- #1
Ene Dene
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I'm having trouble understanding the folowing:
(I'll write h for h/(2Pi))
Momentum in x direction is represented by operator
[tex]p=-ih\frac{d}{dx}[/tex].
So comutator
[tex][x,p]=xp-px=x(-ih)\frac{d}{dx}-(-ih)\frac{dx}{dx}=0+ih*1=ih.[/tex]
Now here comes the part that I don't understand. I'll calculate [x,p^2]:
[tex][x,p^2]=xp^2-p^2x=xpp-ppx[/tex],
but from:
[tex][x,p]=xp-px=ih[/tex]
I have:
[tex][x,p^2]=xpp-ppx=(ih+px)p-p(-ih+xp)=2ih+pxp-pxp=2ih[/tex]
As it should be. But why can't I say:
[tex]p^2=pp[/tex],
[tex](-ih)\frac{d}{dx}*(-ih)\frac{d}{dx}=-h^2\frac{d^2}{dx^2}[/tex]
And write:
[tex][x,p^2]=xp^2-p^2x=x*(-h^2\frac{d^2}{dx^2})+h^2\frac{d^2}{dx^2}(x)=0+0=0[/tex]
(I'll write h for h/(2Pi))
Momentum in x direction is represented by operator
[tex]p=-ih\frac{d}{dx}[/tex].
So comutator
[tex][x,p]=xp-px=x(-ih)\frac{d}{dx}-(-ih)\frac{dx}{dx}=0+ih*1=ih.[/tex]
Now here comes the part that I don't understand. I'll calculate [x,p^2]:
[tex][x,p^2]=xp^2-p^2x=xpp-ppx[/tex],
but from:
[tex][x,p]=xp-px=ih[/tex]
I have:
[tex][x,p^2]=xpp-ppx=(ih+px)p-p(-ih+xp)=2ih+pxp-pxp=2ih[/tex]
As it should be. But why can't I say:
[tex]p^2=pp[/tex],
[tex](-ih)\frac{d}{dx}*(-ih)\frac{d}{dx}=-h^2\frac{d^2}{dx^2}[/tex]
And write:
[tex][x,p^2]=xp^2-p^2x=x*(-h^2\frac{d^2}{dx^2})+h^2\frac{d^2}{dx^2}(x)=0+0=0[/tex]