In Schwartz's book Quantum Field Theory and the Standard Model Exercise 7.2 is to show that in $2\to 4$ scattering there is no interference between 6-point vertex diagrams and the disconnected diagrams with two three point vertex.

When I solve for amplitudes I get a delta function of the type $$\delta^4(p_1+p_2-p_3-p_4-p_5-p_6)$$ for the 6 point vertex and the following for the two 3 point vertex diagrams: $$\delta^4(p_1-p_3-p_4) \delta^4(p_2-p_5-p_6).$$ How to show that these lead to zero interference?


1 Answer 1


Bit of a late response but I was wondering about this myself. I found this thread on PF that explains it quite nicely.

To paraphrase, the idea is that with a total $S$-matrix of $$S=-i\lambda \ \delta^4(\sum_{\text{total}} p) + -ig\delta^4(\sum_{\text{subset}} p)\delta^4(\sum_{\text{subset}} p),$$ the second term is only non-zero on a subset of the phase space with measure zero. More precisely the first term is non-zero on an 8-dimensional subset of the phase space while the latter only on a 4-dimensional subset. Thus for almost all points in the final phase space only the first term contributes.

In the (measure zero) regions where the second term is non-zero however, the double delta function translates to a factor of $(VT)^2$ whereas the first term is only proportional to $VT$. Thus whenever the second term contributes it is infinitely larger than the first term which can be ignored.

What this all means is that in the limit of large $T$ and $V$, only one term in $S$ ever matters and so there is never any interference. In practice what this says is that we only need to consider the fully connected diagram except in regions of the phase space where a disconnected process is allowed, in which case only the disconnected process contributes.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.