# What is the symmetry of the pion triplet ($\pi^{-}, \pi^{0}, \pi^{+}$)?

Under the entry "Isospin" in Wikipedia, it states:

The pions are assigned to the triplet (the spin-1, $$\mathbf{3}$$, or adjoint representation) of $$SU(2)$$

Why is the symmetry not $$SU(3)$$ since there are three particles? And in what circumstance do we have an $$SU(3)$$ symmetry?

• Seems like this was answered here: physics.stackexchange.com/q/351812 Jul 29, 2018 at 18:59
• Isospin SU(2) has a doublet representation, (u,d), a triplet representation, the 3 πs, an isoquartet representation, the 4 Δs, and so on... Do you get the formal connection to angular momentum now? Jul 29, 2018 at 19:15
• @Cosmas Zachos - I don't know how this is connected with angular momentum. Can you explain more clearly?
– Shen
Jul 30, 2018 at 17:43
• SU(2) ~ SO(3) is also the group of angular momentum, except here in isospace, an abstract notional space. The spin doublets, spin 1/2, correspond to isodoublets here, u,d quarks. The spin triplets, spin 1, like 3 vectors, correspond to isotriplets, the pions. The spin quartets, spin 3/2, correspond to the four $\Delta$ baryons, etc.... All you need do is recall the representation theory of angular momentum, otherwise the language would be opaque. Jul 30, 2018 at 18:44
• My sense is you are confusing the dimensionality of the representation with the dimension of the Lie algebra, namely the number of independent generators involved. The pions can be recast into a real 3-vector, so SO(3) ~ SU(2). But fermions cannot, being intrinsically complex spinors, so you need SU(2) for 2 flavors of quark and SU(3) for 3 such. Jul 31, 2018 at 15:35

$$\newcommand{\BK}[3]{\left|{#1},{#2}\right\rangle_{#3}} \newcommand{\BKB}[3]{\mathbf{\left|{#1},{#2}\right\rangle_{\boldsymbol{#3}}}} \newcommand{\FR}[2]{{\textstyle \frac{#1}{#2}}} \newcommand{\BoldExp}[2]{{#1}^{\boldsymbol{#2}}} \newcommand{\CMRR}[2] { \begin{bmatrix} #1 \\ #2 \end{bmatrix} } \newcommand{\MM}[4] { \begin{bmatrix} #1 & #2\\ #3 & #4 \end{bmatrix} } \newcommand{\MMM}[9] { \begin{bmatrix} #1 & #2 & #3 \\ #4 & #5 & #6 \\ #7 & #8 & #9 \\ \end{bmatrix} } \newcommand{\CMRRRR}[4] { \begin{bmatrix} #1 \\ #2 \\ #3 \\ #4 \end{bmatrix} } \newcommand{\CMRRR}[3] { \begin{bmatrix} #1 \\ #2 \\ #3 \end{bmatrix} } \newcommand{\RMCC}[2] { \begin{bmatrix} #1 & #2 \end{bmatrix} } \newcommand{\RMCCC}[3] { \begin{bmatrix} #1 & #2 & #3 \end{bmatrix} } \newcommand{\RMCCCC}[4] { \begin{bmatrix} #1 & #2 & #3 & #4 \end{bmatrix} } \newcommand{\OSS}[1] {\overset{\boldsymbol{\sim}}{#1}} \newcommand{\BoldSub}[2]{{#1}_{\boldsymbol{#2}}} \newcommand{\OSB}[1] {\overset{\boldsymbol{-\!\!\!\!\!-}}{#1}}$$

These pions are mesons, composite particles of a quark $$\boldsymbol{\lbrace}\boldsymbol{u},\boldsymbol{d}\boldsymbol{\rbrace}$$ and an antiquark $$\boldsymbol{\lbrace}\OSB{\boldsymbol{u}},\overline{\boldsymbol{d}}\boldsymbol{\rbrace}$$ :
$$$$\begin{array}{cccccccc} &\boldsymbol{\lbrace}\boldsymbol{u},\boldsymbol{d}\boldsymbol{\rbrace} \!\!\!\!\!&\boldsymbol{\otimes}& \!\!\!\!\boldsymbol{\lbrace}\OSB{\boldsymbol{u}},\overline{\boldsymbol{d}}\boldsymbol{\rbrace} & \!\!\boldsymbol{=}\!\! & \boldsymbol{\lbrace}\boldsymbol{\omega}\boldsymbol{\rbrace}& \!\!\!\!\boldsymbol{\oplus}\!\!&\boldsymbol{\lbrace}\BoldExp{\boldsymbol{\pi}}{-},\BoldExp{\boldsymbol{\pi}}{0},\BoldExp{\boldsymbol{\pi}}{+}\boldsymbol{\rbrace} & \\ & \boldsymbol{2}\!\!\!\!\! & \boldsymbol{\otimes} & \!\!\!\!\OSB{\boldsymbol{2}} & \!\!\boldsymbol{=}\!\!&\boldsymbol{1}&\!\!\!\!\boldsymbol{\oplus}\!\!&\boldsymbol{3}& \end{array} \tag{01}\label{eq01}$$$$ \begin{align} &\left\{ \boldsymbol{\omega} = \sqrt{\tfrac{1}{2}}\left(\boldsymbol{u}\OSB{\boldsymbol{u}}+\boldsymbol{d}\overline{\boldsymbol{d}} \right)\hphantom{=\,}\right\} \quad \,\text{the singlet }\boldsymbol{1} \tag{02.1}\label{eq02.1}\\ &\left. \begin{cases} \BoldExp{\boldsymbol{\pi}}{-} =\boldsymbol{d}\OSB{\boldsymbol{u}} \\ \BoldExp{\boldsymbol{\pi}}{0} =\sqrt{\tfrac{1}{2}}\left(\boldsymbol{u}\OSB{\boldsymbol{u}}-\boldsymbol{d}\overline{\boldsymbol{d}} \right)\\ \BoldExp{\boldsymbol{\pi}}{+} =\boldsymbol{u}\overline{\boldsymbol{d}} \end{cases}\right\}\quad \text{the triplet }\boldsymbol{3} \tag{02.2}\label{eq02.2} \end{align} The subspaces $$\;\boldsymbol{1},\boldsymbol{3}\;$$ are invariant under the isospin group $$\;SU(2)$$.

EDIT

responds to a comment by the OP owner :

This explanation is fine. But I still have a puzzlement. While the three pions ($$\pi^{-}, \pi^{0}, \pi^{+}$$) have an $$SU(2)$$ symmetry, why do the three quarks ($$u,d,s$$) have an $$SU(3)$$ [not $$SU(2)$$] symmetry? More generally, given three similar particles, how do we know whether they have an $$SU(2)$$ symmetry or an $$SU(3)$$ symmetry?

