# Time average value of Spin operator

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

$$S_{avg}=\frac{(S.J)J}{J^2}$$

How to prove this?

kuruman
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Griffiths gives the standard argument in the vector model for the atom that when ##S## precesses rapidly about ##J##, the tranverse components time-average to zero and the operator can be replaced with a time-averaged operator which is the projection of ##S## on ##J##. Now if you have two regular old vectors, ##A## and ##B## with angle ##\theta## between them, you would write the projection of ##A## on ##B## as $$A_B=A\cos\theta=\frac{(\vec A \cdot \vec B)}{AB}A=\frac{(\vec A \cdot \vec B)}{B^2}B.$$

Griffiths gives the standard argument in the vector model for the atom that when ##S## precesses rapidly about ##J##, the tranverse components time-average to zero and the operator can be replaced with a time-averaged operator which is the projection of ##S## on ##J##. Now if you have two regular old vectors, ##A## and ##B## with angle ##\theta## between them, you would write the projection of ##A## on ##B## as $$A_B=A\cos\theta=\frac{(\vec A \cdot \vec B)}{AB}A=\frac{(\vec A \cdot \vec B)}{B^2}B.$$
Im satisfied with the Griffith's explanation for the above expression, but out of curiosity I am looking for the mathematical proof of the same expression. While searching internet about this question, I saw "Wigner Eckart Theorem" could be used to find this expectation value, but I don't know how? Any idea how to do that?

kuruman
I believe that ##\vec S_{avg}## is an operator, not an expectation value. If by "mathematical proof" you mean "Starting with an expression for the time-averaged spin operator, use the Wigner-Eckart theorem to show that $$\vec{S}_{avg}=\frac{(\vec S \cdot \vec J)\vec J}{J^2}$$ in the weak field approximation", the answer is "no I don't have an idea how to do that."
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