Projected spin, orbital angular momentum and gauge-aware generators

Fix a Cartesian orthonormal basis $e_1,e_2,e_3$ of ${\mathbb{C}}^3$ and $\sigma \in \{+1,-1\}$. Since $P_{\sigma }(\mathbf{n})$ has rank one, at least one of $P_{\sigma }(\mathbf{n})e_1,P_{\sigma }(\mathbf{n})e_2,P_{\sigma }(\mathbf{n})e_3$ is nonzero. Define $j_{\sigma }(\mathbf{n})$ to be the least such index. The matrix fields $P_{\sigma }$ are smooth, so each set $\{\mathbf{n}:j_{\sigma }(\mathbf{n})=j\}$ is Borel: it is the intersection of $\{\mathrm{\parallel }{P}_{\sigma }{e}_j\mathrm{\parallel }>0\}$ with the finitely many preceding zero sets. On that set define $e_{\sigma }(\mathbf{n})=P_{\sigma }(\mathbf{n})e_j\mathrm{/}\mathrm{\parallel }{P}_{\sigma }(\mathbf{n})e_j\mathrm{\parallel }$. The denominator is positive on its defining Borel set, so the piecewise formula is Borel, has unit norm, and lies in the range of $P_{\sigma }(\mathbf{n})$.

At each $\mathbf{n}$, the vectors $e_{+}(\mathbf{n}),e_{-}(\mathbf{n})$ are orthonormal because the projectors are orthogonal, and they span $P(\mathbf{n}){\mathbb{C}}^3$ because their ranges are the two rank-one summands of $P(\mathbf{n})$. For every transverse representative $f(\mathbf{k})$, the fiberwise expansion is $f(\mathbf{k})={\sum }_{\sigma =\pm 1}{e}_{\sigma }(\mathbf{n}(\mathbf{k}))(\overline{e_{\sigma }(\mathbf{n}(\mathbf{k}))}\cdot f(\mathbf{k}))$ and $\mathrm{\mid }f(\mathbf{k}){\mathrm{\mid }}^2={\sum }_{\sigma =\pm 1}\mathrm{\mid }(Tf)(\mathbf{k},\sigma ){\mathrm{\mid }}^2$. Borel measurability of $e_{\sigma }$ makes $Tf$ measurable, and integration gives $\mathrm{\parallel }Tf{\mathrm{\parallel }}_{L^2(X)}=\mathrm{\parallel }f{\mathrm{\parallel }}_{{\mathcal{H}}_T^{\nu }}$. Conversely, for any $g\in L^2(X)$ the field $f(\mathbf{k})={\sum }_{\sigma =\pm 1}{e}_{\sigma }(\mathbf{n}(\mathbf{k}))g(\mathbf{k},\sigma )$ is measurable, transverse and square-integrable, and $Tf=g$. Hence $T$ is unitary.

On each fiber $h{e}_{\sigma }=\sigma e_{\sigma }$ and $P_{\tau }{e}_{\sigma }={\mathbf{1}}_{\{\tau =\sigma \}}{e}_{\sigma }$. Substituting the fiberwise expansion from step 2.1 gives $Thf(\mathbf{k},\sigma )=\sigma Tf(\mathbf{k},\sigma )$ and $T{P}_{\tau }f(\mathbf{k},\sigma )={\mathbf{1}}_{\{\tau =\sigma \}}Tf(\mathbf{k},\sigma )$ for every transverse field class. These are precisely the two multiplication-operator identities in the Statement.

Fix $\mathbf{n}\in S^2$ and $v\in P(\mathbf{n}){\mathbb{C}}^3$, so $\mathbf{n}\cdot v=0$. Under [A1], the rotation supplier [F2] provides the stated total operator on the common domain. Decompose $e_i=n_i\mathbf{n}+(e_i-n_i\mathbf{n})$. The cross product of the two vectors $e_i-n_i\mathbf{n}$ and $v$, both perpendicular to $\mathbf{n}$, is parallel to $\mathbf{n}$, so projection by $P$ removes it; hence $P(e_i\times v)=n_i(\mathbf{n}\times v)$. Since $S_{i}v=i(e_i\times v)$ by the stated Levi-Civita convention, $P{S}_{i}Pv=i{n}_i(\mathbf{n}\times v)=n_{i}h(\mathbf{n})v$. Therefore the two displayed multiplication formulas for $S_i^{\mathrm{\prime }}$ agree on every transverse fiber.

