A spatial instability of the uniform polarization source

The positive scalar-current square has a growing channel even when its normal field occupies a complete isotropic spinor multiplet. The result holds for every positive current coefficient and both positive occupations. Adding nonnegative constant Dirac gaps moves the unstable momentum interval without removing it. This is a statement about a uniform source on ordinary noncompact flat space. It does not classify localized sources or a derived discrete spatial spectrum.

Action and occupied background

The target is a doubled spinor matrix $\Psi$. Each normal polarization $B_A$ now has a complete partner multiplet of the same rank $r$. The scalar current is $C=\Psi^\dagger\beta\Psi$, with

$$F_q(C)^2=qr\operatorname{Tr}(C^2)+(1-q)(\operatorname{Tr}C)^2.$$

Choose a Hilbert-Schmidt orthonormal Hermitian basis with $E_1=I/\sqrt r$ and traceless $E_A$ for $A>1$. For $q>0$, the real current coordinates

$$c_1=\operatorname{Tr}C,\qquad c_A=\sqrt{qr}\,\operatorname{Tr}(E_A C)\quad(A>1)$$

satisfy $|c|^2=F_q(C)^2$. The associated arrow matrix is $\begin{pmatrix}0&c^T\\c&0\end{pmatrix}$; its positive eigenvector defines the stated occupied normal polarization. Label its distinguished component by 0 and its remaining components by $A=1,\ldots,r^2$. Now put $t_A=2\operatorname{Re}\operatorname{Tr}(B_0^\dagger\beta B_A)$. The interaction and optional constant gap terms are

$$U=\frac g2|c+t|^2 +m_a\operatorname{Tr}(\Psi^\dagger\beta\Psi) +m_b\sum_{A=0}^{r^2}\operatorname{Tr}(B_A^\dagger\beta B_A), \qquad g>0,\quad m_a,m_b\geq0. \tag{1}$$

All fields retain their canonical Dirac kinetic terms on the same ordinary spatial coordinates. The gaps in (1) are comparison parameters, not derived particle masses. Polarization labels remain hypothetical same-field modes.

Write $a>0,b>0$ for the uniform target and normal canonical densities. The occupied upper target matrix is $\sqrt{a/r}I_r$. The upper normal multiplet is $I_r/\sqrt r$ times the positive arrow eigenvector of norm $b$. All lower components vanish in this background. Define

$$s=g(a+b),\qquad L_a=2(m_a+s),\qquad L_b=2m_b+s. \tag{2}$$

The background phase frequencies are $m_a+s$ and $m_b+s$.

An invariant channel with every omitted component checked

Choose a unit Hermitian traceless partner matrix. Its target upper variation, its lower Clifford partner along the momentum, and the corresponding normal dark upper and lower variations span four complex components. Retain their conjugate components as well. After a sign change on the conjugate lower components, the complete channel generator is

$$\mathcal B=\begin{pmatrix}A&C_1\\-C_1&-A\end{pmatrix}, \qquad H_-=A-C_1,\quad H_+=A+C_1, \tag{3}$$

where

$$H_-= \begin{pmatrix} 0&p&0&0\\p&-L_a&0&0\\ 0&0&-s&p\\0&0&p&-L_b \end{pmatrix}, \tag{4}$$

$$H_+= \begin{pmatrix} 2gqa&p&g\sqrt{2qab}&0\\ p&-L_a&0&0\\ g\sqrt{2qab}&0&-ga&p\\ 0&0&p&-L_b \end{pmatrix}. \tag{5}$$

The squared frequencies are eigenvalues of $H_-H_+$, equivalently $H_+H_-$. This reduction follows by adding and subtracting the ordinary and conjugate equations in (3). The channel embedding is orthonormal in the actual canonical norm. Its invariance is checked by applying the full source derivative to every column, rather than by discarding matrix entries outside the proposed channel.

A nonempty interval necessarily contains growth

Set $y=p^2$. Direct determinants and elimination of the lower rows give

$$\det H_-=-p^2(sL_b-p^2),\qquad \det H_+=L_aL_b\,P(y), \tag{6}$$

$$P(y)=\left(2gqa+\frac y{L_a}\right) \left(-ga+\frac y{L_b}\right)-2g^2qab. \tag{7}$$

Its constant coefficient is $-2g^2qa(a+b)<0$, while its quadratic coefficient is positive. It therefore has exactly one positive root $y_*$. At the zero of the other determinant,

$$P(sL_b)=\frac{sL_b}{L_a}\,gb>0. \tag{8}$$

Consequently $0<y_*<sL_b$. Throughout the nonempty interval

$$\boxed{\sqrt{y_*}<|p|<\sqrt{sL_b},} \tag{9}$$

the product $\det(H_-H_+)$ is negative. Its real characteristic polynomial has positive leading coefficient and even degree four. At zero it is negative, while it is positive at sufficiently negative real argument. It has a negative real eigenvalue $\Omega^2$, which gives an exponentially growing solution. This is sufficient to establish instability; additional growing momenta can occur outside (9).

For zero added gaps, put $\alpha=a/(a+b)$, $u_*=y_*/s^2$. Then

$$u_*^2+\alpha(4q-1)u_*-4q\alpha=0, \quad u_*=\frac{\alpha(1-4q)+ \sqrt{\alpha^2(4q-1)^2+16q\alpha}}2, \quad 0<u_*<1. \tag{10}$$

Increasing a constant gap changes the endpoints in (9); it cannot make them coincide under the stated positive-density assumptions. A spatial confinement law or a different interaction requires a separate analysis.

Verification

isotropic_channel_checks.py passes 314 conditions in isotropic-channel-20260913T200207Z.json. It verifies all stationary components, independent energy and source variations, the full ordinary and conjugate channel embedding at depths two, four and six, and the growing eigenmode. The depth-six test applies the exact full derivative without allocating its large dense Hessian. One hundred additional positive-parameter controls check the interval and determinant sign.

These tests verify the specified field calculation. They neither identify the homogeneous source as a physical particle nor establish instability of the earlier localized inverse-action solutions.

Related research

• Spatial mass operator: results and remaining dynamics — connects this calculation to its premises, later corrections and the complete source question.