I know that mesons are bosons made up of quark-antiquark pairs. But when I see the list of mesons, I can see that the makeup of neutral pions and eta mesons are noted in a strange way.
$$\pi^0=(u\bar{u}-d\bar{d})/\sqrt{2}$$
$$\eta^0=(u\bar{u}+d\bar{d}-2s\bar{s})/\sqrt{6}$$
How am I supposed to understand their compositions?
Interpretation 1: a neutral pion should be understand as a quantum superposition and is actually composed of 2 pairs, sometimes appearing as an up pair, some other times as a down pair.
Interpretation 2: a neutral pion can be an up pair or a down pair. Both compositions lead to mesons with the exact same characteristics and behaviours.
What is the meaning of those square roots? If it's too complicated to be explained within a few lines, can anyone recommend me a website or a book?
Best Answer
Note that in the 3-dimensional complex space spanned by basis $\boldsymbol{\lbrace}\boldsymbol{u}\overline{\boldsymbol{u}},\boldsymbol{d}\overline{\boldsymbol{d}},\boldsymbol{s}\overline{\boldsymbol{s}}\boldsymbol{\rbrace}$, this basis is replaced by $\boldsymbol{\lbrace}\boldsymbol{\pi^{0},\boldsymbol{\eta},\boldsymbol{\eta}^{\prime}}\boldsymbol{\rbrace}$ through a special unitary transformation $\mathrm{V}\in SU(3)$,
\begin{equation} \begin{bmatrix} \boldsymbol{\pi^{0}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{\eta} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{\eta}^{\prime} \vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \sqrt{\tfrac{1}{2}} & \boldsymbol{-} \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0 \vphantom{\tfrac{a}{\tfrac{a}{b}}}\\ \sqrt{\tfrac{1}{6}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{6}} & \boldsymbol{-}\sqrt{\tfrac{2}{3}} \vphantom{\tfrac{a}{\tfrac{a}{b}}}\\ \sqrt{\tfrac{1}{3}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{3}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{3}} \vphantom{\tfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \begin{bmatrix} \boldsymbol{u}\overline{\boldsymbol{u}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{d}\overline{\boldsymbol{d}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{s}\overline{\boldsymbol{s}} \vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} =\mathrm{V} \begin{bmatrix} \boldsymbol{u}\overline{\boldsymbol{u}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{d}\overline{\boldsymbol{d}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{s}\overline{\boldsymbol{s}} \vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \tag{1}\label{1} \end{equation} see Figure.
$ \newcommand{\FR}[2]{{\textstyle \frac{#1}{#2}}} \newcommand{\BK}[3]{\left|{#1},{#2}\right\rangle_{#3}} \newcommand{\BoldExp}[2]{{#1}^{\boldsymbol{#2}}} \newcommand{\BoldSub}[2]{{#1}_{\boldsymbol{#2}}} \newcommand{\MM}[4] {\begin{bmatrix} #1 & #2\\ #3 & #4\\ \end{bmatrix}} \newcommand{\MMM}[9] {\textstyle \begin{bmatrix} #1 & #2 & #3 \\ #4 & #5 & #6 \\ #7 & #8 & #9 \\ \end{bmatrix}} \newcommand{\CMRR}[2] {\begin{bmatrix} #1 \\ #2 \end{bmatrix}} \newcommand{\CMRRR}[3] {\begin{bmatrix} #2 \\ #3 \end{bmatrix}} \newcommand{\CMRRRR}[4] {\begin{bmatrix} #1 \\ #2 \\ #3 \\ #4 \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}} $
