Dirac matrix identities and orthogonal energy projections

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.