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Commutator Problem

  1. Oct 17, 2006 #1
    Question:
    If g(p) can be Taylor expanded in polynomials, then prove that:

    [tex] \left[x, g\left(p\right)\right] = i\hbar \frac{dg}{dp} [/tex]

    To start, I multiply the wave function [itex] \Psi [/itex] and expand the commutator:

    [tex] \left( xg\left(p\right)-g\left(p\right)x \right)\Psi [/tex]

    then expand g(p) using Taylor:

    [tex] x\left(g\left(a\right) + g'\left(a\right)\left(p-a\right)\right)\Psi + \left(g\left(a\right) + g'\left(a\right)\left(p-a\right)\right)x\Psi [/tex]

    Then I'm stuck... I'm nowhere near getting [itex]i\hbar[/itex] or [itex]\frac{dg}{dp}[/itex]

    Any help would be appreciated. Thank you!
    -Rick
     
  2. jcsd
  3. Oct 17, 2006 #2

    StatusX

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    Prove that's true for g(p)=p^n. Then write out the taylor series of g and apply the commutator to each term.
     
  4. Oct 17, 2006 #3
    That was my previous question... I already proved that

    [tex] \left[x, p^{n}\right]=ihnp^{n-1} [/tex]

    But I don't see how that is helpful or how I'm going to use that to do my original question.
     
  5. Oct 17, 2006 #4

    Hurkyl

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    g(a) + g'(a) (p - a) is almost never the Taylor expansion of g(p).
     
  6. Oct 17, 2006 #5

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    [a,b+c]=[a,b]+[a,c]. Use this to show the formula is true for any polynomial. Then take the limit to show it's true of the taylor series (ie, [itex]g(p)=a_0 + a_1 p + a_2 p^2 + ...[/itex]). This part might be a little tricky, and I don't know how rigorous you need to be, but to show two operators are equal, you need to show they agree on every function.
     
  7. Oct 18, 2006 #6
    Solution

    So here's what I have:

    Taylor expansion of g(p) can be expressed as the following:
    [tex]g\left(p\right) \approx a_{o}+a_{1}p+a_{2}p^{2}+\cdots+a_{n}p^{n} [/tex]
    So applying the commutator:
    [tex]\left[x, g\left(p\right)] = [x, a_{o}+a_{1}p+a_{2}p^{2}+\cdots+a_{n}p^{n} = \left[x, a_{o}\right]+\left[x, a_{1}p\right]+\left[x, a_{2}p^{2}\right]+\cdots+\left[x, a_{n}p^{n}\right] [/tex]
    [tex] = 0 + i\hbar a_{1} \frac{dp}{dx} + i\hbar a_{2} \frac{dp^{2}}{dx} + \cdots + i\hbar a_{n} \frac{dp^{n}}{dx} [/tex]
    since [itex] \left[x, p^{n}\right] = i\hbar\frac{dp^{n}}{dx}[/itex].

    So:
    [tex]\left[x, g\left(p\right)\right] = i\hbar\frac{d}{dx}\left(a{o}+a_{1}p+a_{2}p^{2}+\cdots+a_{n}p^{n}\right)=i\hbar\frac{dg\left(p\right)}{dx} [/tex]

    I think I did that okay. What do you think?
     
  8. Oct 18, 2006 #7

    dextercioby

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    Evaluate the Poisson bracket between the 2 functions and then quantize the result via Dirac's prescription.

    Daniel.
     
  9. Oct 18, 2006 #8
    I thought I had it, but it turned out that I was getting the wrong answer. Here's what I did:

    [tex] \left[x, p^{n}\right]\Psi = \left(xp^{n} - p^{n}x\right)\Psi = x\left(-i\hbar\frac{d}{dx}\right)^{n}\Psi - \left(-i\hbar\right)^{n} \left(\frac{d^{n}x}{dx^{n}}\Psi + x\frac{d^{n}}{dx^{n}}\Psi\right) [/tex]
    [tex] = -\left(-i\hbar\right)^{n}\frac{d^{n-1}}{dx^{n-1}}\Psi = i\hbar\left(-i\hbar\frac{d}{dx}\right)^{n-1}\Psi [/tex]

    But then I get [itex] i\hbar p^{n-1} [/itex] as an answer, and I should be getting [itex]i\hbar np^{n-1} [/itex] . Where did I go wrong?

    Thanks!
     
  10. Oct 18, 2006 #9
    This works too, but if you want to do it purely in the framework of QM, I'd recommend trying to do this in the momentum representation, where [tex]x = - i \hbar \partial_x[/tex] and have your trial function sitting up front.

    Commutators are invariant under unitary changes of coordinates.
     
  11. Oct 18, 2006 #10

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    The idea is to start from the fundamental commutation relation [itex][x,p]=-i\hbar[/itex] and not worry about what p is. Then there is a rule that [A,BC] = [A,B]C + B[A,C], so you can use induction to get [itex][x,p^n]=-i\hbar n p^{n-1}[/itex]. Finally, apply this to the taylor series, and you'll end up with what they want you to prove. But this shows that [itex]x=-i\hbar\partial/\partial p[/itex], and symmetrically, [itex]p=-i\hbar\partial/\partial x[/itex], so from that simple commutation relation you've found what p "is".
     
    Last edited: Oct 18, 2006
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