$\boldsymbol{\S\:}\textbf{A. Mesons from three quarks}$ $\boldsymbol{u},\boldsymbol{d},\boldsymbol{s} : \boldsymbol{3}\boldsymbol{\otimes}\overline{\boldsymbol{3}}\boldsymbol{=}\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{8}$
Suppose we know the existence of three quarks only : $\boldsymbol{u}$, $\boldsymbol{d}$ and $\boldsymbol{s}$. Under full symmetry these are the basic states, let
\begin{equation} \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{001}\label{001} \end{equation} 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 \begin{equation} \boldsymbol{\xi}=\xi_u\boldsymbol{u}+\xi_d\boldsymbol{d}+\xi_s\boldsymbol{s}= \begin{bmatrix} \xi_u\\ \xi_d\\ \xi_s \end{bmatrix} \qquad \xi_u,\xi_d,\xi_s \in \mathbb{C} \tag{002}\label{002} \end{equation} For a quark $\boldsymbol{\zeta} \in \mathbf{Q}$ \begin{equation} \boldsymbol{\zeta}=\zeta_u\boldsymbol{u}+\zeta_d\boldsymbol{d}+\zeta_s\boldsymbol{s}= \begin{bmatrix} \zeta_u\\ \zeta_d\\ \zeta_s \end{bmatrix} \tag{003}\label{003} \end{equation} the respective antiquark $\overline{\boldsymbol{\zeta}}$ is expressed by the complex conjugates of the coordinates \begin{equation} \overline{\boldsymbol{\zeta}}=\overline{\zeta}_u \overline{\boldsymbol{u}}+\overline{\zeta}_d\overline{\boldsymbol{d}}+\overline{\zeta}_s\overline{\boldsymbol{s}}= \begin{bmatrix} \overline{\zeta}_u\\ \overline{\zeta}_d\\ \overline{\zeta}_s \end{bmatrix} \tag{004}\label{004} \end{equation} with respect to the basic states
\begin{equation} \overline{\boldsymbol{u}}= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} \qquad \overline{\boldsymbol{d}}= \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix} \qquad \overline{\boldsymbol{s}}= \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \tag{005}\label{005} \end{equation} the antiquarks of $\boldsymbol{u},\boldsymbol{d}$ and $\boldsymbol{s}$ respectively. The antiquarks belong to a different space, the space of antiquarks $\overline{\mathbf{Q}}\equiv \mathbb{C}^{\boldsymbol{3}}$.
Since a meson is a quark-antiquark pair, we'll try to find the product space \begin{equation} \mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\: \left(\equiv \mathbb{C}^{\boldsymbol{9}}\right) \tag{006}\label{006} \end{equation} Using the expressions \eqref{002} and \eqref{004} 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}$ \begin{equation} \begin{split} \mathrm{X}=\boldsymbol{\xi}\boldsymbol{\otimes}\overline{\boldsymbol{\zeta}}=&\xi_u\overline{\zeta}_u \left(\boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_u\overline{\zeta}_d \left( \boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_u\overline{\zeta}_s \left(\boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right)+ \\ &\xi_d\overline{\zeta}_u \left(\boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_d\overline{\zeta}_d \left( \boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_d\overline{\zeta}_s \left(\boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right)+\\ &\xi_s\overline{\zeta}_u \left(\boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_s\overline{\zeta}_d \left( \boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_s\overline{\zeta}_s \left(\boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right) \end{split} \tag{007}\label{007} \end{equation} In order to simplify the expressions, the product symbol $"\boldsymbol{\otimes}"$ is omitted and so \begin{equation} \begin{split} \mathrm{X}=\boldsymbol{\xi}\overline{\boldsymbol{\zeta}}=&\xi_u\overline{\zeta}_u \boldsymbol{u}\overline{\boldsymbol{u}}+\xi_u\overline{\zeta}_d \boldsymbol{u}\overline{\boldsymbol{d}}+\xi_u\overline{\zeta}_s \boldsymbol{u}\overline{\boldsymbol{s}}+ \\ &\xi_d\overline{\zeta}_u \boldsymbol{d}\overline{\boldsymbol{u}}+\xi_d\overline{\zeta}_d \boldsymbol{d}\overline{\boldsymbol{d}}+\xi_d\overline{\zeta}_s \boldsymbol{d}\overline{\boldsymbol{s}}+\\ &\xi_s\overline{\zeta}_u \boldsymbol{s}\overline{\boldsymbol{u}}+\xi_s\overline{\zeta}_d \boldsymbol{s}\overline{\boldsymbol{d}}+\xi_s\overline{\zeta}_s \boldsymbol{s}\overline{\boldsymbol{s}} \end{split} \tag{008}\label{008} \end{equation} Due to the fact that $\mathbf{Q}$ and $\overline{\mathbf{Q}}$ are of the same dimension, it's convenient to represent the meson states in the product 9-dimensional complex space $\:\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\:$ by square $3 \times 3$ matrices instead of row or column vectors \begin{equation} \mathrm{X}=\boldsymbol{\xi}\overline{\boldsymbol{\zeta}}= \begin{bmatrix} \xi_u\overline{\zeta}_u & \xi_u\overline{\zeta}_d & \xi_u\overline{\zeta}_s\\ \xi_d\overline{\zeta}_u & \xi_d\overline{\zeta}_d & \xi_d\overline{\zeta}_s\\ \xi_s\overline{\zeta}_u & \xi_s\overline{\zeta}_d & \xi_s\overline{\zeta}_s \end{bmatrix}= \begin{bmatrix} \xi_u\vphantom{\overline{\zeta}_u}\\ \xi_d\vphantom{\overline{\zeta}_u}\\ \xi_s\vphantom{\overline{\zeta}_u} \end{bmatrix} \begin{bmatrix} \overline{\zeta}_u \\ \overline{\zeta}_d \\ \overline{\zeta}_s \end{bmatrix}^{\mathsf{T}} = \begin{bmatrix} \xi_u\vphantom{\overline{\zeta}_u}\\ \xi_d\vphantom{\overline{\zeta}_u}\\ \xi_s\vphantom{\overline{\zeta}_u} \end{bmatrix} \begin{bmatrix} \overline{\zeta}_u & \overline{\zeta}_d & \overline{\zeta}_s \end{bmatrix} \tag{009}\label{009} \end{equation} The product space $\:\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\:$ is created by completion of the set of states \eqref{008} with arbitrary complex coefficients \begin{equation} \begin{split} \mathrm{X}=&\mathrm{x}_{_{11}}\boldsymbol{u}\overline{\boldsymbol{u}}+\mathrm{x}_{_{12}} \boldsymbol{u}\overline{\boldsymbol{d}}+\mathrm{x}_{_{13}} \boldsymbol{u}\overline{\boldsymbol{s}}+ \\ &\mathrm{x}_{_{21}}\boldsymbol{d}\overline{\boldsymbol{u}}+\mathrm{x}_{_{22}} \boldsymbol{d}\overline{\boldsymbol{d}}+\mathrm{x}_{_{23}} \boldsymbol{d}\overline{\boldsymbol{s}}+ \qquad \mathrm{x}_{_{ij}} \in \mathbb{C}\\ &\mathrm{x}_{_{31}} \boldsymbol{s}\overline{\boldsymbol{u}}+\mathrm{x}_{_{32}} \boldsymbol{s}\overline{\boldsymbol{d}}+\mathrm{x}_{_{33}} \boldsymbol{s}\overline{\boldsymbol{s}} \end{split} \tag{010}\label{010} \end{equation} that is \begin{equation} \mathrm{X}= \begin{bmatrix} \mathrm{x}_{_{11}} & \mathrm{x}_{_{12}} & \mathrm{x}_{_{13}}\\ \mathrm{x}_{_{21}} & \mathrm{x}_{_{22}} & \mathrm{x}_{_{23}}\\ \mathrm{x}_{_{31}} & \mathrm{x}_{_{32}} & \mathrm{x}_{_{33}} \end{bmatrix} \:, \qquad \mathrm{x}_{_{ij}} \in \mathbb{C} \tag{011}\label{011} \end{equation} So $\:\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\:$ is identical to $\mathbb{C}^{\boldsymbol{9}}$ with base states \begin{align} &\boldsymbol{u}\overline{\boldsymbol{u}}= \begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} \quad \boldsymbol{u}\overline{\boldsymbol{d}}= \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} \quad \boldsymbol{u}\overline{\boldsymbol{s}}= \begin{bmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} \tag{012a}\label{012a}\\ &\boldsymbol{d}\overline{\boldsymbol{u}}= \begin{bmatrix} 0 & 0 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} \quad \boldsymbol{d}\overline{\boldsymbol{d}}= \begin{bmatrix} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix} \quad \:\boldsymbol{d}\overline{\boldsymbol{s}}= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix} \tag{012b}\label{012b}\\ &\boldsymbol{s}\overline{\boldsymbol{u}}= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 1 & 0 & 0 \end{bmatrix} \quad \:\boldsymbol{s}\overline{\boldsymbol{d}}= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 1 & 0 \end{bmatrix} \quad \:\boldsymbol{s}\overline{\boldsymbol{s}}= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end{bmatrix} \tag{012c}\label{012c} \end{align} This basis is represented symbolically by a $3\times 3$ array \begin{equation} \mathcal{F}_{\mathbf{M}}= \begin{bmatrix} \boldsymbol{u}\overline{\boldsymbol{u}} & \boldsymbol{u}\overline{\boldsymbol{d}} & \boldsymbol{u}\overline{\boldsymbol{s}}\\ \boldsymbol{d}\overline{\boldsymbol{u}} & \boldsymbol{d}\overline{\boldsymbol{d}} & \boldsymbol{d}\overline{\boldsymbol{s}}\\ \boldsymbol{s}\overline{\boldsymbol{u}} & \boldsymbol{s}\overline{\boldsymbol{d}} & \boldsymbol{s}\overline{\boldsymbol{s}} \end{bmatrix} \tag{013}\label{013} \end{equation} In this Hilbert space the usual inner product between states \begin{equation} \mathrm{X}= \begin{bmatrix} \mathrm{x}_{_{11}} & \mathrm{x}_{_{12}} & \mathrm{x}_{_{13}}\\ \mathrm{x}_{_{21}} & \mathrm{x}_{_{22}} & \mathrm{x}_{_{23}}\\ \mathrm{x}_{_{31}} & \mathrm{x}_{_{32}} & \mathrm{x}_{_{33}} \end{bmatrix} \:, \qquad \mathrm{Y}= \begin{bmatrix} \mathrm{y}_{_{11}} & \mathrm{y}_{_{12}} & \mathrm{y}_{_{13}}\\ \mathrm{y}_{_{21}} & \mathrm{y}_{_{22}} & \mathrm{y}_{_{23}}\\ \mathrm{y}_{_{31}} & \mathrm{y}_{_{32}} & \mathrm{y}_{_{33}} \end{bmatrix} \tag{014}\label{014} \end{equation} is \begin{equation} \begin{split} \langle \mathrm{X},\mathrm{Y}\rangle \equiv &\mathrm{x}_{_{11}}\overline{\mathrm{y}}_{_{11}}+\mathrm{x}_{_{12}}\overline{\mathrm{y}}_{_{12}}+\mathrm{x}_{_{13}}\overline{\mathrm{y}}_{_{13}}+\\ &\mathrm{x}_{_{21}}\overline{\mathrm{y}}_{_{21}}+\mathrm{x}_{_{22}}\overline{\mathrm{y}}_{_{22}}+\mathrm{x}_{_{23}}\overline{\mathrm{y}}_{_{23}}+\\ &\mathrm{x}_{_{31}}\overline{\mathrm{y}}_{_{31}}+\mathrm{x}_{_{32}}\overline{\mathrm{y}}_{_{32}}+\mathrm{x}_{_{33}}\overline{\mathrm{y}}_{_{33}} \end{split} \tag{015}\label{015} \end{equation} which, using the $3\times 3$ matrix representation of states, is the trace of the matrix product $\mathrm{X}\BoldExp{\mathrm{Y}}{*}$
