122. Where are quarks in the Wilson loop?

An anonymous reader from Spain asks in comments to my “Wilson loop – physical introduction” post:

Why do you interpret a mathematical expresion that displays a gluon field (Amu) as a qqbar loop? Where are the q fields? Why they don’t appear in the Wilson loop but you still interpret they’re there? And where did the gluon go?

I think, these questions are such that the answer to them deserves a separate post, so thanks a lot for asking!

1. Small introduction: S-matrix

Let me start from a somewhat different topic, since my thought went along the Polyakov line first in this direction (but you will soon see the connection of this staff to the Polyakov loop).

When we discuss high energy physics and its description in terms of a quantum field theory, we are usually interested to know vacuum-vacuum matrix elements of some operators corresponding to observables:

.

Why vacuum S-matrix elements? The reason is that typically actual interaction between particles happens in the very small volume (at small length scale) compared to the volume of the system or distance that particle passed before the collision. Outside interaction between particles can be neglected, and we can consider it as switching on adiabatically while the particle approaches the region of collision and than switching it off adiabatically while the particle leaves this region. Due to the procedure of adiabatic switching the quantum state does not change: if it was vacuum in the beginning, at , it will be vacuum in the end, at . But where is the effect of interaction? The answer is that due to the interaction process the vacuum has acquired the phase, and this phase is directly related to S-matrix of the interacting theory. In a sense, what we study in QFT is behaviour of these phases.

2. Where are the q fields? Why they don’t appear in the Wilson loop but you still interpret they’re there?

The q fields are in the expression for the Wilson loop, although nobody likes to write them. What we are actually calculating is the vacuum expectation value of the operator

.

If you are taking trace of the VEV of this operator, you can put to the very right using the cycling property of the trace. Therefore, what you have on the right is the operator creating the quark from vacuum and then annihilating it.

Another reason why nobody likes to write quarks in the Wilson loop is that they are infinitely heavy and therefore can be considered (quasi)classical – they do not influence any physics except the only one (albeit important) thing: their wavefunction may acquire phase in the external gluon field.

This is what you have in electrodynamics, too – recall the Aharonov-Bohm effect. Consider a heavy electron in the external field. The electron is quasiclassical, that is, its wavefunction

is a very rapidly oscillating functional of the electron’s trajectory, such that only classical trajectory gives contribution into the functional integral describing the amplitude for the electron to propagate from the point to the point .

However, there is essentially quantum effect present even for this heavy quasiclassical electron – if you put it into the external field, its wavefunction will acquire a phase depending on values of the vector potential along the electron’s trajectory. The additional phase factor will be

,

i.e., exactly the one we have in the definition of the Wilson loop. The only difference with electrodynamics is that now the theory is non-abelian and we want to construct a colorless operator, that is why the trace over color indeces appears.

3. And where did the gluon go?

You are calculating the vacuum expectation value, so no gluons. There are vacuum fluctuations of the gluon field however – and vacuum responds to the introduction of the source (heavy quark ).

By the way, actually the presence of quark operators in the definition of the Wilson loop is directly related to the solution of the Exercise 2 in the technical part of the Wilson loop discussion. Would you want to try and guess what happens if instead of a spin 1/2 particle one will take spin 1 particle?

I hope it is a bit clearer now. If not – you are very welcome to keep asking questions

Update: And there is actually a place where the presence of a quark operator is explained nicely – in the book by Smilga I have mentioned (you need the last section on quark confinement). It is also sanely explained in the Polyakov’s book, but the explanation is somewhat technical. One could also find there the solution for the Exercise 2 mentioned above if one feels a bit lazy, but it would be much nicer to figure it out yourself – the derivation did give me a couple of pleasant moments