Dirac Clifford matrices and energy projections

Let $u=\mathrm{\Lambda }{e}_0=(u^0,\mathbf{u})$. By [F1]–[F2], $u^0>0$ and $-(u^0{)}^2+\mathrm{\mid }\mathbf{u}{\mathrm{\mid }}^2=-1$, so $u^0=\sqrt{1+\mathrm{\mid }\mathbf{u}{\mathrm{\mid }}^2}$. If $\mathbf{u}=0$, then $u=e_0$; choose any unit spatial vector ${\mathbf{n}}_B$ and $\chi =0$, so $B_{{\mathbf{n}}_B}(0)=I$. Otherwise ${\mathbf{n}}_B=\mathbf{u}\mathrm{/}\mathrm{\mid }\mathbf{u}\mathrm{\mid }$ is a unit vector and $u^0>1$. By the range and strict increase in [F3], there is a unique $\chi >0$ with $\cosh \chi =u^0$; then $\sinh \chi >0$ and ${\sinh }^2\chi =(u^0{)}^2-1=\mathrm{\mid }\mathbf{u}{\mathrm{\mid }}^2$, so $\sinh \chi =\mathrm{\mid }\mathbf{u}\mathrm{\mid }$. Define $B_{{\mathbf{n}}_B}(\chi )$ by the block $\left(\begin{array}{cc}\cosh \chi & \sinh \chi \\ \sinh \chi & \cosh \chi \end{array}\right)$ on $\mathrm{span}(e_0,{\mathbf{n}}_B)$ and the identity on its Euclidean orthogonal complement. The hyperbolic identity in [F3] verifies metric preservation and the block determinant is $1$. For a future timelike $v=(v^0,\mathbf{v})$, the signature gives $v^0>\mathrm{\mid }\mathbf{v}\mathrm{\mid }$, and [F5] gives $\mathrm{\mid }\mathbf{v}\cdot {\mathbf{n}}_B\mathrm{\mid }\le \mathrm{\mid }\mathbf{v}\mathrm{\mid }$. Hence $v^{\mathrm{\prime }0}=\cosh \chi v^0+\sinh \chi (\mathbf{v}\cdot {\mathbf{n}}_B)\ge (\cosh \chi -\sinh \chi )v^0>0$ by [F6]. Thus $B\in S{O}^{+}(1,3)$ and $B{e}_0=u$.

Set $R=B^{-1}\mathrm{\Lambda }$. Since both $B$ by step 1.1 and $\mathrm{\Lambda }$ by hypothesis belong to the proper orthochronous Lorentz group, its inverse and composition closure in [F2] give $R\in S{O}^{+}(1,3)$; also $R{e}_0=e_0$. If $v$ is spatial, then ${\eta }_0(Rv,e_0)={\eta }_0(Rv,R{e}_0)={\eta }_0(v,e_0)=0$, so $R$ preserves the spatial orthogonal complement. Its restriction $Q$ there is an orthogonal $3\times 3$ matrix with determinant $+1$ because $\det R=1$ and its time block is $1$. Hence $Q\in SO(3)$.

The real polynomial $p(\lambda )=\det (\lambda I_3-Q)$ is continuous (it is a finite sum of products of affine functions) and has opposite signs for sufficiently large positive and negative $\lambda$, since its leading term is ${\lambda }^3$. The intermediate value theorem [F3] gives a real eigenvalue $\lambda$. Orthogonality implies $\mathrm{\mid }\lambda \mathrm{\mid }=1$, so $\lambda =\pm 1$. If $-1$ is the only real eigenvalue and the other two eigenvalues are nonreal, they are a conjugate pair whose product is positive; then $\det Q=-1$, contrary to $\det Q=1$. If all eigenvalues are real, their product $1$ means the number of $-1$ eigenvalues is even, so at least one eigenvalue is $+1$. Thus $Q$ has a unit fixed vector ${\mathbf{n}}_R$. Choose a standard spatial basis vector $e_j$ not parallel to ${\mathbf{n}}_R$ (one exists because the spatial basis spans ${\mathbb{R}}^3$), and set $f_1=(e_j-(e_j\cdot {\mathbf{n}}_R){\mathbf{n}}_R)\mathrm{/}\mathrm{\mid }{e}_j-(e_j\cdot {\mathbf{n}}_R){\mathbf{n}}_R\mathrm{\mid }$ and $f_2={\mathbf{n}}_R\times f_1$. By [F7], $f_1,f_2$ are an orthonormal basis of ${\mathbf{n}}_R^{\perp }$. The restriction of $Q$ to this plane is orthogonal with determinant $+1$, so in this basis its matrix has form $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$ with $a^2+b^2=1$. By [F4] there is $\theta$ with $a=\cos \theta$, $b=\sin \theta$. Therefore $Q$ is the rotation $R_{{\mathbf{n}}_R}(\theta )$ about ${\mathbf{n}}_R$.

