Does Yang-Mills have free gauge bosons? Is there any physical problem with a propagating non-Abelian gauge boson? That is, a plane wave mode $A_\alpha^\mu e^{ikx}$, where $A_\alpha^\mu=b_\alpha\epsilon^\mu$, with $b_\alpha$ a constant adjoint-representation gauge vector, and $\epsilon^\mu$ a polarization vector orthogonal to $k^\mu$?
We usually say that gluons are confined due to their strong interactions. But it seems like a single boson wouldn't trigger any interactions at all. The YM Lagrangian is $-\frac{1}{4}F_\alpha^{\mu\nu}F_{\alpha\mu\nu}$, where $F_{\alpha\mu\nu}=\partial_\mu A_{\alpha\nu}-\partial_\nu A_{\alpha\mu}+C_{\alpha\beta\gamma}A_{\beta\mu}A_{\gamma\nu}$. As long as all of the spacetime components of  $A_\alpha^\mu$ are parallel to a single adjoint-representation gauge vector, the $C_{\alpha\beta\gamma}A_{\beta\mu}A_{\gamma\nu}$ term vanishes, and the field equation reduces to the wave equation, which is satisfied.
In fact it seems that such a boson would not even emit virtual bosons, so it wouldn't generate a "Coulomb" field. Two wavepackets of the above form could pass near each other without feeling any attraction or repulsion. And if the two of them shared the same $b_\alpha$, then they could even pass through each other with no scattering, just as photons do.
This seems to conflict with the common statement that non-Abelian gauge bosons are "charged". But if we interpret "charge" to mean the timelike components of the conserved currents $J_\alpha^\mu$, then we find that the currents in pure YM are $J_\alpha^\mu=C_{\alpha\beta\gamma}F_\beta^{\mu\nu}A_{\gamma\nu}$. So once again, as long as everything is parallel to a constant gauge vector $b_\alpha$, there is no charge and no current. When we say gauge bosons are "charged", we probably need to clarify that we just mean "transform nontrivially under the gauge group", not that they behave like charged particles.
Once we add in quarks, then all bets are off: quark pair production will happen at a fast rate, leading to more and more interactions. I assume this is why we don't see unconfined gluons in real life. But am I correct that they do exist in YM, or am I making a basic error?
 A: Yes, classical plane wave solutions do exist for non-Abelian gauge theory. They were studied by Sidney Coleman; my example is equivalent to one of his under a gauge transformation. They are sometimes used as a "classical background" in research on the strong-field regime of QCD (but I don't know how the issue of quark pair production plays out). Coleman discusses plane waves, i.e., those with a single direction of propagation, whose field strength is constant in the orthogonal directions. But I see no problem with building a finite wavepacket, as long as we keep everything parallel to a single adjoint-representation vector $b_\alpha$.
As for whether gauge bosons are "charged", it's true that in these plane-wave solutions the currents, as classical observables, are zero. Their quantum expectation values also vanish (at least for one-particle states). But the current operators do have nonzero matrix elements between different one-particle states of this form. [EDIT: Of course, all talk of matrix elements for non-gauge-invariant operators such as the currents requires some gauge-fixing procedure. I think that a physical gauge choice, such as axial gauge $A^3_\alpha=0$, is simplest for my purpose here.]
So the most we can say, in the quantum theory, is that we have found a basis of one-particle states in which the currents have zero expectation. The existence of such a basis should not be surprising; we could always create a basis with this property, if desired, by taking superpositions of states with opposite charges. [EDIT: Actually, the fact that the expectation values for all of the  $N^2-1$ charges vanish in the same basis does seem fairly surprising. My point was just that using this to call the bosons "uncharged" would be highly unjustified.]
The statement that gauge bosons are charged is fully justified by looking at each charge operator in its diagonal basis (that is, diagonal when restricted to the one-particle subspace). The matrices $A^\mu_\alpha$ in those basis states will be of rank two or more, meaning they will have different (probably orthogonal) adjoint-representation vectors for the different spatial directions, giving a nonzero current.  In particular, the W bosons, which directly flip the weak isospin of fermions, do obviously have a nonzero value for one of the $SU(2)$ charges.
