Periodic neutral classical Coulomb constraint model

In the adopted action, ${\rho }_q:=q_{\mathrm{S}\mathrm{I}}{\psi }^{†}\psi$ is the coefficient of the scalar-potential coupling. For the Dirac part, $x^0=ct$ and $A_0=-\phi \mathrm{/}c$ give $D_0\psi =c^{-1}(\dot{\psi }+i{q}_{\mathrm{S}\mathrm{I}}\phi \psi \mathrm{/}\mathrm{\hslash })$ and $D_0\bar{\psi }=c^{-1}(\dot{\bar{\psi }}-i{q}_{\mathrm{S}\mathrm{I}}\phi \bar{\psi }\mathrm{/}\mathrm{\hslash })$. Using [F5], the time part of $\mathrm{\hslash }c{\mathcal{L}}_{\mathrm{n}\mathrm{a}\mathrm{t}}$ is the canonical term $i\mathrm{\hslash }({\psi }^{†}\dot{\psi }-{\dot{\psi }}^{†}\psi )\mathrm{/}2$ minus $q_{\mathrm{S}\mathrm{I}}\phi {\psi }^{†}\psi$. The symmetric spatial derivative, integrated by parts on the periodic torus, is $i\mathrm{\hslash }c{\psi }^{†}\mathit{\alpha }\cdot \mathrm{\nabla }\psi$: the integral of $\mathrm{\nabla }\cdot ({\psi }^{†}\mathit{\alpha }\psi )$ vanishes because opposite periodic faces cancel. The spatial vector-coupling term in the Lagrangian is $q_{\mathrm{S}\mathrm{I}}c{\psi }^{†}\mathit{\alpha }\cdot \mathbf{A}\psi$, and the mass term is $-m_{\mathrm{S}\mathrm{I}}{c}^2{\psi }^{†}\beta \psi$. The first-order canonical momenta are $i\mathrm{\hslash }{\psi }^{†}\mathrm{/}2$ and $-i\mathrm{\hslash }\psi \mathrm{/}2$, so their pairing with the time derivatives cancels the canonical term. Negating the remaining Dirac Lagrangian terms in the Legendre transform gives the matter Hamiltonian density ${\psi }^{†}[c\mathit{\alpha }\cdot (-i\mathrm{\hslash }\mathrm{\nabla }-q_{\mathrm{S}\mathrm{I}}\mathbf{A})+\beta m_{\mathrm{S}\mathrm{I}}{c}^2+q_{\mathrm{S}\mathrm{I}}\phi ]\psi$, whose scalar term is $\phi {\rho }_q$. On stationary fields this source agrees with [F4].

For the finite-mode branch [A4], interpret the local density in [A1] on the torus by its periodic lift and integrate over one spatial fundamental cell. Since $x^0=ct$, the physical Lagrangian is $L(t)=\mathrm{\hslash }c{\int }_{{\mathbb{T}}_L^3}{\mathcal{L}}_{\mathrm{n}\mathrm{a}\mathrm{t}}{d}^3\mathbf{x}$. Using $\bar{\psi }{\gamma }^0={\psi }^{†}$, the part containing time derivatives of the spinor is $\frac{i\mathrm{\hslash }}{2}{\int }_{{\mathbb{T}}_L^3}({\psi }_N^{†}{\dot{\psi }}_N-{\dot{\psi }}_N^{†}{\psi }_N)d^3\mathbf{x}\mathrm{.}$ The $A_0$ coupling and all spatial and mass terms contain no ${\dot{z}}_j$. Substitution of ${\psi }_N={\sum }_j{z}_j{u}_j$ and the stated orthonormality give exactly $L_{\mathrm{k}\mathrm{i}\mathrm{n}}=\frac{i\mathrm{\hslash }}{2}{\sum }_j({\bar{z}}_j{\dot{z}}_j-{\dot{\bar{z}}}_j{z}_j)$. In the canonical normalization [F8], each $u_j$ has unit ${\mathrm{m}}^{-3\mathrm{/}2}$ and each $z_j$ is dimensionless; since $[\mathrm{\hslash }]=\mathrm{J}\mathrm{s}$ and $[{\dot{z}}_j]={\mathrm{s}}^{-1}$, the displayed $L_{\mathrm{k}\mathrm{i}\mathrm{n}}$ has unit joule. This restriction supplies the finite-mode kinetic term but does not assert that the selected mode subspace is invariant under the field equations.

In the real canonical coordinates of the Statement, $\{x_j,P_k^x\}=\{y_j,P_k^y\}={\delta }_{jk}$ and the mixed coordinate-momentum brackets vanish. Substitution of $z=x+iy$ and $p=(P^x-i{P}^y)\mathrm{/}2$ therefore gives the displayed complex canonical brackets. Writing $L_{\mathrm{k}\mathrm{i}\mathrm{n}}=\mathrm{\hslash }{\sum }_j(y_j{\dot{x}}_j-x_j{\dot{y}}_j)$ gives real canonical momenta $P_j^x=\mathrm{\hslash }{y}_j$ and $P_j^y=-\mathrm{\hslash }{x}_j$, hence $p_j=i\mathrm{\hslash }{\bar{z}}_j\mathrm{/}2$ and ${\bar{p}}_j=-i\mathrm{\hslash }{z}_j\mathrm{/}2$. These are precisely the primary constraints in the Statement. The verified finite-dimensional hypotheses of [L2] then give its displayed constraint matrix and inverse, the second-class conclusion, and reduced bracket $\{z_j,{\bar{z}}_k{\}}_D=-i{\delta }_{jk}\mathrm{/}\mathrm{\hslash }$. This is a commuting classical coordinate calculation and does not derive fermionic CAR.

