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Translation of Momentum

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove the following:

    [tex]|\vec{p}\rangle = \exp\left[i\vec{p} \cdot \hat{\vec{x}} / \hbar\right] |\vec{p}=\vec{0}\rangle[/tex]

    Where [tex]|\vec{p}\rangle[/tex] is any eigenstate of the momentum operator [tex]\hat{\vec{p}}[/tex].


    2. Relevant equations

    See above.

    3. The attempt at a solution

    I guess one has to follow the same argument as when one shows that momentum is the generator of translation. Would this mean that position is the generator of momentum?

    When following the argument, one ends up with:

    [tex]|\vec{p} + \vec{p}'\rangle = e^{-i \hat{\vec{K}} \cdot \vec{p}}' |\vec{p}[/tex]

    I just chose the 'translation' operator as [tex]\hat{K}[/tex] again, not to be confused with the one when working with translation of position. I don't really know if this argument will carry on to prove the result or how to do it. Any tips would be appreciated.
     
  2. jcsd
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