Expectation value of total angular momentum $\langle J \rangle$ [I am working with Griffiths Introduction to Quantum Mechanics, 3rd Edition.  My problem is general but if you want to look I am reading from ch 4.1 in which the weak-field Zeeman Effect is being calculated when I got stuck.]
We want to calculate $E _{z} = e/2m * \vec B_{ext} \cdot < \vec J + \vec S>$
we work it out so that all we need to find is $<\vec J>$.
I know that  $\vec J = \vec L +\vec S$, and thus $ |\vec J|^2= |\vec L|^2+|\vec S|^2+\vec L \cdot \vec S$  where $|\vec L|^2 = \hbar l(l+1)$ and $|\vec s|^2 = \hbar s(s+1)$ in the eigenstates of the Hydrogen atom but griffiths does not appear to use any of these facts and (after stating that the $z$-axis will ie along $\vec B _{ext}$ states 
$ \vec B \cdot<\vec J> = \hbar m  _{j}$ 
Maybe I'm just confused about what J is, but how do we goes from one to the other.  
 A: I'm not quite sure what your specific question is, so I'll try to better explain what Griffiths is doing in his book.
In first-order perturbation theory, the Zeeman correction to the energy is:
$$
\begin{align*}
E_{Z}^{1} & = \langle n \; l \; j \; m_{j} \vert H_{Z}^{\prime} \vert n \; l \; j \; m_{j} \rangle \\
& = \langle n \; l \; j \; m_{j} \vert \frac{e}{2m}\left( L + 2S \right) \cdot B_{\text{ext}} \vert n \; l \; j \; m_{j} \rangle \\
& = \frac{e}{2m}B_{\text{ext}} \cdot \langle n \; l \; j \; m_{j} \vert \left(L + 2S \right) \vert n \; l \; j \; m_{j} \rangle \\
& = \frac{e}{2m}B_{\text{ext}} \cdot \langle L + 2S \rangle
\end{align*}
$$
But since $J = L + S$, then $L + 2S$ can be written as $L = J + S$. Since the total angular momentum, $J$, is constant and $L$ and $S$ precess around $J$, we can work out the time average value of $S$ by calculating its projection on $J$:
$$
S_{\text{ave}} = \frac{\left( S \cdot J \right)}{J^{2}}J
$$
So now we need to find out what $S \cdot J$ is, which is not immediately obvious. But consider the following:
$$
\begin{align*}
L^{2} & = \left( J - S \right) \left( J - S \right) = J \cdot J - 2J \cdot S + S \cdot S \\
& = J^{2} + S^{2} - 2J \cdot S
\end{align*}
$$
And so if we re-arrange this, we obtain an expression for $S \cdot J$:
$$
S \cdot J = \frac{1}{2}\left(J^{2} + S^{2} - L^{2} \right)
$$
But we know that $J^{2} = j\left(j + 1\right)h^{2}$, and similarly with $S^{2}$ and $L^{2}$; so our expression becomes:
$$
\begin{align*}
S \cdot J & = \frac{1}{2}\left[j\left(j + 1\right)\hbar^{2} + s\left(s + 1\right)\hbar^{2} - l\left(l + 1\right)\hbar^{2} \right] \\
& = \frac{\hbar^{2}}{2}\left[j\left(j + 1\right) + s\left(s + 1\right) - l\left(l + 1\right) \right]
\end{align*}
$$
And so it follows that:
$$
\begin{align*}
\langle L + 2S \rangle & = \langle J + S \rangle \\
& = \langle \left( 1 + \frac{S \cdot J}{J^{2}} \right)J \rangle \\
& = \left[ 1 + \frac{\frac{\hbar^{2}}{2}\left[j\left(j + 1 \right) + s\left(s + 1\right) - l\left(l + 1\right) \right]}{j\left(j + 1\right)\hbar^{2}}\right]\langle J \rangle \\
& = \left[ 1 + \frac{\left[j\left(j + 1 \right) + s\left(s + 1\right) - l\left(l + 1\right) \right]}{2j\left(j + 1\right)}\right]\langle J \rangle \\
& = g_{J}\langle J \rangle
\end{align*}
$$
where $g_{J}$ is the Landé g-factor.
Recall our expression for the first-order correction to the energy:
$$
E_{Z}^{1} = \frac{e}{2m}B_{\text{ext}}\cdot \langle L + 2S \rangle
$$
We just showed that $\langle L + 2S \rangle = g_{J} \langle J \rangle$, so we have:
$$
E_{Z}^{1} = \frac{e}{2m}B_{\text{ext}} \cdot g_{J} \langle J \rangle
$$
At this point, we can choose the z-axis to lie along the direction of $B_{\text{ext}}$. In this case, $B_{\text{ext}} \cdot \langle J \rangle = B_{\text{ext}} \langle J_{z} \rangle$. Of course, the expectation value $\langle J_{z} \rangle = \hbar m_{j}$, and so we have:
$$
\begin{align*}
E_{Z}^{1} & = \frac{\hbar e}{2m}B_{\text{ext}}g_{J}m_{j} \\
& = \mu_{B} B_{\text{ext}}g_{J}m_{j}
\end{align*}
$$
where $\mu_{B}$ is the Bohr magneton.
