When we encounter a problem in physics which can be expressed in terms of matrices or tensors, why do we decompose the tensor in terms of its symmetric and antisymmetric or trace components? What is the physics motivation behind doing so?

  • 1
    $\begingroup$ Typically, commutators of matrices project out their traceful (identity) pieces, so the split between traceless and traceful parts in a stable one under Lie-algebraic manipulations. Likewise, splitting S and A parts showcases simple parities easy to track off in multiplication and commutation. $\endgroup$ Aug 27, 2017 at 13:40

3 Answers 3


A very interesting example where the invariance of symmetry or antisymmetry is used successfully is 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
\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} Let take 2 more quarks in order to construct baryons from 3 quarks \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} \:, \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{B-03} \end{equation} A baryon state $\:T\:$ in the product space \begin{equation} \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{B-04} \end{equation} is the product of the states of above 3 quarks \begin{equation} T=\boldsymbol{\xi}\boldsymbol{\otimes}\boldsymbol{\eta}\boldsymbol{\otimes}\boldsymbol{\zeta} \tag{B-05} \end{equation} so is represented by a tensor with three indices \begin{equation} T_{ijk}=\xi_{i}\eta_{j}\zeta_{k}\:,\qquad i,j,k \in \left\lbrace 1,2,3\right\rbrace \tag{B-06} \end{equation} Now, under a unitary transformation $\;U=u_{\ell m} \in SU(3)\;$ in the 3-dimensional space of quarks $\;\mathbf{Q}\;$, that is \begin{equation} \boldsymbol{\xi}^{\prime}=U\boldsymbol{\xi}\:,\qquad\boldsymbol{\eta}^{\prime}=U\boldsymbol{\eta}\:,\qquad \boldsymbol{\zeta}^{\prime}=U\boldsymbol{\zeta} \tag{B-07} \end{equation} or by components \begin{equation} \xi^{\prime}_{\rho}=u_{\rho i}\xi_{i}\:,\qquad \eta^{\prime}_{\sigma}=u_{\sigma j}\eta_{j}\:,\qquad \zeta^{\prime}_{\tau}=u_{\tau k}\zeta_{k} \tag{B-08} \end{equation} for the transformation of the baryon state we have \begin{equation} T^{\prime}_{\rho\sigma\tau}=\xi^{\prime}_{\rho}\eta^{\prime}_{\sigma}\zeta^{\prime}_{\tau} =\left(u_{\rho i}\xi_{i}\right)\left(u_{\sigma j}\eta_{j} \right)\left(u_{\tau k}\zeta_{k}\right)=\left(u_{\rho i}u_{\sigma j}u_{\tau k}\right)\xi_{i}\eta_{j}\zeta_{k} \tag{B-09} \end{equation} so[Note A] \begin{equation} T^{\prime}_{\rho\sigma\tau}=\left(u_{\rho i}u_{\sigma j}u_{\tau k}\right)T_{ijk} \tag{B-10} \end{equation}
In order to find the invariant subspaces or in other terms the irreducible representations, we'll try to find the invariant properties of the law (B-10). So, we note that if $\:T_{ijk}\:$ is symmetric (+) or anti-symmetric (-) with respect to a pair of indices, let $\:(i,j)\:$, so does the transformed state $\:T^{\prime}_{\rho\sigma\tau}\:$ with respect to the respective pair of indices, in our case $\:(\rho,\sigma)$, since \begin{equation} T_{ijk}=\;\pm\;T_{jik} \Longrightarrow T^{\prime}_{\rho\sigma\tau}=\left(u_{\rho i}u_{\sigma j}u_{\tau k}\right)T_{ijk} =\;\pm\;\left(u_{\sigma j}u_{\rho i}u_{\tau k}\right)T_{jik}=\;\pm\; T^{\prime}_{\sigma\rho\tau} \tag{B-11} \end{equation} If a tensor $\:T_{i_{1}i_{2}\cdots i_{r}\cdots i_{s}\cdots i_{p}}\:$ with $\:p\:$ indices is symmetric (antisymmetric) with respect to any pair of them $\:\left( i_{r},i_{s}\right)$, then we call it totally symmetric (antisymmetric) or simply symmetric (antisymmetric).