We must not confuse the number $$\;n\;$$ of the symmetry group $$\;SU(n)\;$$ with the number $$\;m\;$$ of the resulting $$\;m-$$plets (singlets,doublets,triplets,...nonets, etc).

In the following three examples the number $$\;n\;$$ of the symmetry group $$\;SU(n)\;$$ is the number of the $$\;n\;$$ independent $$\;n-$$dimensional systems we put together to build a composite system.

$$\color{blue}{\textbf{Example A :}}$$ If we put together a particle $$\;\alpha\;$$ of spin angular momentum $$\;j_{\alpha}=\frac12\;$$ with a particle $$\;\beta\;$$ of spin angular momentum $$\;j_{\beta}=\frac12\;$$ then the resulting multiplets is a singlet of angular momentum $$\;j_{1}=0\;$$ and a triplet of angular momentum $$\;j_{2}=1\;$$ $$$$\boldsymbol{2}\boldsymbol{\otimes}\boldsymbol{2}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3} \tag{ed-01}\label{eqed-01}$$$$ Now let apply the following $$\;SU(2)\;$$ transformations to the systems $$\;\alpha,\beta\;$$ (particles) respectively \begin{align} ^{\bf 2}U_{\bf \alpha} & = \MM{\hphantom{\boldsymbol{-}}g_{\bf \alpha}}{h_{\bf \alpha}}{\vphantom{h^{\boldsymbol{*}}_{\bf \beta}}\boldsymbol{-}h^{\boldsymbol{*}}_{\bf \alpha}}{g^{\boldsymbol{*}}_{\bf \alpha}}_{\bf a} \,,\quad g_{\bf \alpha}g^{\boldsymbol{*}}_{\bf \alpha}\boldsymbol{+}h_{\bf \alpha}h^{\boldsymbol{*}}_{\bf \alpha}=1 \tag{ed-02a}\label{eqed-02a}\\ ^{\bf 2}U_{\bf \beta} & = \MM{\hphantom{\boldsymbol{-}}g_{\bf \beta}}{h_{\bf \beta}}{\boldsymbol{-}h^{\boldsymbol{*}}_{\bf \beta}}{g^{\boldsymbol{*}}_{\bf \beta}}_{\bf b} \,,\quad g_{\bf \beta}g^{\boldsymbol{*}}_{\bf \beta}\boldsymbol{+}h_{\bf \beta}h^{\boldsymbol{*}}_{\bf \beta}=1 \tag{ed-02b}\label{eqed-02b} \end{align} In the composite system this is a $$\;SU(4)\;$$ transformation, the product of the two ones above

$$$$^{\bf 4}U_{ f} = \left(^{\bf 2}U_{\bf \alpha}\right)\boldsymbol{\otimes}\left(^{\bf 2}U_{\bf \beta}\right) = \MM{\hphantom{\boldsymbol{-}}g_{\bf \alpha}}{h_{\bf \alpha}}{\vphantom{h^{\boldsymbol{*}}_{\bf \beta}}\boldsymbol{-}h^{\boldsymbol{*}}_{\bf \alpha}}{g^{\boldsymbol{*}}_{\bf \alpha}}_{\bf a}\!\!\! \boldsymbol{\otimes} \MM{\hphantom{\boldsymbol{-}}g_{\bf \beta}}{h_{\bf \beta}}{\boldsymbol{-}h^{\boldsymbol{*}}_{\bf \beta}}{g^{\boldsymbol{*}}_{\bf \beta}}_{\bf b}\!\!\! = \begin{bmatrix} \hphantom{\boldsymbol{-}}g_{\bf \alpha}g_{\bf \beta} & \hphantom{\boldsymbol{-}}g_{\bf \alpha}h_{\bf \beta} & \hphantom{\boldsymbol{-}}h_{\bf \alpha}g_{\bf \beta} & h_{\bf \alpha}h_{\bf \beta} \\ \boldsymbol{-}g_{\bf \alpha}h^{\boldsymbol{*}}_{\bf \beta} & \hphantom{\boldsymbol{-}}g_{\bf \alpha}g^{\boldsymbol{*}}_{\bf \beta} & \boldsymbol{-}h_{\bf \alpha}h^{\boldsymbol{*}}_{\bf \beta} & h_{\bf \alpha}g^{\boldsymbol{*}}_{\bf \beta} \\ \boldsymbol{-}h^{\boldsymbol{*}}_{\bf \alpha}g_{\bf \beta} & \boldsymbol{-}h^{\boldsymbol{*}}_{\bf \alpha}h_{\bf \beta} & \hphantom{\boldsymbol{-}}g^{\boldsymbol{*}}_{\bf \alpha}g_{\bf \beta} & g^{\boldsymbol{*}}_{\bf \alpha}h_{\bf \beta} \\ \hphantom{\boldsymbol{-}}h^{\boldsymbol{*}}_{\bf \alpha}h^{\boldsymbol{*}}_{\bf \beta} & \boldsymbol{-}h^{\boldsymbol{*}}_{\bf \alpha}g^{\boldsymbol{*}}_{\bf \beta} & \boldsymbol{-}g^{\boldsymbol{*}}_{\bf \alpha}h^{\boldsymbol{*}}_{\bf \beta} & g^{\boldsymbol{*}}_{\bf \alpha}g^{\boldsymbol{*}}_{\bf \beta} \end{bmatrix}_{\bf e} \tag{ed-03}\label{eqed-03}$$$$

But the $$\;SU(2)\;$$ transformations in \eqref{eqed-02a},\eqref{eqed-02b} represent rotations in the real space $$\;\mathbb{R}^{3}\;$$ wherein both particles live, so they must be identical (we would not rotate one system differently from the other) $$$$^{\bf 2}U_{\bf \alpha} =\,^{\bf 2}U_{\bf \beta}=\, ^{\bf 2}U = \MM{\:\:g}{h}{\boldsymbol{-}h^{\boldsymbol{*}}}{\:\:g^{\boldsymbol{*}}} \,,\quad gg^{\boldsymbol{*}}\boldsymbol{+}hh^{\boldsymbol{*}}=1 \tag{ed-04}\label{eqed-04}$$$$ so that \eqref{eqed-03} yields $$$$^{\bf 4}U_{ f} = \left(^{\bf 2}U_{\bf \alpha}\right)\boldsymbol{\otimes}\left(^{\bf 2}U_{\bf \beta}\right) =\left(^{\bf 2}U\right)^{\boldsymbol{\otimes}2} = \begin{bmatrix} \:g^{2} & \:\:gh & \:hg & \!\!\!h^{2} \\ \boldsymbol{-}gh^{\boldsymbol{*}} & \hphantom{\boldsymbol{-}}gg^{\boldsymbol{*}} & \boldsymbol{-}hh^{\boldsymbol{*}} & hg^{\boldsymbol{*}}\\ \boldsymbol{-}h^{\boldsymbol{*}}g & \,\boldsymbol{-}h^{\boldsymbol{*}}h & \hphantom{\boldsymbol{-}}g^{\boldsymbol{*}}g & g^{\boldsymbol{*}}h \\ \hphantom{\boldsymbol{-}}h^{\boldsymbol{*}2} & \:\:\boldsymbol{-}h^{\boldsymbol{*}}g^{\boldsymbol{*}} & \:\:\boldsymbol{-}g^{\boldsymbol{*}}h^{\boldsymbol{*}} & g^{\boldsymbol{*}2} \end{bmatrix}_{\bf e} \tag{ed-05}\label{eqed-05}$$$$

