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Problem with commutation

  1. Dec 3, 2007 #1
    1. The problem statement, all variables and given/known data

    where B=×A , p=i[tex]\bar{h}[/tex] and q is constant

    2. Relevant equations

    3. The attempt at a solution

    if the x component is: px-qAx
    and the y component is py-qAy

    Then the x and y components shouldn't commute [px-qAx,py-qAy] <>0. This is where the extra term q[tex]\bar{h}[/tex]σB comes from.

    However, it seems that these values do commute:

    Am I missing something here that would cause these values not to commute?
  2. jcsd
  3. Dec 4, 2007 #2


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    Ay can be a function of x, so it does not commute with px (and vice versa).
  4. Dec 4, 2007 #3
    so what would [px-qAx,py-qAy] evaluate as?
  5. Dec 4, 2007 #4


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    In one dimension,
    [tex][p,f(x)]=-i\hbar f'(x)[/tex]
  6. Dec 4, 2007 #5
    oops, I asked the question wrong. I was already given what it evaluates as, I mean to ask why.
    [p,f(x)]=pf(x)-f(x)p... I guess I'm not seeing how the f' and ih terms are coming.
  7. Dec 4, 2007 #6


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    Consider f(x)=x^n. Using [A,BCD...]=[A,B]CD... +B[A,C]D... + AC[B,D]... + ...,
    [tex][p,x^n] = [p,x]x^{n-1}+x[p,x]x^{n-2}+\ldots+x^{n-1}[p,x]x+[p,x]x^n[/tex]
    In each term, [itex][p,x]=-i\hbar[/itex], and this is a number that can be pulled out front. There are n terms, and each has n-1 powers of x, so we get
    [tex][p,x^n] = -i\hbar nx^{n-1}[/tex]
    This is [itex]-i\hbar[/itex] times the derivative of x^n with respect to x. So to compute [p,f(x)], we do a Taylor expansion, and get the derivative term by term.
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