Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Wave packet propagating in one dimension

  1. Nov 14, 2006 #1
    I am working on a problem that goes
    "Show that for a wave packet propagating in one dimension, for a free particle Hamiltonian"

    m d<x^2>/dt = <xp> + <px>

    What I think I want to do.

    Use
    d<A>/dt = i/hbar * <[H,A]>

    Which leads me to

    d<x^2>/dt = i/hbar * <[H,x^2]>

    For the free particle H = p^2/2m

    Noting the relationship [x,p] = xp - px = ih => xp = ih + px
    note in the below I have not typed the bar behind the h to indicate that it is indeed hbar.
    [H,X^2] = [P^2/2m,x^2] for the free particle
    [H,X^2] = 1/2m (ppxx - xxpp)
    [H,X^2] = 1/2m (ppxx - (x(ih + px)p))
    [H,X^2] = 1/2m (ppxx - (ihx + xpxp))
    [H,X^2] = 1/2m (ppxx - (ihx + (ih + px)(ih + px))
    [H,X^2] = 1/2m (ppxx - (ihx + ihih + 2ihpx + (px)(px))
    [H,X^2] = 1/2m (ppxx - (ihx -h^2+ 2ihpx + ppxx)
    [H,X^2] = 1/2m (ppxx -ihx +h^2 - 2ihpx - ppxx)
    [H,X^2] = 1/2m ( -ihx +h^2 - 2ihpx)
    or something like this.

    This is where I get stuck, as a matter of fact I am not sure of any of the above is correct, I just tried to follow a similar example that goes

    m d<x>/dt = i/hbar <[P^2/2m,x]>
    [H,X] = 1/2m (ppx - xpp)
    [H,X] = 1/2m (ppx - (ih + px)p)
    [H,X] = 1/2m (ppx - (pxp + ihp))
    [H,X] = 1/2m (ppx - (p(ih + px) + ihp))
    [H,X] = 1/2m (ppx - (ihp + ppx + ihp))
    [H,X] = -ihp/m

    Any help on this would be appreciated.
     
  2. jcsd
  3. Nov 15, 2006 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Using the Schroedinger picture

    [tex] m\frac{d}{dt}\langle \hat{x}^{2} \rangle_{|\psi\rangle} =...= \frac{1}{2i\hbar} \langle \psi|[\hat{x}^{2},\hat{H}]_{-}|\psi\rangle =...= \langle \psi|\hat{x}\hat{p}_{x}+\hat{p}_{x}\hat{x}|\psi\rangle [/tex]

    Daniel.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook