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Commutation relation of the position and momentum operators

  1. Feb 1, 2007 #1
    1. The problem statement, all variables and given/known data

    I've just initiated a self-study on quantum mechanics and am in need of a little help.

    The position and momentum operators do not commute. According to my book which attemps to demonstrate this property,
    (1) [tex]\hat{p} \hat{x} \psi = \hat{p} x \psi = -i \hbar \frac{\partial}{\partial x}(x \psi)[/tex]

    OK, so far I was following. I'm not understaing how they get to this next equation from the previous one...
    (2) [tex] = -i \hbar \left( \psi + x \frac{\partial \psi}{\partial x}\right)[/tex]

    To be more precise, where did the second psi symbol come from and where did the + symbol come from in equation 2?
    2. Relevant equations

    [tex] \hat{x} = x [/tex]
    [tex] \hat{p} = -i \hbar \frac{\partial}{\partial x} [/tex]


    3. The attempt at a solution

    When I attempt to multiply the momentum and position operators, I get:
    (3) [tex] \hat{p} \hat{x} \psi = \hat{p} x \psi = \left( -i \hbar \frac{\partial}{\partial x} \right) x \psi [/tex]

    But how do I go from equation 3 and arrive at equation 2? Or can I?
     
    Last edited: Feb 1, 2007
  2. jcsd
  3. Feb 1, 2007 #2

    Meir Achuz

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    psi also depends on x. This leads to the second derivative.
    You may need a live teacher to learn QM.
     
  4. Feb 1, 2007 #3
    OK, I understand the first part, but I am not following why it "leads to the second derivative." Could you please explain in more detail? Or preferably, could you show me the next few steps after equation (1)?
     
  5. Feb 1, 2007 #4

    dextercioby

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    What book are you using ? Throw it away. One postulates the commutation relations, they cannot be proved.
     
  6. Feb 1, 2007 #5
    The author of the book is demonstrating that [tex] \hat{p} \hat{x} \psi does NOT equal \hat{x} \hat{p} \psi [/tex] .

    Sorry if I miscommunicated this point in the first post.
     
  7. Feb 1, 2007 #6
    I understand the solution to why [tex] \hat{x} \hat{p} \psi [/tex] equals what it equals. My question just has to do with how the author gets the solution that he does for the [tex] \hat{p} \hat{x} \psi [/tex]
     
  8. Feb 1, 2007 #7

    cristo

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    Are you asking how you get from the left to the right hand side of this? [tex]\left( -i \hbar \frac{\partial}{\partial x} \right) x \psi = -i \hbar \left( \psi + x \frac{\partial \psi}{\partial x}\right)[/tex].

    If so, then note that [itex]x\psi[/itex] is a product, each of which is dependent on x, and so we must use the product rule. So, starting from the left [tex]\left( -i \hbar \frac{\partial}{\partial x} \right) x \psi =-i\hbar\frac{\partial}{\partial x}(x\psi)=-i\hbar(x\frac{\partial\psi}{\partial x}+\psi)[/tex]
     
    Last edited: Feb 2, 2007
  9. Feb 1, 2007 #8

    nrqed

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    It's simply the product rule. If [itex] f (x) [/itex] is a function of x, then [tex] \frac{d}{dx} ( x f(x) ) = 1 \times f(x) + x \times \frac{df(x)}{dx} [/tex]
    , right?

    What book are you using? I would highly recommend Shankar and, as a supplement, Griffiths.
     
  10. Feb 2, 2007 #9
    Great. Thanks a LOT cristo and nrqed!

    I love this site! :)

    I am currently using two books: "Quantum Mechanics for Undergraduates" by Norbury as well as Griffiths (both books use the same template and complement each other strongly). I also have Shankar, however, after going through the first chapter, I found it was too math heavy... perhaps it is aimed at graduates?
     
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