As a first step we extract from $\:T_{ijk}\:$ a totally symmetric part $\:S_{ijk}\:$ and a totally anti-symmetric part $\:A_{ijk}\:$ \begin{equation} T_{ijk}= S_{ijk}+A_{ijk}+R_{ijk} \tag{B-12} \end{equation} and check after that the properties of the rest part $\:R_{ijk}.$ Now, $\:T_{ijk}\:$ is a $\:3^{3}\!=\!27\!-\!$dimensional quantity and for our purpose we classify the set of its 27 elements into 10 classes as in the following scheme : \begin{equation} \begin{matrix} 01)\left[300\right]: & T_{111} \\ 02)\left[210\right]: & T_{112},T_{121},T_{211} \\ 03)\left[201\right]: & T_{113},T_{131},T_{311} \\ 04)\left[120\right]: & T_{122},T_{212},T_{221} \\ 05)\left[111\right]: & T_{123},T_{231},T_{312},T_{321},T_{213},T_{132} \\ 06)\left[102\right]: & T_{133},T_{313},T_{331}\\ 07)\left[030\right]: & T_{222} \\ 08)\left[021\right]: & T_{223},T_{232},T_{322} \\ 09)\left[012\right]: & T_{332},T_{323},T_{233} \\ 10)\left[003\right]: & T_{333} \end{matrix} \tag{B-13} \end{equation} The code number of each class is the string $\;\left[ x_{1}x_{2}x_{3}\right]\;$ which denotes that the 3 indices of the elements of the class contain $x_{1}$ times the number $1$, $x_{2}$ times the number $2$ and $x_{3}$ times the number $3$. The number of classes is 10 because[Note B]: so many are the ordered triads $\;\left(x_{1},x_{2},x_{3}\right)\;$ of nonnegative integers by which the integer 3 ($\equiv$ the number of indices) could be split :
\begin{equation} x_{1}+x_{2}+x_{3}=3 \tag{B-14} \end{equation} Let arrange the 27 components of tensor $\:T_{ijk}\:$ on a cube. In this case to the 10 classes of its elements, see equation (B-13), there correspond 10 geometrical figures : \begin{equation} \begin{matrix} \left[300\right],\left[030\right],\left[003\right] & : & \textbf{3} & \textbf{points} \text{ on the main diagonal} \hphantom{===============}\\ \left[210\right],\left[201\right],\left[120\right],\left[102\right],\left[021\right],\left[012\right] & : & \textbf{6} & \textbf{equilateral triangles} \text{ normal to the main diagonal}\hphantom{==}\\ \left[111\right] & : & \textbf{1} & \textbf{regular hexagon} \text{ normal to the main diagonal}\hphantom{====} \end{matrix} \tag{B-13a} \end{equation} This representation is given in the Figure below. Each one of these 10 figures is invariant with respect to any pair of indices of its elements. So, if to the elements of each figure we attach a variable independent of the 9 variables attached to the rest 9 figures we produce a totally symmetric tensor $\:S_{ijk}$. This tensor, by the method of its production, is 10-dimensional. Such a production is given in equation (B-15). See each one of the 10 figures separately in larger size here : 10 invariant figures.
enter image description here So, we would extract from $\:T_{ijk}\:$ a 10-dimensional totally symmetric part $\:S_{ijk}\:$ as follows : \begin{equation} \begin{split} 1)\left[300\right]: & S_{111}=T_{111} \\ 2)\left[210\right]: & S_{112}=S_{121}=S_{211}=\dfrac{T_{112}+T_{121}+T_{211}}{3}\\ 3)\left[201\right]: & S_{113}=S_{131}=S_{311}=\dfrac{T_{113}+T_{131}+T_{311}}{3}\\ 4)\left[120\right]: & S_{122}=S_{212}=S_{221}=\dfrac{T_{122}+T_{212}+T_{221}}{3}\\ 5)\left[111\right]: & S_{123}=S_{231}=S_{312}=S_{321}=S_{213}=S_{132}=\\ & \dfrac{\left(T_{123}+T_{231}+T_{312}\right)+\left(T_{321}+T_{213}+T_{132}\right)}{6}\\ 6)\left[102\right]: & S_{133}=S_{313}=S_{331}=\dfrac{T_{133}+T_{313}+T_{331}}{3}\\ 7)\left[030\right]: & S_{222}=T_{222} \\ 8)\left[021\right]: & S_{223}=S_{232}=S_{322}=\dfrac{T_{223}+T_{232}+T_{322}}{3}\\ 9)\left[012\right]: & S_{332}=S_{323}=S_{233}=\dfrac{T_{332}+T_{323}+T_{233}}{3}\\ 10)\left[003\right]: & S_{333}=T_{333} \end{split} \tag{B-15} \end{equation} that is taking $^{\prime\prime}$mean value$^{\prime\prime}$ in each class.

