Quantum Chromodynamics and Group Theory

Question

I know nothing of QCD, but I was watching this youtube video and pondered on whether the additive structure described is a group, and if so, which is it?

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As of now, I know the group must contain the elements $\{R,G,B,-R,-G,-B,W\}$, where

1. $W$ is the identity.
2. $X$ is the inverse of $-X$ (for $X\in\{R,G,B\}$).
3. Addition is associative and commutative.
4. The following equation holds: $$R+G+B = W.$$

Points $1$ to $3$ make me certain that group is hidden somewhere, but I do not have enough knowledge of the remaining properties of the structure to pinpoint which group it is.

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• Intuitively the group is generated by the seven mentioned members, and thus by just $\{R,G,B\}$, and in fact by any two of these.

• If the group $G$ consists only of the mentioned seven members, then $G=\mathbb{Z}_7$. I find that unlikely, as it would imply that $$nR := R + \dots + R = -R$$ for some $n\in[2,7]$, which contradicts some vague intuition I have that "a lot of red can only be cancelled by a lot of anti-red". If the elements $nR$ for $n\in\mathbb{N}$ are all different, we would be dealing with an infinite group.

Answer 1

You have been barking up the wrong tree... r,b,g are component labels (basically, 1,2,3) of complex 3-vectors ψ that the infinite group elements of SU(3) act on to transform in the triplet (fundamental) representation.

SU(3) is an infinite-dimensional (Lie) group, generated by eight generators, Lie algebra elements. In this, 3, representation, they are the 8 independent hermitian traceless 3×3 matrices provided.

Every unitary matrix U can be written in the form $U=e^{iH}$, where H is hermitian, so the elements of SU(3) in this representation can be expressed as $$U=e^{i\sum{a_k\lambda_k}},$$ where $\lambda_k$ are the Gell-Mann basis of the 8 linearly independent hermitian traceless matrices mentioned. So the group is unitary, but also simple (unimodular): $\det(e^H)=e^{\operatorname{tr} (H)}=1$ since H is traceless. There is a separate gluon corresponding to each λ matrix, as displayed in the video in Dirac notation.

The "poetic" color analogy of your video evokes "colorless" combinations of quarks, meaning SU(3) invariants, SU(3) singlets, where no color sticks out: any group elements do not affect a singlet, so your speaker analogizes it to white, taking the optical analogy further than he should, color additivity nonsense and all. You can trivially see $\bar \psi U^\dagger U \psi= \bar \psi \psi$ is a singlet, and all the rest.

The WP link provided is hugely more informative than the video rant.

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Geeky

In any case, if you truly wish to mathematically connect the two, it appears from his 13:16-13:40 video diagrams ("visualizing" the root diagram of WP in terms of Grassmann's laws ), that, up to normalizations, in his peculiar labelling scheme,

1. The $\bar G\to G$ axis characterizes the eigenvalues of the 3rd components of isospin, $I_3\sim \lambda_3$, α in the root diagram ;

2. his $ \bar R\to R$ axis the signed eigenvalues of $U_3\sim\sqrt{3} \lambda_8-\lambda_3$, β in the WP root diagram;

3. and his $B\to\bar B$ axis the eigenvalues of $V_3\sim \sqrt{3} \lambda_8 +\lambda_3$, α+β in the superbly more mainstream root diagram.

You may wish to convince yourself of the suitable properties of this confusing scheme. The two (Cartan subalgebra) generators in the center have zero 3rd component of I,U, and V spin eigenvalues , but they are not colorless, of course.

Matt's Grassmann color scheme would really, really, really, lead a clueless meticulous listener astray. For instance, if you identify his two hexagons (eightfold ways) with $R=\lambda_1, ~B=\lambda_5, ~G=\lambda_6$, $\bar R=\lambda_2, ~\bar B=\lambda_4, ~\bar G=\lambda_7$, and the Cartan generators $\lambda_3$ and $\sqrt{3} \lambda_8$ with his black and white dots, you certainly cannot add R+B+G, as you do, and get anything meaningful—certainly not 0, nor the identity, etc. O'Dowd's white is a feature of r,b,g composites and not a generator, which should have null action on it, as above. I wouldn't go there...