Is colour the conserved charge of global $SU(3)$ color symmetry?

Question

Consider the Lagrangian consisting of three Dirac fields

$$ \mathcal{L} = \sum_{a=1}^3 \bar{\psi}_a ( i \gamma^\mu \partial_\mu - m ) \psi_a$$

where $a$ is an internal index labelling the colour degree of freedom. This is symmetric under global $SU(3)$ transformations

$$ \psi(x) \rightarrow U \psi(x)$$

where $U \in SU(3)$ is a unitary matrix that acts only on the colour indices. Applying Noether's theorem to this will give me a set of eight conserved charges. However, the Cartan subalgebra is two-dimensional so I can label my eigenstates with only two charges.

My question

Now I would expect that this means that an $SU(3)$ invariant system has two conserved charges. However, I have come across quite a few contradicting answers here:

1. I have read here that I can also combine this symmetry with the $U(1)$ symmetry to get a set of three conserved charges that commute. Under a change of basis, these three charges can be interpreted as the three colours.
2. On the other hand, this answer says that there are no conserved charges for $SU(3)$!
3. Finally, this answer suggests says that there are two charges (which I would expect) but within the 2D charge space, we can define three colours.

I am quite confused by these contradicting answers. Why do some people state that $SU(3)$ charges are colour (or at least related to colour), whilst others say there is no conserved charge?

Answer 1

There is no contradiction in the three answers you are citing, just loose language. Only the third answer deals with your spare globally symmetric Lagrangian. The other two deal with QCD, its gauged (local) extension, involving gluons, which most texts write down: it involves an extra fermion bilinear/ gluon linear term, and a kinetic term for the gluons. It too is also globally, beyond being locally invariant, but the first two answers fuss the analog Noether's theorem for gauge theories which the preamble to your question excludes.

Noether's (first) theorem deals with the eight Lie-algebraic generators $F^\alpha, ~~~~\alpha=1,2,...,8$, which you may think of as eight linearly independent 3×3 matrices acting on the three a labels of the states you wrote down. (I assume you are not interested in the specific basis thereof, etc.), $$ \psi_a\to \Bigl (\exp (i\theta^\alpha F_\alpha)\Bigr )_{ab} \psi_b . $$ Summation over repeated indices is implied.

There is one "color rotation angle" parameter $\theta^\alpha$ for each of the eight generators written. For infinitesimally small such, the generic increment of the 3-spinors your wrote down is $$ \delta \psi_a = i\theta^\alpha (F_\alpha)_{ab} \psi_b. $$

These yield eight "color charges" through Noether's (first) theorem, $$ Q^\alpha =i\int\!\! dx~~ \bar \psi_a (F^\alpha)_{ab} \gamma_0\psi_b, $$ which you evidently checked are on-shell (~through the application of the equations of motion) time-conserved.

Commuting these eight charges with the 3-spinors yields the above $\delta \psi$ increments. Indeed, the two charges corresponding to the two Cartan generators, together with the Baryon charge, $\propto i\int\!\! dx~~ \bar \psi_a \gamma_0\psi_a$, also commuting with those, suffice to uniquely specify the independent components of each triplet $\psi_a$, some people call r,g,b for convenience. So: three colors, eight independent rotations thereof (conserved charges).

If/when you add gluon gauge fields, which transform among themselves through 8×8 (not 3×3) matrices, the expanded Lagrangian (QCD) is also globally SU(3) invariant, but through the magic of the gauge construction, it is, in addition, locally (gauge) invariant an issue outside the ambit of your question, if not your confusion.

The way the 8 Lie algebra generators and corresponding color charges compose among themselves (commutation) is best described through two roots, corresponding to two Cartan generators; you may think about them as commutator motions among these 8 charges.

So, in a way, your three questions harmonize, after all: it's just that they describe three different aspects of the elephant to the three blind men touching three different parts of her...

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Clarification for comment question: Yes it does, only up to language. Algebraically, flavor SU(3) and color SU(3) are identical, so one uses labels and matrices of the two interchangeably, trusting the reader cannot be confused.

