An electron with a retained quantum oscillator source

For a smooth test in the electron coordinate,

The integration by parts uses $\operatorname{div}\widehat r=2/|r|$ and extends by Hardy to $H^1$. Thus $h+1\geq p_r^2/4$. Young's inequality gives $gYd_R\geq-\Omega Y^2/4-g^2\|d_R\|_\infty^2/\Omega$, and $H_s-\Omega Y^2/4\geq H_s/2$. In particular

The true common form domain is $\mathcal V=H^1(\mathbb R^3_r\times\mathbb R_Y) \cap\{Yu\in L^2\}$. Coulomb is infinitesimally Laplacian bounded by Hardy and interpolation. The independent source oscillator satisfies $\|Yu\|\leq\sqrt{2/\Omega}\|H_s^{1/2}u\|$; hence the bounded-profile coupling is infinitesimally operator bounded against $H_{\rm c}=h+H_s$. After a shift the two central operators are positive and strongly commute, so $D(H_{\rm c})=D(h)\cap D(H_s)$. This is the common first domain of every $H(t)$.

Use the tangent/source estimate operator $\mathcal A=1+J_e^2+N_Y$, $J_e=r\times(-i\nabla_r)$. It strongly commutes with $H_{\rm c}$. Angular commutation with multiplication by $d_R$ produces a bounded angular derivative and at most one angular generator; source commutation gives $[N_Y,Y]=-\partial_Y$ and $[N_Y,\partial_Y]=-Y$. Thus each $\operatorname{ad}_{\mathcal A}^{\ell}(Yd_R)$, $\ell\leq3$, has tangent/source degree at most $\ell+1\leq2\ell$ and uses at most six profile jets. Angular representation and oscillator ladder bounds give (6.1), while four differentiated jets and integrable $g'$ give (6.2). The source $Y$ is unbounded but its degree has been paid explicitly.

The entrance lies in every $\mathcal A$ graph and in the true first $H$ domain. Its central part is an eigenvector, and $Yd_R f$ has only electron angular degree one and source occupations zero and two. These have square-integrable radial factors, so $H(0)f\in D(\mathcal A^2)$. The test core crosses $r=0$ in the operator domain; a collision-deleted core is used only for form density. The spectral projections of $\mathcal A$ commute with the shifted central form and converge on its joint core. All hypotheses of Theorem 6.1 are thereby verified. In particular the weakly closed identity is

On compact electron annuli and bounded source charts, subtract the angular/source terms, controlled by $\mathcal A^3\psi$, to obtain the genuine radial second-order equation for $\mathcal A^2\psi$. There are three tangent coordinates, the two electron angles and $Y$. Equations (6.8)–(6.9) with $\theta>3/8$ give joint spacetime continuity and local full $H^2$ away from the collision. The argument has not required a high power of the full Coulomb Hamiltonian.

Testing the physical form against real multipliers yields the full Cartesian conservation law $\partial_t\rho+\operatorname{div}_rj_r+\partial_Yj_Y=0$ on all four coordinates, with no deleted-collision boundary source. The coercive floor and first graph give finite current length/action and logarithmic costs by Proposition 3.4. Hardy in the electron coordinate gives

The smooth logarithmic collision test used in (8.3) excludes every-time collision encounters for the reference paths. The initial source node is a different issue; Lemma 3.2 applies to the proved continuous full wave and finite log/action costs, without deleting that node. The deterministic theorem now gives the conclusion. Since the entrance density is positive almost everywhere except the measure-zero source nodal plane, any original joint Lebesgue-AC law is also AC relative to it. Singular entrance mass on that plane is outside this assertion.

$Hf=(E_0+gYd_R)f$, where $f$ is a real spatial factor times a fixed internal vector. Both $f$ and $Hf$ belong to the physical form domain: the cusp is $H^1$, the profile is smooth, and all oscillator polynomial factors have finite first form. Spectral calculus for the constant parent therefore gives differentiability at zero in the form norm with $\dot\psi(0)=-iHf$. Differentiating the current bilinear, whose form-to-$L^1$ continuity follows by Cauchy, cancels the two terms containing $(E_0+gYd_R)f\nabla f$. The remaining term is $-\rho_0\nabla(gYd_R)$, with the appropriate kinetic coefficient, which gives (9.4). The profile $d_R$ is odd under $z\mapsto-z$ and the entrance electron density is even, so its electron integral vanishes. At a generic configuration $d_R\rho_0\neq0$. Replacing $Y$ by its mean would erase this complete-coordinate response. The result is a first derivative for constant entrance coupling, not an asserted $O(t^2)$ remainder for an arbitrary merely AC control.