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Operators in Quantum mechanics: can one swap \Psi and \Psi^*

  1. Sep 14, 2015 #1
    1. The problem statement, all variables and given/known data
    The demonstration for the momentum operator in Quantum Mechanics goes something like this

    <v>=\frac{d}{dt}<x>=\frac{d}{dt} \int x \Psi^* \Psi dx

    and then one ends up with

    <p>=m<v>=\int \Psi^* (-i \hbar \frac{d}{dx}) Psi dx

    however, if you swap the congugates you get
    <p>=m<v>=\int \Psi (i \hbar \frac{d}{dx}) Psi^* dx

    Can somebody confirm that this is true? When one works with operators the order of the \Psi and \Psi* is important?

    Thanks!



    2. Relevant equations


    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Sep 14, 2015 #2

    blue_leaf77

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    $$
    -i\hbar \int_{-\infty}^{\infty} \psi^* \frac{\partial \psi}{\partial x} dx = -i \hbar \left(|\psi|^2 \big|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \psi \frac{\partial \psi^*}{\partial x} dx \right)
    $$
    $$
    = i\hbar \int_{-\infty}^{\infty} \psi \frac{\partial \psi^*}{\partial x} dx
    $$
    where we have assumed that ##\psi## is square integrable. This is no surprise since momentum is supposed to be Hermitian.
     
  4. Sep 14, 2015 #3

    micromass

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    But why does ##\psi## vanish at ##+\infty## and ##-\infty##? Surely not every square-integrable function has that property?
     
  5. Sep 15, 2015 #4

    blue_leaf77

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    What might be an example of a square-integrable function which does not vanish at infinities?
     
  6. Sep 15, 2015 #5

    micromass

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    [tex]f(x) = x^2 \text{exp}(-x^8\sin^2(x))[/tex]
     
  7. Sep 15, 2015 #6

    blue_leaf77

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    Why does this function not vanish for x approaches infinity, I think by simple inspection it should? What am I missing here?

    EDIT: After seeing the post in http://math.stackexchange.com/quest...ction-tend-to-0-as-its-argument-tends-to-infi, I come to agree with you. It's the infinitely zero width but finite height that may make a function square-integrable although it does not necessarily vanish at infinities. Thanks anyway.
     
    Last edited: Sep 15, 2015
  8. Sep 15, 2015 #7

    micromass

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    If ##x=\pi n ## for some integer ##n##, then ##\sin^2(x) = 0##. Hence for these ##x##, we have ##f(x) = x^2=\pi^2 n^2##. This does not converge to ##0## if ##n## gets large. In fact, it becomes unbounded.
     
    Last edited: Sep 15, 2015
  9. Sep 15, 2015 #8

    micromass

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  10. Sep 15, 2015 #9

    blue_leaf77

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    CORRECTION to post #2:
    The requirement "where we have assumed that ##\psi## is square integrable" should be narrowed down to ##\psi## which goes to zero when ##|x|## goes to infinity.
     
  11. Sep 16, 2015 #10
    Thank you all for your useful answers.

    $$\Psi$$
     
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