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Consider a $\varphi^3$ theory: $$ Z_1(J) \propto \exp\left[\frac{i}{6} Z_g g\int \mathrm{d}^4 x \left(\frac{1}{i}\frac{\delta}{\delta J}\right)^3\right] Z_0(J), $$ where $$ Z_0(J) = \exp\left[\frac{i}{2} \int \mathrm{d}^4 x \mathrm{d}^4 x' J(x)\Delta(x-x')J(x')\right]. $$ That is $$ Z_1(J) \propto \sum_{V=0}^\infty \frac{1}{V!}\left[\frac{i}{6} Z_g g \int \mathrm{d}^4 x \, \left(\frac{1}{i}\frac{\delta}{\delta J}\right)^3\right]^V \times \sum_{P=0}^\infty \frac{1}{P!}\left[\frac{i}{2} \int \mathrm{d}^4 y \, \,\mathrm{d}^4 z\, J(y)\Delta(y-z)J(z)\right]^P. $$ In particular, we can consider the term when $V=2, P=3$. Calculation shows that \begin{equation} - i \frac{1}{2!}\frac{1}{3!} \frac{(Z_g g)^2}{6^2*2^3} \left[\int \mathrm{d}^4\, x_1 \, \mathrm{d}^4 x_2 \, \left(\frac{\delta}{\delta J(x_1)}\right)^3 \left(\frac{\delta}{\delta J(x_2)}\right)^3\right] \left[\int \mathrm{d}^4 y \, \mathrm{d}^4 z \, J(y)\Delta(y-z)J(z)\right]^3\\= - i \frac{1}{2!}\frac{1}{3!} \frac{(Z_g g)^2}{6^2*2^3} \int \mathrm{d}^4 x_1 \, \mathrm{d}^4 x_2 \, \left[3^3*2^4 \Delta(x_1-x_1)\Delta(x_1-x_2)\Delta(x_2-x_2) + \\ 3^2\times 2^5\Delta(x_1-x_2)\Delta(x_1-x_2)\Delta(x_1-x_2)\right]\\= -i(Z_g g)^2 \int \mathrm{d}^4 x_1 \, \mathrm{d}^4 x_2\, \times \left[\frac{1}{2^3} \Delta(x_1-x_1)\Delta(x_1-x_2)\Delta(x_2-x_2) + \\ \frac{1}{2\times 3!}\Delta(x_1-x_2)\Delta(x_1-x_2)\Delta(x_1-x_2)\right], \end{equation} where $\Delta(x_1-x_1)\Delta(x_1-x_2)\Delta(x_2-x_2)$ and $\Delta(x_1-x_2)\Delta(x_1-x_2)\Delta(x_1-x_2)$ correspond to their Feynman diagram. Then the question is why $\frac{1}{2^3}$ and $\frac{1}{2\times 3!}$ are just the reciprocal of symmetry factors of the corresponding Feynman diagram respectively?

In the general case of $V, P$, why the coefficients of the terms in the result of calculation are just the reciprocal of symmetry factors of the corresponding Feynman diagram respectively?

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  • $\begingroup$ I applaud you for writing that much LaTeX, but please use the properly scaled brackets for clarity next time, e.g. 'left[', '\right]' and '\left(', '\right).' $\endgroup$ – JamalS May 11 '14 at 14:39
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This is exactly the point of the symmetry factor.

Let's call the term in $Z$ that we're considering $T$.

Without considering the symmetric exchanges that produce the symmetry factor, the contribution of each diagram to $T$ is simply its associated term without any numerical factor in front (a factor of 1). This is because when we count every possible exchange of vertices, propagators, derivatives, etc. that leaves the Feynman diagram invariant, this number neatly cancels out the factorials in the Taylor expansion and our choice of 1/6 and 1/2 in the field Lagrangian. If the symmetry factor for a diagram is 1, each of these exchanges gives rise to an identical term in the $T$.

When a diagram has a symmetry factor that is not 1, some of these exchanges mentioned above no give rise to additional terms. Hence the contribution of that particular diagram must be divide by the symmetry factor $S$.

This is a confusing topic, I wrote a note specifically on this kind of counting here

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  • $\begingroup$ Thanks! Could you explain the last paragraph in details? The key point may be there. $\endgroup$ – Eden Harder May 15 '14 at 2:11
  • $\begingroup$ It would help if you tell me exactly which part you do not understand. $\endgroup$ – zzz May 15 '14 at 2:55
  • $\begingroup$ Why if 'some of these exchanges mentioned above no give rise to additional terms' then 'the contribution of that particular diagram must be divide by the symmetry factor $S$'? $\endgroup$ – Eden Harder May 15 '14 at 4:36
  • $\begingroup$ So let's consider your example of V=2 and P=3, you can only draw 2 distinct Feynman diagrams for thus case, this tells you that there are two types of terms, I.e. Any term you get from doing the P=3 V=2 integral is identical to one of the Feynman diagrams, the question left is then how many identical terms are there that correspond to each diagram $\endgroup$ – zzz May 15 '14 at 13:20
  • $\begingroup$ And by counting arguments (By the notation I'm assuming you're using Srednicki, he outlines these counting arguments), we can show that if every exchange of vertices, propagators, etc that leaves the diagram invariant correspond to an identical term that contributes, the number of identical terms cancel the Taylor coefficients. $\endgroup$ – zzz May 15 '14 at 13:23

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