# Why do we factor out a momentum-conserving delta function in a 1-loop 1PI diagram in $\phi^4$ theory?

I do not understand the meaning of factoring out $$(2\pi)^4 \delta ^4(p_1 - p_2)$$, surely the delta function will have argument zero ($$p_1 = p_2$$) and should consume one integration and $$(2\pi)^4$$ should also cancel the $$(2\pi)^4$$ factor in the denominator (which comes due to loop integration of internal momenta).

Please tell me where I am incorrect and how to do this.

Due to spacetime translation symmetry, each Feynman diagram $$i{\cal M} ~(2\pi)^4\delta^4(\sum_e p_e)$$ is proportional to a delta function containing overall momentum conservation of external legs (with appropriate sign conventions). For this reason, it is in practice convenient to just list/handle/discuss the non-trivial part $${\cal M}$$.

References:

1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; eq. (4.103).

It just depends on what you want to compute. In the text they are focused on the loop, which is the part that contains divergences, the external legs have been amputated if you want. All they mean by the sentence you highlight is: "we are not computing the exact amplitude for the diagram, but just extracting the divergent part". Which is totally fine in order to study the divergence alone. However once you want to deal with a particular process, or renormalize the theory you will have to properly deal with the full diagram. The full amplitude just looks like

$$\langle f | S | i \rangle = (2\pi)^4 \delta^4(p_i-p_f) I$$

but they want to study $$I$$ not $$\langle f | S | i \rangle$$, that is all.

• Thankyou for the detailed answer. – sawan kt May 14 at 9:52