For the totally anti-symmetric part $\:A_{ijk}\:$ : if an element has the same value to at least two indices then is zero. It's clear that the elements $\:A_{123}, A_{231}, A_{312}\:$ must have the same value, say $\:b_{0}\:$, as also $\:A_{321},A_{213},A_{132}\:$ must do and moreover[Note C] \begin{equation} A_{123}=A_{231}=A_{312}=\;b_{0}\;=-A_{321}=-A_{213}=-A_{132} \tag{B-16} \end{equation} Although it makes no sense, the elements of $\:A_{ijk}\:$ are classified in the above-mentioned 10 classes and the component $\:b_{0}\:$ of equation (B-16 ) is defined by \begin{equation} b_{0}=\dfrac{\left(T_{123}+T_{231}+T_{312}\right)-\left(T_{321}+T_{213}+T_{132}\right)}{6}\\ \tag{B-17} \end{equation} \begin{equation} \begin{split} 1)\left[300\right]: & A_{111}=0 \\ 2)\left[210\right]: & A_{112}=A_{121}=A_{211}=0\\ 3)\left[201\right]: & A_{113}=A_{131}=A_{311}=0\\ 4)\left[120\right]: & A_{122}=A_{212}=A_{221}=0\\ 5)\left[111\right]: & A_{123}=A_{231}=A_{312}=-A_{321}=-A_{213}=-A_{132}=\\ &\dfrac{\left(T_{123}+T_{231}+T_{312}\right)-\left(T_{321}+T_{213}+T_{132}\right)}{6}\\ 6)\left[102\right]: & A_{133}=A_{313}=A_{331}=0\\ 7)\left[030\right]: & A_{222}=0 \\ 8)\left[021\right]: & A_{223}=0\\ 9)\left[012\right]: & A_{332}=A_{323}=A_{233}=0\\ 10)\left[003\right]: & A_{333}=0 \end{split} \tag{B-18} \end{equation} The component $\:b_{0}\:$ of the 1-dimensional $\:A_{ijk}\:$ may be any arbitrary complex number. The above choice, see equation (B-17), is necessary to feed the rest of the tensor \begin{equation} R_{ijk} \equiv T_{ijk}-S_{ijk}-A_{ijk} \tag{B-19} \end{equation} with the invariant property \begin{equation} R_{ijk}+R_{jki}+R_{kij}=0\;, \quad \text{cyclic permutation of the indices } \;i,j,k \tag{B-20} \end{equation} That the property in (B-20) remains invariant under the transformation law (B-10) is proved as follows \begin{equation} \begin{split} R^{\prime}_{\rho\sigma\tau}+R^{\prime}_{\sigma\tau\rho}+R^{\prime}_{\tau\rho\sigma}&=\left(u_{\rho i}u_{\sigma j}u_{\tau k}\right)R_{ijk}+\left(u_{\sigma j}u_{\tau k}u_{\rho i}\right)R_{jki}+\left(u_{\tau k}u_{\rho i}u_{\sigma j}\right)R_{kij}\\ &=\left(u_{\rho i}u_{\sigma j}u_{\tau k}\right)\underbrace{\left(R_{ijk}+R_{jki}+R_{kij}\right)}_{0}=0 \end{split} \nonumber \end{equation} The components of 16-dimensional tensor $\: R_{ijk}\:$ are given below \begin{equation} \begin{split} 1)\left[300\right]: & R_{111}=0 \\ 2)\left[210\right]: & R_{112}=\dfrac{2T_{112}-\left(T_{121}+T_{211}\right)}{3}, R_{121}=\dfrac{2T_{121}-\left(T_{211}+T_{112}\right)}{3}, R_{211}=\dfrac{2T_{211}-\left(T_{112}+T_{121}\right)}{3}\\ 3)\left[201\right]: & R_{113}=\dfrac{2T_{113}-\left(T_{131}+T_{311}\right)}{3}, R_{131}=\dfrac{2T_{131}-\left(T_{311}+T_{113}\right)}{3}, R_{311}=\dfrac{2T_{311}-\left(T_{113}+T_{131}\right)}{3}\\ 4)\left[120\right]: & R_{122}=\dfrac{2T_{122}-\left(T_{212}+T_{221}\right)}{3}, R_{212}=\dfrac{2T_{212}-\left(T_{221}+T_{122}\right)}{3}, R_{221}=\dfrac{2T_{221}-\left(T_{122}+T_{212}\right)}{3}\\ 5)\left[111\right]: & R_{123}=\dfrac{2T_{123}-\left(T_{231}+T_{312}\right)}{3}, R_{231}=\dfrac{2T_{231}-\left(T_{312}+T_{123}\right)}{3}, R_{312}=\dfrac{2T_{312}-\left(T_{123}+T_{231}\right)}{3}\\ & R_{321}=\dfrac{2T_{321}-\left(T_{213}+T_{132}\right)}{3}, R_{213}=\dfrac{2T_{213}-\left(T_{132}+T_{321}\right)}{3}, R_{132}=\dfrac{2T_{132}-\left(T_{321}+T_{213}\right)}{3}\\ 6)\left[102\right]: & R_{133}=\dfrac{2T_{133}-\left(T_{313}+T_{331}\right)}{3}, R_{313}=\dfrac{2T_{313}-\left(T_{331}+T_{133}\right)}{3}, R_{331}=\dfrac{2T_{331}-\left(T_{133}+T_{313}\right)}{3}\\ 7)\left[030\right]: & R_{222}=0\\ 8)\left[021\right]: & R_{223}=\dfrac{2T_{223}-\left(T_{232}+T_{322}\right)}{3}, R_{232}=\dfrac{2T_{232}-\left(T_{322}+T_{223}\right)}{3}, R_{322}=\dfrac{2T_{223}-\left(T_{232}+T_{322}\right)}{3}\\ 9)\left[012\right]: & R_{233}=\dfrac{2T_{233}-\left(T_{323}+T_{332}\right)}{3}, R_{323}=\dfrac{2T_{323}-\left(T_{332}+T_{233}\right)}{3}, R_{332}=\dfrac{2T_{332}-\left(T_{233}+T_{323}\right)}{3}\\ 10)\left[003\right]: & R_{333}=0 \end{split} \tag{B-21} \end{equation} The tensor $\: R_{ijk}\:$ is not symmetric or anti-symmetric relative to any pair of indices. So, it is further reduced if we split it in a symmetric $\: X_{ijk}\:$ and an anti-symmetric part $\: Y_{ijk}\:$ relative to one and only one pair of indices, say $\:(i,j)\:$ \begin{equation} R_{ijk}=X_{ijk}+Y_{ijk}\;, \quad X_{ijk}=\;+\;X_{jik}\;, \quad Y_{ijk}=\;-\;Y_{jik} \tag{B-22} \end{equation}
