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In on-shell (OS) 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?

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  • $\begingroup$ @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" $\endgroup$
    – Jia Yiyang
    Commented May 8, 2013 at 15:51
  • $\begingroup$ Is the question "why does it behave as a free field propagator?" or "why is it important that it should behave as a free field propagator?" $\endgroup$
    – MannyC
    Commented May 3, 2020 at 22:07

3 Answers 3

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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...

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  • $\begingroup$ 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. $\endgroup$
    – Jia Yiyang
    Commented May 9, 2013 at 5:10
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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 $1$.

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  • $\begingroup$ 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. $\endgroup$
    – Jia Yiyang
    Commented May 9, 2013 at 5:16
  • $\begingroup$ 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. $\endgroup$
    – Siva
    Commented May 9, 2013 at 8:36
  • $\begingroup$ That's a nice answer. Related, I think, to the need to add factors of $\sqrt{R}$ to external lines in other schemes. $\endgroup$
    – innisfree
    Commented May 9, 2013 at 9:11
  • $\begingroup$ @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? $\endgroup$
    – Jia Yiyang
    Commented May 11, 2013 at 12:26
  • $\begingroup$ 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 $X$. $\endgroup$
    – Siva
    Commented May 11, 2013 at 16:44
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  1. It should be stressed that the fact that the propagator has a pole at the physical mass $k^2=-m_{\rm ph}^2$ is the very definition of the physical mass, i.e. it does not depend on the renormalization scheme, cf. e.g. Ref. 1. The on-shell (OS) renormalization scheme equates the renormalized mass $m\equiv m_r$ and the physical mass $m_{\rm ph}$.

  2. The unit residue condition (in the $k^2$ variable) is strictly speaking just a natural/convenient convention in the OS renormalization scheme. It is in general incompatible with other renormalization schemes, such as, e.g. the $\overline{MS}$ renormalization scheme.

References:

  1. M. Srednicki, QFT, 2007; eq. (27.7). A prepublication draft PDF file is available here.
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