Dirac Fock covariance, vacuum cyclicity and spectrum

For translations the identities $S\mathrm{\Gamma }(p)S^{-1}=\mathrm{\Gamma }(\mathrm{\Lambda }p)$ and $S\beta S^{†}=\beta$ hold because $S=I$ and $\mathrm{\Lambda }=I$. They hold for each explicit rotation and boost generator by [F2]. They are preserved under products: if they hold for $({\mathrm{\Lambda }}_1,S_1)$ and $({\mathrm{\Lambda }}_2,S_2)$, then $(S_2{S}_1)\mathrm{\Gamma }(p)(S_2{S}_1{)}^{-1}=\mathrm{\Gamma }({\mathrm{\Lambda }}_2{\mathrm{\Lambda }}_{1}p)$ and $(S_2{S}_1)\beta (S_2{S}_1{)}^{†}=\beta$. They are also preserved under inverses, since $S^{-1}\mathrm{\Gamma }(r)S=\mathrm{\Gamma }({\mathrm{\Lambda }}^{-1}r)$ follows by setting $p={\mathrm{\Lambda }}^{-1}r$ in the first identity, and $S^{-1}\beta S^{-†}=\beta$ follows by multiplying the second identity by $S^{-1}$ and $S^{-†}$. Thus both identities hold for every element of the subgroup $\tilde{G}$ generated by the stated triples. Now for $p=(\omega ,\mathbf{k})$ the projection identity in [F1] gives $P_{+}(\mathbf{k})\beta =(\mathrm{\Gamma }(p)+\mu )\mathrm{/}(2\omega )$. For $q=(\omega ,-\mathbf{k})$ it gives $P_{-}(\mathbf{k})\beta =(\mathrm{\Gamma }(q)-\mu )\mathrm{/}(2\omega )$. Thus, using $d{\nu }_{\mu }=d^3\mathbf{k}\mathrm{/}((2\pi {)}^{3}2\omega )$, the separate kernels in [F4] are $C_{+}(z)=\int (\mathrm{\Gamma }(p)+\mu )\beta e^{-ip\circ z}d{\nu }_{\mu }(p),C_{-}(z)=\int (\mathrm{\Gamma }(q)-\mu )\beta e^{+iq\circ z}d{\nu }_{\mu }(q),$ where $p\circ z=p^0{z}^0-\mathbf{p}\cdot \mathbf{z}$ and both $p,q$ range over the future mass shell. The matrix relations now established for every $\tilde{g}\in \tilde{G}$, together with $S^{-1}\beta =\beta S^{†}$, give $S(\mathrm{\Gamma }(p)\pm \mu )\beta S^{†}=(\mathrm{\Gamma }(\mathrm{\Lambda }p)\pm \mu )\beta$. In each shell integral, change variables $r=\mathrm{\Lambda }p$ (or $r=\mathrm{\Lambda }q$); invariance of $d{\nu }_{\mu }$ [F3] and $(\mathrm{\Lambda }p)\circ z=p\circ ({\mathrm{\Lambda }}^{-1}z)$ show $C_{\pm }(\mathrm{\Lambda }z)=S{C}_{\pm }(z)S^{†}$. The coefficients grow at most linearly on the shell, so these Fourier integrals define tempered distributions and their pairings with Schwartz tests converge.

For $F,G\in \mathcal{S}({\mathbb{R}}^4;{\mathbb{C}}^4)$, [F4] identifies $⟨P_{+}{u}_F,P_{+}{u}_G⟩$ and $⟨P_{-}{u}_F,P_{-}{u}_G⟩$ with the respective kernel pairings $\int F(x{)}^{†}{C}_{\pm }(x-y)G(y)d^{4}x{d}^{4}y$. Set $F^{\mathrm{\prime }}={\tau }_{\tilde{g}}F$ and $G^{\mathrm{\prime }}={\tau }_{\tilde{g}}G$. In each pairing, the spin factors are $S^{-1}$ and $S^{-†}$; changing variables $x=a+\mathrm{\Lambda }{x}^{\mathrm{\prime }}$, $y=a+\mathrm{\Lambda }{y}^{\mathrm{\prime }}$ has unit Jacobian, cancels the translation in $x-y$, and the kernel identity of step 1.1 cancels both spin factors. Hence both polarized one-particle pairings are preserved, and the maps $P_{\pm }{u}_F\mapsto P_{\pm }{u}_{F^{\mathrm{\prime }}}$ are well-defined isometries on their test-generated subspaces. Applying the inverse group element proves they are onto after completion. For the conjugate negative-energy sector, conjugating the same preserved pairing and reversing its arguments gives the corresponding isometry on $\overline{{\mathcal{H}}_{-}}$.

The test-generated subspaces are dense. For $u\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3\setminus \{0\};{\mathbb{C}}^4)$, $P_{\pm }u$ is smooth and compactly supported away from the massless apex; such projected vectors are dense in ${\mathcal{H}}_{\pm }$ by [F1] and [F6], since removing the single point does not change an $L^2$ class. Choose $g\in C_c^{\mathrm{\infty }}(\mathbb{R})$ with $\int g(t)dt=1$ and set $\widehat{F}(t,\mathbf{k})=g(t)W_D(-t,\mathbf{k})u(\mathbf{k})$. This is smooth and compactly supported in $(t,\mathbf{k})$; its inverse spatial Fourier transform is Schwartz by repeated integration by parts. The definition in [F4] then gives $u_F=u$, and hence $P_{\pm }{u}_F=P_{\pm }u$.

