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Counting and Grouping Feynman Diagrams

  1. Oct 17, 2010 #1
    I'm working through Srednicki's QFT text, and I'm continuously vexed by the various numerical factors in diagrams and vertices, as well as the grouping of diagrams. For example, in Chapter 10 (pg 75) Srednicki treats basic [tex]\phi\phi\rightarrow\phi\phi[/tex] scattering processes in [tex]\phi^3[/tex]. He claims the 4 sources on the interaction yield 4!=24 different diagrams (alright), collected in 3 different groups (hmm) with 8 graphs each (which "neatly cancels the symmetry factor"?).

    Each of those statements seems logical, and the first two make some sense (I know the basic digram, and have read elsewhere you can "twist" diagrams to get the 3 groups he comes up with). Mathematically or intuitively, though, how can I look at this problem and see that there are indeed 3 kinds of diagram AND know that they each have 8-fold degeneracy?

    It's this final bit that I cannot quite grasp; any help would be appreciated.
  2. jcsd
  3. Oct 18, 2010 #2
    I don't quite understand what you mean by "twisting" the diagrams to get the three groups, but afaik, the three groups roughly correspond to the "s", "t" and "u" channels familiar from QED. And to see to which group a given connected diagram belongs, you simply have to look at the four-momentum of the internal line: whether it is s=p1+p2, t=p3-p1 or u=p4-p1 (where the in-coming particles have momenta p1 and p2 and the out-going p3 and p4).

    This may be trivial for you, I cannot unfortunately give a precise mathematical statement how to see that there are these three groups of diagrams.

    And to tell the truth, I cannot see how he gets simply 4! diagrams. Perhaps I should review my QFT...

    EDIT: No wait, I think I get it slowly. There are 3*3=9 ways to connect the vertices together, after which there are 4! different ways to connect the external points to the vertices. And then... what?
    Last edited: Oct 18, 2010
  4. Oct 18, 2010 #3
    If the ends of the external propogators are unlabelled you can do various symmetry operations without changing the diagram. Swapping the two functional derivs on left hand can be duplicated by swapping the two external props at the same time for example (indicating when you counted the number of such diagrams from your expansion initially you over counted by a factor of 2. If you don't remember counting then this was the reason for the factors like 1/V! that "neatly cancelled"). Similarly the right hand side external props can be swapped=> another factor of 2. Finally you can exchange the set of external props on the left for the set on the right and we can duplicate this by swapping the vertices and flipping the internal prop.
    This gives S=8 for the diagram in question.

    Now to get the 4 point correlation func, you apply the functional derivitive in the way Srednicki specifies in 10.4, but you are only interested in the fully connected diagrams, which amounts basically to taking all the diagrams with 4 sources, removing those sources and applying the labels 1,2,1',2' in every possible way. There are 24 possible ways of doing this, (at the moment we are considering permutations were order matters)4x3x2x1=24. Just think about the 4 corners lined up in a row say each as a box |_|_|_|_| and you need to work out number of ways you can arrange 4 distinct things, 1,1',2,2' into those.

    OK but all of these 24 diagrams are not "topologically distinct", you can rotate some into each other as you say. What are these "rotations" they are just the symmetries of the diagram we discussed earlier, if you label the external lines any way you like, you can get to another equivalent diagram by swapping both left hand props say: this should be obviously equivalent (it's just a different perspective on the same scattering process). Sim swap right hand side props, or swap both sets of props etc. So you have 3 groups of 8 within your 24. So it's no coincidence this cancels your 1/8 symmetry factor from earlier.

    These 3 groups are usually called the s,u and t channels.
  5. Oct 18, 2010 #4
    This may confuse someone, but here's my way to think about the prefactors. There are 3*3=9 ways to connect the vertices together. After this, there are 4! ways to connect the external points to the vertices (since there are 2+2 free fields left in the vertices). This means that there are 9*4!=216 Wick contractions that give rise to connected diagrams. For each channel (s,t and u) there are 216/3=72 Wick contractions.

    (To take the concrete example of the s-channel, this factor could be obtained directly by noting that when e.g. phi(x1) is connected to one of the remaining four fields (4 ways to do this), phi(x2) has to be connected to the same vertex, by the definition of the s-channel. phi(x3) can then be connected to the remaining two fields and for phi(x4) there is only one field left. Thus for the s-channel, there are 9*4*2=72 Wick contractions.)

    The pre-factor in the S-matrix expansion is 1/(2!3!3!)=1/72, which means that for each channel, the prefactor is simply 1.

    There are three groups of diagrams, since once you have contracted, say, phi(x1) with any vertex, there are three different external field terms that you can connect to this same vertex, and each choice gives rise to different dynamics (since the internal line has different momenta). I cannot directly see how this reasoning could be generalized to higher orders in perturbation theory, it is just a nice property of the four-particle correlation function in the second order of perturbation theory.
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