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Commutators and wave function

  1. Aug 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Using the results of the previous problem, find [x,p2 ] and from that determine [x,p2 ][itex]\psi[/itex](x)

    2. Relevant equations

    The solution to the previous problem was [A,BC]=[A,B]C+B[A,C]

    3. The attempt at a solution

    As I'm suppose to use the results of the previous problem I think what I need to do is

    [x,pp]=[x,p]p+p[x,p]

    But that doesn't really seem right to me... shouldn't this lead me to an identity (it doesn't say but usually in a QM problem p is momentum and this would seem to be leading to a wave equation in momentum space). If I really am approaching this the correct way, then how should I treat [x,p2 ][itex]\psi[/itex](x)?
     
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  3. Aug 3, 2011 #2

    George Jones

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    What is [x,p]?
     
  4. Aug 3, 2011 #3

    George Jones

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    Yes, you are on the right track. I should have started my previous post with this.
     
  5. Aug 3, 2011 #4
    My first instinct is to answer [x,p]=-[p,x]...but I don't think that was what you were asking?
     
  6. Aug 3, 2011 #5

    George Jones

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    While this true, you need to evaluate [x,p] explicitly. The answer is something simple, and will have been covered earlier in your text/course.
     
  7. Aug 3, 2011 #6
    Am I correct in thinking that x is position and p momentum?
     
  8. Aug 3, 2011 #7
    The canonical commutation? [x,p]=i [itex]\hbar[/itex]
     
  9. Aug 3, 2011 #8

    George Jones

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    Yes; now, use this in the expression at the end of your first post.
     
  10. Aug 4, 2011 #9
    (i [itex]\hbar[/itex])p+p(i [itex]\hbar[/itex]) ?
     
  11. Aug 4, 2011 #10

    George Jones

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    Right, but remember that [itex]\left[ x , p \right] = i\hbar[/itex] really means [itex]\left[ x , p \right] = i\hbar I[/itex], where [itex]I[/itex] is the identity operator.
     
  12. Aug 4, 2011 #11
    So in the first part I then end up with a result of

    (i [itex]\hbar[/itex] I)p + p(i [itex]\hbar[/itex] I)

    Do the standard rules of algebra apply here allowing me to combine this?

    Then for the second part is my result just the combined result with the [itex]\psi[/itex](x)? (a complex wave equation in momentum space?)
     
  13. Aug 4, 2011 #12

    George Jones

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    Yes, [itex]i\hbar I[/itex] commutes with all operators, and [itex]IA = A[/itex] for all operators [itex]A[/itex].
    [itex]\psi \left( x \right)[/itex] is a complex wave function (not equation) in position space (since [itex]\psi[/itex] is a function of position [itex]x[/itex]). What is the position space representation of the momentum operator, i.e., [itex]p = ?[/itex] This also will have been covered earlier in your text/course.
     
  14. Aug 5, 2011 #13
    I think I'm starting to grasp this. It seems like the final solution is heading towards something similar to px = i [itex]\hbar[/itex] [itex]\partial[/itex]/[itex]\partial[/itex]x .
     
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