Suppose I have a very simple QCD diagram, where a quark emits a gluon and continues on. Can I describe such process through "usual" quantum mechanics? More specifically, can I use the concept of quantum state (colour state in this case) in any meaningful way?

E.g. we have the set of colours $r$, $g$ and $b$. Suppose the colour state of the initial quark was $|r\rangle$. As I understand, the quark after the gluon emission should change its colour state to either $|g\rangle$ or $|b\rangle$ (or it might remain in the state $|r\rangle$). Is that correct? And the gluon accordingly would be in a colour state $|r\bar{g}\rangle$ or $|r\bar{b}\rangle$ (or $|r\bar{r}\rangle$ or any of the two other "colourless" ones in case the quark remained the same). Wiki says that's not entirely valid and one should look for superposition states for gluons - is it relevant in this particular case? What exactly is affected by this discrepancy?

And how would the final state for the whole system look like? Initially we've had just $|r\rangle$, but in the end, I suppose, we'd have an entangled state of this quark and a gluon, right? So maybe it would look something like this: $\frac{1}{\sqrt{3}}|r\rangle\left(|r\bar{r}\rangle+|g\bar{g}\rangle+|b\bar{b}\rangle\right)+|g\rangle|r\bar{g}\rangle+|b\rangle|r\bar{b}\rangle$?

Is this correct?

Does any of this make sense at all?

  • $\begingroup$ A gluon has one of 3 colors and one of 3 anti-colors. So the total number of different gluons, 3 multiplied by 3 equals ... 8. $\endgroup$
    – safesphere
    Aug 14, 2017 at 22:59

1 Answer 1


Re: The wikipedia comment regarding superposition of colors.

A quark in a meson/baryon is not going to be |r>, it is going to be in superposition such that the combined (anti)quarks are in a color singlet.

As an analogy, consider the deuteron. It's S-state (symmetric) and spin-1 (symmetric), so to be overall antisymmetric, the isospin state is singlet: $\frac{1}{\sqrt{2}}(np - pn) $. What that means is that there not "a proton" and "a neutron" inside, rather, the nucleons are mixed, and if you address one of them, it's both proton and neutron--as a superposition.

Likewise with the red quark--it just can't be "just red".

As far as the rest of your question: consider the nucleon emitting a virtual pion, as in $p\rightarrow \pi^+n$. Does the $n$ part require $n\rightarrow \pi^-p$?

Go with whatever conserves charge (color) at the vertex.

  • $\begingroup$ thanks for this. It has struck me recently that the term "charge" for color is misleading, because charge is an immovable characteristic of the particle definition. the electron does not play ball with its charge. I am now looking for articles on "quark color charge and compositeness" $\endgroup$
    – anna v
    Aug 15, 2017 at 3:42
  • $\begingroup$ Thank you for the answer. And what if I'm dealing with "free" quark (under perturbative QCD conditions)? Would it make more sense to describe it the way I did? And sorry, I honestly did not understand your point with p and n decay. Could you please elaborate a bit more on that? $\endgroup$ Aug 15, 2017 at 6:17
  • $\begingroup$ That wasn't decay. A bound nucleon exchanging virtual pions will change it's charge in the same way a bound quark will change it's color. Sometimes understanding the effective field theory, quantum hadrodynamics, can help in making the jump to the full theory: QCD. $\endgroup$
    – JEB
    Aug 16, 2017 at 12:47

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