One quartic current law for conditional self frequency and pair response

The rank-independent one-current excess preserves isolated occupied sources but begins at eighth order in the field. The localized action already contains several Lorentz-scalar currents of the same field. Comparing two of them gives a quartic field law with a direct spatial-pattern response. This is an inverse candidate. Its spatial carrier, mode weights and physical normalization have not been derived from the master interaction.

One same-field quartic and its complete source

Retain the doubled field and partner indices of the localized action. At each retained oscillator order $k$, let $C_k$ be the current with $B_0=\sigma_z\otimes I$ and let $K_k$ use $B_2=I\otimes\beta$, including the same spectator trace in both. Both are Hermitian Lorentz-scalar matrices in the stated Dirac representation. Put $h_{k,d}=\binom{k+d-1}{d-1}$. A single quartic comparison is

$$\boxed{U_q=g_d\sum_{k\ge0}h_{k,d} \left[\operatorname{Tr}(C_k^2) -q\operatorname{Tr}(K_k^2)\right].} \tag{1}$$

The retained-current maps and their adjoints are part of this equation. If one suppresses those maps to display the local matrix variation, the source is

$$\frac{\partial U_q}{\partial\Psi^\dagger} =2g_d\sum_k h_{k,d} \left[B_0\Psi C_k-qB_2\Psi K_k\right], \tag{2}$$

with each matrix lifted back into its original oscillator grade. This formula includes both complex partner and conjugate variations. At $q=-1$, the two current squares have equal positive coefficients: $U_{-1}=g_d\sum h_{k,d} [\operatorname{Tr}C_k^2+\operatorname{Tr}K_k^2]\ge0$. It is smooth even when a current vanishes. Equal coefficients are a specified symmetry choice, not yet a consequence of the master field.

Conditional self frequency and its norm

At a spatial point, take an upper-only coherent field $\phi=\sqrt{\rho}\,U$, where $U$ is unitary in the complete partner frame, and put $B=U^\dagger\beta U$. Then $B^2=I$ and at every occupied order

$$C_k=\rho I_r\otimes P_{z,k},\qquad K_k=\rho B\otimes P_{z,k}. \tag{3}$$

Both current squares act on the occupied field as $\rho^2$. The exact occupied source from (2) is $2g_d(1-q)\rho\,D_N\phi$, where $D_N=\sum_{k=0}^{N}h_{k,d}=\binom{N+d}{d}$ and the maintained particle index is $n=N+1$. Thus a spatial zero-mode field with constant $\rho$ has the conditional frequency

$$\boxed{\omega_N=2g_d(1-q)\rho D_N,\qquad q=-1:\ \omega_N=4g_d\rho D_N.} \tag{4}$$

This is mass from a self-coupling term in the standing equation, and the same term supplies the pair response below. No frequency coefficient is fitted separately for two grades at one depth. At common canonical norm $Q$ and physical spatial volume $\mathcal V_d$, this square-unitary ansatz has $\rho=Q/(r_d\mathcal V_d)$. The frequency unit is therefore $4g_d Q/(r_d\mathcal V_d)$ for the equal-coefficient law. The volume, common norm, cross-depth relation and suitable spatial zero modes remain physical inputs. A nonconstant upper-only density makes the right-hand side of (4) position dependent, so the earlier localized core is not a solution of (1). It does not supply the square-unitary current pattern or the simplex enhancement required here.

There is an exact energy normalization issue. Let $T_{\rm sp}$ be the quadratic spatial kinetic energy and $U$ the quartic (1). For a stationary field, homogeneity gives $\omega Q=T_{\rm sp}+2U$, while its canonical Hamiltonian is $E=T_{\rm sp}+U=\omega Q-U$. On a spatial zero mode,

$$\boxed{E=\frac{\omega Q}{2}.} \tag{5}$$

At the working one-particle norm $Q=1$, this energy is half the frequency-equivalent rest energy. Setting $Q=2$ would remove the factor algebraically on that branch, but no norm quantization or full boost stress has selected it. The finite frozen controls below do not resolve this inertial-mass identification.

Reciprocal prepared-pair law

For two same-depth upper fields, let their local densities be $\rho,\sigma$, their spatial partner involutions $B_A=U_A^\dagger\beta U_A$ and $B_B=U_B^\dagger\beta U_B$, and their coherent oscillator-ray overlap be $s=|z^\dagger w|^2$. For a shared order $k$, direct substitution into (1) and subtraction of both isolated energies gives

$$\boxed{V_{AB,k} =2g_d\rho\sigma\left[r_d-q\operatorname{Tr}(B_A B_B)\right]s^k.} \tag{6}$$

The expression is symmetric in $A,B$ and varies both fields. For $q=-1$ it becomes

$$V_{AB,k} =g_d\rho\sigma\|B_A+B_B\|_{\rm HS}^2s^k\ge0. \tag{7}$$

Compatible $B_A=B_B$ patterns have the largest prepared cross energy; opposite $B_A=-B_B$ patterns have zero. This is a shape-sensitive energy, not an observed interaction rate. The six-depth quartic calculation checks (2), isolated sources and (6) with complete complex field variations. The actual retained grade-zero tower gives the same pair formula and reciprocal derivatives at all six maintained depths, with pair formula errors below $7.2\times10^{-15}$.

