I'm a mathematician trying to get some very basic physical intuition on gauge theories, so I apologize if what follows is really naive. My first super elementary question is:
Am I right to think that in magneto-static, the Wilson loop for a
simple curve is nothing but the exponential of the magnetic flux
through any surface bounded by that curve ?
This seems obvious but I've never seen it explicitly written so I fear I'm missing something obvious. Now of course this is special to electromagnetism because the gauge group is abelian (hence holonomy of the one form $A$ describing the magnetic potential is exp of the integral of $A$ along the curve, which by Stokes theorem is the integral of $dA$ over any surface bounded by the curve), and as far as I understand classical non-abelian Yang-Mills theories do not quite describe any "real" physics, still:
Do classical Wilson loops have some sort of physical meaning?
Wilson loops are still well-defined for self-intersecting loops. But if I understand correctly, when going to the quantum picture using the Feynman integral formulation there is some "regularization" issue, and one way to deal with it is to equip the curve at hand with a framing, which in particular forces to consider only simple loops. This is beautifully explained in Witten's Jones polynomial paper, and in a way is the very reason why Chern-Simons knot invariants are non-trivial (which as you may have guessed is the reason why I'm asking those questions).
Is there some intuitive/layman physical explanation as to why quantum Wilson loops are only well defined on simple curves? I would already be interested in an answer in the abelian case, perhaps something related to the fact that in that case it's only for simple curves that the classical ones are related to the magnetic flux?
Edit I realize my question was unclear, mostly because I don't really know how to use the jargon accurately. By "quantum Wilson loop" I really meant its vacuum expectation value. Is there a simple explanation of what this quantity means ? Now this has, I believe, a divergence when you take the limit as the size of the loop goes to 0. For the classical Wilson loop for a given $A$, this basically gives the field strength at the point which is the limit of the loop. As far as I understand, the reason why the VEV is divergent in that situation, is that it doesn't quite make sense physically to measure the field at one point. Rather, all we can do is measure the flux going through a very small loop near that point, and somehow quantum physics is sensitive to that. Does that make sense ? Is there a similar explanation as to why the VEV of a WIlson loop diverge for self-intersecting loops ?
Best Answer
I'm a physicist in training who doesn't know any math, so hopefully I'll have good naive answers for your naive questions.
Yes, for a purely spatial Wilson loop that's perfectly correct. For a temporal Wilson loop, such as one lying in the $zt$ plane, you instead get $$\int E(z, t) \, dz dt$$ which, roughly speaking, integrates the potential difference between the two temporal sides of the loop over time.
The quantum Wilson loop yields the phase a quantum particle picks up upon going around the loop, but the quantum phase is nothing but the classical on-shell action. So the spatial Wilson loop tells you the extra action cost for a spatial path when there's a magnetic field around. It's trickier to talk about a temporal Wilson loop since a single classical particle can't go back in time, but you can use it to compare the actions of two classical particles that start and end in the same place.
Upon quantization, the spatial case gives you the Aharanov-Bohm phase, while the temporal case simply gives you the dynamical $e^{-iEt/\hbar}$ phase where the energy $E$ includes the electric potential.
If you're asking whether you can understand this classically, I think the answer is no. Witten states that this restriction has to do with regularization of inherently quantum effects. I don't think the nonapplicability of Stokes' theorem really has anything to do with it, as you can still apply it if you split the curve into simple pieces.