From [F1]–[F3], $S_0^2={\mu }_0\mathrm{\hslash }c$ and $F_{\mathrm{S}\mathrm{I}}=S_{0}f$, so the physical Maxwell Lagrangian density is $-F_{ab}^{\mathrm{S}\mathrm{I}}{F}_{\mathrm{S}\mathrm{I}}^{ab}\mathrm{/}(4{\mu }_0)$. The tensor conventions [F2] give $F_{ab}^{\mathrm{S}\mathrm{I}}{F}_{\mathrm{S}\mathrm{I}}^{ab}=2(\mathrm{\mid }\mathbf{B}{\mathrm{\mid }}^2-\mathrm{\mid }\mathbf{E}{\mathrm{\mid }}^2\mathrm{/}{c}^2)$; using ${\epsilon }_0=({\mu }_0{c}^2{)}^{-1}$ from [F1], this becomes ${\mathcal{L}}_M={\epsilon }_0\mathrm{\mid }\dot{\mathbf{A}}+\mathrm{\nabla }\phi {\mathrm{\mid }}^2\mathrm{/}2-\mathrm{\mid }\mathbf{B}{\mathrm{\mid }}^2\mathrm{/}(2{\mu }_0)$. Differentiating with respect to $\dot{\mathbf{A}}$ gives ${\mathit{\pi }}_A={\epsilon }_0(\dot{\mathbf{A}}+\mathrm{\nabla }\phi )=-{\epsilon }_0\mathbf{E}$; there is no $\dot{\phi }$ term, so ${\pi }_{\phi }=0$. Since $\dot{\mathbf{A}}={\mathit{\pi }}_A\mathrm{/}{\epsilon }_0-\mathrm{\nabla }\phi$, the Maxwell Legendre transform is $\mathrm{\mid }{\mathit{\pi }}_A{\mathrm{\mid }}^2\mathrm{/}(2{\epsilon }_0)+\mathrm{\mid }\mathbf{B}{\mathrm{\mid }}^2\mathrm{/}(2{\mu }_0)-{\mathit{\pi }}_A\cdot \mathrm{\nabla }\phi$. Integrating the last term by parts on the torus gives $+\phi \mathrm{\nabla }\cdot {\mathit{\pi }}_A$; opposite periodic faces cancel. Adding the matter Hamiltonian from step 1.1 gives exactly the Hamiltonian in the Statement, including $\phi (\mathrm{\nabla }\cdot {\mathit{\pi }}_A+{\rho }_q)$.

Varying the unrestricted multiplier $\phi$ in step 2.1 imposes $\mathrm{\nabla }\cdot {\mathit{\pi }}_A+{\rho }_q=0$. Substituting ${\mathit{\pi }}_A=-{\epsilon }_0\mathbf{E}$ gives ${\epsilon }_0\mathrm{\nabla }\cdot \mathbf{E}={\rho }_q$. In Coulomb gauge, $\mathrm{\nabla }\cdot \mathbf{E}=-\mathrm{\Delta }\phi$ because $\mathrm{\nabla }\cdot \dot{\mathbf{A}}=0$, so the constraint becomes $-\mathrm{\Delta }\phi ={\rho }_q\mathrm{/}{\epsilon }_0$. This is the Gauss sign in [F3]. The multiplier equation holds for every smooth periodic source; neutrality is needed only for a periodic solution.

Under the neutral-branch hypothesis in the Statement, the smoothness, periodicity, and zero-mean hypotheses of [L1] hold at each $t\in I$. It supplies the unique zero-mean Fourier solution and energy identity. Each summand is nonnegative because ${\epsilon }_0>0$ by [F1] and $\mathrm{\mid }2\pi \mathbf{n}\mathrm{/}L{\mathrm{\mid }}^2>0$ for $\mathbf{n}\mathrm{\ne }0$; hence the longitudinal energy is nonnegative. This remains conditional on the current draft solver and Countable Choice and asserts no solution for nonneutral data.

On the neutral branch [A3], [F1] and step 1.1 give ${\rho }_q=-e_{\mathrm{S}\mathrm{I}}{\psi }^{†}\psi$ and ${\int }_{{\mathbb{T}}_L^3}{\psi }^{†}\psi d^3\mathbf{x}=0$ because $e_{\mathrm{S}\mathrm{I}}>0$. The integrand is continuous and nonnegative. If it were positive at any point, continuity would make it positive on a ball of positive volume, forcing a positive integral. Thus it vanishes everywhere and $\psi =0$. The finite-mode calculation of step 1.2 is logically separate and does not assert a solution in this neutral branch.