A smooth cubic current excess for prepared pair coupling

The scalar-moment law joins a nonzero coherent mass source to a stable prepared-pair alignment, but its degree-one current formula has no differentiable source at full current cancellation. This calculation asks whether a smooth local polynomial can retain those two useful properties. It does not derive the master mass operator or a measured interaction rate.

Current excess and its complete derivative

Use the same Lorentz-scalar Hermitian retained current $C=\Psi^\dagger B_0\Psi$, including its indefinite values. Let $r$ be the partner rank, $T=\operatorname{Tr}C$, and $t=T/r$. Define the cubic excess

$$\boxed{H_r(C)=\operatorname{Tr}[C(C-tI)^2] =p_3-\frac{2T}{r}p_2+\frac{T^3}{r^2},} \quad p_j=\operatorname{Tr}C^j. \tag{1}$$

This polynomial is smooth at $C=0$ and for indefinite currents. For $C\ge0$, it is nonnegative because it is a sum of $\lambda_i(\lambda_i-t)^2$. It changes sign under $C\mapsto-C$, so it is not a globally positive energy on all Hermitian currents. Its full noncommuting matrix gradient is

$$\boxed{\nabla H_r(C)=3C^2-4tC+ \left(3t^2-\frac{2p_2}{r}\right)I.} \tag{2}$$

The derivative of (2) in any Hermitian direction $X$ includes $3(CX+XC)$, the trace variation of $t$, and the variation of $p_2$. The joint script executes matrix, oscillator-tower, frozen-field and parent-embedding controls in one run. Its report records random indefinite noncommuting gradient checks at ranks 2, 4, 8 and 32, with largest relative gradient-derivative error below $1.9\times10^{-11}$.

An inverse action trial repeats one expression at each retained order:

$$U_{\eta,d}=\omega_{0,d}\sum_{k\ge0} \binom{k+d-1}{d-1} \bigl[\operatorname{Tr}C_k+\eta H_{r_d}(C_k)\bigr]. \tag{3}$$

The coefficient $\eta>0$ has units of inverse current squared. It has not been derived or fixed by (1). The shell weights, frequency unit, canonical norm, and separate nonlinear spatial binding term also remain inverse inputs. Writing (3) as one bracket does not establish a single physical origin for its two terms.

Exact isolated source and prepared pair

On an isolated coherent current $C_0=\rho I_r\otimes P$, with $P$ unit rank, $H_r(C_0)=0$ and $(\nabla H_r(C_0))P=0$. Thus (3) gives exactly the previous occupied source and conditional simplex mass-frequency $\omega_N=\omega_{0,d}\binom{N+d}{d}$, with maintained particle index $n=N+1$. A fixed-trace partner split $\rho(1\pm\varepsilon)$ costs $2\rho^3\varepsilon^2+O(\varepsilon^3)>0$ in $H_r$.

For a prepared same-depth pair at order $k$, take $C_k=I_r\otimes(\rho P_z+\sigma P_w)$, $s=|z^\dagger w|^2$, and $L=\rho+\sigma$. The two nonzero ray eigenvalues obey $\lambda_+\lambda_-= ho\sigma(1-s^k)$. Using their quadratic identity in (1) gives

$$\boxed{H_r(C_k)=rL\rho\sigma(1-s^k)\ge0.} \tag{4}$$

For $s=\cos^2\theta$, the coefficient of $\theta^2$ is $rL\rho\sigma k>0$ when $k>0$. Both isolated cores have nonzero mass frequency, while their aligned prepared pair has zero cubic excess and is an angular energy minimum. With spatially varying densities the initial potential from (3) is

$$V_\eta(s)=\eta\omega_{0,d}r \int(\rho+\sigma)\rho\sigma \sum_{k=1}^{\min(N,M)} \binom{k+d-1}{d-1}(1-s^k)\,d^dx. \tag{5}$$

Equation (5) is a reciprocal prepared-pair energy, not a scattering cross section. It assumes both currents use the same partner frame and retained depth. If the spatial patterns and orientations vary during an encounter, the complete field equation must evolve both. Internal complex ray phases enter through $s$, while a common overall phase cancels in these current projectors.

