not a different wavefunction from $\psi(x)$

Since an arbitrary phase makes no difference, you can propose a new wavefunction $\psi^\prime(x) = e^{i\alpha}\psi(x)$ ... then the boundary condition becomes: $$\psi^\prime(x+L) = \psi^\prime(x)$$

Last edited by a moderator:
DrDu
Note that while a constant phase factor is indeed unimportant in QM, a position or angle dependent factor is not. Specifically,
changing $\psi(\phi)\to \exp(i\alpha(\phi))\psi(\phi)$ is a gauge transformation which has to be compensated by changing $L_z=-i\partial /\partial \phi$ to $L'_z=L_z-\partial \alpha/\partial \phi$. Writing $\alpha=\int A d\phi$ you can interpret this as the effect of a magnetic vector potential A due to a magnetic flux line in the center of your ring. So you rediscovered the Aharonov Bohm effect.

hmm I understand. So what goes wrong in my calculation, where I get an energy dependent on α?

DrDu
Your phase factor is definitively dependent on angle $\phi$, as $\alpha(0)\neq \alpha(2\pi)$.

I don't understand. I let the the periodic boundary condition care an arbitrary phase factor. But shouldn't that be allowed when the phase of the wave function carries no physical significance?

Simon Bridge
Homework Helper
Was the phase factor you added a constant phase?

DrDu
As Simon explained before: If the phase factor is constant, the new boundary conditions are $\exp(i\alpha) \psi(2\pi)=\exp(i\alpha)\psi(0)$. So you see that a constant phase factor won't change the original boundary conditions.

ShayanJ
Gold Member
Last edited by a moderator:
vanhees71
Gold Member
2019 Award
This is a very subtle issue. A long time ago, I thought about this in connection with a preprint on the arXiv, which however obviously never made it into a publication in a peer reviewed journal. Of course, there are no truely infinite potential wells and thus the problem is rather academic, but it's fun to think about the fundamental difference between Hermitean and essentially self-adjoint operators. Here are my thoughts on this problem in connection with thermodynamics:

http://fias.uni-frankfurt.de/~hees/tmp/box.pdf [Broken]

Last edited by a moderator:
Simon Bridge
Homework Helper
Hmmm ... perhaps OP does not see why the phase factor added was not of a constant phase?
But without feedback we cannot really tell for sure.

sorry yes. I thought it was a constant phase? What is a constant phase if not the one i added?

Simon Bridge