Why is color conserved in QCD?

Question

According to Noether's theorem, global invariance under $SU(N)$ leads to $N^2-1$ conserved charges. But in QCD gluons are not conserved; color is. There are N colors, not $N^2-1$ colors. Am I misunderstanding Noether's theorem?

My only guess (which is not made clear anywhere I can find) is that there are $N_R^2-1$ conserved charges, where $N_R$ is the dimension of the representation of SU(N) that the matter field transforms under.

EDIT:

I think I can answer my own question by saying that eight color combinations are conserved which do correspond to the colors carried by gluons. Gluon number is obviously not conserved, but the color currents of each gluon type are conserved. An arbitrary number of gluons can be created from the vacuum without violating color conservation because color pair production {$r,\bar{r}$}, {$g,\bar{g}$}, {$b,\bar{b}$} does not affect the overal color flow. Lubos or anyone please correct me if this is wrong, or if you want to clean it up and incorporate it into your answer Lubos I will accept your answer.

Answer 1

Global invariance under $SU(N)$ is equivalent to the conservation of $N^2-1$ charges – these charges are nothing else than the generators of the Lie algebra ${\mathfrak su}(N)$ that mix some components of $SU(N)$ multiplets with other components of the same multiplets. These charges don't commute with each other in general. Instead, their commutators are given by the defining relations of the Lie algebra, $$ [\tau_i,\tau_j] = f_{ij}{}^k \tau_k $$ But these generators $\tau_i$ are symmetries because they commute with the Hamiltonian, $$[\tau_i,H]=0.$$ None of these charges may be interpreted as the "gluon number". This identification is completely unsubstantiated not only in QCD but even in the simpler case of QED. What is conserved in electrodynamics because of the $U(1)$ symmetry is surely not the number of photons! It's the electric charge $Q$ which is something completely different. In particular, photons don't carry any electric charge.

Similarly, this single charge $Q$ – generator of $U(1)$ – is replaced by $N^2-1$ charges $\tau_i$, the generators of the algebra ${\mathfrak su}(N)$, in the case of the $SU(N)$ group.

Also, it's misleading – but somewhat less misleading – to suggest that the conserved charges in the globally $SU(N)$ invariant theories are just the $N$ color charges. What is conserved – what commutes with the Hamiltonian – is the whole multiplet of $N^2-1$ charges, the generators of ${\mathfrak su}(N)$.

Non-abelian algebras may be a bit counterintuitive and the hidden motivation behind the OP's misleading claim may be an attempt to represent $SU(N)$ as a $U(1)^k$ because you may want the charges to be commuting – and therefore to admit simultaneous eigenstates (the values of the charges are well-defined at the same moment). But $SU(N)$ isn't isomorphic to any $U(1)^k$; the former is a non-Abelian group, the latter is an Abelian group.

At most, you may embed a $U(1)^k$ group into $SU(N)$. There's no canonically preferred way to do so but all the choices are equivalent up to conjugation. But the largest commuting group one may embed into $SU(N)$ isn't $U(1)^N$. Instead, it is $U(1)^{N-1}$. The subtraction of one arises because of $S$ (special, determinant equals one), a condition restricting a larger group $U(N)$ whose Cartan subalgebra would indeed be $U(1)^N$.

For example, in the case of $SU(3)$ of real-world QCD, the maximal commuting (Cartan) subalgebra of the group is $U(1)^2$. It describes a two-dimensional space of "colors" that can't be visualized on a black-and-white TV, to use the analogy with the red-green-blue colors of human vision. Imagine a plane with hexagons and triangles with red-green-blue and cyan-purple-yellow on the vertices.

But grey, i.e. color-neutral, objects don't carry any charges under the Cartan subalgebra of $SU(N)$. For example, the neutron is composed of one red, one green, one blue valence quark. So you could say that it has charges $(+1,+1,+1)$ under the "three colors". But that would be totally invalid. A neutron (much like a proton) actually carries no conserved QCD "color" charges. It is neutral under the Cartan subalgebra $U(1)^2$ of $SU(3)$ because the colors of the three quarks are contracted with the antisymmetric tensor $\epsilon_{abc}$ to produce a singlet. In fact, it is invariant under all eight generators of $SU(3)$. It has to be so. All particles that are allowed to appear in isolation must be color singlets – i.e. carry vanishing values of all conserved charges in $SU(3)$ – because of confinement!

