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Quantum operator problem

  1. Jan 12, 2008 #1
    1. The problem statement, all variables and given/known data

    I got stuck to this problem:

    To prove that (x p)^2 is not equal to (x)^2 (p)^2

    where x and p are position and mometum operator in QM.


    2. Relevant equations
    3. The attempt at a solution

    I approached this way:

    Two operators A and B are equal iff Af=Bf for all f

    So,here {(x p)^2 - (x)^2 (p)^2}f=0

    Since this is valid for all f, we must have {(x p)^2 - (x)^2 (p)^2}=0

    In other words, [(x p)^2 , (x)^2 (p)^2 ]=0

    But I could not disprove this commutator.Calculations are big and after some steps I doubt whether this might at all be the correct way.

    Can anyone please give some hint?
     
  2. jcsd
  3. Jan 12, 2008 #2

    malawi_glenn

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    (xp)^2 = xpxp

    if xpxp = xxpp, then what can we say about [x,p] = xp - px ?
     
  4. Jan 12, 2008 #3

    kdv

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    You can indeed do it with commutators (the identity [AB,CD] = A[B,C]D + etc would be useful) or here it might be simpler to work with explicit differential operators. Just replace the operator p by its usual expression and apply the two expressions to a test function. Obviously, X^2 p^2 f is simply [tex]- \hbar^2 x^2 \frac{d^2f}{dx^2} [/tex]. Calculating the other one won't be too hard and you will see if you get the same result (you won't).
     
  5. Jan 12, 2008 #4

    kdv

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    But xpxp is NOT equal to xxpp!
     
  6. Jan 12, 2008 #5

    kdv

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    Oh, I see your point. yes, that's a nice way to see right away that the two expressions can't be equal because they would imply that x and p commute. That's nice and quick indeed.

    Regards
     
  7. Jan 12, 2008 #6

    malawi_glenn

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    Now we solved the problem for the OP, which is not so good. So I propose we delete our latest posts :)

    EDIT: too late ;)

    I should have taken the discussion with you kdv via PM ;)
     
  8. Jan 12, 2008 #7
    I got glenn's point. Really, it is lot easier.

    I also tried in the way kdv indicated.But that did not work.This is really much more simple.
     
  9. Jan 13, 2008 #8

    malawi_glenn

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    Doing it the other wat should aslo work, remember that an operator operates on EVERYTHING to the right of it, for example

    [tex] px\psi (x)= p(x \psi (x) ) = \frac{d}{dx}(x \psi (x)) [/tex]

    where the last thing is evaluated according to the leibniz product rule for derivatives. Here i have omitted hbar and the imaginary unit from the p-operator just to make things clearer.
     
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