As discussed in the paragraph next to the transformation law (B-10), the symmetric and anti-symmetric properties of $\: X_{ijk}\:$ and $\: Y_{ijk}\:$ respectively, relatively to the pair indices $\:(i,j)$, remain invariant under aforementioned law.
From (B-22) \begin{equation} X_{ijk}=\dfrac{R_{ijk}+R_{jik}}{2}\;, Y_{ijk}=\dfrac{R_{ijk}-R_{jik}}{2} \tag{B-23} \end{equation} The tensor $\: X_{ijk}\:$ is symmetric with respect only to one pair of indices, it's characterized as mixed symmetric (MS), it is 8-dimensional and its elements are given in (B-24) \begin{equation} \begin{split} 1)\left[300\right]: & X_{111}=0 \\ 2)\left[210\right]: & X_{112}=-2X_{121}=-2X_{211}=\dfrac{2T_{112}-\left(T_{121}+T_{211}\right)}{3}\\ 3)\left[201\right]: & X_{113}=-2X_{131}=-2X_{311}=\dfrac{2T_{113}-\left(T_{131}+T_{311}\right)}{3}\\ 4)\left[120\right]: & X_{221}=-2X_{122}=-2X_{212}=\dfrac{2T_{221}-\left(T_{122}+T_{212}\right)}{3}\\ 5)\left[111\right]: & X_{123}=X_{213}=\dfrac{2\left(T_{123}+T_{213}\right)-\left(T_{231}+T_{312}+T_{132}+T_{321}\right)}{6}\\ & X_{231}=X_{321}=\dfrac{2\left(T_{231}+T_{321}\right)-\left(T_{312}+T_{123}+T_{213}+T_{132}\right)}{6}\\ & X_{312}=X_{132}=-\left( X_{123}+X_{231}\right) =\dfrac{2\left(T_{312}+T_{132}\right)-\left(T_{123}+T_{231}+T_{321}+T_{213}\right)}{6}\\ 6)\left[102\right]: & X_{331}=-2X_{133}=-2X_{313}=\dfrac{2T_{331}-\left(T_{133}+T_{313}\right)}{3}\\ 7)\left[030\right]: & X_{222}=0\\ 8)\left[021\right]: & X_{223}=-2X_{232}=-2X_{322}=\dfrac{2T_{223}-\left(T_{232}+T_{322}\right)}{3}\\ 9)\left[012\right]: & X_{332}=-2X_{233}=-2X_{323}=\dfrac{2T_{332}-\left(T_{233}+T_{323}\right)}{3}\\ 10)\left[003\right]:& X_{333}=0 \end{split} \tag{B-24} \end{equation} The tensor $\: Y_{ijk}\:$ is anti-symmetric with respect only to one pair of indices, it's characterized as mixed anti-symmetric (MA), it is also 8-dimensional and its elements are given in (B-25) \begin{equation} \begin{split} 1)\left[300\right]: & Y_{111}=0 \\ 2)\left[210\right]: & Y_{112}=0,Y_{121}=-Y_{211}=\dfrac{T_{121}-T_{112}}{2}\\ 3)\left[201\right]: & Y_{113}=0,Y_{131}=-Y_{311}=\dfrac{T_{131}-T_{113}}{2}\\ 4)\left[120\right]: & Y_{221}=0,Y_{122}=-Y_{212}=\dfrac{T_{122}-T_{212}}{2}\\ 5)\left[111\right]: & Y_{123}=-Y_{213}=\dfrac{2\left(T_{123}-T_{213}\right)-\left(T_{231}+T_{312}-T_{132}-T_{321}\right)}{6}\\ &Y_{231}=-Y_{321}=\dfrac{2\left(T_{231}-T_{321}\right)-\left(T_{312}+T_{123}-T_{213}-T_{132}\right)}{6}\\ & Y_{312}=-Y_{132}=-\left(Y_{123}+Y_{231}\right)=\dfrac{2\left(T_{312}-T_{132}\right)-\left(T_{123}+T_{231}-T_{321}-T_{213}\right)}{6}\\ 6)\left[102\right]: & Y_{331}=0,Y_{133}=-Y_{313}=\dfrac{T_{133}-T_{313}}{2}\\ 7)\left[030\right]: & Y_{222}=0\\ 8)\left[021\right]: & Y_{223}=0,Y_{232}=-Y_{322}=\dfrac{T_{232}-T_{223}}{2}\\ 9)\left[012\right]: & Y_{332}=0,Y_{323}=-Y_{233}=\dfrac{T_{323}-T_{332}}{2}\\ 10)\left[003\right]:& Y_{333}=0 \end{split} \tag{B-25} \end{equation} The irreducible representation of $\:\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}\:$ is shown in (B-26) below. The subscripts used are $\:A=A$nti-symmetric (totally), $S=S$ymmetric (totally), $MS=M$ixed $S$ymmetric and $MA=M$ixed $A$nti-symmetric. \begin{equation} \boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}= \boldsymbol{1}_{A}\boldsymbol{\oplus}\boldsymbol{10}_{S}\boldsymbol{\oplus} \boldsymbol{8}_{MS}\boldsymbol{\oplus}\boldsymbol{8}_{MA} \tag{B-26} \end{equation} while its relation with the above-mentioned tensors is shown in (B-27) \begin{equation} \begin{matrix} \underbrace{T_{ijk}}_{\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}}\quad = \underbrace{A_{ijk}}_{\boldsymbol{1}_{A}} +\underbrace{S_{ijk}}_{\boldsymbol{10}_{S}} +\underbrace{ X_{ijk}}_{\boldsymbol{8}_{MS}}+\underbrace{ Y_{ijk}}_{\boldsymbol{8}_{MA}} \end{matrix} \tag{B-27} \end{equation} Equation (B-27) is an alternative expression without subscripts. \begin{equation} \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{B-28} \end{equation} The connection of any tensor independent component with a baryon is achieved by :

