The average value of S operator

In summary, while discussing the fine-structure of hydrogen and spin-orbit coupling, the author explains that the average value of the S operator is the projection of S onto J. This is based on the fine-structure Hamiltonian and the commutator [H, S]. Using the Ehrenfest theorem, it can be shown that the time-averaged value of the perpendicular component of S is zero, indicating that it precesses like a spinning top.
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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?
 
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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.
 
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Thank you all for useful replies.
 

Related to The average value of S operator

1. What is the S operator?

The S operator, also known as the spin operator, is a mathematical operator used in quantum mechanics to describe the spin of a particle. It is represented by the symbol S and has a magnitude of ½.

2. How is the average value of S operator calculated?

The average value of the S operator is calculated by taking the sum of all possible spin values and their corresponding probabilities, weighted by the probability of each spin state occurring.

3. What is the physical significance of the average value of S operator?

The average value of the S operator represents the expected value of the spin of a particle in a given state. It provides information about the orientation and magnitude of the spin of a particle.

4. Can the average value of S operator be negative?

Yes, the average value of the S operator can be negative. This indicates that the spin of the particle is in a negative direction, relative to the chosen axis of measurement.

5. How is the average value of S operator related to the uncertainty principle?

The average value of the S operator is related to the uncertainty principle in that it is impossible to know the exact spin of a particle at any given moment. The uncertainty principle states that the more precisely one property of a particle is known, the less precisely the other property can be known. In the case of the S operator, this means that the more precisely the spin of a particle is known, the less precisely its position can be known, and vice versa.

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