**SECTION A :** *What remains invariant for a complex $\:3\times 3\:$ tensor depends upon its transformation law under* $\:U \in SU(3)\:$

CASE 1 : $\:\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}=\boldsymbol{6}\boldsymbol{\oplus}\overline{\boldsymbol{3}}\:$

The transformation law for the complex $\:3\times 3\:$ tensor $\:\mathrm{X}\:$ in this case is
\begin{equation}
\mathrm{X }^{\prime}=U\mathrm{X}U^{\mathsf{T}}\quad
\tag{A-01}
\end{equation}
Here the symmetry (+) or antisymmetry (-) is invariant since
\begin{equation}
\mathrm{X}^{\mathsf{T}}=\pm\:\mathrm{X} \Longrightarrow
{(\mathrm{X }^{\prime})}^{\mathsf{T}}=(U\mathrm{X}U^{\mathsf{T}})^{\mathsf{T}}=
{(U^{\mathsf{T}})}^{\mathsf{T}}\mathrm{X}^{\mathsf{T}}U^{\mathsf{T}}=U(\pm\:\mathrm{X})U^{\mathsf{T}}=\pm\:\mathrm{X }^{\prime}
\tag{A-02}
\end{equation}

In this case *it makes sense to split the tensor in its symmetrical and anti-symmetrical parts*
\begin{equation}
\mathrm{\Psi}=\dfrac{1}{2} \left(\mathrm{X}+\mathrm{X}^{\mathsf{T}}\right)\:, \quad\mathrm{\Omega}=\dfrac{1}{2} \left(\mathrm{X}-\mathrm{X}^{\mathsf{T}}\right)
\tag{A-03}
\end{equation}
The symmetrical part $\:\mathrm{\Psi}\:$ depends on 6 parameters, so is identical to a complex $\:6$-vector $\:\boldsymbol{\psi}\:$ which belongs to a *complex 6-dimensional invariant subspace* and is transformed under a special unitary transformation $\:W \in SU(6)\:$

\begin{equation}
\boldsymbol{\psi}^{\prime}=W\boldsymbol{\psi}\:, \quad W \in SU(6)
\tag{A-04}
\end{equation}

while, on the other hand, the anti-symmetrical part $\:\mathrm{\Omega}\:$ depends on 3 parameters, so is identical to a complex $\:3$-vector $\:\boldsymbol{\omega}\:$ which belongs to a *complex 3-dimensional invariant subspace* and is transformed under the special unitary transformation $\:\overline{U} \in SU(3)\:$

\begin{equation}
\boldsymbol{\omega}^{\prime}=\overline{U}\boldsymbol{\omega}\:, \quad \overline{U} \in SU(3)
\tag{A-05}
\end{equation}
That's why the symmetrical and anti-symmetrical parts give rise to the terms $\:\boldsymbol{6}\:$ and $\:\overline{\boldsymbol{3}}\:$ in the right hand of equation $\:\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}=\boldsymbol{6}\boldsymbol{\oplus}\overline{\boldsymbol{3}}\:$ respectively.

CASE 2 : $\:\boldsymbol{3}\boldsymbol{\otimes}\overline{\boldsymbol{3}}=\boldsymbol{8}\boldsymbol{\oplus}\boldsymbol{1}\:$

The transformation law for the complex $\:3\times 3\:$ tensor $\:\mathrm{X}\:$ in this case is
\begin{equation}
\mathrm{X }^{\prime}=U\mathrm{X}U^{\boldsymbol{*}}=U\mathrm{X}U^{-1}
\tag{A-06}
\end{equation}
For those interested, this is proved in **SECTION B**, motivated by the adventure of explaining the structure of mesons under the quark theory.

Here the symmetry (+) or antisymmetry (-) is **NOT** invariant
\begin{equation}
\mathrm{X}^{\mathsf{T}}=\pm\:\mathrm{X} \Longrightarrow
{(\mathrm{X }^{\prime})}^{\mathsf{T}}=(U\mathrm{X}U^{\boldsymbol{*}})^{\mathsf{T}}=
{(U^{\boldsymbol{*}})}^{\mathsf{T}}\mathrm{X}^{\mathsf{T}}U^{\mathsf{T}}=\overline{U}(\pm\:\mathrm{X})U^{\mathsf{T}} \ne\pm\:\mathrm{X }^{\prime}
\tag{A-07}
\end{equation}

So *it makes NO SENSE to split the tensor in its symmetrical and anti-symmetrical parts.*

To the contrary :

(1) if $\:\mathrm{X}\:$ is a constant tensor, that is a *scalar* multiple of the identity, $\:\mathrm{X}=z\mathrm{I}\:$ ($\:z \in \mathbb{C}\:$) , then is invariant $\:\mathrm{X }^{\prime}= U\mathrm{X}U^{-1}=U\left(z\mathrm{I}\right)U^{-1}=z\mathrm{I}=\mathrm{X}\:$

or

(2) since the transformation (A-06) is a similarity transformation, *it preserves the Trace* (=sum of the elements on the main diagonal) of $\:\mathrm{X}\:$, that is $\:Tr \left(\mathrm{X}^{\prime}\right)=Tr\left(\mathrm{X}\right)\:$. So a traceless tensor remains traceless.

