Summary: Eliminating spatial modes dresses both the interaction vertex and its kinetic normalization, with exact finite-frequency conservation identities and nonzero responses that a bare projection can miss.

Carry the source through the same reduction as the dynamics

The projected Dirac current shows why a current can lose required terms under direct mode compression. This note calculates those terms by eliminating the omitted modes of the same field. It supplies the physical normalization and the complete source matrix for a specified linear Hermitian problem.

All modes describe Ψ∞\Psi_\infty on the one nested manifold of ordinary macroscopic spatial dimensions. Retained and omitted components are a mathematical decomposition of that field. They are not extra dimensions, independent interaction fields or mediator particles.

Let a fixed orthogonal mode projector define a block decomposition

H=(ABB†C),H=H†.(1) H=\begin{pmatrix}A&B\\B^\dagger&C\end{pmatrix}, \qquad H=H^\dagger. \tag{1}

The formulas apply to a finite regulator or to operators with compatible domains and the stated inverses. Work at a real spectral parameter EE outside the spectrum of CC. A continuum or a resonance requires the appropriate boundary-value resolvent and flux normalization; the positive norm identities below concern this Hermitian, real-energy setting. The full real-linear Dirac fluctuation problem may also require its ordinary and conjugate blocks before this reduction is applicable.

Exact reconstruction and the norm of a reduced amplitude

The omitted row of (H−E)Ψ=0(H-E)\Psi=0 gives

R(E)=(E−C)−1,Ψ=Ω(E)u,Ω(E)=(IR(E)B†).(2) R(E)=(E-C)^{-1},\qquad \Psi=\Omega(E)u,\qquad \Omega(E)=\begin{pmatrix}I\\R(E)B^\dagger\end{pmatrix}. \tag{2}

The retained equation and its inverse operator are

Heff(E)=A+BR(E)B†,K(E)=EI−Heff(E),K(E)u=0.(3) H_{\rm eff}(E)=A+BR(E)B^\dagger, \qquad K(E)=EI-H_{\rm eff}(E),\qquad K(E)u=0. \tag{3}

This is the effective-operator construction explained in Compression and effective dynamics. It obeys the useful exact identity

(E−H)Ω(E)=(K(E)0).(4) (E-H)\Omega(E)=\begin{pmatrix}K(E)\\0\end{pmatrix}. \tag{4}

The full field norm is not generally u†uu^\dagger u. Since ∂ER=−R2\partial_E R=-R^2,

Z(E)=Ω†Ω=I+BR(E)2B†=I−∂EHeff(E),∥Ψ∥2=u†Z(E)u.(5) \boxed{Z(E)=\Omega^\dagger\Omega =I+BR(E)^2B^\dagger =I-\partial_EH_{\rm eff}(E),\qquad \|\Psi\|^2=u^\dagger Z(E)u.} \tag{5}

ZZ is positive and at least II in this setting. For a selected energy, b=Z1/2ub=Z^{1/2}u gives a normalized physical amplitude and Ω^=ΩZ−1/2\widehat\Omega=\Omega Z^{-1/2} is an isometry. Energy dependence makes this a mode normalization; replacing Z(E)Z(E) by one constant over an entire broad spectrum is an extra approximation.

Differentiate the entire effective operator

Let H(a)=H+aV+O(a2)H(a)=H+aV+O(a^2), where aa is a specified source perturbation amplitude and V=V†V=V^\dagger is its full spatial vertex. The same-field interpretation requires that amplitude and vertex to be obtained from the adopted action. This calculation does not identify aa with a physical photon in advance.

At fixed EE and fixed mode projector, direct differentiation gives

∂aHeff=VPP+VPQRB†+BRVQP+BRVQQRB†=Ω†VΩ.(6) \begin{aligned} \partial_aH_{\rm eff}={}&V_{PP} +V_{PQ}RB^\dagger+BRV_{QP} +BRV_{QQ}RB^\dagger\\ ={}&\Omega^\dagger V\Omega. \end{aligned} \tag{6}

The final term follows from differentiating the inverse: ∂aR=RVQQR\partial_aR=RV_{QQ}R. The middle terms retain the probe's effect on the coupling to the omitted modes. Their signs and coefficients are fixed by the same full operator.

Consequently the normalized static source vertex is

V^(E)=Z(E)−1/2Ω(E)†VΩ(E)Z(E)−1/2.(7) \boxed{\widehat V(E)=Z(E)^{-1/2}\Omega(E)^\dagger V\Omega(E)Z(E)^{-1/2}.} \tag{7}

For an isolated eigenvalue E(a)E(a) and reduced eigenvector uu, differentiating K(E(a),a)u=0K(E(a),a)u=0 yields

dEda=u†(∂aHeff)uu†Zu.(8) \frac{dE}{da} =\frac{u^\dagger(\partial_aH_{\rm eff})u}{u^\dagger Zu}. \tag{8}

This equals the full-field expectation of VV in the normalized eigenstate. Both the source and its normalization are therefore required to predict an energy shift. A probe-dependent projector adds its own variation terms; (6) is explicitly the fixed-projector formula.

For a transition between two retained spatial modes, the corresponding vertex is

Γfi=Ωf†VfiΩi,Γ^fi=Zf−1/2ΓfiZi−1/2.(9) \Gamma_{fi}=\Omega_f^\dagger V_{fi}\Omega_i, \qquad \widehat\Gamma_{fi}=Z_f^{-1/2}\Gamma_{fi}Z_i^{-1/2}. \tag{9}

The two lifts include both mode preparations. If the full vertex satisfies Vif=Vfi†V_{if}=V_{fi}^\dagger, then Γ^if=Γ^fi†\widehat\Gamma_{if}=\widehat\Gamma_{fi}^\dagger. Polarization and shape enter through both full spatial states and the source operator; this relation does not replace either preparation by an isotropic average.

