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A Annihilation creation operators in case of half H.oscillator

  1. Sep 12, 2016 #1
    I'm trying to check if Ehrenfest theorem is satisfied for this wave function,
    |Y>=(1/sqrt(2))*(|1>+|3>),
    where |1> and |3> are the ground and 1st exited state wave functions of a half harmonic oscillator. When I'm calculating the expectation values of x and p using annihilation creation operators and trying to check the Ehrenfest theorem I'm getting
    m*(time derivative of <x>)=2*<p>
    which violates the Ehrenfest theorem. So, my question is whether annihilation creation operators are valid in case of a half harmonic oscillator.
     
  2. jcsd
  3. Sep 12, 2016 #2

    Demystifier

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    If ##a## and ##a^{\dagger}## are destruction and creation operators for the full harmonic oscillator, then the destruction and creation operators for the half harmonic oscillator are
    $$A=aa$$
    $$A^{\dagger}=a^{\dagger}a^{\dagger}$$
    The operators ##a## and ##a^{\dagger}## are not defined (do not exist) in the space of states of the half harmonic oscillator.
     
  4. Sep 12, 2016 #3

    A. Neumaier

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    What is this?
     
  5. Sep 12, 2016 #4

    BvU

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    Come on ! you can google it too !
     
  6. Sep 12, 2016 #5

    A. Neumaier

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    There are two natural boundary conditions one can place at ##x=0##. It is odd that http://www.physicspages.com/2012/08/18/half-harmonic-oscillator/ only describes the odd half with Dirichlet boundary conditions; the even half with Neumann boundary conditions is not even mentioned.
     
    Last edited: Sep 12, 2016
  7. Sep 13, 2016 #6
    Thanks a lot.
     
  8. Sep 13, 2016 #7

    Demystifier

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    This is the standard practice in QM (see e.g. a textbook treatment of infinite well potential) to require that ##\psi## (not ##\psi'##) vanishes at the points where the potential ##V## is infinite. The textbooks also give a physical reason for that. I has to do with the fact energy should be finite and that Schrodinger equation has ##\psi## (and ##\psi''##) but not ##\psi'##.

    More generally, Neumann boundary conditions in physics usually correspond to ends that can oscillate freely. It should be intuitively clear that ends at infinite ##V## are far from being "free". Think, e.g., of a guitar string.
     
    Last edited: Sep 13, 2016
  9. Sep 13, 2016 #8
    Sorry, but I'm still getting the same $$2<p>$$result when I replaced ##a## and ##a^{\dagger}## by ##A## and ##A^{\dagger}## in expression for x and p and did the calculations. Why this "2" is coming in the expression? Is Ehrenfest theorem also not defined in this case?
     
  10. Sep 13, 2016 #9

    Demystifier

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    I did not say that such a replacement is legitimate. You should not express ##x## and ##p## in terms of creation/destruction operators. Ehrenfest theorem is still valid, but it is more subtle.
     
  11. Sep 14, 2016 #10
    So, I have to take the Hermite polynomials and calculate expectation values in integration method? Or, is there any other easier method to do this? So that I can verify Ehrenfest theorem.
     
  12. Sep 15, 2016 #11

    Demystifier

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    The approach with the Heisenberg picture should be the easiest. The direct calculation with Hermite polynomials should work too, but make sure to first renormalize them because now they are defined only for x>0.
     
  13. Sep 15, 2016 #12
    I'll try that. Thanks again.
     
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