5
$\begingroup$

I was studying about quarkonia systems and reached this page at CERN Courier. Here, I came across the following text:

While the failure to reproduce an experimental observable that is perturbatively calculable in the electroweak or strong sector would be interpreted as a sign of new physics, phenomena requiring a non-perturbative treatment – such as those related to the long-distance regime of the strong force – are less likely to trigger an immediate reaction.

What does this exactly mean? Can somebody roughly sum it up, using some example, if possible?

$\endgroup$
5
$\begingroup$

Simply speaking because non-perturbative QCD is affected by far larger systematic errors than perturbative QCD. The only way to handle non-perturbative QCD from first principles is lattice QCD but only a small fractions of observables can be computed. All other computational methods are based on some semi-phenomenological methods, chiral perturbation theory, heavy quark effective theory, non-relativistic QCD, and 1/N expansion being the main ones, which all have their share of uncertainties. For example, in chiral perturbation theory, at each order of perturbation, there is a coefficient that must be determined from experiments (simplifying a lot).

An interesting example relevant to the discussion is the study of CP violation in B meson decays. One is interested in extracting the parameters of the Standard Model responsible for mixing quark flavours (they are gathered in a matrix called the CKM matrix). In practice, one seek to measure a phase angle originating in the coefficients of this matrix. But QCD can contribute an extra phase, and some of that is non-perturbative. As a result, the "golden" channels are those where QCD does not contribute any phase. For example, $b\to c\bar{c}s$. So if one would see a strong deviation from the Standard Model in this golden channel, people would be very confident that there is new physics. But on the contrary, a deviation in a channel with a sizeable QCD contribution, such as $b\to c\bar{c}d$, would raise less of an eyebrow.

$\endgroup$
  • $\begingroup$ Why are lattice computations not feasible? They were using such methods 30 years ago and today they have computers that are enormously more powerful than what they had then. Also there has been some progress in methods such as finite size scaling allowing one to extrapolate from finite lattice results. $\endgroup$ – Count Iblis Jul 1 '17 at 18:57
  • $\begingroup$ There are various computational hurdles that can't be easily dwarfed as I understand it. The biggest problem is the evaluation of the so-called fermion determinant. Most of the flops comes from inverting a matrix using iterative methods and the computing cost is then proportional to the condition number, which is horrible because the smallest eigenvalue is the mass $m$ of the quark, i.e. it scales as $1/m$. $\endgroup$ – user154997 Jul 1 '17 at 19:09
  • $\begingroup$ @CountIblis Exponential scalings can do wonders in making huge increases in computational power mean very little. I'm unsure how it applies to QCD in practice, but it can make life very hard for cold-atoms-on-a-lattice simulations if you have strong interactions. $\endgroup$ – Emilio Pisanty Jul 1 '17 at 21:00
  • $\begingroup$ @EmilioPisanty There is that, indirectly. If $N$ is the number of lattice steps in each dimension, then a naive computation of the fermion determinant I talked about scales as $N^{12}$. A cost to be paid at each iteration of a Monte-Carlo integration. But a clever workaround avoid this cost, leading to the point I made above. $\endgroup$ – user154997 Jul 1 '17 at 22:17
4
$\begingroup$

My interpretation of the text - which is all any individual other than the author may offer - is that it rightly states that an observation not in alignment with perturbative calculations would be suggestive of new physics.

For example, in quantum field theory, in perturbation theory, one thing we can compute is the branching ratio for a process, which roughly means what proportion is to occur relative to other processes that produce the same final state. If we did this calculation and found in experiment radically different branching ratios, it would hint that the theory is missing something.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.