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What is the symmetry factor for the following Feynman diagram if we assume that the external points are held fixed?enter image description here

Please ignore the arrows in the diagram. I am referring to the second diagram on the third row of Figure 9.7, page 62, of Srednicki's book, but, in contrast to Srednicki, I want the external points to be held fixed.

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In this case, the symmetry factor is 2, coming from the loop in the upper part of the diagram. You can either exchange both propagators or the derivatives at the vertices, producing overcounting.

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  • $\begingroup$ Apart from swapping the propagators in the loop in the upper part of the diagram, are we also allowed to swap the two vertices in the loop in the upper part of the diagram, giving us a symmetry factor of $4$? $\endgroup$
    – user127054
    Commented Jun 21, 2014 at 12:16
  • $\begingroup$ @user127054: Not if you hold the external points fixed. $\endgroup$
    – Frederic Brünner
    Commented Jun 21, 2014 at 16:33
  • $\begingroup$ One more question: if we set the external momenta equal to zero, does the symmetry factor change? $\endgroup$
    – user127054
    Commented Jun 22, 2014 at 20:44
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    $\begingroup$ @user127054: Since the symmetry factor does not say anything about the precise value of the momenta, I would say that the answer is no. $\endgroup$
    – Frederic Brünner
    Commented Jun 22, 2014 at 20:46
  • $\begingroup$ @FredericBrünner Even if we hold the external points fixed, we are allowed to swap the two vertices in the loop in the upper part of the diagram because they only connect to internal points, right? What am I missing? $\endgroup$
    – user87745
    Commented Apr 29, 2021 at 10:19

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