This matrix expressed in the basis of the irreducible direct sum \eqref{eqed-01} is $$$$^{\bf 4}\OSS{U}_{ f}= \begin{bmatrix} \begin{array}{c|ccc} \:\: 1 \:\: &\rule [0ex]{20pt}{0.0ex}&\rule [-2.5ex]{0pt}{6.0ex} \rule [0ex]{16pt}{0ex}& \rule [0ex]{16pt}{0ex}\\ \hline \rule [-3ex]{0pt}{6ex}&g^{2}& \sqrt{2} g h & h^{2} \\ \rule [-3ex]{0pt}{6ex}& -\sqrt{2} g h^{\boldsymbol{*}} & \left(g g^{\boldsymbol{*}}-h h^{\boldsymbol{*}}\right) & \sqrt{2} g^{\boldsymbol{*}} h \\ \rule [-3ex]{0pt}{6ex}& \left(h^{\boldsymbol{*}}\right)^{2} & - \sqrt{2}g^{\boldsymbol{*}} h^{\boldsymbol{*}} & \left(g^{\boldsymbol{*}}\right)^{2} \end{array} \end{bmatrix}_{\:\mathbf{f}} = \begin{bmatrix} \begin{array}{c|ccc} ^{\mathbf{1}}U_{\boldsymbol{\left[1\right]}}&\rule [0ex]{20pt}{0.0ex}&\rule [-2.5ex]{0pt}{6.0ex} \rule [0ex]{16pt}{0ex}& \rule [0ex]{16pt}{0ex}\\ \hline \rule [-3ex]{0pt}{6ex}&\rule [0.0ex]{50pt}{0.0ex}& \rule [0.0ex]{50pt}{0.0ex} &\rule [0.0ex]{50pt}{0.0ex}\\ \rule [-3ex]{0pt}{6ex}& & ^{\mathbf{3}}U_{\boldsymbol{\left[2\right]}} & \\ \rule [-3ex]{0pt}{6ex}& & & \end{array} \end{bmatrix}_{\:\mathbf{f}} \tag{ed-06}\label{eqed-06}$$$$ where $$\:^{\mathbf{1}}U_{\boldsymbol{\left[1\right]}}\:$$ and $$\:^{\mathbf{3}}U_{\boldsymbol{\left[2\right]}}\:$$ are special unitary matrices in the spaces of the singlet and of the triplet respectively given by $$$$^{\mathbf{1}}U_{\boldsymbol{\left[1\right]}}= \begin{bmatrix} 1 \end{bmatrix} \quad \in SU(1)\equiv \{1\} \tag{ed-07}\label{eqed-07}$$$$

$$$$^{\mathbf{3}}U_{\boldsymbol{\left[2\right]}}= \begin{bmatrix} g^{2}& \sqrt{2} g h & h^{2} \rule [-3ex]{0pt}{6ex}\\ -\sqrt{2} g h^{\boldsymbol{*}} & \left(g g^{\boldsymbol{*}}-h h^{\boldsymbol{*}}\right) & \sqrt{2} g^{\boldsymbol{*}} h \rule [-3ex]{0pt}{6ex}\\ \left(h^{\boldsymbol{*}}\right)^{2} & - \sqrt{2}g^{\boldsymbol{*}} h^{\boldsymbol{*}} & \left(g^{\boldsymbol{*}}\right)^{2} \rule [-3ex]{0pt}{6ex} \end{bmatrix} \quad \in SU(3) \tag{ed-08}\label{eqed-08}$$$$ So if we apply the $$\;SU(2)\;$$ transformation $$\:^{\bf 2}U\:$$ of \eqref{eqed-04} on both spaces in the product of the lhs of \eqref{eqed-01} then the spaces of the terms of the direct sum of the rhs side of the same equation remain invariant, the singlet \eqref{eq02.1} invariant under $$\;SU(1)\;$$ (more exactly unchanged) and the triplet \eqref{eq02.2} transformed under $$\;SU(3)\;$$ remaining in its invariant space.

We say that the symmetry group is $$\;SU(2)$$, NOT $$\;SU(1)\;$$ or $$\;SU(3)\;$$ of the resulting multiplets.

Reference link : Total spin of two spin-1/2 particles.

