B The average value of S operator

Viona
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Why the average value of S operator is considered to be the projection of S onto J ?
While reading in the book of Introduction to Quantum Mechanics by David Griffith in the section of Fine structure of Hydrogen: spin- orbit coupling, he said that the average value of S operator is considered to be the projection of S onto J. I could not understand why he assumed that. please help me to understand.
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Given the fine-structure Hamiltonian ##H = A\vec{L} \cdot \vec{S}##, try to calculate the commutator ##[H,\vec{S}]##. What do you get, in vector form?
 
What I was getting at is that if you do the algebra, I get $$[H,\vec{S}] = iA\vec{S} \times \vec{L} = iA\vec{S} \times \vec{J}$$ where the last step is true because ##\vec{J} = \vec{L} + \vec{S}##. Use the Ehrenfest theorem on this to get $$\frac{d\langle \vec{S} \rangle}{dt} = \frac{A}{\hbar} \langle \vec{S} \times \vec{J} \rangle $$

##\vec{S}## can be split into two components: ##S_{||}## which is parallel to ##\vec{J}## and ##\vec{S}_{\perp}## which is perpendicular to ##\vec{J}##. Since the cross product cancels the parallel part, only the perpendicular part has time dependence (it precesses). The equation for ##\langle \vec{S}_{\perp} \rangle## is the equation for a spinning top (precession). The time averaged value of ##\vec{S}_{\perp}## is zero.
 
Thank you all for useful replies.
 
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