1
$\begingroup$

I am trying to understand how to enumerate higher-order Feynman diagrams.

In his book on Elementary Particle Physics, Griffiths considers a simple "ABC toy theory" which has:

  • three (scalar, massive) particles A, B, and C (which are their own antiparticles), and
  • only one allowed interaction vertex ABC.

He counts explicitly the number of higher-order diagrams for the process $A+A\to B+B$ at order $g^4$, i.e. with 4 vertices.

What confuses me is that his result changes in the 2nd edition:

In the first edition of his book (p. 207) he finds that there are 15 because each additional line can start at one of the 5 lines of the original $A+A\to B+B$ diagram and end on the same or another line, so enumerating

  • $1\to1$, $1\to2$, ..., $1\to5$,
  • $2\to2$, ..., $2\to5$,
  • ...
  • $5\to5$,

we end up with 15. $\Rightarrow$ This makes sense naively.

However, in the second, revised edition, he talks about 8 diagrams only (p. 217): 5 self-energy diagrams ($i\to i$, a line "sprouts a loop"), two vertex corrections ("a vertex becomes a triangle") and one box diagram.

My question: How many diagrams are there? 15 or 8? Where have the other 7 gone, e.g. a line connecting an initial and a final-state particle? Was it wrong to include them? If so why? Are they equivalent to another diagram or ruled out for some reason?

$\endgroup$
  • 1
    $\begingroup$ And please add a comment if you downvote so I can improve the question! :/ $\endgroup$ – fuenfundachtzig May 4 '15 at 9:35
1
$\begingroup$

The 8 is correct. The diagrams where you connect one outgoing and one incoming line are equivalent to the ones, where you connect one outgoing line and the internal line, as both lead to the vertex-correction diagram.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.