$$\color{blue}{\textbf{Example B :}}$$ The quark model of baryons consisting of three quarks. So, suppose we know the existence of three quarks only : $$\boldsymbol{u}$$, $$\boldsymbol{d}$$ and $$\boldsymbol{s}$$. Under full symmetry (the same mass) these are the basic states, let
$$$$\boldsymbol{u}= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} \qquad \boldsymbol{d}= \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix} \qquad \boldsymbol{s}= \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \tag{ed-09}\label{eqed-09}$$$$ of a 3-dimensional complex Hilbert space of quarks, say $$\mathbf{Q}\equiv \mathbb{C}^{\boldsymbol{3}}$$. A quark $$\boldsymbol{\xi} \in \mathbf{Q}$$ is expressed in terms of these basic states as $$$$\boldsymbol{\xi}=\xi_1\boldsymbol{u}+\xi_2\boldsymbol{d}+\xi_3\boldsymbol{s}= \begin{bmatrix} \xi_1\\ \xi_2\\ \xi_3 \end{bmatrix} \qquad \xi_1,\xi_2,\xi_3 \in \mathbb{C} \tag{ed-10}\label{eqed-10}$$$$ Let take 2 more quarks in order to construct baryons from 3 quarks $$$$\boldsymbol{\eta}=\eta_1\boldsymbol{u}+\eta_2\boldsymbol{d}+\eta_3\boldsymbol{s}= \begin{bmatrix} \eta_1\\ \eta_2\\ \eta_3 \end{bmatrix} \:, \qquad \boldsymbol{\zeta}=\zeta_1\boldsymbol{u}+\zeta_2\boldsymbol{d}+\zeta_3\boldsymbol{s}= \begin{bmatrix} \zeta_1\\ \zeta_2\\ \zeta_3 \end{bmatrix} \tag{ed-11}\label{eqed-11}$$$$ A baryon state $$\:T\:$$ in the product space $$$$\mathbf{B}=\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}=\mathbf{Q}\boldsymbol{\otimes}\mathbf{Q}\boldsymbol{\otimes}\mathbf{Q}\equiv \mathbb{C}^{\boldsymbol{3}}\boldsymbol{\otimes}\mathbb{C}^{\boldsymbol{3}}\boldsymbol{\otimes}\mathbb{C}^{\boldsymbol{3}}=\mathbb{C}^{\boldsymbol{27}} \tag{ed-12}\label{eqed-12}$$$$ is the product of the states of above 3 quarks $$$$T=\boldsymbol{\xi}\boldsymbol{\otimes}\boldsymbol{\eta}\boldsymbol{\otimes}\boldsymbol{\zeta} \tag{ed-13}\label{eqed-13}$$$$ The final result of a full analysis is $$$$\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}= \boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{10}\boldsymbol{\oplus} \boldsymbol{8}^{\boldsymbol{\prime}}\boldsymbol{\oplus}\boldsymbol{8} \tag{ed-14}\label{eqed-14}$$$$ that is the space of states of a baryon is the direct sum of a singlet $$\;\boldsymbol{1}$$, a decuplet $$\;\boldsymbol{10}$$, a mixed symmetric octet $$\;\boldsymbol{8'}$$ and a mixed anti-symmetric octet $$\;\boldsymbol{8}$$.

Now applying a $$\;SU(3)\;$$ transformation $$\;^{\bf 3}U\;$$ on the 3-dimensional space $$\mathbf{Q}\equiv \mathbb{C}^{\boldsymbol{3}}$$ results in a $$\;SU(27)\;$$ transformation $$\;^{\bf 27}U\;$$ on the 27-dimensional space $$\;\mathbf{B}\;$$ of equation \eqref{eqed-12} $$$$^{\bf 27}U = \left(^{\bf 3}U\right)\boldsymbol{\otimes}\left(^{\bf 3}U\right)\boldsymbol{\otimes}\left(^{\bf 3}U\right) =\left(^{\bf 3}U\right)^{\boldsymbol{\otimes}3} \tag{ed-15}\label{eqed-15}$$$$ The space of each $$\;m-$$plet remains invariant and a state in this $$\;m-$$plet is transformed under a $$\;SU(m)\;$$ transformation, where $$\;m=1,10,8,8$$. But

We say that the symmetry group is $$\;SU(3)$$, NOT $$\;SU(1)\;$$ or $$\;SU(10)\;$$ or $$\;SU(8)\;$$ of the resulting multiplets.

Reference link : Symmetry in terms of matrices.

• ω , a vector meson, in the same breath as pseudscalars, just because the η is messier? Does it clarify anything? Jul 29, 2018 at 19:37
• This explanation is fine. But I still have a puzzlement. While the three pions ($\pi^{-}, \pi^{0}, \pi^{+}$) have an $SU(2)$ symmetry, why do the three quarks ($u, d, s$) have an $SU(3)$ [not $SU(2)$] symmetry? More generally, given three similar particles, how do we know whether they have an $SU(2)$ symmetry or an $SU(3)$ symmetry?
– Shen
Jul 30, 2018 at 17:28
• @frobenius LPT: instead of typing \boldsymbol{8}^{\boldsymbol{\prime}}\boldsymbol{\oplus}\boldsymbol{8}, you could just type \boldsymbol{8'\oplus8} for the exact same output. Similarly for the rest of your formulas. Jul 30, 2018 at 23:57
• It's always a pleasure to see posts so well-formatted.
– Nat
Aug 2, 2018 at 12:48
• @Nat : Thanks, I always try to do the best for format and Figures. Aug 2, 2018 at 12:52

As per urging by @rob, here is the concise answer:

Isospin SU(2) has a doublet representation, (u,d); a triplet representation, the 3 πs; an isoquartet representation, the 4 Δs; and so on... You already know this from angular momentum, since, SU(2) ~ SO(3) is also the group of rotations/angular momentum, except here in isospace, an abstract notional space: The spin doublets, spin 1/2, correspond to isodoublets here, u,d quarks. The spin triplets, spin 1, like 3-vectors, correspond to isotriplets, the pions. The spin quartets, spin 3/2, correspond to the four Δ baryons, etc... All SU(2) irreps are real (in a slightly technical sense... even the spinors).

Now, unlike SU(2), flavor SU(3) has a truly complex representation, a triple (u,d,s); a real octet representation; a complex decuplet, etc...

Now you consider a real triplet of pions, thus a real 3-vector. You know this vector transforms under SO(3) ~ SU(2), just like rotations of real vectors, so the group is the isospin SU(2) as stated.

However, if it were a spinor, instead, a complex triplet, it would have to transform under an SU(3): you could not restrict the number of independent transformations of its components to SO(3), and you'd be stuck with SU(3), eight independent transformation directions.

This is what dictates SU(3) for a complex triplet of quarks, (u,d,s); even though, historically, the logic went backwards: the complex triplet was suggested by the fundamental representation of flavor SU(3), inferred by the real meson octet!

• In case you were stumped by the complex $\pi^{\pm}\equiv (\pi^1\pm i\pi^2)/\sqrt{2}$, this is just the spherical vector rewriting of the Cartesian components $\pi^1,\pi^2$, so group theoretically the pion is still a real triplet.
• just want to draw your attention to this physics.stackexchange.com/questions/96440/… . My vague understanding is that the operators take care of this difference, but searching on google I found somebody answering that the field is spllt into ψ+ and ψ- . It might be good if you answered this. thanks Sep 12, 2018 at 3:51
• @anna ... i think the answer there is just fine... yes, the particle & antiparticle are encoded in the same field, in different components.... Dirac thought of them as holes, but the QFT interpretation is smoother... QFT books hash it out: different Fourier components of the same field tag the particle vs antiparticle pieces.... Sep 12, 2018 at 8:42
• Thanks, do you recomment a particular book? . Sep 12, 2018 at 9:14
• Maybe Peskin & Schroeder, p 29? Sep 12, 2018 at 14:40

