Full spatial response of a scalar-null plane

Summary: A specified scalar-null Dirac plane has an exact full spatial spectrum, a sharp density-dependent linear stability threshold and a directional scalar response that cannot be replaced by a universal static overlap.

Exact background and complete perturbation

Use the local scalar equation on flat ordinary spatial $\mathbb R^d$, $d\ge2$, with $\hbar=c=1$:

$$i\partial_t\Psi=-i\alpha\cdot\nabla\Psi+ g(\Psi^\dagger\beta\Psi)\beta\Psi+J\beta\Psi. \tag{1}$$

All $\alpha_j,\beta$ are Hermitian anticommuting involutions on a minimal module of complex dimension $n=2^{\lfloor(d+1)/2\rfloor}$. $J$ is a real scalar probe of the same field. The following constant-density plane is an unbound control per unit volume. It is not an identified physical vacuum or particle.

Choose $p,\rho>0$, $|e|=1$, $\alpha(e)w=w$, and $w^\dagger w=1$. Then

$$\Psi_0=\sqrt\rho\,w\,e^{ip e\cdot x-ipt},\qquad w^\dagger\beta w=0,\qquad \nu=g\rho \tag{2}$$

solves (1) at $J=0$ for every real $g$. The scalar vanishes because $\beta w$ has the opposite eigenvalue of $\alpha(e)$. This background exists in both even and odd $d$; the invariant linear classes answer the stronger question of arbitrary spatial patterns.

In the carrier frame use the doubled perturbation $\zeta_q=(\eta_q,\overline{\eta_{-q}})^{\mathsf T}$. With $v=\beta w$, the full Fourier symbol is

$$\mathcal H(q)= \begin{pmatrix} A(q)&B\\-\overline B&-\overline{A(-q)} \end{pmatrix},\qquad A(q)=\alpha\cdot(pe+q)-pI+\nu vv^\dagger,\quad B=\nu vv^{\mathsf T}. \tag{3}$$

The opposite Fourier mode and its conjugate are required by the real scalar variation. They introduce no second physical field. The pointwise source rank alone would miss the spatial kinetic terms in (3).

Exact four-dimensional response block

Write $q_\parallel=e\cdot q$, $q_\perp=q-q_\parallel e$ and $k=|q_\perp|$. For $k>0$ put $t=\alpha(q_\perp/k)v$. The columns $W=(v,t)$ are orthonormal, and $T=\operatorname{diag}(W,\overline W)$ is an isometric invariant frame for (3). In this order,

$$\boxed{H_4= \begin{pmatrix} \nu-2p-q_\parallel&k&\nu&0\\ k&q_\parallel&0&0\\ -\nu&0&-\nu+2p-q_\parallel&k\\ 0&0&k&q_\parallel \end{pmatrix}.} \tag{4}$$

For $k=0$ choose any physical transverse direction to define $t$. No transverse restriction has been placed on the perturbations. The remaining $2n-4$ dimensions have the free frequencies

$$-p\pm|pe+q|,\qquad p\pm|pe-q|, \tag{5}$$

each with multiplicity $n/2-1$. The source-generated block therefore includes all scalar response, while the complete spectrum retains every other spatial polarization.

Put $a=\sqrt{p(p-\nu)}$, allowing a complex square root. The four remaining frequencies are

$$\boxed{\omega_{s,t}=s a+ t\sqrt{(q_\parallel-s a)^2+k^2},\qquad s,t\in\{-1,1\}.} \tag{6}$$

Equivalently their characteristic equation is

$$(\omega^2-|q|^2)^2=4p(p-\nu)(\omega-q_\parallel)^2. \tag{7}$$

For example, reorder (4) as $(v,\bar v,t,\bar t)$. It becomes

$$\begin{pmatrix}-q_\parallel I+M&kI\\kI&q_\parallel I\end{pmatrix}, \qquad M=\begin{pmatrix}\nu-2p&\nu\\-\nu&2p-\nu\end{pmatrix}, \qquad M^2=4p(p-\nu)I. \tag{8}$$

Its two $M$ eigenvalues give (6), and continuity gives the degenerate cases without discarding any modes.

