2
$\begingroup$

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?

$\endgroup$

1 Answer 1

3
$\begingroup$

It actually makes sense to perform the resummation of all the lowest order diagrams only. Doing this just turns the propagator into a more useful form where the shifted pole tells us the renormalized mass, also to lowest order. Of course, the renormalized mass is definitely not correct to the second order, because as you have noticed we are not adding any of the other 1PI diagrams. However, as we are only working to first order, this doesn't matter.

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.

By the way, my question was a bit wrong because I originally had the same misunderstanding as you. It does make sense to sum all the tadpole diagrams in a geometric series, as this just turns the exact propagator into a more useful form and allows us to find the renormalized mass to first order. However, adding only the Saturn diagrams in a geometric series, which is what Maggiore then goes on to do, doesn't make sense, as there is not really any meaning to the propagator in this case without also including the tadpole.

$\endgroup$
5
  • 1
    $\begingroup$ I have now corrected the original question as well to be in line with this answer. $\endgroup$ Jul 22, 2023 at 14:19
  • $\begingroup$ Thanks a lot! So in other words: If we only consider the first order, we can add and subtract higher-order contributions as we like, because by design, these won't change the first-order result. $\endgroup$ Jul 22, 2023 at 14:21
  • 1
    $\begingroup$ Yes, that sounds right to me $\endgroup$ Jul 22, 2023 at 14:24
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 22, 2023 at 14:31
  • $\begingroup$ Yes, this makes perfect sense to me now. It's just very unusual that an expression can be brought into a more convenient form by including higher-order terms. Usually it's the other way around :D $\endgroup$ Jul 22, 2023 at 15:06

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.