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I will briefly summarize what I know and then ask my questions. If you spot mistakes in my summary, please tell me.

The idea of flavor symmetry is that massless QCD is invariant under SU(6) transformations on the 6-dimensional flavor space for quarks. Since up- and down-type quarks are treated different in electroweak theory, speaking about flavor symmetry only makes sense when speaking about strong interactions/QCD.

The typical energy scale of QCD is the proton mass. After including quark masses through electroweak symmetry breaking, flavor symmetry is therefore still an approximate symmetry for a subset of quarks with mass differences that are negligible compared to the proton mass. It turns out that one has an almost exact SU(2) flavor symmetry for $\{u,d\}$ and an acceptable SU(3) flavor symmetry for $\{u,d,s\}$. Flavor symmetries with heavier quarks are broken so badly that there is no sense in talking about them.

Formally, one can arrange the quark flavors in a SU(n) fundamental representation. One can then do tensorproducts for flavor and spin to construct other representations like baryons and mesons. An example: $2\otimes 2 = 3\oplus 1$ for spin SU(2) gives scalar- and vector mesons, $3\otimes \bar{3} = 8\oplus 1$ for flavor SU(3) gives the eightfold way for both scalar and vector mesons.

Why do these tensor methods predict the right hadrons? One also finds hadron multiplets for flavor SU(4), which is badly broken. As I got it, there is no sense in doing tensorproducts in representations of broken symmetries. What did I miss?

Furthermore, one can attach flavor quantum numbers $I_3, S, C, B, T$ to SU(n) flavor symmetry. Quantum numbers are defined to be conserved for exact symmetries, so these quantum numbers should not be conserved since flavor symmetry is broken. But there are no flavor changing processes in QCD, so surprisingly these quantum numbers are conserved in QCD. Why are flavor quantum numbers conserved in QCD even though flavor symmetry is broken?

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@anna's answer gives you what you really want to know in physics, but I will address some of your formal anxieties. A major theme is the sharp distinction between degeneracy symmetries (the Lie algebras of operators that commute or almost commute with the Hamiltonian) and spectrum-generating symmetries (the Lie algebras of operators that do not commute with the hamiltonian, and, in fact, move you from one rung of the spectrum to others).

For the quantum oscillator, the Heisenberg algebra $[a,a^\dagger]=1$ does not commute with the number operator Hamiltonian: it ladders you up and down to non-degenerate states. For the Hydrogen atom, the spectrum-generating symmetries so(4,1) and so(4,2) connect states of different energy, as the Hamiltonian is not a function of their Casimir invariants, but, instead, contains "ladder" pieces moving eigenstates of it to different, non-degenerate, eigenstates. When one turns off such pieces, the SGA collapses to a basically dull degeneracy algebra.

Recall how su(3) works. On the one hand, in the limit of equal quark masses, it is a good degeneracy symmetry. But, we are far away from this limit. In fact, the strange quark mass differs from the u,d masses by more than $\Lambda_{QCD}$, or the constituent quark mass, a third of the proton mass. The genius of flavor su(3) is first it tabulates all states made up of these quarks, a pretty tabulation. The su(4) pyramid does this as well.

But, importantly, second, it tells you how this symmetry is broken, by U-spin and V-spin operators, in a systematic, predictable, way: it is such amplitudes, couplings, Clebsches, etc... that expedite an awful lot of the heavy lifting involved in hadron interactions. (Doing this sort of thing with constituent quark wave functions is a godawful mess... you want to know how it's done, and it makes sense, but in all likelihood you won't use it in all but simple estimates, like magnetic moments.)

You may well do the same for flavor su(6), but our visual intuition is lacking in 5 dimensions, so I don't know anyone doing this. In a way, they do, when they segregate the 3 light quarks off the 3 heavy ones, in "WIsgur" stunts, and connect the "brown muck" QCD effects of each.

QCD is blind to all such structures: it couples the same way to all quarks, of whatever mass or flavor, but its effects vary with their masses. It doesn't not alter flavor.

As the other answer points out, such flavor groups are also broken by the EW symmetry, which does alter flavor, adding another layer of systematic complication to the picture.

It is fair to say that "symmetry operators" is an imperfect physics synonym for "Lie algebra generators", whose currents are not always even close to being conserved, as you observe. Lie theory, however, is so powerful it helps a lot even when it appears lost.

Now flavor numbers. These are mere tags reminding you what quark you are discussing. They correspond to independent rephasing of each flavor quark separately, and their currents are conserved and do nothing. QCD, unlike the weak interactions, does not mutate flavor, same as electromagnetism (which can still tell the difference of their charges).

As a result, flavor charges, like, e.g., S, are strictly conserved, outside the realm of the weak interactions. They are not traceless su(3) generators, obviously, and same for su(2), su(4)... So nothing breaks them, and QCD treats them all equally. They are not part of your "broken flavor symmetry" canard...

