# Why does it make sense to do Dyson Resummation with first-order 1PI-diagrams?

This is somewhat related to This question.

The way I understand renormalisation (e.g of the mass) is that we consider the loop-corrections order by order in the coupling constant. If we then consider e.g. $$\phi^4$$ theory, the first-order 1PI-diagram is given by $$-\frac{ig}{2}\Delta(0)$$ where $$\Delta(p)$$ is the Feynman Propagator. The Lecture notes I'm studying then go on to write the 2-point connected Green's function as $$W(p_1, p_2) = (2\pi)^4 \delta(p_1+p_2) \left[ \frac{i}{p_1^2-m^2} +\frac{i}{p_1^2-m^2} \frac{\frac{g}{2} \Delta(0)}{i} \frac{i}{p_1^2-m^2} + \dots \right].$$ Then they go on to perform the Dyson Resummation and find that the mass gets shifted by $$m^2 + \frac{g}{2}\Delta(0) := m_r^2.$$ Conceptually, I don't understand what's going on here. Sure, we can only consider the 1PI-diagrams up to leading order. That means, we should neglect all diagrams of order $$g^2$$. But then, when we do Dyson Resummation, the third term is already of order $$g^2$$. Thus, it seems to me that we only consider some of the higher-order diagrams (i.e. the ones that are reducible). But there are also irreducible order $$g^2$$ contributions (for example the one where you have two loops stacked on top of each other) that simply don't appear here.

How is this consistent? In the question I linked above, the consensus seems to be that it doesn't make sense to do Dyson Resummation with only some if the 1PI diagrams. But if that's true, then how is it at all possible to do renormalisation order by order in the coupeling constant? Because in the lecture notes, renormalisation is always performed on the Vertex functions, and these are essentially obtained from Dyson resummation (at least in the 2-point case), correct?

This is perhaps more easily seen by looking at the exact propagator $$\frac{i}{p^2-m^2-\Sigma(p)}\tag{2}$$ where $$-i\Sigma(p)$$ is the sum of all 1PI diagrams (using the language from my question). If we include in this only the tadpole, we get the same result your lecture notes found to lowest order, and it doesn't matter that we didn't include the Saturn diagram to this order.
• Also note for example that the Taylor expansion of $\frac{1}{x-a}$ is just $\frac{1}{x}+\frac{1}{x}a\frac{1}{x}$ to first order. So the effect of adding on all the other higher order terms in a geometric series is just for illustrative purposes and they don't actually do anything to the result. Commented Jul 22, 2023 at 14:31