By steps 2.1 and 2.2 the particle and conjugate antiparticle test maps extend to unitaries on ${\mathcal{H}}_{+}$ and $\overline{{\mathcal{H}}_{-}}$. Their direct sum $T(\tilde{g})$ is unitary. Define $U(\tilde{g})={⨁}_{n\ge 0}T(\tilde{g}{)}^{{\otimes }_{a}n}$ on the antisymmetric Fock sectors. Each tensor power is unitary by its isometric action on elementary tensors and inverse tensor power; it commutes with the signed permutation average and therefore restricts to a unitary on the antisymmetric sector. The direct sum is unitary and fixes $\mathrm{\Omega }$. On elementary wedge tensors the insertion/contraction formulas of [F5] give $U{c}^{†}(h)U^{-1}=c^{†}(Th)$ and $Uc(h)U^{-1}=c(Th)$; density extends these identities to the finite-particle domain. Thus $U(\tilde{g})\mathrm{\Psi }(F)U(\tilde{g}{)}^{-1}=\mathrm{\Psi }({\tau }_{\tilde{g}}F)$. Direct substitution in the affine and spinor formulas gives ${\tau }_{{\tilde{g}}_2}{\tau }_{{\tilde{g}}_1}={\tau }_{{\tilde{g}}_2{\tilde{g}}_1}$: the coordinate argument is $({\mathrm{\Lambda }}_2{\mathrm{\Lambda }}_1{)}^{-1}(x-a_2-{\mathrm{\Lambda }}_2{a}_1)$ and the matrix factor is $(S_2{S}_1{)}^{-†}=S_2^{-†}{S}_1^{-†}$. By [F8] the Lorentz factors compose within $S{O}^{+}(1,3)$; by the definition of $\tilde{G}$ its triples are closed under this product. Hence $T$ and its Fock lift $U$ obey the same product law and $U$ is a representation.

The map $\tilde{g}\mapsto {\tau }_{\tilde{g}}F$ is continuous in Schwartz seminorms for each $F$, since it is composition with affine maps and finite matrices whose entries vary continuously. By the continuous test estimate of [F4], the one-particle representation is continuous on the dense test-generated subspaces of step 2.2; its operators have norm one, so approximation extends strong continuity to all one-particle vectors. On an elementary $n$-tensor, the difference of $T(g{)}^{\otimes n}$ and $T(g_0{)}^{\otimes n}$ is a sum of $n$ terms, each bounded by the norm of one factor difference times the norms of unchanged factors. It tends to zero as $g\to g_0$. Density and the unit norm bound extend this to each completed antisymmetric sector; finite-sector truncation then extends strong continuity to their direct sum.

By step 2.2 the ranges of $F\mapsto P_{+}{u}_F$ and $F\mapsto \overline{P_{-}{u}_F}$ are dense in ${\mathcal{H}}_{+}$ and $\overline{{\mathcal{H}}_{-}}$, respectively. Moreover, $\mathrm{\Psi }(F{)}^{†}\mathrm{\Omega }=c^{†}(P_{+}{u}_F)\mathrm{\Omega }$ and $\mathrm{\Psi }(F)\mathrm{\Omega }=c^{†}(\overline{P_{-}{u}_F})\mathrm{\Omega }$, so their created one-particle components range densely over both summands. For any ordered list of $n$ such fields or adjoints, the all-creation term in their product on $\mathrm{\Omega }$ is the wedge of their $n$ creation components. Expand the product into creation and annihilation terms and use the CAR to move every annihilator to the right: a term with a remaining annihilator kills $\mathrm{\Omega }$, while each contraction leaves a product with at most $n-2$ creation factors. By induction, those lower-degree wedge vectors are in the polynomial orbit span, so subtracting them isolates the desired degree-$n$ wedge. The degree-zero base also lies in the orbit span: step 2.2 gives a test $F$ with $u_F\mathrm{\ne }0$, and the CAR formula of [F4] gives $\{\mathrm{\Psi }(F),\mathrm{\Psi }(F{)}^{†}\}=\mathrm{\parallel }{u}_F{\mathrm{\parallel }}^{2}I$, hence $I$ belongs to the generated algebra. Wedges of vectors from the dense one-particle ranges are dense in each antisymmetric sector by the Hilbert tensor completion, and finite-sector truncations are dense in Fock space. Hence $\mathrm{\Omega }$ is cyclic for the global graded field algebra.

For a pure translation, $\widehat{F(\cdot -a)}(t,\mathbf{k})=\chi (-\mathbf{k}\cdot \mathbf{a})\widehat{F}(t-a^0,\mathbf{k})$, so $u_{F(\cdot -a)}=W_D(a^0,\mathbf{k})\chi (-\mathbf{k}\cdot \mathbf{a})u_F$. On ${\mathcal{H}}_{+}$ this is the phase $\chi (\omega a^0-\mathbf{k}\cdot \mathbf{a})$; after conjugation on $\overline{{\mathcal{H}}_{-}}$ it is $\chi (\omega a^0+\mathbf{k}\cdot \mathbf{a})$. Thus the induced Fock translations are precisely those of [F7], whose joint spectrum is in the forward cone, whose zero-energy eigenspace is $\mathbb{C}\mathrm{\Omega }$, and whose positive-$\mu$ gap is at least $\mu$. By [F9], $S_R(2\pi )=-I_4$ and $S_R(4\pi )=I_4$; hence a $2\pi$ rotation acts as $-I$ on each one-particle summand and $(-1{)}^n$ on sector $n$, while a $4\pi$ rotation is the identity. The first-quantized $h_D$ has eigenvalues $\pm \omega$, while the Fock translation time generator in [F7] multiplies both particle and antiparticle excitations by $+\omega$; their domains and Hilbert spaces differ.