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A The Optical Theorem for Feynman Diagrams

  1. Jun 10, 2018 #1
    In Peskin's textbook section 7.3 The Optical Theorem for Feynman Diagrams(Page233), he said it is easy to check that the corresponding t- and u-channel diagrams have no branch cut singularities for s above threshold.

    But I can't figure out how to prove it. Can angone help me? Thanks!
     
  2. jcsd
  3. Jun 12, 2018 #2

    CAF123

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    For the case of equal mass scattering, one may derive in the centre of mass frame the following relations for the Mandelstam invariants ##s,t,u##:
    $$s = 4(\mathbf p^2 + m^2) ,\,\,\,\,t=-2\mathbf p^2 (1- \cos \theta), \,\,\,\text{and} \,\,\,\,u = -2\mathbf p^2 (1+\cos \theta) $$ As the physical domain of the variables is such that ##\mathbf p^2 \geq 0## and ## -1 \leq \cos \theta \leq 1##, means that ##s \geq 4m^2, t \leq 0, u \leq 0## with the equalities holding when the relative momentum vector ##\mathbf p = 0##.

    In particular, and coming to your question now, if ##s## is above threshold, meaning that ##s## is higher than the ##4m^2##, then ##u## and ##t## are strictly less than zero. This all collectively maintains the constraint ##s+t+u = 4m^2##. So, the ##t## and ##u## channel diagrams proportional to ##1/t## and ##1/u## respectively, at the amplitude level, do not admit singular structures as ##t## and ##u## may not be nullified when ##s## is above ##4m^2.##
     
  4. Jun 13, 2018 #3
    Thank you.
    ##1.## You mean that the Mandelstam variables ##t## and ##u## correspond to the branch cut singularities of ##t##- and ##u##-channel diagrams respectively? Why?
    ##2.## I can't understand your last sentence.
    ##3.## ##t=-2\mathbf p^2 (1- \cos \theta)^2, \,\,\,\text{} \,\,\,\,u = -2\mathbf p^2 (1+\cos \theta)^2 ##. You lost ##^2## above the ##(1- \cos \theta)## and ##(1+\cos \theta)##?
     
  5. Jun 13, 2018 #4

    CAF123

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    Actually I confess that those statements were made with the tree level diagrams in mind and I implicitly assumed the propagator was massless. But the argument goes through with a massive propagator because then the t and u diagrams are proportional to 1/(t-m2) and 1/(u-m2) respectively and for s>4m2 strictly t,u<0 so one never encounters a pole at tree level. This is a general feature under 'S-matrix analyticity' .

    What happens at loop level?

    One finds that the t,u diagrams may contribute an imaginary part because of their proportionality to ##\sqrt{1-4m^2/q^2}##, where q2 stands for either t or u. Given that s>4m2 again we must have t,u<0 by the above. So the argument of the square root may never be negative and, as such, we are away from the branch cut and in the domain of parameter space where the diagrams are real.

    I think there might be another argument without using the explicit form of the diagrams (indeed P&S make their statement before starting their computation of the s channel diagram) but I don't see it yet.
    Does the above make it better?
    Hmm I don't think so. What makes you think there is a square?
     
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