If I had 2 pions that were identical, except one was comprised of a red and anti-red, and the other was comprised of a green and anti-green, would I be able to perform an experiment that distinguishes between them?


2 Answers 2


Color charge in the sense of "being blue, red, green" is not a quantum mechanical observable because the $\mathrm{SU}(3)$ gauge transformations mix the colors. This means it is meaningless to say "We have a blue particle", because we can perform a gauge transformation and then we "have a red particle". Since physical descriptions related by gauge transformations are equivalent, there is no difference between "having a red particle" and "having a blue particle". You cannot, even in principle, determine the "color" of an object in this sense.

The popular phrasings of "red, blue and green" quarks are actually meaningless. They give a nice heuristic because the "color language" gives a way to draw many intuitive conclusions about otherwise unintuitive group theory, but "red, blue or green" quarks do not exist. Objects in the theory that are related by a gauge transformation are literally the same, there is no difference between a "red quark" and a "green quark" - a quark is a quark is a quark.

What we are able to say (if you manage to deconfine the color-charged stuff, since confinement means that we only see colorless objects) is "I have a color-charged particle", and specify "which kind of color charge" it has, i.e. whether it has merely a color (like quarks), or a color-anticolor (like gluons), or color-anticolor-anticolor (like nothing we know), and so on. (These correspond formally to different $\mathrm{SU}(3)$ representations) This - "color/color-anticolor/color-anticolor-anticolor/..." - is the proper generalization of the $\mathrm{U}(1)$ electric charge to non-Abelian gauge theories.

  • 1
    $\begingroup$ It should be pointed out that a gauge transformation affects the gluons too and not just the quarks. That's what stops you from comparing two quarks at different points by bringing them to the same point---the gluon field conspires to ensure the quarks' colours change as they move. $\endgroup$
    – Brian Bi
    Commented Apr 17, 2015 at 4:23
  • $\begingroup$ Is this also true about electric charge? If so is that why we never talk about an object being in a superposition of positive and negitive charge? Or have I never heard of a superposition of positive and negitive charge? $\endgroup$ Commented Jan 10, 2016 at 2:00
  • $\begingroup$ @Shane: No, positive and negative charge are not the analogue to the colors red, green, blue, but to the representations "color", "color-anticolor", and so on. The electromagnetic gauge transformations just "rotate a phase", they do not mix positive and negative. $\endgroup$
    – ACuriousMind
    Commented Jan 10, 2016 at 14:35
  • $\begingroup$ So what about SU(2), also non-Abelian? If weak isospin/I3 isn't the analogue to positive and negative charge, what is? And why can different weak isospins be isolated and observed, while colors can't? $\endgroup$ Commented Dec 7, 2018 at 8:44
  • $\begingroup$ @alexchandel The third component of weak isospin is part of the unbroken $\mathrm{U}(1)$ of electromagnetism and hence observable. Note that no one claims that particles have definite first or second components of weak isospin. $\endgroup$
    – ACuriousMind
    Commented Dec 8, 2018 at 13:04

Usually, the charge we refer to in QFT means the Noether charge of some global (i.e physical) symmetry. For example, the Noether charge associated with a global $U(1)$ transformation in QED is called electric charge. One must be careful with this $U(1)$ transformation because lots of people confused it with the $U(1)$-gauge invariance in QED, i.e the redundancy under a local $U(1)$-transformation.

Noether's theorems only applies for global symmetries. You can try applying it to a gauge (i.e local) transformation and find a conserved quantity, but it is not an physical observable because it is not gauge invariant.

More specifically, the conserved quantity you have for a $U(1)$-gauge transformation $A^{\mu}\rightarrow A^{\mu}+\partial^{\mu}\Lambda$ is $$J^\mu=\frac{\partial\mathcal L}{\partial(\partial_\mu A_\nu)}\delta A_\nu=-\frac12 F^{\mu\nu}\partial_\nu\Lambda,$$

which is indeed conserved, but is not gauge invariant.

The same thing goes to $SU(3)$-gauge theory. The colors of quarks are conserved, but are not gauge invariant. In QED, one can essentially view the global U(1) symmetry of a charged fermion as the "global part" of the $U(1)$-gauge invariance. However, in the non-Abelian case, one can easily see that the "global part" of $SU(3)$ is its center, which is a discrete subgroup. In other words, one should not expect a conserved current associated with it.

Instead, given the Lagrangian of a $SU(3)$-gauge theory, $$\mathcal{L}=-\frac{1}{4}\mathrm{Tr}(F_{\mu\nu}F^{\mu\nu})+\bar{\Psi}(iD\!\!\!\!/-m)\Psi,$$

one finds a current $$J^{\mu}=\sum_{a=1}^{N}\left(\bar{\Psi}\gamma^{\mu}T_{a}\Psi\right)T_{a}$$

which is covariantly conserved, i.e $D_{\mu}J^{\mu}=0$, following its equations of motion. Notice that this current is not locally conserved because of the covariant derivative. In other words, it is not a Noether current.

