# What is the significance of the branch cut in renormalization group logarithms?

What is the physical significance of the branch cut in renormalization group logarithms?

(Is this just an avatar of the optical theorem, or is there something to be understood about these logarithms from the point of view of taking the scale as a complex parameter? Usually non-analyticity is a signal of something of physical importance, right?)

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Please make the question more precise/concrete. There are many things to be said about RG logarithms, so it would be nice to know what exactly you are interested in. This topic ranges from the whole issue of analytic structure in QFT to the technical re-sumamtion of large logs via the RG trick... –  Zohar Ko Nov 4 '11 at 22:00
Hi Zohar, thanks. I'm primarily interested in the analytic structure of QFT---is there any significance to the branch cut as a function of the renormalization scale mu, or is this the same thing as the optical theorem telling me that the discontinuity as a function of the energy scale of the process has to do with on-shell virtual particles? (Sorry, I'm new to stackexchagne---should I also modify the original question above?) –  Flip Nov 5 '11 at 2:25
Well, a good question. I would like to see some more complete answer as well. The branch cut from $\ln(E)$ surely means that singular physics we understand takes place at $E=0$ as well as $E=\infty$ and you're not allowed to circumvent these points. It's somewhat questionable whether all the functions with $\ln(E)$ have natural extrapolations to the complex plane. Also, note that a log may be written as an integral of $1/(E-E_0)$ so the log branch cut comes from a continuous union of many ordinary first-order poles. In $N=2$ SUSY QFTs (Seiberg-Witten etc.), there are many sharper facts on logs –  Luboš Motl Nov 5 '11 at 15:41

So if one promotes all momenta in a scattering process to complex variables, then the complex function one obtains is not single valued, in other words, the amplitude lives in a multi-cover of some $C^n$. All the branch cuts must live on the real axis. The significance of the coefficients of the brach cuts is that some "virtual particles can become on shell," or more abstractly, that some subset of the external legs can create an on-shell state with the prescribed kinematics. (These are the famous unitarity cuts.) There could be poles on the real axis too, but these should be thought of as a special case of a branch cut, one where the discontinuity is a delta function. This means that the state in the Hilbert space is “isolated” and it is thus interpreted as a bound state.
RG is quite a different set of ideas altogether. Perhaps it is best to illustrate why we need the equations of RG by an example. Consider some two point function of currents in a QCD-like theory. One removes the divergences with dimensional regularization and so the answer should be $\sim g^2$, where $g$ is the gauge coupling. However, since there are virtual particles in the loop this cannot be the right answer because there must be a branch cut. And indeed, one finds $g^2\log(p^2/\mu^2)$ where $\mu$ is an arbitrary fixed scale and $p$ is the momentum. The branch cut is independent of $\mu$ because the imaginary part must be finite, scheme independent. A result like $g^2\log(p^2/\mu^2)$ is highly unsatisfying practically because the expansion in $g$ breaks down at large enough momentum. This is where the RG equations come in. Basically to understand the large $p^2$ regime (which is ought to be weakly coupled in AF theories!!) one needs to re-sum perturbation theory, but one needs to worry only about the leading large logarithms which endanger the $g$ expansion. One finds that the answer can be written as $g^2(p^2)$ and $g^2(p^2)$ satisfies the RG equation, which, in QCD, guarantees that, at large $p$, $g^2(p^2)$ is small. So perturbation theory is saved. Another way to think about RG is that there is a family of possible regularizations of the theory, parameterized by some scale $\mu$, and we can compute any process we want with any choice of $\mu$ , these are all equivalent, but to change $\mu$ to $\mu'$ one needs to redefine the coupling constants. And indeed, to avoid large logarithms it is best to choose $\mu$ around the momentum scale. This is why $g^2(p^2)$ effectively becomes $g^2(\mu)$.