# Coefficients in Feynman Diagrams

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:

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?

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 $$8\times3\times4\times3\times2$$.