# Why is color conserved in QCD?

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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.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
retagged Apr 11, 2015
Related question by OP: physics.stackexchange.com/q/56866/2451

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user Qmechanic
@DJBunk, drawing diagrams to convince myself, it would seem to me that it is the three primary colors that are conserved, hence my confusion.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
Excellent question! It's definitely color charge that is conserved (gluon number is not). But I don't know what the resolution of this issue is.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user David Z
Dear @David, I am a sort of a fan of yours but this basic confusion of yours came as a big surprise to me.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user Luboš Motl
@Luboš what confusion are you talking about? If you mean the fact that I don't know how to answer the question, it hardly seems like the sort of thing a grad student should be expected to know off the top of their head (except for one who specializes in group theory).

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user David Z
@lubos, it would be extremely helpful if you would compare what I say in my edit to what wikipedia says (en.wikipedia.org/wiki/Gluon#Eight_gluon_colors), where they They specifically associate the 8 gluon color combinations with the gell-mann matrices. This is common in other texts. Either I'm right, or everybody else is stupid. It would be great if, instead of continuing to call me stupid, if you would actually try to address this discrepancy, and in so doing, my question.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247

This seems to be a rather elementary confusion, it's like asking since Lorentz transformations are 4 by 4 matrices and act on column vectors with 4 independent entries, how can there be more than 4 conserved quantities originating from Lorentz invariance?

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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.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user Luboš Motl
answered Mar 14, 2013 by (10,268 points)
So it is just a coincidence that there are $N^2-1$ conserved charges and $N^2-1$ gauge fields? What are the conserved charges called? Can you point to a color combination or any specific example of one of the eight conserved charges? For example, are the three SU(2) weak charges the three components of weak isospin?

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
"For example, baryons would have to contain 4 quarks to be color-neutral." -- wait, are you saying that in SU(2) weak theory particles must be "color neutral", where now there are two colors?

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
What I mean about the "coincidence" is that the generators of SU(N) are both the conserved charges, and also correspond to the gauge fields, apparently. But the gauge fields are not conserved, the charges are. How can they both correspond to the generators, but only one is conserved?

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
For SU(2) the generators are weak isospin T1,T2,T3. Shouldn't all three be conserved? Is only T3 conserved because of electroweak symmetry breaking?

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
@user1247 QCD is in a strong coupling confining phase, while weak theory is in a weak coupling Higgs phase, so the low energy phenomenology of the two theories is completely different. If you wrote down a confining SU(2) analogy to QCD (not the physical electroweak theory) then you would have colour neutral combinations of two colour quarks.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user Michael Brown
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