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Solving [H,x] =

  1. Oct 26, 2013 #1
    I'm trying to practice some operator algebra..

    Solve [H,x] where H is the Hamiltonian operator, x is position operator, and assuming one dimensional potential energy, U(x).

    I know the commutator comes out as -ih_bar(p_op)/m

    Here is my work so far.

    [H,x] = Hx - xH

    note: H = [ -ih_bar(p/2m) + U(x) ]

    so.. plug it in.. [ -ih_bar(p/2m) + U(x) ] x - x [ -ih_bar(p/2m) + U(x) ]

    ( x U(x) and -x U(x) cancel)

    now.. -(ih_bar/2m)(p x) + (ih_bar/2m)(x p)

    -(ih_bar/2m)[ p x - x p ]

    x and p are both operators.. so I know they don't cancel.. I'm kind of lost at this point.
  2. jcsd
  3. Oct 27, 2013 #2
    Well, [itex]px - xp[/itex] is just another way of writing [itex][p,x][/itex], which by definition is [itex]i\hbar[/itex]. However, this doesn't give you the answer you're looking for, because your Hamiltonian isn't right. The standard nonrelativistic Hamiltonian is [itex]\frac{\hat{p}^2}{2m} + U(\hat{x})[/itex], which should give you the commutation relations you're looking for.
  4. Oct 27, 2013 #3
    Oh yeah.. haha thanks.
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