Orbital angular momentum eigenfunctions: the low spherical harmonics of the massive model

[Low polynomials] Substituting the constants of [F1] gives $h_{00}=2^{-1\mathrm{/}2}(2\pi {)}^{-1\mathrm{/}2}=(4\pi {)}^{-1\mathrm{/}2}$, $h_{10}=\sqrt{3\mathrm{/}2}(2\pi {)}^{-1\mathrm{/}2}{x}_3=\sqrt{3\mathrm{/}(4\pi )}{x}_3$, and $h_{1,\pm 1}=\mp \sqrt{3\mathrm{/}4}(2\pi {)}^{-1\mathrm{/}2}(x_1\pm i{x}_2)=\mp \sqrt{3\mathrm{/}(8\pi )}(x_1\pm i{x}_2)$, which are the displayed polynomials of (i); each is harmonic and homogeneous of the stated degree by [F1].

[General sphere identity and the spatial generator domain] The general sphere identity $L_3^{\mathrm{s}\mathrm{p}\mathrm{h}}{Y}_{lm}=\mathrm{\hslash }m{Y}_{lm}$ is [F3]. Put $\psi (x)=R(\mathrm{\mid }x\mathrm{\mid })h_{lm}(x)w$, extended by zero at $0$. Its support lies in a fixed compact annulus, so it is smooth there and zero near $0$. Rotations preserve $\mathrm{\mid }x\mathrm{\mid }$, hence by [F4], $U_i(t)\psi (x)=R(\mathrm{\mid }x\mathrm{\mid })h_{lm}(R_i(-t)x)w$. In the fixed finite-dimensional space of homogeneous degree-$l$ polynomials, the coefficients of $h_{lm}(R_i(-t)x)$ are smooth functions of $t$; differentiating the rotation matrices at $0$ gives the polynomial derivative $-D_i{h}_{lm}$. Each fixed polynomial times $R(\mathrm{\mid }x\mathrm{\mid })w$ belongs to $H_s$ by [F4], since it is bounded and compactly supported. Thus coefficientwise convergence of the finitely many difference quotients gives convergence in $H_s$ to $-R(\mathrm{\mid }x\mathrm{\mid })D_i{h}_{lm}(x)w$. The domain definition in [F4] proves $\psi \in D(L_i)$ and $L_i\psi =-i\mathrm{\hslash }R(\mathrm{\mid }x\mathrm{\mid })D_i{h}_{lm}(x)w$. This establishes the needed full spatial action directly for these vectors, using the assigned operators without a general spectral or Stone construction.

[Sphere norms and orthogonality] By [F2] and step 1.1, ${\int }_{S^2}\mathrm{\mid }{Y}_{00}{\mathrm{\mid }}^{2}d\sigma =\frac{1}{4\pi }\cdot 4\pi =1$; ${\int }_{S^2}\mathrm{\mid }{Y}_{10}{\mathrm{\mid }}^{2}d\sigma =\frac{3}{4\pi }\int {\mathrm{\Omega }}_3^{2}d\sigma =\frac{3}{4\pi }\cdot \frac{4\pi }{3}=1$; and ${\int }_{S^2}\mathrm{\mid }{Y}_{1,\pm 1}{\mathrm{\mid }}^{2}d\sigma =\frac{3}{8\pi }\int ({\mathrm{\Omega }}_1^2+{\mathrm{\Omega }}_2^2)d\sigma =\frac{3}{8\pi }\cdot \frac{8\pi }{3}=1$. The mixed products with $Y_{00}$ integrate to zero: ${\mathrm{\Omega }}_1$ and ${\mathrm{\Omega }}_2$ have zero $\phi$-integrals, while ${\int }_0^{\pi }\cos \theta \sin \theta d\theta =0$ for ${\mathrm{\Omega }}_3$, by the angle formula and the trigonometric primitives in [F2]. The mixed products of $Y_{10}$ with $Y_{1,\pm 1}$ vanish by $\int {\mathrm{\Omega }}_3{\mathrm{\Omega }}_{1}d\sigma =\int {\mathrm{\Omega }}_3{\mathrm{\Omega }}_{2}d\sigma =0$ from [F2]. Finally, $Y_{11}\overline{Y_{1,-1}}=-\frac{3}{8\pi }({\mathrm{\Omega }}_1^2-{\mathrm{\Omega }}_2^2+2i{\mathrm{\Omega }}_1{\mathrm{\Omega }}_2)$ has zero integral because [F2] gives equal second moments for ${\mathrm{\Omega }}_1,{\mathrm{\Omega }}_2$ and a zero cross moment; the reversed mixed integrals are their complex conjugates. Hence the four functions are orthonormal.

