Symmetry factors (Srednicki ch9)

[LAHLH]

Hi,

I'm not sure if I understand symmetry factors correctly or not. Looking at the second diagram in Srednicki's fig 9.1.

The way I understand things is this corresponds to a number of terms in the expansion, that are algebraically different somehow, e.g perhaps one has a propagator like $$\Delta(x_1-y_1)$$ whilst another term has $$\Delta(x_2-y_1)$$ etc. But nevertheless, all terms corresponding to that diagram each have 3 propagators, no J's, 2 lot's of $$iZ_{g}g\int d^4x$$.

So to count number of terms corresponding to this diagram say, we have 3 functional derivatives for each of our two vertices that can be permuted, so 3!x3!, we can also swap the two vertices so we get another factor of 2!. We can swap the two sources at the ends of each propagator (basically swap the ends of each propagator round on the diagram), which gives us 2!^3. Finally we can permute the 3 propagators amongst themselves for a further 3! terms. So we might guess at no terms=[3!^2]*[2!]*[2!^3]*3!=36*2*8*6=3456.

But we will find we have over counted, since sometimes a certain permutation may result in exactly the same term (not just same diagram) as another permutation.

So for example when swapping our three propagators around here in the 3! ways, we would also find these same terms cropping up when we permute functional derivatives in our 3!*3! ways, so we can cancel out the 3! from the prop permuting, and say we've already included it in the functional derivative permutations.

Similary swapping the two vertices in the 2 ways we have, can be duplicated in one of our (2!)^3 ways of swapping endpoints of propagators around (namely the one where each endpoint of all three is swapped simultaneously), so we can cancel out those 2 ways of swapping the vertices.

However there is no way to replicate swapping of the end points of the propagators now, by just permuting the functional derivatives of each vertex about (since they would still be connect to the "wrong ends" of the propagators). So we can't cancel any more

Therefore S=2*3!. Our number of terms corresponding to this diagram is then:
[3!^2]*[2!^3]=288

I would be really grateful if someone could tell me if I am thinking about this correctly, and if my answer is right.

Thanks

 

[eljose79]

Hi there,

Your understanding of symmetry factors is mostly correct. The way I like to think about it is that symmetry factors account for the different ways in which a particular diagram can be constructed from its basic building blocks (vertices, propagators, etc.). In this case, the symmetry factor of 2*3! means that there are 2 ways to construct the diagram by swapping the vertices and 3! ways to construct it by permuting the functional derivatives.

However, your calculation of the number of terms may be slightly off. The correct formula is actually [3!^2]*[2!^3]/S, where S is the symmetry factor. In this case, that would give us 288/2=144 terms. This is because we need to account for the fact that some of the terms will be identical due to the symmetry of the diagram. So, in your example, we would have 144 unique terms, each with a symmetry factor of 2*3!, giving a total of 288 terms.

I hope this helps clarify things for you. Keep in mind that symmetry factors can be tricky to calculate and it's always a good idea to double check your work. Good luck with your studies!