The spatial restriction of $R$ is the rotation from step 3.1, so $R=R_{{\mathbf{n}}_R}(\theta )$ on all of ${\mathbb{R}}^4$ because $R{e}_0=e_0$. Since $\mathrm{\Lambda }=BR$, this is the required boost-rotation factorization.

Direct multiplication gives ${\sigma }_j^{†}={\sigma }_j$, ${\sigma }_j^2=I_2$, ${\sigma }_1{\sigma }_2=i{\sigma }_3$, ${\sigma }_2{\sigma }_3=i{\sigma }_1$, and ${\sigma }_3{\sigma }_1=i{\sigma }_2$; reversing each distinct pair changes the sign. Thus ${\sigma }_i{\sigma }_j+{\sigma }_j{\sigma }_i=2{\delta }_{ij}{I}_2$. The block formulas then give ${\alpha }_i^{†}={\alpha }_i$, ${\beta }^{†}=\beta$, ${\alpha }_i{\alpha }_j+{\alpha }_j{\alpha }_i=2{\delta }_{ij}{I}_4$, and ${\alpha }_i\beta +\beta {\alpha }_i=0$. Consequently $h_D^{†}=h_D$ and $h_D^2=(\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2+{\mu }^2)I_4={\omega }^2{I}_4$.

Where $\omega >0$, the self-adjoint matrix $h_D\mathrm{/}\omega$ squares to $I_4$. Hence $P_{\pm }^{†}=P_{\pm }$, $P_{\pm }^2=P_{\pm }$, $P_{+}{P}_{-}=0$, and $P_{+}+P_{-}=I_4$. Since $\mathrm{tr}(h_D)=0$, $\mathrm{tr}(P_{\pm })=2$; a finite-dimensional orthogonal projection has eigenvalues only $0,1$, so its trace equals its rank and each rank is two. The chosen values at the massless apex have the same projection and rank properties. The matrix fields are measurable and have operator norm at most one, so their multiplication operators are bounded everywhere-defined projections; their ranges are closed and give ${\mathcal{H}}_D={\mathcal{H}}_{+}\oplus {\mathcal{H}}_{-}$.

The block products also give ${\gamma }^0=\beta$, ${\gamma }^{j†}=-{\gamma }^j$, $\{{\gamma }^0,{\gamma }^j\}=0$, and $\{{\gamma }^i,{\gamma }^j\}=-2{\delta }_{ij}{I}_4$. With $\eta =\mathrm{diag}(-1,1,1,1)$ these are exactly $\{{\gamma }^a,{\gamma }^b\}=-2{\eta }^{ab}{I}_4$. Moreover $i{\gamma }^a{\mathrm{\partial }}_a-\mu =0$ is equivalent, after multiplication by the invertible $\beta$, to $i{\mathrm{\partial }}_0=(-i{\sum }_j{\alpha }_j{\mathrm{\partial }}_j+\beta \mu )$; for the Fourier convention $\widehat{F}(\mathbf{k})=\int e^{-i\mathbf{k}\cdot \mathbf{x}}F(\mathbf{x})d^3\mathbf{x}$, the right side has multiplier $h_D(\mathbf{k})$.

