Section 10 9 October 2026

The actual compact-profile electron–source parent

Jeremy Rodgers · Independent Researcher
doi:10.5281/zenodo.23259571

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10 The actual compact-profile electron–source parent

\chi_R(R+t)=1- \frac{e^{-1/t}}{e^{-1/t}+e^{-1/(1-t)}}.

For $t\leq1/2$ its derivative is bounded by $e^{2-1/t}(t^{-2}+4)\leq8$ ; reflection treats the other half. Set

d_R(r)=r_z\chi_R(|r|),\quad d_0=\|d_R\|_\infty\leq R+1,\qquad H=h+H_s+gYd_R ,

(10.1)

with real constant $g$ . This is a prescribed compact-profile scalar parent. The finite annular electrostatic density $-\epsilon_0\Delta d_R$ does not by itself construct its quantized source, field energy or laboratory realization.

Lemma 10.1 (Physical domains and forcing tail)

The parent (10.1) is self-adjoint on $D(h)\cap D(H_s)$ and has form domain $H^1(\mathbb R^3\times\mathbb R)\cap\{Y\psi\in L^2\}$ . It satisfies

H+2+\frac{g^2d_0^2}{\Omega} \geq I+\tfrac14p_r^2+\tfrac12H_s .

(10.2)

For $f_R=d_R\phi$ and $R\geq40$ ,

\begin{align}\|f_R-f\|^2&\leq\epsilon_R^2:= e^{-2R}\left(\tfrac23R^4+\tfrac43R^3+2R^2+2R+1\right), \tag{10.3}\\ \|(h+1)^{1/2}(f_R-f)\| &\leq\sqrt{41.5+R^{-2}}\,\epsilon_R . \tag{10.4}\end{align}

At $R=40$ these upper bounds are respectively $6\cdot10^{-15}$ in norm and $4\cdot10^{-14}$ in first-form norm.

Proof

Hardy and interpolation make Coulomb infinitesimally Laplacian bounded. The independent electron and oscillator operators commute; after a common positive shift their sum has domain $D(h)\cap D(H_s)$ . Since $\|Y\psi\|\leq\sqrt{2/\Omega}\|H_s^{1/2}\psi\|$ and $d_R$ is bounded, $gYd_R$ is infinitesimally operator bounded relative to this sum. This proves the operator assertion and the corresponding common closed form.

The square $\frac14\|\nabla u+2\widehat r\,u\|^2\geq0$ gives $h+1\geq p_r^2/4$ . Young's inequality gives $gYd_R\geq-\Omega Y^2/4-g^2d_0^2/\Omega$ and $H_s-\Omega Y^2/4\geq H_s/2$ , proving (10.2).

The omitted forcing is confined to $r\geq R$ and has magnitude at most that of $u_f$ . Integrating $(4/3)r^4e^{-2r}$ gives (10.3). For the radial error $w=(\chi_R-1)u_f$ , $|u_f'|\leq u_f$ on this tail and therefore $|w'|\leq9u_f$ . The positive upper bound on the remaining radial form potential is $1+r^{-2}$ . Thus

\langle w,(h_1+1)w\rangle \leq\left(\tfrac{81}2+1+R^{-2}\right) \int_R^\infty|u_f|^2\,dr,

which is (10.4). Reciprocal positive Taylor sums for $e^{80}$ enclose the stated numerical tails.

□

The cutoff is essential to the actual source parent. With an uncut $gYr_z$ , completing the source square leaves $-g^2r_z^2/(2\Omega)$ . Electron packets translated to $r_z=L$ and source packets translated to $Y=-gL/\Omega$ have energy tending to $-\infty$ . Thus no stable uncut source Hamiltonian is being approximated here. Equations (10.3)–(10.4) compare forcing vectors, not complete actual parents.

10.1 Exact blocks, including source feedback

Let

P=|\phi\rangle\langle\phi|\otimes I,\qquad Q=I-P,\qquad \Psi=\phi p+q,\quad q=Q\Psi .

