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Scattering channels, stationary resolvents and time delay

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-10-22 (gpt-6.1-sol)Open item page →

Wave, scattering and channel operators

Definition

Assume the Axiom of Choice. Let H0,H be separable complex Hilbert spaces, H0 and H self-adjoint operators on them, and J:H0→H a bounded linear map. Write EH0 and EH for their spectral projection valued measures. The absolutely continuous subspace Hac(H0) consists of vectors u for which the scalar measure B↦∥EH0(B)u∥2 is absolutely continuous with respect to Lebesgue measure; Pac(H0) is its orthogonal projection. Define the two-space wave operators, when the pointwise strong limits exist on H0, by

Ω±(H,H0;J):=s-lim⁡t→±∞eitHJe−itH0Pac(H0),

where a strong limit means that for every u∈H0 the vector-valued expression converges in H. The two signs are separate limits. When J=I and H0=H, write Ω±(H,H0). A wave operator is complete when its range is all of Hac(H). When both wave operators exist and are isometries on Hac(H0), the scattering operator on that subspace is

S(H,H0;J):=Ω+∗Ω−.

Whenever both wave operators exist, are isometries on Hac(H0), and are complete with range Hac(H), the scattering operator S(H,H0;J) is unitary. For a fixed integer r≥1 and an open energy interval I⊂R, finite-channel data consists of the standard space Cr and a matrix field S:I→U(r), where

U(r):={U∈Mr(C):U∗U=UU∗=Ir}.

Here U∗ is conjugate transpose, tr⁡U=∑j=1rUjj, and det⁡U is the usual alternating determinant. This matrix field is separate specified data; the definition does not assert that it is obtained from the abstract wave operators above.

Channel diagram

Hac(H0)HHac(H0)¬¡S¬¤+

The upper arrow is Ω− restricted to the free absolutely continuous subspace. The right arrow is the full adjoint Ω+∗:H→Hac(H0); its codomain restriction follows from the factor Pac(H0) in the definition of Ω+. Whenever both limits exist, the diagonal arrow is their composite; if both are also isometries, this is the scattering operator S=Ω+∗Ω− defined above. The diagram records this composition only and asserts neither existence nor completeness.

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-10-22 (gpt-6.1-sol)Open item page →

Continued resolvent matrix elements and resonance poles

Definition

Let T be self-adjoint on a complex Hilbert space H whose inner product is linear in the first variable, use the resolvent convention RT(z)=(T−zI)−1, and let X⊆H be a specified vector subspace. For f,h∈X, the scalar resolvent matrix element is

rf,h(z):=⟨RT(z)f,h⟩,Im⁡z>0.

Given an open real interval I, a point E0∈I, and a disk D centered at E0, a local analytic continuation of these matrix elements through E0 is a specified family of holomorphic functions r~f,h:D→C agreeing with rf,h on D∩{Im⁡z>0}, for the stated f,h∈X. A meromorphic continuation on D is instead a specified family r~f,h agreeing with rf,h on D∩{Im⁡z>0}, holomorphic on D except for isolated poles. At every crossing point where no pole occurs it is a regular analytic continuation on a smaller neighborhood. For such a meromorphic continuation, if z0∈D is nonreal and for some f,h∈X the continued function satisfies

r~f,h(z)=(z−z0)−kq(z)

on a punctured neighborhood of z0, for an integer k≥1 and a holomorphic q with q(z0)≠0, then z0 is a pole of order k of that continued matrix element. Such a pole is called a resonance matrix-element pole for the specified pair (T,X) and continuation. This definition asserts neither that a continuation exists nor that a pole is an eigenvalue of T; the continuation and the chosen test vectors are part of the data.

