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Concerning the momentum operator

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data

    If [tex]U[/tex] is an operator so [tex]U\Psi(x)[/tex] = [tex]\Psi(x-a)[/tex].

    How can I show that [tex]exp(-iaP/h) = U[/tex]

    where P is the momentum operator [tex]P = -ih(d/dx)[/tex]

    2. Relevant equations

    Not sure

    3. The attempt at a solution

    What I do know that if I have a function F of an operator then

    [tex]F(P)\psi[/tex] = [tex]$\sum_{i} c_iF(\lambda_i)\psi_i[/tex]

    where [tex]\lambda_i[/tex] are the eigenvalues of [tex]P[/tex]

    and [tex]c_i = <\psi_i,\psi>[/tex]

    can I somehow relate all of this to the operator [tex]U[/tex]
  2. jcsd
  3. Sep 22, 2009 #2


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    Homework Helper
    Gold Member

    I think what you'll want to do is combine the fact that (by the definition of the exponential of an operator)

    [tex]e^{-iaP/\hbar}=\sum_{n=0}^{\infty}\frac{(-1)^na^nP^n}{\hbar^n n!}[/tex]

    with your expression for [itex]P[/itex] in the x-basis to calculate [itex]e^{-iaP/\hbar}\Psi(x)[/itex]....your result should remind you of a Taylor series!:wink:
  4. Sep 22, 2009 #3
    Ok so:

    [tex]e^{-iaP/\hbar}= e^{-aD_x} = 1 - aD_x + \frac{a^2D_x^2}{2!} - \frac{a^3D_x^3}{3!} + ... [/tex]

    and applying this to [tex]\psi(x)[/tex] I can´t really say that I recognise the series, but I suppose it is the series for shifted [tex]\psi(x)[/tex] about a
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