# How to understand the makeup of neutral pi and eta mesons?

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?

• Interpretation 1 is right I think, but it would be good to see an answer explaining the relative signs. – jacob1729 Jun 7 '19 at 15:52
• The square roots are normalization factors. – G. Smith Jun 7 '19 at 16:17

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 $$\mathrm{V}\in SU(3)$$,

$$$$\begin{bmatrix} \boldsymbol{\pi^{0}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{\eta} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{\eta}^{\prime} \vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \sqrt{\tfrac{1}{2}} & \boldsymbol{-} \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0 \vphantom{\tfrac{a}{\tfrac{a}{b}}}\\ \sqrt{\tfrac{1}{6}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{6}} & \boldsymbol{-}\sqrt{\tfrac{2}{3}} \vphantom{\tfrac{a}{\tfrac{a}{b}}}\\ \sqrt{\tfrac{1}{3}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{3}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{3}} \vphantom{\tfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \begin{bmatrix} \boldsymbol{u}\overline{\boldsymbol{u}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{d}\overline{\boldsymbol{d}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{s}\overline{\boldsymbol{s}} \vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} =\mathrm{V} \begin{bmatrix} \boldsymbol{u}\overline{\boldsymbol{u}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{d}\overline{\boldsymbol{d}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{s}\overline{\boldsymbol{s}} \vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \tag{1}\label{1}$$$$ see Figure.


$$\boldsymbol{\S\:}\textbf{A. Mesons from three quarks}$$ $$\boldsymbol{u},\boldsymbol{d},\boldsymbol{s} : \boldsymbol{3}\boldsymbol{\otimes}\overline{\boldsymbol{3}}\boldsymbol{=}\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{8}$$