\begin{equation} \langle \mathrm{X},\mathrm{Y}\rangle =\mathrm{Tr}\left[\mathrm{X}\BoldExp{\mathrm{Y}}{*}\right] \tag{016}\label{016} \end{equation} given that $\BoldExp{\mathrm{Y}}{*}$ is the complex conjugate of the transpose of $\mathrm{Y}$ \begin{equation} \BoldExp{\mathrm{Y}}{*}\equiv \BoldExp{ \begin{bmatrix} \mathrm{y}_{_{11}} & \mathrm{y}_{_{12}} & \mathrm{y}_{_{13}}\\ \mathrm{y}_{_{21}} & \mathrm{y}_{_{22}} & \mathrm{y}_{_{23}}\\ \mathrm{y}_{_{31}} & \mathrm{y}_{_{32}} & \mathrm{y}_{_{33}} \end{bmatrix}} {*} = \overline{\begin{bmatrix} \mathrm{y}_{_{11}} & \mathrm{y}_{_{12}} & \mathrm{y}_{_{13}}\\ \mathrm{y}_{_{21}} & \mathrm{y}_{_{22}} & \mathrm{y}_{_{23}}\\ \mathrm{y}_{_{31}} & \mathrm{y}_{_{32}} & \mathrm{y}_{_{33}} \end{bmatrix}^{\mathsf{T}}} = \begin{bmatrix} \overline{\mathrm{y}}_{_{11}} & \overline{\mathrm{y}}_{_{21}} & \overline{\mathrm{y}}_{_{31}}\\ \overline{\mathrm{y}}_{_{12}} & \overline{\mathrm{y}}_{_{22}} & \overline{\mathrm{y}}_{_{32}}\\ \overline{\mathrm{y}}_{_{13}} & \overline{\mathrm{y}}_{_{23}} & \overline{\mathrm{y}}_{_{33}} \end{bmatrix} \tag{017}\label{017} \end{equation} Now, under a unitary transformation $\;W \in SU(3)\;$ in the 3-dimensional space of quarks $\;\mathbf{Q}\;$, we have \begin{equation} \BoldExp{\boldsymbol{\xi}}{'} = W\boldsymbol{\xi} \tag{018}\label{018} \end{equation} so in the space of antiquarks $\overline{\mathbf{Q}}\;$, since $\;\BoldExp{\boldsymbol{\zeta}}{'}=W \boldsymbol{\zeta}\;$ \begin{equation} \overline{\BoldExp{\boldsymbol{\zeta}}{'}}= \overline{W}\;\overline{\boldsymbol{\zeta}} \tag{019}\label{019} \end{equation} and for the meson state \begin{align} \BoldExp{\mathrm{X}}{'} & =\BoldExp{\boldsymbol{\xi}}{'}\boldsymbol{\otimes}\overline{\BoldExp{\boldsymbol{\zeta}}{'}}=\left(W\boldsymbol{\xi}\vphantom{\overline{W}\overline{\boldsymbol{\zeta}} }\right)\left(\overline{W}\overline{\boldsymbol{\zeta}} \right) = \Biggl(W\begin{bmatrix} \xi_u\vphantom{\overline{\zeta}_u}\\ \xi_d\vphantom{\overline{\zeta}_u}\\ \xi_s\vphantom{\overline{\zeta}_u} \end{bmatrix}\Biggr) \Biggl(\overline{W}\begin{bmatrix} \overline{\zeta}_u\\ \overline{\zeta}_d\\ \overline{\zeta}_s \end{bmatrix}\Biggr)^{\mathsf{T}} \nonumber\\ & = W\Biggl(\begin{bmatrix} \xi_u\vphantom{\overline{\zeta}_u}\\ \xi_d\vphantom{\overline{\zeta}_u}\\ \xi_s\vphantom{\overline{\zeta}_u} \end{bmatrix} \begin{bmatrix} \overline{\zeta}_u & \overline{\zeta}_d & \overline{\zeta}_s \end{bmatrix}\Biggr)\overline{W}^{\mathsf{T}} =W\left(\boldsymbol{\xi}\boldsymbol{\otimes}\overline{\boldsymbol{\zeta}}\right)\BoldExp{W}{*}=W\;\mathrm{X}\;\BoldExp{W}{*} \nonumber \tag{020}\label{020} \end{align} that is \begin{equation} \BoldExp{\mathrm{X}}{'} = W\;\mathrm{X}\;\BoldExp{W}{*} \tag{021}\label{021} \end{equation} Above equation \eqref{021} is the transformation law of meson states in the 9-dimensional space $\;\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\;$ induced by a unitary transformation $\;W \in SU(3)\;$ in the 3-dimensional space of quarks $\mathbf{Q}$.
