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Homework Help: 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
    So comutator
    Now here comes the part that I don't understand. I'll calculate [x,p^2]:
    but from:
    I have:
    As it should be. But why can't I say:
    And write:
  2. jcsd
  3. Aug 9, 2007 #2


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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


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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


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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:


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


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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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