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
$$$$\boldsymbol{u}= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} \qquad \boldsymbol{d}= \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix} \qquad \boldsymbol{s}= \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \tag{001}\label{001}$$$$ of a 3-dimensional complex Hilbert space of quarks, say $$\mathbf{Q}\equiv \mathbb{C}^{\boldsymbol{3}}$$. A quark $$\boldsymbol{\xi} \in \mathbf{Q}$$ is expressed in terms of these basic states as $$$$\boldsymbol{\xi}=\xi_u\boldsymbol{u}+\xi_d\boldsymbol{d}+\xi_s\boldsymbol{s}= \begin{bmatrix} \xi_u\\ \xi_d\\ \xi_s \end{bmatrix} \qquad \xi_u,\xi_d,\xi_s \in \mathbb{C} \tag{002}\label{002}$$$$ For a quark $$\boldsymbol{\zeta} \in \mathbf{Q}$$ $$$$\boldsymbol{\zeta}=\zeta_u\boldsymbol{u}+\zeta_d\boldsymbol{d}+\zeta_s\boldsymbol{s}= \begin{bmatrix} \zeta_u\\ \zeta_d\\ \zeta_s \end{bmatrix} \tag{003}\label{003}$$$$ the respective antiquark $$\overline{\boldsymbol{\zeta}}$$ is expressed by the complex conjugates of the coordinates $$$$\overline{\boldsymbol{\zeta}}=\overline{\zeta}_u \overline{\boldsymbol{u}}+\overline{\zeta}_d\overline{\boldsymbol{d}}+\overline{\zeta}_s\overline{\boldsymbol{s}}= \begin{bmatrix} \overline{\zeta}_u\\ \overline{\zeta}_d\\ \overline{\zeta}_s \end{bmatrix} \tag{004}\label{004}$$$$ with respect to the basic states
$$$$\overline{\boldsymbol{u}}= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} \qquad \overline{\boldsymbol{d}}= \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix} \qquad \overline{\boldsymbol{s}}= \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \tag{005}\label{005}$$$$ the antiquarks of $$\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 $$$$\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\: \left(\equiv \mathbb{C}^{\boldsymbol{9}}\right) \tag{006}\label{006}$$$$ Using the expressions \eqref{002} and \eqref{004} of the quark $$\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}$$ $$$$\begin{split} \mathrm{X}=\boldsymbol{\xi}\boldsymbol{\otimes}\overline{\boldsymbol{\zeta}}=&\xi_u\overline{\zeta}_u \left(\boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_u\overline{\zeta}_d \left( \boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_u\overline{\zeta}_s \left(\boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right)+ \\ &\xi_d\overline{\zeta}_u \left(\boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_d\overline{\zeta}_d \left( \boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_d\overline{\zeta}_s \left(\boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right)+\\ &\xi_s\overline{\zeta}_u \left(\boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_s\overline{\zeta}_d \left( \boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_s\overline{\zeta}_s \left(\boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right) \end{split} \tag{007}\label{007}$$$$ In order to simplify the expressions, the product symbol $$"\boldsymbol{\otimes}"$$ is omitted and so $$$$\begin{split} \mathrm{X}=\boldsymbol{\xi}\overline{\boldsymbol{\zeta}}=&\xi_u\overline{\zeta}_u \boldsymbol{u}\overline{\boldsymbol{u}}+\xi_u\overline{\zeta}_d \boldsymbol{u}\overline{\boldsymbol{d}}+\xi_u\overline{\zeta}_s \boldsymbol{u}\overline{\boldsymbol{s}}+ \\ &\xi_d\overline{\zeta}_u \boldsymbol{d}\overline{\boldsymbol{u}}+\xi_d\overline{\zeta}_d \boldsymbol{d}\overline{\boldsymbol{d}}+\xi_d\overline{\zeta}_s \boldsymbol{d}\overline{\boldsymbol{s}}+\\ &\xi_s\overline{\zeta}_u \boldsymbol{s}\overline{\boldsymbol{u}}+\xi_s\overline{\zeta}_d \boldsymbol{s}\overline{\boldsymbol{d}}+\xi_s\overline{\zeta}_s \boldsymbol{s}\overline{\boldsymbol{s}} \end{split} \tag{008}\label{008}$$$$ Due to the fact that $$\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$$ matrices instead of row or column vectors $$$$\mathrm{X}=\boldsymbol{\xi}\overline{\boldsymbol{\zeta}}= \begin{bmatrix} \xi_u\overline{\zeta}_u & \xi_u\overline{\zeta}_d & \xi_u\overline{\zeta}_s\\ \xi_d\overline{\zeta}_u & \xi_d\overline{\zeta}_d & \xi_d\overline{\zeta}_s\\ \xi_s\overline{\zeta}_u & \xi_s\overline{\zeta}_d & \xi_s\overline{\zeta}_s \end{bmatrix}= \begin{bmatrix} \xi_u\\ \xi_d\\ \xi_s \end{bmatrix} \begin{bmatrix} \overline{\zeta}_u \\ \overline{\zeta}_d \\ \overline{\zeta}_s \end{bmatrix}^{\mathsf{T}} = \begin{bmatrix} \xi_u\\ \xi_d\\ \xi_s \end{bmatrix} \begin{bmatrix} \overline{\zeta}_u & \overline{\zeta}_d & \overline{\zeta}_s \end{bmatrix} \tag{009}\label{009}$$$$ The product space $$\:\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\:$$ is created by completion of the set of states \eqref{008} with arbitrary complex coefficients $$$$\begin{split} \mathrm{X}=&\mathrm{x}_{_{11}}\boldsymbol{u}\overline{\boldsymbol{u}}+\mathrm{x}_{_{12}} \boldsymbol{u}\overline{\boldsymbol{d}}+\mathrm{x}_{_{13}} \boldsymbol{u}\overline{\boldsymbol{s}}+ \\ &\mathrm{x}_{_{21}}\boldsymbol{d}\overline{\boldsymbol{u}}+\mathrm{x}_{_{22}} \boldsymbol{d}\overline{\boldsymbol{d}}+\mathrm{x}_{_{23}} \boldsymbol{d}\overline{\boldsymbol{s}}+ \qquad \mathrm{x}_{_{ij}} \in \mathbb{C}\\ &\mathrm{x}_{_{31}} \boldsymbol{s}\overline{\boldsymbol{u}}+\mathrm{x}_{_{32}} \boldsymbol{s}\overline{\boldsymbol{d}}+\mathrm{x}_{_{33}} \boldsymbol{s}\overline{\boldsymbol{s}} \end{split} \tag{010}\label{010}$$$$ that is $$$$\mathrm{X}= \begin{bmatrix} \mathrm{x}_{_{11}} & \mathrm{x}_{_{12}} & \mathrm{x}_{_{13}}\\ \mathrm{x}_{_{21}} & \mathrm{x}_{_{22}} & \mathrm{x}_{_{23}}\\ \mathrm{x}_{_{31}} & \mathrm{x}_{_{32}} & \mathrm{x}_{_{33}} \end{bmatrix} \:, \qquad \mathrm{x}_{_{ij}} \in \mathbb{C} \tag{011}\label{011}$$$$ So $$\:\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$$ array $$$$\mathcal{F}_{\mathbf{M}}= \begin{bmatrix} \boldsymbol{u}\overline{\boldsymbol{u}} & \boldsymbol{u}\overline{\boldsymbol{d}} & \boldsymbol{u}\overline{\boldsymbol{s}}\\ \boldsymbol{d}\overline{\boldsymbol{u}} & \boldsymbol{d}\overline{\boldsymbol{d}} & \boldsymbol{d}\overline{\boldsymbol{s}}\\ \boldsymbol{s}\overline{\boldsymbol{u}} & \boldsymbol{s}\overline{\boldsymbol{d}} & \boldsymbol{s}\overline{\boldsymbol{s}} \end{bmatrix} \tag{013}\label{013}$$$$ In this Hilbert space the usual inner product between states $$$$\mathrm{X}= \begin{bmatrix} \mathrm{x}_{_{11}} & \mathrm{x}_{_{12}} & \mathrm{x}_{_{13}}\\ \mathrm{x}_{_{21}} & \mathrm{x}_{_{22}} & \mathrm{x}_{_{23}}\\ \mathrm{x}_{_{31}} & \mathrm{x}_{_{32}} & \mathrm{x}_{_{33}} \end{bmatrix} \:, \qquad \mathrm{Y}= \begin{bmatrix} \mathrm{y}_{_{11}} & \mathrm{y}_{_{12}} & \mathrm{y}_{_{13}}\\ \mathrm{y}_{_{21}} & \mathrm{y}_{_{22}} & \mathrm{y}_{_{23}}\\ \mathrm{y}_{_{31}} & \mathrm{y}_{_{32}} & \mathrm{y}_{_{33}} \end{bmatrix} \tag{014}\label{014}$$$$ is $$$$\begin{split} \langle \mathrm{X},\mathrm{Y}\rangle \equiv &\mathrm{x}_{_{11}}\overline{\mathrm{y}}_{_{11}}+\mathrm{x}_{_{12}}\overline{\mathrm{y}}_{_{12}}+\mathrm{x}_{_{13}}\overline{\mathrm{y}}_{_{13}}+\\ &\mathrm{x}_{_{21}}\overline{\mathrm{y}}_{_{21}}+\mathrm{x}_{_{22}}\overline{\mathrm{y}}_{_{22}}+\mathrm{x}_{_{23}}\overline{\mathrm{y}}_{_{23}}+\\ &\mathrm{x}_{_{31}}\overline{\mathrm{y}}_{_{31}}+\mathrm{x}_{_{32}}\overline{\mathrm{y}}_{_{32}}+\mathrm{x}_{_{33}}\overline{\mathrm{y}}_{_{33}} \end{split} \tag{015}\label{015}$$$$ which, using the $$3\times 3$$ matrix representation of states, is the trace of the matrix product $$\mathrm{X}\BoldExp{\mathrm{Y}}{*}$$
$$$$\langle \mathrm{X},\mathrm{Y}\rangle =\mathrm{Tr}\left[\mathrm{X}\BoldExp{\mathrm{Y}}{*}\right] \tag{016}\label{016}$$$$ given that $$\BoldExp{\mathrm{Y}}{*}$$ is the complex conjugate of the transpose of $$\mathrm{Y}$$ $$$$\BoldExp{\mathrm{Y}}{*}\equiv \BoldExp{ \begin{bmatrix} \mathrm{y}_{_{11}} & \mathrm{y}_{_{12}} & \mathrm{y}_{_{13}}\\ \mathrm{y}_{_{21}} & \mathrm{y}_{_{22}} & \mathrm{y}_{_{23}}\\ \mathrm{y}_{_{31}} & \mathrm{y}_{_{32}} & \mathrm{y}_{_{33}} \end{bmatrix}} {*} = \overline{\begin{bmatrix} \mathrm{y}_{_{11}} & \mathrm{y}_{_{12}} & \mathrm{y}_{_{13}}\\ \mathrm{y}_{_{21}} & \mathrm{y}_{_{22}} & \mathrm{y}_{_{23}}\\ \mathrm{y}_{_{31}} & \mathrm{y}_{_{32}} & \mathrm{y}_{_{33}} \end{bmatrix}^{\mathsf{T}}} = \begin{bmatrix} \overline{\mathrm{y}}_{_{11}} & \overline{\mathrm{y}}_{_{21}} & \overline{\mathrm{y}}_{_{31}}\\ \overline{\mathrm{y}}_{_{12}} & \overline{\mathrm{y}}_{_{22}} & \overline{\mathrm{y}}_{_{32}}\\ \overline{\mathrm{y}}_{_{13}} & \overline{\mathrm{y}}_{_{23}} & \overline{\mathrm{y}}_{_{33}} \end{bmatrix} \tag{017}\label{017}$$$$ Now, under a unitary transformation $$\;W \in SU(3)\;$$ in the 3-dimensional space of quarks $$\;\mathbf{Q}\;$$, we have $$$$\BoldExp{\boldsymbol{\xi}}{'} = W\boldsymbol{\xi} \tag{018}\label{018}$$$$ so in the space of antiquarks $$\overline{\mathbf{Q}}\;$$, since $$\;\BoldExp{\boldsymbol{\zeta}}{'}=W \boldsymbol{\zeta}\;$$ $$$$\overline{\BoldExp{\boldsymbol{\zeta}}{'}}= \overline{W}\;\overline{\boldsymbol{\zeta}} \tag{019}\label{019}$$$$ and for the meson state \begin{align} \BoldExp{\mathrm{X}}{'} & =\BoldExp{\boldsymbol{\xi}}{'}\boldsymbol{\otimes}\overline{\BoldExp{\boldsymbol{\zeta}}{'}}=\left(W\boldsymbol{\xi}\right)\left(\overline{W}\overline{\boldsymbol{\zeta}} \right) = \Biggl(W\begin{bmatrix} \xi_u\\ \xi_d\\ \xi_s \end{bmatrix}\Biggr) \Biggl(\overline{W}\begin{bmatrix} \overline{\zeta}_u\\ \overline{\zeta}_d\\ \overline{\zeta}_s \end{bmatrix}\Biggr)^{\mathsf{T}} \nonumber\\ & = W\Biggl(\begin{bmatrix} \xi_u\\ \xi_d\\ \xi_s \end{bmatrix} \begin{bmatrix} \overline{\zeta}_u & \overline{\zeta}_d & \overline{\zeta}_s \end{bmatrix}\Biggr)\overline{W}^{\mathsf{T}} =W\left(\boldsymbol{\xi}\boldsymbol{\otimes}\overline{\boldsymbol{\zeta}}\right)\BoldExp{W}{*}=W\;\mathrm{X}\;\BoldExp{W}{*} \nonumber \tag{020}\label{020} \end{align} that is $$$$\BoldExp{\mathrm{X}}{'} = W\;\mathrm{X}\;\BoldExp{W}{*} \tag{021}\label{021}$$$$ Above equation \eqref{021} is the transformation law of meson states in the 9-dimensional space $$\;\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 $$$$\langle \BoldExp{\mathrm{X}}{'},\BoldExp{\mathrm{Y}}{'}\rangle =\mathrm{Tr}\left[\BoldExp{\mathrm{X}}{'}\BoldExp{\BoldExp{\mathrm{Y}}{'}}{*}\right]=\mathrm{Tr}\Bigl[\left(W\mathrm{X}\BoldExp{W}{*}\right) \BoldExp{\left(W\mathrm{Y}\BoldExp{W}{*}\right)}{*}\Bigr]=\mathrm{Tr}\Bigl[W \left( \mathrm{X}\BoldExp{Y}{*}\right)\BoldExp{W}{*}\Bigr]=\mathrm{Tr}\Bigl[\mathrm{X}\BoldExp{Y}{*}\Bigr] \tag{022}\label{022}$$$$ The last equality in above equation \eqref{022} is valid since under the transformation law \eqref{021} the trace remains invariant. More generally, for unitary $$\;W \in SU(n)\;$$ and $$\;A\;$$ a $$\;n \times n\;$$ complex matrix the transformation
$$$$\BoldExp{\mathrm{A}}{'} = W\;\mathrm{A}\;\BoldExp{W}{*} \tag{023}\label{023}$$$$
if expressed in terms of elements, yields (we use the Einstein summation convention) $$$$\BoldExp{a_{ij}}{'} = w_{i\rho}a_{\rho\sigma}\BoldExp{w_{\sigma j}}{*} \tag{024}\label{0242}$$$$ so $$$$\mathrm{Tr}\Bigl[\BoldExp{\mathrm{A}}{'}\Bigr]=\BoldExp{a_{ii}}{'} = w_{i\rho}a_{\rho\sigma}\BoldExp{w_{\sigma i}}{*}=(\BoldExp{w_{\sigma i}}{*}w_{i\rho})a_{\rho\sigma}=\delta_{\sigma\rho}a_{\rho\sigma}=a_{\rho\rho}=\mathrm{Tr}\Bigl[A\Bigr] \tag{025}\label{025}$$$$ proving the invariance of inner product under the transformation law \eqref{021} $$$$\langle \BoldExp{\mathrm{X}}{'},\BoldExp{\mathrm{Y}}{'}\rangle =\langle \mathrm{X},\mathrm{Y}\rangle \tag{026}\label{026}$$$$

