Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A question on eigenstates and operators

  1. Dec 26, 2009 #1
    Why can we say that:

    [tex]<x'|e^{i\hat{x}}|x>=e^{ix'}\delta(x'-x) [/tex]

    where where [tex]\hat{x}[/tex] is an operator?

    I mean if

    [tex]\hat{x}|x>=x|x> [/tex]

    we may write [tex]<x'|\hat{x}|x>=x<x'|x>=x\delta(x'-x)[/tex]

    but in the expression at the top, we have an exponential operator (something I've never come across before) - is |x> an eigenstate of this operator (it seems to be), and why is the eigenvalue of the operator exactly the same form as the operator?

  2. jcsd
  3. Dec 26, 2009 #2
    You can prove it by using power series expansion

    [tex]e^{i\hat{x}} = 1 + i\hat{x} - \hat{x}^2/2! + \ldots [/tex]

  4. Dec 26, 2009 #3
    Aaahh thanks Eugene. So in general if [tex]{\hat{Z}}[/tex] is any operator and |Z> the eigenfunctions of this operator, we can also say that |Z> are also the eigenfunctions of [tex]e^{i\hat{Z}}[/tex].
  5. Dec 26, 2009 #4
    That's right. |Z> is an eigenfunction of any function [tex]f(\hat{Z})[/tex] of the operator [tex]\hat{Z}[/tex]. The corresponding eigenvalue is [tex]f(Z)[/tex].

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook