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Old Quantum Theory = Oscillations?

  1. Feb 8, 2009 #1
    Hello again,

    I am having this problem understanding Quantum Theory, when it comes to calculations and applications. I'll give you an example.

    When using the Bohr Sommerfeld quantization rule we use calculus in...what? Is it an oscillating particle?? That's what I can't seem to understand. Or (this is from some solved exercises I was working on before) when using the WKBJ semiclassical approximation, after putting momentum under the integral, I noticed that the solution had a number multiplying the integral, either 2 or 4 (sometimes there wasn't any). So I have concluded that the WKBJ integral is multiplied by the number of times we want that specific part of the full oscillation of the particle we are studying. Is that really it? I hope you can understand what I am talking about! I also think that it all has to do about the potential Energy U(x), or even something about a potential well.

    I really need some guidance over the matter. I would really appreciate it if you could guide me!
  2. jcsd
  3. Feb 8, 2009 #2
    what potentials are you looking at?
  4. Feb 9, 2009 #3
    Well actually even this little question sounds a bit tricky to me.

    You see in this book of mine, there are exercises that talk about Potentials V(x), that I think are mistaken for U(x) Potential energies. This is another one of my queries.

    As for the potentials the particle moves in, there are different ones like [itex]V(x)=1/2 mw^2x_o ^2[/itex] or [itex]V(x)=b|x|[/itex].

    The integrals for each of the cases mentioned are as follows:

    1st: [itex]\int_{-x_o}^{x_0}dx=(n+1/2)\pi * h-bar[/itex]

    and the 2nd Integral: [itex]2\int_{0}^{x_1}dx=(n+1/2)\pi * h-bar[/itex]

    I just want to ask, what is the meaning of the multiplier in front of the integral in an oscillation level. Does it mean, as I said, parts of one full oscillation? Because I know that xo and -x0 are the limits of one oscillation. Or am I wrong?

    (BTW: If this thread goes to a deeper more Homework level, feel free to move it admins)
  5. Feb 9, 2009 #4
    Your integrals look a bit funny, since you're not integrating anything. Try to be a bit more specific in what the integral really is (also, the latex code for h-bar is \hbar).

    Anyways, I think that the factor in front is related to the symmetry of the function. If the function is symmetric, then an integral from [itex]-x_0[/itex] to [itex]x_0[/itex] is the same as 2 times the integral from [itex]0[/itex] to [itex]x_0[/itex]. If the wavefunction is antisymmetric, then it would automatically zero. Your potentials are indeed symmetric, so the energy eigenstates you encounter are either symmetric or antisymmetric. Hence the simplification.
  6. Feb 9, 2009 #5
  7. Feb 9, 2009 #6
    To xepma: Yes I'm sorry for the funny integrals, I thought that by saying WKBJ or Wilsony-Bohr-Sommerfeld quantization you would understand what I am talking about!

    To The Dagda: Thanks very much for the link, so dumm of me to jump the search-in-physics-forums step! Well that last thing for the (I think it's called Contour) Integral and the separate parts are really useful.

    So one last thing to conclude a series of questions : The oscillations described (from what I've understood are oscillations) are Simple Harmonic Oscillations with [tex]x_0[/tex] and [tex]-x_0[/tex] edges? Or are they circular orbits around let's say a nucleus?

    The Potentials I've described are actually Potential energy? From what I've understood they're the potential energy when the particle stands in [tex]x_max = x_0 [/tex], right where Kinetic energy is zero, or something like that. Please excuse my potential misunderstanding since it is somewhat difficult for someone my age not to confuse oscillations combined with quantum mechanics!

    (PS. You guys are really helpful! One would think you're secretly paid for this :P)
  8. Feb 10, 2009 #7
    According to that link they are elliptical orbits, they also reflect the kinetic energy of the system too in discrete quanta. Kinetic energy is only 0 at t=0.
    Last edited: Feb 10, 2009
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