Charge normalization and a finite-frequency identity

If the adopted action already supplies one conserved generator qIqI on the full mode family, its normalized equal-energy density vertex is exactly

Z−1/2Ω†(qI)ΩZ−1/2=qI.(10) Z^{-1/2}\Omega^\dagger(qI)\Omega Z^{-1/2}=qI. \tag{10}

A small retained fraction cannot turn this supplied charge into a neutral particle. This preserves a given charge; it does not assign qq to any sector or identify the physical electromagnetic symmetry.

For two different energies at the same spatial momentum, the resolvent identity gives

R(Ef)−R(Ei)=−(Ef−Ei)R(Ef)R(Ei). R(E_f)-R(E_i)=-(E_f-E_i)R(E_f)R(E_i).

Hence the complete density vertex obeys

Ωf†Ωi=I−Heff(Ef)−Heff(Ei)Ef−Ei.(11) \boxed{\Omega_f^\dagger\Omega_i =I-\frac{H_{\rm eff}(E_f)-H_{\rm eff}(E_i)}{E_f-E_i}.} \tag{11}

Its coincident limit is (5). Equation (11) ties the energy dependence of the dynamics to the finite-frequency density response, so they cannot be chosen independently.

The full spatial current retains continuity

For an affine spatial symbol

H(p)=M+∑aαapa,M=M†,αa=αa†,(12) H(p)=M+\sum_a\alpha_a p_a, \qquad M=M^\dagger,\quad\alpha_a=\alpha_a^\dagger, \tag{12}

use the same fixed retained projector at incoming and outgoing momenta. The minimally varied density and spatial-current matrices, without their common supplied charge factor, are

Γfi0=Ωf†Ωi,Γfia=Ωf†αaΩi.(13) \Gamma^0_{fi}=\Omega_f^\dagger\Omega_i, \qquad \Gamma^a_{fi}=\Omega_f^\dagger\alpha_a\Omega_i. \tag{13}

Apply (4) on the left and right of the exact full-operator difference. Writing ω=Ef−Ei\omega=E_f-E_i and k=pf−pik=p_f-p_i gives

ωΓfi0−kaΓfia=K(Ef,pf)−K(Ei,pi).(14) \boxed{\omega\Gamma^0_{fi}-k_a\Gamma^a_{fi} =K(E_f,p_f)-K(E_i,p_i).} \tag{14}

Between actual incoming and outgoing solutions, the right side has zero matrix element. This is the reduced continuity identity. It remains valid when MM and the velocities mix the retained and omitted modes strongly, as long as the stated reconstruction exists.

For a nonlinear spatial symbol, the finite-momentum current must instead be obtained by varying its actual coupling to the probe. Replacing that vertex by a derivative at one momentum need not satisfy the finite-difference identity. A transverse photon would contract the correctly derived spatial vertex with its physical polarization; continuity fixes the longitudinal relation but does not supply the transverse interaction by itself.

A bare zero can miss a real source

Consider the completely specified two-mode model

H=(0ttΔ),V=(0110),Δ>0.(15) H=\begin{pmatrix}0&t\\t&\Delta\end{pmatrix}, \qquad V=\begin{pmatrix}0&1\\1&0\end{pmatrix}, \qquad \Delta>0. \tag{15}

Retain the first component. Its bare vertex is VPP=0V_{PP}=0, but Heff=t2/(E−Δ)H_{\rm eff}=t^2/(E-\Delta) and ∂aHeff=2t/(E−Δ)\partial_aH_{\rm eff}=2t/(E-\Delta) when the probe changes tt to t+at+a. The full lower eigenvalue gives

dE−da∣a=0=−2tΔ2+4t2.(16) \frac{dE_-}{da}\bigg|_{a=0} =-\frac{2t}{\sqrt{\Delta^2+4t^2}}. \tag{16}

At t=0.4t=0.4, Δ=2\Delta=2, the physical vertex is about −0.37139-0.37139. Equation (7) reproduces it exactly. This is a source through the components of the same field which the bare projection omitted.

Even a truly invariant retained space at zero probe can retain a nonlinear response. Set t=0t=0 in (15). The first derivative vanishes, but

E−(a)=Δ−Δ2+4a22=−a2Δ+O(a4),E−′′(0)=−2Δ.(17) E_-(a)=\frac{\Delta-\sqrt{\Delta^2+4a^2}}2 =-\frac{a^2}{\Delta}+O(a^4),\qquad E_-''(0)=-\frac2\Delta. \tag{17}

The induced source is linear in a small applied amplitude even though the unperturbed source expectation is zero. Thus a neutral charge, a vanishing linear transition vertex, and absence of polarizability are separate tests. The geometric response examples require this same distinction.

Application to the neutral spatial modes

Spatial transport of a neutral doublet shows that a fixed spinor projection can remove the first-order Dirac action of perpendicular spatial momenta. Equations (2)–(14) provide a way to retain that contribution in both the dynamics and the source before identifying a physical coupling. They also prevent a missing projected current from being mistaken for derived electromagnetic neutrality.

The required next inputs are the full spatial particle solution, the actual photon inclusion or collective map, and the action which defines VV. The descending photon source specifies the propagation-preserving output projection; its complete current must be carried through the same reduction.

Verification

dressed_coupling_checks.py compares the source derivative with independent finite differences of the full effective operator, reconstructs full eigenstates, and checks their energy shifts against direct diagonalization. Independent incoming and outgoing momenta test (14), including on-shell matrix elements. The two-mode examples verify both the nonzero dressed source and quadratic response. All conditions remain active under Python optimization.