Time average value of Spin operator

Question

From the book Introduction to Quantum Mechanics by Griffiths,. In the section 6.4.1 (weak field zeeman effect) Griffiths tells that the time average value of $S$ operator is just the projection of $S$ onto $J$ while finding the expectation value of $J+S$

$$\vec{S}_{avg}=\frac{(\vec{S}.\vec{J})\vec{J}}{J^2}$$ How to prove this?

Answer 1

You can use the Wigner-Eckart corollary where the operator $\vec{S}$ is used:

$\langle JM_J|\vec{S}|JM_J \rangle = \langle J||\vec{S}||J| \rangle \langle JM_J|\vec{J}|JM_J \rangle ;$

$\langle JM_J|\vec{J}\cdot\vec{S}|JM_J \rangle =\langle J||\vec{S}||J \rangle \langle JM_J|\vec{J}^2|JM_J \rangle;$

From this one,

$\langle J||\vec{S}||J\rangle=\frac{ \langle JM_J|\vec{J}\cdot\vec{S}|JM_J\rangle}{\langle JM_J|\vec{J}^2|JM_J\rangle}$

Now, by inserting it in the first equation,

$\langle JM_J|\vec{S}|JM_J \rangle = \frac{ \langle JM_J|\vec{J}\cdot\vec{S}|JM_J \rangle}{\langle JM_J|\vec{J}^2|JM_J \rangle} \langle JM_J|\vec{J}|JM_J \rangle $

Generally we have

$ \vec{S}_{average} = \frac{\vec{S}\cdot\vec{J}}{\vec{J}^2} \vec{J}$