# How to avoid the ordinary Coulomb solution in QCD?

To see where QCD starts to differ from the behavior of EM fields, we might begin by looking at the classical field. A search brings up [question 339978] and [question 360061] but no answer is found there. With the usual [QCD Lagrangian] definition: \begin{align} \mathcal L_{\text{QCD}} &= \bar\psi(i{\large \not}D-m)\psi -{ \frac{_1}{^4}} G^a_{\mu\nu}G_a^{\mu\nu} \\[6pt] D_\mu &= \partial_\mu - i g\ T_a A_\mu^a \\[6pt] G^a_{\mu\nu} &= \partial_\mu A_\nu^a - \partial_\nu A_\mu^a + g f_{abc} A_\mu^b A_\nu^c \end{align} we can use Euler-Lagrange, $${\small \partial_\mu}\frac{{\large \delta} \mathcal L}{{\large \delta}\partial_\mu A_\nu^a} = \frac{{\large \delta} \mathcal L}{{\large \delta} A_\nu^a}$$, to find field equations: $$\partial_\mu G^{\mu\nu}_a = J^\nu_a + g f_{abc} G^{\nu\rho}_b A_\rho^c$$ where $$J$$ is the source current from the matter field only (the second term shows how the field also acts as its own source). My question is: what goes wrong if I assume that the matter field gives us only static color charge density of one kind, say $$a=1$$? So the only component of the source current is $$J^0_1$$. It seems that we then get exactly the Coulomb solution for the color-electric field $$G_1^{\,0i}$$ with $$i=1..3$$, and we can even express it simply using an electrostatic potential: $$V \equiv A_0^1, \quad\text{with}\ \ \nabla^2 V = -J^0_1, \ \ E_i^1 \equiv G_{i0}^1= -\nabla_i V, \quad\text{etc.}$$

As long as there is only a component of $$A$$ for one value of the $$SU(3)$$ index, the terms $$f_{abc} A_\mu^b A_\nu^c$$ will always be $$0$$ because the structure constants $$f_{abc}$$ are antisymmetric. So we get ordinary electrostatics back and there is no confinement! (Likewise, if we would make it non-static, we would get electrodynamics back if there are only the components $$J^\mu_1$$.)

Is this perhaps impossible because the fermion field can never create a charge density $$\bar\psi \,T_a\,\psi$$ for only one value of $$a$$? If $$\psi$$ is in the $${\bf 3}$$ representation of $$SU(3)$$ that seems plausible, since it has only 3 (complex) degrees of freedom and can never create any arbitrary point in the 8-dimensional space of $$SU(3)$$ generators. But what about the $${\bf 8}$$ or the other higher representations? Or is there something else we have to change? Do we have to go to QM? The question then seems to become: what is the simplest (but not simpler) analysis that we can use to see confinement.

• Confinement is not a classical property. The simple analysis you can do to see confinement is in the 82nd section of web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf (Srednicki book). Commented Jul 14 at 15:48
• What makes QCD very different from QED is the fact that, while photons are charge neutral, the gluons carry color charge and therefore interact with each other. As a result one would not get a Coulomb potential with gluons. Commented Jul 15 at 4:06
• @flippiefanus With only the $a=1$ component the nonlinear terms would vanish, as the example tries to explain. So does that mean there cannot be an "$a=1$ only" case, or do we need some additional reasoning to show that there always is confinement? That was the question... Commented Jul 15 at 8:30
• When you set $a=1$ for a quark, the interactions among the gluons will generate the other colors. If you force those to be $a=1$ as well, you get something that does not make sense. Not sure what you can learn from such an exercise. Commented Jul 16 at 3:51
• Getting "$a=1$ only" current from the quark field $\psi$ is not even possible, at least 2 of the Gell-Mann matrices will give non-zero result, whatever you choose. And then the nonlinear field equations will indeed add at least a 3rd component. That's all well and good, but is that enough to prove confinement? Or do we need another essential step? And can the current come from another kind of field which does allow an "$a=1$ only" case? And would that not give confinement?! All these these questions seem to be mathematically traceable, and interesting. Do you know the answers? Commented Jul 16 at 6:32

• Yes any charge or current will change under SU(3) gauge transform, that's why it's color charge. This actually helps (thanks for the tip!) to see which current can be realized by the matter field (just one case to be analyzed, then transformed to any possible other case). But doing so only confirms that a pure $a=1$ current is not possible, we always end up in at least an $SU(2)$ generated subspace. So then the classical non-linear terms could still give confinement (for sufficiently large $g$ of course). How can we see that they don't do that?! Commented Jul 15 at 8:26