Angular momentum and rotations

[Kashmir]

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?

 

[vanhees71]

Unitarity of $R \equiv R_{e_z}$ means
$$R R^{\dagger}=(1-\mathrm{i} \mathrm{d} \alpha J_z)(1+\mathrm{i} \mathrm{d} \alpha) J_z^{\dagger} = 1 -\mathrm{i} \mathrm{d} \alpha (J_z - J_z^{\dagger}) + \mathcal{O}(\mathrm{d} \alpha^2) \stackrel{!}{=} 1 + \mathcal{O}(\mathrm{d} \alpha^2) \; \Rightarrow \; J_z=J_z^{\dagger}.$$

 

[Kashmir]

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

 

[Omega0]

... 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.."?

 

[PeroK]

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.