106. Criteria for confinement. Wilson loop – getting more technical

Last time we have discussed a bit the behavior of the Wilson loop expected in the confinement and deconfinement phases and have concluded from simple physical considerations that the first one corresponds to the area law, while the second – to the perimeter law. Let us now show directly that the Wilson loop VEV satisfies the area law for a large rectangular contour. This derivation will allow us to get familiar with several interesting features of the Wilson loop variable.

For simplicity we will choose the gauge and the contour with the length along the spacelike direction and the length along the timelike direction. The Wilson loop VEV is given by

The vector potential operator is the matrix (in the fundamental representation of the group ), so is the Wilson loop operator

,

understood as the usual exponent of the matrix. By the way, it is clear why we have to introduce -ordering: the operator exponents taken at different points of the spacetime do not commute with each other, since is the non-abelian group.

Exercise. What is the value of the commutator of two such exponents taken at different points?

We can rewrite the trace above in the form

, (1)

where the operator

.

Next, we can relate the operator to with the help of evolution operator. We have

.

At the next step we insert the complete set of excited states between the operators
and to find

.

Since ,

we conclude that

, (2)

where is the matrix element

.

(Basically, we wrote the expansion for the propagator of the complex field over the complete set of eigenstates of the Hamiltonian .)

What do we learn from the exression (2)? There are actually several important lessons. First of all, let us consider a Euclidean version of the theory (that can be obtained by exchanging in formulae above). The expression for the VEV of the Wilson loop is given by

.

For very long contours with the only important contribution is given by the ground state:

,

where

,

i.e., trace of the VEV squared of the Wilson line, connecting heavy quark and antiquark. The vacuum energy is clearly given by the interquark potential
– recall that the quarks are infinitely heavy and therefore static, and the only contribution into the overall energy of the system comes from their potential energy. It is worth noting in this respect that we can only talk about the interquark potential, if the quarks are non-relativistic (even better – static).

If the potential is growing linearly, we find that the VEV of the Wilson loop satisfies the area law, famous criterion of confinement first proposed by Kenneth Wilson.

The second lesson is that area law does not hold for short contours, since excited states start to give larger contribution into (2). It is physically clear what happens at very small . The gluodynamics is an asymptotically free theory, and the effective coupling gets small for – the theory becomes effectively free or, as we often say, enters the Coulomb phase.

Exercise 1. Try to estimate the order of magnitude of the gap in the spectrum of the Hamiltonian .

The third, very important, lesson is that we don’t actually expect area law for the Wilson law if we take two infinitely heavy bosons charged under SU(N)!

Exercise 2. Think a bit why it is so. If you will be unable to figure it out, then I shall explain this in the one of the next posts. But you see… from my point of view deriving things yourself (or at least being able to derive them) is equivalent to having the best fun ever Theoretical physics (and QCD in particular!) may become a huge source of fun in this respect

And if you want a hint, consider representations for various spins and figure out what the trace really means in the Wilson loop.

To be continued.