Axial symmetry constrains spin-to-orbital conversion

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.