Weights and roots of $SU(3)$

Question

I am self-studying group theory from Lie Algebras in Particle Physics by H. Georgi and I am having trouble following some of his arguments. In section 7.2 titled Weights and roots of $SU(3)$ he starts out by finding the weights of the generators \begin{equation} T_3 = \begin{pmatrix} \frac{1}{2} & 0 & 0\\ 0 & -\frac{1}{2} & 0\\ 0 & 0 & 0 \end{pmatrix} \hspace{1cm} T_{8} = \frac{\sqrt{3}}{6} \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -2 \end{pmatrix}. \end{equation} The eigenvectors and the associated weights are given to be \begin{align} \begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix} & \longrightarrow (1/2,\sqrt{3}/6)\\ \begin{pmatrix} 0\\1\\0 \end{pmatrix} &\longrightarrow (-1/2,\sqrt{3}/6)\\ \begin{pmatrix} 0\\0\\1 \end{pmatrix} &\longrightarrow (0,-\sqrt{3}/3) \end{align} So far so good. What I don't understand is how Georgi finds the roots. He says that "The roots are going to be differences of weights, because the corresponding generators take us from one weight to another." I do not understand this statement, can someone please elaborate?

Next he says "It is not hard to see that the corresponding generators are those that have only one off-diagonal entry" \begin{align} \frac{1}{\sqrt{2}}(T_1\pm i T_2) &= E_{\pm 1,0}\\ \frac{1}{\sqrt{2}}(T_4\pm i T_5) &= E_{\pm 1/2,\pm\sqrt{3}/2}\\ \frac{1}{\sqrt{2}}(T_6\pm i T_7) &= E_{\mp 1/2,\pm\sqrt{3}/2} \end{align} I thought the generators were the $T_i$'s. Can someone please explain? I find the book to be a bit terse sometimes and lacking elaborate explanations.

Answer 1

By definition, the non-zero roots are the ladder operators of the algebra. The zero roots (at the center of the root diagram) are the diagonal operators.

Now, whether you use $T_1$ or $T_1\pm i T_2$ depends on how you define $T_1$ and $T_2$. In this case, it seems they are defined as hermitian so, like $L_x$ and $L_y$, it is the combinations $L_\pm = L_x\pm i L_y$ that are the ladder operators. Their matrix representation (as $2\times 2$ matrices acting on a 2d Hilbert space) contains zeroes everywhere except for a single $1$ somewhere off-diagonal.

In general, the ladder operators satisfy $[h_i,T_\alpha]=\alpha(i) T_\alpha$ where $\alpha$ is a root and $\alpha(i)$ is a component of the root vector. This simply generalizes $[L_z,L_\pm]=\pm L_\pm$. The normalization of the diagonal operators is usually chosen so the roots span a regular lattice in $\mathbb{R}^r$, where $r$ is the rank of the algebra. Moreover, the weights of a representation, when placed on the lattice, all differ by an integer linear combination of roots.

Thus, just like $L_\pm$ move you between $su(2)$ states $\vert \ell,m\rangle$ of different weights, the non-zero roots will move you between states of different weights. It is almost tautological to see that applying various roots to a state of a given weight will produce a state with a weight that is exactly the linear combination related to the application of the roots: in other words, if $$ E_\alpha E_\beta \vert \boldsymbol{m}\rangle $$ then the resulting state will have weight $\boldsymbol{m}+\alpha+\beta$.

It must follow that $E_\beta\vert\boldsymbol{m}\rangle$ has weight $\boldsymbol{m}+\beta$ because \begin{align} h_iE_\beta\vert\boldsymbol{m}\rangle&= [h_i,E_\beta]\vert\boldsymbol{m}\rangle + E_\beta\,h_i\vert \boldsymbol{m}\rangle\\ &=\beta(i)E_\beta\vert\boldsymbol{m}\rangle+ m_i E_\beta\vert\boldsymbol{m}\rangle\, ,\\ &= (\beta(i)+m_i)\vert\boldsymbol{m}\rangle \end{align} or, in other words, $E_\beta\vert\boldsymbol{m}\rangle$ has weight component $m_i+\beta_i$ as it is an eigenstate of $h_i$ with this eigenvalue.