# Weights and roots of $SU(3)$

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 $$$$T_3 = \begin{pmatrix} \frac{1}{2} & 0 & 0\\ 0 & -\frac{1}{2} & 0\\ 0 & 0 & 0 \end{pmatrix} \hspace{1cm} T_{8} = \frac{\sqrt{3}}{6} \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -2 \end{pmatrix}.$$$$ 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.

• I think you have a typo in the normalization of $T_8$
– lcv
Commented Jan 27 at 11:42
• @lcv thanks, edited. Commented Jan 27 at 11:59
• WP might illuminate the picture for you. Presumably you have mastered this. Commented Jan 27 at 17:03
• One should never use Georgi to learn this topic. See this post for alternatives. Commented Jan 28 at 4:31
• @ZeroTheHero thanks. I feel like the book is so condensed that it serves better as a refresher. Commented Jan 28 at 5:48

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.