# The average value of S operator

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• Viona
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.
Viona
TL;DR Summary
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.

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?

Viona
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.

Viona
Thank you all for useful replies.

## 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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