Now, obviously the meson state represented by the identity matrix $$$$\mathrm{I}= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \tag{027}\label{027}$$$$ is unchanged under the transformation \eqref{021} and if normalized yields $$$$\BoldSub{\mathrm{F}}{0}=\sqrt{\tfrac{1}{3}} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} =\sqrt{\tfrac{1}{3}}\left(\boldsymbol{u}\overline{\boldsymbol{u}}+\boldsymbol{d}\overline{\boldsymbol{d}}+\boldsymbol{s}\overline{\boldsymbol{s}} \right)\equiv \BoldExp{\boldsymbol{\eta}}{\prime} \tag{028}\label{028}$$$$ that is, it represents the $$\;\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 $$$$\mathrm{X}\perp \boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace}\Leftrightarrow\langle \mathrm{X},\BoldSub{\mathrm{F}}{0}\rangle =0\Leftrightarrow\mathrm{Tr}\left[\mathrm{X}\BoldSub{\mathrm{F}}{0}^{\boldsymbol{*}}\right]=0\Leftrightarrow\mathrm{Tr}\left[\mathrm{X}\right]=0 \tag{029}\label{029}$$$$
So, the 8-dimensional linear subspace of all meson states with traceless matrix representation is the orthogonal complement of the 1-dimensional subspace $$\;\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}\;$$ is any basis which spans this space then $$$$\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{1},\BoldSub{\mathrm{F}}{2},\cdots,\BoldSub{\mathrm{F}}{8}\boldsymbol{\rbrace}=\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace}^{\boldsymbol{\perp}}=\Bigl\{ \mathrm{X} \in \mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\; :\; \mathrm{Tr}\left[X\right]=0 \; \Bigr\} \tag{030}\label{030}$$$$
This space is invariant under the transformation \eqref{021}. There are arbitrary many choices of the basis $$\;\left(\BoldSub{\mathrm{F}}{1},\BoldSub{\mathrm{F}}{2},\cdots,\BoldSub{\mathrm{F}}{8}\right)\;$$ but a proper one must correspond to mesons in the real world and be orthonormal if possible. So, the normalized traceless meson state $$$$\BoldSub{\mathrm{F}}{3}=\sqrt{\tfrac{1}{2}} \begin{bmatrix} 1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0\\ 0 & \boldsymbol{-}1 & \hphantom{\boldsymbol{-}}0\\ 0 & \hphantom{\boldsymbol{-}} 0 & \hphantom{\boldsymbol{-}}0 \end{bmatrix} =\sqrt{\tfrac{1}{2}}\left(\boldsymbol{u}\overline{\boldsymbol{u}}-\boldsymbol{d}\overline{\boldsymbol{d}} \right)\equiv \BoldExp{\boldsymbol{\pi}}{0} \tag{031}\label{031}$$$$
represents of course the $$\;\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 $$$$\begin{split} &\mathcal{F}_{\mathbf{M}}=\Bigl(\tfrac{1}{3}\mathrm{Tr}\left[\mathcal{F}_{\mathbf{M}}\right]\Bigr)\mathcal{I}+\Bigl[\mathcal{F}_{\mathbf{M}}-\Bigl(\tfrac{1}{3}\mathrm{Tr}\left[\mathcal{F}_{\mathbf{M}}\right]\Bigr)\mathcal{I}\Bigr]\\ &=\begin{bmatrix} \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}} & \mathbf{0} & \mathbf{0}\\ \mathbf{0} & \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}} & \mathbf{0}\\ \mathbf{0} & \mathbf{0} & \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}} \end{bmatrix} + \begin{bmatrix} \dfrac{\left(2\boldsymbol{u}\overline{\boldsymbol{u}}-\boldsymbol{d}\overline{\boldsymbol{d}}-\boldsymbol{s}\overline{\boldsymbol{s}}\right) }{3}{\rule[0ex]{-10pt}{0ex}} & \boldsymbol{u}\overline{\boldsymbol{d}} & \boldsymbol{u}\overline{\boldsymbol{s}}\\ \boldsymbol{d}\overline{\boldsymbol{u}} & \dfrac{\left(-\boldsymbol{u}\overline{\boldsymbol{u}}+2\boldsymbol{d}\overline{\boldsymbol{d}}-\boldsymbol{s}\overline{\boldsymbol{s}}\right) }{3} & \boldsymbol{d}\overline{\boldsymbol{s}} \\ \boldsymbol{s}\overline{\boldsymbol{u}} & \boldsymbol{s}\overline{\boldsymbol{d}} & {\rule[-2ex]{-10pt}{6ex}} \dfrac{\left(-\boldsymbol{u}\overline{\boldsymbol{u}}-\boldsymbol{d}\overline{\boldsymbol{d}}+2\boldsymbol{s}\overline{\boldsymbol{s}}\right)}{3} \end{bmatrix} \end{split} \tag{032}\label{032}$$$$

