Should the quark propagator vanish because of confinement?

Nicolasrll
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Hi everyone! Something has been bothering me lately. Consider the quark propagator:

\langle 0|\psi_a(x)\psi_a(0) |0\rangle

For a given color a. Now let's say we insert 1 = \sum |n \rangle \langle n| between the two quark fields, where the sum is over a complete set of energy eigenstates. We then have a sum over a bunch of terms of the form \langle n|\psi_a(0)|0\rangle

Here then is what I'm worried about: confinement, as I understand it, tells us that all energy eigenstates must be color singlets. But unless I'm mistaken, \psi_a(0)|0\rangle is not at all a color singlet, and therefore I would expect all the terms of the form shown above to be zero. This, obviously, clashes hideously with the quark propagator used in perturbative QCD. So what am I missing?

Thank you for your insight!
 
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Confinement is more than just color neutrality or color singulet states. Color confinement means that in addition to color neutrality color charges cannot be separated from each other.

Color neutrality can be derived from the Gauss law constraint "algebraically", whereas confinement requires for a dynamical explanation, which is still missing.
 
Right, but I'm not sure how that resolves the issue. I mean, whether we call it color neutrality or color confinement, the end result seems to be that there are no colored eigenstates, so \langle n|\psi_a(0)|0\rangle would seem to vanish, along with the quark propagator. I don't think this can be right, so how do I get out this bind?
 
I am not sure how to derive the quark propagator in this picture. It depends on what |0> is, the (trival) Fock vacuum or the physical vacuum. In order to enforce color neutrality one has to restrict the physical states to the kernel of the Gauss law Ga(x)|phys> = 0; but this is a gauge-dependend constraint derived for A°=0. And of course the propagator itself is gauge-dependent, too.

Of course due to confinement the quark propagator (in a certain regime) needs to vanish somehow, but thus is a non-perturbative statement and cannot be derived from the perturbative vacuum.
 
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