(a) replacing every $\:T_{ijk}\:$ in the expression of the component by its set of indices
$\:ijk\:$ and in this set substituting the numbers $\:1,2,3\:$ by quarks $\:\boldsymbol{u},\boldsymbol{d},\boldsymbol{s}\:$ respectively \begin{equation} 1\longrightarrow \boldsymbol{u}\;,\quad 2\longrightarrow \boldsymbol{d}\;,\quad 3\longrightarrow \boldsymbol{s} \tag{B-29} \end{equation}
(b) normalizing the resulting baryon basic state. In the following the symbol $\:\left[q_{1},q_{2}\right]\:$ is used for the anti-symmetric expression $\:q_{1}q_{2}-q_{2}q_{1}\:$ \begin{equation} \left[q_{1},q_{2}\right]\equiv q_{1}q_{2}-q_{2}q_{1} \tag{B-30} \end{equation} So :
(A) Singlet $\:\boldsymbol{1}\:$ $\:\Lambda^{0}_{1}\:$ from $\:A_{ijk}.$
To the total anti-symmetric 1-dimensional tensor $\:A_{ijk}\:$, see (B-18), and to the component \begin{equation} A_{123}=\dfrac{\left(T_{123}+T_{231}+T_{312}\right)-\left(T_{321}+T_{213}+T_{132}\right)}{6} \tag{B-31} \end{equation} there corresponds a singlet, the baryon \begin{equation} \Lambda^{0}_{1}\sim\dfrac{\left(uds+dsu+sud\right)-\left(sdu+dus+usd\right)}{6} \tag{B-32} \end{equation} which normalized and after use of the symbol (B-30) is \begin{equation} \Lambda^{0}_{1}=\frac{1}{\sqrt{6}}\Bigl(\left[d,s\right]u+\left[s,u\right]d+\left[u,d\right]s\Bigr) \tag{B-33} \end{equation} (B) Decuplet $\:\boldsymbol{10}\:$ $\:\lbrace \Delta^{++},\Delta^{+},\Delta^{0},\Delta^{-},{\Sigma^{*}}^{^{\boldsymbol{+}}},{\Sigma^{*}}^{^{\boldsymbol{0}}},{\Sigma^{*}}^{^{\boldsymbol{-}}},{\Xi^{*}}^{^{\boldsymbol{0}}},{\Xi^{*}}^{^{\boldsymbol{-}}},\Omega^{-}\rbrace\:$ from $\:S_{ijk}.$ To the total symmetric 10-dimensional tensor $\:S_{ijk}\:$, see (B-15), there corresponds component by component the following decuplet \begin{equation} \begin{split} 1.\quad S_{111}: & \qquad \Delta^{++}=uuu \\ 2.\quad S_{112}: & \qquad \Delta^{+}=\frac{1}{\sqrt{3}}\left(uud+udu+duu\right)\\ 3.\quad S_{122}: & \qquad \Delta^{0}=\frac{1}{\sqrt{3}}\left(udd+dud+ddu\right)\\ 4.\quad S_{222}: & \qquad \Delta^{-}=ddd\\ 5.\quad S_{113}: & \qquad {\Sigma^{*}}^{^{\boldsymbol{+}}}=\frac{1}{\sqrt{3}}\left(uus+usu+suu\right)\\ 6.\quad S_{123}: & \qquad {\Sigma^{*}}^{^{\boldsymbol{0}}}=\frac{1}{\sqrt{6}}\left(uds+dsu+sud+sdu+dus+usd\right)\\ 7.\quad S_{223}: & \qquad {\Sigma^{*}}^{^{\boldsymbol{-}}}=\frac{1}{\sqrt{3}}\left(dds+dsd+sdd\right)\\ 8.\quad S_{133}: & \qquad {\Xi^{*}}^{^{\boldsymbol{0}}}=\frac{1}{\sqrt{3}}\left(uss+sus+ssu\right)\\ 9.\quad S_{233}: & \qquad {\Xi^{*}}^{^{\boldsymbol{-}}}=\frac{1}{\sqrt{3}}\left(ssd+sds+dss\right)\\ 10.\quad S_{333}: & \qquad \Omega^{-}=sss \end{split} \tag{B-34} \end{equation} (C) Octet $\:\boldsymbol{8}\:$ $\:\:\lbrace p,n,\Sigma^{+},\Sigma^{0},\Sigma^{-},\Lambda^{0},\Xi^{0},\Xi^{-}\rbrace\:$ from $\:Y_{ijk}.$ To the mixed anti-symmetric 8-dimensional tensor $\:Y_{ijk}\:$, see (B-25), there corresponds component by component the following octet \begin{equation} \begin{split} 1.\quad Y_{121}: & \qquad p=\frac{1}{\sqrt{2}}\left(udu-uud\right)\\ 2.\quad Y_{122}: & \qquad n=\frac{1}{\sqrt{2}}\left(udd-dud\right)\\ 3.\quad Y_{131}: & \qquad \Sigma^{+}=\frac{1}{\sqrt{2}}\left(usu-uus\right)\\ 4.\quad Y^{\prime}_{231}: & \qquad \Sigma^{0}=\dfrac{1}{2}\Bigl(\left[d,s\right]u+\left[u,s\right]d\Bigr)\\ 5.\quad Y_{232}: & \qquad \Sigma^{-}=\frac{1}{\sqrt{2}}\left(dsd-dds\right)\\ 6.\quad Y_{123}: & \qquad \Lambda^{0}=\frac{1}{\sqrt{12}}\left[ 2\left(uds-dus\right) -\left( dsu+sud-usd-sdu\right)\right]\\ & \qquad \quad =\frac{1}{\sqrt{12}}\Bigl(2\left[u,d\right]s-\left[d,s\right]u-\left[s,u\right]d\Bigr)\\ 7.\quad Y_{133}: & \qquad \Xi^{0}=\dfrac{1}{\sqrt{2}}\left(uss-sus\right)=\dfrac{1}{\sqrt{2}}\left[u,s\right]s\\ 8.\quad Y_{323}: & \qquad \Xi^{-}=\dfrac{1}{\sqrt{2}}\left(sds-ssd\right)\\ \end{split} \tag{B-35} \end{equation} Note that for the formation of the $\:\Sigma^{0}\:$ baryon as eighth independent component in place of $\: Y_{231}\:$ is used the linear combination \begin{equation} Y^{\prime}_{231} \equiv Y_{231}-Y_{312}=\dfrac{\left(T_{231}-T_{321}\right)+\left(T_{132}-T_{312}\right)}{2} \tag{B-36} \end{equation}