The 3rd diagonal element of the traceless component of $$\mathcal{F}_{\mathbf{M}}$$, if opposed and normalized, yields $$$$\BoldSub{\mathrm{F}}{8}=\sqrt{\tfrac{1}{6}} \begin{bmatrix} 1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0\\ 0 & \hphantom{\boldsymbol{-}}1 & \hphantom{\boldsymbol{-}}0\\ 0 & \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}2 \end{bmatrix} =\sqrt{\tfrac{1}{6}}\left(\boldsymbol{u}\overline{\boldsymbol{u}}+\boldsymbol{d}\overline{\boldsymbol{d}}-2\boldsymbol{s}\overline{\boldsymbol{s}} \right)\equiv \boldsymbol{\eta} \tag{033}\label{033}$$$$
that is, it represents the $$\;\boldsymbol{\eta}\;$$ meson.

(to be continued in $$\boldsymbol{\S\:}\textbf{B}$$)

• Doesn't this beg the question of how the unitary transformation is constructed? Can't, for example, the entry -$\sqrt{\frac{2}{3}}$ be $\sqrt{\frac{2}{3}}$? – Deschele Schilder Jun 8 '19 at 15:01
• @descheleschilder You are absolutely right. I gave here a final conclusion and to answer this in details demands space greater than that of a PSE answer (30.000 characters). – Frobenius Jun 8 '19 at 17:28
• Thanks for taking the time to give the full details, even giving some maths reminder along the way for old skunks like me! But I still have 2 questions: - Why do we search for states invariance as stated in (026)? - It looks like you didn't give an answer to @descheleschilder remark in (033). So (1/6;1/6;-2/3) and (-1/6;-1/6;2/3) are the same and we chose one over the other by convention? – OOEngineer Jun 18 '19 at 19:09
• @OOEngineer "So $(\sqrt{\frac{1}{6}},\sqrt{\frac{1}{6}},\boldsymbol{-}\sqrt{\frac{2}{3}})$ and $(\boldsymbol{-}\sqrt{\frac{1}{6}},\boldsymbol{-}\sqrt{\frac{1}{6}},\sqrt{\frac{2}{3}})$ are the same and we chose one over the other by convention?" Yes, precisely. Since a constant multiple of a state vector represents the same state. But I think that descheleschilder asks why we don't choose $(\sqrt{\frac{1}{6}},\sqrt{\frac{1}{6}},\boldsymbol{+}\sqrt{\frac{2}{3}})$. Because this is an other state not corresponding to a meson in the real world. – Frobenius Jun 18 '19 at 20:55
• I see it's more clear. Then why states representing actual particles must be invariant following the way you defined between (022) and (026)? – OOEngineer Jun 18 '19 at 21:12


$$\boldsymbol{\S\:}\textbf{B. continued from }\boldsymbol{\S\:}\textbf{A}$$

Now, we have pions $$\BoldExp{\boldsymbol{\pi}}{+}$$,$$\BoldExp{\boldsymbol{\pi}}{-}$$ $$$$\BoldSub{\mathrm{F}}{1}= \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} =\boldsymbol{u}\overline{\boldsymbol{d}}\equiv \BoldExp{\boldsymbol{\pi}}{+} \tag{034}\label{034}$$$$

$$$$\BoldSub{\mathrm{F}}{2}= \begin{bmatrix} 0 & 0 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} =\boldsymbol{d}\overline{\boldsymbol{u}}\equiv \BoldExp{\boldsymbol{\pi}}{-} \tag{035}\label{035}$$$$

The rest four(4) basic meson states define mesons called kaons $$$$\BoldSub{\mathrm{F}}{4}= \begin{bmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} =\boldsymbol{u}\overline{\boldsymbol{s}}\equiv \BoldExp{\mathbf{K}}{+} \tag{036}\label{036}$$$$

