What potential between charges leads to confinement

Question

In (2+1)-d, instanton effect leads to a linear potential between charges. If we have two particles with opposite charges in this case, since linear potential diverges when the distance between the two charges goes to infinity, it is energetically favorable to produce two more charged particles to neutralize the original two particles, instead of letting them separate as far as possible. This is basically the argument on confinement of (2+1)-d EM field.

It looks the essential element in this argument is the presence of a potential between charged particles which diverges when the distance goes to infinity, so does logarithmic potential lead to confinement? If so, we know that in (2+1)-d charges interact via a logarithmic potential, then why cannot we conclude that (2+1)-d EM field is confined without considering instanton effect?

I think the answer is no, so how strong should the potential be in order to have confinement?

Answer 1

It's not in 2+1d linear EM field that there are instantons, rather in 2+1d nonlinear gauge theory. The 2+1 d free EM field is not confining, but has logarithmically rising potential, and a force that falls off as 1/r. The 1+1 d linear EM field is confining, this is the Schwinger model, but this is not a surprise as in 1d, the potential is automatically linear.

The logarithmic potential in 2+1d does mean that particles can't separate, but it is not the technical definition of a confining potential, which is that the particles should have a behavior similar to that of the 1+1d Schwinger model, that they are linked by a string with tension.

That's not a full answer, you presumably want to know what the spectrum of 2+1d QED looks like, is it all logarithmically confined positron states. I don't know the detailed spectrum of this theory, but it can be calculated, so it's a good question, but I think the question should be worded more clearly, so that it doesn't seem to be talking about instanton effects in QED. But +1, because I know what you mean.