$\newcommand{\BK}[3]{\left|{#1},{#2}\right\rangle_{#3}} \newcommand{\BKB}[3]{\mathbf{\left|{#1},{#2}\right\rangle_{\boldsymbol{#3}}}} \newcommand{\FR}[2]{{\textstyle \frac{#1}{#2}}} \newcommand{\BoldExp}[2]{{#1}^{\boldsymbol{#2}}} \newcommand{\CMRR}[2] { \begin{bmatrix} #1 \\ #2 \end{bmatrix} } \newcommand{\MM}[4] { \begin{bmatrix} #1 & #2\\ #3 & #4 \end{bmatrix} } \newcommand{\MMM}[9] { \begin{bmatrix} #1 & #2 & #3 \\ #4 & #5 & #6 \\ #7 & #8 & #9 \\ \end{bmatrix} } \newcommand{\CMRRRR}[4] { \begin{bmatrix} #1 \\ #2 \\ #3 \\ #4 \end{bmatrix} } \newcommand{\CMRRR}[3] { \begin{bmatrix} #1 \\ #2 \\ #3 \end{bmatrix} } \newcommand{\RMCC}[2] { \begin{bmatrix} #1 & #2 \end{bmatrix} } \newcommand{\RMCCC}[3] { \begin{bmatrix} #1 & #2 & #3 \end{bmatrix} } \newcommand{\RMCCCC}[4] { \begin{bmatrix} #1 & #2 & #3 & #4 \end{bmatrix} } \newcommand{\OSS}[1] {\overset{\boldsymbol{\sim}}{#1}} \newcommand{\BoldSub}[2]{{#1}_{\boldsymbol{#2}}} \newcommand{\OSB}[1] {\overset{\boldsymbol{-\!\!\!-}}{#1}}$

$\color{blue}{\textbf{Example C :}}$ The quark model of mesons consisting of two quarks (relevant to the question). So, suppose we know the existence of two quarks only: $\boldsymbol{u}$ and $\boldsymbol{d}$. Under full symmetry these are the basic states, let
$$\boldsymbol{u}= \begin{bmatrix} \:1\:\vphantom{\dfrac12}\\ \:0\:\vphantom{\dfrac12} \end{bmatrix} \qquad \boldsymbol{d}= \begin{bmatrix} \:0\:\vphantom{\dfrac12}\\ \:1\:\vphantom{\dfrac12} \end{bmatrix} \tag{ed-16}\label{eqed-16}$$ of a 2-dimensional complex Hilbert space of quarks, say $\boldsymbol{\lbrace}\boldsymbol{u},\boldsymbol{d}\boldsymbol{\rbrace}=\mathbf{Q}\equiv \mathbb{C}^{\boldsymbol{2}}$. A quark $\boldsymbol{\xi} \in \mathbf{Q}$ is expressed in terms of these basic states as $$\boldsymbol{\xi}=\xi_u\boldsymbol{u}+\xi_d\boldsymbol{d}= \begin{bmatrix} \:\xi_u\:\vphantom{\dfrac12}\\ \:\xi_d\:\vphantom{\dfrac12} \end{bmatrix} \qquad \xi_u,\xi_d \in \mathbb{C} \tag{ed-17}\label{eqed-17}$$ For a quark $\boldsymbol{\zeta} \in \mathbf{Q}$ $$\boldsymbol{\zeta}=\zeta_u\boldsymbol{u}+\zeta_d\boldsymbol{d}= \begin{bmatrix} \:\zeta_u\:\vphantom{\dfrac12}\\ \:\zeta_d\:\vphantom{\dfrac12} \end{bmatrix} \tag{ed-18}\label{eqed-18}$$ the respective antiquark $\overline{\boldsymbol{\zeta}}$ is expressed by the complex conjugates of the coordinates $$\overline{\boldsymbol{\zeta}}=\overline{\zeta}_u \OSB{\boldsymbol{u}}+\overline{\zeta}_d\overline{\boldsymbol{d}}= \begin{bmatrix} \:\overline{\zeta}_u\:\vphantom{\dfrac12}\\ \:\overline{\zeta}_d\:\vphantom{\dfrac12}\\ \end{bmatrix} \tag{ed-19}\label{eqed-19}$$ with respect to the basic states
$$\OSB{\boldsymbol{u}}= \begin{bmatrix} \:1\:\vphantom{\dfrac12}\\ \:0\:\vphantom{\dfrac12} \end{bmatrix} \qquad \overline{\boldsymbol{d}}= \begin{bmatrix} \:0\:\vphantom{\dfrac12}\\ \:1\:\vphantom{\dfrac12} \end{bmatrix} \tag{ed-20}\label{eqed-20}$$ the antiquarks of $\boldsymbol{u},\boldsymbol{d}$ respectively. The antiquarks belong to a different space, the space of antiquarks $\boldsymbol{\lbrace}\OSB{\boldsymbol{u}},\overline{\boldsymbol{d}}\boldsymbol{\rbrace}=\overline{\mathbf{Q}}\equiv \mathbb{C}^{\boldsymbol{2}}$.

Since mesons here are quark-antiquark pairs, they belong to the product space $$\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\: \left(\equiv \mathbb{C}^{\boldsymbol{4}}\right) \tag{ed-21}\label{eqed-21}$$ Using the expressions \eqref{eqed-17} and \eqref{eqed-19} of the quark $\boldsymbol{\xi} \in \mathbf{Q}$ and the antiquark $\overline{\boldsymbol{\zeta}} \in \overline{\mathbf{Q}}$ respectively we have for the product meson state $\mathrm{X} \in \mathbf{M}$