It would sound not very well, but in this case the invariants are the "tracelessness" and the "scalarness".

In this case *it makes sense to split the tensor in a traceless and in a scalar part* :
\begin{equation}
\mathrm{\Phi}=\mathrm{X}-\left[\dfrac{1}{3}Tr\left(\mathrm{X}\right)\right]\cdot\mathrm{I}\:, \quad \mathrm{\Upsilon}=\left[\dfrac{1}{3}Tr\left(\mathrm{X}\right)\right]\cdot\mathrm{I}
\tag{A-08}
\end{equation}
The traceless part $\:\mathrm{\Phi}\:$ depends on 8 (=3x3-1) parameters, so is identical to a complex $\:8$-vector $\:\boldsymbol{\phi}\:$ which belongs to a *complex 8-dimensional invariant subspace* **NOT FURTHER REDUCED TO INVARIANTS SUBSPACES** and is transformed under a special unitary transformation $\:V \in SU(8)\:$
\begin{equation}
\boldsymbol{\phi}^{\prime}=V\boldsymbol{\phi}\:, \quad V \in SU(8)
\tag{A-09}
\end{equation}
while, on the other hand, the scalar part $\:\mathrm{\Upsilon}\:$ depends on 1 parameter, so is identical to a complex $\:1$-vector $\:\boldsymbol{\upsilon}\:$ which belongs to a *complex 1-dimensional invariant subspace* (identical to the set of complex numbers $\:\mathbb{C}\:$) and is transformed under the special unitary transformation $\:\mathrm{I} \in SU(1)\:$

(identical to the identity)
\begin{equation}
\boldsymbol{\upsilon}^{\prime}=\mathrm{I}\boldsymbol{\upsilon}=\boldsymbol{\upsilon}
\tag{A-10}
\end{equation}
Note that $\:SU(1)\equiv \{\:\mathrm{I}\:\}\:$, that is the group $\:SU(1)\:$ has only one element, the identity $\:\mathrm{I}\:$, while $\:U(1)\equiv\{\:U\::\:U=e^{i\theta}\mathrm{I}\:, \quad \theta \in \mathbb{R} \}\:$, that is mathematically identical to the unit circle in $\:\mathbb{C}\:$.

================================================================================

**SECTION B :** *Mesons from three quarks*

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

\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{B-01}
\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_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{B-02}
\end{equation}
For a quark $\boldsymbol{\eta} \in \mathbf{Q}$
\begin{equation}
\boldsymbol{\eta}=\eta_1\boldsymbol{u}+\eta_2\boldsymbol{d}+\eta_3\boldsymbol{s}=
\begin{bmatrix}
\eta_1\\
\eta_2\\
\eta_3
\end{bmatrix}
\tag{B-03}
\end{equation}

the respective antiquark $\overline{\boldsymbol{\eta}}$ is expressed by the complex conjugates of the coordinates

\begin{equation}
\overline{\boldsymbol{\eta}}=\overline{\eta}_1 \overline{\boldsymbol{u}}+\overline{\eta}_2\overline{\boldsymbol{d}}+\overline{\eta}_3\overline{\boldsymbol{s}}=
\begin{bmatrix}
\overline{\eta}_1\\
\overline{\eta}_2\\
\overline{\eta}_3
\end{bmatrix}
\tag{B-04}
\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{B-05}
\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{B-06}
\end{equation}

