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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?

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    $\begingroup$ 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
    Commented Dec 24, 2016 at 9:44
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    $\begingroup$ 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
    Commented Dec 24, 2016 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
    Commented Dec 24, 2016 at 11:17
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    $\begingroup$ @LubošMotl That comment (possibly slightly expanded) would make a perfectly fine answer ;) $\endgroup$
    – ACuriousMind ♦
    Commented Dec 24, 2016 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
    Commented Dec 24, 2016 at 13:46

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In this context, I think Peskin and Schroeder are just commenting on the fact that the multi-particle states contribute a continuum of propagators to the Fourier Transform of the interaction two-point function. This can be seen in eq (7.9) of Peskin and Schroeder:

$\int d^{4}x\, e^{ipx}\,\langle \Omega | T \phi(x)\phi(0)|\rangle$ $= \frac{iZ}{p^{2}-m^{2}+i\epsilon} + \int_{\sim 4m^{2}}^{\infty}\frac{dM^{2}}{2\pi}\rho(M^{2})\frac{i}{p^{2}-M^{2}+i\epsilon}$

where $m$ is the particle rest mass. Since this contribution is singular for all $p^{2} = M^{2} > 4m^{2}$ on the real-axis, it demands a branch-cut for any potential contour integrals of this quantity. I believe this is what they mean by "branch-cut singularity".

This differs from the contribution of the single-particle / bound states, which contribute isolated poles.

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