We know that nuclear magnetic moment can be expressed in terms of the expected value for nuclear spin as:
$$\langle\mu\rangle =[g_lj+(g_s-g_l)\langle s_z\rangle]\frac{\mu_N}{\hbar}$$
(Cf. Krane), where $\vec{j}$ is the total angular momentum, $\vec{l}+\vec{s}$.
How does the expected $\langle s_z\rangle$ value relate to the $\vec{j}$-component of spin, $\langle s_j\rangle$? Krane mentions that only that value is needed, given that it remains constant.
Best Answer
From the magnetic moment $$\mathbf{\mu}=\mathbf{\mu_L}+ \mathbf{\mu_S}=(g_l \mathbf{L}+g_s\mathbf{S})\frac{\mu_N}{\hbar}\tag1$$ take the scalar product with $\bf{J}$ $$\mathbf{\mu_J} ·\mathbf{J}=(1/2(g_l +g_s)\mathbf J^2+1/2(g_l -g_s)(\mathbf L^2-\mathbf S^2))\frac{\mu_N}{\hbar} \tag2$$ so with commutator relation $$\mu=(1/2(g_l +g_s)j+1/2(g_l -g_s)\frac{(l-s)(l+s+1)}{j+1}) \mu_N \tag3$$
since $s=1/2$ and $j= l\pm1/2$ you end with two possible values for $\mu$ $$\mu=(jg_l-1/2(g_l -g_s)) \mu_N \quad\text{for }j= l+1/2\\ \mu=(jg_l+(g_l -g_s)\frac{j}{2j+1}) \mu_N \quad\text{for }j= l-1/2 \tag4$$
this is your $⟨s⟩$ projection in the $j$ direction.
For $s\neq 1/2$ equation, $(3)$ still holds. But for $s>l$ the factor must be changed to the general form $$\frac{l(l+1)- s (s+1)}{j+1}.$$