LemmaStatement: AI-adaptedProof: AI-generatedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Explicit Gaussian continuation of a free resolvent matrix element

Statement

Assume the Axiom of Choice and Countable Choice. On H=L2(R3,dp;C) with ⟨f,h⟩=∫R3f(p)h(p)‾ dp, let H0=M∣p∣2 on its maximal domain {f:∣p∣2f∈L2}, let g(p)=π−3/4e−∣p∣2/2, and put

m(z):=⟨(H0−zI)−1g,g⟩,Im⁡z>0.

This is a dimensionless model; no physical units are assigned to p or z.

For w∈C, define the entire function

E(w):=2π∫Γwe−ζ2 dζ,Γw={w+t:t≥0},

where Γw is the horizontal ray oriented from w toward +∞+w. For any E0>0, take a disk D centered at E0 with 0∉D, and the holomorphic square root k:D→C with k(E0)=E0>0. Then m has the local analytic continuation through E0

mcont(z)=2+2iπ k(z)e−zE(−ik(z)),z∈D.

Facts & Assumptions

Given: The Hilbert space, multiplication operator, vector g, scalar resolvent matrix element, and local continuation definition in the Statement.

[A1]

Assume the Axiom of Choice and Countable Choice as required by the self-adjoint multiplication and integration suppliers (The Axiom of Choice, The Axiom of Countable Choice (ACω)).

[F1]

Real multiplication by a measurable finite real function on its maximal domain is self-adjoint; in particular H0=M∣p∣2≥0 on the domain stated above (Real multiplication operators and free plus bounded-potential Schrödinger operators).

[F3]

For a radial integrand in R3, dp=4πr2 dr after angular integration (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

[F4]

Nonnegative iterated integrals may be interchanged by Tonelli's theorem on sigma-finite product measure spaces (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F5]

Dominated convergence applies to the finite integrals and locally uniform parameter bounds used below (Dominated convergence).

[F6]

Holomorphic functions agreeing on a set with an accumulation point in a connected domain agree throughout that domain (Identity theorem for holomorphic functions).

Proof

Given: [A1], the Hilbert space, multiplication operator, normalized vector g, scalar resolvent element, and the disk and branch data in the Statement.

1.1A1F1F2F3F4F5

By [F1], H0 is self-adjoint and nonnegative. The Gaussian integral in [F2], applied in each coordinate and combined with Tonelli [F4], gives ∫R3e−∣p∣2 dp=π3/2, so ∥g∥=1. Polar coordinates [F3] and the resolvent multiplier therefore give m(z)=4π∫0∞r2e−r2r2−z dr for Im⁡z>0. The denominator is nonzero on the integration ray, and the integral is locally uniformly dominated on the upper half-plane, so it defines the stated resolvent matrix element.

1.2F2F4algebra

For Re⁡a>0, put I(a)=∫0∞e−r2(r2+a2)−1 dr. When a>0 is real, the Laplace representation of (r2+a2)−1 and Tonelli [F4] give I(a)=∫0∞e−a2s∫0∞e−(1+s)r2 dr ds=π2∫0∞e−a2s1+s ds by [F2]. The substitutions u=1+s and v=au then give I(a)=πea2∫1∞e−a2u2 du=πea2a∫a∞e−v2 dv=π2aea2E(a), using the horizontal-ray definition of E for real a>0.

