Skip to main content

Section 4.3 \(\bmit{\su(3)}\)

The unitary group \(\SU(3)\) consists of all \(3\times3\) unitary matrices with determinant \(1\text{,}\) that is

\begin{equation} \SU(3) = \{M\in\CC^{3\times3}:M^\dagger M=\one, |M|=1\} .\tag{4.3.1} \end{equation}

The group \(\SU(3)\) is the smallest of the unitary groups to be unrelated to the orthogonal groups; it's something new. As is the case for \(\su(2)\text{,}\) the Lie algebra \(\su(3)\) consists of all \(3\times3\) tracefree, anti-Hermitian matrices, that is

\begin{equation} \su(3) = \{A\in\CC^{3\times3}:A^\dagger+A=0, \tr(A)=0\} .\tag{4.3.2} \end{equation}

The standard basis for the Lie algebra \(\su(3)\otimes\CC\) consists of the Gell-Mann matrices  1 

\begin{equation} \begin{aligned} \lambda_1 \amp= \begin{pmatrix} 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \end{pmatrix} ,\qquad \lambda_2 = \begin{pmatrix} 0 \amp -i \amp 0 \\ i \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \end{pmatrix} ,\nonumber\\ \lambda_3 \amp= \begin{pmatrix} 1 \amp 0 \amp 0 \\ 0 \amp -1 \amp 0 \\ 0 \amp 0 \amp 0 \end{pmatrix} ,\qquad \lambda_4 = \begin{pmatrix} 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \\ 1 \amp 0 \amp 0 \end{pmatrix} ,\nonumber\\ \lambda_5 \amp= \begin{pmatrix} 0 \amp 0 \amp i \\ 0 \amp 0 \amp 0 \\ -i \amp 0 \amp 0 \end{pmatrix} ,\qquad \lambda_6 = \begin{pmatrix} 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \end{pmatrix} ,\nonumber\\ \lambda_7 \amp= \begin{pmatrix} 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp -i \\ 0 \amp i \amp 0 \end{pmatrix} ,\qquad \lambda_8 = \frac1{\sqrt{3}}\begin{pmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -2 \end{pmatrix} .\end{aligned}\tag{4.3.3} \end{equation}

As with the Pauli matrices, the Gell-Mann matrices are Hermitian; unlike the Pauli matrices, they do not square to \(\pm1\text{.}\) However, they are again orthonormal with respect to the Killing form (with overall normalization \(2\)). We can obtain an anti-Hermitian basis of \(\su(3)\) itself (that is, not complexified) by using the matrices

\begin{equation} \mu_m = -i \lambda_m\tag{4.3.4} \end{equation}

Alternatively, we can work with the real subset of these matrices, and study the real form \(\sl(3,\RR)\) of \(\su(3)\text{,}\) that is

\begin{equation} \sl(3,\RR) = \langle \lambda_1, \mu_2, \lambda_3, \lambda_4, \mu_5, \lambda_6, \mu_7, \lambda_8 \rangle\tag{4.3.5} \end{equation}

where \(\langle...\rangle\) denotes the span of the given elements, that is, the set of all linear combinations of these elements. By inspection, \(\sl(3,\RR)\) contains 5 boosts and 3 rotations.

The given form of these matrices makes clear that \(\lambda_3\) and \(\lambda_8\) commute with each other, and that no larger set of basis elements will do so. The advantage of working with \(\sl(3,\RR)\) is that these real symmetric matrices have real eigenvalues; at the Lie algebra level, their real eigenvectors will lie inside the algebra, without the need for complexification.

Explicitly, we have the commutation relations

\begin{equation} \begin{aligned} \lbrack\lambda_3,\lambda_8\rbrack \amp= 0 ,\nonumber\\ \left\lbrack\lambda_3,\frac12(\lambda_1\mp\mu_2)\right\rbrack \amp= \pm\frac22(\lambda_1\mp\mu_2) ,\quad \left\lbrack\lambda_8,\frac12(\lambda_1\mp\mu_2)\right\rbrack = 0 ,\nonumber\\ \left\lbrack\lambda_3,\frac12(\lambda_4\mp\mu_5)\right\rbrack \amp= \mp\frac12(\lambda_4\mp\mu_5) ,\quad \left\lbrack\lambda_8,\frac12(\lambda_4\mp\mu_5)\right\rbrack = \mp\frac{\sqrt{3}}2(\lambda_4\mp\mu_5) ,\nonumber\\ \left\lbrack\lambda_3,\frac12(\lambda_6\mp\mu_7)\right\rbrack \amp= \mp\frac12(\lambda_6\mp\mu_7) ,\quad \left\lbrack\lambda_8,\frac12(\lambda_6\mp\mu_7)\right\rbrack = \pm\frac{\sqrt{3}}{2}(\lambda_6\mp\mu_7) .\end{aligned}\tag{4.3.6} \end{equation}
Figure 4.3.1. The root diagram of \(\su(3)\text{.}\)

Regarding the eigenvalues as vectors in \(\RR^2\text{,}\) we can identify our basis elements with their eigenvalues, as follows:

\begin{equation} \begin{aligned} \frac12(\lambda_1\mp\mu_2) \amp\longleftrightarrow (\pm2,0) \nonumber ,\\ \frac12(\lambda_4\mp\mu_6) \amp\longleftrightarrow (\mp1,\mp3) \nonumber ,\\ \frac12(\lambda_6\mp\mu_7) \amp\longleftrightarrow (\mp1,\pm3) ,\end{aligned}\tag{4.3.7} \end{equation}

and both \(\lambda_3\) and \(\lambda_8\) correspond to \((0,0)\text{.}\) As with \(\su(2)\text{,}\) we recover almost all of the structure of the Lie algebra by plotting these points. The result is shown in Figure 4.3.1, and is called the root diagram of \(\su(3)\). Each family of parallel lines represents the action of one of the three pairs of eigenvectors on the other eigenvectors; again, the eigenvectors can be thought of as raising and lowering operators. It is a useful exercise to work out all the commutators, and to compare the result with the root diagram.

Our definition of \(\lambda_5\) differs by an overall minus sign from the standard definition, in order to correct a minor but annoying lack of cyclic symmetry in the original definition.