What are particle multiplets in the Standard Model?

Question

The particles of the standard model are often displayed in groupings known as multiplets. I know that this somehow relates to the underlying symmetries of the standard model, which can be viewed as tranforming these particles into each other. I have still struggled to pin down exactly what is meant by these multiplet groupings however, and to get a good 'feel' of the concept. Could somebody elaborate please?

Answer 1

It was the observation of symmetries in studying nuclear physics that led to the use of group theory and multiplets. It was experimentally found that the behavior of nuclei did not depend to first order on the number of protons and neutrons, but on the number of nucleons ( either protons or neutrons). In addition it was found that a spin assignment on the nucleon of +1/2 for the proton and -1/2 for the neutron would describe economically interactions observed . This was called isotopic spin and could be organized with the group SU(2).

Then scatterings of elementary particles gave rise to the representations in multiples., because more quantum numbers were found and the symmetries found could be described with multiplets of SU(3).

The meson octet. Particles along the same horizontal line share the same strangeness, s, while those on the same diagonals share the same charge, q.

Isospin on the x axis and strangeness quantum number on the y describe the baryon octet. The masses to first order of the isospin multiplets of the x axis are the same.

The organization into multiplets for all resonances and excitations had predictive behavior, as with the prediction of the omega- , the tip of the decouplet.

The first Omega baryon discovered was the Ω−, made of three strange quarks, in 1964.3 The discovery was a great triumph in the study of quark processes, since it was found only after its existence, mass, and decay products had been predicted by American physicist Murray Gell-Mann in 1962 and independently by Israeli physicist Yuval Ne'eman.

The symmetries observed led to the quark model of elementary particles to start with, and to the extensive use of group symmetries in the proposed theories, leading to the standard model with the $SU(3)\times SU(2)\times U(1)$ symmetries.

Answer 2

Put simply, a particle multiplet is a combination of particles that transform into each other under a symmetry transformation.

In order to describe a system you need two main ingredients:

1. Symmetry groups
2. Field content

If the system is invariant under a symmetry then the fields must be of the form of multiplets (otherwise its impossible to form an combination of fields). There are many examples of such multiplets. These field combinations are just the most convenient way to describe the system with a symmetry.

A particularly rudimentary example is from spin-spin interactions in Quantum Mechanics. We assume the system is invariant under a spin symmetry, $SU(2)$. Then if we suppose the system has two spin 1/2 particles then the four possible states are the singlet and triplet states: \begin{equation} \left|0,0 \right\rangle , \quad \left| 1,1 \right\rangle , \left| 1, 0 \right\rangle , \left| 1 , - 1 \right\rangle \end{equation} Under an SU(2) rotation we have, \begin{align} & \psi ^{ singlet} \rightarrow \psi ^{ singlet} \\ & \psi ^{ triplet} _i \rightarrow U _{ ij} \psi ^{ triplet} _{ j} \end{align} In other words transformations produce rotations between the particles in the multiplet but never take a field outside that multiplet (a triplet can't rotate into a singlet).

Another important example is brought up by AnnaV. The SM is approximately invariant under an $ SU(3) $ chiral symmetry, under which the up, down, and strange quark transform into one another. To see this consider the QCD Lagrangian at energies well below the charm mass such that we can effectively ignore it as well as the bottom and top: \begin{equation} {\cal L} _{ QCD} = \sum _{ i = u,d ,c }\bar{\psi} _i ( i D _\mu \gamma ^\mu - m _i ) \psi _i - \frac{1}{4} G _{ \mu \nu } ^a G ^{ \mu \nu } _{ a} \end{equation} This Lagrangian is not invariant under the flavor $ SU(3) $ transformation, \begin{equation} \psi _i \rightarrow U _{ ij} \psi _i \end{equation} since the mass term in not invariant. But if we work well above the strange mass (but still below the charm mass) then we approximately have, \begin{equation} {\cal L} _{ QCD} \approx \sum _{ i = u,d ,c }\bar{\psi} _i ( i D _\mu \gamma ^\mu ) \psi _i - \frac{1}{4} G _{ \mu \nu } ^a G ^{ \mu \nu } _{ a} \end{equation} which is approximately invariant under the flavor symmetry.

Hadrons obtain their masses primarily due to non-perturbative interactions between the quarks. It turns out that QCD becomes non-perturbative around, \begin{equation} \Lambda _{ QCD} \approx 200 \mbox{MeV} \end{equation} while the charm mass is $ \approx 1000 \mbox{MeV} $ and the strange mass is $ 100 \mbox{MeV} $. Thus hadrons masses can be approximately described the massless Lagrangian above. Since the Lagrangian has an additional symmetry, the particles must form multiplets of the symmetry. While we can't calculate their masses directly, they should approximately exhibit such a symmetry in their masses. This is why we expect the hadron masses to be arranged into flavor multiplets.

