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Symmetry factor of diagrams

  1. Aug 17, 2010 #1
    I was suddenly confused by the calculation of symmetry factors of Feynmann diagrams.
    For example, in Peskin's text book, as the attached pdf file,
    Below eq(4.45), he calculated the symmetry factor in detail,
    however, I was confused by the last 1/2 factor.

    I'm trying to realize this factor as for the double-counting when we connect two of four legs of [tex]u[/tex]-vertex. But, I am not quite convinced by this above argument.
    If we label the four legs of [tex]u[/tex]-vertex with 1,2,3,4.
    Connecting legs 1,2 to form a loop, and then connecting legs 3,4 to [tex]w-[/tex]vertex is different from connecting legs 3,4 to form a loop and then connecting 1,2 to [tex]w-[/tex]vertex, right? In this way there is no double-counting!?

    Where does that 1/2 factor come from?
    Thanks.
     

    Attached Files:

  2. jcsd
  3. Aug 17, 2010 #2
    Oh I figured it out.
    When we form a loop at vertex [tex]u[/tex], we have 6 ways.
    But when we connect two legs, say 1,2, of vertex [tex]u[/tex] to, say 1,2, of vertex [tex]w[/tex], we have only 2 possible ways. That is, (11)(22) and (12)(12).
    However, we treated the contraction between vertices [tex]u[/tex] and [tex]w[/tex] as four different combination. So the double counting occurs here.
    :shy:
     
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