Sharp linear stability boundary

For $\nu<p$, the real matrix

$$G=\begin{pmatrix}2p-\nu&-\nu\\-\nu&2p-\nu\end{pmatrix} \tag{9}$$

has eigenvalues $2p$ and $2(p-\nu)$, both positive, and $GM=M^{\mathsf T}G$. Thus $\operatorname{diag}(G,G)$ symmetrizes (8). Normalize it by $2p$ and extend by identity on the free orthogonal complement. The result gives the uniform-in-momentum bound

$$\boxed{\|\zeta(t)\|_{L^2}\le \sqrt{\frac{\max(1,1-\nu/p)}{\min(1,1-\nu/p)}} \|\zeta(0)\|_{L^2},\qquad \nu<p.} \tag{10}$$

This proves bounded linear evolution for the specified background. It is stronger than merely finding real frequencies, but is not a nonlinear stability or particle-binding theorem.

At $\nu=p$, $M$ is nonzero and nilpotent. At $q=0$ the normalized initial vector $(i,0,-i,0)/\sqrt2$ has evolved norm $\sqrt{1+4p^2t^2}$. The threshold has secular growth and fails (10).

For $\nu>p$, write $a=ib$, $b=\sqrt{p(\nu-p)}$. At $k=0$ there is an exponentially growing frequency $-q_\parallel+2ib$. Continuity gives growth on an open set of nearby momenta, so the instability is available to spatial wavepackets. Its largest possible growth rate is $2b$. To see the bound, write $r=q_\parallel$ and use

$$\left|\operatorname{Im}\sqrt{(r-ib)^2+k^2}\right|^2 =\frac{\sqrt{(r^2+k^2-b^2)^2+4r^2b^2} -(r^2+k^2-b^2)}2\le b^2. \tag{11}$$

Squaring the last inequality leaves $4b^2k^2\ge0$. The threshold depends on frequency, density and the written coefficient; it contains no special spatial depth.

For $\nu<p$, each cone in (6) has group speed one away from its apex. Near $q=0$ the low branches behave as $\omega=q_\parallel\pm k^2/(2a)+O(|q|^3/a^2)$. The transverse low-frequency dispersion is quadratic there.

Scalar susceptibility and the static limit

The force and scalar readout in (4) are $\sqrt\rho(1,0,-1,0)^{\mathsf T}$ and $\sqrt\rho(1,0,1,0)$, respectively. Solving the resolvent gives

$$\boxed{\chi_{\rm null}(z,q)= \frac{-4p\rho(z-q_\parallel)^2} {(z^2-|q|^2)^2-4p(p-\nu)(z-q_\parallel)^2}.} \tag{12}$$

One direct derivation eliminates both transverse partners and sets $F=(z^2-|q|^2)/(z-q_\parallel)$. The remaining block is $FI-M$, with forcing $\sqrt\rho(1,-1)$ and readout $\sqrt\rho(1,1)$. Its numerator is $-4p\rho$; multiplying by $(z-q_\parallel)^2$ gives (12). The rational expression supplies the continuous limits where this intermediate elimination divides by zero.

For stable backgrounds take the retarded boundary value $z=\omega+i0$. For unstable backgrounds the initial-value response uses a Laplace contour above the growing poles. An unqualified $+i0$ prescription would omit that condition. The local first-order equation retains the causal domain of dependence described in causal scalar response.

At zero momentum the removable factors cancel:

$$\chi_{\rm null}(z,0)= -\frac{4p\rho}{z^2-4p(p-\nu)},\qquad \chi_{\rm null}(0,0)=\frac{\rho}{p-\nu}\quad(\nu<p). \tag{13}$$

In contrast, a nonzero purely transverse static momentum has $\chi_{\rm null}(0,q)=0$. General static poles satisfy $|q|^2=2a|q_\parallel|$: two elastic-scattering spheres centered at $\pm ae$, each of radius $a$. These directional limits and resonances exclude treating this background's response as one universal positive elliptic overlap coefficient.

Nonlinear comparison and scope

The nonnull moving branches emerge at the same threshold and have a continuously connecting scalar response. Neither family is a localized particle or a discrete mass selection.

scalar_plane_response_checks.py compares (4) with the complete spinor symbol in spatial dimensions two through six and ten, with complex polarizations and general momenta. Independent Fourier differences of the nonlinear source check both ordinary and conjugate blocks. The diagnostic verifies the full spectrum, positive metric, evolution bound, critical Jordan mode and scalar resolvent.

It also evolves the original eight-component nonlinear plane at $p=\rho=1$ and $\nu=0.3,1,1.4$. Halving the initial perturbation from $10^{-5}$ to $5\times10^{-6}$ approximately halves the error divided by that perturbation amplitude, as required for a first variation. Over $0\le t\le4$, the selected linear perturbations amplify by approximately $1.016$, $8.062$ and $157.53$, respectively. Positive norm and the complete kinetic-plus-scalar energy remain conserved. These are dynamical checks of the stated controls; they do not measure a physical particle coupling.