Where does $SU(2)$ isospin symmetry come from? Strong (nuclear) interaction does not distinguish between neutrons and protons. I know that neutron and proton form an Isopin $SU(2)$ doublet. But I have a doubt. Strong interaction has $SU(3)_C$ color symmetry between the quark field and not $SU(2)$. Then where does this $SU(2)$ group arise? Moreover is it not at the level of quarks? Does the SU(2)-isopin symmetry of up and down quarks result in the isopin symmetry between neutrons and protons?
SU(2) Isospin Symmetry – Origins and Explanation
group-theoryisospin-symmetrynuclear-physicsparticle-physicsstandard-model
Related Solutions
In the standard model of particle physics, there exist elementary particles out of which all other matter is composed.
elementary particles
The proton and the neutron are composed out of up and down quarks
the proton is (u u d) and the neutron (u d d). If you notice in the table the quarks have different charges. In the hypothetical case of no charges, there will still be the symmetry to which the strong interaction ties the quarks, so a proton and a neutron will still be different, occupying the isotopic spin +1/2 and -1/2 respectively in the representation. They will have the same strong force effects as composites of quarks , following the symmetries.
This is just a statement to demonstrate the isotopic spin symmetry that arises from the strong interactions. Charges are fundamental in nature because we cannot turn them off.
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.
Flavor Symmetry
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.
Approximate Symmetry
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:
$$(\text{actual result}) = (\text{result with exact isospin}) \times \left[1+ \mathcal O\left(\frac{m_u-m_d}{\Lambda_\text{QCD}}\right)\right]$$
(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.
Notes
Why do we focus on isospin and not flavor symmetry?
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.
What is flavor symmetry good for?
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.)
Isn't flavor broken all over the place?
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.
Where did this symmetry come from?
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, $\langle q\bar q\rangle$.) 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.
The other quarks
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.
Best Answer
DomDoe's answer is the historical answer, but I suspect that mithusengupta123 may really be asking something like:
Given our understanding of the Standard Model of particle physics, how is it that the low-energy physics of hadrons has an approximate isospin symmetry?
1. Isospin acting on quarks
Historically, isospin historically proposed as a symmetry between protons and neutrons. A nucleon is a field $N = (p, n)^T$ that is a doublet under SU(2) isospin. Equivalently, we may define isospin on the light quarks, placing the up and down into an isospin doublet, $q = (u, d)^T$.
By addition of spin, a state with two up quarks and a down quark has net isospin $I^3 = +1/2$, and so this is consistent with the proton being a $+1/2$ state. Similarly with the neutron having isospin $-1/2$.
2. Isospin is not exact
Isospin is not an exact symmetry. It is an approximate symmetry. It is broken by:
The masses of the quarks. The up quark has a different mass than the down quark. Similarly, the proton has a different mass than the neutron.
It is also broken my electromagnetism. The up and down quarks have different electric charge. Similarly, the proton is charged whereas the neutron is neutral.
In this sense, isospin is not a symmetry. It's almost a symmetry. What do we mean by almost? We mean that there is a small, dimensionless parameter that we can expand about. When we're talking about low-energy hadronic physics, this parameter is something like $(m_u - m_d)/\Lambda_\text{QCD}$, where $\Lambda_\text{QCD}$ is the confinement scale (alternatively you could put in the proton mass, which is roughly the same order of magnitude). The other expansion parameter is $\alpha = 1/137$. Both of these expansion parameters are around the percent level.
This means that if we have a result that is true in the exact limit of isospin symmetry, then the actual result in nature is the same thing up to percent-level corrections. Further, we can use techniques like perturbation theory to solve for these corrections order by order.
3. Isospin in practice
How do we use isospin? One simple example are the pions. We know that pions are bound states of two light quarks. That is, they are in an isospin representation that comes from the product of two doublets. (There's a subtlety here because it's really a quark--anti-quark pair, see e.g. this question).) We know that the combination of two SU(2) doublets gives a triplet and a singlet.
