Controlled paraxial spin–orbital labels and nonparaxial corrections

By [F4], the Axiom of Choice in [A1] implies Countable Choice, so the countable-choice hypotheses in [F1] and [F2] hold. The vectors ${\mathbf{e}}_{\theta }$ and ${\mathbf{e}}_{\phi }$ are orthonormal, transverse to $\mathbf{n}$, and satisfy $\mathbf{n}\times {\mathbf{e}}_{\theta }={\mathbf{e}}_{\phi }$ and $\mathbf{n}\times {\mathbf{e}}_{\phi }=-{\mathbf{e}}_{\theta }$. Thus $e_{\sigma }$ has unit norm and $h{e}_{\sigma }=\sigma e_{\sigma }$; multiplication by the phase $e^{i\sigma \phi }$ preserves both properties. By [F2], the resulting north gauge extends smoothly through $+e_z$, while the stated transition prevents treating it as a global frame.

For $\mathrm{\ell }>0$, $(\sin \theta {)}^{\mathrm{\mid }\mathrm{\ell }\mathrm{\mid }}{e}^{i\mathrm{\ell }\phi }=(n_x+i{n}_y{)}^{\mathrm{\ell }}$, and for $\mathrm{\ell }<0$ it equals $(n_x-i{n}_y{)}^{\mathrm{\mid }\mathrm{\ell }\mathrm{\mid }}$; for $\mathrm{\ell }=0$ the factor is $1$. These are smooth polynomials in the Cartesian coordinates. Since $b$ is smooth in $(r,n_z)$ and $r>0$, the specified coefficient is smooth across the axis. Away from the axis, the azimuthal phase is smooth wherever its envelope is smooth.

Since $R_z(t){\mathbf{e}}_{\theta }(\theta ,\phi -t)={\mathbf{e}}_{\theta }(\theta ,\phi )$ and $R_z(t){\mathbf{e}}_{\phi }(\theta ,\phi -t)={\mathbf{e}}_{\phi }(\theta ,\phi )$, the local frame obeys $R_z(t)e_{\sigma }^N(\theta ,\phi -t)=e^{-i\sigma t}{e}_{\sigma }^N(\theta ,\phi )$. This identity fixes the phase contribution of the chosen gauge under the rotation convention in [F3].

For any nonempty bounded open interval $(u,v)$, choose $c$ and $\epsilon >0$ with $[c-\epsilon ,c+\epsilon ]\subset (u,v)$. Apply [F2] in ${\mathbb{R}}^3$ with radii $\epsilon \mathrm{/}2<\epsilon$, obtaining a smooth $\rho$ equal to $1$ on the smaller ball and supported in the larger. The restriction $g(x)=\rho ((x-c)e_1)$ is smooth, equals $1$ on a neighborhood of $c$, and is supported in $(u,v)$. Applying this construction to $({\kappa }_{-},{\kappa }_{+})$ and $({\theta }_1,{\theta }_2)$ gives the required nonzero functions $a$ and $b$.

The integrand $\mathrm{\mid }a(r){\mathrm{\mid }}^2\mathrm{\mid }b(\theta ){\mathrm{\mid }}^2$ is bounded and supported in a finite radial annulus and a finite-measure spherical chart, so it is integrable. By [F1], [F3], and [F4], $\mathrm{\parallel }{f}_{\sigma ,\mathrm{\ell }}^0{\mathrm{\parallel }}_{\nu }^2=\frac{1}{2}({\int }_{{\kappa }_{-}}^{{\kappa }_{+}}r\mathrm{\mid }a(r){\mathrm{\mid }}^{2}dr)({\int }_{S^2}\mathrm{\mid }b(\theta ){\mathrm{\mid }}^{2}d\mathrm{\Omega })=\pi AB\mathrm{.}$ The first factor $A$ is finite and positive because $a$ is bounded, compactly supported, and equals $1$ on an interval. The spherical chart gives ${\int }_{S^2}\mathrm{\mid }b(\theta ){\mathrm{\mid }}^{2}d\mathrm{\Omega }=2\pi B$, so $B$ is finite and positive for the same reason. Thus $0<\mathrm{\parallel }{f}_{\sigma ,\mathrm{\ell }}^0{\mathrm{\parallel }}_{\nu }^2=\pi AB<\mathrm{\infty }$, and the stated normalization is valid.