(D ) Octet $\:\boldsymbol{8'}\:$ from $\:X_{ijk}.$ To the mixed symmetric 8-dimensional tensor $\:X_{ijk}\:$, see (B-24), there corresponds component by component the following octet \begin{equation} \begin{split} 1.\quad X_{121}: & \qquad \frac{1}{\sqrt{6}}\left(udu+duu-2uud\right)\\ 2.\quad X_{131}: & \qquad \frac{1}{\sqrt{6}}\left(usu+suu-2uus\right)\\ 3.\quad X_{122}: & \qquad \frac{1}{\sqrt{6}}\left(udd+dud-2ddu\right)\\ 4.\quad X_{123}: & \qquad \frac{1}{\sqrt{12}}\left(2uds+2dus-usd-sud-dsu-sdu\right)\\ 5.\quad X'_{231}: & \qquad \frac{1}{\sqrt{2}}\left(sdu+dsu-sud-usd\right)\\ 6.\quad X_{133}: & \qquad \frac{1}{\sqrt{6}}\left(uss+sus-2ssu\right)\\ 7.\quad X_{232}: & \qquad \frac{1}{\sqrt{6}}\left(dsd+sdd-2dds\right)\\ 8.\quad X_{233}: & \qquad \frac{1}{\sqrt{6}}\left(dss+sds-2ssd\right)\\ \end{split} \tag{B-37} \end{equation} Note that for the formation of the member 5. baryon as eighth independent component in place of $\: X_{231}\:$ is used the linear combination \begin{equation} X^{\prime}_{231} \equiv X_{231}-X_{312}=\dfrac{\left(T_{321}-T_{312}\right)+\left(T_{231}-T_{132}\right)}{2} \tag{B-38} \end{equation}