$$$$\BoldSub{\mathrm{F}}{5}= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 1 & 0 & 0 \end{bmatrix} =\boldsymbol{s}\overline{\boldsymbol{u}}\equiv \BoldExp{\mathbf{K}}{-} \tag{037}\label{037}$$$$ $$$$\BoldSub{\mathrm{F}}{6}= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix} =\boldsymbol{d}\overline{\boldsymbol{s}}\equiv \BoldExp{\mathbf{K}}{0} \tag{038}\label{038}$$$$ $$$$\BoldSub{\mathrm{F}}{7}= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 1 & 0 \end{bmatrix} =\boldsymbol{s}\overline{\boldsymbol{d}}\equiv \BoldExp{\overline{\mathbf{K}}}{0} \tag{039}\label{039}$$$$ The octet $$\;\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{1},\BoldSub{\mathrm{F}}{2},\BoldSub{\mathrm{F}}{3},\BoldSub{\mathrm{F}}{4},\BoldSub{\mathrm{F}}{5},\BoldSub{\mathrm{F}}{6},\BoldSub{\mathrm{F}}{7},\BoldSub{\mathrm{F}}{8}\boldsymbol{\rbrace}\;$$ or in terms of meson states the octet $$$$\boldsymbol{\lbrace}\BoldExp{\boldsymbol{\pi}}{+},\BoldExp{\boldsymbol{\pi}}{-},\BoldExp{\boldsymbol{\pi}}{0},\BoldExp{\mathbf{K}}{+},\BoldExp{\mathbf{K}}{-},\BoldExp{\mathbf{K}}{0},\BoldExp{\overline{\mathbf{K}}}{0},\boldsymbol{\eta}\boldsymbol{\rbrace} \nonumber$$$$ is a complete orthonormal basis of the 8-dimensional subspace of traceless meson states.

Note that the six mesons $$$$\BoldExp{\boldsymbol{\pi}}{+},\BoldExp{\boldsymbol{\pi}}{-},\BoldExp{\mathbf{K}}{+},\BoldExp{\mathbf{K}}{-},\BoldExp{\mathbf{K}}{0},\BoldExp{\overline{\mathbf{K}}}{0} \nonumber$$$$ are represented by matrices having zeros on the main diagonal, see equations \eqref{034}-\eqref{039}. This is due to the fact that these states are orthogonal to the space spanned by the three mesons $$$$\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0},\BoldSub{\mathrm{F}}{3},\BoldSub{\mathrm{F}}{8}\boldsymbol{\rbrace} \equiv \boldsymbol{\lbrace}\BoldExp{\boldsymbol{\eta}}{\prime},\BoldExp{\boldsymbol{\pi}}{0},\boldsymbol{\eta}\boldsymbol{\rbrace} \nonumber$$$$ Indeed, for any state $$$$\mathrm{X}= \begin{bmatrix} \mathrm{x}_{_{11}} & \mathrm{x}_{_{12}} & \mathrm{x}_{_{13}}\\ \mathrm{x}_{_{21}} & \mathrm{x}_{_{22}} & \mathrm{x}_{_{23}}\\ \mathrm{x}_{_{31}} & \mathrm{x}_{_{32}} & \mathrm{x}_{_{33}} \end{bmatrix} \tag{040}\label{040}$$$$ we have

1. From orthogonality $$\mathrm{X}\perp \BoldSub{\mathrm{F}}{0}\equiv \BoldExp{\boldsymbol{\eta}}{\prime}$$ $$$$\mathrm{Tr}\left[\mathrm{X}\right]=\mathrm{x}_{_{11}}+\mathrm{x}_{_{22}}+\mathrm{x}_{_{33}}=0 \tag{041}\label{041}$$$$

2. From orthogonality $$\mathrm{X}\perp \BoldSub{\mathrm{F}}{3}\equiv \BoldExp{\boldsymbol{\pi}}{0}$$
$$$$\mathrm{x}_{_{11}}-\mathrm{x}_{_{22}}=0 \tag{042}\label{042}$$$$

3. From orthogonality $$\mathrm{X}\perp \BoldSub{\mathrm{F}}{8}\equiv \boldsymbol{\eta}$$
$$$$\mathrm{x}_{_{11}}+\mathrm{x}_{_{22}}-2\mathrm{x}_{_{33}}=0 \tag{043}\label{043}$$$$ and so by equations \eqref{041}-\eqref{043}: $$$$\mathrm{x}_{_{11}}=\mathrm{x}_{_{22}}=\mathrm{x}_{_{33}}=0 \nonumber$$$$

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 $$\boldsymbol{\lbrace}\BoldExp{\boldsymbol{\pi}}{0},\boldsymbol{\eta},\BoldExp{\boldsymbol{\eta}}{\prime}\boldsymbol{\rbrace}$$ through a special unitary transformation $$\mathrm{V}\in SU(3)$$ $$$$\begin{bmatrix} \boldsymbol{\pi^{0}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{\eta} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{\eta}^{\prime} \vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \sqrt{\tfrac{1}{2}} & \boldsymbol{-} \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0 \vphantom{\tfrac{a}{\tfrac{a}{b}}}\\ \sqrt{\tfrac{1}{6}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{6}} & \boldsymbol{-}\sqrt{\tfrac{2}{3}} \vphantom{\tfrac{a}{\tfrac{a}{b}}}\\ \sqrt{\tfrac{1}{3}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{3}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{3}} \vphantom{\tfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \begin{bmatrix} \boldsymbol{u}\overline{\boldsymbol{u}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{d}\overline{\boldsymbol{d}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{s}\overline{\boldsymbol{s}} \vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} =\mathrm{V} \begin{bmatrix} \boldsymbol{u}\overline{\boldsymbol{u}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{d}\overline{\boldsymbol{d}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{s}\overline{\boldsymbol{s}} \vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \tag{044}\label{044}$$$$

