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I am reading Cheng and Li. On page 9, it is written that the coefficient $\frac{1}{2 \cdot (4!)^2}$ for the second order term of the four point function becomes just $\frac{1}{2}$ for the following diagram:

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I am not getting that answer. Call the vertices in the diagram $y_1$ and $y_2$. Imagine $y_1$ and $y_2$ as a point with four lines coming out. (As it is $\phi^4$) My reasoning is as follows: $x_1$ can be connected to any one of the four lines emanating, after that $x_4$ has three choices for joining to one of the vertices. Ways of joining $x_1$ and $x_4$ is $4 \times 3$. Similarly for $x_2$ and $x_3$. Now $y_1$ and $y_2$ have two lines left which can be joined in a loop in two ways So total number of ways is $4 \times 3 \times 4 \times 3 \times 2$ . So I am getting the coeffecient as $\frac{4 \times 3 \times 4 \times 3 \times 2}{2 \times (4!)^2}=\frac{1}{4}$.

Where am I going wrong?

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The first thing when you start connecting is that x_1 has 8 lines to connect because otherwise you are leaving the combinantion in which y_1 and y_2 could be exchanged because they are arbitrary.

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