The tower component inserts (1) into the complete grade-two and grade-one current maps at every maintained depth 2, 3, 4, 5, 6 and 10. The largest isolated source error is below $4.2\times10^{-16}$ relative, and the largest pair-formula error is $4.3\times10^{-14}$. These are finite coefficient controls at one spatial point.

Why a globally nonnegative smooth correction is flatter

Compare the globally nonnegative quartic current polynomial $H_{4,r}=\operatorname{Tr}[C^2(C-tI)^2]$. On the same prepared pair, its exact value is

$$\boxed{H_{4,r}=2r(\rho\sigma)^2(1-s^k)^2,} \tag{6}$$

which begins at order $\theta^4$. More generally, suppose a smooth function of every Hermitian current is nonnegative and zero on every coherent rank-$r$ current. Its gradient vanishes there. Its Hessian along a tangent rotation of the coherent projector also vanishes, since its value is identically zero on that coherent manifold. The prepared-pair path has the same first-order tangent at alignment. Its angular excess therefore has no $\theta^2$ term. The cubic (1) escapes this restricted result because it is nonnegative only on the positive-current cone; a coherent current lies on that cone's boundary in the full Hermitian space.

This is a restriction on an excess that vanishes on all coherent currents. It is not a no-go for every smooth full action or a reason to impose global energy positivity on an indefinite Dirac carrier.

Frozen response and nested-sector control

At a constant unit-density coherent core with one oscillator column, the traceless partner-current Hessian of (3) equals the earlier trace-current model with $q=2\eta$. Its exact high-momentum block has real bounded frequencies for $0<q<1$, a secular endpoint at $q=1$, and growth tending to $\omega\sqrt{q-1}$ for $q>1$. For a core with local density $\rho$, the equivalent parameter is $q_{\rm loc}=2\eta\rho^2$. Thus $2\eta\rho(x)^2<1$ is a necessary local bound for this partner block, not a full spatial stability theorem.

The frozen component retains complete real and conjugate spinor perturbations at depths 2, 4 and 6, one or two oscillator columns, core-gap ratios 0 and 1, and momenta $0.1,1,10$ in frequency units. At unit current, $\eta=0.05$ and $0.25$ have no resolved growth above $1.3\times10^{-14}$; $\eta=1$ reaches growth $0.98076$ at momentum 10, approaching the analytic unit limit. The sampled stable values are controls, not a proof for the spatially varying core or untested oscillator modes.

There is also a cross-depth condition. If a rank-$s$ lower-sector current is zero padded into a parent current of rank $r>s$, then even the isolated lower state has

$$H_r(\rho P_s)=s\rho^3(1-s/r)^2,\qquad (\nabla H_r)_{\rm occupied}=3\rho^2(1-s/r)^2. \tag{7}$$

The second quantity changes its local isolated mass source. The embedding component checks every nested pair among the six maintained depths and rotates the embedded projector independently. This result excludes simple zero padding as a transparent common-parent coupling implementation. It does not rule out a derived embedding with additional same-field structure. Sector nesting alone does not specify the field map between particle patterns.

Physical status

The cubic excess is a smooth algebraic candidate for same-depth prepared coupling. It leaves the conditional simplex mass branch intact and supplies a reciprocal alignment cost with one coefficient per current rule. It is not yet the coupling law sought for actual moving particles: $\eta$'s physical scale, the cross-depth map, the permitted polarization dynamics, and encounter rates are not derived. The mass branch still uses stipulated shell weights and separate spatial binding. The inherited coherent upper-only core has the constrained energetic-saddle obstruction even though finite frozen frequencies can be real. The master mass operator and common physical normalization therefore remain open.