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I was reading this stackexchange question, and found the answer to my question not totally answered. Clearly there is color and anti-color in analogy to electric charge, and color charge clearly cannot vary from color to anti-color. However can color (or anti-color) continuously vary between a red green and blue basis, or is it like wavelengths in atomic orbitals where in order to go from one color to another you have emit the exact amount of color charge required?

Qmechanic
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Craig
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1 Answers1

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Naively, color can vary continuously between the colors according to a gauge transformation $\psi\mapsto \mathrm{e}^{\mathrm{i}\epsilon^a T^a}$ for some $\mathfrak{su}(2)$-valued object $\epsilon$, this is precisely the same as saying that a particle with electrical charge $e$ can vary continuously in phase according to $\psi\mapsto \mathrm{e}^{\mathrm{i}e\phi}\psi$.

However, there is a crucial difference: The $\mathrm{U}(1)$ symmetry of electromagnetism is Abelian, and so all transformations with constant $\phi$ are global symmetry transformations that have no gauge character, since the gauge field does not change under such transformations. In contrast, the $\mathrm{SU}(3)$ symmetry group is non-Abelian, and even constant $\epsilon$ change the gauge field, unless they commute with it. The set of elements of a non-Abelian group that commute with all others is called the center, and the center of $\mathrm{SU}(N)$ is the discrete group $\mathbb{Z}_n$.

So while electrically charged matter retains a continuous $\mathrm{U}(1)$ symmetry even after eliminating the gauge, color-charged matter retains only a discrete $\mathbb{Z}_3$ symmetry. That is, if you eliminate the gauge (which, in general, we cannot do: Gribov ambiguities prevent us even in principle from doing so globally, and even then, we will face a loss of covariance) you will end up with a particular set of red/blue/green particles that no longer can transform into each other. In this gauge-fixed world, you could think of color as a fixed property of each object, but this is not a useful intuitive picture to have. We describe the world through gauge theories precisely because the gauge-less description is not tractable.

However, that there is a discrete global $\mathbb{Z}_3$ symmetry is a valuable insight, as this is what is actually broken in the Higgs mechanism, as explained in this answer by Dominic Else.

ACuriousMind
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    That's interesting, as I've always heard this situation described as spontaneous breaking of global $SU(3)$ symmetry, not global $\mathbb{Z}_3$ symmetry. Is this because you're choosing to interpret transformations with constant gauge parameter as redundancies? Is this convention also taken by the usual QFT texts? – knzhou Dec 21 '18 at 02:10
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    @knzhou It's not a "choice". By the very nature of a gauge theory, configurations of the gauge field related by a gauge transformation (regardless of the space-time dependence of the transformation parameter!) are physically equivalent. Therefore, the only transformations between physically distinct configurations can be those that leave the gauge field invariant. But I'll grant that many texts do not ponder this subtlety at all, since they follow the strange pedagogy of thinking of a gauge theory as arising from "gauging" a global symmetry. – ACuriousMind Dec 21 '18 at 02:32
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    (As supplementary eivdence that my claims are correct, consider that no text I've ever read spends any time to dwell on the conserved quantities associated to the global variant of a non-Abelian gauge symmetry. That's because the global symmetry is discrete, and Noether's theorem doesn't apply to discrete symmetries) – ACuriousMind Dec 21 '18 at 02:35
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    Hmm, I'm still not sure. How does your second comment square with your answer here? – knzhou Dec 21 '18 at 02:40
  • @knzhou Sorry, that comment was ill thought-out. Noethers theorem as a mathematical tool applies of course to all non-spacetime-dependent continuous symmetries. But the "conserved quantity" for the "global continuous " part of a non-Abelian gauge symmetry is still gauge-variant, hence not observable. – ACuriousMind Dec 21 '18 at 03:20
  • Just to be clear, you are saying the 8 conserved currents naively associated to global symmetry do not actually exist as operators in the theory? I have never heard this claim before so I'm interested. – octonion Dec 21 '18 at 04:04
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    @octonion I was using "observable" in the meaning of "physically measurable" there. However, now that I think about it, this local/global/global continuous distinction is a can of worms and I'm not sure my understanding is exactly right. I might ask a more detailed question soon. – ACuriousMind Dec 21 '18 at 12:40
  • Yes I would be interested in a question along these lines. I had convinced myself thinking about fiber bundles at one point that there is always a real global symmetry associated to a gauge symmetry, but now that I think about it I'm not so sure either. – octonion Dec 21 '18 at 12:52
  • @ACuriousMind in general, will different gauge fixes give us different "fixed" colors to deal with? As well, I think you have $\mathfrak(su(2))$ instead of $\mathfrak(su(3))$ – Craig Dec 21 '18 at 20:03
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    @Craig The gauge group of the strong force to which the colors belong is $\mathrm{SU}(3)$. As for "different colors", I don't quite understand what you mean by that: There's a residual $\mathbb{Z}_3$ symmetry that swaps three colors around. These colors have no other "identity", "red", "green" and "blue" are just labels. You cannot actually say what makes "red" different from "blue", so it's not clear how you could compare them against another set of colors. – ACuriousMind Dec 21 '18 at 20:22
  • Alas, I have more reading to do – Craig Dec 21 '18 at 20:47