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Consider the 4-point correlation function, $G(x_1,x_2,x_3,x_4)$, in $\phi^4$ theory. Let us consider the term which is $\propto \lambda^2$, represented by the following Feynman's diagram:

enter image description here

According to symmetry rules, I should get a $1/2$ symmetry factor, due to the middle lines. Hence, I would suppose that

$$G_2(x_1,x_2,x_3,x_4) = -\frac{\lambda}{2}\int\int d^4x_5 d^4x_6\Delta_F(x_1-x_5)\Delta_F(x_2-x_5)\Delta_F(x_4-x_6)\Delta_F(x_3-x_6)\left(\Delta_F(x_5-x_6)\right)^2$$

where $x_5$ and $x_6$ denote de vertices. However, if I carry out the calculations by hand, I should have something like $\frac{1}{2\times 24}\times 4 \times 3 \times 2 \times 6 \times 2 = 6 \neq \frac{1}{2}$. What am I doing wrong here?

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The factor is $\frac{1}{2!} \times \frac{1}{4!} \times \frac{1}{4!} \times 8 \times 3 \times 4 \times 3 \times 2 = \frac{1}{2}$.

The $\frac{1}{2!}$ comes from expanding $e^x$ to second order. The two factors of $\frac{1}{4!}$ come from the interaction term (which is $\frac{\lambda}{4!}$). The remaining factors come from counting contractions so lets look at that. The fields to be contracted are $$ \phi_1 \phi_2 \phi_3 \phi_4 \, \, \phi_x\phi_x\phi_x\phi_x \, \, \phi_y\phi_y\phi_y\phi_y $$ First $\phi_1$ can contract with 8 different fields. Everything is completely symmetrical, so lets assume it contracts with $\phi_x$. $\phi_2$ must then contract with $\phi_x$ also which happens in 3 different ways. $\phi_3$ contracts with $\phi_y$ in 4 different ways and $\phi_4$ contracts with another $\phi_y$ in 3 different ways. Finally, the remaining un-contracted fields are $\phi_x \phi_x \,\, \phi_y \phi_y$. The first $\phi_x$ can contract with a $\phi_y$ in 2 different ways and the final contraction is then fixed. Multiplying all the numbers in this paragraph, we find that the number of contractions is $8\times3\times4\times3\times2$.

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  • $\begingroup$ Ahhhh! This makes much more sense than what I was thinking. Thank you so much. $\endgroup$ – miniplanck May 2 at 21:17

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