It is a well known fact that the location of the pole of a propagator (in QFT) can be interpreted as the physical mass.

Is there an interpretation for the residue of the propagator?

Note: I´m thinking of generalised propagator, not necessarily a propagator of a fundamental field.

  • $\begingroup$ +1, interesting question. The residue theorem is a special type of Stokes theorem, are you sure that the pole is interpreted as the mass and not the residue? The pole is an infinity, but the residue gives the pole a value (for integrable poles), or any closed closed loop the value of the pole. Intuitively it would seem to give a sort of measure of the mass? I'm not an expert in QFT or QM however... $\endgroup$
    – daaxix
    Dec 31, 2012 at 22:51
  • $\begingroup$ @daaxix it's the location of the pole, not the value of the pole, that gives the mass of the particle. $\endgroup$
    – David Z
    Dec 31, 2012 at 23:27
  • $\begingroup$ @DavidZaslavsky, ah, ok. Well the residue gives the value of the pole itself. Is there a difference between integrable and non-integrable poles in QFT? Are they all integrable? $\endgroup$
    – daaxix
    Dec 31, 2012 at 23:32
  • $\begingroup$ There are sometimes branch cuts in propagators, but I couldn't tell you that much about them offhand. Other than that, I can't think of any examples of nonintegrable poles. Generally, a propagator is of the form $\frac{\text{stuff}}{p^2 - m^2}$, and the residue depends on the stuff in the numerator, so it doesn't necessarily correspond to a particular physical quantity that I know of. $\endgroup$
    – David Z
    Jan 1, 2013 at 0:08
  • $\begingroup$ Related: physics.stackexchange.com/q/63809/2451 $\endgroup$
    – Qmechanic
    Oct 18, 2019 at 9:31

1 Answer 1


As you say, the location of the pole gives the mass. Higher-order diagrams shift the location of the pole, causing a (often infinite!) renormalization of the mass.

The residue at the pole location simply gives the normalization of the wave function, which, as far as I remember, is just absorbed into scale of the field.

  • $\begingroup$ Thanks Tom. "Happy New year" to you too. But, please don't post these in answers. You could leave a comment below your own post though you can't leave comments in others' post. BTW, please don't leave your user signature. Your user card has your name, rep. score, badges, etc. :-) $\endgroup$ Jan 1, 2013 at 5:13
  • $\begingroup$ Thank you Tom. Thus, just for clarification, the shift of the location of the pole renormalise the mass, whilst the shift to the residue is reabsorbed in the renormalisation of the field. Cheers, and Happy new year. $\endgroup$
    – Dox
    Jan 2, 2013 at 9:31

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