The projection $P$ does not reduce $H$ . The odd profile has $Pd_RP=0$ , so the exact equations are

\begin{align}i\dot q&=Aq+Lp,& A&=QHQ,& Lp&=gf_R\otimes Yp, \tag{10.5}\\ i\dot p&=(H_s-\tfrac12)p+ gY\langle\phi,d_Rq\rangle_r . \tag{10.6}\end{align}

The subscript on the last inner product means integration in the electron coordinate only. The actual vector $p(Y,t)$ retains its source phase and every finite internal coefficient. It is not prescribed as a free source waveform.

On $Q$ put

A_0=Q(h+H_s)Q,\qquad K_0=A_0+1,\qquad K=A+1,\qquad T_{\rm ph}=I+\tfrac12p_r^2+H_s .

(10.7)

Here $A_0$ is the self-adjoint restriction of $h+H_s$ to $Q\mathcal H$ . The spectral projection $Q$ preserves the common operator and form domains. More precisely, $A$ denotes the self-adjoint operator $A_0+gYQd_RQ$ on $D(A_0)$ : the electronic factor $Qd_RQ$ is bounded and commutes with the source coordinate, so the same infinitesimal relative-bound argument used in Lemma 10.1 applies. Thus $QHQ$ denotes this proved realization, rather than an assumed self-adjoint compression.

Since the complete excited electronic floor is $-1/8$ ,

K_0\geq7/8+H_s,\qquad K_0\geq p_r^2/4+H_s,\qquad T_{\rm ph}\leq(22/7)K_0 .

(10.8)

For the last inequality take $4/11$ of the first lower bound and $7/11$ of the second, then multiply by $22/7$ . This gives the exact coefficients $1$ and $1/2$ of the identity and electron kinetic terms; the source coefficient is larger than required.

The unbounded source interaction has the bounded form sandwich

\|K_0^{-1/2}gYQd_RQK_0^{-1/2}\| \leq\theta:=|g|d_0\sqrt{\frac{2}{\Omega(7/8)}}.

(10.9)

Indeed use $\|u\|\leq(7/8)^{-1/2}\|K_0^{1/2}u\|$ on one factor, and $\|Yv\|\leq\sqrt{2/\Omega}\|K_0^{1/2}v\|$ on the other. This bounds the full sesquilinear form and hence its operator. If $\theta<1$ , then

(1-\theta)K_0\leq K\leq(1+\theta)K_0 .

(10.10)

All these inequalities are physical form inequalities on the complete $Q$ space. They neither truncate the oscillator nor delete $QYd_RQ$ .

10.2 The spectral supremum and its correction

For $\zeta=E+i\eta$ , $\eta>0$ , the spectral theorem gives

\begin{align}J_{\rm spec}(E,\eta) &:=\|K_0^{1/2}(A_0-\zeta)^{-1}K_0^{1/2}\| =\sup_{a\in\sigma(A_0)}\frac{a+1}{|a-\zeta|} \notag\\ &\leq\overline J(E,\eta):= \sup_{a\geq-1/8+\Omega/2}\frac{a+1}{|a-\zeta|}. \tag{10.11}\end{align}

The inequality can be strict. The half-line in the last expression contains spectral gaps and cannot generally replace $\sigma(A_0)$ in an equality.

Example 10.2 (A strict gap in the half-line bound)

Take $\Omega=.01$ , $E=-.117$ , $\eta=.0001$ . The lowest spectral points of $A_0$ are $-.120$ and $-.110$ ; the $n\geq3$ electronic ladders begin above $-.051$ , and the continuum begins at $.005$ . At $a=E$ , the half-line expression is $8830$ . On the actual spectrum the maximum is

J_{\rm spec}=\frac{.88}{\sqrt{.003^2+.0001^2}}<294.

Indeed the first point gives a value above $280$ , while for every other spectral point $a\geq-.110$ the quotient is at most $.89/.007<128$ . Thus the half-line equality fails by a large factor even for this concrete parent.

Theorem 10.3 (An earned low-window inverse for the actual block)

If $\overline J\theta<1$ , then

\|K_0^{1/2}(A-\zeta)^{-1}LH_s^{-1/2}\| \leq\frac{|g|}{1-\overline J\theta} \left\{\frac2\Omega\sup_{s\geq\Omega/2} \int\frac{e+s+1}{|e+s-\zeta|^2}\,\mu_{f_R}(de)\right\}^{1/2}.

