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Harmonic potential energy well

  1. Jun 11, 2009 #1
    1. A particle in a harmonic potential energy well is in a state described by the initial wave function

    Ψ(x, 0) = (1/√2) (ψ1(x)+ iψ3(x))

    where ψ1(x)and ψ3(x) are real normalized energy eigenfunctions of the harmonic oscillator with quantum numbers n =1 and n = 3 respectively.

    (a)
    Write down an expression for Ψ(x, t) that is valid for all t> 0. Express your answer in terms of ψ1(x), ψ3(x)and ω0, the classical angular frequency of the oscillator.

    (b)
    Find an expression for the probability density function at any time t> 0. Express your answer in terms of ψ1(x), ψ3(x), ω0 and t.Use the symmetry of this function to show that the expectation value,<x> = 0 at all times.

    2. Relevant equations



    3. The attempt at a solution


    I have reached Ψ(x, t) = (1/√2) (ψ1(x)exp(-3iw[tex]_{}0[/tex]t/2+ iψ3(x)exp(-7iw[tex]_{}0[/tex]t/2)


    for part (b) I'm not sure how to calculate the probability density function at any time t>0 ?

    I know that the probability density needs to be even function of x and so therefore being symmetrical about the centre of the well at all times.

    Then using the sandwich integral to calculate <x>, it will yield zero as the integrand is an odd function.
     
  2. jcsd
  3. Jun 11, 2009 #2

    Cyosis

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    What is the definition of the probability density function?
     
  4. Jun 11, 2009 #3
    it's just the |Ψ|^2 is it not? I don't know how to do the math
     
  5. Jun 11, 2009 #4

    Cyosis

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    That's correct, do you now that [itex]\psi^* \psi=|\psi|^2[/itex], with [itex]\psi^*[/itex] the conjugated wave function?
     
  6. Jun 11, 2009 #5
    so the conjugate of Ψ(x, t) is (1/√2)(ψ1(x)exp(+3iwt/2) + iψ3(x)exp(7iwt/2)) ?
     
  7. Jun 11, 2009 #6
    The exponentials disappear and then |Ψ|^2 = (1/2)(ψ1(x) - ψ3(x)).

    Am I on the right tracks here?
     
  8. Jun 11, 2009 #7

    Cyosis

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    That doesn't look correct to me. Write out the whole expression and show your steps, start with showing what you got for[itex]\psi^*[/itex].
     
  9. Jun 11, 2009 #8
    i get [tex]\Psi[/tex][tex]\ast[/tex]=(1/√2)(ψ1(x)exp(+3iwt/2) + iψ3(x)exp(7iwt/2)) for
     
  10. Jun 11, 2009 #9

    Cyosis

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    That is not correct. Note that there is an i in front of [itex]\psi_3[/itex].
     
  11. Jun 11, 2009 #10
    [tex]\Psi[/tex][tex]\ast[/tex]=(1/√2)(ψ1(x)exp(+3iwt/2) - iψ3(x)exp(7iwt/2))
     
  12. Jun 11, 2009 #11

    Cyosis

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    Looking good. Now multiply and be careful when working out the brackets.
     
  13. Jun 12, 2009 #12
    Right so I should have

    |[tex]\Psi[/tex]|^2 = 1/2 |[tex]\psi[/tex]1|^2 + |[tex]\psi[/tex]3|^2

    + 2[tex]\psi[/tex]1[tex]\psi[/tex]3sin(2wot)
     
  14. Jun 12, 2009 #13

    Cyosis

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    Almost correct, don't forget that the entire expression is multiplied by 1/2 not just the first term.

    Now write down the expression for <x>.
     
  15. Jun 12, 2009 #14
    that should have been

    |[tex]\Psi[/tex]|^2 = [1/2][|[tex]\psi[/tex]1|^2 + |[tex]\psi[/tex]3|^2

    + 2[tex]\psi[/tex]1[tex]\psi[/tex]3sin(2wot)]
     
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