2.1step 1.2F2F5F6algebra

Fix a0 with Re⁡a0>0 and a closed disk K about a0 contained in that half-plane. Choose η>0 and A<∞ so that Re⁡a≥η and ∣a∣≤A on K. For r≥0, ∣r2+a2∣=∣a+ir∣∣a−ir∣≥η2. For nonzero complex δ small enough that a0+δ∈K, the exact difference quotient of the integrand of I is e−r2δ(1r2+(a0+δ)2−1r2+a02)=−(2a0+δ)e−r2(r2+(a0+δ)2)(r2+a02). For ∣δ∣≤1, its modulus is at most (2A+1)η−4e−r2, an integrable majorant by [F2]. The quotients are measurable in r, and tend pointwise to −2a0e−r2/(r2+a02)2. Applying [F5] along every sequence of nonzero complex increments tending to zero proves that I is complex differentiable at a0, with derivative the integral of this limit. Thus I is holomorphic on the right half-plane; this conclusion uses the actual difference quotients, not an assumption that dominated convergence is itself a differentiation theorem. For E, put F(w,t)=e−(w+t)2. Its exact segment identity is F(w+δ,t)−F(w,t)δ=∫01−2(w+sδ+t)e−(w+sδ+t)2 ds. For completeness this identity follows by expanding the exponential series on the finite segment, differentiating each polynomial in s, and integrating it: the series and its derivative converge uniformly there, since for ∣w+sδ+t∣≤B they are dominated respectively by ∑B2n/n! and ∣δ∣∑n≥12nB2n−1/n!, both finite. Hence passage through the finite segment integral is justified by the uniform remainder bound. For ∣w∣≤R, ∣δ∣≤1 and 0≤s≤1, write w+sδ=u+iv; then ∣u∣,∣v∣≤R+1 and (t+u)2≥t2/2−u2. Consequently the segment derivative is bounded in modulus by CR(1+t)e−t2/2, independently of w,δ,s. This majorant is integrable: (1+t)e−t2/4 is bounded, and the remaining scaled Gaussian is integrable by [F2]. The original integrand is bounded the same way without the factor 1+t. The measurable difference quotients therefore have one integrable majorant and tend pointwise, as complex δ→0, to −2(w+t)e−(w+t)2. Apply [F5] to every such sequence to obtain the complex derivative of E at every w. It is entire. Both sides of step 1.2 are therefore holomorphic on the connected half-plane Re⁡a>0; their agreement on the positive real axis has an accumulation point inside this domain, so [F6] extends the identity throughout the half-plane.

3.1step 2.1F2F5algebra∎

For Im⁡z>0, choose the square root a=−z with Re⁡a>0. Since r2/(r2−z)=1+z/(r2−z), m(z)=2+4zπI(a); substituting the identity from step 2.1 and z=−a2 gives m(z)=2−2π ae−zE(a). On D∩{Im⁡z>0}, the root a equals −ik(z): both square to −z, and Re⁡(−ik(z))>0. Substitution gives m(z)=2+2iπ k(z)e−zE(−ik(z)) there. The right hand side is holomorphic throughout D, so it is the required local continuation through E0.

TheoremStatement: AI-adaptedProof: AI-generatedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Compact-support short-range scattering closure

Statement

Assume the Axiom of Choice and Countable Choice. Let V∈L∞(R3;R) and suppose there is a compact set K⊂R3 such that V=0 almost everywhere outside K. Let H=L2(R3;C), and set H0=−Δ and H=H0+MV on D(H0)=D(H)=H2(R3;C). Use

f^(p)=(2π)−3/2∫R3e−ip⋅xf(x) dx,Ω±=s-lim⁡t→±∞eitHe−itH0,S=Ω+∗Ω−.

Here MVf=Vf, and the strong limits act on every f∈H. Then:

(a) The wave operators exist, are isometries, and satisfy

ran⁡Ω+=ran⁡Ω−=Hac(H)=Hc(H),Hsc(H)={0}.

For u∈H, write μu(B)=∥EH(B)u∥2. The subspace Hac(H) consists of vectors with μu≪dE; Hsc(H) is the subspace with atomless scalar spectral measures singular with respect to Lebesgue measure; Hpp(H) is the closed span of eigenspaces, and Hc(H)=Hpp(H)⊥. These are the canonical spectral-type subspaces of the PVM.

(b) The scattering operator S is unitary on H and commutes with e−itH0 for every t∈R.

Use time in the corresponding ℏ=1 evolution convention and the standard scaled Schrödinger normalization in which the kinetic coefficient ℏ2/(2m) is set to 1. When length is dimensional, E and V then have units L−2 and k=E has units L−1; the proof is an operator deduction in this normalization.

