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!

Expectation value of the time evolution operator

  1. Nov 19, 2013 #1
    This problem pertains to the perturbative expansion of correlation functions in QFT.

    1. The problem statement, all variables and given/known data

    Show that [itex]\langle0|T\left[exp\left(i\int_{-t}^{t}dt' H_{I}^{'}(t')\right)\right]|0\rangle = \left(\langle0|T\left[exp\left(-i\int_{-t}^{t}dt' H_{I}^{'}(t')\right)\right]|0\rangle\right)^{-1}[/itex]

    2. Relevant equations

    [itex]H_{I}[/itex] satisfies [itex]i \frac{\partial U}{\partial t} = H_{I}U(t)[/itex] and is the Hamiltonian in the interaction picture for the [itex]\varphi_{in}[/itex] fields, which was used in the perturbative expansion. Also, [itex]|0\rangle\langle0| = 1[/itex]

    3. The attempt at a solution

    [itex]1=\langle0|1|0\rangle = \langle0|T\left[exp\left((i-i)\int_{-t}^{t}dt' H_{I}^{'}(t')\right)\right]|0\rangle[/itex]

    [itex]1=\langle0|T\left[exp\left(i\int_{-t}^{t}dt' H_{I}^{'}(t')\right)\right]|0\rangle\langle0|T\left[exp\left(-i\int_{-t}^{t}dt' H_{I}^{'}(t')\right)\right]|0\rangle[/itex]

    From there it follows that [itex]\langle0|T\left[exp\left(i\int_{-t}^{t}dt' H_{I}^{'}(t')\right)\right]|0\rangle = \left(\langle0|T\left[exp\left(-i\int_{-t}^{t}dt' H_{I}^{'}(t')\right)\right]|0\rangle\right)^{-1}[/itex] simply by dividing...

    I can't see what's wrong with my solution... but I have the feeling that there is something wrong.
     
    Last edited: Nov 19, 2013
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Expectation value of the time evolution operator
Loading...