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?


2 Answers 2


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:

  1. 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.

  2. 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:

  1. 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.

  2. 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.

  • $\begingroup$ In part 4, you said $U(6)_{D}$ is broken to $U(1)^{6}$ because of different masses of each quark, but the mass splitting between the up and the down quark is relatively modest... so the up and the down have an approximate $SU(2)$ symmetry left over. Why is the leftover not $U(2)$ but $SU(2)$? Also, in part 5, why did you say "the $SU(2)_{L} \times SU(2)_{R}$ subgroup of $U(6)_{L} \times U(6)_{R}$" instead of the $U(2)_{L} \times U(2)_{R}$ subgroup of $U(6)_{L} \times U(6)_{R}$? $\endgroup$
    – Shen
    Commented Dec 10, 2021 at 13:41
  • $\begingroup$ @Shen: at the level of the discussion above, the leftover piece is indeed U(2). We focus on the SU(2) since that was the initial question. The U(1) factors that distinguish U(2) from SU(2) go into: (1) rephrasing the mass terms to make them real, (2) unbroken baryon (-lepton) number, and (3) the anomalous axial symmetry. You may find a discussion in Section 2 of arXiv:1711.03624 $\endgroup$ Commented Dec 15, 2021 at 2:14

Isospin was introduced before the quark-model was developed. Since up- and down-quarks have about the same mass and the same coupling to the strong force, this symmetry works pretty well for nucleons (proton $uud$, neutron $udd$). $SU(2)$ was proposed as an analogy to regular spin. It treats the proton and the neutron as just two different states (spin up, spin down) of the same particle (hence the $SU(2)$).

After the introduction of the quark model (where it was already noticed that there are much heavier quarks than up/down), all other quarks were handed an isospin $0$-charge to make it solely a symmetry of up/down quarks ($I_{3_{u/d}}=\pm \frac{1}{2}$).

  • $\begingroup$ What about SU(3) flavour symmetry? Is it there for the QCD Lagrangian at the quark level? $\endgroup$ Commented Aug 14, 2017 at 10:17
  • $\begingroup$ @mithusengupta123 Have you looked it up? $\endgroup$ Commented Aug 14, 2017 at 12:34
  • $\begingroup$ $SU(3)_F$ is only an approximate symmetry in the quark model, since anything above the strange quark mass (charm, bottom, top) breaks this symmetry badly. $\endgroup$
    – DomDoe
    Commented Aug 14, 2017 at 13:37
  • 1
    $\begingroup$ In general, when one has two particles (things)one can try to model them by an SU(2) symmetry, i.e. that they are interacting exactly the same way quantum mechanically except for a "spin" difference which will enter algebraically in the calculations. Then wone checks if the symmetry holds. If three particles SU(3) can be tried as a symmetry of the system, and checked with data. for n it is SU(n) $\endgroup$
    – anna v
    Commented Aug 15, 2017 at 3:34
  • $\begingroup$ @annav - If one has two particles, why does one not model them by U(2) or SO(2) but SU(2)? $\endgroup$
    – Shen
    Commented Dec 10, 2021 at 12:13

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