$$\mathrm{X}=\boldsymbol{\xi}\boldsymbol{\otimes}\overline{\boldsymbol{\zeta}}=\xi_u\overline{\eta}_u \left(\boldsymbol{u}\boldsymbol{\otimes}\OSB{\boldsymbol{u}}\right)+\xi_u\overline{\zeta}_d \left( \boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_d\overline{\zeta}_u\left( \boldsymbol{d}\boldsymbol{\otimes}\OSB{\boldsymbol{u}}\right)+\xi_d\overline{\zeta}_d\left( \boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right) \nonumber \tag{ed-22}\label{eqed-22}$$ In order to simplify the expressions, the product symbol $"\boldsymbol{\otimes}"$ is omitted and so $$\mathrm{X}=\boldsymbol{\xi}\overline{\boldsymbol{\zeta}}=\xi_u\overline{\zeta}_u \boldsymbol{u}\OSB{\boldsymbol{u}}+\xi_u\overline{\zeta}_d \boldsymbol{u}\overline{\boldsymbol{d}}+\xi_d\overline{\zeta}_u \boldsymbol{d}\OSB{\boldsymbol{u}}+\xi_d\overline{\zeta}_d \boldsymbol{d}\overline{\boldsymbol{d}} \tag{ed-23}\label{eqed-23}$$ or in one column matrix form $$\mathrm{X}= \begin{bmatrix} \begin{array}{rrrr} \xi_u\overline{\zeta}_u\vphantom{\dfrac12}\\ \xi_u\overline{\zeta}_d\vphantom{\dfrac12}\\ \xi_d\overline{\zeta}_u\vphantom{\dfrac12}\\ \xi_d\overline{\zeta}_d\vphantom{\dfrac12} \end{array} \end{bmatrix}_{\mathbf{e}} \tag{ed-24}\label{eqed-24}$$ This representation is with respect to the basis
$$\boldsymbol{e_1}=\boldsymbol{u}\OSB{\boldsymbol{u}}= \begin{bmatrix} 1\\ 0\\ 0\\ 0 \end{bmatrix}\,, \quad \boldsymbol{e_2}=\boldsymbol{u}\overline{\boldsymbol{d}}= \begin{bmatrix} 0\\ 1\\ 0\\ 0 \end{bmatrix}\,, \quad \boldsymbol{e_3}=\boldsymbol{d}\OSB{\boldsymbol{u}}= \begin{bmatrix} 0\\ 0\\ 1\\ 0 \end{bmatrix}\,, \quad \boldsymbol{e_4}=\boldsymbol{d}\overline{\boldsymbol{d}}= \begin{bmatrix} 0\\ 0\\ 0\\ 1 \end{bmatrix} \tag{ed-25}\label{eqed-25}$$ The final result of a full analysis is $$\boldsymbol{2}\boldsymbol{\otimes} \OSB{\boldsymbol{2}} \boldsymbol{=}\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3} \tag{ed-26}\label{eqed-26}$$ that is the space of states of a meson is the direct sum of a singlet $\;\boldsymbol{1}\equiv\boldsymbol{\lbrace}\boldsymbol{\omega}\boldsymbol{\rbrace}$ and a triplet $\;\boldsymbol{3}\equiv\boldsymbol{\lbrace}\BoldExp{\boldsymbol{\pi}}{-},\BoldExp{\boldsymbol{\pi}}{0},\BoldExp{\boldsymbol{\pi}}{+}\boldsymbol{\rbrace}$.

Now, if we apply a $\;SU(2)\;$ transformation in the space $\;\boldsymbol{2}=\boldsymbol{\lbrace}\boldsymbol{u},\boldsymbol{d}\boldsymbol{\rbrace}=\mathbf{Q}$ represented with respect to the basis \eqref{eqed-16} of this space by the matrix $$^{\boldsymbol{2}}U \equiv \begin{bmatrix} \:\:\:g\hphantom{\boldsymbol{-}}\vphantom{\dfrac12} & h \vphantom{\dfrac12}\:\:\:\\ \!\!\!\!\!\boldsymbol{-}\overline{h} & \OSB{g}\vphantom{\dfrac12}\:\: \end{bmatrix}_{\mathbf{ud}}\;, \qquad g\OSB{g}+h\overline{h}=\vert g \vert ^{2} + \vert h \vert ^{2} = 1 \tag{ed-27}\label{eqed-27}$$ then we must apply in the space $\OSB{\boldsymbol{2}}=\boldsymbol{\lbrace}\OSB{\boldsymbol{u}},\overline{\boldsymbol{d}}\boldsymbol{\rbrace}=\overline{\mathbf{Q}}\equiv \mathbb{C}^{\boldsymbol{2}}$ its complex conjugate represented with respect to the basis \eqref{eqed-20} of this space by the matrix $$^{\boldsymbol{2}}\OSB{U} \equiv \begin{bmatrix} \:\:\:\OSB{g}\hphantom{\boldsymbol{-}}\vphantom{\dfrac12} & \overline{h} \vphantom{\dfrac12}\:\:\:\\ \!\!\!\!\!\boldsymbol{-}h & g\vphantom{\dfrac12}\:\: \end{bmatrix}_{\mathbf{\OSB{\boldsymbol{u}}\overline{\boldsymbol{d}}}} \tag{ed-28}\label{eqed-28}$$
In the composite system $\;\boldsymbol{2}\boldsymbol{\otimes} \OSB{\boldsymbol{2}}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\;$ this is a $\;SU(4)\;$ transformation, the product of these transformations above, represented with respect to the basis \eqref{eqed-25} of this space by the matrix $$^{\boldsymbol{4}}U=\left(^{\boldsymbol{2}}U\vphantom{^{\boldsymbol{2}}\OSB{U}}\right)\boldsymbol{\otimes}\left(^{\boldsymbol{2}}\OSB{U}\right) = \begin{bmatrix} \:\:\:g\hphantom{\boldsymbol{-}}\vphantom{\dfrac12} & h \vphantom{\dfrac12}\:\:\:\\ \!\!\!\!\!\boldsymbol{-}\overline{h} & \OSB{g}\vphantom{\dfrac12}\:\: \end{bmatrix} \boldsymbol{\otimes} \begin{bmatrix} \:\:\:\OSB{g}\hphantom{\boldsymbol{-}}\vphantom{\dfrac12} & \overline{h} \vphantom{\dfrac12}\:\:\:\\ \!\!\!\!\!\boldsymbol{-}h & g\vphantom{\dfrac12}\:\: \end{bmatrix} = \begin{bmatrix} \:g\OSB{g} & \:\:g\overline{h} & \:h\OSB{g} & \!\!\!h\overline{h} \\ \boldsymbol{-}gh & \hphantom{\boldsymbol{-}}g^{2} & \boldsymbol{-}h^{2} & hg\\ \boldsymbol{-}\overline{h}\OSB{g} & \,\boldsymbol{-}\overline{h}^{2} & \hphantom{\boldsymbol{-}}\OSB{g}^{2} & \OSB{g}\overline{h} \\ \hphantom{\boldsymbol{-}}\overline{h}h & \:\:\boldsymbol{-}\overline{h}g & \:-\OSB{g}h & \:\OSB{g}g \end{bmatrix}_{\bf e} \tag{ed-29}\label{eqed-29}$$ We change from the old basis $\;\boldsymbol{\lbrace e_k\rbrace}$, see equation \eqref{eqed-25}, to this new one