(10.12)

For the static row

\Omega=1/100,\quad |g|\leq10^{-6},\quad R=40,\quad d_0\leq41, \qquad E\leq-.47,\quad\eta>0,

(10.13)

the complete physical sandwich satisfies

\|T_{\rm ph}^{1/2}(A-E-i\eta)^{-1}LH_s^{-1/2}\|^2 <4.035635\cdot10^{-9}<4.05\cdot10^{-9}.

(10.14)

Proof

Let $V_0=K_0^{-1/2}gYQd_RQK_0^{-1/2}$ and $B_0=K_0^{1/2}(A_0-\zeta)^{-1}K_0^{1/2}$ . The bounded form pencil of $A-\zeta$ has inverse

K_0^{1/2}(A-\zeta)^{-1}K_0^{1/2} =(I+B_0V_0)^{-1}B_0.

The order is fixed: the pencil is $B_0^{-1}+V_0=B_0^{-1}(I+B_0V_0)$ . The Neumann bound is $\|(I+B_0V_0)^{-1}\|\leq(1-\overline J\theta)^{-1}$ . This is the concrete form-inverse construction of Theorem 5.1; no Hilbert-norm boundedness of $YQd_RQ$ has been assumed.

In the baseline squared norm, integrate out only the electron spectral measure. The remaining positive source operator is

H_s^{-1/2}YF(H_s)YH_s^{-1/2},\qquad F(s)=\int\frac{e+s+1}{|e+s-\zeta|^2}\,\mu_{f_R}(de).

Bound $F$ by its supremum, then use $\|YH_s^{-1/2}\|\leq\sqrt{2/\Omega}$ . This proves (10.12). Moving $Y$ through $F(H_s)$ or through the resolvent would not prove that estimate.

For (10.13), $\theta<.00062$ and $\overline J\leq88/35$ . To see the latter, first lower every denominator to $a+.47$ and use $a\geq-.12$ ; the resulting $(a+1)/(a+.47)$ decreases and has value $88/35$ at the floor. Thus the correction in (10.11) leaves the low-window bound unchanged.

The source supremum also needs its own argument. First dominate the integral kernel at the fixed endpoint $E=-.47,\eta=0$ , since $e+s\geq-.12$ . At this endpoint the derivative of $(e+s+1)/(e+s+.47)^2$ with respect to $s$ has sign $-(e+s+1.53)<0$ . It is consequently enough to use $s=1/200$ . This monotonicity was not asserted for every negative $E$ before making the endpoint domination.

For the uncut forcing, isolate the exact $2p$ mass $w_2=32768/59049$ . All remaining spectral mass lies at $e\geq-1/18$ , and the same endpoint kernel is decreasing in $e$ . Its complete bound-plus-continuum integral is therefore

F_f(1/200)\leq w_2\frac{352}{49}+(1-w_2)\frac{123048}{22801}<6.4 .

(10.15)

This analytic ceiling requires neither a continuum quadrature nor a truncation of the Rydberg series.

The compact forcing changes the baseline weighted inverse norm, before multiplication by $|g|$ , by at most

\overline J\,\epsilon_R\sqrt{\frac2{\Omega(7/8)}}.

This follows by inserting $K_0^{-1/2}$ and applying its lower form floor to $f_R-f$ and the ordered source factor. Use (10.8), $\epsilon_R<6\cdot10^{-15}$ , $\sqrt{1280}<35.778$ and $\sqrt{1600/7}<16$ . The resulting entirely rational upper bound is

\frac{22}{7}\,10^{-12} \frac{\big[35.778+(88/35)(6\cdot10^{-15})16\big]^2} {[1-(88/35)(.00062)]^2} <4.035635\cdot10^{-9}.

This proves (10.14), retaining the actual excited-block feedback in the inverse.

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The result is a uniform frequency-window statement with source weight $H_s^{1/2}$ . It does not say that the actual time-dependent input $p$ has Fourier support in this window. Such a use requires the temporal band and complementary-frequency terms of the window theorem. We next give a different all-frequency bound with the stronger, explicitly earned source weight $H_s$ .