Put $k_{\parallel }=\mathbf{n}\cdot \mathbf{k}$ and ${\mathbf{k}}_{\perp }=\mathbf{k}-k_{\parallel }\mathbf{n}$, so ${\mathbf{k}}_{\perp }\perp \mathbf{n}$. The Pauli products in step 1.1 give ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{\sum }_k{ϵ}_{ijk}{\sigma }_k$, where $ϵ$ is the alternating symbol with ${ϵ}_{123}=1$; by [F6] this implies ${\mathrm{\Sigma }}_{\mathbf{n}}{\alpha }_{\mathbf{n}}={\alpha }_{\mathbf{n}}{\mathrm{\Sigma }}_{\mathbf{n}}$ and ${\mathrm{\Sigma }}_{\mathbf{n}}(\mathit{\alpha }\cdot {\mathbf{k}}_{\perp })=i\mathit{\alpha }\cdot (\mathbf{n}\times {\mathbf{k}}_{\perp })$. Expanding with $S_R=cI-is{\mathrm{\Sigma }}_{\mathbf{n}}$ and $S_R^{-1}=cI+is{\mathrm{\Sigma }}_{\mathbf{n}}$ gives $S_R{\alpha }_{\mathbf{n}}{S}_R^{-1}={\alpha }_{\mathbf{n}}$ and $S_R(\mathit{\alpha }\cdot {\mathbf{k}}_{\perp })S_R^{-1}=(c^2-s^2)\mathit{\alpha }\cdot {\mathbf{k}}_{\perp }+2cs\mathit{\alpha }\cdot (\mathbf{n}\times {\mathbf{k}}_{\perp })=\mathit{\alpha }\cdot (\cos \theta {\mathbf{k}}_{\perp }+\sin \theta \mathbf{n}\times {\mathbf{k}}_{\perp })$. This is exactly the rotation action in the Statement. Since ${\mathrm{\Sigma }}_{\mathbf{n}}$ is Hermitian with square $I$, $S_R$ is unitary; it commutes with $\beta$, so $S_R\beta S_R^{-1}=S_R\beta S_R^{†}=\beta$. Therefore $S_R\mathrm{\Gamma }(p)S_R^{-1}=\mathrm{\Gamma }(R_{\mathbf{n}}(\theta )p)$.

Put ${\alpha }_{\mathbf{n}}=\mathit{\alpha }\cdot \mathbf{n}$ and, for each spatial $\mathbf{v}$, ${\gamma }_{\mathbf{v}}=\beta (\mathit{\alpha }\cdot \mathbf{v})$. Then ${\alpha }_{\mathbf{n}}^2=I$, $\{{\alpha }_{\mathbf{n}},\beta \}=0$, $[{\alpha }_{\mathbf{n}},{\gamma }_{\mathbf{m}}]=0$ for $\mathbf{m}\perp \mathbf{n}$, and $[{\alpha }_{\mathbf{n}},\beta ]=-2{\gamma }_{\mathbf{n}}$, $[{\alpha }_{\mathbf{n}},{\gamma }_{\mathbf{n}}]=-2\beta$. Expanding $S_B=\cosh (\chi \mathrm{/}2)I+\sinh (\chi \mathrm{/}2){\alpha }_{\mathbf{n}}$ and its inverse gives $S_B\beta S_B^{-1}=\beta \cosh \chi -{\gamma }_{\mathbf{n}}\sinh \chi$ and $S_B{\gamma }_{\mathbf{n}}{S}_B^{-1}={\gamma }_{\mathbf{n}}\cosh \chi -\beta \sinh \chi$, while perpendicular $\gamma$ components are unchanged. Therefore $S_B\mathrm{\Gamma }(p)S_B^{-1}=\beta (p^0\cosh \chi +p_{\parallel }\sinh \chi )-{\gamma }_{\mathbf{n}}(p^0\sinh \chi +p_{\parallel }\cosh \chi )-{\gamma }_{{\mathbf{p}}_{\perp }}=\mathrm{\Gamma }(B_{\mathbf{n}}(\chi )p)$. Since $\beta S_B=S_B^{-1}\beta$ and $S_B^{†}=S_B$, $S_B\beta S_B^{†}=\beta$.