Since the same generators are involved, one automatically maps the three diagonal matrices' $\lambda_8; \lambda_3; {\mathbb I}$ diagonals, suitably normalized, to the state vectors $r=(1,0,0); g=(0,1,0); b=(0,0,1)$, virtually a change of basis. The idea is people learn their SU(3) from flavor (the eightfold way), and then instantly translate to color. As generators, they are conserved charges (the baryon number is the identity in that space of triplets), but not all conserved color charges.

Answer 2

For a classical gauge charge, there is a charge component $\left(e_a: a = 0,1,⋯,N-1\right)$ for each of the $N$ dimensions of the gauge group. For $SU(3)$, $N = 8$. The charge precesses under Wong's Equation $$\frac{de_a}{dt} = \sum_{0≤b,c<N} f_{ab}^c \left(φ^b - 𝐯·𝐀^b\right) e_c,$$ where $f_{ab}^c = -f_{ba}^c$ are the structure coefficients of the gauge group's Lie algebra, $\left(\left(φ^b, 𝐀^b\right): b = 0,1,⋯,N-1\right)$ the $N$ copies of the field's potentials and $𝐯$ the charge's velocity.

This entails that the quadratic invariant $$α = ½ \sum_{0≤a,b<N}k^{ab} e_a e_b,$$ is conserved, where $k^{ab} = k^{ba}$ is the inverse of the gauge group's metric, since we may write: $$\frac{dα}{dt} = \sum_{0≤a,d<N} k^{ad} \frac{de_a}{dt} e_d = \sum_{0≤a,b,c,d<N} k^{ad} f_{ab}^c \left(φ^b - 𝐯·𝐀^b\right) e_c e_d = 0,$$ since the metric is assumed to be "adjoint invariant" so that it $k_{ab}$ and its inverse $k^{ab}$ satisfy the relations: $$\sum_{0≤d<N} \left(f_{ab}^d k_{dc} + f_{ac}^d k_{db}\right) = 0, \hspace 1em \sum_{0≤a<N} \left(k^{ac} f_{ab}^d + k^{ad} f_{ab}^c\right) = 0.$$

For simple gauge groups, the adjoint-invariant metric is given uniquely - up to a constant factor - by the Killing metric $κ_{ab} = \sum_{0≤c,d<N} f_{ac}^d f_{bd}^c$. In the usual matrix representations of simple gauge groups, this metric is diagonal so that the invariant amounts to a sum of squares of the components $\left(e_a: 0≤a<N\right)$. More generally: the coefficients are (up to a factor) the same as the coefficients for the quadratic Casimir invariant of the Lie algebra.

As a point of interest, when $SU(3)$ charge is quantized, then two components - whose operators are mutually commuting - are selected out to define the eigen-space for the charge. They play a role analogous to "chromaticity coordinates". Call them $U$ and $V$. In the usual matrix representation, they're associated with $Λ_3$ and $Λ_8$.

If you run through the entire standard model spectrum, labeling them $x$ "white" (anti-lepton); $o$ "black" (lepton); $r$ "red", $g$ "green", $b$ "blue" (quark); $c$ "cyan" = "anti-red", $m$ "magenta" = "anti-green", $a$ "amber" = "anti-blue" (anti-quark), then their assignments - up to proportion - can be written down as: $$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline · & x & o & r & g & b & c & m & a \\ \hline \\ U & 0 & 0 & +1 & 0 & -1 & -1 & 0 & +1 \\ \hline \\ V & 0 & 0 & -1 & +2 & -1 & +1 & -2 & +1 \\ \hline \\ B-L & +3 & -3 & +1 & +1 & +1 & -1 & -1 & -1 \\ \hline \end{array}$$ where I also added the assignments for the "baryon minus lepton" number $B - L$.

The quark and anti-quark colors span a space, with $U$ and $V$ as its coordinates, analogous to the Newton color wheel, at 6 cardinal points that form a hexagon.

There is no quadratic invariant involving $U$ and $V$ that covers the full spectrum of fundamental fermions. However, as you can clearly see, there is a quadratic invariant, if you also include $B-L$ in the list, namely: $$6U^2 + 2V^2 + \left(B-L\right)^2 = 9.$$

The spectrum, here, with the inclusion of $B-L$, is now analogous to the color cube and the cardinal points are the 8 vertices of the cube. That's actually a characteristic pattern of $SO(6)$ and or $SU(4)$, which both have the same Lie algebra.