Answer 3

There are 5 standard model (SM) multiplets per generation of fermions.

The SM gauge group is $\mathcal{G}_\text{SM} = SU(3)_C \times SU(2)_L \times U(1)_Y$. Various multiplets can then be written as $\mathcal{G}_\text{SM} \ni x = (C,T)_{(Y)}$, where $C$ denotes colour multiplet, $T$ weak isospin multiplet and $Y$ hypercharge value. Multiplets (1st generation) are then

$$ Q = (3,2)_{(1/3)} = \begin{pmatrix} u_r & u_g & u_b \\ d_r & d_g & d_b \end{pmatrix}, \quad \text{quark multiplet}, $$ $$ L = (1,2)_{(-1)} = \begin{pmatrix} \nu_e \\ e \end{pmatrix}, \quad \text{leptonic doublet}, $$ $$ u^c = (\bar{3},1)_{(-4/3)} = \begin{pmatrix} u^c_{\bar{r}} & u^c_{\bar{g}} & u^c_{\bar{b}} \end{pmatrix}, \quad \text{anti-up quarks}, $$ $$ d^c = (\bar{3},1)_{(2/3)} = \begin{pmatrix} d^c_{\bar{r}} & d^c_{\bar{g}} & d^c_{\bar{b}} \end{pmatrix}, \quad \text{anti-down quarks}, $$ $$ e^c = (1,1)_{(2)}, \quad \text{positron}. $$ Anti-neutrino (right-handed component -- you need it in case of massive neutrinos) is a SM singlet, so it doesn't transform under $\mathcal{G}_\text{SM}$, meaning the multiplet could be written as $$ \nu^c = (1,1)_{(0)} $$ The reason, why $e^c$ is included as a SM ''multiplet", and $\nu^c$ isn't, is because it has a non-zero hypercharge, and participates in the $U(1)_Y$ interactions.

You have these 5 multiplets for each generation: $Q_1$ for example contains $u$ and $d$ quarks, $Q_2$ contains $c$ and $s$ quarks and so on. The number within that multiplet notation means which (irreducible) representation the multiplet belongs to. This why $u^c$ and $d^c$ belong to the $\overline{\mathbf{3}}$ representation and have anti-colours in the subscripts.

From here you can see that for example $Q$ transforms as a triplet under $SU(3)_C$, that it transforms as a doublet under $SU(2)_L$ and that it belongs to a non-trivial representation under $U(1)_Y$ (meaning it has a non-zero hypercharge). This means it will interact via all three fundamental interactions. Leptons for example will only interact via the electroweak interactions (in the multiplet notation it's obvious they don't carry colour).

For $u^c$ and $d^c$ you see they don't interact under $SU(2)_L$. However, gauge bosons $W^\pm$ and $Z$ don't belong to the $SU(2)_L$ directly: they're linear combinations of $SU(2)_L$ and $U(1)_Y$ generators. The "weak" interaction via $Z$ and $W^\pm$ isn't strictly weak interaction, in the sense, that your multiplet could be a $SU(2)_L$ singlet, and as long it has a non-zero hypercharge it will interact with those bosons. True $SU(2)_L$ generators are also massless, $Z$ and $W^\pm$ are just what's left after you break the symmetry via the Higgs mechanism at low enough energies (around $M_Z$). In the same sense the photon isn't the $U(1)_Y$ generator, but what's left of the $SU(2)_L\times U(1)_Y$ generators, and mediates the $U(1)_\text{em}$ (electromagnetic) interaction. So yes, photon is also a linear combination of the $SU(2)_L$ and $U(1)_Y$ generators.

As to 'why do you need them' ... you don't really, unless you're doing Grand Unified Theories (GUTs). There all the SM particless are typically in one or two multiplets for generation. For example, under $SO(10)$ GUT, entire generation of particles then belongs to only one representation (multiplet), and that is $\mathbf{16}$: $$ \mathbf{16} = (Q, u^c, d^c, L, e^c, \nu^c). $$ It's then easier to identify SM particles in a GUT if you already know how they transform. For example, you see that after you break $SO(10)$ to $\mathcal{G}_\text{SM}$ you find that the first 6 components of this vector transform as $(3,2)_{(1/3)}$, so you can identify them with the quark multiplet $Q$.