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Quantum Pendulum question

  1. Jul 11, 2014 #1
    Hi guys,
    I've been studying the problem of the simple, one-dimensional quantum mechanical pendulum of length [itex] \ell [/itex] and mass [itex] m [/itex]. We first apply the small-angle approximation which of course reduces the problem to the simple harmonic oscillator. This part is easy enough.

    However, we then are asked to use perturbation theory to solve for the first order energy correction due to the invalidity of the small-angle approximation. My confusion is this: a true eigenfunction on a ring must have periodic boundary conditions imposed. However, the small-angle approximation in this problem yields eigenfunctions which certainly aren't periodic on the circle. I guess this is okay, after all, we're looking only at small angles. However, perturbation theory requires integrals over the *entire* configuration space of the coordinate. So in my case, do I integrate from 0 to 2pi, even though the un-perturbed eigenfunctions aren't really eigenfunctions over this space? I'm tempted to integrate over the real line instead, but this also doesn't sound quite okay.
     
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  3. Jul 12, 2014 #2

    Simon Bridge

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    What is the configuration space of the coordinate?
     
  4. Jul 12, 2014 #3
    well, I just mean that for the familiar 1d oscillator, when computing expectation values, say, you integrate over all of [itex] \mathbb{R} [/itex] whereas for a particle on a ring, you integrate from 0 to 2pi. So which do I use in my case? Because, the approximate un-perturbed eigenfunctions certainly aren't well-defined, periodic functions on the circle.
     
  5. Jul 12, 2014 #4

    Simon Bridge

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    Walk through it one step at a time...
    So what are you integrating?
    How are the wavefunctions defined exactly?
    What is the operation you need to do?
     
  6. Jul 12, 2014 #5
    Ok so when you write out the full Hamiltonian for the quantum pendulum you get something like:

    [tex] H=\frac{-\hbar^{2}}{2m \ell^{2}} \frac{d^{2}}{d \phi^{2}}-mg \cos \phi [/tex]

    which we can then apply the small-angle approximation to get:

    [tex] H \psi = \frac{-\hbar^{2}}{2m \ell^{2}} \frac{d^{2}\psi}{d \phi^{2}}+\frac{1}{2} m g \ell \phi^{2} \psi = E \psi [/tex]

    This is just the simple harmonic oscillator with energies [itex] E_{n}= \hbar \sqrt{\frac{g}{\ell}} (n +1/2) [/itex] and the ground state eigenfunction is the familiar oscillator eigenfunction in the variable [itex] \ell \phi [/itex]:

    [tex] u_{0}( \ell \phi)=\bigg(\frac{m}{\hbar \pi}\sqrt{\frac{g}{\ell}}\bigg)^{1/4}e^{-\frac{m}{2\hbar}\sqrt{\frac{g}{\ell}}(\ell \phi)^{2}}[/tex]

    So my confusion is that there is not a chance this is an acceptable eigenfunction on a ring because it is not periodic. This is okay since we're only looking in the small-angle regime, but what happens when you need to do a computation that involves inserting a complete set of states? For example, what if we think of the next highest term in the cosine expansion as a small perturbation? Then we would have a small perturbing Hamiltonian of the form:

    [tex] H_{1}=-\frac{m g \ell}{24} \phi^{4} [/tex]

    Then, the first order energy correction to the ground state would entail a matrix element of the form:

    [tex] E_{0}^{(1)}= \langle 0 | H_{1} | 0 \rangle [/tex]

    But then evaluating this would involve inserting two complete sets of states in the coordinate [itex] \phi [/itex] (or [itex] \ell \phi [/itex]??). It is tempting to write:

    [tex] I=\int_{0}^{2 \pi} d \phi |\phi \rangle \langle \phi | [/tex]

    but this doesn't seem right since our eigenfunctions from the small-angle approximation certainly aren't complete on the circe. What should I do with computations like this?

    Thanks in advance!
     
  7. Jul 12, 2014 #6

    mfb

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    The perturbation you use is a valid approach only if the considered states correspond to small angles - where a correction might be useful, but you can still neglect angles larger than pi/2. Just integrate up to this (or pi if you like, or infinity, should all give the same result), for large angles your potential is completely wrong and won't matter anyway.
     
  8. Jul 12, 2014 #7

    Simon Bridge

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    One of the things that can happen at this level is that you get to a point in a calculation where you cannot see where it is going to end up and you kind-of recoil at that.

    You have to learn to keep going - this level of physics is getting more and more like exploring a dense jungle: you can easily feel lost, it happens a lot to everyone. Keep going. This is usually OK as long as you are careful about what the math you've done so far is saying... which is what mfb was clearing up for you.

    Later on you will find yourself starting calculations where you have little or no idea where you'll end up or even if the approach you have chosen is valid. That is normal - enjoy it. You have to keep going until you get some sort of outcome or you won't learn anything.
     
  9. Jul 13, 2014 #8

    Vanadium 50

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    Let's sweep away some brush.

    Consider two systems - a ball that is sliding (but not rolling) in a parabolic bowl, and a pendulum. Classically, at the bottom, for small amplitudes, the ball and the bob move together. They have the same potentials, and that means they have the same equations of motion.

    Now, when I quantize these systems, again, they have the same potentials and the same equations of motion, so therefore have the same wavefunctions. You are right to notice that the integration bounds differ in these two systems. The correct conclusion is that it doesn't matter whether I integrate from -pi/2 to pi/2, or -pi to pi, or -infinity to infinity: I will get essentially the same answer. (Classically, this is a statement that the potential where the test particle isn't makes no difference; quantum mechanically this is a statement that when the wavefunction is negligible the effects of the wavefunction are as well.)

    So, which one do you choose? The one that makes your calculation easiest. In this case, it will be integrating from -infinity to infinity. Now, the smart way to do this is not to use the actual integral at all. Think "complete sets of states", "orthogonal wavefunctions" and "raising and lowering operators" and you should be able to answer this without any explicit integration.
     
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