(c) Put h=L2(S2,dσ), where dσ(S2)=4π, and define the normalized free energy transform

(F0f)(E,ω)=E1/42f^(E ω),F0:H→L2((0,∞),dE;h).

Fix a bounded ball B containing K. For E>0, k=E, define bounded maps Z0(E):L2(B)→h and Z0(E)∗:h→L2(B) by

(Z0(E)f)(ω)=k4π3/2∫Be−ikω⋅xf(x) dx,(Z0(E)∗a)(x)=k4π3/2∫S2eikω⋅xa(ω) dσ(ω).

Let RH(E+i0):L2(B)→L2(B) be the operator-norm limit lim⁡ϵ↓01B(H−(E+iϵ)I)−11B; existence and continuity for E>0 are part of the stated supplier input. The products below are these bounded maps on B. Then F0 is unitary and there is a measurable unitary field Sstat(E)∈B(h) such that

F0SF0∗=MSstat(E),Sstat(E)=Ih−2πi Z0(E)[V−VRH(E+i0)V]Z0(E)∗,E>0.

The formula uses only positive energies; it makes no claim at the threshold E=0, and no long-range Coulomb potential is included.

Facts & Assumptions

Given: AC and CC; a real bounded compactly supported V; the free Fourier convention, operators and wave limits in the Statement.

[A1]

Assume AC (The Axiom of Choice).

[F1]

The real multiplier and bounded-potential theorem gives H0=−Δ self-adjoint on H2, H0≥0, and H=H0+MV self-adjoint on the same domain (Real multiplication operators and free plus bounded-potential Schrödinger operators).

[F2]

For the exact hypotheses V∈L∞(R3;R), compact support, H0=−Δ, and H=H0+MV on H2, the compact-support completeness supplier proves that Ω± exist by Cook's construction, their individual ranges equal Hac(H)=Hc(H), the singular-continuous subspace is zero, and S=Ω+∗Ω− is unitary and commutes with the free group (Compact-support asymptotic completeness).

[F3]

For the same real bounded compactly supported V, the stationary/dynamical supplier uses f^(p)=(2π)−3/2∫e−ip⋅xf(x) dx, F0f(E,ω)=E1/4f^(E ω)/2, and proves that F0 is unitary, that Z±=F0Ω±∗, and that the positive-energy stationary matrix is

I−2πi Z0(E)[V−VRH(E+i0)V]Z0(E)∗.

It proves this matrix is unitary for every E>0, with the local resolvent and compact-region adjoint conventions in the Statement (Stationary and dynamical scattering in the full energy-angular representation).

[F4]

The definitions of Ω±, completeness, and S=Ω+∗Ω− are those of Wave, scattering and channel operators.

[F5]

The canonical spectral-type decomposition defines the absolutely continuous, singular-continuous, and pure-point subspaces from the spectral PVM and scalar measures μu(B)=∥EH(B)u∥2 (Canonical decomposition into pure point, absolutely continuous and singular continuous parts).

Proof

Given: [A1], [A2], [F1]–[F5], and the operators and conventions stated above.

1.1A1A2F1F2algebra

The Fourier convention in [F1] is f^2π(ξ)=∫e−2πix⋅ξf(x) dx; with p=2πξ, its multiplier 4π2∣ξ∣2 is exactly ∣p∣2, and the unitary momentum transform is (2π)−3/2∫e−ip⋅xf(x) dx. Thus the H0 and domain H2 in [F2] agree with those in the Statement. Since the potential is the same real bounded compactly supported multiplication operator, [F1] also gives the self-adjoint realization required by [F2].

1.2F2F4F5algebra

By [F2], the two stated pointwise strong limits exist with the exact sign convention eitHe−itH0, are isometries, and have the stated individual ranges. That same supplier proves the absent singular-continuous subspace; [F5] fixes the PVM meaning of these spectral subspaces. In the degenerate case V=0, the approximants are eitH0e−itH0=I, so Ω±=I and S=I. By [F4], the supplier's scattering operator is exactly the S defined in the Statement, so its unitarity and free-group commutation give (b).