The support lies in ${\kappa }_{-}<r<{\kappa }_{+}$ and ${\theta }_1<\theta <{\theta }_2$, a compact region away from $\mathbf{k}=0$ and the azimuthal axis after taking the compact supports of $a,b$. The frame $e_{\sigma }^N$ is smooth and transverse with $P_{\sigma }{e}_{\sigma }^N=e_{\sigma }^N$ by [F1]; because $\mathrm{\ell }$ is integral, $e^{i\mathrm{\ell }\phi }$ is single-valued and smooth on this region. Extension by zero is smooth since $a,b$ have compact support in the open intervals. Hence $f_{\sigma ,\mathrm{\ell }}^0\in \mathcal{D}$, and dividing by its positive norm preserves membership and helicity while giving norm one.

Under $R_z(-t)$, $r$ and $n_z$ stay fixed and $\phi$ becomes $\phi -t$. Therefore [F1] gives $U(R_z(t))f_{\sigma ,\mathrm{\ell }}=e^{-i\mathrm{\ell }t}{e}^{-i\sigma t}{f}_{\sigma ,\mathrm{\ell }}=e^{-i(\mathrm{\ell }+\sigma )t}{f}_{\sigma ,\mathrm{\ell }}$. Since $f_{\sigma ,\mathrm{\ell }}\in \mathcal{D}\subseteq D(J_z)$ by [F5], differentiating at $t=0$ in the Stone convention of [F5] yields $J_z{f}_{\sigma ,\mathrm{\ell }}=(\mathrm{\ell }+\sigma )\mathrm{\hslash }{f}_{\sigma ,\mathrm{\ell }}$.

Under the Choice assumptions [A1], [F2] gives $⟨f,S_z^{\mathrm{\prime }}f{⟩}_{\nu }=\mathrm{\hslash }\sigma {\int }_M\cos \theta d\rho$, where $d\rho =\mathrm{\mid }f{\mathrm{\mid }}^{2}d\nu$ is a probability measure because $\mathrm{\parallel }f{\mathrm{\parallel }}_{\nu }=1$. Since $\mathrm{\mid }\sigma \mathrm{\mid }=1$, $\mid ⟨f,S_z^{\mathrm{\prime }}f{⟩}_{\nu }-\mathrm{\hslash }\sigma \mid =\mathrm{\hslash }(1-{\int }_M\cos \theta d\rho )$; the right side is nonnegative by the range bound in [F7].

On the support, $0\le \theta \le {\theta }_{\max }<\pi \mathrm{/}2$. The monotonicity in [F7] gives $0\le 1-\cos \theta \le 1-\cos {\theta }_{\max }$ pointwise. Integrating against the probability measure $\rho$ in step 1.1 proves the first upper bound.

By [F8], $1-\cos {\theta }_{\max }=2{\sin }^2({\theta }_{\max }\mathrm{/}2)$. Applying [F6] with $u={\theta }_{\max }\mathrm{/}2$ gives $2{\sin }^2({\theta }_{\max }\mathrm{/}2)\le 2({\theta }_{\max }\mathrm{/}2{)}^2={\theta }_{\max }^{}\mathrm{/}2$. Combining this with step 2.1 proves the spin estimate.

Under [A1], the preceding lemma [F5] gives $J_{z}f=(\mathrm{\ell }+\sigma )\mathrm{\hslash }f$ and $f\in D(J_z)$ in the rotation convention [F4]. By [F3], $L_z^{\mathrm{\prime }}=J_z-S_z^{\mathrm{\prime }}$ on that domain, so $⟨f,L_z^{\mathrm{\prime }}f{⟩}_{\nu }=(\mathrm{\ell }+\sigma )\mathrm{\hslash }-⟨f,S_z^{\mathrm{\prime }}f{⟩}_{\nu }$. Substituting [F2] gives the displayed exact orbital expectation; subtracting $\mathrm{\ell }\mathrm{\hslash }$ and applying step 3.1 gives its stated bound.