Let $\mathcal{R}$ be the countable real rational-box family from [F3]. For $u\in L^2({\mathbb{R}}^3;\mathbb{C})$, its real and imaginary parts are in real-valued $L^2$, since $\mathrm{\mid }\mathrm{Re}u\mathrm{\mid },\mathrm{\mid }\mathrm{Im}u\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }$, and $\mathrm{\parallel }u-(f+ig){\mathrm{\parallel }}_{L^2(\mathbb{C})}^2=\mathrm{\parallel }\mathrm{Re}u-f{\mathrm{\parallel }}_{L^2(\mathbb{R})}^2+\mathrm{\parallel }\mathrm{Im}u-g{\mathrm{\parallel }}_{L^2(\mathbb{R})}^2\mathrm{.}$ Approximating these two parts by $f,g\in \mathcal{R}$ shows that $\mathcal{R}+i\mathcal{R}$ is dense in complex scalar $L^2$; it is countable by enumerating pairs of members of $\mathcal{R}$ in increasing order of the sum of their indices. Let ${\mathcal{V}}_D=(\mathcal{R}+i\mathcal{R}{)}^4$. Repeated pairing shows it is countable, and componentwise density shows it is dense for the norm of ${\mathcal{H}}_D$; hence ${\mathcal{H}}_D$ is separable. A closed subspace of a separable Hilbert space is separable: intersect a countable base of open balls with the subspace and choose one point from each nonempty intersection; Countable Choice supplies these choices. Thus ${\mathcal{H}}_{+}$ and ${\mathcal{H}}_{-}$ are separable.

Write $T_{+}$ and $T_{-}$ for multiplication by $+\omega$ on ${\mathcal{H}}_{+}$ and by $-\omega$ on ${\mathcal{H}}_{-}$, each with domain $D(T_{\pm })=\{u\in {\mathcal{H}}_{\pm }:\omega u\in {\mathcal{H}}_{\pm }\}$. Since $P_{\pm }$ are bounded projections and ${\mathcal{V}}_D$ from step 3.1 is dense in ${\mathcal{H}}_D$, the sets $P_{\pm }{\mathcal{V}}_D$ are dense in ${\mathcal{H}}_{\pm }$. Each element of ${\mathcal{V}}_D$ is supported on a bounded finite union of rational boxes; multiplication by $P_{\pm }$ preserves that support, and $\omega$ is bounded there. Thus $P_{\pm }{\mathcal{V}}_D\subseteq D(T_{\pm })$, so both maximal domains are dense. Pointwise, [step 2.1] gives $h_D=+\omega I$ on ${\mathcal{H}}_{+}$ and $h_D=-\omega I$ on ${\mathcal{H}}_{-}$. By [step 1.1], $\mathrm{\mid }{h}_{D}u\mathrm{\mid }=\omega \mathrm{\mid }u\mathrm{\mid }$ almost everywhere, so the domain in Dirac Clifford matrices and energy projections is exactly $\{u:\omega u\in {\mathcal{H}}_D\}$. For either sign, $T_{\pm }$ is symmetric because its multiplier is real. If $g\in D(T_{\pm }^{\ast })$ has adjoint value $q$, then for each integer $N\ge 1$ the vector $v_N=(\pm \omega g-q){\mathbf{1}}_{\{\omega \le N\}}$ lies in ${\mathcal{H}}_{\pm }$ and in $D(T_{\pm })$; the cutoff preserves the projected subspace and makes the multiplier bounded. The adjoint identity gives $0=⟨T_{\pm }{v}_N,g⟩-⟨v_N,q⟩={\int }_{\{\omega \le N\}}\mathrm{\mid }\pm \omega g-q{\mathrm{\mid }}^2{d}^3\mathbf{k}\mathrm{/}(2\pi {)}^3$. Since $\omega$ is finite everywhere, these sets increase to ${\mathbb{R}}^3$, so $q=\pm \omega g$ almost everywhere and $\omega g\in {\mathcal{H}}_{\pm }$. Conversely every vector in the maximal domain satisfies the adjoint identity because the multiplier is real; hence $T_{\pm }=T_{\pm }^{\ast }$. Testing the adjoint of $T_{+}\oplus T_{-}$ on vectors supported in each summand shows $(T_{+}\oplus T_{-}{)}^{\ast }=T_{+}^{\ast }\oplus T_{-}^{\ast }$ with domain $D(T_{+}^{\ast })\oplus D(T_{-}^{\ast })$. Therefore $h_D=T_{+}\oplus T_{-}$ is self-adjoint on $\{u:\omega u\in {\mathcal{H}}_D\}$.

