Simplifying Feynman Diagram for Nasty 10 Point Green's Function?

[vertices]

Why is the Feynman diagram for the following nasty 10 point Green's function so simple: I mean it only has two external points, one vertex, and one loop:

Here is the offending function:

\int d^4y_1 d^4y_2 <0|T[\phi (x_1) \phi (x_2) \phi^4 (y_1) \phi^4 (y_2)]|0>

which I am assuming is simply equal to:

\int d^4y_1 d^4y_2 <0|T[\phi (x_1) \phi (x_2) \phi (y_1) \phi (y_1)\phi (y_1) \phi (y_1) \phi (y_2)\phi (y_2) \phi (y_2) \phi (y_2)]|0>?

I mean this expression is very complicated - let's see:

F(\phi (x_1) \phi (x_2))F(\phi (y_1) \phi (y_1) ) F(\phi (y_1) \phi (y_1) ) F(\phi (y_2) \phi (y_2))F(\phi (y_2) \phi (y_2)) +<br /> F(\phi (x_1) \phi (y_1))F( \phi (x_2) \phi (y_1))F(\phi (y_1) \phi (y_1) ) F(\phi (y_2) \phi (y_2))F(\phi (y_2) \phi (y_2)) +...

(where F( ) is a contraction of operators).

Is there any way to simply this horrendous expression?

Thanks...

 

[azntoon]

Yes, there is a way to simplify this expression. First, note that the Feynman diagram for this nasty 10-point Green's function will only have two external points, one vertex, and one loop. This means that all of the terms in the expression can be grouped together into a single diagram by using Wick's theorem. This theorem states that any product of fields can be expressed as a sum of contractions of pairs of operators. Therefore, each term in the expression can be represented as a single Feynman diagram with two external points, one vertex, and one loop. The final result is that the nasty 10-point Green's function can be represented as a single Feynman diagram with two external points, one vertex, and one loop. This is much simpler than the original expression and is much easier to work with.