By the hypotheses on $U$, ${\psi }_{\mathrm{o}\mathrm{u}\mathrm{t}}\in D(J_z)$ and $J_z{\psi }_{\mathrm{o}\mathrm{u}\mathrm{t}}=U{J}_z\psi$. Unitarity and the second-variable-linear Hilbert pairing give $⟨{\psi }_{\mathrm{o}\mathrm{u}\mathrm{t}},J_z{\psi }_{\mathrm{o}\mathrm{u}\mathrm{t}}{⟩}_{\nu }=⟨U\psi ,U{J}_z\psi {⟩}_{\nu }=⟨\psi ,J_z\psi {⟩}_{\nu }$.

If $Uh=hU$, then $U^{\ast }hU=h$ and unitarity gives $⟨U\psi ,hU\psi {⟩}_{\nu }=⟨\psi ,h\psi {⟩}_{\nu }$ for every $\psi$. Conversely, if these expectations agree for every unit vector, homogeneity gives equality for every vector. The bounded self-adjoint operator $A=U^{\ast }hU-h$ therefore has $⟨x,Ax{⟩}_{\nu }=0$ for every $x$. Expanding this identity at $x+y$, $x-y$, $x+iy$, and $x-iy$ shows both real and imaginary parts of $⟨x,Ay{⟩}_{\nu }$ vanish for every $x,y$, so the polarization identity gives $A=0$ and hence $Uh=hU$.

Under the Choice assumptions [A1], choose the two packets $f_{+}=f_{+1,0}$ and $f_{-}=f_{-1,2}$ from [F4]. They are normalized and orthogonal because $P_{+}{P}_{-}=0$ by [F1]. Both have $J_z$ eigenvalue $\mathrm{\hslash }$, while $h{f}_{+}=f_{+}$ and $h{f}_{-}=-f_{-}$. Let $K=\mathrm{span}\{f_{+},f_{-}\}$, let $U_0$ exchange $f_{+}$ and $f_{-}$ on $K$, and let $U_0$ be the identity on $K^{\perp }$. Its matrix on the ordered orthonormal basis $(f_{+},f_{-})$ is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, which equals its adjoint and has square equal to the identity; thus $U_0$ is unitary.

By [F2], $L_z^{\mathrm{\prime }}=J_z-S_z^{\mathrm{\prime }}$ on the domain containing both input and output. Subtracting the finite expectations in step 1.1 gives $\mathrm{\Delta }⟨L_z^{\mathrm{\prime }}⟩=\mathrm{\Delta }⟨J_z⟩-\mathrm{\Delta }⟨S_z^{\mathrm{\prime }}⟩=-\mathrm{\Delta }⟨S_z^{\mathrm{\prime }}⟩$. If the input also satisfies the fixed-helicity and support hypotheses of [F3], that fact bounds its input spin expectation. Applying the bound to the output would require the corresponding output hypotheses; the exact balance requires neither these packet restrictions nor spin conservation.

The finite-dimensional subspace $K$ lies in the $\mathrm{\hslash }$ eigenspace of the self-adjoint $J_z$. Its orthogonal projection $P_K$ preserves $D(J_z)$ and satisfies $J_z{P}_K\eta =\mathrm{\hslash }{P}_K\eta$. If $\eta \in D(J_z)\cap K^{\perp }$, self-adjointness gives $⟨v,J_z\eta {⟩}_{\nu }=⟨J_{z}v,\eta {⟩}_{\nu }=0$ for every $v\in K$, so $J_z\eta \in K^{\perp }$. Thus both $K$ and $K^{\perp }$ reduce $J_z$, and $U_0$ preserves $D(J_z)$ and commutes with $J_z$ there.

But $U_{0}h{f}_{+}=f_{-}$ whereas $h{U}_0{f}_{+}=-f_{-}$. Hence $U_{0}h\mathrm{\ne }h{U}_0$ although $U_0$ is unitary and commutes with $J_z$; by step 1.2 it changes mean helicity for an input. This proves that axial symmetry alone does not imply helicity conservation.