The multiplier $n_{i}h(\mathbf{n})$ is self-adjoint on the transverse fiber because $n_i$ is real and $h^{\ast }=h$, and its norm is at most $1$ because $\mathrm{\mid }{n}_i\mathrm{\mid }\le 1$ and $\mathrm{\parallel }h\mathrm{\parallel }=1$ there. Integrating the fiberwise Hermitian identity shows that $S_i^{\mathrm{\prime }}$ is symmetric on the whole Hilbert space; since it is bounded and defined everywhere, the adjoint definition gives $S_i^{\mathrm{\prime }\ast }=S_i^{\mathrm{\prime }}$. Its norm is at most $\mathrm{\hslash }$. Its matrix coefficients are smooth on $M$; multiplication therefore preserves smoothness and compact support away from $\mathbf{k}=0$, and its values remain transverse by [F1]. Consequently $S_i^{\mathrm{\prime }}\mathcal{D}\subseteq \mathcal{D}$, and $L_i^{\mathrm{\prime }}=J_i-S_i^{\mathrm{\prime }}$ on $D(J_i)$ is well-defined and restricts to an operator from $\mathcal{D}$ to ${\mathcal{H}}_T^{\nu }$.

It suffices first to consider the axis $e_3$. The $z$-axis and the positive-$x$ branch cut, with the origin removed, are $\nu$-null by [F8]: on each annulus $1\mathrm{/}m\le \mathrm{\mid }\mathbf{k}\mathrm{\mid }\le m$, the mass-shell density is bounded and both sets lie in Lebesgue-null coordinate hyperplanes. To apply the Borel change-of-variables formula [F4] to an arbitrary $f\in {\mathcal{H}}_T^{\nu }$, take a finite-valued representative and apply [F11] to the Lebesgue-measurable sublevel sets of each of its six real components at rational levels. Countable Choice lets us select the corresponding Borel sets; their symmetric differences lie in a countable union of Borel null sets. Outside that union, each component equals the infimum of the rational levels whose selected Borel sets contain $\mathbf{k}$; defining each infimum to be $0$ on the union gives Borel real-valued representatives, hence a Borel vector representative $f_0=f$ almost everywhere. Since $d\nu =(2\mathrm{\mid }\mathbf{k}\mathrm{\mid }{)}^{-1}d{\lambda }_3$ has strictly positive finite density on $M$, Lebesgue-null and $\nu$-null sets there coincide. Thus $f_0$ represents the same Hilbert-space class, and [F4] applies to its coordinate pullback; the same formula shows the pullback is independent of the representative. Use the cylindrical coordinates of [F4], put $r=\sqrt{{\rho }^2+z^2}$ and ${\mathbf{n}}_0(\rho ,z)=(\rho ,0,z)\mathrm{/}r$, and define $g(\rho ,z,\varphi )=R_{e_3}(-\varphi )f_0(\rho \cos \varphi ,\rho \sin \varphi ,z)$. The rotation matrices preserve the Hermitian norm, so the measure identity and its inverse coordinate map $f_0(\mathbf{k})=R_{e_3}(\varphi )g(\rho ,z,\varphi )$ make this a unitary identification with the subspace of $L^2(\mathrm{\Omega }\times \mathbb{T},wd\rho dzd\varphi \mathrm{/}(2\pi );{\mathbb{C}}^3)$ satisfying ${\mathbf{n}}_0(\rho ,z)\cdot g(\rho ,z,\varphi )=0$. Direct substitution gives $U(R_{e_3}(t))g(\rho ,z,\varphi )=g(\rho ,z,\varphi -t)$.

