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I Solving for <E^2> of a non-stationary state of the QSHO

  1. Jul 14, 2017 #1
    I have the following non-stationary state of the QSHO:

    Ψ = [​IMG][ gif&s=25.gif gif&s=25.gif gif&s=62.gif e −3iωt/2 + gif&s=62.gif [ gif&s=49.gif ][​IMG]e −5iωt/2]

    where β = mω/ħ

    in calculating <E2>

    The answer I see in the textbook is 6.17 ħ2ω2.

    This answer suggests that in calculating <E2> = Ψ*Ĥ2Ψ dx

    where Ψ is composed of the above two terms Ψ1 and Ψ2 (a linear combination of E1 and E2 of the QSHO) the cross terms need to be zero to eliminate the time dependence as reflected in the answer in the text book.

    This leaves us with the inside and outside terms

    Ψ12Ψ1 dx + Ψ22Ψ2 dx and calculating the sum of these inside and outside terms producing the answer in the text book.

    However, as I calculated the cross terms Ψ12Ψ2 dx is zero as it is an odd function, but Ψ22Ψ1 dx is not zero as it is an even function. This would not eliminate the time dependence and would not result in a non oscillating result as reflected in the text book.

    Am I missing something here?

    Attached Files:

  2. jcsd
  3. Jul 14, 2017 #2
    Sorry part of Ψ did not load

    Ψ = upload_2017-7-13_22-17-2.png [ upload_2017-7-13_22-20-37.png βx upload_2017-7-13_22-21-56.png e −3iωt/2 + upload_2017-7-13_22-23-56.png [ upload_2017-7-13_22-23-37.png ] upload_2017-7-13_22-17-30.png e −5iωt/2]
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