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SymmetryFactor Calculation

  1. Sep 23, 2014 #1

    ChrisVer

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    Can someone check the symmetry factor I've found in the following diagram and verify it?
    Do you know if there is any way to determine the SF of a diagram fast?

    So I have [itex]\phi_x[/itex] which can be contracted with 4 [itex]\phi_w[/itex] : 4
    Then [itex]\phi_y[/itex] which can be contracted with 3 [itex]\phi_w[/itex] : 3
    Then I have 2 [itex]\phi_w[/itex] which can be contracted with 4 [itex] \phi_u [/itex]: 2x4
    Then I have 3 [itex]\phi_u[/itex] which can be contracted with 4 [itex]\phi_z[/itex]: 3x4
    The rest [itex]\phi_u[/itex] get contracted together (only 2 left), whereas the [itex]\phi_z[/itex] we get a factor of 2 since there are 2 possibilities to contract 3 fields.
    Finally the last contraction gives just a factor 1 (no possible alternative choices).

    So the result from contractions is 4x3x2 x4 x3 x2 x 4 = 4! x 4! x 4
    The diagram is of order 3 (3 vertices) so there is a factor 1/3! and also the 1/4! from the coupling constant for [itex]\phi^4[/itex] theory.
    So is it correct to say that the symmetry factor is afterall SF=16?
     

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  2. jcsd
  3. Sep 24, 2014 #2

    nrqed

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    Your factor of 2 (in blue) should be 3. From three fields, there are three possible distinct pairs to construct (AB, AC and BC).
    But there is also a second point: there should be a factor of 1/4! for each vertex, so you should have 1/(4!)^3 instead of 1/4!
    As for a shortcut: usually people drop the denominator and simply calculate the symmetry factor of the graph (the number of automorphisms in mathspeak). I think there is a nice discussion in Peskin and Schroeder if I recall correctly.
     
  4. Sep 24, 2014 #3

    ChrisVer

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    In Peskin they say that they drop the 1/4! factor and write the vertex factor as [itex]\int d^4 z (-i \lambda)[/itex] because they say that to a generic vertex has four lines coming in from four different places so the various contractions of [itex] \phi \phi \phi \phi[/itex] is 4!...Also the n! from the Taylor expansion will cancel because of interchanging of vertices...
    But I don't understand either of these explanations...
    So in case I used (1/4!)3 I would have obtained the correct result?

    [itex]A= \frac{16}{4! 4!}= \frac{1}{3! 3!} = \frac{1}{36} [/itex]
    So the symmetry factor is: 36?
     
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