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Symmetry factors (Srednicki ch9)

  1. Feb 18, 2010 #1
    Hi,

    I'm not sure if I understand symmetry factors correctly or not. Looking at the second diagram in Srednicki's fig 9.1.

    The way I understand things is this corresponds to a number of terms in the expansion, that are algebraically different somehow, e.g perhaps one has a propagator like [tex] \Delta(x_1-y_1) [/tex] whilst another term has [tex] \Delta(x_2-y_1) [/tex] etc. But nevertheless, all terms corresponding to that diagram each have 3 propagators, no J's, 2 lot's of [tex] iZ_{g}g\int d^4x[/tex].

    So to count number of terms corresponding to this diagram say, we have 3 functional derivatives for each of our two vertices that can be permuted, so 3!x3!, we can also swap the two vertices so we get another factor of 2!. We can swap the two sources at the ends of each propagator (basically swap the ends of each propagator round on the diagram), which gives us 2!^3. Finally we can permute the 3 propagators amongst themselves for a further 3! terms. So we might guess at no terms=[3!^2]*[2!]*[2!^3]*3!=36*2*8*6=3456.

    But we will find we have over counted, since sometimes a certain permutation may result in exactly the same term (not just same diagram) as another permutation.

    So for example when swapping our three propagators around here in the 3! ways, we would also find these same terms cropping up when we permute functional derivatives in our 3!*3! ways, so we can cancel out the 3! from the prop permuting, and say we've already included it in the functional derivative permutations.

    Similary swapping the two vertices in the 2 ways we have, can be duplicated in one of our (2!)^3 ways of swapping endpoints of propagators around (namely the one where each endpoint of all three is swapped simultaneously), so we can cancel out those 2 ways of swapping the vertices.

    However there is no way to replicate swapping of the end points of the propagators now, by just permuting the functional derivatives of each vertex about (since they would still be connect to the "wrong ends" of the propagators). So we can't cancel any more

    Therefore S=2*3!. Our number of terms corresponding to this diagram is then:
    [3!^2]*[2!^3]=288

    I would be really grateful if someone could tell me if Im thinking about this correctly, and if my answer is right.

    Thanks
     
  2. jcsd
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