1.3F3algebra

The Fourier scaling and energy change of variables in [F3] use the same momentum transform and dσ(S2)=4π as in the Statement. Hence the local F0, Z0(E), resolvent boundary value, and compact-region generalized adjoint agree term-for-term with [F3]; no independent normalization or sign change is introduced.

2.1A1A2F1F2F3F4step 1.1step 1.2step 1.3∎

Applying [F3] with the identical H0,H,V now proves that F0 is unitary, that F0SF0∗ is multiplication by Sstat(E), and that the displayed stationary formula is unitary for every E>0. The result is conditional on AC, CC and the compact-support model and makes no threshold or Coulomb assertion. This proves (a)–(c).

TheoremStatement: AI-adaptedProof: AI-generatedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Wigner–Smith matrix for differentiable finite channels

Statement

Let I⊂R be an open energy interval with fixed finite channel number r≥1, let S:I→U(r) be C1 in the finite-channel convention of Wave, scattering and channel operators, and let ℏ>0. Define

Q(E):=−iℏ S(E)∗S′(E),E∈I.

Then Q(E) is Hermitian and

tr⁡Q(E)=−iℏ (det⁡S)′(E)det⁡S(E)=−iℏ ddElog⁡det⁡S(E),

where the last expression means the logarithmic derivative (det⁡S)′/det⁡S, independent of a choice of logarithm branch. If the incoming and outgoing bases are changed by energy-independent unitary matrices Uin and Uout, so that S♯=Uout∗SUin, then

Q♯=Uin∗QUin.

If E is measured in joules and ℏ in joule-seconds, Q has units of seconds. This is an algebraic scattering-matrix result; it does not by itself establish an operational arrival-time or dwell-time interpretation.

Facts & Assumptions

Given: The open interval I, fixed r, C1 unitary matrix field S(E), and ℏ>0.

[F1]

A finite-channel scattering matrix is a matrix field S(E)∈U(r); S∗ is conjugate transpose, while trace and determinant are the usual finite-dimensional trace and alternating determinant (Wave, scattering and channel operators).

[F2]

The finite-dimensional derivative, product and trace operations are entrywise operations on C1 scalar functions; determinant is the alternating multilinear polynomial in the matrix entries (Wave, scattering and channel operators).

Proof

Given: [F1], [F2] and the data in the Statement.

1.1givenF1F2algebra

Differentiating S(E)∗S(E)=Ir entrywise gives S′(E)∗S(E)+S(E)∗S′(E)=0. Thus A:=S∗S′ satisfies A∗=−A, and Q∗=(−iℏA)∗=iℏA∗=−iℏA=Q.

2.1step 1.1F1F2algebra

Since S is unitary, S−1=S∗ and det⁡S≠0. Multilinearity of the alternating determinant gives det⁡(Ir+hA+o(h))=1+htr⁡(A)+o(h) for A=S(E)−1S′(E): terms linear in h are exactly the r choices of one diagonal entry, while every nonidentity permutation or two off-diagonal factors contributes o(h). Applying this to S(E+h)=S(E)(Ir+hS(E)−1S′(E)+o(h)) yields (det⁡S)′(E)=det⁡S(E)tr⁡(S(E)−1S′(E))=det⁡S(E)tr⁡(S(E)∗S′(E)). Multiplication by −iℏ/det⁡S(E) proves the trace formula.

3.1step 1.1givenalgebra∎

For constant unitary Uin,Uout, differentiation gives (S♯)′=Uout∗S′Uin, while (S♯)∗=Uin∗S∗Uout. Therefore (S♯)∗(S♯)′=Uin∗S∗S′Uin, proving the stated conjugation rule. Energy-dependent basis changes introduce derivatives of the basis matrices, so they are not covered by this claim.

