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

Peskin equation (7.51)

  1. Jan 3, 2012 #1
    This is trivially true on the real line where s<s_0. Then he analytically continued [itex]M(s)[/itex] to the entire complex plane and then made use of (7.51) off the real line. But how one can be sure (7.51) holds on entire complex plane?
  2. jcsd
  3. Jan 4, 2012 #2


    User Avatar
    Science Advisor

    They already assumed that the S-matrix element [itex]M(s)[/itex] is an analytic function of the complex variable [itex]s=E_{cm}^2[/itex].

    It is a simple result in complex analysis that if [itex]f(z)[/itex] is complex-analytic, then so is [itex]\left[f(z^*)\right]^*[/itex].

    Then, the statement

    M(s) ~=~ \left[M(s^*)\right]^*
    for [itex]s < s_0[/itex] on the real line, says that two analytic functions coincide on an open subset of the real line. There is a theorem about "analytic continuation" that if two analytic functions (with different domains in general) coincide on an open subset of there common domain, each can be considered as an analytic continuation of the other into the other's domain. If it's possible to do this along two different paths from one domain to the other such that no poles are enclosed by these paths, then such an analytic continuation is unique and it makes sense to think of the whole as a single analytic function on the combined domain.
    (See any textbook on complex analysis for more discussion of this.)

    A (more direct) application of this is as follows (taken from Schaum's Complex Variables, problem 10.1):

    Theorem: Let F(z) be analytic in a region R and suppose that F(z)=0 at all points on an arc PQ inside R. Then F(z)=0 throughout R.

    Hopefully it's obvious how to apply this to the current case of [itex]M(s) - \left[M(s^*)\right]^* = 0[/itex] ?
  4. Jan 4, 2012 #3
    Thanks, this is precisely what I assumed he's using, but I never learnt complex analysis systematically so I didn't know.

    EDIT:Emm, turns out not hard at all to prove this using Cauchy-Riemann equation, but shamefully I didn't even try.
    Last edited: Jan 4, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook