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


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?

  • $\begingroup$ But ... why not? $\endgroup$ – Kostya May 8 '13 at 15:18
  • $\begingroup$ @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 May 8 '13 at 15:51

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...

  • $\begingroup$ 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 May 9 '13 at 5:10

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 $1$.

  • $\begingroup$ 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 May 9 '13 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 May 9 '13 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 May 9 '13 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 May 11 '13 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 May 11 '13 at 16:44

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