B The average value of S operator

Viona
Messages
49
Reaction score
12
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.
Untitled.png
 
Physics news on Phys.org
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.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
According to recent podcast between Jacob Barandes and Sean Carroll, Barandes claims that putting a sensitive qubit near one of the slits of a double slit interference experiment is sufficient to break the interference pattern. Here are his words from the official transcript: Is that true? Caveats I see: The qubit is a quantum object, so if the particle was in a superposition of up and down, the qubit can be in a superposition too. Measuring the qubit in an orthogonal direction might...
Back
Top