Write $g_m(\rho ,z)={\int }_{\mathbb{T}}g(\rho ,z,\varphi )e^{-im\varphi }d\varphi \mathrm{/}(2\pi )$. By [F5], Parseval identifies the squared norm with ${\sum }_m\mathrm{\parallel }{g}_m{\mathrm{\parallel }}_{L^2(\mathrm{\Omega },w)}^2$, and the rotation difference quotient has Fourier coefficients $(e^{-imt}-1)g_m\mathrm{/}t$. If ${\sum }_m{m}^2\mathrm{\parallel }{g}_m{\mathrm{\parallel }}^2<\mathrm{\infty }$, then $\mathrm{\mid }(e^{-imt}-1)\mathrm{/}t\mathrm{\mid }\le \mathrm{\mid }m\mathrm{\mid }$ and dominated convergence in the square-summable coefficient space gives convergence to $(-im{g}_m{)}_m$. Conversely, if the difference quotients converge in norm, their $m$th coefficients converge to $-im{g}_m$; Fatou applied to finite partial sums and then their increasing limit gives ${\sum }_m{m}^2\mathrm{\parallel }{g}_m{\mathrm{\parallel }}^2<\mathrm{\infty }$. By [F10], this is exactly $D(J_{e_3})$; the generator acts as $J_{e_3}=-i\mathrm{\hslash }{\mathrm{\partial }}_{\varphi }$ in the rotating coordinates, and the squared graph norm is ${\sum }_{m\in \mathbb{Z}}(1+{\mathrm{\hslash }}^2{m}^2)\mathrm{\parallel }{g}_m{\mathrm{\parallel }}_{L^2(\mathrm{\Omega },w)}^2$.

Let $g$ represent any vector in $D(J_{e_3})$. For $n\ge 2$ use the explicit cutoffs ${\chi }_n$ from [F7]. They lie in $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$, satisfy $0\le {\chi }_n\le 1$, and converge pointwise to $1$. Since ${\chi }_n$ depends only on $(\rho ,z)$, it commutes with the $\varphi$-derivative. Dominated convergence [F9] therefore gives ${\chi }_{n}g\to g$ in the graph norm of [F3]. Its Fourier truncations $g^{(N)}={\sum }_{\mathrm{\mid }m\mathrm{\mid }\le N}{\chi }_n{g}_m{e}^{im\varphi }$ converge in that graph norm by the weighted Parseval sum in step 2.1.

Fix $n,N$ and put ${\eta }_n={\chi }_{2n}$; [F7] makes ${\eta }_n=1$ on the support of ${\chi }_n$ and gives compact support inside $\mathrm{\Omega }$. Each of the finitely many coefficients ${\chi }_n{g}_m$ is supported in $\mathrm{supp}{\chi }_n$ and lies in ${\mathbf{n}}_0^{\perp }$. Extend it by zero to ${\mathbb{R}}^2$. On the compact support of ${\eta }_n$, the weight in [F4] is bounded above and below by positive constants, so [F6] gives a smooth compactly supported approximation $a_m^0$ in the weighted norm. Replace it by $a_m={\eta }_{n}P({\mathbf{n}}_0)a_m^0$. Because $P({\mathbf{n}}_0){\chi }_n{g}_m={\chi }_n{g}_m$ and $\mathrm{\parallel }P\mathrm{\parallel }\le 1$, this is transverse and does not increase the approximation error. Choose the finitely many coefficient errors so that ${\sum }_{\mathrm{\mid }m\mathrm{\mid }\le N}(1+{\mathrm{\hslash }}^2{m}^2)\mathrm{\parallel }{a}_m-{\chi }_n{g}_m{\mathrm{\parallel }}^2$ is arbitrarily small. Then ${\sum }_{\mathrm{\mid }m\mathrm{\mid }\le N}{a}_m(\rho ,z)e^{im\varphi }$ is smooth, transverse, and compactly supported away from $\rho =0$; after applying $R_{e_3}(\varphi )$ and returning to $\mathbf{k}$ coordinates it is an element of $\mathcal{D}$, since both factors are $2\pi$-periodic and their derivatives match across the angular seam. Combining these finite coefficient approximations with the cutoff and Fourier truncation from step 3.1 gives graph convergence to $g$, so $\mathcal{D}$ is a core for $J_{e_3}$.

