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What makes a system strongly coupled?

by physlad
Tags: coupled, makes, strongly
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physlad
#1
May31-09, 01:33 PM
P: 22
I was reading about AdS/CFT and its possible description of superconductors. It is said that superconductors may exhibit a strong coupling behavior. In general, AdS/CFT is thought to be a good tool to study strong coupling dynamics. Now, my question is what are the properties that define a strongly coupled system? In my mind QCD is strongly coupled because of the confinement of quarks and gluons. But it seems that this may not be the case. Can anybody explain?
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genneth
#2
Jun1-09, 01:01 AM
P: 980
Strong coupling is whenever a perturbative expansion fails to converge.
Parlyne
#3
Jun1-09, 07:17 AM
P: 546
Quote Quote by genneth View Post
Strong coupling is whenever a perturbative expansion fails to converge.
By this definition, all interacting quantum field theories would be strongly coupled, since the radius of convergence in coupling constant space is always strictly zero. (There is always at least one coupling constant which must have a positive real part.)

Strong coupling is the situation where the assumptions going into the perturbation expansion break down. This essentially means that the potential energy involved in the interaction should is not small compared to the kinetic plus rest energy.

Haelfix
#4
Jun1-09, 05:51 PM
Sci Advisor
P: 1,685
What makes a system strongly coupled?

Said another way, the dimensionless parameter (coupling constant or other) that you use to perturb around is of order 1. So for instance QCD's coupling (which is a function of energy) becomes of order 1 in the IR and you can no longer rely on perturbation theory.
genneth
#5
Jun2-09, 03:52 AM
P: 980
Quote Quote by Parlyne View Post
By this definition, all interacting quantum field theories would be strongly coupled, since the radius of convergence in coupling constant space is always strictly zero. (There is always at least one coupling constant which must have a positive real part.)

Strong coupling is the situation where the assumptions going into the perturbation expansion break down. This essentially means that the potential energy involved in the interaction should is not small compared to the kinetic plus rest energy.
Indeed. I should have been more careful and say that it's when the expansion fails to converge even asymptotically.
genneth
#6
Jun2-09, 03:56 AM
P: 980
Quote Quote by Haelfix View Post
Said another way, the dimensionless parameter (coupling constant or other) that you use to perturb around is of order 1. So for instance QCD's coupling (which is a function of energy) becomes of order 1 in the IR and you can no longer rely on perturbation theory.
But it's not quite that simple. For instance, in a crystal, atoms are pretty strongly coupled (strong used informally, as in the coupling constant is in some sense greater than 1 in the appropriate units), but with a suitable transformation, we can turn it into a (mathematically) weakly coupled theory of phonons.

I'm fairly sure that the colloquial usage is simply that we can't do perturbation theory. Bear in mind that lots of "non-interacting" theories also fall into this domain, such as non-linear sigma models.
Bob_for_short
#7
Jun2-09, 01:29 PM
P: 1,160
I think a strong coupling reduces to bound state appearance.

By the way, there are methods of non-linear series summation that permit to use the perturbative expansions (asymptotic, divergent, whatever) in practical calculations.

Bob_for short.


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