Moreover, the matrix $$\mathrm{V}$$ is real with $$\det\left({\mathrm{V}}\right) =+1$$. It would represent a rotation in the 3-dimensional real space. Its inverse is equal to its transpose, $$\BoldExp{\mathrm{V}}{-1}=\BoldExp{\mathrm{V}}{\mathrm{T}}$$

$$$$\begin{bmatrix} \boldsymbol{u}\overline{\boldsymbol{u}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{d}\overline{\boldsymbol{d}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{s}\overline{\boldsymbol{s}} \vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} =\BoldExp{\mathrm{V}}{-1} \begin{bmatrix} \boldsymbol{\pi^{0}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{\eta} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{\eta}^{\prime} \vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} = \begin{bmatrix} \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{6}} & \sqrt{\tfrac{1}{3}} \vphantom{\tfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{-}\sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{6}} & {\rule[0ex]{+8pt}{0ex}}\sqrt{\tfrac{1}{3}} \vphantom{\tfrac{a}{\tfrac{a}{b}}}\\ \hphantom{\boldsymbol{-}} 0& \boldsymbol{-}\sqrt{\tfrac{2}{3}} & {\rule[0ex]{+8pt}{0ex}}\sqrt{\tfrac{1}{3}} \vphantom{\tfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \begin{bmatrix} \boldsymbol{\pi^{0}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{\eta} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \boldsymbol{\eta}^{\prime} \vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \tag{045}\label{045}$$$$ so the expressions \begin{align} \boldsymbol{u}\overline{\boldsymbol{u}} & = \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}}+\dfrac{\BoldExp{\boldsymbol{\pi}}{0}}{\sqrt{2}}+\dfrac{\boldsymbol{\eta}}{\sqrt{6}} \tag{046}\label{046}\\ \boldsymbol{d}\overline{\boldsymbol{d}} & = \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}}-\dfrac{\BoldExp{\boldsymbol{\pi}}{0}}{\sqrt{2}}+\dfrac{\boldsymbol{\eta}}{\sqrt{6}} \tag{047}\label{047}\\ \boldsymbol{s}\overline{\boldsymbol{s}} & = \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}}-\dfrac{2\boldsymbol{\eta}}{\sqrt{6}} \tag{048}\label{048} \end{align} Inserting above in place of the diagonal elements of $$\mathcal{F}_{\mathbf{M}}$$ in equation \eqref{013} and using expressions \eqref{034}-\eqref{039} for the off-diagonal elements we have $$$$\begin{split} \mathcal{F}_{\mathbf{M}}&= \begin{bmatrix} \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}}+\dfrac{\BoldExp{\boldsymbol{\pi}}{0}}{\sqrt{2}}+\dfrac{\boldsymbol{\eta}}{\sqrt{6}} & \BoldExp{\boldsymbol{\pi}}{+} & \BoldExp{\mathbf{K}}{+} \\ \BoldExp{\boldsymbol{\pi}}{-} & \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}}-\dfrac{\BoldExp{\boldsymbol{\pi}}{0}}{\sqrt{2}}+\dfrac{\boldsymbol{\eta}}{\sqrt{6}} & \BoldExp{\mathbf{K}}{0} {\rule[-4.5ex]{0pt}{10ex}}\\ \BoldExp{\mathbf{K}}{-} & \BoldExp{\overline{\mathbf{K}}}{0} & {\rule[0ex]{+12pt}{0ex}}\dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}}-\dfrac{2\boldsymbol{\eta}}{\sqrt{6}}{\rule[0ex]{+12pt}{0ex}} \end{bmatrix}\\ &=\begin{bmatrix} \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}} & \mathbf{0} & \mathbf{0}\\ \mathbf{0} & {\rule[0ex]{+12pt}{0ex}}\dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}}{\rule[0ex]{+12pt}{0ex}} & \mathbf{0} {\rule[-4.5ex]{0pt}{10ex}}\\ \mathbf{0} & \mathbf{0} & \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}} \end{bmatrix} + \begin{bmatrix} \dfrac{\BoldExp{\boldsymbol{\pi}}{0}}{\sqrt{2}}+\dfrac{\boldsymbol{\eta}}{\sqrt{6}} & \BoldExp{\boldsymbol{\pi}}{+} & \BoldExp{\mathbf{K}}{+} \\ \BoldExp{\boldsymbol{\pi}}{-} & -\dfrac{\BoldExp{\boldsymbol{\pi}}{0}}{\sqrt{2}}+\dfrac{\boldsymbol{\eta}}{\sqrt{6}} & \BoldExp{\mathbf{K}}{0} {\rule[-4.5ex]{0pt}{10ex}}\\ \BoldExp{\mathbf{K}}{-} & \BoldExp{\overline{\mathbf{K}}}{0} & {\rule[0ex]{+12pt}{0ex}}-\dfrac{2\boldsymbol{\eta}}{\sqrt{6}}{\rule[0ex]{+12pt}{0ex}} \end{bmatrix} \end{split} \tag{049}\label{049}$$$$ That the 9-dimensional product space $$\;\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\;$$ is identical to the direct sum of the invariant subspaces $$\;\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace}\;$$ and $$\;\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{1},\BoldSub{\mathrm{F}}{2},\BoldSub{\mathrm{F}}{3},\BoldSub{\mathrm{F}}{4},\BoldSub{\mathrm{F}}{5},\BoldSub{\mathrm{F}}{6},\BoldSub{\mathrm{F}}{7},\BoldSub{\mathrm{F}}{8}\boldsymbol{\rbrace}\;$$ with dimensions 1 and 8 respectively, that is $$$$\begin{split} \mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}&= \boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace}\boldsymbol{\oplus}\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{1},\BoldSub{\mathrm{F}}{2},\BoldSub{\mathrm{F}}{3},\BoldSub{\mathrm{F}}{4},\BoldSub{\mathrm{F}}{5},\BoldSub{\mathrm{F}}{6},\BoldSub{\mathrm{F}}{7},\BoldSub{\mathrm{F}}{8}\boldsymbol{\rbrace}\\ &= \boldsymbol{\lbrace}\BoldExp{\boldsymbol{\eta}}{\prime}\boldsymbol{\rbrace}\boldsymbol{\oplus}\boldsymbol{\lbrace}\BoldExp{\boldsymbol{\pi}}{+},\BoldExp{\boldsymbol{\pi}}{-},\BoldExp{\boldsymbol{\pi}}{0},\BoldExp{\mathbf{K}}{+},\BoldExp{\mathbf{K}}{-},\BoldExp{\mathbf{K}}{0},\BoldExp{\overline{\mathbf{K}}}{0},\boldsymbol{\eta}\boldsymbol{\rbrace} \end{split} \tag{050}\label{050}$$$$
is expressed by the identity $$$$\boldsymbol{3}\boldsymbol{\otimes}\overline{\boldsymbol{3}}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{8} \tag{051}\label{051}$$$$