On the other hand, in QCD there are still global symmetries, i.e. Baryon number conservation (i.e global $U(1)$ symmetry) and flavour number conservation etc.

Since @octonian mentioned this, it should be emphasized that here the current $$J^{\mu}=\sum_{a=1}^{N}\left(\bar{\Psi}\gamma^{\mu}T_{a}\Psi\right)T_{a}$$

for non-Abelian gauge theory is not gauge invariant. First of all, the current comes from equations of motion $$D_{\mu}F^{\mu\nu}=J^{\nu} \tag{1}$$ $$(iD\!\!\!\!/-m)\Psi=0 \tag{2}$$

where equation (1) is the Yang-Mills equation, which is the non-Abelian version of the inhomogeneous pair of the Maxwell equations. Under a $SU(3)$-gauge transformation, one has $$F^{\prime}=UFU^{-1},\quad D^{\prime}=UDU^{-1},\quad\mathrm{and}\quad J^{\prime}=UJU^{-1},$$

which implies that the equations of motion is invariant under the gauge transformation, i.e $$D^{\prime}\star F^{\prime}=J^{\prime}.$$

The covariant-conservation of $J$ can be easily checked:

\begin{align} D\star J&=DD\star F \\ &=F\wedge\star F-\star F\wedge F \\ &=F^{a}\wedge\star F^{b}[T_{a},T_{b}] \\ &=F^{a}\wedge\star F^{b}f_{ab}^{\,\,\,\,c}T_{c} \\ &=\left\langle F^{a},F^{b}\right\rangle f_{ab}^{\,\,\,\,c}T_{c} \\ &=0, \end{align}

where in the last line the fact that $f_{ab}^{\,\,\,\,c}$ is anti-symmetric wrt $a$ and $b$ has been used.

To avoid any further misunderstandings, please notice that here the curvautre 2-form is $F=dA+A\wedge A$. Since it was mentioned by @octonion in the comment section, it should be emphasized that $F=0$ does not imply flat connection $A=0$!This is true even in Abelian gauge theory. This is easy t understand if one calculates the Christoffel symbols in spherical coordinates of Minkowski spacetime. That the current $J$ is covariantly conserved is a special property for non-Abelian gauge theory. In contrast, in Abelian gauge theory the non-homogeneous pair of the Maxwell equations read $$\star d\star F=j,$$

and the current $j$ is always locally conserved, i.e $d\star j=0$, regardless of whether the curvature $F$ vanishes or not.

  • $\begingroup$ So that $J$ current is not conserved in much the same way that the energy-momentum tensor in curved spacetime is not conserved. It still seems to me that both are very conservation law like. $\endgroup$
    – octonion
    Commented May 7, 2022 at 14:23
  • $\begingroup$ @octonion The current $J$ here is in flat spacetime. The covariant derivative here is also in flat spacetime. Here, $D_{\mu}=\partial_{\mu}+A_{\mu}$. $\endgroup$
    – Valac
    Commented May 7, 2022 at 15:49
  • $\begingroup$ @octonion Here the current $J$ is not gauge invariant. Under a generic $SU(3)$-gauge transformation $U$, the current $J$ transforms as $J\rightarrow UJU^{-1}$. On the other hand, the stress energy-momentum tensor is gauge invariant. So it makes no sense to compare the two. $\endgroup$
    – Valac
    Commented May 7, 2022 at 15:52
  • 1
    $\begingroup$ It makes perfect sense to compare the two. Both involve a conservation law extended by a covariant derivative. The biggest difference is just that general relativity is nearly always considered as a classical field theory, and non-Abelian gauge theories as a QFT. If the curvature vanishes in GR then we can pick a flat connection and the energy-momentum tensor is strictly conserved. If the curvature 2-form vanishes in a non-Abelian gauge theory then we can pick a flat connection and those 8 $J$ currents are strictly conserved. $\endgroup$
    – octonion
    Commented May 7, 2022 at 20:07
  • $\begingroup$ @octonion Sorry but your last comment makes no sense. Even the curvature 2 form vanishes, that does not make it Abelian. As long as it is non-Abelian, the current always covariantly conserve, and is not locally conserved. Let's assume $F=0$ as you said. Can you write down the Lagrangian of the theory? Notice that here $F=dA+A\wedge A$. Setting $F=0$ does not guarantee that $A=0$. $\endgroup$
    – Valac
    Commented May 7, 2022 at 22:53

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