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Momentum and Translation operator.

  1. Oct 12, 2011 #1
    In the concept of translation operator, there's a line said that

    pUIINX.gif

    where gif.latex?T(0)=1.gif and pQzIHQ.gif

    How could they interpret that this parameter ρ is really a momentum operator?
     
    Last edited by a moderator: Apr 14, 2017
  2. jcsd
  3. Oct 12, 2011 #2

    dextercioby

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    In the position representation, the momentum along axis x is proportional to the derivative operator wrt x. This is a consequence of the Stone-von Neumann uniqueness theorem.

    Since the infinitesimal generator of translations along x axis is proportional to the first derivative, again wrt x, it follows simply that the momentum along x is the generator of translations along the same axis, i.e. x.
     
  4. Oct 12, 2011 #3

    vanhees71

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    I don't understand your notation. Anyway, in position representation, the translation operator is given by

    [tex]\hat{U}(\vec{\xi})=\exp(\mathrm{i} \hat{\vec{p}} \cdot \vec{\xi}),[/tex]

    where I've set [itex]\hbar=1[/itex] for simplification of notation. By definition it acts on a wave function in the domain of the momentum operator by

    [tex]\hat{U}(\vec{\xi}) \psi(\vec{x}):=\psi(\vec{x}+\vec{\xi}).[/tex]

    Comparing this to the Taylor-expansion formula,

    [tex]\psi(\vec{x}+\vec{\xi})=\exp(\vec{\xi} \cdot \vec{\nabla}) \psi(\vec{x}),[/tex]

    immediately gives the well-known expression

    [tex]\hat{\vec{p}}=-\mathrm{i} \vec{\nabla}[/tex]

    for the momentum operator in the position representation.
     
  5. Oct 12, 2011 #4
    @dextercioby: That's exactly what I want as an answer. Thanks.
    @vanhees71: Thanks for that reply :)
     
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