For any unit $\mathbf{a}$, choose $R\in SO(3)$ with $R{e}_3=\mathbf{a}$. The rotation representation gives $U(R)U(R_{e_3}(t))U(R{)}^{\ast }=U(R_{\mathbf{a}}(t))$; Stone uniqueness then gives $J_{\mathbf{a}}=U(R)J_{e_3}U(R{)}^{\ast }$ with the conjugated domain. Also $U(R)\mathcal{D}=\mathcal{D}$, because rotations preserve smoothness, compact support away from zero, and transversality. Applying the unitary $U(R)$ to the graph-dense approximation in step 4.1 proves that $\mathcal{D}$ is a core for $J_{\mathbf{a}}$.

The domain $\mathcal{D}$ is invariant under each $J_i$ and $S_i^{\mathrm{\prime }}$ by their smooth differential and multiplication formulas. Write $X_i=(e_i\times \mathbf{k})\cdot {\mathrm{\nabla }}_{\mathbf{k}}$ and $C_{i}v=e_i\times v$, so $L_i=-i\mathrm{\hslash }{X}_i$ and $\mathrm{\hslash }{S}_i=i\mathrm{\hslash }{C}_i$. The product rule for the linear vector fields gives $[X_i,X_j]f=(e_j\times (e_i\times \mathbf{k})-e_i\times (e_j\times \mathbf{k}))\cdot \mathrm{\nabla }f=-{\sum }_k{ϵ}_{ijk}{X}_{k}f$. The vector triple-product identity gives $[C_i,C_j]={\sum }_k{ϵ}_{ijk}{C}_k$, while constant matrices commute with the scalar differential coefficients, so $[L_i,\mathrm{\hslash }{S}_j]=0$. Therefore $[L_i,L_j]=i\mathrm{\hslash }{\sum }_k{ϵ}_{ijk}{L}_k$ and $[\mathrm{\hslash }{S}_i,\mathrm{\hslash }{S}_j]=i\mathrm{\hslash }{\sum }_k{ϵ}_{ijk}\mathrm{\hslash }{S}_k$. Adding these identities yields $[J_i,J_j]=i\mathrm{\hslash }{\sum }_k{ϵ}_{ijk}{J}_k$ on $\mathcal{D}$.

Since $J_i$ is self-adjoint and $-S_i^{\mathrm{\prime }}$ is bounded self-adjoint, it is $J_i$-bounded with relative-bound coefficient $a=0<1$ and finite constant $b=\mathrm{\parallel }{S}_i^{\mathrm{\prime }}\mathrm{\parallel }$. Apply [F4] to $A=J_i$ and $B=-S_i^{\mathrm{\prime }}$ to conclude $L_i^{\mathrm{\prime }}=J_i-S_i^{\mathrm{\prime }}$ is self-adjoint on exactly $D(J_i)$. For every $\psi \in D(J_i)$, $\mathrm{\parallel }{L}_i^{\mathrm{\prime }}\psi \mathrm{\parallel }\le \mathrm{\parallel }{J}_i\psi \mathrm{\parallel }+\mathrm{\hslash }\mathrm{\parallel }\psi \mathrm{\parallel }$ and $\mathrm{\parallel }{J}_i\psi \mathrm{\parallel }\le \mathrm{\parallel }{L}_i^{\mathrm{\prime }}\psi \mathrm{\parallel }+\mathrm{\hslash }\mathrm{\parallel }\psi \mathrm{\parallel }$. These inequalities make the two graph norms equivalent. Hence [F3] implies that $\mathcal{D}$ is also a core for $L_i^{\mathrm{\prime }}$.

The operators $S_i^{\mathrm{\prime }}$ are multiplication by the commuting matrix fields $\mathrm{\hslash }{n}_{i}h(\mathbf{n})$. Thus on every transverse fiber $S_i^{\mathrm{\prime }}{S}_j^{\mathrm{\prime }}={\mathrm{\hslash }}^2{n}_i{n}_j{h}^2=S_j^{\mathrm{\prime }}{S}_i^{\mathrm{\prime }}$, so $[S_i^{\mathrm{\prime }},S_j^{\mathrm{\prime }}]=0$ as bounded operators, in particular on $\mathcal{D}$.

