4
$\begingroup$

Peskin & Schröder say on page 216:

The poles in $p^2$ come only from one-particle intermediate states, while multiparticle intermediate states give weaker branch cut singularities.

In order to figure out what "branch cut singularities" are in this context, I have come across many terms like "branch poins", "branch cut" or "branch point which is a singularity", but I am still confused about what a branch cut singularity is.

I know what a branch cut is - a curve that you can not perform a integral around - but I don't understand how a branch cut could be a singularity. Furthermore, why should we examine its properties in a physical context? What is the physical meaning of such a branch cut singularity?

$\endgroup$
  • 1
    $\begingroup$ All these terms are exactly the same thing. en.wikipedia.org/wiki/Branch_point - The branch point or branch point singularity is at the point of the multi-valued function where the arbitrarily small loop monodromy is nontrivial. Because of the multi-valuedness, there has to be a branch cut (=branch cut singularity) coming from the branch point on which the function jumps from one value to another. Around branch points, the function is continuous but it's still "singular" according to physics jargon - it can't be Taylor-expanded there, for example. $\endgroup$ – Luboš Motl Dec 24 '16 at 9:44
  • 2
    $\begingroup$ In physics perspective, branch points are singular - not regular - because they're special. One can't consider the patches around branch points to be smoothly differentiable patches of manifolds. $\endgroup$ – Luboš Motl Dec 24 '16 at 9:46
  • $\begingroup$ @LubošMotl Thanks a lot for your explanation! By saying that Taylor-expansion can not be performed there, it is now, practically, much clear to me in understanding the singular property of branch cuts. Still for another question, if I may ask, which role does the discontinuity(value) of the branch cut play in QFT? $\endgroup$ – Patrick Dec 24 '16 at 11:17
  • $\begingroup$ @LubošMotl That comment (possibly slightly expanded) would make a perfectly fine answer ;) $\endgroup$ – ACuriousMind Dec 24 '16 at 12:38
  • $\begingroup$ Tx, aCuriousMind, but "it's a synonym" isn't really what I consider a full-fledged answer. ;-) @Patrick - the location of the branch cut is a matter of convention in general but we normally place it on the real axis of $p^2$ etc. The discontinuity of various Green's functions is equal to expressions proportional to on-shell propagators of some intermediate particles. Search for "Cutkosky's cutting rules". It's in textbooks or e.g. here cds.cern.ch/record/334567/files/9709423.pdf - but I can't guarantee it's the best presentation of them. $\endgroup$ – Luboš Motl Dec 24 '16 at 13:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.