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Half-harmonic oscillator potential

  1. Feb 21, 2005 #1
    I have to find the allowed energies of this potential:

    \frac1{2}m\omega^2x^2 & \text{for } x > 0\\
    \infty & \text{for } x < 0

    My suggestion is that all the odd-numbered energies (n = 1, 3, 5...) in the ordinary harmonic osc. potential are allowed since [tex]\psi(0) = 0[/tex] in the corresponding wave functions and this is consistent with the fact that [tex]\psi(x)[/tex] has to be 0 where the potential is infinite.

    In the assignment it says that it takes some careful thought to reach this result and it took me 10 seconds to figure this out. In other words; somethings is wrong :tongue2:

    PS. I am new to quantum mechanics so please don't use any obscure notation :wink:
  2. jcsd
  3. Feb 21, 2005 #2

    Tom Mattson

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    Ha Ha, no thing is wrong with your solution. You're just a smart cookie. :smile:
  4. Feb 21, 2005 #3
    Wow, that was fast! Thanks. :smile:
  5. Sep 25, 2006 #4
    question: how do you prove that odd N numbered solutions to the harmonic oscillator are odd? It's not necessary to solve this problem, but i'm trying to remember what my teacher said about it. I assume that you can prove that [tex]A_+ \psi_0 [/tex] is odd, while [tex]A_+^2 \psi_0 [/tex] is even.
  6. Sep 25, 2006 #5
    You can f.e. write

    [itex] \phi_n=d(a_+)^n \phi_0 [/itex], while [itex] \phi_0 = c e^{-(\frac{x}{2x_0})^2} [/itex].

    Knowing something like [itex] a_+ = (\frac{x}{x_0} - x_0 \frac{d}{dx}) [/itex] (not completely sure about this) you can see that [itex]a_+[/itex] adds or removes one x to/from every term every time you apply it, so the resulting function must be odd/even.

    edit: I remember I also did the "half-oscillator" this way, but my professor said that you also have to consider functions which are not normalized in the harmonic oszillator potential, and prove that there is no even function with [itex] \phi_n(0) = 0 [/itex].
    Last edited: Sep 25, 2006
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