TheoremStatement: AI-adaptedProof: AI-generatedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Resonance poles in an analytic rank-one model

Statement

Assume the Axiom of Choice and Countable Choice. Let H0 be a self-adjoint operator on a complex Hilbert space H with inner product linear in its first variable, bounded below, let g∈H, let λ∈R∖{0}, and put

Pgf=⟨f,g⟩g,Hλ=H0+λPgon D(H0),R0(z)=(H0−zI)−1.

Adopt as an explicit conditional input the rank-one resolvent interface: Hλ is self-adjoint, and whenever z∈ρ(H0) and d(z):=1+λm(z)≠0, where m(z)=⟨R0(z)g,g⟩, one has

(Hλ−zI)−1=R0(z)−λR0(z)PgR0(z)d(z).

Spectral variables use the energy unit of H0. If ∥g∥ is dimensionless, m(z) has inverse-energy units and λ has energy units; in general the coupling is chosen so that λm(z) is dimensionless.

Let E⋆∈R, take an open disk D centered at E⋆, and let E⋆ lie in an open interval I on which the spectral PVM EH0(⋅) of H0 is absolutely continuous: EH0(N)=0 for every Lebesgue-null Borel set N⊆I. Suppose m and, for specified test vectors f,h in a subspace X⊆H, the free matrix elements

⟨R0(z)f,h⟩,af(z):=⟨R0(z)f,g⟩,bh(z):=⟨R0(z)g,h⟩

have holomorphic continuations through E⋆ to D. Suppose dcont(z)=1+λmcont(z) has a zero z0∈D with Im⁡z0<0 of finite order k≥1, meaning that on some neighborhood of z0 one has dcont(z)=(z−z0)kq(z) for a holomorphic function q with q(z0)≠0, and suppose af(z0)bh(z0)≠0. Then the continued matrix element ⟨(Hλ−zI)−1f,h⟩ has a resonance pole of order k. In the simple-zero case its residue is

−af(z0)bh(z0)mcont′(z0).

In particular, for the Gaussian model H0=M∣p∣2, g(p)=π−3/4e−∣p∣2/2, and X=span⁡{g}, take E⋆>0, an open disk D centered at E⋆ with 0∉D, and the holomorphic square-root branch kD:D→C normalized by kD(E⋆)=E⋆>0. (The subscript distinguishes this branch from the zero order k.) The continuation in the Gaussian lemma with precisely these disk and branch data supplies mcont; every interval I⊂(0,∞) is absolutely continuous by the radial spectral measure formula. Every finite-order zero of 1+λmcont in the lower half of such a disk gives a pole of the continued scalar element ⟨(Hλ−zI)−1g,g⟩. These poles are not nonreal eigenvalues of the self-adjoint Hλ.

Statement references

Facts & Assumptions

Given: AC and CC; the self-adjoint lower-semibounded H0, g, real nonzero λ, Hλ and R0 in the Statement; the specified test space and local continuations.

[A1]

The rank-one resolvent theorem is used at its exact current interface: under AC, self-adjoint lower-semibounded H0, g∈H, real λ, Pgf=⟨f,g⟩g, and Hλ on D(H0), it gives self-adjointness and the displayed resolvent identity when 1+λ⟨R0(z)g,g⟩≠0 (Rank-one resolvent and bound-state equation). The complete current proof and hash-current Step 1 ready record were inspected; this use matches both exactly.

[F1]

The continuation and pole terms have the precise matrix-element meaning of Continued resolvent matrix elements and resonance poles.

[F2]

For the Gaussian model, E⋆>0, a disk D centered at E⋆ with 0∉D, and the holomorphic branch kD2=z on D with kD(E⋆)=E⋆>0, the explicit horizontal-contour formula supplies a holomorphic mcont agreeing with m on D∩{Im⁡z>0}. These are the exact branch and disk hypotheses of Explicit Gaussian continuation of a free resolvent matrix element.