For $R\in SO(3)$, the cross-product identity $(R\mathbf{n})\times (Rv)=R(\mathbf{n}\times v)$ gives $P(R\mathbf{n})=RP(\mathbf{n})R^{-1}$ and $h(R\mathbf{n})=Rh(\mathbf{n})R^{-1}$. Also $R{S}_j{R}^{-1}={\sum }_m{R}_{mj}{S}_m$, since rotations carry $e_j$ to ${\sum }_m{R}_{mj}{e}_m$. Substituting these identities in $S_j^{\mathrm{\prime }}(\mathbf{n})=\mathrm{\hslash }P(\mathbf{n})S_{j}P(\mathbf{n})$ yields $U(R)S_j^{\mathrm{\prime }}U(R{)}^{\ast }={\sum }_m{R}_{mj}{S}_m^{\mathrm{\prime }}$. For $R=R_{e_i}(t)$, differentiating the right side at zero gives ${\sum }_m{ϵ}_{ijm}{S}_m^{\mathrm{\prime }}$. For $\psi \in \mathcal{D}$, both $\psi$ and $S_j^{\mathrm{\prime }}\psi$ lie in $D(J_i)$ by [F1]–[F2], so the product rule for the left side gives ${\frac{d}{dt}\mid }_{0}U(R_{e_i}(t))S_j^{\mathrm{\prime }}U(R_{e_i}(t){)}^{\ast }\psi =-\frac{i}{\mathrm{\hslash }}[J_i,S_j^{\mathrm{\prime }}]\psi$. Equating the two derivatives proves $[J_i,S_j^{\mathrm{\prime }}]=i\mathrm{\hslash }{\sum }_m{ϵ}_{ijm}{S}_m^{\mathrm{\prime }}$ on $\mathcal{D}$.

Using $L_i^{\mathrm{\prime }}=J_i-S_i^{\mathrm{\prime }}$ and invariance of $\mathcal{D}$, expand on $\mathcal{D}$: $[L_i^{\mathrm{\prime }},S_j^{\mathrm{\prime }}]=[J_i,S_j^{\mathrm{\prime }}]-[S_i^{\mathrm{\prime }},S_j^{\mathrm{\prime }}]=i\mathrm{\hslash }{\sum }_k{ϵ}_{ijk}{S}_k^{\mathrm{\prime }}$, and $[L_i^{\mathrm{\prime }},L_j^{\mathrm{\prime }}]=[J_i,J_j]-[J_i,S_j^{\mathrm{\prime }}]-[S_i^{\mathrm{\prime }},J_j]+[S_i^{\mathrm{\prime }},S_j^{\mathrm{\prime }}]=i\mathrm{\hslash }{\sum }_k{ϵ}_{ijk}(J_k-2S_k^{\mathrm{\prime }})=i\mathrm{\hslash }{\sum }_k{ϵ}_{ijk}(L_k^{\mathrm{\prime }}-S_k^{\mathrm{\prime }})$. Here $[S_i^{\mathrm{\prime }},J_j]=-[J_j,S_i^{\mathrm{\prime }}]=i\mathrm{\hslash }{\sum }_k{ϵ}_{ijk}{S}_k^{\mathrm{\prime }}$ because ${ϵ}_{jik}=-{ϵ}_{ijk}$. Along with step 1.3 and step 1.1, these are all stated identities. The term $-S_k^{\mathrm{\prime }}$ in the orbital commutator prevents $L_i^{\mathrm{\prime }}$ from satisfying an independent $\mathfrak{s}\mathfrak{o}(3)$ algebra; the commuting $S_i^{\mathrm{\prime }}$ also do not satisfy the nonzero $\mathfrak{s}\mathfrak{o}(3)$ brackets.

