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Referring to the quantum observables of quarks, is it true that the color $SU(3)$ unitary representation commutes with all observables?

I am not referring here to the local $SU(3)$ gauge theory, I am just considering the global color $SU(3)$ symmetry.

My problem is that, if it were the case, it would not possible to distinguish a pair of apparently different (pure) states represented by unit vectors $\Psi$, $\Psi'$ if $\Psi' = U\Psi$ when $U$ is such a gauge transformation, by means of any sort of measurement. Indeed if $A$ is an observable $$\langle \Psi'|A \Psi' \rangle = \langle U\Psi|A U\Psi \rangle = \langle \Psi|U^*A U\Psi \rangle = \langle \Psi|A U^*U\Psi \rangle = \langle \Psi|A \Psi \rangle\:.$$ In other words $U$ plays the exact role of phases in QM when the algebra of observables is the whole class of selfadjoint operators. In this respect, states should be re-defined as equivalence classes $[\Psi]$ with respect to the equivalence under the action of color $SU(3)$.

Conversely, I have read that the color degrees of freedom have been introduced (also) to fix problems with the spin statistic principle.

There are states $\Psi$ of hadrons composed of three quarks which seem to be symmetric under swap of particles in spite of the fact that quarks are fermions. Adding an antisymmetric color part $\Phi_{color}$ to the state, the complete state $\Psi \otimes \Phi_{color}$ satisfies the spin statistic principle. Here the color $SU(3)$ representation just acts on the part $\Phi_{color}$ of the state.

Assuming I have understood well, I am not sure to agree with this idea because as the part $\Phi_{color}$ is a priori unobservable it cannot play any physical role and it cannot be used to fix problems with Pauli principle.

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When the gauge theory is quantized in the proper way, you cannot even meaningfully talk about the action of an $\mathrm{SU}(3)$ transformation on the space of states because the quantization of a gauge theory essentially requires you to quotient out the gauge transformations (all of them, including the global ones), so that they are "do nothing" transformations on the actual physical states. In particular, the question of whether the gauge transformations commute with all observables becomes trivial as the gauge transformations are the identity on the space of physical states.

Now, the introduction of color solves the spin-statistic problem by building the state of space fully before quotienting out the transformations. On the gauge-variant "pre-physical states", you can build the well-known colorless state consisting of three quarks. Since it is gauge-invariant as a colorless pre-physical state, quotienting out the transformations when passing the true physical Hilbert space does nothing to it, and it is therefore still there after that.

A bit more technically (I might be cheating here because you really need to carry out the BRST procedure to get the physical space of states), on the pre-physical space you have three different fermionic creation operators $a^\dagger_b,a^\dagger_g,a^\dagger_r$ associated to the triplet-valued fermionic fields. The spin-statistic theorem essentially says that each of these anticommutes with itself, and hence gives zero when squared, but it doesn't forbid applying each of them once to the vacuum. After quotienting out the gauge transformations, these operators don't descend to well-defined operator on the quotient exactly because they are not invariant under the group action we quotiented out, and nothing forces that the image of the colorless pre-physical three-quark state $a^\dagger_b a^\dagger_g a^\dagger_r \lvert 0\rangle$ in the quotient should vanish - and indeed it doesn't.

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  • $\begingroup$ I do not understand well. Is the space of states a Hilbert space, after taking the quotient? If the answer is positive, is a true vector state of quarks antisymmetric under interchange of identical particles? $\endgroup$ – Valter Moretti Jul 21 '16 at 16:56
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    $\begingroup$ @ValterMoretti: Yes, the quotient is there to make it a Hilbert space, actually. The space before the quotient has zero norm states as well as negative norm "ghost states" in it, which we get rid of by the quotient and turn the pre-physical pseudo-inner product space into an actual Hilbert space. The formal procedure is called BRST quantization and is rather subtle to get right in full generality. $\endgroup$ – ACuriousMind Jul 21 '16 at 17:01
  • $\begingroup$ I understand, it is similar to what happens in QED with Gupta-Bleuler formalism. $\endgroup$ – Valter Moretti Jul 21 '16 at 17:03
  • $\begingroup$ So it is false a claim, sometimes stated in the literature, that QCD is an example where the von Neumann algebra of observables is not $B(H)$ but a smaller factor with non-Abelian commutant! $\endgroup$ – Valter Moretti Jul 21 '16 at 17:05
  • $\begingroup$ @ValterMoretti: Yes indeed, Gupta-Bleuler is the version of that that works only for Abelian gauge theories. $\endgroup$ – ACuriousMind Jul 21 '16 at 17:08

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