- #1
krakes
- 1
- 0
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.
"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.