Coherent pair deformation

Summary: A common positive-density quartic drives two coherent mode patterns into shape deformations, with reciprocal changes in contact fixed by conserved energy and rotation moments.

The specified component action

At a common projective mode fibre of dimension $d$, retain complete bands $\mathcal H_N=\operatorname{Sym}^N(\mathbb C^{d+1})$ and $\mathcal H_M=\operatorname{Sym}^M(\mathbb C^{d+1})$, with $N,M\ge1$. Set $D_R=\binom{R+d}{d}$ and use normalized coherent transforms

$$s_a(u)=\sqrt{D_N}\langle u;N|a\rangle, \qquad s_b(u)=\sqrt{D_M}\langle u;M|b\rangle.$$

Assume the two retained families are orthogonal bundle components of the same field, with the declared positive density $\rho=|s_a|^2+|s_b|^2$ and canonical coefficient kinetic form. The action is

$$S=\int dt\left[ \frac i2(a^\dagger\dot a-\dot a^\dagger a +b^\dagger\dot b-\dot b^\dagger b) -\frac\kappa2\int\rho^2d\nu_{\rm FS}\right]. \tag{1}$$

The line bundles of degrees $N$ and $M$ need not be identified to add their Hermitian densities. Orthogonality and the kinetic metric are assumptions of this component action; spatial nesting alone supplies neither. A coherent scalar sum with interference is a different current. The positive density here is also different from the indefinite scalar Dirac bilinear $\bar\Psi\Psi$.

This calculation retains all complex coefficients of its two families. It tests mode-pattern deformation in a specified action, rather than claiming a full particle-pair interaction on the physical base. The projective fibre adds no physical spatial dimensions. Different physical depths additionally require actual normal-bundle maps and spatial kinetic normalization. No independent field or mediator particle is introduced.

A single symmetric product gives both reciprocal forces

Let $T_{NM}:\mathcal H_N\otimes\mathcal H_M\to\mathcal H_{N+M}$ orthogonally symmetrize every tensor index. In normalized occupations,

$$T_{NM}(|\alpha\rangle\otimes|\beta\rangle) =\sqrt{\frac{\binom N\alpha\binom M\beta} {\binom{N+M}{\alpha+\beta}}}\,|\alpha+\beta\rangle. \tag{2}$$

The multi-index Vandermonde identity gives $T_{NM}T_{NM}^\dagger=I$. Define

$$C_{NM}=\frac{D_ND_M}{D_{N+M}},\qquad q=T_{NM}(a\otimes b),\qquad A_bh=T_{NM}(h\otimes b),\qquad B_ak=T_{NM}(a\otimes k).$$

The cross-density overlap is exactly

$$\boxed{I(a,b)=\int|s_a|^2|s_b|^2d\nu_{\rm FS} =C_{NM}\|q\|^2.} \tag{3}$$

Its two derivatives are $C_{NM}A_b^\dagger q$ and $C_{NM}B_a^\dagger q$. If $F_N$ is the complete unit-coupling single-family quartic gradient, (1) gives

$$\boxed{ i\dot a=\kappa\bigl(F_N(a)+C_{NM}A_b^\dagger q\bigr),\qquad i\dot b=\kappa\bigl(F_M(b)+C_{NM}B_a^\dagger q\bigr).} \tag{4}$$

Both responses, their relative coefficient and their self-interactions come from the same $\kappa$. No pair-specific force coefficient has been chosen. Independent component phases conserve $J_a=\|a\|^2$ and $J_b=\|b\|^2$.

Coherent overlap and the missing shape directions

For unit coherent preparations $a=z^{\otimes N}$, $b=w^{\otimes M}$, let $\tau=|z^\dagger w|^2$. Symmetrizing the two products gives the finite polynomial

$$\boxed{I(z^{\otimes N},w^{\otimes M}) =\frac{C_{NM}}{\binom{N+M}{N}} \sum_{j=0}^{\min(N,M)}\binom Nj\binom Mj\tau^j.} \tag{5}$$

One derivation groups permutations by the number $j$ of indices exchanged between the two tensor blocks. Those terms have overlap $\tau^j$ and multiplicity $\binom Nj\binom Mj$; their total number at $\tau=1$ is $\binom{N+M}{N}$ by Vandermonde. Equation (5) is an overlap identity, not a proof that the coherent subset is dynamically invariant. The graph-ray benchmark demonstrates the constant-overlap precession obtained in a specified two-ray reduction.

To test invariance, use common mode-frame covariance and component phases to set $z=e_0$ and $w=ce_0+se_1$, where $c,s>0$ and $c^2+s^2=1$. The component of the cross-gradient on $|N-j,j\rangle$ is

$$\begin{aligned} G_j={}&\frac{D_ND_M\sqrt{\binom Nj}\,d!}{(N+M+d)!} \sum_{r=0}^{M-j}\binom Mr\binom M{r+j} c^{2M-2r-j}s^{2r+j}\\ &\hspace{27mm}\cdot(N+M-r-j)!\,(r+j)!, \qquad 0\le j\le\min(N,M). \end{aligned} \tag{6}$$

Components with $j>M$ vanish. Expanding the two powers of $w$ in the projected current and using the FS monomial integral proves (6). Every summand is positive in the stated nonaligned preparation. In particular, $G_2>0$ whenever $N,M\ge2$.

At a coherent tensor, its phase and norm occupy $j=0$, and its coherent ray tangents occupy $j=1$. Therefore the $j=2$ force lies outside that tangent space. The self-gradient contributes only the carrier direction, so it cannot cancel this force. The reciprocal calculation gives the same conclusion for the other pattern.