I think your interpretation 1 is correct. Here is a way to understand the linear combinations:

A meson made initially from just $$u$$ and $$\bar{u}$$ will not stay that way for long, because the quarks can annihilate and then reappear as $$d\bar{d}$$ or $$s\bar{s}$$. However, certain superpositions of $$u\bar{u}$$, $$d\bar{d}$$ and $$s\bar{s}$$ will remain constant over time. These are the linear combinations you listed, and they are the $$eigenvectors$$ of the system's Hamiltonian.

They can be derived by writing down the Hamiltonian in the $${u\bar{u}, d\bar{d}, s\bar{s}}$$ basis: $$H = \begin{bmatrix}2m + A & A & A\\A & 2m+A & A\\A & A & 2m+A\end{bmatrix},$$

where $$m$$ is the mass of a quark, and $$A$$ is the coupling between the basis states - it is the amplitude for a pair to annihilate and reappear as a different (or the same) pair. (We are assuming here that all quarks have the same mass and the same annihilation amplitudes. This is known as SU(3) flavor symmetry.)

You can check that the eigenvectors are the two you listed (with eigenvalue $$2m$$), along with $$(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$$, with eigenvalue $$2m + 3A$$, which corresponds to the $$\eta^{\prime}$$ meson.

Note: As alluded to in the other answers, this whole discussion is in the context of the simple static quark model, which is a big simplification of the real physics.

• I like your straightforward explanation for this. Since the only eigenvector of $H$ with eigenvalue $2m+3A$ is $\frac{1}{\sqrt{3}}(1,1,1)$, it must be the $\eta'$. However, the eigenspace for eigenvalue $2m$ is 2-degenerate, so there are many orthonormal combinations we could choose. How can you justify that the $pi^0$ and $\eta$ are the specific linear combinations given, rather than any other? – Jordan Aug 21 '19 at 3:40
• Good question. You are right that any linear combination of $\pi^0$ and $\eta^0$ is a valid eigenvector, in the limit of exact $SU(3)$ symmetry. In the real world $SU(3)$ is not exact, mainly because the strange quark mass is bigger than the up and down quark masses. (There are other, smaller effects as well.) So a more accurate Hamiltonian would have $2(m+\Delta) + A$ in place of $2m + A$ in the lower right corner of the above matrix. You can check that that modified Hamiltonian still has $\pi^0$ as one of its eigenvectors ... – Paul G Aug 21 '19 at 17:02
• ... (And the other one is a linear combination of $\eta^0$ and $\eta^{\prime}$.) That's the reason why people usually state that $\pi^0$ is an eigenvector of the $SU(3)$-symmetric Hamiltonian, rather than some other equally valid linear combination of $\pi^0$ and $\eta^0$: It's because $\pi^0$ remains an eigenvector when $SU(3)$ breaking is turned on. – Paul G Aug 21 '19 at 17:03

One has to keep in mind that all these constituents of the hadrons are elementary particles, i.e.quantum entities. All mathematical expressions follow the rules of quantum mechanics.

Quantum entitities are expressed with normalized wave functions . So one should read the expression as the wavefunction of a pi0, and the wavefunction of the eta. Wavefunctions

will give the probability of finding an up quark or a down quark when scattering off a pion ,and a strange quark when scattering off an eta. The square roots come so that the probability is normalized to one.

Hadrons are more complicated than the valence quarks that characterize their symmetries. The proton does not have just the valence quarks but a sea of quarks and gluons due to the strong interaction between quarks.

It is not easy to scatter off a pion :), but the proton has been extensively studied . These scatters are what come up in the parton distribution functions within hadrons .

Figure 1: Overview of the CTEQ6M proton parton distribution at Q = 2 GeV (Pumplin et al. 2002).

As you see it is much more complicated sinc there are not only the valence quarks but also the sea of quarks and gluons with which incoming leptonic probes can scatter. The valence quarks are important in the assignment of the symmetry groups,

The meson octet. Particles along the same horizontal line share the same strangeness, s, while those on the same left-leaning diagonals share the same charge, q (given as multiples of the elementary charge).

The eightfold way symmetries clinched the existence of quarks by the prediction and subsequent discovery of the omega-.

A lot enters when one is really studying hadrons.

• "all these are elementary particles" Could you please clarify what you mean by "all these"? My understanding is pions are not considered elementary anymore? – akhmeteli Jun 8 '19 at 5:33
• @akhmeteli I mean the contents displayed in the formulas – anna v Jun 8 '19 at 6:29
• So to sum up, both of my interpretations were wrong. Neutral pions, eta and neutral kaons are the only hadrons where their makeup in valence quarks cannot be nicely simplified as a quark-antiquark pair and thus show a bit of the underlying QCD mathematics. Am I right? – OOEngineer Jun 8 '19 at 11:52