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Quantum pendulum

  1. Oct 18, 2014 #1
    1. The problem statement, all variables and given/known data
    A particle is in the ground state of a pendulum of length ##L##. ##L## suddenly increases to ##4L##. Find the probability that the particle will be in the ground state of the new pendulum.

    2. Relevant equations


    3. The attempt at a solution

    What's this question even asking? So the particle has energy ##E=\frac{1}{2} \hbar \omega##, where ##\omega=\sqrt{\frac{g}{L}}##, then ##E \rightarrow \frac{1}{2} E##. Isn't it still going to be in the ground state? I don't get what the problem is trying to ask.
     
  2. jcsd
  3. Oct 19, 2014 #2

    BvU

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    Note that E of the particle stays the same. But it's no longer the energy of the ground state of the 4L pendulum. That means that the wave function of the particle has components of higher energy states of the 4L.
     
  4. Oct 19, 2014 #3
    Okay, that makes more sense. The energy of the particle is the same, but how do I get a wave function when transforming? I might be grossly misunderstanding, but if we're given an energy level we don't necessarily know the particle's wave function, right?
     
  5. Oct 19, 2014 #4

    vela

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    I don't think you can assume E stays the same. What you can assume is that the wave function immediately before and after the transition are the same.
     
  6. Oct 19, 2014 #5
    So the probability will be 0 then, correct? Because if the wave function remains the same, and it's the wave function for the ground state of the L pendulum, which is higher than the ground state of the 4L pendulum.

    Feels wrong.
     
  7. Oct 19, 2014 #6
    Or wait, I would simply take the inner product of the two wave functions, right?
     
  8. Oct 19, 2014 #7

    vela

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    Because it is wrong. ;)

    Right.
     
  9. Oct 20, 2014 #8

    BvU

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    I have a hard time getting my head around this one. Math wise I could do the work and re-calculate E, but a) too lazy and b) prefer to think of the physics instead.

    I may have jumped on ##E \rightarrow \frac{1}{2} E## and claimed that E stays the same a little too quickly, but on the other hand: the full wave function ##|\psi(x,t)>## and its first derivatives are supposed to be continuous. And the time derivative has E in it. An argument to expect E stays the same ?

    Now the exercise has the particle in a well specified state ##|\psi_{E_0}(x,t)> = \phi_{E_0} \; \exp (-i\hbar E_0/t) ##. Potential function changes from ##{1\over 2} {mg\over L} x^2## to ##{1\over 8} {mg\over L} x^2##, giving the impression that, indeed, E changes ?

    By now, John must have 'simply' done the ##\ <\phi_{E_0, 4L} | \phi_{E_0, L} >\ ## inner product (answer ?), but I wonder what happened to the difference between ##\ E_{0, L}\ ## and ##\ \sum_{i = 0}^\infty \ <\phi_{E_{2i}, 4L} | \phi_{E_0, L} > ##

    Perhaps Vela can shed some light on this ? All I can come up with is that time derivative doesn't have to be continuous under such drastic instantaneous changes.
     
  10. Oct 20, 2014 #9

    mfb

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    We change the potential so fast the wavefunction does not change significantly during the adjustment. Afterwards, the time-derivative is different and E changes. Actually, it is not in an energy eigenstate any more, and I think even <E> changes as we change the potential.
     
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