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Free particle on a ring

  1. Jan 27, 2015 #1
    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.
     
  2. jcsd
  3. Jan 28, 2015 #2

    Simon Bridge

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    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: Jan 29, 2015
  4. Jan 28, 2015 #3

    DrDu

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    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.
     
  5. Jan 28, 2015 #4
    hmm I understand. So what goes wrong in my calculation, where I get an energy dependent on α?
     
  6. Jan 28, 2015 #5

    DrDu

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    Your phase factor is definitively dependent on angle ##\phi##, as ##\alpha(0)\neq \alpha(2\pi)##.
     
  7. Jan 28, 2015 #6
    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?
     
  8. Jan 29, 2015 #7

    Simon Bridge

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    Was the phase factor you added a constant phase?
     
  9. Jan 29, 2015 #8

    DrDu

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    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.
     
  10. Jan 29, 2015 #9

    ShayanJ

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    Last edited by a moderator: May 7, 2017
  11. Jan 29, 2015 #10

    vanhees71

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    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: May 7, 2017
  12. Jan 30, 2015 #11

    Simon Bridge

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    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.
     
  13. Feb 4, 2015 #12
    sorry yes. I thought it was a constant phase? What is a constant phase if not the one i added?
     
  14. Feb 5, 2015 #13

    Simon Bridge

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    Look back at post #2: for the phase to be constant, you have to add it to the x+L wavefunction as well as the x wavefunction. These are actually the same wavefunction evaluated at two different positions. Otherwise the added phase is different for x and x+L: i.e. not constant.
     
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