\begin{align} \OSS{\boldsymbol{e}}_{\bf 1} & =\sqrt{\tfrac{1}{2}}\left(\boldsymbol{e_1}+\boldsymbol{e_4} \right) = \sqrt{\tfrac{1}{2}}\left(\boldsymbol{u}\OSB{\boldsymbol{u}}+\boldsymbol{d}\overline{\boldsymbol{d}} \right)=\boldsymbol{\omega} \tag{ed-30.1}\label{eqed-30.1}\\ \OSS{\boldsymbol{e}}_{\bf 2} & =\boldsymbol{e_2} =\boldsymbol{u}\overline{\boldsymbol{d}} =\BoldExp{\boldsymbol{\pi}}{+} \tag{ed-30.2}\label{eqed-30.2}\\ \OSS{\boldsymbol{e}}_{\bf 3} & =\sqrt{\tfrac{1}{2}}\left(\boldsymbol{e_1}-\boldsymbol{e_4} \right)=\BoldExp{\boldsymbol{\pi}}{0} \tag{ed-30.3}\label{eqed-30.3}\\ \OSS{\boldsymbol{e}}_{\bf 4} & =\boldsymbol{e_3} =\boldsymbol{d}\OSB{\boldsymbol{u}} =\BoldExp{\boldsymbol{\pi}}{-} \tag{ed-30.4}\label{eqed-30.4} \end{align} Formally $$\begin{bmatrix} \OSS{\boldsymbol{e}}_{\bf 1}\vphantom{\sqrt{\tfrac{1}{2}}}\\ \OSS{\boldsymbol{e}}_{\bf 2}\vphantom{\sqrt{\tfrac{1}{2}}} \\ \OSS{\boldsymbol{e}}_{\bf 3}\vphantom{\sqrt{\tfrac{1}{2}}} \\ \OSS{\boldsymbol{e}}_{\bf 4}\vphantom{\sqrt{\tfrac{1}{2}}} \end{bmatrix} = \begin{bmatrix} \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0 &\hphantom{\boldsymbol{-}} 0 & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{2}}\\ 0 & \hphantom{\boldsymbol{-}}1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0\vphantom{\sqrt{\tfrac{1}{2}}}\\ \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}} 0 & \hphantom{\boldsymbol{-}}0 & -\sqrt{\tfrac{1}{2}}\\ 0 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}1 & \hphantom{\boldsymbol{-}}0\vphantom{\sqrt{\tfrac{1}{2}}} \end{bmatrix} \begin{bmatrix} \boldsymbol{e_1}\vphantom{\sqrt{\tfrac{1}{2}}}\\ \boldsymbol{e_2}\vphantom{\sqrt{\tfrac{1}{2}}}\\ \boldsymbol{e_3}\vphantom{\sqrt{\tfrac{1}{2}}}\\ \boldsymbol{e_4}\vphantom{\sqrt{\tfrac{1}{2}}} \end{bmatrix} =\mathrm{K} \begin{bmatrix} \boldsymbol{e_1}\vphantom{\sqrt{\tfrac{1}{2}}}\\ \boldsymbol{e_2}\vphantom{\sqrt{\tfrac{1}{2}}}\\ \boldsymbol{e_3}\vphantom{\sqrt{\tfrac{1}{2}}}\\ \boldsymbol{e_4}\vphantom{\sqrt{\tfrac{1}{2}}} \end{bmatrix} \tag{ed-31}\label{eqed-31}$$ where $\;\mathrm{K}\;$ the following $\;4\times 4\;$ real orthogonal matrix $$\mathrm{K} = \begin{bmatrix} \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0 &\hphantom{\boldsymbol{-}} 0 & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{2}}\\ 0 & \hphantom{\boldsymbol{-}}1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0\vphantom{\sqrt{\tfrac{1}{2}}}\\ \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}} 0 & \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}\sqrt{\tfrac{1}{2}}\\ 0 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}1 & \hphantom{\boldsymbol{-}}0\vphantom{\sqrt{\tfrac{1}{2}}} \end{bmatrix} \tag{ed-32}\label{eqed-32}$$ with property $$\mathrm{K}^{\boldsymbol{-1}} = \begin{bmatrix} \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0 &\hphantom{\boldsymbol{-}} \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0\\ 0 & \hphantom{\boldsymbol{-}}1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0\vphantom{\sqrt{\tfrac{1}{2}}}\\ 0 & \hphantom{\boldsymbol{-}} 0 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}1\vphantom{\sqrt{\tfrac{1}{2}}}\\ \sqrt{\tfrac{1}{2}}& \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}\sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0 \end{bmatrix} =\mathrm{K}^{\boldsymbol{\top}} \tag{ed-33}\label{eqed-33}$$ The matrix $\;^{\boldsymbol{4}}U$, see equation \eqref{eqed-29}, representing the $\;SU(4)\;$ transformation with respect to the basis $\;\boldsymbol{\lbrace \boldsymbol{e}_k\rbrace}$ has with respect to the new basis $\;\boldsymbol{\lbrace \OSS{\boldsymbol{e}}_k\rbrace}$, see equations \eqref{eqed-30.1}-\eqref{eqed-30.4}, the following form \begin{align} & ^{\boldsymbol{4}}\OSS{U}=\mathrm{K}\left(^{\boldsymbol{4}}U\right)\mathrm{K}^{\boldsymbol{-1}} \nonumber\\ & = \begin{bmatrix} \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0 &\hphantom{\boldsymbol{-}} 0 & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{2}}\\ 0 & \hphantom{\boldsymbol{-}}1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0\vphantom{\sqrt{\tfrac{1}{2}}}\\ \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}} 0 & \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}\sqrt{\tfrac{1}{2}}\\ 0 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}1 & \hphantom{\boldsymbol{-}}0\vphantom{\sqrt{\tfrac{1}{2}}} \end{bmatrix} \begin{bmatrix} \:g\OSB{g} & \:\:g\overline{h} & \:h\OSB{g} & \!\!