1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    gabbagabbahey

    User Avatar
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Concerning the momentum operator
  1. Momentum operator (Replies: 7)

  2. Momentum operators (Replies: 4)

  3. Momentum operator (Replies: 3)

Loading...