I Angular momentum and rotations

Kashmir
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Cohen tannoudji. Vol 1.pg 702"Now, let us consider an infinitesimal rotation ##\mathscr{R}_{\mathbf{e}_z}(\mathrm{~d} \alpha)## about the ##O z## axis. Since the group law is conserved for infinitesimal rotations, the operator ##R_{\mathbf{e}_z}(\mathrm{~d} \alpha)## is necessarily of the form: $$ R_{\mathbf{e}_z}(\mathrm{~d} \alpha)=1-\frac{i}{\hbar} \mathrm{d} \alpha J_z $$ where ##J_z## is a Hermitian operator since ##R_{\mathbf{e}_z}\left(\mathrm{~d} \alpha\right.## ) is unitary (cf. Complement ##\mathrm{C}_{\mathrm{II}}, \S 3## ). This relation is the definition of ##J_z##."

Why is it that; Since the group law is conserved for infinitesimal rotations, the operator ##R_{\mathbf{e}_z}(\mathrm{~d} \alpha)## is necessarily of the form: $$ R_{\mathbf{e}_z}(\mathrm{~d} \alpha)=1-\frac{i}{\hbar} \mathrm{d} \alpha J_z $$ where ##J_z## is a Hermitian operator?
 
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Unitarity of ##R \equiv R_{e_z}## means
$$R R^{\dagger}=(1-\mathrm{i} \mathrm{d} \alpha J_z)(1+\mathrm{i} \mathrm{d} \alpha) J_z^{\dagger} = 1 -\mathrm{i} \mathrm{d} \alpha (J_z - J_z^{\dagger}) + \mathcal{O}(\mathrm{d} \alpha^2) \stackrel{!}{=} 1 + \mathcal{O}(\mathrm{d} \alpha^2) \; \Rightarrow \; J_z=J_z^{\dagger}.$$
 
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vanhees71 said:
Unitarity of ##R \equiv R_{e_z}## means
$$R R^{\dagger}=(1-\mathrm{i} \mathrm{d} \alpha J_z)(1+\mathrm{i} \mathrm{d} \alpha) J_z^{\dagger} = 1 -\mathrm{i} \mathrm{d} \alpha (J_z - J_z^{\dagger}) + \mathcal{O}(\mathrm{d} \alpha^2) \stackrel{!}{=} 1 + \mathcal{O}(\mathrm{d} \alpha^2) \; \Rightarrow \; J_z=J_z^{\dagger}.$$
I was trying to ask about why does
The group law being conserved for infinitesimal rotations imply that

##R_{\mathbf{e}_z}(\mathrm{~d} \alpha)=1-\frac{i}{\hbar} \mathrm{d} \alpha J_z## . Why does it necessarily have this form
 
... because this is the infinitesimal generator relative to an virtual z axis? Is your question like "why is the Taylor expansion of the e function is at it is.."?
 
Kashmir said:
I was trying to ask about why does
The group law being conserved for infinitesimal rotations imply that

##R_{\mathbf{e}_z}(\mathrm{~d} \alpha)=1-\frac{i}{\hbar} \mathrm{d} \alpha J_z## . Why does it necessarily have this form
Every operator parameterized by an infinitesimal has that form. The group law implies ##J_z## is Hermitian. That's the point.
 
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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
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
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