For $\omega >0$, write $W_D(t)=\cos (\omega t)I+i\sin (\omega t)h_D\mathrm{/}\omega$. Using $h_D^2={\omega }^{2}I$ and $h_D^{†}=h_D$ from step 1.1, the product $W_D(t{)}^{†}{W}_D(t)=({\cos }^2(\omega t)+{\sin }^2(\omega t))I=I$. The sine and cosine addition formulas give $W_D(t+s)=W_D(t)W_D(s)$, and differentiating gives ${\mathrm{\partial }}_t{W}_D=i{h}_D{W}_D$. At the massless apex the assigned value is the identity and does not affect the multiplication operator.

For every integer $N\ge 2$, repeated integration by parts in $\mathbf{x}$ gives $(1+\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2{)}^N\widehat{F}(t,\mathbf{k})=\widehat{(1-{\mathrm{\Delta }}_{\mathbf{x}}{)}^{N}F}(t,\mathbf{k})$. Since $F$ is Schwartz, the right side is bounded by a finite Schwartz seminorm times $(1+\mathrm{\mid }t\mathrm{\mid }{)}^{-2}$. Thus $\mathrm{\mid }\widehat{F}(t,\mathbf{k})\mathrm{\mid }\le C_N(F)(1+\mathrm{\mid }t\mathrm{\mid }{)}^{-2}(1+\mathrm{\mid }\mathbf{k}\mathrm{\mid }{)}^{-2N}$. The matrix multiplier $W_D(t,\mathbf{k})$ is pointwise unitary by [F2]; integrating this bound in $t$ yields $\mathrm{\mid }{u}_F(\mathbf{k})\mathrm{\mid }\le C_N^{\mathrm{\prime }}(F)(1+\mathrm{\mid }\mathbf{k}\mathrm{\mid }{)}^{-2N}$. This is square-integrable in three dimensions for $N\ge 2$, with norm bounded by the displayed Schwartz seminorm, so $u_F\in {\mathcal{H}}_D$ continuously.

The equal-time CAR expansion gives $\{{\mathrm{\Psi }}_t(u),{\mathrm{\Psi }}_t(v{)}^{†}\}=\int \widehat{u}(\mathbf{k}{)}^{†}(P_{+}+P_{-})\widehat{v}(\mathbf{k})d^3\mathbf{k}\mathrm{/}(2\pi {)}^3$. By [F2] the sum is $I_4$, and [F4] turns the integral into $\int u^{†}v{d}^3\mathbf{x}$. The same-type CAR vanish by [F3].

The first field term is antilinear in $F$ because $u_F$ is linear and $c$ is conjugate-linear; the second is antilinear because complex conjugation into $\overline{{\mathcal{H}}_{-}}$ is conjugate-linear and $c^{†}$ is linear. Expanding the anticommutators with [F3], the only nonzero terms in $\{\mathrm{\Psi }(F),\mathrm{\Psi }(G{)}^{†}\}$ are the particle pairing $⟨P_{+}{u}_F,P_{+}{u}_G⟩$ and the conjugate negative-space pairing $⟨\overline{P_{-}{u}_G},\overline{P_{-}{u}_F}⟩=⟨P_{-}{u}_F,P_{-}{u}_G⟩$. Orthogonality and $P_{+}+P_{-}=I$ make their sum $⟨u_F,u_G⟩$. The same-type anticommutator is zero because same-type CAR vanish within each species and the two species are orthogonal. Boundedness follows from $\mathrm{\parallel }\mathrm{\Psi }(F)\mathrm{\parallel }\le \mathrm{\parallel }{P}_{+}{u}_F\mathrm{\parallel }+\mathrm{\parallel }{P}_{-}{u}_F\mathrm{\parallel }\le \sqrt{2}\mathrm{\parallel }{u}_F\mathrm{\parallel }$.

