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Momentum Operator Proof

  1. Apr 21, 2008 #1
    Does anyone have a momentum operator proof? The book I'm using is skipping a lot of steps .
     
  2. jcsd
  3. Apr 21, 2008 #2

    siddharth

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    Could you be more specific? Proof of what?
     
  4. Apr 21, 2008 #3
    sorry i re-read my post and realized it was really vague. The momentum operator used to determine the momentum of a particle described by schrodingers wave equation.
     
  5. Apr 21, 2008 #4

    siddharth

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    Yes, what about it? I don't understand what proof you have in mind. Maybe, if you post the exact problem statement, and where you are stuck, it'll be easier to help.
     
  6. Apr 22, 2008 #5

    malawi_glenn

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    I am also stunned, are you saying that you want a proof that

    [tex] -i\hbar \nabla [/tex]

    gives the momentum of a state?

    Be specific, and write down for us where you get stucked.
     
  7. Apr 22, 2008 #6

    JK423

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    There is no *proof*, but there is a "classical" way to calculate it. But still it`s not a *proof*.
    Assume that this relation for the mean values is valid:
    <P>=m*d<x>/dt.
    Then substitute <x>=(Ψ,xΨ), do the math and you`ll get the well known result:
    <P>=(Ψ,pΨ), where p is the momentum`s operator.
     
  8. Apr 23, 2008 #7

    malawi_glenn

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    No JK423, that is not what you do. You want to show that the p operator is the generator of translations and that it fulfills certain commutation relations.
     
  9. Apr 23, 2008 #8

    JK423

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    I mentioned a way to calculate momentum`s operator in {x} representation from a known classical equation. What`s wrong about that?
    What do you mean by
    "You want to show that the p operator is the generator of translations and that it fulfills certain commutation relations" ?
    You can show these these things without knowing what operator p equals to?
     
  10. Apr 23, 2008 #9

    malawi_glenn

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    Well what you wrote is ok :-)
    but is not a proof, as you said.



    but I know what p equals, so I can proove that it is the generator of translation and fulfills certain commutator relations. I did not say that i was about to derive the momentum operator, which is done by doing the steps backwards. I.e I demand translations symmetry and that it fulfills certain commutator relations.

    It is quite simple.
     
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