[F3]

A self-adjoint Hλ has real spectrum; every nonreal z is in its resolvent set (Resolvent of a self-adjoint operator: nonreal resolvents and the estimate).

[F4]

The spectral PVM of H0=M∣p∣2 acts by multiplication by 1{∣p∣2∈B}, and its scalar measure for u∈L2(R3) is ∫{∣p∣2∈B}∣u(p)∣2 dp (Spectral theorem for unbounded self-adjoint operators (PVM form), Projection valued measure, Integral of a measurable function against a projection-valued measure).

[F5]

Polar integration on R3 gives dp=r2 dr dσ(ω), with finite unit-sphere measure (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

[F6]

Since Lebesgue measure on R3 is the completion of its Borel restriction, every Lebesgue-measurable function agrees almost everywhere with a Borel-measurable function; apply this to ∣u∣2 (L(Rn) is exactly the completion of the restriction of λn to the Borel sets, A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra).

Proof

Given: [A1], [F1]–[F6], and the free matrix-element continuations in the Statement.

1.1A1F3givenalgebra

Fix z with Im⁡z>0. Self-adjointness of H0 puts z in ρ(H0); self-adjointness of Hλ in [A1] puts it in ρ(Hλ). Also d(z)≠0: if d(z)=0, then g≠0, u=R0(z)g≠0, and (Hλ−zI)u=g+λ⟨R0(z)g,g⟩g=0, contradicting z∈ρ(Hλ). Thus [A1] applies on the upper half-plane.

1.2A1givenF1algebra

Pairing the rank-one identity [A1] against f,h and using the first-variable-linear convention gives ⟨Rλ(z)f,h⟩=⟨R0(z)f,h⟩−λ af(z)bh(z)d(z). Each numerator term has the holomorphic continuation assumed in the Statement, so the right side defines the corresponding meromorphic continuation on D.

2.1step 1.2givenF1algebra

By the finite-order zero hypothesis, dcont(z)=(z−z0)kq(z) near z0, where q is holomorphic and q(z0)≠0. The product afbh is holomorphic and nonzero at z0, so its quotient by dcont has a pole of order exactly k. The first term ⟨R0(z)f,h⟩cont is holomorphic and cannot cancel that pole. Thus the continued Hλ matrix element has the asserted resonance pole. If k=1, its residue is −λaf(z0)bh(z0)/dcont′(z0)=−af(z0)bh(z0)/mcont′(z0).

3.1F2F4F5F6step 2.1algebra

In the Gaussian model take exactly the positive center E⋆, disk excluding zero, and normalized holomorphic branch kD specified in [F2]. Its formula therefore supplies the assumed holomorphic continuation on this D, not a continuation across the threshold zero or a differently signed sheet. Choose a Borel representative of ∣u∣2 by [F6]. The PVM formula [F4] and polar integration [F5] then give, for every u∈L2(R3), μu(B)=∫{∣p∣2∈B}∣u(p)∣2 dp=∫B∩(0,∞)E2∫S2∣u(E ω)∣2 dσ(ω) dE. The substitution E=r2 gives r2 dr=(E/2) dE. Thus EH0(N)=0 for every Lebesgue-null N⊂(0,∞), so any open I⊂(0,∞) meets the Statement's absolute-continuity condition. For f=h=g, one has ag=bg=mcont. At a zero of 1+λmcont, mcont(z0)=−1/λ≠0. Hence the noncancellation condition of step 2.1 holds at every finite-order zero, proving the stated Gaussian specialization.

4.1A1algebra∎

If a nonzero vector u∈D(Hλ) satisfied Hλu=z0u with Im⁡z0<0, self-adjointness in [A1] would make ⟨Hλu,u⟩ real, but it equals z0∥u∥2, which is nonreal. This contradiction proves that the continued pole is not a nonreal eigenvalue.

5 · Examples, experiments and counterexamples

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Sources