# Renormalization condition: why must be the residue of the propagator be 1

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In on-shell scheme, one of the renormalization conditions is that the propagator, say, a scalar theory

$$\frac{1}{p^2+m^2-\Sigma(p^2)-i\epsilon}$$

must have a unit residue at the pole of physical mass $p^2=-m^2$. Some textbooks say this is to make sure the propagator behaves like a free field propagator near the pole. But why?

UPDATE(27-Jun-2015): I was just randomly checking my old questions, now I fully agree with innisfree's answer.

This post imported from StackExchange Physics at 2014-04-24 02:31 (UCT), posted by SE-user Jia Yiyang

edited Jun 27, 2015
But ... why not?

This post imported from StackExchange Physics at 2014-04-24 02:31 (UCT), posted by SE-user Kostya
@Kostya: I'd agree that it will reduce to free field propagator if the interaction is somehow turned off, but I can't see any connection between "$p^2\to-m^2$" and "interaction being turned off"

This post imported from StackExchange Physics at 2014-04-24 02:31 (UCT), posted by SE-user Jia Yiyang

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The OS condition that $$\frac{\partial\Sigma}{\partial p^2}|_{p^2=-m^2} = 0$$ implies that the residue in the propagator remains equal to one.

Suppose that we used a different renormalization scheme in which our counter-terms contain no finite parts (e.g. MS scheme). In the OS scheme, we removed finite parts which were logarithmic in our artificial regularization scale $\mu$. In our new choice, the propagator might have a residue, say $R$.

This residue manifests itself in an irritating way; the field will be re-normalized such that $\phi = \frac{1}{\sqrt{R}} \phi_B$. In the LSZ formula, however, external lines contribute factors $R$ (from the KG equation cancelling the propagators). So external scalar lines contribute a factor $\sqrt{R}$ in the MS scheme.

So, whilst this choice in the OS scheme is somewhat arbitrary, it's convenient, because external scalar lines contribute a factor 1.

I'm trying to learn these points myself, so hopefully someone can expand/correct my answer where necessary...

This post imported from StackExchange Physics at 2014-04-24 02:31 (UCT), posted by SE-user innisfree
answered May 8, 2013 by (295 points)
So you are saying it's more of a mathematical convenience? I haven't read MS scheme so I'll keep this in mind for now.

This post imported from StackExchange Physics at 2014-04-24 02:31 (UCT), posted by SE-user Jia Yiyang

@JiaYiyang: It is a consensus of the physics community, hence a cultural convenience, not a mathematical one. The inverse of the Green's function is a differential operator perturbed in the non-free case by an integral operator called the self-energy, and asymptotically it equals the Fourier transform of the differential operator defining the asymptotic states. For your choice of the Minkowski metric, this should give $p^2+m^2$, and not $a(p^2+m^2)$ with some $a\ne 1$.

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The pole corresponds to an on-shell particle going from one point to another. Then, the residue effectively tells you how many of those particles are being transmitted. Since in your physical/renormalized theory, the propagator should correspond to $1$ quantum of the renormalized field being transmitted, you set the residue at the pole to $1$.

This post imported from StackExchange Physics at 2014-04-24 02:31 (UCT), posted by SE-user Siva
answered May 8, 2013 by (720 points)
But why does the residue equal (instead of just being proportional to) the number of particles? Can you give a reasoning without reference to free field case? In the sense that I don't want a reasoning like “because in free field propagator they are equal”, since this would be logically no difference with the textbook argument.

This post imported from StackExchange Physics at 2014-04-24 02:31 (UCT), posted by SE-user Jia Yiyang
Well, what if the start and the end point were the same. Then you'd expect to get the vev of $|\phi*(x) \phi(x)$ which you expect to give you the particle density at that point... and you want a single operator insertion to create 1 particle.

This post imported from StackExchange Physics at 2014-04-24 02:31 (UCT), posted by SE-user Siva
That's a nice answer. Related, I think, to the need to add factors of $\sqrt{R}$ to external lines in other schemes.

This post imported from StackExchange Physics at 2014-04-24 02:31 (UCT), posted by SE-user innisfree
@Siva：I don't quite get you, isn't the vev of $\phi^*(x)\phi(x)$ just 0, since particle density is 0 for vacuum? And by "single operator insertion to create 1 particle" what kind of insertion do you mean exactly?

This post imported from StackExchange Physics at 2014-04-24 02:31 (UCT), posted by SE-user Jia Yiyang
If you expand out $\phi$ and $\phi^*$ ino creation and annihilation operators, you will notice that one term will survive (and three will vanish due to normal ordering). So you'll have $<\phi^* \phi> = <a^{\dagger}a> = 1$. By "operator insertion at X", I just mean that a field operator acts acting at point $X$.

This post imported from StackExchange Physics at 2014-04-24 02:31 (UCT), posted by SE-user Siva
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I propose to look at this from the coordinate representation. The Green's function (propagator) may be built from the equation solutions, so the question is in fact about solutions $\phi$ of an "interacting" theory. Let's simplify the theory to a non-relativistic scattering of a scalar particle with mass $m$ from a static potential. The exact solution is not reduced to a free one, but has asymptotics at large distances describing free physical particles with the same mass. Similar property must be fulfilled by the Green's function. So, if one wants to build an "interacting" theory for physical particles, one should guess the interaction correctly. It should not modify the particle mass, but should deviate its path from a free one (scatter, in other words). In the occupation number representation it means change of occupation numbers of physical particles. The on-shell renormalization condition is such a subtraction way that removes the self-mass from the initial "interaction". Renormalization (subtraction) "finalizes" our initially wrongly guess of the interaction of physical particles.

The propagator residue is indeed directly connected with the wave function normalization condition in a trivial way.

UPDATE of 07 august 2021:

I would like to clarify once and forever the statement "... I can't see any connection between "$p^2→−m^2$" and "interaction being turned off" ".

Actually there are many (wrong) speculations about "bare particles" parameters, as if they existed indeed, but were non observable due to absence of interactions. In fact we always proceed from the equations of motions (good or bad) of interacting observable particles. That is why we know their physical parameters like mass, charge, spin, etc. However there exist a regime of very high kinetic energy giving the solutions numerically nearly identical to the solutions of free equations. Thus arose a notion of a "free particles". It is not a regime of coupling tending to zero; the coupling may stay the same (permanent for dressed particles or decaying at infinity or at shorter range for scattering processes), but the kinetic energy may prevail effects of those interactions in our numerically acceptable accuracy for this high energy particle (the interaction effects like radiation or excitations are also governed with some other equations, which have non trivial physical solutions). So we (may) study a free particle equations with physical mass at different $\vec{p}$. That is why in the propagators stays always the same physical mass.

The situation is slightly more complicated for dressed particles (infra-particles, for example), however in principle it is the same - our accuracy requirements are fulfilled for "free" equation solutions, so we content ourselves with these solutions. But one should never forget - the origin of them is our "full-fledged equations".

You may read my papers on arXiv about dressing examples and integration of short distance "physics".

answered Jun 29, 2015 by (102 points)
edited Sep 10, 2021

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