I am working on a problem that goes(adsbygoogle = window.adsbygoogle || []).push({});

"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.

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# Homework Help: Wave packet propagating in one dimension

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