1. Not finding help here? Sign up for a free 30min 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!

Expectation values as a phase space average of Wigner functions

  1. Apr 12, 2017 #1
    Hi. I'm trying to prove that

    [tex][\Omega] = \int dq \int dp \, \rho_{w}(q,p)\,\Omega_{w}(q,p) [/tex]


    where

    [tex]\rho_{w}(q,p) = \frac{1}{2\pi\hbar} \int dy \, \langle q-\frac{y}{2}|\rho|q+\frac{y}{2}\rangle\,\exp(i\frac{py}{\hbar})[/tex]
    is the Wigner function, being \rho a density matrix. On the other hand
    [tex]\Omega_{w}(q,p) = \frac{1}{2\pi\hbar} \int dy \, \langle q-\frac{y}{2}|\Omega|q+\frac{y}{2}\rangle\,\exp(i\frac{py}{\hbar})[/tex]
    is the Wigner representation of the operator I'm interested in. The expectation value of an operator can be calculated from

    [tex][\Omega] = Tr(\rho\Omega) = \int dp \, \langle p|\rho\Omega|p\rangle[/tex]
    [tex]= \int dp^{\prime} \int dp \, \langle p|\rho|p^{\prime}\rangle\langle p^{\prime}|\Omega|p\rangle[/tex]


    Now, the matrix elements [tex]\langle p | \rho | p^{\prime} \rangle \,\,\, \mathrm{and} \,\,\, \langle p | \Omega| p^{\prime} \rangle[/tex]

    Should lead to the Wigner function and the Wigner representation of the operator but they only do so if
    [tex]p=p^{\prime}[/tex]

    Do you know how can I solve this?

    Thank you.
     
  2. jcsd
  3. Apr 17, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Expectation values as a phase space average of Wigner functions
Loading...