Second-quantized total rotations and helicity counts

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.

If $P_{\sigma }f=f$, then $hf=\sigma f$ by [F4]. By [F6], $S_z^{\mathrm{\prime }}f(\mathbf{k})=\mathrm{\hslash }{n}_z(\mathbf{k})h(\mathbf{n}(\mathbf{k}))f(\mathbf{k})=\mathrm{\hslash }\sigma \cos \theta (\mathbf{k})f(\mathbf{k})$. Taking the Hilbert pairing with $f$, using $\mathrm{\parallel }f{\mathrm{\parallel }}_{\nu }=1$ and $n_z=k_z\mathrm{/}\mathrm{\mid }\mathbf{k}\mathrm{\mid }=\cos \theta$, gives the displayed expectation identity. The multiplier is bounded, so the integral is finite.