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!

Klein-Gordon operator on a time-ordered product

  1. Nov 4, 2014 #1
    1. The problem statement, all variables and given/known data
    Hey guys,

    So here's the problem I'm faced with. I have to show that

    [itex] (\Box + m^{2})<|T(\phi(x)\phi^{\dagger}(y))|>=-i\delta^{(4)}(x-y) [/itex],

    by acting with the quabla ([itex]\Box[/itex]) operator on the following:

    [itex]T(\phi(x)\phi^{\dagger}(y))=\theta(x_{0}-y_{0})\phi(x)\phi^{\dagger}(y)+\theta(y_{0}-x_{0})\phi^{\dagger}(y)\phi(x)[/itex]

    2. Relevant equations
    [itex]\partial_{0}\theta(x_{0}-y_{0})=\delta(x_{0}-y_{0})[/itex]

    3. The attempt at a solution
    So I've split the quabla into its time and spatial derivatives: [itex]\Box = \partial_{0}^{2}-\nabla^{2}[/itex] and I'm applying the time derivative first, using the product rule:

    [itex]
    \partial_{0}T(\phi(x)\phi^{\dagger}(y))
    =\delta(x_{0}-y_{0})\phi(x)\phi^{\dagger}(y) +\theta(x_{0}-y_{0})\dot{\phi}(x)\phi^{\dagger}(y)\\
    +\theta(x_{0}-y_{0})\phi(x)\dot{\phi}^{\dagger}(y) -\delta(x_{0}-y_{0})\phi^{\dagger}(y)\phi(x)
    +\theta(y_{0}-x_{0})\dot{\phi}^{\dagger}(y)\phi(x) +\theta(y_{0}-x_{0})\phi^{\dagger}(y)\dot{\phi}(x)
    [/itex]

    However, the two delta terms vanish as the commutator of two fields is 0. so I'm left with

    [itex]
    \partial_{0}T(\phi(x)\phi^{\dagger}(y))
    =\theta(x_{0}-y_{0})\dot{\phi}(x)\phi^{\dagger}(y) +\theta(x_{0}-y_{0})\phi(x)\dot{\phi}^{\dagger}(y)\\
    +\theta(y_{0}-x_{0})\dot{\phi}^{\dagger}(y)\phi(x) +\theta(y_{0}-x_{0})\phi^{\dagger}(y)\dot{\phi}(x)
    [/itex]

    At this point I'm meant to be using the equal-time commutation relation: [itex] [\phi(x),\dot{\phi}^{\dagger}(y)] = i\delta^{(3)}(x-y)[/itex] but all my signs are positive...so what do I do?

    Thanks guys...
     
  2. jcsd
  3. Nov 9, 2014 #2
    Thanks for the post! 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?
     
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: Klein-Gordon operator on a time-ordered product
Loading...