For $R\in SO(3)$, $U(R{)}^{\otimes n}$ commutes with the permutation action on ${\mathcal{H}}_T^{\widehat{\otimes }n}$, so it preserves the symmetric sector and is unitary there; set its vacuum-sector value to $1$. The tensor and direct-sum definitions in [F2] give $\mathrm{\Gamma }(U(R_1))\mathrm{\Gamma }(U(R_2))=\mathrm{\Gamma }(U(R_1{R}_2))$. On an elementary tensor, the telescoping estimate $\mathrm{\parallel }{⨂}_{r}U(R){\psi }_r-{⨂}_{r}U(R_0){\psi }_r\mathrm{\parallel }\le {\sum }_j\mathrm{\parallel }U(R){\psi }_j-U(R_0){\psi }_j\mathrm{\parallel }{\prod }_{r\mathrm{\ne }j}\mathrm{\parallel }{\psi }_r\mathrm{\parallel }$ shows strong continuity from that of $U$. Density extends it to each symmetric sector; finite-particle vectors are dense by [F2], and approximation by such vectors with unitarity extends continuity to the full Fock direct sum. For each axis, Stone's theorem [F7] supplies the unique self-adjoint generator $J_{\mathbf{a}}^{\mathrm{F}}$. For ${\psi }_1,\dots ,{\psi }_n\in D(J_{\mathbf{a}})$, differentiating the finite tensor product at $t=0$ by the product rule gives the sum of the one-particle derivatives $-i{J}_{\mathbf{a}}{\psi }_r\mathrm{/}\mathrm{\hslash }$; symmetrization commutes with the tensor action. Hence the Stone generator acts on these finite symmetric tensors by the displayed $d\mathrm{\Gamma }(J_{\mathbf{a}})$ sum.

Apply the unitary $T$ of [F4] and its Fock lift $\mathrm{\Gamma }(T)={⨁}_{n\ge 0}{T}^{{\otimes }_{s}n}$. Each sector map is unitary because tensor powers preserve inner products and commute with permutations, so $\mathrm{\Gamma }(T)$ is unitary. Put $X=M\times \{+1,-1\}$ and $\mathit{\omega }(\mathbf{k},\sigma )=(\mathrm{\mid }\mathbf{k}\mathrm{\mid },{\mathbf{1}}_{\{\sigma =+1\}},{\mathbf{1}}_{\{\sigma =-1\}})$. The measure $\nu \otimes \mathrm{\#}$ is sigma-finite by the bundle definition in [F3], and $\mathit{\omega }$ is finite-valued and measurable. The three coordinates in [F5] therefore give the self-adjoint Fock multipliers $H_0=d\mathrm{\Gamma }(\mathrm{\mid }\mathbf{k}\mathrm{\mid })$, $N_{+}$, and $N_{-}$ on their exact domains; the third linear combination $H_{(0,1,-1)}$ is the self-adjoint sector multiplier ${\sum }_r{\sigma }_r=d\mathrm{\Gamma }(h)$. On each finite-particle sector, ${\mathbf{1}}_{\{\sigma =+1\}}-{\mathbf{1}}_{\{\sigma =-1\}}=\sigma$, so this operator agrees there with the difference $N_{+}-N_{-}$. Its full domain is the exact $D(H_{(0,1,-1)})$ supplied by [F5], not an informal difference of unbounded operators.

The space ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}(\mathcal{D})$ is dense because $\mathcal{D}$ is dense in ${\mathcal{H}}_T^{\nu }$ by [F3], algebraic tensors are dense in each Hilbert tensor completion, and finite-particle truncations are dense in the direct sum by [F2]. It is invariant under the sector sums $d\mathrm{\Gamma }(J_i)$ because $J_i\mathcal{D}\subseteq \mathcal{D}$ by [F3]. On an elementary $n$-tensor, expanding each sector sum shows that terms acting on distinct tensor factors commute, while terms on the same factor give $[J_i,J_j]$; hence $[d\mathrm{\Gamma }(J_i),d\mathrm{\Gamma }(J_j)]={\sum }_{r=1}^n[J_i,J_j{]}^{(r)}=i\mathrm{\hslash }{\sum }_k{ϵ}_{ijk}d\mathrm{\Gamma }(J_k)$ on the symmetrized tensor core by [F6]. This is the asserted Fock total rotation algebra.

In the scalarized one-particle representation, $V_t$ is multiplication by $e^{-it\mathrm{\mid }\mathbf{k}\mathrm{\mid }}$ on both helicity components, so its Fock lift is the phase group $e^{-it{H}_0}$ from [F5]. The joint spectral measure in [F5] makes this group commute with every Borel function of the helicity coordinates, in particular with $H_{(0,1,-1)}=d\mathrm{\Gamma }(h)$; the joint multiplier also preserves its exact domain. Thus $\mathrm{\Gamma }(V_t)D_h=D_h\mathrm{\Gamma }(V_t)$ for every real $t$, proving the conservation statement.