The spatial Fourier transform of $(i{\mathrm{\partial }}_0-h_D(-i\mathrm{\nabla }))F$ is $i{\mathrm{\partial }}_t\widehat{F}-h_D(\mathbf{k})\widehat{F}$. Since $F$ is Schwartz, $W_D(t,\mathbf{k})\widehat{F}(t,\mathbf{k})$ tends to zero as $t\to \pm \mathrm{\infty }$. Integration by parts and ${\mathrm{\partial }}_t{W}_D=i{h}_D{W}_D$ give $\int W_D(t,\mathbf{k})i{\mathrm{\partial }}_t\widehat{F}dt=\int h_D{W}_D(t,\mathbf{k})\widehat{F}dt$, which cancels the second term in the definition of $u$. Hence $u_{(i{\mathrm{\partial }}_0-h_D(-i\mathrm{\nabla }))F}=0$. The matrix identity in [F2] identifies this with the stated tested Dirac equation.

The unsmeared CAR expansion gives the two separate kernels $C_{+}$ and $C_{-}$ displayed in the Statement: the positive term has coefficient $P_{+}$ and phase $e^{-i\omega \tau +i\mathbf{k}\cdot \mathbf{x}}$, while the negative term has coefficient $P_{-}$ and phase $e^{+i\omega \tau +i\mathbf{k}\cdot \mathbf{x}}$. Their sum is $\int (P_{+}(\cos (\omega \tau )-i\sin (\omega \tau ))+P_{-}(\cos (\omega \tau )+i\sin (\omega \tau )))e^{i\mathbf{k}\cdot \mathbf{x}}{d}^3\mathbf{k}\mathrm{/}(2\pi {)}^3=W_D(-\tau ,\mathbf{k})$ in Fourier form. Since $h_D^2={\omega }^2{I}_4$, this equals ${\mathrm{\partial }}_{\tau }{\mathrm{\Delta }}_{\mu }{I}_4-i{h}_D(-i\mathrm{\nabla }){\mathrm{\Delta }}_{\mu }$, using ${\mathrm{\partial }}_{\tau }(\sin \omega \tau \mathrm{/}\omega )=\cos \omega \tau$. The equality is distributional: the shell weights grow at most polynomially and the test Fourier transforms decay rapidly, so the pairings converge absolutely and Fubini [F7] applies. Differential operators do not enlarge distributional support; [F5] therefore gives $\mathrm{supp}R\subseteq \{\mathrm{\mid }\mathbf{x}\mathrm{\mid }\le \mathrm{\mid }\tau \mathrm{\mid }\}$.

If compact supports $K,L$ are spacelike separated, [F6] places the compact difference set $K-L$ outside $\mathrm{supp}R$. Pairing $R(x-y)$ with the product of any two supported spinor tests therefore gives $\{\mathrm{\Psi }(F),\mathrm{\Psi }(G{)}^{†}\}=0$. The field-field anticommutator vanishes for all tests independently: its same-type CAR terms vanish by [F3], and its mixed terms pair the orthogonal summands ${\mathcal{H}}_{+}$ and $\overline{{\mathcal{H}}_{-}}$ of $\mathcal{K}$, so their CAR scalar products are zero. Taking adjoints and interchanging $F,G$ gives the remaining cross anticommutators. Thus all field generators in the two spacelike-separated regions anticommute pairwise. Moving each of $r$ factors past each of $s$ factors contributes $(-1{)}^{rs}$; when both monomials have even degree this sign is $+1$. Finite sums of even monomials therefore commute, and norm limits preserve commutation because the generating fields are bounded.