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Momentum operator and comutators

  1. Aug 9, 2007 #1
    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]
     
  2. jcsd
  3. Aug 9, 2007 #2

    Dick

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    I don't know how you are reducing the last line to zero. [x,p^2] in this representation is an operator. It needs to operate on something. Work out [x,p^2](f(x)) for an arbitrary function f(x).
     
  4. Aug 9, 2007 #3

    malawi_glenn

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    try to apply the comutator on a arbitrary wavefunction[tex]\psi [/tex]. Remember that [itex]p^2[/itex] is not the p-operator squared, it is performed twice on a wavefunction:

    [tex]p^2 \psi=-ih\frac{d}{dx} \left(-ih\frac{d}{dx} \psi \right) [/tex]

    Edit: Dick was 5 seconds before me =)
     
  5. Aug 9, 2007 #4
    Yes, of course, I understand now! Thank you!
     
  6. Aug 10, 2007 #5

    olgranpappy

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    the third equal sign is the above equation is incorrect (you didn't distribute the p correctly). The equation should read:

    [tex]
    [x,p^2]=xpp-ppx=(ih+px)p-p(-ih+xp)=2ihp+pxp-pxp=2ihp
    [/tex]

    In general
    [tex]
    [x,f(p)]=i\hbar f'(p)
    [/tex]
    and the above is a special case with f(p)=p^2.:wink:
     
    Last edited: Aug 10, 2007
  7. Aug 10, 2007 #6

    CompuChip

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    You better also had done the first one that way, as not to get confused by what the differential operator acts on.

    Let f denote a function of x, then
    [tex][x,p]f = xpf-pxf=x(-ih)\frac{df}{dx}-(-ih)\frac{dxf}{dx}= -i h x f' + i h (f + x f') = i h f[/tex]
    so dropping the function we see [tex][x, p] = i h[/tex].

    And this method will give you the correct result for any commutator (just remember, the differentiation operator works as far as possible to the right, so [itex]\frac{d}{dx} x f [/itex] really is [itex]\frac{d (x f)}{dx}[/itex] and not [itex]\left( \frac{dx}{dx} \right) \cdot f[/itex]).
     
    Last edited: Aug 10, 2007
  8. Jan 28, 2011 #7
    How come there is no h^2 term after we do it this way? Using the definition, I don't see why I only get an h and not h^2.
     
    Last edited: Jan 28, 2011
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