Experimentally, we can identify the triplet and the singlet as the three pions and the $\eta$, respectively. The three pions are related to each other by isospin symmetry, whereas the $\eta$ is its own object. Indeed, the $\eta$ is about four times heavier than the pions, which are roughly the same mass.
On the other hand, the pions do not have the same exact mass. The charged pions are 140 MeV, while the neutral pion is 135 MeV. This few percent correction to the exact isospin limit is precisely the result of the mass splitting of the quarks and the electromagnetic discrimination between the charged and neutral states.
4. Where did isospin come from?
Now to the crux of the question: if we know the Standard Model, how do we understand that at low energies there is an approximate isospin symmetry? How is this related to any of the other symmetries of the Standard Model?
The answer is that isospin symmetry is the result of chiral symmetry breaking.
Imagine writing down all of the Standard Model particles without any interactions. There is a symmetry among the quarks. $U(6)_L\times U(6)_R$, which rotates the six left-handed quarks separately from the six right-handed quarks. (Recall from representation theory that left-chiral and right-chiral fields are, a priori, completely different things which can have different charges.) This is actually
This symmetry is broken by:
The electroweak force, which distinguishes between left- and right-chiral quarks. Further, it places left-handed quarks into doublets and distinguishes between right-handed quarks with up-type and down-type charges.
The Yukawa interactions with the Higgs, which (upon electroweak symmetry breaking) gives different masses to vectorlike combinations of left- and right-chiral quarks. This combines Weyl fermions into Dirac fermions.
Let us ignore the electroweak force---these effects come with electroweak couplings which we know are relatively small. Certainly at low energies when they are either mediated by a photon ($\alpha = 1/137$) or a $W/Z$ boson (suppressed contributions at low energies because they're heavy).
Then the mass terms pair left- and right-chiral fermions. These mass terms look like $m_q \bar q_L q_R+\text{h.c.}$. First, let us pretend that all of the quarks have the same mass. That is, $m_q$ is universal. Then this means that our original $U(6)_L\times U(6)_R$ symmetry is broken to $U(6)_D$, where the $D$ means diagonal. If you rotate between the six left-handed quarks, you have to do a compensating rotation among the six right-handed quarks in order for the mass term to remain invariant.
Once you turn on the different masses of each quark, then this $U(6)_D$ is broken further to $U(1)^6$, which is basically the rephasing of each type of quark.
We know that there's a big hierarchy in the quark masses, so for the most part, this breaking down to $U(1)^6$ symmetry is pretty rigorous. Each U(1) represents the conservation of up-ness, down-ness, strangeness, charm-ness, etc. (We know that the interactions of the $W$ boson violate these, but for now we're ignoring the electroweak interactions.) However, the mass splitting between the up and the down quark is relatively modest... so in fact, the up and the down have an approximate $SU(2)$ symmetry left over. (I'm being sloppy with the U(1) factors, you can repackage them into overall baryon number conservation and other conservation laws.) This $SU(2)$ symmetry is precisely isospin.
5. What does this buy us?
Why is this useful from the fundamental point of view?
We know how to deal with spontaneous symmetry breaking. In particular, we know that the breaking of the $SU(2)_L\times SU(2)_R$ subgroup of $U(6)_L\times U(6)_R$ is a spontaneous breaking of an approximate symmetry. Thus we can describe the interactions of the Goldstone bosons of this breaking by the nonlinaer sigma model, up to corrections.
The power of this point of view is that the pions are identified with the Goldstone bosons of $SU(2)_L\times SU(2)_R \to SU(2)_D$. Their interactions with each other are predicted by the non-linear sigma model once one defines the scale at which this symmetry is broken. (This is the pion decay constant, which is related to the chiral symmetry breaking scale by the QCD chiral condensate.)
One can extend this and say that the strange quark is also reasonably close in mass to the up and down. Then one can talk about an approximate $SU(3)_L\times SU(3)_R \to SU(3)_D$ breaking. In the limit where the masses of all three light quarks are degenerate, then one can describe the Goldstone bosons as an octet of pions and kaons. The interactions between these particles are all predicted by the nonlinear sigma model, up to corrections that are now a little bigger than in the pure SU(2) isospin case.