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Harmonic Oscillator

  1. Mar 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Using the normalization constant A and the value of a, evaluate the probability to find an oscillator in the ground state beyond the classical turning points ±x0. Assume an electron bound to an atomic-sized region (x0 = 0.1 nm) with an effective force constant of 1.0 eV/nm2.

    2. Relevant equations

    [itex]\psi(x)=Ae^{-ax^{2}}[/itex], where [itex]A=(\frac{m\kappa}{\pi^{2}\hbar^{2}})^{1/8}[/itex] and [itex]a=\sqrt{m\kappa}/2\hbar[/itex]

    3. The attempt at a solution

    How do I find m?
     
    Last edited: Mar 5, 2012
  2. jcsd
  3. Mar 5, 2012 #2

    vela

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    The mass of the electron? You look it up.
     
  4. Mar 5, 2012 #3

    tiny-tim

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    or you could weigh one …

    if you have one on you o:)
     
  5. Mar 5, 2012 #4
    I will try to read the question more carefully next time.
     
  6. Mar 5, 2012 #5
    Am I supposed to find probabilities at both tails of the distribution?
     
  7. Mar 5, 2012 #6

    vela

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    Yes.
     
  8. Mar 7, 2012 #7
    Apparently there is an inconsistency in the question and there are two interpretations of the question. What would be the other one?
     
  9. Mar 8, 2012 #8

    vela

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    Beats me.
     
  10. Mar 10, 2012 #9
    My interpretation: find probabilities at both tails of the distribution i.e. [itex]\int^{0}_{-\infty}\psi^{2}(x)\;dx[/itex] and [itex]1-\int^{0.1}_{-\infty}\psi^{2}(x)\;dx[/itex].
     
  11. Mar 11, 2012 #10

    vela

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    I'd say the problem is asking for one number, P(|x|>x0). The upper limit on your first integral is wrong.
     
  12. Mar 19, 2012 #11
    Do you mean it is wrong according to your interpretation or it is generally wrong? Should it be 0.1?
     
    Last edited: Mar 19, 2012
  13. Mar 20, 2012 #12

    vela

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    It is generally wrong. Why would the upper limit be 0?
     
  14. Mar 20, 2012 #13
    It should be -0.1, right?
     
    Last edited: Mar 20, 2012
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