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What is the most general Feynman diagram?

Srednicki, in his Quantum Field Theory book, says:

The most general diagram consists of a product of several connected diagrams. Let $C_I$ stand for a particular connected diagram, including its symmetry factor. A general diagram $D$ can then be expressed as $$ D = \frac{1}{S_D} \prod_I (C_I)^{n_I}$$ where $n_I$ is an integer that counts the number of $C_I$ ’s in $D$, and $S_D$ is the additional symmetry factor for $D$ (that is, the part of the symmetry factor that is not already accounted for by the symmetry factors already included in each of the connected diagrams). [see eqn (9.12)]

Can anyone please explain this for me? It would be very helpful if you exemplify this.

I have another question too. Can you give me a reference where phi-cubed theory is treated as I can use it as a reference while reading Srednicki.

Lorenzo B.
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rainman
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    Possibly the phrase "the most general" is tripping you up. There isn't a singular "the most general" Feynman diagram - obviously any given diagram is a specific thing. Try substituting "general" with "arbitrary," "typical," or "generic." He is just saying that any old diagram is made of one or more connected pieces. – Michael Jun 10 '13 at 15:59
  • @MichaelBrown: Thanks :( But do you know any other references where phi-cubed theory is discussed? I have found Srednicki very terse :( – rainman Jun 10 '13 at 16:02
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    Not off hand sorry. $\phi^3$ isn't discussed very often since it is a rather unphysical theory: the Hamiltonian is unbounded below and the perturbative vacuum is unstable. You'll find a lot more material about $\phi^4$ theory. But, if you are finding Srednicki terse you might have more luck with Tong. – Michael Jun 10 '13 at 16:19
  • But that is also written in $\phi^4$ theory :'( – rainman Jun 10 '13 at 16:24
  • What are $\phi^3$ used for, in theory as to model the real world they have some issues as @MichaelBrown says? I too see only the $\phi^4$ everywere, so I am curious to what theoretical issues could be sturdied with a $\phi^3$? – Dilaton Jun 10 '13 at 22:15

1 Answers1

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To explain what Srednicki is doing:

$C_i$ labels the connected diagrams with symmetry factors associated with them (individual diagrams) included, $n_i$ represents the number of diagrams $C_i$ present in the disconnected diagram $D$ and $S_D$ is an extra symmetry factor for the entire disconnected diagram due to interchange of lines between different connected graphs. Note that because we have $n_i$ copies of $C_i$ in our disconnected diagram, not only can we interchange lines in the individual diagrams due to symmetries (which has already been taken into account), but we can also change lines between diagrams of the same type, this contributes to the overall symmetry factor of the diagram. As we have $n_i$ of each $C_i$ we find $S_D = \prod_i n_i!$, so each diagram $D$ can be written as: $$D = \prod_{i}\frac{1}{n_i!}{C_i}^{n_i}$$

Will
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  • Can you please tell me why we are using symmetry factors here? – rainman Jun 11 '13 at 07:59
  • Do you understand how symmetry factors work? Think about it for a while. If you're still confused ask again and I can give a detailed account. – Will Jun 11 '13 at 14:30
  • Please give a detailed account. I have just started studying QFT and feeling confusion at many places. – rainman Jun 11 '13 at 14:45
  • Maybe you should ask this as a new question and explain where you are confused. There isn't enough space in the comment section if you want a general description with examples. Also, thinking about how you are confused might help you understand what's going on. – Will Jun 11 '13 at 16:10
  • Let, for specific values of $V$ and $P$ from eqn (9.11) we get some terms. One of them is a disconnected diagram consisted of two connected diagrams $C_1$ and $C_2$. The disconnected diagrams symmetry factor is, say, $S$. Now we write the term for disconnected diagram according to the eqn (9.12) given above. In this is this true: $ S = \frac{1}{n_1!} \times \frac{1}{n_2 !} \times C_1$'s symmetry factor $\times C_2$'s symmetry factor? – rainman Jun 11 '13 at 16:49