[Note A] : The transformation $\:\left(u_{\rho i}u_{\sigma j}u_{\tau k}\right)\:$ in (B-10) is the product \begin{equation} u_{\rho i}u_{\sigma j}u_{\tau k} \longrightarrow U\!\boldsymbol{\otimes}\!U\!\boldsymbol{\otimes}\!U=U^{\boldsymbol{\otimes}3} \tag{N-01} \end{equation} [Note B] : Let a set of complex numbers be represented by a mathematical quantity $T_{i_{1}i_{2}\cdots i_{p-1}i_{p}}$ with $p$ indices. These indices take values in the set $\left\lbrace 1,2,3, \cdots ,d\!-\!1,d \right\rbrace$. Here $\:p\:$ and $\:d\:$ are positive integers, so
\begin{equation} \begin{split} T_{i_{1}i_{2}\cdots i_{p-1}i_{p}} \in \mathbb{C}\;, \qquad & i_{k}\in \left\lbrace 1,2,3, \cdots ,d-1,d \right\rbrace\\ & p, d \in \mathbb{N}=\left\lbrace 1,2,\cdots\right\rbrace \end{split} \tag{N-02} \end{equation} The positive integer $\:d\:$ usually represents the dimension of a linear space.
The general tensor $\:T_{i_{1}i_{2}\cdots i_{p-1}i_{p}}\:$ defined by equation (N-02) has $\:d^{p}\:$ linearly independent elements classified in $\:N\!\left(p,d\right)\:$ classes. Each class has a code or name $\:\left[ x_{1}x_{2}\cdots x_{d}\right]\:$, where in this class $x_{1}$ indices take the value $1$, $x_{2}$ indices take the value $2$ and so on $x_{d}$ indices take the value $d$. So, the number of the classes $\:N\!\left(p,d\right)\:$ is the number of the ordered solutions $\:\mathbf{x}=\left(x_{1}, x_{2},\cdots x_{d}\right)\:$ of the equation
\begin{equation} x_{1}+x_{2}+\cdots + x_{d}=p \tag{N-03} \end{equation} where all $x_{k}$ are nonnegative integers. From combinatorics this number is \begin{equation} N\!\left(p,d\right)=\binom{p\!+\!d\!-\!1}{d\!-\!1} \tag{N-04} \end{equation} The number of elements in class $\:\left[ x_{1}x_{2}\cdots x_{d}\right]\:$ is \begin{equation} \text{number of elements in class }\:\left[ x_{1}x_{2}\cdots x_{d}\right] =\frac{p!}{x_{1}!x_{2}!\cdots x_{d}!} \tag{N-05} \end{equation} so \begin{equation} \sum_{x_{1}\!+\!x_{2}\cdots \!+\!x_{d}=p}\frac{p!}{x_{1}!x_{2}!\cdots x_{d}!}=d^{p} \tag{N-06} \end{equation} If $\:T_{i_{1}i_{2}\cdots i_{p-1}i_{p}}\:$ is symmetric with respect to any pair of indices, called totally symmetric, then taking one and only one element of each class we form the set of its $\:N\!\left(p,d\right)\:$ linearly independent elements. For example, if $\:p=3\:$ and $\:d=3\:$ then \begin{equation} N\!\left(p,d\right)=N\!\left(3,3\right)=\binom{3\!+\!3\!-\!1}{3\!-\!1}=\binom{5}{2}= 10 \tag{N-07} \end{equation} In the Figure below the 27 components of a tensor $\mathrm{a}_{ijk}$ are arranged on a cube. If this tensor is totally symmetric then its linear independent components are 10. A choice of such a $^{\prime\prime}$decuplet$^{\prime\prime}$ of elements is the 10 elements on the pyramid shown (green balls). This pyramid is formed from two diagonal planes : a diagonal plane which separates the elements of the cube because of the symmetry with respect to the pair of indices $\left(k,i\right)$ and a diagonal plane which separates the elements of the cube because of the symmetry with respect to the pair of indices $\left(j,k\right)$. Then automatically there exists symmetry with respect to the third pair of indices $\left(i,j\right)$.
enter image description here See a 3D version of above Figure here : Totally Symmetric Matrix

[Note C] : In the Figure below a totally antisymmetric tensor $\mathrm{b}_{ijk}$ is shown. The tensor is 1-dimensional.

enter image description here

See a 3D version of above Figure here : Totally Antisymmetric Matrix


Usually this is done because the various pieces in the decomposition transform "nicely". For instance, the pieces that are symmetric will usually only transform amongst themselves, and likewise for the antisymmetric pieces. This decomposition thus often facilitates bookkeeping in calculations.

The process is similar in spirit to decomposing many-particle spin 1/2 states into states of definite total net spin: in this case states of a given net total spin $S$ transform amongst themselves.

A practical byproduct of such decompositions is that some terms may cancel out "by symmetry", just like - say - some integrals are obviously $0$ since the integrand is odd but the integration interval is symmetric.

As another application, selection rules might also allow or disallow some processes to occur based only on symmetry arguments, in which case only the properly symmetrized pieces remain.


OP is basically asking:

Why do we decompose (reducible) group representations in irreducible group representations?

Partial answer:

  1. To classify the (reducible) representation.

  2. Irreducible representations can not be further truncated without destroying the group symmetry.

  3. Because certain irreducible sub-representations of the given (reducible) representation may be forbidden by e.g. selection rules, other physical principles, etc, and this is always useful information.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.