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I find myself very confused by the usage of spin terminology to other quantum numbers. A singlet state is the only spinless state of the system. Now, if we consider the possible colour states $|r\rangle$, $|g\rangle$ and $|b\rangle$ of a quark, what does the term "singlet state" actually mean in this context? Presumably it means the only colourless state of the system. In the context of mesons this would be

$$|\psi\rangle = \frac{1}{\sqrt{3}} (|r\bar{r}\rangle+|b\bar{b}\rangle+|g\bar{g}\rangle).$$

This is intuitively colourless since colour and anti-colour presumably cancel to no net colour, and its designation as a singlet also makes intuitive sense since there is no other state that is colourless (given the constraint that each meson is a pairing of a quark and an anti-quark). NB the state does not have anti-symmetry or symmetry but there is no need for it because the quark and anti-quark are distinguishable.

In the context of baryons, the singlet state is given as

$$|\psi\rangle = \frac{1}{\sqrt{6}} (|rgb\rangle-|rbg\rangle+|brg\rangle-|bgr\rangle+|gbr\rangle-|grb\rangle).$$

How is this state colourless? Do the red, green and blue colours somehow cancel each other out? Even if they do, symmetric colourless states would exist such as $|rgb\rangle$, so this would not be the only colourless state. Why, therefore, is this called a singlet state?

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    $\begingroup$ The spin language is used since that is a similar example you are familiar with - for instance you say $|\psi\rangle$ is intuitively colourless, but is the state $|\phi\rangle = |\uparrow \downarrow\rangle +|\downarrow \uparrow \rangle$ spinless? $\endgroup$
    – jacob1729
    Commented Jan 5, 2021 at 11:33
  • $\begingroup$ @Pancake_Senpai To help people write answers at the appropriate level: Are you familiar with the idea that each particle transforms as $|k\rangle\to \sum_n U_{kn}|n\rangle$ or as $|\bar k\rangle\to \sum_n U^*_{nk}|\bar n\rangle$, for the unbarred and barred versions respectively, where each index runs over the $N$ colours and $U$ is an $N\times N$ unitary matrix with determinant $1$? (The ordinary-spin case and the three-colour case are both special cases of this, and saying that a multiparticle state is a singlet means it doesn't mix with other states under this transformation.) $\endgroup$
    – Chiral Anomaly
    Commented Jan 5, 2021 at 14:58
  • $\begingroup$ Singlet, scalar, colorless, etc... refer to the same thing: invariance under color transformations. You should be able to prove this for your "meson" states yourself, but your "baryon" one, your last formula is invariant by dint of combinatoric properties of the Levi-Civita symbol. Your text should be covering it, if any good. $\endgroup$
    – Cosmas Zachos
    Commented Jan 5, 2021 at 23:20
  • $\begingroup$ related: physics.stackexchange.com/q/563495/226902 $\endgroup$
    – Quillo
    Commented May 8, 2022 at 14:10

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Singlet, scalar, colorless, etc... all refer to the same thing: invariance under color transformations.

That is, parameterizing your silly popular science r,b,g by a 3-valued index i, your mesons are basically $q^{\dagger~i} q^i$. This is invariant under the generic SU(3) simple unitary ($U^\dagger U=1\!\! 1$ ) transformation $q^i \mapsto U_{ij}q^j$, $$ q^{\dagger~i}U^{\dagger}_{ij}U_{jk} q^k =q^{\dagger~i} q^i, $$ so it is a scalar, colorless, etc. The two qs may have differing "flavor" indices, but I suppress them to simplify the demonstration.

Your second, baryon, formula can be written compactly by use of the fully antisymmetric Levi-Civita symbol, $$\epsilon_{ijk} q^iq^jq^k, $$ and it too is invariant, singlet, color-neutral, etc, by dint of $$ \epsilon_{ijk}U_{im} q^m U_{jn}q^n U_{ks}q^s=\det (U) ~ \epsilon_{ijk} q^iq^jq^k= \epsilon_{ijk} q^iq^jq^k, $$ since U is unimodular (unit determinant) by construction. The first equality follows from the neat identity $\epsilon_{ijk} U_{im} U_{jn} U_{ks} = \det(U)~\epsilon_{mns} .$

NB Your alternative state $q^1q^2q^3$ is not invariant, unless it is fully antisymmetrized by the Levi-Civita symbol. The above determinant stunt would fail. It is not enough to have one color of each kind to have an invariant combination.

NB In the language of Young tableaux that physicists use, these two points are self-evident by inspection. A box under two boxes is a 3 box column, so, then, a singlet, which is also the fully antisymmetrized "baryon" combination we did last.

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