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In $\phi^3$ theory, the generating functional for interacting field theory is given by: $$ Z_1(J) = \sum_{V=0}^{\infty} \frac{1}{V!} \Big[ \frac{iZ_g g}{6} \int \Big( \frac{1}{i}\frac{\delta}{\delta J}\Big)^3 d^4 x \Big]^V \times \sum_{P=0}^{\infty} \frac{1}{P!} \Big[ \frac{i}{2} \int J(y) \Delta(y-z) J(z) \, d^4 y \, d^4z \Big]^P $$

[Reference: Srednicki: eqn. (9.11)]

Let, for specific values of $V$ and $P$ we get some terms from it. One of them is a disconnected diagram consisted of two connected diagrams $C_1$ and $C_2$. The disconnected diagrams symmetry factor is, say, $S$; that is the term for disconnected diagram has a numerical coefficient: $\frac{1}{S}$. Now we write the term for disconnected diagram according to the eqn (9.12): $$ D = \frac{1}{S_D} \prod_I (C_I)^{n_I}$$

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 $D$. Here, $S_D = \prod_I n_I !$

In this case 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?

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Yes, it is true. For $V=2$, $P=4$ and $E = 2$ it can be proved.

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    $\begingroup$ You know you can accept your own answer if you solved the problem yourself. Also, it's great you figured it out, but if you want this to be useful for future readers you may want to flesh this out with a bit of explanation. Up to you. $\endgroup$ – Michael Brown Jun 14 '13 at 14:52

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