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Proof that the linear momentum operator is hermitian

  1. May 23, 2014 #1
    hello
    i have to proof that Px (linear momentum operator ) is hermitian or not
    i have added my solution in attachments

    please look at my solution and tell me if its correct


    thank you all
     

    Attached Files:

  2. jcsd
  3. May 23, 2014 #2

    Matterwave

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    There's a sign error in the last expression of the first page, but you "fixed" it in the next expression.

    This looks OK to me, but of course you should explain why ##\left.\psi_b^*\psi_a\right|_{-\infty}^\infty=0##.
     
  4. May 23, 2014 #3

    micromass

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    What is the domain of the linear momentum operator that you're looking at? You can't just look at arbirary wave functions, because then ##\psi_b^*\psi_a\vert_{-\infty}^{+\infty}## will not be zero. You need to restrict to special wave function which do have that property.

    Another solution is that you use "distributional differentiation".
     
  5. May 24, 2014 #4
    thank you for your answers
    if i take a special function which satisfy the property then is my solution ok?
     
  6. May 24, 2014 #5

    Matterwave

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    Yea, it looks OK to me.
     
  7. May 24, 2014 #6
    ok thank you very much
     
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