Using the expressions (B-02) and (B-04) of the quark $\boldsymbol{\xi} \in \mathbf{Q}$ and the antiquark $\overline{\boldsymbol{\eta}} \in \overline{\mathbf{Q}}$ 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{\eta}}=&\xi_1\overline{\eta}_1 \left(\boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_1\overline{\eta}_2 \left(\boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_1\overline{\eta}_3 \left(\boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right)+ \\
&\xi_2\overline{\eta}_1 \left(\boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_2\overline{\eta}_2 \left( \boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_2\overline{\eta}_3 \left(\boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right)+\\
&\xi_3\overline{\eta}_1 \left(\boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_3\overline{\eta}_2 \left(\boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_3\overline{\eta}_3 \left(\boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right)
\end{split}
\tag{B-07}
\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{\eta}}=&\xi_1\overline{\eta}_1 \left(\boldsymbol{u}\overline{\boldsymbol{u}}\right)+\xi_1\overline{\eta}_2 \left(\boldsymbol{u}\overline{\boldsymbol{d}}\right)+\xi_1\overline{\eta}_3 \left(\boldsymbol{u}\overline{\boldsymbol{s}}\right)+ \\
&\xi_2\overline{\eta}_1 \left(\boldsymbol{d}\overline{\boldsymbol{u}}\right)+\xi_2\overline{\eta}_2 \left( \boldsymbol{d}\overline{\boldsymbol{d}}\right)+\xi_2\overline{\eta}_3 \left(\boldsymbol{d}\overline{\boldsymbol{s}}\right)+\\
&\xi_3\overline{\eta}_1 \left(\boldsymbol{s}\overline{\boldsymbol{u}}\right)+\xi_3\overline{\eta}_2 \left(\boldsymbol{s}\overline{\boldsymbol{d}}\right)+\xi_3\overline{\eta}_3 \left(\boldsymbol{s}\overline{\boldsymbol{s}}\right)
\end{split}
\tag{B-08}
\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{\eta}}=
\begin{bmatrix}
\xi_1\overline{\eta}_1 & \xi_1\overline{\eta}_2 & \xi_1\overline{\eta}_3\\
\xi_2\overline{\eta}_1 & \xi_2\overline{\eta}_2 & \xi_2\overline{\eta}_3\\
\xi_3\overline{\eta}_1 & \xi_3\overline{\eta}_2 & \xi_s\overline{\eta}_3
\end{bmatrix}=
\begin{bmatrix}
\xi_1\\
\xi_2\\
\xi_3
\end{bmatrix}
\begin{bmatrix}
\overline{\eta}_1 \\
\overline{\eta}_2 \\
\overline{\eta}_3
\end{bmatrix}^{\mathsf{T}}
=
\begin{bmatrix}
\xi_1\\
\xi_2\\
\xi_3
\end{bmatrix}
\begin{bmatrix}
\overline{\eta}_1 & \overline{\eta}_2 & \overline{\eta}_3
\end{bmatrix}
\tag{B-09}
\end{equation}

Now, under a unitary transformation $\;U \in SU(3)\;$ in the 3-dimensional space of quarks $\;\mathbf{Q}\;$, we have
\begin{equation}
\boldsymbol{\xi}^{\prime}= U\boldsymbol{\xi}
\tag{B-10}
\end{equation}
so in the space of antiquarks $\overline{\mathbf{Q}}\;$, since $\;\boldsymbol{\eta}^{\prime}=U\boldsymbol{\eta}\;$
\begin{equation}
\overline{\boldsymbol{\eta}^{\prime}}= \overline{U}\;\overline{\boldsymbol{\eta}}
\tag{B-11}
\end{equation}
and for the meson state
\begin{equation}
\mathrm{X}^{\prime}=\boldsymbol{\xi}^{\prime}\boldsymbol{\otimes}\overline{\boldsymbol{\eta}^{\prime}}=\left(U\boldsymbol{\xi}\right)\left(\overline{U}\overline{\boldsymbol{\eta}}\right)
=
\Biggl(U\begin{bmatrix}
\xi_1\\
\xi_2\\
\xi_3
\end{bmatrix}\Biggr)
\Biggl(\overline{U}\begin{bmatrix}
\overline{\eta}_1\\
\overline{\eta}_2\\
\overline{\eta}_3
\end{bmatrix}\Biggr)^{\mathsf{T}}
\\=
U\Biggl(\begin{bmatrix}
\xi_1\\
\xi_2\\
\xi_3
\end{bmatrix}
\begin{bmatrix}
\overline{\eta}_1 & \overline{\eta}_2 & \overline{\eta}_3
\end{bmatrix}\Biggr)\overline{U}^{\mathsf{T}}
=
U\left(\boldsymbol{\xi}\boldsymbol{\otimes}\overline{\boldsymbol{\eta}}\right)U^{*}=U\;\mathrm{X}\;U^{*}
\tag{B-12}
\end{equation}

$===================\text{end of answer}=======================$

This post imported from StackExchange Physics at 2015-06-05 21:02 (UTC), posted by SE-user diracpaul