Does gluon oscillation violate color conservation?

[Zarathustra0]

Since the actual mass-eigenstate gluons are not the simple red-antired, red-antigreen, etc. but rather linear combinations thereof, is color charge still absolutely conserved? It seems that if we (perhaps naïvely) treat a gluon as simply fluctuating from one of the color-anticolor combinations of which it is a superposition to another, this would result in the possibility that, e.g., a red up quark and antigreen up antiquark could annihilate into a red-antigreen gluon, which could then oscillate into a green-antired gluon and produce a green quark and antired antiquark, violating color conservation.

If this approach of oscillating gluons is in fact naïve to the point of inaccuracy and we simply imagine this gluon as a superposition of red-antigreen and green-antired in equal proportions (1/√2), ignoring the collapse into one state or the other upon observation, we have a sort of colorless state, with equal parts red and antired and equal parts green and antigreen. Really, regardless of how the intermediate gluon itself is viewed, its being a superposition would seem to allow the transition of a red quark and antigreen antiquark to a green quark and antired antiquark.

 

[Vanadium 50]

There are no gluon color oscillations. A gluon is produced in one of the 8 SU(3) color combinations, and it stays there until it is destroyed.

 

[Zarathustra0]

This I understand, but what I'm wondering is what these superpositions mean in terms of actual interactions. An individual quark or antiquark has a definite color (correct?), so if they annihilate to a gluon, which of the eight gluons do they make? Since the color and anticolor of a gluon are probabilistic upon observation, would the color and anticolor of a quark-antiquark pair produced by this gluon also be probabilistic, and if so, wouldn't this allow for color nonconservation?

 

[Vanadium 50]

Just because something is not in an eigenstate does not mean it is not conserved. All three components of angular momentum are conserved, but you'll never see anything in an Lx, Ly, Lz eigenstate.

 

[tom.stoer]

There are several approaches which demonstrate that color is strictly conserved.

1) Looking at complicated Feynman diagrams one could introduce a cut at constant but arbitrary time. Analyzing the total color of all "particles" defined at this cut (no matter whether "real" or "virtual") it always corresponds to the color of the initial state; this is due to the color coupling rules at the individual vertices.

2) Formulating QCD in a Hilbert space framewerk with gauge fixing and therefore restricting to physical states one finds that these physical states are all color neutral (= in the singulet representation of SU(N)_color); in addition all physical operators (e.g. the Hamiltonian creating time evolution and the S-matrix) are gauge invariant and therefore commute with the gauge symmetry generators.