for general angular momentum state, it is a “ugly”combination of the orbital state and the spin state:

$\left|l,m_l,s,m_s \right> = \left|l,m_l\right>\bigotimes \left|s,m_s \right>$

the above combination only possible when:

$[L_i,S_j]=0$

$L = ( L_x,L_y,L_z)$

$L^2 \left|l,m_l \right> = l(l+1) \left |s,m_s \right>$

$L_z \left|l,m_l \right> = m_l \left |s,m_s \right>$

$[L_x,L_y]= i L_z$

$L_{\pm} \left|l,m_l\right> = \sqrt{ l(l+1) - m(m\pm1)} \left|l,m_l \right>$

the above are basic properties shared by spin angular momentum operator.

$[L_z, L_{\pm}]=\pm L_{\pm}$

$[L^2,L_{\pm}]=0$

and the total angular momentum operator is

$J= L\bigotimes 1 + 1 \bigotimes S$

and the state is

$\left| j, m_j \right> = \left| l, m_l, s, m_s \right>$

$J_z \left| j, m_j \right> = m_j \left| j, m_j \right>$

by $J_z= L_z\bigotimes 1 + 1 \bigotimes S_z$

$m_j =m_l+m_s$

by same on x component and y component:

$J_{\pm} = L_{\pm} + S_{\pm}$

and by $[L_i,S_j]=0$

$[J_x,J_y ] = i J_z$

$J^2 = L^2 + S^2 + 2 L \cdot S$

To find the possible $j$, we can use the highest state withe $J_{\pm}$ and $J_z$. Since the highest state are always symmetric. we have to find the others orthogonal states, which are anti-symmetric.

the result is

$\left| l-s \right| \leq j \leq \left| l + s \right|$