Under this transformation law the inner product of two meson states is invariant because its relation with the trace, equation \eqref{016}, yields \begin{equation} \langle \BoldExp{\mathrm{X}}{'},\BoldExp{\mathrm{Y}}{'}\rangle =\mathrm{Tr}\left[\BoldExp{\mathrm{X}}{'}\BoldExp{\BoldExp{\mathrm{Y}}{'}}{*}\right]=\mathrm{Tr}\Bigl[\left(W\mathrm{X}\BoldExp{W}{*}\right) \BoldExp{\left(W\mathrm{Y}\BoldExp{W}{*}\right)}{*}\Bigr]=\mathrm{Tr}\Bigl[W \left( \mathrm{X}\BoldExp{Y}{*}\right)\BoldExp{W}{*}\Bigr]=\mathrm{Tr}\Bigl[\mathrm{X}\BoldExp{Y}{*}\Bigr] \tag{022}\label{022} \end{equation} The last equality in above equation \eqref{022} is valid since under the transformation law \eqref{021} the trace remains invariant. More generally, for unitary $\;W \in SU(n)\;$ and $\;A\;$ a $\;n \times n\;$ complex matrix the transformation
\begin{equation} \BoldExp{\mathrm{A}}{'} = W\;\mathrm{A}\;\BoldExp{W}{*} \tag{023}\label{023} \end{equation}
if expressed in terms of elements, yields (we use the Einstein summation convention) \begin{equation} \BoldExp{a_{ij}}{'} = w_{i\rho}a_{\rho\sigma}\BoldExp{w_{\sigma j}}{*} \tag{024}\label{0242} \end{equation} so \begin{equation} \mathrm{Tr}\Bigl[\BoldExp{\mathrm{A}}{'}\Bigr]=\BoldExp{a_{ii}}{'} = w_{i\rho}a_{\rho\sigma}\BoldExp{w_{\sigma i}}{*}=(\BoldExp{w_{\sigma i}}{*}w_{i\rho})a_{\rho\sigma}=\delta_{\sigma\rho}a_{\rho\sigma}=a_{\rho\rho}=\mathrm{Tr}\Bigl[A\Bigr] \tag{025}\label{025} \end{equation} proving the invariance of inner product under the transformation law \eqref{021} \begin{equation} \langle \BoldExp{\mathrm{X}}{'},\BoldExp{\mathrm{Y}}{'}\rangle =\langle \mathrm{X},\mathrm{Y}\rangle \tag{026}\label{026} \end{equation}
Now, obviously the meson state represented by the identity matrix \begin{equation} \mathrm{I}= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \tag{027}\label{027} \end{equation} is unchanged under the transformation \eqref{021} and if normalized yields \begin{equation} \BoldSub{\mathrm{F}}{0}=\sqrt{\tfrac{1}{3}} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} =\sqrt{\tfrac{1}{3}}\left(\boldsymbol{u}\overline{\boldsymbol{u}}+\boldsymbol{d}\overline{\boldsymbol{d}}+\boldsymbol{s}\overline{\boldsymbol{s}} \right)\equiv \BoldExp{\boldsymbol{\eta}}{\prime} \tag{028}\label{028} \end{equation} that is, it represents the $\;\BoldExp{\boldsymbol{\eta}}{\prime}\;$ meson. The 1-dimensional subspace $\;\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace}\;$ spanned by this state is invariant. Note that $\;\BoldExp{\boldsymbol{\eta}}{\prime}=\sqrt{3}\cdot \mathrm{Tr}\left[\mathcal{F}_{\mathbf{M}}\right]$.
Any meson state orthogonal to this space, $\mathrm{X}\perp\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace} $, remains orthogonal under the transformation. But \begin{equation} \mathrm{X}\perp \boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace}\Leftrightarrow\langle \mathrm{X},\BoldSub{\mathrm{F}}{0}\rangle =0\Leftrightarrow\mathrm{Tr}\left[\mathrm{X}\BoldSub{\mathrm{F}}{0}^{\boldsymbol{*}}\right]=0\Leftrightarrow\mathrm{Tr}\left[\mathrm{X}\right]=0 \tag{029}\label{029} \end{equation}
So, the 8-dimensional linear subspace of all meson states with traceless matrix representation is the orthogonal complement of the 1-dimensional subspace $\;\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace}\;$ and if $\;\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{1},\BoldSub{\mathrm{F}}{2},\cdots,\BoldSub{\mathrm{F}}{8}\boldsymbol{\rbrace}\;$ is any basis which spans this space then \begin{equation} \boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{1},\BoldSub{\mathrm{F}}{2},\cdots,\BoldSub{\mathrm{F}}{8}\boldsymbol{\rbrace}=\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace}^{\boldsymbol{\perp}}=\Bigl\{ \mathrm{X} \in \mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\; :\; \mathrm{Tr}\left[X\right]=0 \; \Bigr\} \tag{030}\label{030} \end{equation}
This space is invariant under the transformation \eqref{021}. There are arbitrary many choices of the basis $\;\left(\BoldSub{\mathrm{F}}{1},\BoldSub{\mathrm{F}}{2},\cdots,\BoldSub{\mathrm{F}}{8}\right)\;$ but a proper one must correspond to mesons in the real world and be orthonormal if possible. So, the normalized traceless meson state \begin{equation} \BoldSub{\mathrm{F}}{3}=\sqrt{\tfrac{1}{2}} \begin{bmatrix} 1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0\\ 0 & \boldsymbol{-}1 & \hphantom{\boldsymbol{-}}0\\ 0 & \hphantom{\boldsymbol{-}} 0 & \hphantom{\boldsymbol{-}}0 \end{bmatrix} =\sqrt{\tfrac{1}{2}}\left(\boldsymbol{u}\overline{\boldsymbol{u}}-\boldsymbol{d}\overline{\boldsymbol{d}} \right)\equiv \BoldExp{\boldsymbol{\pi}}{0} \tag{031}\label{031} \end{equation}
represents of course the $\;\BoldExp{\boldsymbol{\pi}}{0}\;$ meson (pion).
