Finite-bandwidth vortex packets are rotation eigenvectors

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 $J_z{f}_{\sigma ,\mathrm{\ell }}=(\mathrm{\ell }+\sigma )\mathrm{\hslash }{f}_{\sigma ,\mathrm{\ell }}$.