So as far as the $SU(3)$ charges go, nothing prevents a neutron from decaying to completely neutral final products such as photons. It's only the (half-integral) spin $J$ and the (highly approximately) conserved baryon number $B$ that only allow the neutron to decay into a proton, an electron, and an antineutrino and that make the proton stable (so far) although the proton's decay to completely quark-free final products such as $e^+\gamma$ is almost certainly possible even if very rare.

Answer 2

What Luboš has written is totally right but I also understand that it does not completely anwser to your question. By the statement "color is conserved in QCD" you probably mean that there are three U(1) symmetries corresponding to red, green and blue color. You know it because you have seen a lot of QCD pictures such as this one where the colored lines never end. I find it interesting that this is almost nowhere explained explicitly. Consider my anwser as a continuation to the one of Luboš's rather than an alternative explanation.

As Luboš had written, the $SU(3)$ gauge symmetry implies that there are two commuting operators because Rank[$SU(3)$]=2. We can conventionally choose those operators as $$ \lambda_3=\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}, \quad \lambda_8 = \frac{1}{\sqrt{3}} \begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}. $$ Those operators act on the quarks $q=(q_r, q_g, q_b)^T$ via left multiplication and on the gluons $G_\mu = \sum_{i=1}^8\lambda_i G^{i}_\mu$ via commutation. Apparently, those are not operators of red, green or blue color.

However, as Luboš has also written, there is an additional global $U(1)$ symmetry corresponding to the baryon number conservation. The represantation of its generator, i.e., the operator of baryon number, in the same notation takes the form $$ B = \frac{1}{3} \begin{pmatrix}1&0&0\\0&1&0\\0&0&1 \end{pmatrix}.$$

Hence $\{\lambda_3, \lambda_8, B\}$ spans the trivial Lie algebra corresponding to the $U(1)^3$ symmetry group. Since any basis is as good as any other, we can change the basis by making the linear combinations of the 3 generators to the following one: $$ r= \begin{pmatrix} 1&0&0&\\0&0&0\\0&0&0\end{pmatrix}, \quad g= \begin{pmatrix} 0&0&0&\\0&1&0\\0&0&0\end{pmatrix}, \quad b= \begin{pmatrix} 0&0&0&\\0&0&0\\0&0&1\end{pmatrix} $$ Obviously, those operators deserve to be called the operators of red, green and blue color bacause $$ \psi_r = \begin{pmatrix} 1\\0\\0\end{pmatrix}, \quad \psi_g = \begin{pmatrix} 0\\1\\0\end{pmatrix}, \quad \psi_b = \begin{pmatrix} 0\\0\\1\end{pmatrix} $$ are their simultaneous eigenstates with the ancticipated eigenvalues.

This is why colors are converved. Therefore, it does make some sense to state that neutron has colors (+1,+1,+1), but it is equivalent to the (more elegant) statement that neutron is a colorless baryon.

Answer 3

I think color charge may NOT be a Noether charge. In the context of Yang-Mills theory, color is just the index of the generators of the matrix representation of gauge group.

The statement "colors are conserved" may come from Fierz indentity:

$\sum_a T^a_{ij}T^a_{kl} = \frac{1}{2} \left( \delta_{il}\delta_{jk} - \frac{1}{N} \delta_{ij}\delta_{kl}\right)$

This structure may appear in Feynman diagrams. For example, for $u\bar{d} \to u \bar{d}$ scattering, we have in tree-level: \begin{equation} T_{j i}^{a} T_{k l}^{a}\left(i g_{s}\right)^{2} \bar{u}_{j}\left(p_{2}\right) \gamma^{\mu} u_{i}\left(p_{1}\right) \frac{-i\left[g_{\mu \nu}-(1-\xi) \frac{k_{\mu} k_{\nu}}{k^{2}}\right]}{k^{2}} \bar{v}_{k}\left(p_{3}\right) \gamma^{\nu} v_{l}\left(p_{4}\right) \end{equation} Thus if the incoming "color" $i \neq j$, then we have outgoing color to be $l=i,~ k = j$, which may be interpret as "conservation of color". If $i =j$, then we call it "color singlet", and the final state can be red/anti-red, blue/anti-blue or green/anti-green.