Verify (2.16). To do this, it is useful to form the so-called "step-up" and "step-down" operators
$$
J_{\pm}=J_{1} \pm i J_{2}
$$
First, show that
$$
J_{\pm}|j m\rangle=(C-m(m \pm 1))^{1 / 2}|j, m \pm 1\rangle
$$
that is, $J_{\pm}$step $m$ up and down by one unit, respectively. Show that $C=j(j+1)$
A state $|j m\rangle$ is transformed under a rotation through an angle $\theta$ about the 2-axis into a linear combination of the $2 j+1$ states $\left|j m^{\prime}\right\rangle$, with $m^{\prime}=-j,-j+$ $1, \ldots, j$
$$
e^{-i \theta J_{2}}|j m\rangle=\sum_{m^{\prime}} d_{m^{\prime} m}^{j}(\theta)\left|j m^{\prime}\right\rangle
$$
where the coefficients $d_{m^{\prime} m}^{\prime}$ are written in conventional notation and are frequently called rotation matrices. From $(2.19)$, we see the states having the same $j$ but all possible $m$ values transform among themselves under rotations. In fact, all the $2 j+1$ states are mixed by rotations. They form the basis of a $(2 j+1)=$ dimensional irreducible representation of the rotation group. The set of states is called a multiplet.