  • Bonus problem. Can you see how the su(3) currents for $\lambda_3$ and also $\lambda_8$ are conserved, after all?
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  • $\begingroup$ Thank you very much for caring about my formal anxieties. As I got it, the idea is to distinguish between cartan generators(commute with all generators and can be used to define quantum numbers for states, degeneracy symmetries, $J_3$ for SU(2)) and the other generators(generate transitions between the states, spectrum-generating symmetries, $J_1,J_2$ for SU(2)). Currents corresponding to cartan generators are conserved in flavor symmetry since there are no flavor changes is QCD, but currents corresponding to the other generators are not conserved since quark masses forbid state-transitions. $\endgroup$ – jonas Aug 15 at 8:51
  • $\begingroup$ Since the cartan generator currents are conserved after flavor symmetry breaking, one can still label states with flavor quantum numbers and construct higher flavor representations like hadron multipletts. However, transititions between elements of the multiplet are not allowed since the currents of the corresponding generators are not conserved any more. Please correct me, if I got something wrong. $\endgroup$ – jonas Aug 15 at 8:57
  • $\begingroup$ One last thing: Since states in flavor SU(3) can be labeled by $I_3, S$, I think that they correspond to the generators $\lambda_3, \lambda_8$. However, I dont know how to write down these generators in a nice way. $\endgroup$ – jonas Aug 15 at 9:02
  • $\begingroup$ Basically yes, but, careful, Cartan generators commute with each other, not the rest of the generators!! The currents are $J_i^\mu\propto \bar q \lambda_i \gamma^\mu q$, and the charges are the time integrals thereof. So $J_3^\mu\propto \bar u \gamma^\mu u - \bar d \gamma^\mu d$, ... $\endgroup$ – Cosmas Zachos Aug 15 at 10:14
  • $\begingroup$ Ah, I see. Appreciate this very much! $\endgroup$ – jonas Aug 15 at 10:44
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It is instructive if you understand how the quarks were discovered, at the time of the eightfold way.

Here is the octet

enter image description here

The meson octet. Particles along the same horizontal line share the same strangeness, s, while those on the same left-leaning diagonals share the same charge, q (given as multiples of the elementary charge).

It is the fact that the masses are different, i.e. are differently broken by the electroweak symmetry breaking that allowed experimentally to see the symmetry. The symmetry is there, before breaking too, but we would have had a hard time to see experimentally the weak SU(3) representations that led to the quark model.

Flavor symmetries are broken when you have different masses for the quarks after electroweak breaking. But the quantum numbers are not affected by the electroweak breaking, they were experimentally discovered to be conserved (and the laws governing their changes) after breaking anyway, and it is assumed they are the same before the symmetry breaking mechanism; so maybe you should rewrite this:

Quantum numbers are defined to be conserved for exact symmetries, so these quantum numbers should not be conserved since flavor symmetry is broken.

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  • $\begingroup$ Really nice answer. $\endgroup$ – Árpád Szendrei Aug 14 at 16:04
  • $\begingroup$ Thank you for your answer. I dont see why the eightfold way SU(3) symmetry and the conservation of the flavor quantum numbers should not break down when electroweak symmetry breaking occurs. Essentially, the quark masses, which are also quantum numbers(with respect to the Poincare group), do change during this process. $\endgroup$ – jonas Aug 14 at 16:10
  • $\begingroup$ @jonas that is the way physics theories are formed. mass is not a quantum number afaik, it is an observable, but it is not conserved except classically.. The quark masses arise because of the higgs mechansm and because they are not composite (as our theories assume now) , they do not change $\endgroup$ – anna v Aug 14 at 16:28
  • $\begingroup$ @annav The Wigner classification defines a casimir operator(like commuting observable in QM) $p_\mu p^\mu$ with the generator $p^\mu$ of translation that has the eigenvalues $m^2$. Since it commutes with all other generators of the poincare group, one can label states with $m^2$ and rest mass behaves like a quantum number describing representations of the poincare group. Quark masses change from 0 to finite within the electroweak symmetry breaking. Seems rather random to me, why SU(3) symmetry and flavor quantum numbers should survive. $\endgroup$ – jonas Aug 14 at 16:54
  • $\begingroup$ Using poincare arguments here may be risky since this is not a usual mass term, but generated by higgs mechanism. However, quantum numbers may change if their symmetry is broken. Take QM as an example: In $H=\frac{p^2}{2m}$ I can label states with $p$ since I have p-conservation, if I add a symmetry violating term $\frac{m\omega^2}{2}x^2$, I need other quantum numbers since my p-symmetry is broken. $\endgroup$ – jonas Aug 14 at 17:23

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