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To my understanding, particles in a multiplet have similar properties (similar masses and so on) and slight or symmetric differences (in mass, electric charge, spin, and so on). They can be regarded as the same particle split into different states. For example, Heisenburg proposed that a proton and a neutron are the same particle (neutron) in different states, which form a doublet. Similarly, the three pions ($\pi^{+}, \pi^{0}, \pi^{-}$) form a triplet. Moreover, we have baryon octet ($n, p, \Sigma^{-}, \Sigma^{0}, \Sigma^{+}, \Lambda, \Xi^{-}, \Xi^{0}$), meson octet ($K^{0}, K^{+}, \pi^{-}, \pi^{0}, \pi^{+}, \eta, K^{-}, \overline{K}^{0}$), and meson decuplet ($\Delta^{-}, \Delta^{0}, \Delta^{+}, \Delta^{++}, \Sigma^{*-}, \Sigma^{*0}, \Sigma^{*+}, \Xi^{*-}, \Xi^{*+}, \Omega^{-}$). When particles in a multiplet rotate into each other, the symmetry is invariant. However, $e^{-}$ and $\nu_{e}$ are not similar but conspicuously different. While the former is massive, the latter is massless (in Standard Model). Their difference is neither slight nor symmetric in any sense. Why do we regard ($e^{-}, \nu_{e}$) as a doublet?

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You are contrasting the Weak interactions to the Strong interactions, where the breaking is spontaneous (and large) as contrasted to explicit (and small), respectively.

($e^{-}, \nu_{e}$) are a doublet under weak isospin. This means the the corresponding SM lagrangian is invariant under an SU(2) gauge group -- you could rotate the fields under such a transformation and the Lagrangian would stay the same. This gives these two fermions remarkably many properties in common, such as lepton number, and dictates how they would connect under WI transmutations; this only applies to left-handed chiral components, the ones in this doublet. The right handed components are unconnected.

However, the electroweak v.e.v. in this symmetry is not invariant, and so the symmetry is spontaneously broken, and many mass degeneracies of such doublets are vitiated. In fact, the particular Yukawa coupling giving mass to the electron is different than that giving mass to the neutrino, so their masses are completely disconnected: this is an unsung glory of the Standard Model.

Their charges are different, but related, as well, since the electric charge does not commute with these SU(2) generators, but entangles into their common weak hypercharge.

Their spins are the same.

The multiplets you consider, by contrast, are hadron multiplets "almost degenerate" under flavor SU(3) transformation, which (almost) commute with the Hamiltonian to be sure, but also leave the strong interaction vacuum invariant, unlike the above. So, to a zeroth approximation, their masses are the same ... but you noticed their charges are also different, systematically related by the strong hypercharge Gell-Mann--Nishijima formula. Their spins are the same.

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  • $\begingroup$ To a zeroth approximation, a cow is a sphere :) $\endgroup$ – my2cts Jul 21 '18 at 18:49
  • $\begingroup$ I thought the SU(3) symmetry of the standard model was not broken. Why "(almost)"? $\endgroup$ – Noiralef Jul 21 '18 at 21:19
  • $\begingroup$ You are both asking the same question. The flavor SU(3) symmetry is broken by the quark masses, explicitly, so it is almost good, but pions are not degenerate with kaons. (Technically, quark masses are smaller than $\Lambda_{QCD}$, the scale of strong dynamics.) But, e.g. the charmed quark mass is larger, so flavor SU(4) is not a good symmetry: it is more like a cow to a sphere, as opposed to an apple to a sphere, to slide down the bannister of unstoppable daffy metaphors... $\endgroup$ – Cosmas Zachos Jul 21 '18 at 23:58
  • $\begingroup$ Oh, I somehow thought you were talking about color SU(3), not flavor SU(3)! I will leave my comments up, in case someone else makes the same mistake. $\endgroup$ – Noiralef Jul 22 '18 at 17:16
  • $\begingroup$ @CosmasZachos - Is the flavor symmetry exact only when the masses of the multiplet particles are the same? If so, the $SU(2)$ symmetry of ($e^{-}, \nu_{e}$) is broken because $e^{-}$ and $\nu_{e}$ have quite different masses. Is this right? Then, why do we still think they have an $SU(2)$ symmetry? Is this an approximate $SU(2)$ symmetry? $\endgroup$ – Shen Aug 12 '18 at 18:48

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