For a larger grade $N$ and smaller grade $M$, the full shared-order pair factor is

$$F_{M,d}(s)=\sum_{k=0}^{M}\binom{k+d-1}{d-1}s^k,\qquad F_{M,d}(1)=D_M,\qquad \lim_{M\to\infty}F_{M,d}(s)=(1-s)^{-d}\quad(0\le s<1). \tag{8}$$

Thus imperfect overlap can make pair strength much less sensitive to the occupied mass grade than the self frequency. In the exact maintained depth-six comparison at $s=0.5$, the $n=13$ factor is about $0.95187$ of the $n=35$ factor. At $s=1$ their ratio is $0.0048364$. The law does not select $s$ or ensure it stays away from one. Keeping only order zero would make the pair factor exactly grade independent, but that would be a different contraction rule from (1) and needs its own derivation.

Transverse depth-two control and spatial separation

For normalized transverse Jones vectors $e,f\in\mathbb C^2$, let $B_e=2ee^\dagger-I_2$. These are Hermitian involutions, and the rank-two factor in (7) is exactly

$$2+\operatorname{Tr}(B_eB_f)=4|e^\dagger f|^2. \tag{9}$$

The depth-two control checks linear, right- and left-circular, and elliptical patterns, including their relative phases. Orthogonal linear and opposite circular states give zero, while matched states give the maximum. A common physical rotation of the transverse plane preserves (9); rotating only one linear pattern changes it as $4\cos^2\theta$, while a circular pattern keeps its overlap under such a rotation. This checks the required polarization geometry if those Jones states are occupied photon patterns in the IDWT field. That identification, photon propagation and the $n=0$ massless boundary remain unproved.

The different sign $q=1$ has a useful prepared spatial identity even though it gives zero coherent self frequency. On an upper field $\phi$, decompose its spatial spinor with $P_\pm=(I\pm\beta)/2$ and set $A=\phi^\dagger P_+\phi$, $B=\phi^\dagger P_-\phi$. Then

$$\operatorname{Tr}(C^2)-\operatorname{Tr}(K^2) =4\operatorname{Tr}(AB)\ge0. \tag{10}$$

It vanishes, with zero complete source, on a single square-unitary coherent core. For two such prepared cores, its cross energy is $\rho\sigma s^k\|B_A-B_B\|_{\rm HS}^2$. This provides a local quartic correction to the previous trace mass action but does not itself produce self mass.

For the analytic stereographic upper-core profile of the localized action, the spatial calculation derives the exact displaced-core potential for (10). Write $h=|\delta|/R$, $\rho(x)=A_R^2(1+|x|^2/R^2)^{1-d}$, and $J_d=\pi^{d/2}\Gamma(3d/2)/\Gamma(2d)$. One shared shell gives

$$V_{-,k}(\delta) =4r_d A_R^4R^d\,s^k h^2J_d \,{}_2F_1\!\left(\frac{3d}{2},d;d+\frac12;-\frac{h^2}{4}\right). \tag{11}$$

The pair energy is zero at coincidence, rises to a finite barrier, and decays at large separation. Independent shifted-profile quadrature and Feynman-parameter integration verify (11) at the tested depths and separations. The gapless analytic depth-two profile has divergent canonical norm, so its value here is only a local preparation. Odd-depth entries use the integral formula without a constructed matching Clifford spatial core. For the $q=-1$ positive sum, the same preparation instead has its largest cross energy at coincidence in the sampled separations; that action changes the isolated analytic core, so this is a prepared-pair comparison rather than a force between its solutions.

Frozen control and remaining derivation

The unified batch and report retain all real and conjugate spinor and partner directions in finite uniform-core tests at depths 2, 4 and 6, one or two oscillator columns, and momenta $0.1,1,10$. For the constant-density self-mass action (1), the equal-positive choice $q=-1$ has no resolved imaginary frequency above $3\times10^{-14}$; $q=-0.5$ and $q=0$ also pass these finite samples. The tested $q=0.25,0.5,0.75$ have growth approximately $0.630,1.295,2.608$ respectively at momentum ten. These are frozen controls, not stability of a spatially varying core or every momentum and mode.

The equal-positive quartic is the first simple trial here in which one smooth local expression supplies both a conditional simplex self frequency and a polarization-sensitive pair response. The theorem is conditional: its full spatial state, common finite-volume normalization, physical depth-two photon map, cross-depth sector embedding, particle mode selection and complete response have not been obtained. Equation (5) is a specific unresolved inertial-energy defect at the working $Q=1$. The retained weights and the relation of this quartic to the master field remain assumptions. It is therefore a useful one-law candidate, not yet the physical mass operator or coupling law.