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

Schwinger Quantum action

  1. May 7, 2008 #1
    Given the path integral

    [tex] < -\infty | \infty > = N \int D[\phi ]e^{i \int dx (L+J \phi ) } [/tex]

    then , it would be true that (Schwinger)

    [tex] \frac{\partial < -\infty | \infty >}{\partial J }= < -\infty |\delta \int L d^{4}x | \infty > [/tex]

    If so, could someone provide an exmple with the Kelin-gordon scalar field plus an interaction of the form [tex] \phi ^{4} [/tex]
  2. jcsd
  3. May 7, 2008 #2
    This is a consequence of the fact that
    [tex]\int\!\mathcal{D}\phi\, \frac{\delta}{\delta \phi} \left( \dots \right) = 0[/tex]
    Essentially, one integrates by parts and hopes that the exponential damps out the boundary in function space, if such a boundary exists. Thus, the equations of motion for [tex]\phi[/tex] are obeyed inside the expectation value.

    Anyway, using (space) integration by parts, write your action as
    [tex]\int\!d^4x\, \frac{1}{2}\phi ( - \partial^2 - m^2 ) \phi + \frac{\lambda}{4!} \phi^4[/tex].
    You should now be able to find the variation in this action just like taking ordinary derivatives. If you get stuck, tell us what you've done and we can see about help.

    I hope someone doesn't just blurt out the answer. This forum seems to be full of keeners like that =)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook