= Solution
A fundamental matrix realization compatible with the <root> normalization is
$$
H_1=\frac12\operatorname{diag}(1,-1,0),\qquad
H_2=\frac1{2\sqrt3}\operatorname{diag}(1,1,-2).
$$
The <weights of a representation> on the coordinate states $u,d,s$ are therefore
$$
\boxed{u=\left(\frac12,\frac1{2\sqrt3}\right),\qquad
d=\left(-\frac12,\frac1{2\sqrt3}\right),\qquad
s=\left(0,-\frac1{\sqrt3}\right).}
$$
Each has <weight multiplicity> one. The <weight diagram of the defining SU(3) representation> is the corresponding equilateral triangle. These are $(H_1,H_2)$ coordinates; the familiar vertical <flavor hypercharge> coordinate is $Y=2H_2/\sqrt3$.
For the <adjoint representation of SU(3)>, the states are elements of the <Lie algebra>. The two Cartan generators commute with both $H_i$, giving <representation weight> $(0,0)$ with multiplicity two. Each <root> generator $E_\pm^r$ is an eigenvector of the Cartan adjoint action with <representation weight> $\pm\alpha_r$. Consequently
$$
\boxed{\text{adjoint weights: }\,
\pm(1,0),\quad\pm(-1/2,\sqrt3/2),\quad
\pm(1/2,\sqrt3/2),\quad (0,0)\text{ twice}.}
$$
The six nonzero <representation weights> form a regular hexagon, with a doubly occupied centre. Its <highest weight> is $\alpha_3=\omega_1+\omega_2$, so its <Dynkin labels> are $(1,1)$ and its <dimension> is eight. The triplet and octet panels in the preceding figure give these two <weight diagrams> explicitly.
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