Commutator proof

  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. Hurkyl

    Hurkyl 16,090
    Staff Emeritus
    Science Advisor
    Gold Member

    How do you know that?
     
  4. 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. Hurkyl

    Hurkyl 16,090
    Staff Emeritus
    Science Advisor
    Gold Member

    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. 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 :)
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?