Commutator proof

  1. Mar 27, 2010 #1
    1. The problem statement, all variables and given/known data
    Show [tex]\left[x,f(p)[/tex][tex]\right)][/tex] = [tex]i\hbar\frac{d}{dp}(f(p))\right.[/tex]


    2. Relevant equations

    I can use [tex]\left[x,p^{n}[/tex][tex]\right)][/tex] = [tex]i\hbar\\n\right.[/tex][tex]p^{n}\right.[/tex]
    f(p) = [tex]\Sigma[/tex] [tex]f_{n}[/tex][tex]p^{n}[/tex] (power series expansion)


    3. The attempt at a solution
    I started by expanding f(p) to the power series which makes

    [tex]\left[x,\Sigma\\f_{n}\\p^{n}[/tex][tex]\right)][/tex]

    and I know I must use the commutator identity [A, BC] = [A,B]C + B[A,C]
    but the power series cannot be split up into two products(BC) ? So I'm not sure how to go on
     
  2. jcsd
  3. Mar 27, 2010 #2

    Hurkyl

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    How do you know that?
     
  4. Mar 28, 2010 #3
    In a text book it says it can be shown using that equation

    Trying a different method:

    [x, f(p)] = [x,[tex]\sum_{n}\\f_{n}p^{n}[/tex]] = [x,fnpn + [tex]\sum_{n-1}\\f_{n}p^{n}[/tex]]

    using [A, B+C] = [A,B] + [A,C]

    = [x, fnpn] + [x, [tex]\sum_{n-1}\\f_{n}p^{n}[/tex]]

    using [A, BC] = C[A,B] + B[A,C]

    = fn[x, pn] + pn[x, fn] + [x, [tex]\sum_{n-1}\\f_{n}p^{n}[/tex]]

    using [x, pn] = i[tex]\hbar[/tex]npn-1

    = fni[tex]\hbar[/tex]npn-1 + pn[x, fn] + [x, [tex]\sum_{n-1}\\f_{n}p^{n}[/tex]]

    [x, fn] = 0 as fn is a const.

    = fni[tex]\hbar[/tex]npn-1 + [x, [tex]\sum_{n-1}\\f_{n}p^{n}[/tex]]

    am I on the right track?
     
    Last edited: Mar 28, 2010
  5. Mar 28, 2010 #4

    Hurkyl

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    I'm curious why you used
    [A, rC] = [A,r]C + r[A,C]​
    to pull out a scalar, rather than just using
    [A, rC] = r [A,C]​

    I'm also curious why you stopped using
    [A, B + C] = [A,B] + [A,C]​
    after a single addition.

    But that aside, everything you wrote looks correct. We won't know if you're on the right track until we see where this path leads, though!
     
  6. Mar 28, 2010 #5
    I see what you mean. [x, fnpn] = fn[x, pn] is fine.

    I kept going with the addition and noticed a pattern and managed to solve it. Thanks for the hints :)
     
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