\!h\overline{h}\vphantom{\sqrt{\tfrac{1}{2}}} \\ \boldsymbol{-}gh & \hphantom{\boldsymbol{-}}g^{2} & \boldsymbol{-}h^{2} & hg\vphantom{\sqrt{\tfrac{1}{2}}}\\ \boldsymbol{-}\overline{h}\OSB{g} & \,\boldsymbol{-}\overline{h}^{2} & \hphantom{\boldsymbol{-}}\OSB{g}^{2} & \OSB{g}\overline{h}\vphantom{\sqrt{\tfrac{1}{2}}} \\ \hphantom{\boldsymbol{-}}\overline{h}h & \:\:\boldsymbol{-}\overline{h}g & \:-\OSB{g}h & \:\OSB{g}g\vphantom{\sqrt{\tfrac{1}{2}}} \end{bmatrix} \begin{bmatrix} \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0 &\hphantom{\boldsymbol{-}} \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0\\ 0 & \hphantom{\boldsymbol{-}}1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0\vphantom{\sqrt{\tfrac{1}{2}}}\\ 0 & \hphantom{\boldsymbol{-}} 0 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}1\vphantom{\sqrt{\tfrac{1}{2}}}\\ \sqrt{\tfrac{1}{2}}& \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}\sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0 \end{bmatrix} \nonumber\\ & = \begin{bmatrix} \sqrt{\tfrac{1}{2}} & 0 & 0 & \sqrt{\tfrac{1}{2}} \\ \boldsymbol{-}gh & \hphantom{\boldsymbol{-}}g^{2} & \boldsymbol{-}h^{2} & hg\vphantom{\sqrt{\tfrac{1}{2}}}\\ \sqrt{\tfrac{1}{2}}\left(g\OSB{g}\boldsymbol{-}\overline{h}h\right) & \sqrt{2}g\overline{h} & \sqrt{2}\OSB{g}h & \sqrt{\tfrac{1}{2}}\left(\overline{h}h\boldsymbol{-}g\OSB{g}\right) \\ \boldsymbol{-}\overline{h}\OSB{g} & \,\boldsymbol{-}\overline{h}^{2} & \hphantom{\boldsymbol{-}}\OSB{g}^{2} & \OSB{g}\overline{h}\vphantom{\sqrt{\tfrac{1}{2}}} \end{bmatrix} \begin{bmatrix} \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0 &\hphantom{\boldsymbol{-}} \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0\\ 0 & \hphantom{\boldsymbol{-}}1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0\vphantom{\sqrt{\tfrac{1}{2}}}\\ 0 & \hphantom{\boldsymbol{-}} 0 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}1\vphantom{\sqrt{\tfrac{1}{2}}}\\ \sqrt{\tfrac{1}{2}}& \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}\sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0 \end{bmatrix} \nonumber\\ & = \begin{bmatrix} 1 & 0 & 0 & 0 \vphantom{\sqrt{\tfrac{1}{2}}}\\ 0 & g^{2} & \boldsymbol{-}\sqrt{2}gh &\boldsymbol{-}h^{2}\vphantom{\sqrt{\tfrac{1}{2}}}\\ 0 & \sqrt{2}g\overline{h} & \left(g\OSB{g}\boldsymbol{-}h\overline{h}\right) & \sqrt{2}\OSB{g}h \\ 0 & \,\boldsymbol{-}\overline{h}^{2} & \boldsymbol{-}\sqrt{2}\OSB{g}\overline{h} & \OSB{g}^{2} \vphantom{\sqrt{\tfrac{1}{2}}} \end{bmatrix}_{\OSS{\boldsymbol{e}}} \tag{ed-34}\label{eqed-34} \end{align} so $$^{\bf 4}\OSS{U}= \begin{bmatrix} \begin{array}{c|ccc} \:\: 1 \:\: &\rule [0ex]{20pt}{0.0ex}&\rule [-2.5ex]{0pt}{6.0ex} \rule [0ex]{16pt}{0ex}& \rule [0ex]{16pt}{0ex}\\ \hline \rule [-3ex]{0pt}{6ex}&g^{2} & \boldsymbol{-}\sqrt{2}gh &\boldsymbol{-}h^{2} \\ \rule [-3ex]{0pt}{6ex}& \sqrt{2}g\overline{h} & \left(g\OSB{g}\boldsymbol{-}h\overline{h}\right) & \sqrt{2}\OSB{g}h \\ \rule [-3ex]{0pt}{6ex}& \boldsymbol{-}\overline{h}^{2} & \boldsymbol{-}\sqrt{2}\OSB{g}\overline{h} & \OSB{g}^{2} \end{array} \end{bmatrix}_{\OSS{\boldsymbol{e}}} = \begin{bmatrix} \begin{array}{c|ccc} ^{\mathbf{1}}U_{\boldsymbol{\left[1\right]}}&\rule [0ex]{20pt}{0.0ex}&\rule [-2.5ex]{0pt}{6.0ex} \rule [0ex]{16pt}{0ex}& \rule [0ex]{16pt}{0ex}\\ \hline \rule [-3ex]{0pt}{6ex}&\rule [0.0ex]{50pt}{0.0ex}& \rule [0.0ex]{50pt}{0.0ex} &\rule [0.0ex]{50pt}{0.0ex}\\ \rule [-3ex]{0pt}{6ex}& & ^{\mathbf{3}}U_{\boldsymbol{\left[2\right]}} & \\ \rule [-3ex]{0pt}{6ex}& & & \end{array} \end{bmatrix}_{\OSS{\boldsymbol{e}}} \tag{ed-35}\label{eqed-35}$$ where $\:^{\mathbf{1}}U_{\boldsymbol{\left[1\right]}}\:$ and $\:^{\mathbf{3}}U_{\boldsymbol{\left[2\right]}}\:$ are special unitary matrices in the spaces of the singlet $\;\boldsymbol{1}\equiv\boldsymbol{\lbrace}\boldsymbol{\omega}\boldsymbol{\rbrace}$ and of the triplet $\;\boldsymbol{3}\equiv\boldsymbol{\lbrace}\BoldExp{\boldsymbol{\pi}}{-},\BoldExp{\boldsymbol{\pi}}{0},\BoldExp{\boldsymbol{\pi}}{+}\boldsymbol{\rbrace}$ respectively given by $$^{\mathbf{1}}U_{\boldsymbol{\left[1\right]}}= \begin{bmatrix} 1 \end{bmatrix} \quad \in SU(1)\equiv \{1\} \tag{ed-36}\label{eqed-36}$$

$$^{\mathbf{3}}U_{\boldsymbol{\left[2\right]}}= \begin{bmatrix} g^{2} & \boldsymbol{-}\sqrt{2}gh &\boldsymbol{-}h^{2}\vphantom{\sqrt{\dfrac{1}{2}}}\\ \sqrt{2}g\overline{h} & \left(g\OSB{g}\boldsymbol{-}h\overline{h}\right) & \sqrt{2}\OSB{g}h\vphantom{\sqrt{\dfrac{1}{2}}} \\ \boldsymbol{-}\overline{h}^{2} & \boldsymbol{-}\sqrt{2}\OSB{g}\overline{h} & \OSB{g}^{2} \vphantom{\sqrt{\dfrac{1}{2}}} \end{bmatrix} \quad \in SU(3) \tag{ed-37}\label{eqed-37}$$ results in all respects similar to those in $\color{blue}{\textbf{Example A }}$, see equations (ed-06), (ed-07) and (ed-08).

Again : We say that the symmetry group is $\;SU(2)$, NOT $\;SU(1)\;$ or $\;SU(3)\;$ of the resulting multiplets.

• Is the new basis, eigenbasis? Jul 8, 2022 at 16:30