# Factors for topologically similar Feynman Diagrams

I am reading Dr. Mattuck's A Guide to Feynman Diagrams in the Many-body Problem and found a rule I didn't see in my other QFT textbooks. It states as:

If we are given a diagram, and form a new diagram from it by twisting one or more of its interaction wiggles through 180 degrees, then the new diagram has the same value as the original one. Hence all twisted diagrams may be omitted if we just multiply the amplitude by a factor of $$2^n$$.

where $$n$$ is the number of interaction wiggles in a diagram. For example, the bubble diagrams in the diagram series for the single-particle propagator in interacting Fermi system are

According to the rule quoted here, we need to put a "$$2$$" in front of the LHS diagram to omit the RHS diagram above. So my question is: Why do we bother to add factors in front of topologically similar diagrams since they are essentially telling the same story?

## 1 Answer

Why do we bother to add factors in front of topologically similar diagrams since they are essentially telling the same story?

Diagrams do not "tell a story." They are also not a real time snapshot of a physical process (I thought you might have this doubt judging by the presence of a time axis in your drawings).

Diagrams are simply pieces of an integral. They are bookkeeping devices that allow us to keep track of all the terms. As such, we need to consider all of them. All possible diagrams that can be drawn are terms that will appear in the perturbative expansion of the integral under question.

Since there are symmetries (like twisting the wiggles as you said) many terms give the same numerical value, so we just have to compute them once. But still we need to know how many of them there were.

## Simplifying the concept of Feynman diagram

### (continue reading at your own risk)

I will give you a simplified version of the Feynman diagrams. This is what really happens in QFT, but instead of dealing with an infinite dimensional integral we deal with a one dimensional one. The underlining idea is the same.

Consider the following integral $$\mathcal{I}_{2n} = \int_{-\infty}^\infty \mathrm{d}x \,x^{2n} \exp\left(-\frac1{2g}x^2\right) \equiv \langle x^{2n}\rangle\, \mathcal{I}_0\,.$$ I gave it the suggestive name of a vacuum expectation value of the "operator" $$x^{2n}$$ because that's what it stands for if we are doing an analogy with QFT. There are no operators here, this is just to make the analogy clearer. Then I also normalize it by $$\mathcal{I}_0$$. Now, because this is just a single integral, we know $$\langle x^{2n}\rangle= g^{n}\,(2n-1)!!\,.$$ But let's forget that and try to compute it the "Feynman way." The "partition function," namely the expectation value of $$e^{Jx}$$ is \begin{aligned} Z(J) &\equiv \langle e^{Jx}\rangle = \frac{1}{\sqrt{2\pi g}}\int_{-\infty}^\infty \mathrm{d}x\,\exp\left(-\frac{1}{2g}x^2 + Jx\right) \\ &= e^{\frac{g}2 J^2}\,. \end{aligned} We are defining this in order to do this trick: every time we take a derivative with respect to $$J$$ we bring down an $$x^2$$ in the integrand. So we can write $$\langle x^{2n} \rangle = \frac{\mathrm{d}^{2n}}{\mathrm{d}J^{2n}} Z(J)\big|_{J=0}\,.$$ This derivation will produce a lot of terms: they grow exponentially as $$n$$ grows. And, here is the thing: every term is a Feynman diagram. In the QFT case the same thing happens, we also have some complicated derivative (in an infinite dimensional space) and every term that comes out of it is what we call a Feynman diagram.

But how so, you may ask. Well, let's analyse this theory: the "Lagrangian" is quadratic, so we only have propagators and no vertices. This theory is kind of trivial: if you have $$2n$$ "external points," meaning you are computing $$\langle x^{2n}\rangle$$, then your only Feynman diagram is the one where all external points are joined pairwise. We conclude that the derivative in the previous equation is computed by summing all such Feynman diagrams.

Furthermore, unlike QFT, these Feynman diagrams require no integration nor any difficult computation. They are just the product of all the propagators. Where a propagator is given by $$\mathrm{Propagator} = \langle x^2 \rangle = \frac{\mathrm{d}^{2}}{\mathrm{d}J^{2}} Z(J)\big|_{J=0} = g\,.$$ I'm not drawing the diagrams, but they are just a set of $$n$$ lines joining pairwise $$2n$$ external points.

Ok so, wait a minute. If all we did so far was correct, we would conclude that $$\langle x^{2n}\rangle = \mbox{Product of n propagators} = g^n\,.$$ But this is wrong, we are missing the $$(2n-1)!!$$ factor! Why is that? Because of the simmetries! There are actually many Feynman diagrams that evaluate to the same answer, and we need to count all of them. Otherwise the result would be wrong.

This is a simple combinatorial problem: in how many ways can I join $$2n$$ points pairwise? The first doesn't matter. So I have $$2n-1$$ choices for the one that links to the first. Out of the remaining $$2n-2$$, the first doesn't matter and I have $$2n-3$$ choices for the second, and so on, so $$\mbox{Number of equivalent Feynman diagrams}\equiv N_F= (2n-1)(2n-3)(2n-5)\cdots 3\cdot 1\,.$$ Which leads us to $$\langle x^{2n}\rangle = N_F\, g^n = (2n-1)!!\, g^n\,.$$

### Discussion

This trivial example doesn't have loops. But we could introduce them! We could, for instance, compute the integral $$\int_{-\infty}^\infty \mathrm{d}x \,x^{2n} \exp\left(-\frac1{2g}x^2 + \frac{\lambda}{4!} x^4\right)\,,$$ and that would be very similar to QFT: there will be loops and many diagrams will be equivalent in the sense that they evaluate to the same thing. Nevertheless we need to count them all.

• Wow, this is a very intuitive answer, thank you so much! Mar 29, 2020 at 22:00