Physics Asked on July 25, 2021
There is an approximate $SU(3)_F$ flavour symmetry that exists at the quark level between $u$, $d$ and $s$ quarks. But we often talk about an approximate isospin $SU(2)$ between up and down quark only? Why?
Both SU(3) flavor and SU(2) isospin are approximate symmetries of the Standard Model at low energies. Consider physics below the proton mass, where we can talk about the pions and kaons that are the avatars of these symmetries. At energies this low, it doesn't make sense to talk about the heavy quarks (charm, bottom, top), so we're left with the light quarks: up, down, strange.
In the limit where the quark masses are identical, then it looks like there's nothing to distinguish these three quarks. But wait, you might argue, the down-type quarks have different charge than the up quark! This is true, but at low energies, the quarks are all confined into mesons: so we only ever see quark--anti-quark pairs and are not sensitive to the charges of the constituents.
So in the limit where $m_u = m_d = m_s$, we can talk about an SU(3) symmetry that rotates these quarks into each other. It turns out that this flavor symmetry let us relate the properties of pions and kaons in a nice way. For example, you can experimentally measure the decay constant of a pion---something which is not calculable from perturbation theory---and then invoke SU(3) flavor symmetry to say that a kaon should have the same decay constant.
In reality the pion and kaon decay constants are not the same. This is because in reality, the light quarks do not have degenerate masses and so do not have an exact SU(3) flavor symmetry.
$m_u = 2.3$ MeV
$m_d = 4.8$ MeV
$m_s = 95$ MeV.
This means that SU(3) flavor symmetry is broken by the different quark masses. But we can see from the numerical values that the up and down quarks almost have the same mass. Thus there is almost an SU(2) symmetry where you rotate between up and down. If you have a result which is true in the limit where this SU(2) symmetry were exact, then you can say that the result is true up to corrections that scale like $(m_u-m_d)$. You'll need to write this in terms of a dimensionless ratio, usually in flavor physics the ratio is something like $(m_u-m_d)/Lambda_text{QCD}$. Thus we would write:
$$(text{actual result}) = (text{result with exact isospin}) times left[1+ mathcal Oleft(frac{m_u-m_d}{Lambda_text{QCD}}right)right]$$
(Don't worry too much about the use of $Lambda_text{QCD}approx m_{text{proton}}$ as the heavy scale. The expansion will be something around this value if you're looking light hadrons.)
You can say the same about SU(3) flavor, but now we see that because the strange quark is more than an order of magnitude heavier than the up/down quarks, the symmetry between the strange and the lighter quarks is worse. What we mean by "worse" is that any result that is true in the limit of exact SU(3) symmetry is corrected by something that goes like $(m_s - m_{u,d})/Lambda_text{QCD}$.
Note that $(m_s - m_{u,d})/Lambda_text{QCD}$ is not a bad expansion parameter. So it still makes sense to use SU(3) flavor as an approximate symmetry. But you can see that SU(2) isospin is more precise at the same order of symmetry breaking.
A more practical answer to your question may be pedagogical. When learning how to use approximate symmetries to extract physical predictions, it is useful to start with the simpler SU(2) isospin symmetry since this encodes all of the physics intuition. SU(3) is essentially the same story but with more generators and multiple breaking parameters. Thus if your goal is to learn physics, then SU(2) is often sufficient.
We've noted that quarks are confined, at least at the energy scales at which we would be talking about SU(3) flavor symmetry. In this sense, what good is it to talk about symmetries of the quarks if the quarks are all locked up in hadrons?
The reason is "addition of angular momentum." Well, not angular momentum literally, but the power of using representation theory to decompose product representations. We say that the up and down form an isodoublet. This means that a bound state of two of these particles is in a representation that must be in the decomposition of the product of two SU(2) doublets. We know that this decomposes into an isotriplet and an isosinglet (analogous to the hydrogen atom!). Indeed, these are the pions and the $eta$. ((There's a little subtlety because the bound states are quark--anti-quark pairs rather than spin--spin pairs in the case of the angular momentum of hydrogen.))
So in other words: flavor symmetry is a way to understand and classify the interactions of the hadrons composed of the light quarks based on the approximate symmetry of the constituent quarks. (Historically isospin was originally identified as a relation between protons and neutrons, but it is identical and we don't have to take the historical view.)
If approximate flavor symmetry is something that came from the more fundamental particle theory of quarks, then maybe we go back to saying that it seems to make no sense that there's a symmetry relating up, down, and strange if these have different charges. In other words, flavor symmetry (isospin) is broken by electromagnetism.
Indeed, the reason why the charged pions are heavier than the neutral pion is precisely because of corrections to their self-energy from electromagnetism. Since electromagnetism has a small interaction strength, this is a reasonably modest correction.
One thing you might also be thinking about is where flavor symmetry came from. Flavor symmetry is a subgroup of the chiral symmetry, which is explicitly broken by the quark masses (Yukawa interactions with the Higgs) and the electroweak force (which treats left- and right-handed particles differently). Chiral symmetry is the observation that one can rotate the left-handed quarks between themselves separately from the right-handed quarks.
This symmetry is spontaneously broken by the same process that breaks electroweak symmetry. The Higgs does this and gives masses to the quarks. The mass terms connect a left-chiral quark to a right-chiral quark, and so rotations among the left-handed quarks must be accompanied by rotations of the right-handed quarks in order to remain invariant:
$$m_q bar q_L q_R to m_Q bar q_L U_L^dagger U_R q_R$$
is invariant only when $U_L = U_R$. In other words, chiral symmetry is spontaneously broken to the diagonal group. (A quick aside: chiral symmetry is broken by the Higgs, but also at low energies by QCD from the chiral condensate, $langle qbar qrangle$.) The residual symmetry is precisely what we call SU(3) flavor. The fact that it is approximate is again a relic that this symmetry is explicitly broken "by a small amount" by two effects:
The different masses of the light quarks
The electromagnetism, which can tell up quarks from down/strange quarks.
When we have a spontaneous breaking of global symmetries, we know that we are left with a theory of Goldstone bosons. Because the symmetry is explicitly broken, these Goldstones are "pseudo-Goldstones," which practically means that they have a small mass (that is otherwise protected from large quantum corrections due to the approximate Goldstone shift symmetry).
These small mass states are precisely the light pseudoscalar mesons. We know that the interactions of Goldstone bosons is given by the non-linear sigma model, and so we can write down a theory of pion/kaon interactions in the SU(3) flavor limit. This is quite remarkable when we remember that these are bound states under the strong force, which is non-perturbative at these energies.
This should also give an explanation for why we only talk about flavor symmetries between light quarks. The heavier quarks are heavier than the strong scale, $Lambda_text{QCD}$ and the symmetry between these quarks and the lighter quarks is very poor.
Correct answer by Henry Deith on July 25, 2021
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