Time average value of Spin operator

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SUMMARY

The time average value of the spin operator, denoted as S, is defined as the projection of S onto J when calculating the expectation value of J + S, as outlined in Griffiths' "Introduction to Quantum Mechanics." The formula for the average value is given by Savg = (S·J)J/J2. The discussion highlights the rapid precession of S about J, which causes the transverse components to average to zero, allowing for the replacement with a time-averaged operator. While the Wigner-Eckart theorem is mentioned as a potential method for deriving this expression, it is not necessary to find the expectation value, as Griffiths' equations 6-73 to 6.75 provide a sufficient approach.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly angular momentum.
  • Familiarity with Griffiths' "Introduction to Quantum Mechanics" and its equations.
  • Knowledge of vector projections and their mathematical representations.
  • Basic understanding of the Wigner-Eckart theorem in quantum mechanics.
NEXT STEPS
  • Study Griffiths' equations 6-73 to 6-75 for a detailed understanding of the time-averaged spin operator.
  • Research the Wigner-Eckart theorem and its applications in quantum mechanics.
  • Explore vector projections in quantum mechanics to solidify understanding of angular momentum projections.
  • Investigate the weak field Zeeman effect and its implications in quantum mechanics.
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Students and researchers in quantum mechanics, particularly those focusing on angular momentum, spin operators, and the mathematical foundations of quantum theory.

Muthumanimaran
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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?
 
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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 $$A_B=A\cos\theta=\frac{(\vec A \cdot \vec B)}{AB}A=\frac{(\vec A \cdot \vec B)}{B^2}B.$$
 
kuruman said:
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?
 
Muthumanimaran said:
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?
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."
However, you don't need the Wigner-Eckart theorem to find the expectation value ##<\vec S_{avg}>.~## Just follow Griffiths, equations 6-73 to 6.75.
 
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