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QM: 1D Potential Well Spring - Energy Levels

  1. Nov 27, 2017 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    1D Potential V(x) = mw^2x^2/2, part of a harmonic oscillator.
    Suppose that the spring can only be stretched, so that the potential becomes V=infinity for x<0. What are the energy levels of this system?

    2. Relevant equations


    3. The attempt at a solution

    I argued my way though this problem by the following:
    We know that V(x) = infinity
    V(0) = 0
    V(x) = 0 otherwise

    From our typical energy levels we know E_n = ħw(n+1/2) for n=0,1,2,3,...

    But there is a barrier at x =0. Therefore we need x=0 to have E=0.
    Energy levels are thus:
    E_n = ħw(n+1/2) with n=1,3,5,7,...

    One can see this though the wave function graphs: https://i.stack.imgur.com/rb340.gif

    Is that argued properly? Did I find the right solution?
     
  2. jcsd
  3. Nov 27, 2017 #2

    BvU

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    What are the boundary conditions at ##x=0## ?
     
  4. Nov 27, 2017 #3

    RJLiberator

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    I thought it was V(0) = 0. Did I interpret this incorrectly? Since the spring can only be stretched and not compressed?
     
  5. Nov 27, 2017 #4

    BvU

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    I mean the conditions imposed on the solution . You want to make a distinction between the independent variable ##x## and the solution ##\ \psi(x) \ ## that has to satisfy a second order equation (##\ {\mathcal H}\psi = E\psi\ ## in your case) plus two boundary conditions.
     
  6. Nov 27, 2017 #5

    BvU

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    I don't want to confuse you, though: your line of thinking is correct; it's just the wording that is unfortunate:
    The solution for x < 0 is ##\ \Psi(x) = 0\ ## and at ##\ x=0 \ ## the boundary condition is that the ##\ \Psi \ ## has to be continuous.

    The first derivative does not have to be continuous at ##\ x=0 \ ## because the potential function isn't continuous either (I silently hope a theoretician will improve on this somewhat).

    At the turning point (Where ##\ V(x) = E\ ##) the situation is different and both ##\ \Psi \ ## and its first derivative have to be continuous.
     
  7. Dec 4, 2017 #6

    RJLiberator

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    @BvU sorry for not getting back to you earlier in this thread -- but your help was spot on. It was a much more basic question then I assumed and I got full credit for it, partially thanks to your guidance.
    Cheers.
     
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