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  Termination of Diagrams in Chiral Perturbation Theory

+ 3 like - 0 dislike

In the Chiral perturbation theory scheme ( Scholarpedia article - http://www.scholarpedia.org/article/Chiral_perturbation_theory ), one does, for example, meson-meson scattering, limiting to LO, NLO or at most NNLO. The basic units in the Lagrangian (like a nonlinear Sigma model) are \[U = \exp (i {\vec \pi}\cdot {\vec \tau}/f),\] for instance for pions. However, 

Quoting verbatim from the book Quarks, baryons and chiral symmetry, by Hosaka, Toki: 

"Since there is no small coupling constant, calculation of various amplitudes would require all Feynman diagrams of all orders of perturbation theory, which would make practical calculations formidable. In fact, however, there are examples in which computation of finite numbers of diagrams provides good description of, for instance, meson-meson scatterings. The termination of the diagrams is not,however, justified from a theoretical ground."

If termination is not justified on theoretical grounds, why are calculations terminated to LO or NLO in articles using this method? What if the NNLO contribution comes out to be larger (at least in principle)? Won't we get a divergent series then, just like the usual case if we had calculated using quark fields instead of meson fields?

asked Apr 20, 2015 in Theoretical Physics by Anonymous_Curious (15 points) [ no revision ]

The quote from Hosaka and Toki sounds wrong (but maybe it was made in a slightly different context?). The correct answer is given in the scholarpedia article you link to: Chiral perturbation theory is an effective field theory, and at any order in the small expansion parameter $p^2/\Lambda_\chi^2$ there are a finite number of diagrams.

@Thomas - Thanks, but I guess what I was trying to ask here was - are we absolutely sure that other diagrams are not relevant here (especially since there doesn't seem to a rigorous convergence proof). We should not strike out possibilities unless we are totally convinced they won't make any appreciable difference. Thanks for your time, by the way :)

Yes, this is known as ``Weinberg's power counting theorem'', and described in Vol II of  Weinberg, and in many other text books (Donoghue, Golowich & Holstein, or Manohar & Wise, ...).

Thanks @Thomas for the inputs. (Sorry, I didn't login for a while.) :)

1 Answer

+ 3 like - 0 dislike

There is a hidden dependence on $\hbar$, which is usually set to 1 for computational simplicity. But if the right factors are reinstalled, the loop expansion is an expansion in $\hbar$. Thus one may regard $\hbar$ as the small parameter, though it is not really small in the regime considered. But one also uses large $N$ expansions although $N=3$ is not really large - and it works to some extent. 

In general, one just does the calculations that are feasible and hopes for the best - the quality is decided by the interpretability of the outcome and by comparing with experiment. In quantum field theory, a more sensible error analysis is possible only in simple cases such as QED.

In case of the loop expansion, one finds that. in practice, a low order expansion works reasonably well for many problems, though (except in toy problems such as the anharmonic oscillator) there is no proof of this.

answered Apr 20, 2015 by Arnold Neumaier (15,787 points) [ revision history ]
edited Apr 20, 2015 by Arnold Neumaier
Most voted comments show all comments

I am not sure (to say the least) that one can say "$\hbar$ is small'. One has to compare dimensionless quantities, for example, the de Broglie wavelength and the characteristic length of the potential variation, so the velocity (or energy) is involved too.

@VladimirKalitvianski: I said that it is not really small, and that the argument proves nothing in a rigorous sense. But one can expand in $\hbar$ and hope that (often after some resummation, or through variational perturbation theory) the first neglected term is small compared to the terms kept - this quotient is always dimensionless. If the system is not too far from classicality (which corresponds to $\hbar=0$), the procedure is reasonable.

The "distance" from classicality is the ratio $\lambda/L$ mentioned above.

@Anonymous_Curious: You have the same problem as expanding a function in Taylor series. In the absence of rigor about convergence, you just compare numerically successive terms in the absolute values. If they are decreasing, you may estimate what precision you are getting. Physically, if the initial approximation captures well the physics, the higher orders only modify it numerically. If not, the higher orders may indeed bring "new physics".

By "terminating diagrams" you mean truncating the series, don't you?

Note that this is not just reproduction. If you reproduced a few numbers correctly by some approximation scheme you may be reasonably confident that the other stuff you predict with the same approximation scheme at similar energies will be of similar accuracy. 

It is similar to the situation in QED. Here theory alone is unpredictive, as one needs two numbers $e$ and $m$ from experiment. Once these are known, everything else can be predicted well. 

In applications where the approximations are less favorable than in QED one needs more experimental data that free constants in the theory. But suppose the theory has only 2 constants and you can choose these to reproduce 5 experimental values with 5% accuracy, your scheme is likely to be accurate to about 5% even for the observables where no experimental values are available, as long as the orders of magnitude of the computations remain the same. 

Most recent comments show all comments

Also, a second issue is - since you assert that that NNLO is smaller than NLO can not be rigorously proved in general for ChiPT, how does one know that NNLO is only adding more decimal places to a NLO result which reproduces experimental numbers. What if NNLO coughs up some interesting Physics which we disregard with no justification?

Thanks in advance (and sorry, I am unable to vote up your answer - probably because of insufficient score points.)

Thank you Arnold, and @VladimirKalitvianski for the valuable insights. Many thanks :)

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