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Could somebody explain how one can derive the symmetry factor both by counting possible contractions and by looking at the symmetry of a diagram. Consider for example this diagramenter image description here

in $\phi^4$-theory with $\mathcal{L}_\text{int} = -\frac{\lambda}{4!}\phi^4$.

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    $\begingroup$ More on symmetry factors of Feynman diagrams. $\endgroup$
    – Qmechanic
    Dec 2, 2014 at 21:17
  • $\begingroup$ I've seen the other three questions. None of them helped me to understand how to tackle a diagram such as the above. $\endgroup$
    – Janosh
    Dec 2, 2014 at 21:39
  • $\begingroup$ Usually you never get anything beyond a two for a symmetry factor for a Feynman diagram. That does not answer your question, I know. So here is a site that will help you. I will give you a site to go to that explains it better than I could in the little space I have. Google, Feynman diagrams Radovan Dermisek, he explains it two different ways. He uses Srednicki's book, but a draft of that book is online for free. After you see it, you will understand why it is not so easy to explain in one post. I think he does a pretty good job of explaining it. Hope it helps! $\endgroup$
    – Reddy15
    Dec 3, 2014 at 13:44
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    $\begingroup$ I assume, you are referring to these lecture notes. I read parts of them. However, I find that often 'powerpoint style' notes aren't very informative without the person who created them explaining their contents. After all, their main purpose is to back up the person's lecture and allow listeners to write down notes in context during the talk. $\endgroup$
    – Janosh
    Dec 4, 2014 at 12:12
  • $\begingroup$ I will give Srednicki's book a closer look this weekend. $\endgroup$
    – Janosh
    Dec 4, 2014 at 12:13

2 Answers 2

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The procedure is quite simple: the symmetry factor counts how many of the same diagrams you would be counting if you didn't consider that the single particles in your diagrams are identical. To do this, ask yourself "how many of this same diagram can I make?" where "same diagram" means "same external particles, same connections, and same vertices". To do this you want to

  • draw only the external particles and the vertices, leaving other particle lines "unconnected"
  • pick one of the unconnected lines: in how many ways can you connect it so that you have the desired diagram? Count them, and then connect it in one of the ways you just found (doesn't matter which one)
  • continue for the other unconnected lines, minding that now some of the earlier possible connections will be "taken"
  • the product of all the numbers you found is your symmetry factor, and its inverse is to be multiplied by your diagram in the perturbation series

To work with your example in $\lambda\phi^4$ theory: enter image description here

  • external particles and vertices drawn
  • I picked the upper left line, and to make the desired diagram I can connect it either in the blue or orange way: so my number is two, and I write it down. Now I pick the blue connection (but I could've picked the orange)
  • now I move on to the lower left line, but there is only one possible connection to be drawn: the green one, so I write one
  • the symmetry factor is therefore $2\times 1=2$

Or yet another example: enter image description here

  • external particles and vertices drawn
  • I picked the upper left line, and to make the desired diagram I can connect it either in the blue, orange, or green way: so my number is three, and I write it down. Now I pick the blue connection (but I could've picked the orange or green)
  • now I move on to the center left line, but there are two possible connections to be drawn, so I write two
  • ...
  • the symmetry factor is therefore $3\times 2 \times 1=3!=6$
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    $\begingroup$ But how does this work if we want to determine the symmetry factor of a double loop (8)? $\endgroup$
    – Silas
    Nov 16, 2022 at 21:01
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    $\begingroup$ @Silas what do you mean by "double loop (8)"? $\endgroup$ Nov 16, 2022 at 21:46
  • $\begingroup$ For the vacuum process $\int\mathrm{d}^4xD_F(x-x)D_F(x-x)$ (looks like 8 or infinity) we have only one vertex but I don‘t know how to get the symmetry factor $S=8$ with this technique. $\endgroup$
    – Silas
    Nov 16, 2022 at 22:14
  • $\begingroup$ Sorry for bothering you again but I haven't found any book that deals with this method. Is it possible to determine the symmetry factor of a loop diagram with this method or does it only works for connected diagrams? $\endgroup$
    – Silas
    Jan 10, 2023 at 19:05
  • $\begingroup$ It would be 8, right? $\endgroup$
    – Guy
    Jul 2, 2023 at 19:01
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Keep the external points $x_i$ fixed and look at the exchanging of propagators for the virtual particles. In this case there the factor is $2$ since you can only exchange them once and get the same diagram.

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