[Degree-one eigenvalues] By step 1.1, direct differentiation gives $D_3{h}_{1m}=(x_1{\mathrm{\partial }}_2-x_2{\mathrm{\partial }}_1)h_{1m}=im{h}_{1m}$ for $m=-1,0,1$, hence $-i\mathrm{\hslash }{D}_3{h}_{1m}=\mathrm{\hslash }m{h}_{1m}$ as a polynomial identity; restricting to the sphere and using [F3] gives $L_3^{\mathrm{s}\mathrm{p}\mathrm{h}}{Y}_{1m}=\mathrm{\hslash }m{Y}_{1m}$. The Casimir identity of [F3] with $l=1$ gives ${\sum }_i(L_i^{\mathrm{s}\mathrm{p}\mathrm{h}}{)}^{2}F=2{\mathrm{\hslash }}^{2}F$ for $F\in V_1$. For $l=0$, each $D_i{h}_{00}=0$, so [F3] gives $L_i^{\mathrm{s}\mathrm{p}\mathrm{h}}{Y}_{00}=0$ and therefore ${\sum }_i(L_i^{\mathrm{s}\mathrm{p}\mathrm{h}}{)}^2{Y}_{00}=0$. This proves (ii).

[Radial norm and Casimir form] By homogeneity, $h_{lm}(r\mathrm{\Omega })=r^l{Y}_{lm}(\mathrm{\Omega })$, and $-i\mathrm{\hslash }{D}_i{h}_{lm}$ is the homogeneous polynomial representative of $L_i^{\mathrm{s}\mathrm{p}\mathrm{h}}{Y}_{lm}$ by [F3]. Polar integration [F4] and $\mathrm{\parallel }w\mathrm{\parallel }=\mathrm{\parallel }{Y}_{lm}\mathrm{\parallel }=1$ therefore give $\mathrm{\parallel }\psi {\mathrm{\parallel }}^2=A_R$ and $\mathrm{\parallel }{L}_i\psi {\mathrm{\parallel }}^2=A_R\mathrm{\parallel }{L}_i^{\mathrm{s}\mathrm{p}\mathrm{h}}{Y}_{lm}{\mathrm{\parallel }}^2$, where $A_R={\int }_0^{\mathrm{\infty }}\mathrm{\mid }R(r){\mathrm{\mid }}^2{r}^{2l+2}dr<\mathrm{\infty }$. Since $R$ is nonzero and continuous, it is bounded below in modulus by a positive constant on some nondegenerate interval inside $(0,\mathrm{\infty })$, so $A_R>0$. The sphere generators preserve $V_l$ and are self-adjoint there; hence [F3] gives ${\sum }_i\mathrm{\parallel }{L}_i^{\mathrm{s}\mathrm{p}\mathrm{h}}{Y}_{lm}{\mathrm{\parallel }}^2=⟨{\sum }_i(L_i^{\mathrm{s}\mathrm{p}\mathrm{h}}{)}^2{Y}_{lm},Y_{lm}⟩={\mathrm{\hslash }}^{2}l(l+1)$. Dividing the spatial norm identities by $A_R$ proves ${\sum }_i\mathrm{\parallel }{L}_i(\psi \mathrm{/}\mathrm{\parallel }\psi \mathrm{\parallel }){\mathrm{\parallel }}^2={\mathrm{\hslash }}^{2}l(l+1)$, the rest of (iii). Together with steps 1.1, 2.1 and 2.2 this completes (i)-(iii), under the stated Countable Choice hypothesis.