Thus the coherent product manifold is not invariant for a generic nonaligned pair with $N,M\ge2$. A calculation that varies only the two rays discards genuine solutions of the retained coefficient equations. When $\min(N,M)=1$, the cross-force has only $j=0,1$ in either adapted frame. In that special case the coherent pair manifold is invariant. Aligned and orthogonal coherent preparations are also stationary rays.

Changing contact follows from conservation

Write $A(a)=\int|s_a|^4$, $B(b)=\int|s_b|^4$, and $\lambda_R=D_R^2/D_{2R}$. The single-family sharp bounds give

$$A(a)\le\lambda_NJ_a^2, \qquad B(b)\le\lambda_MJ_b^2,$$

with equality exactly for coherent tensors. Prepare both patterns coherently and evolve the complete equations (4), with $\kappa\ne0$. Conservation of $E=\kappa(A+B)/2+\kappa I$ then gives

$$\boxed{I(t)-I(0) =\frac{\lambda_NJ_a^2-A(a(t)) +\lambda_MJ_b^2-B(b(t))}{2}\ge0.} \tag{7}$$

Loss of individual coherence is balanced by increased cross-density contact. Equation (7) is an exact bound relative to this particular initial preparation. It does not say that $I(t)$ increases monotonically.

The first time derivative vanishes initially. Resolve the initial coefficient velocities into each coherent frame and let $v_{a,j}$, $v_{b,j}$ denote the full vectors of degree-$j$ transverse components. The complete single-family Hessian gives

$$\boxed{I''(0)=2\sum_{j\ge2}\left[ \lambda_NJ_a(1-2r_{N,j})\|v_{a,j}\|^2 +\lambda_MJ_b(1-2r_{M,j})\|v_{b,j}\|^2\right],} \tag{8}$$

where $r_{R,j}=(R)_{\underline j}/(2R)_{\underline j}$ and absent components are zero. Since $1-2r_{R,j}>0$ for $j\ge2$, (6) makes (8) strictly positive for a generic nonaligned coherent pair. The actual velocities include $\kappa$, so this curvature is proportional to $\kappa^2$.

Other preparations can evolve toward less contact. In the real occupation convention, the Hamiltonian obeys $F(\bar a,\bar b)=\overline{F(a,b)}$. Complex-conjugating a later state therefore initiates its reversed trajectory, along which the earlier contact increase becomes a decrease. This is a mathematical time-reversal test of (4); it assigns no physical meaning to antiparticles.

Rotation moments give a second compatibility balance

Let $\rho_a,\rho_b$ be the positive one-index contractions of $|a\rangle\langle a|$ and $|b\rangle\langle b|$, with traces $J_a,J_b$. Common $U(d+1)$ covariance of the mode action gives

$$S=N\rho_a+M\rho_b,\qquad \dot S=0. \tag{9}$$

This mode symmetry contains the physical rotation subgroup but also mixes the homogenizing coefficient with oscillator directions. It is part of the specified isotropic completion, not an additional physical rotation group inferred from spatial depth.

For the same coherent initial preparation, conservation of $\operatorname{tr}S^2$ gives

$$\boxed{ \operatorname{tr}(\rho_a(t)\rho_b(t))-J_aJ_b\tau_0 =\frac{N^2[J_a^2-\operatorname{tr}\rho_a(t)^2] +M^2[J_b^2-\operatorname{tr}\rho_b(t)^2]}{2NM}\ge0.} \tag{10}$$

The inequality uses positivity and the fixed traces. This one-index compatibility is distinct from the full density overlap $I$; both obey preparation-dependent balances. If both patterns remain coherent, (10) forces their relative ray overlap to remain $\tau_0$. Full shape changes are precisely what allow this restricted conclusion to change.

Exact active-subspace reduction for this preparation

Initially, $S$ in (9) has support on $V=\operatorname{span}\{z,w\}$, of complex dimension at most two. For $x\perp V$, conservation gives $0=x^\dagger Sx=Nx^\dagger\rho_ax+Mx^\dagger\rho_bx$. Both terms are nonnegative, hence both vanish. The tensors therefore remain in

$$\operatorname{Sym}^N(V)\oplus\operatorname{Sym}^M(V). \tag{11}$$

This is an invariant consequence of the complete equations for the stated preparation. All $N+1$ and $M+1$ coefficients on $V$ must still evolve; it is not a two-ray truncation. The ambient normalization factors $D_R(d)$, $\lambda_R(d)$ and $C_{NM}(d)$ remain those of the original fibre. Replacing them by their dimension-one values changes the action.

For example, the degree pair $(12,34)$ at $d=6$ can be evolved with 48 complex coefficients while retaining the ambient band factors. The diagnostic uses unit norms, initial $\tau_0=0.95$ and time measured in units of $[|\kappa|\max(\lambda_N,\lambda_M,C_{NM})]^{-1}$. Over 100 such units its sampled contact rises from approximately $2225.8413$ to $2275.4$, while the first pattern's one-index purity falls below $0.947$. These are dimensionless conditional action results, not measured lepton interaction strengths.

The invariant-subspace argument does not establish stability against arbitrary outside perturbations. The separate aligned-pair calculation supplies a complete ambient Hessian and finite-mode orbital stability for that stationary set. Neither result replaces the actual spatial polarization, transport and Dirac current needed for the physical compatibility hypothesis.

Verification

coherent_pair_variation_checks.py uses independent FS quadrature for the energy and both reciprocal forces, all linear and conjugate Hessian blocks, exact monomial leakage formulas, arbitrary complex mode-frame rotations, conserved moments, full nonlinear evolution and reversed trajectories. It also checks the degree-one invariant case and large-band formulas at depths six and ten. coherent_pair_algebra.py implements the normalized products and complete derivatives.