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In Peskin and Schroeder's QFT book, on page 219, there is the following equation:enter image description here

The heading to the equation is: "The Fourier transform of the two-point function can now be written as".

Could someone help me formally prove this formula? I don't really get why the perturbation expansion decomposes into 1PI's.

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The goal is to find the single particle propagator in the presence of interactions. This propogator will be the sum of all diagrams which have two external vertices.

This sum of diagrams would be difficult to compute, but it turns out it easy to write this big sum of diagrams in terms of a sum of a smaller set of diagrams: the set of "one particle irreducible" (1PI) diagrams. A 1PI diagram is a diagram that cannot be split into two disjoint diagrams by cutting a single propagator edge.

To see why this is true, we realize that any diagram at all can be uniquely "untangled" into a finite sequence of 1PI diagrams. This means that the set of all diagrams is contained in the set of sequences of 1PI diagrams.

On the other hand it is clearly true that any sequence of 1PI diagrams is a diagram. Thus the set of all sequences of 1PI diagrams is in the set of all diagrams.

Thus we conclude that looking at all finite sequences of 1PI diagrams is the same as looking at all diagrams.

Now that we are looking at the sums of 1PI diagrams we notice that the sum factors: suppose for the sake of argument that $x$, $y$, and $z$ were the only 1PI diagrams. Then our sum of sequences of 1PI diagrams would look like $1+x+y+z+xx+xy+xz+yx+yy+yz+zx+zy+zz+xxx+\cdots$. Now notice this factors into $1+(x+y+z)+(x+y+z)^2+\cdots$. The key point here is that the sum of all length $n$ sequences of 1PI diagrams is the $n$th power of the sum of all 1PI diagrams. This is because when you construct a length $n$ sequence of 1PI diagrams, you just choose a 1PI diagram for the first and another for the second and so on. This is the same thing that happens when you compute a power of the sum of 1PI diagrams.

The next step is basically to recognize that $1+(x+y+z)+(x+y+z)^2+\cdots$ is a geometric series. After you sum this series, you have a formula for the interacting propagator in terms of a sum of just the 1PI diagrams, as desired. This is Eq. 7.23 in Peskin

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  • $\begingroup$ Small comment: You didn't mention tadpole diagrams. Any diagram at all can be uniquely "untangled" into a finite sequence of 1PI diagrams and tadpole contributions (which are typically zero). $\endgroup$ – JeffDror Jan 30 '14 at 16:34

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