The basis $\mathcal{F}_{\mathbf{M}}$ may be expressed symbolically as sum of a diagonal and a traceless component \begin{equation} \begin{split} &\mathcal{F}_{\mathbf{M}}=\Bigl(\tfrac{1}{3}\mathrm{Tr}\left[\mathcal{F}_{\mathbf{M}}\right]\Bigr)\mathcal{I}+\Bigl[\mathcal{F}_{\mathbf{M}}-\Bigl(\tfrac{1}{3}\mathrm{Tr}\left[\mathcal{F}_{\mathbf{M}}\right]\Bigr)\mathcal{I}\Bigr]\\ &=\begin{bmatrix} \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}} & \mathbf{0} & \mathbf{0}\\ \mathbf{0} & \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}} & \mathbf{0}\\ \mathbf{0} & \mathbf{0} & \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}} \end{bmatrix} + \begin{bmatrix} \dfrac{\left(2\boldsymbol{u}\overline{\boldsymbol{u}}-\boldsymbol{d}\overline{\boldsymbol{d}}-\boldsymbol{s}\overline{\boldsymbol{s}}\right) }{3}{\rule[0ex]{-10pt}{0ex}} & \boldsymbol{u}\overline{\boldsymbol{d}} & \boldsymbol{u}\overline{\boldsymbol{s}}\\ \boldsymbol{d}\overline{\boldsymbol{u}} & \dfrac{\left(-\boldsymbol{u}\overline{\boldsymbol{u}}+2\boldsymbol{d}\overline{\boldsymbol{d}}-\boldsymbol{s}\overline{\boldsymbol{s}}\right) }{3} & \boldsymbol{d}\overline{\boldsymbol{s}} \\ \boldsymbol{s}\overline{\boldsymbol{u}} & \boldsymbol{s}\overline{\boldsymbol{d}} & {\rule[-2ex]{-10pt}{6ex}} \dfrac{\left(-\boldsymbol{u}\overline{\boldsymbol{u}}-\boldsymbol{d}\overline{\boldsymbol{d}}+2\boldsymbol{s}\overline{\boldsymbol{s}}\right)}{3} \end{bmatrix} \end{split} \tag{032}\label{032} \end{equation}
The 3rd diagonal element of the traceless component of $\mathcal{F}_{\mathbf{M}}$, if opposed and normalized, yields \begin{equation} \BoldSub{\mathrm{F}}{8}=\sqrt{\tfrac{1}{6}} \begin{bmatrix} 1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0\\ 0 & \hphantom{\boldsymbol{-}}1 & \hphantom{\boldsymbol{-}}0\\ 0 & \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}2 \end{bmatrix} =\sqrt{\tfrac{1}{6}}\left(\boldsymbol{u}\overline{\boldsymbol{u}}+\boldsymbol{d}\overline{\boldsymbol{d}}-2\boldsymbol{s}\overline{\boldsymbol{s}} \right)\equiv \boldsymbol{\eta} \tag{033}\label{033} \end{equation}
that is, it represents the $\;\boldsymbol{\eta}\;$ meson.
(to be continued in $\boldsymbol{\S\:}\textbf{B}$)