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Resonance poles in an analytic rank-one model

Statement

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

$P_{g} f = ⟨ f, g ⟩ g, H_{λ} = H_{0} + λ P_{g} on D (H_{0}), R_{0} (z) = (H_{0} - z I)^{- 1} .$

Adopt as an explicit conditional input the rank-one resolvent interface: $H_{λ}$ is self-adjoint, and whenever $z \in ρ (H_{0})$ and $d (z) : = 1 + λ m (z) \neq 0$ , where $m (z) = ⟨ R_{0} (z) g, g ⟩$ , one has

$(H_{λ} - z I)^{- 1} = R_{0} (z) - \frac{λ R_{0} (z) P_{g} R_{0} (z)}{d (z)} .$

Spectral variables use the energy unit of $H_{0}$ . If $∥ g ∥$ is dimensionless, $m (z)$ has inverse-energy units and $λ$ has energy units; in general the coupling is chosen so that $λ m (z)$ is dimensionless.

Let $E_{⋆} \in R$ , take an open disk $D$ centered at $E_{⋆}$ , and let $E_{⋆}$ lie in an open interval $I$ on which the spectral PVM $E_{H_{0}} (\cdot)$ of $H_{0}$ is absolutely continuous: $E_{H_{0}} (N) = 0$ for every Lebesgue-null Borel set $N \subseteq I$ . Suppose $m$ and, for specified test vectors $f, h$ in a subspace $X \subseteq H$ , the free matrix elements

$⟨ R_{0} (z) f, h ⟩, a_{f} (z) : = ⟨ R_{0} (z) f, g ⟩, b_{h} (z) : = ⟨ R_{0} (z) g, h ⟩$

have holomorphic continuations through $E_{⋆}$ to $D$ . Suppose $d_{c o n t} (z) = 1 + λ m_{c o n t} (z)$ has a zero $z_{0} \in D$ with $Im z_{0} < 0$ of finite order $k \geq 1$ , meaning that on some neighborhood of $z_{0}$ one has $d_{c o n t} (z) = (z - z_{0})^{k} q (z)$ for a holomorphic function $q$ with $q (z_{0}) \neq 0$ , and suppose $a_{f} (z_{0}) b_{h} (z_{0}) \neq 0$ . Then the continued matrix element $⟨ (H_{λ} - z I)^{- 1} f, h ⟩$ has a resonance pole of order $k$ . In the simple-zero case its residue is

$- \frac{a_{f} (z_{0}) b_{h} (z_{0})}{m_{c o n t}^{'} (z_{0})} .$

In particular, for the Gaussian model $H_{0} = M_{∣ p ∣^{2}}$ , $g (p) = π^{- 3 / 4} e^{- ∣ p ∣^{2} / 2}$ , and $X = span {g}$ , take $E_{⋆} > 0$ , an open disk $D$ centered at $E_{⋆}$ with $0 \notin D$ , and the holomorphic square-root branch $k_{D} : D \to C$ normalized by $k_{D} (E_{⋆}) = \sqrt{E_{⋆}} > 0$ . (The subscript distinguishes this branch from the zero order $k$ .) The continuation in the Gaussian lemma with precisely these disk and branch data supplies $m_{c o n t}$ ; every interval $I \subset (0, \infty)$ is absolutely continuous by the radial spectral measure formula. Every finite-order zero of $1 + λ m_{c o n t}$ in the lower half of such a disk gives a pole of the continued scalar element $⟨ (H_{λ} - z I)^{- 1} g, g ⟩$ . These poles are not nonreal eigenvalues of the self-adjoint $H_{λ}$ .

Statement references

The Axiom of Choice

The Axiom of Countable Choice ( ${A C}_{ω}$ )

Continued resolvent matrix elements and resonance poles

Projection valued measure

Integral of a measurable function against a projection-valued measure

Explicit Gaussian continuation of a free resolvent matrix element

Real multiplication operators and free plus bounded-potential Schrödinger operators

Resolvent of a self-adjoint operator: nonreal resolvents and the estimate

Spectral theorem for unbounded self-adjoint operators (PVM form)

Rank-one resolvent and bound-state equation

Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma

$L (R^{n})$ is exactly the completion of the restriction of $λ_{n}$ to the Borel sets

A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra

Facts & Assumptions

Given: AC and CC; the self-adjoint lower-semibounded $H_{0}$ , $g$ , real nonzero $λ$ , $H_{λ}$ and $R_{0}$ 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 $H_{0}$ , $g \in H$ , real $λ$ , $P_{g} f = ⟨ f, g ⟩ g$ , and $H_{λ}$ on $D (H_{0})$ , it gives self-adjointness and the displayed resolvent identity when $1 + λ ⟨ R_{0} (z) g, g ⟩ \neq 0$ (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 $0 \notin D$ , and the holomorphic branch $k_{D}^{2} = z$ on $D$ with $k_{D} (E_{⋆}) = \sqrt{E_{⋆}} > 0$ , the explicit horizontal-contour formula supplies a holomorphic $m_{c o n t}$ agreeing with $m$ on $D \cap {Im z > 0}$ . These are the exact branch and disk hypotheses of Explicit Gaussian continuation of a free resolvent matrix element.

[F3]

A self-adjoint $H_{λ}$ 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 $H_{0} = M_{∣ p ∣^{2}}$ acts by multiplication by $1_{{∣ p ∣^{2} \in B}}$ , and its scalar measure for $u \in L^{2} (R^{3})$ is $\int_{{∣ p ∣^{2} \in B}} ∣ u (p) ∣^{2} d p$ (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 $R^{3}$ gives $d p = r^{2} d r d σ (ω)$ , with finite unit-sphere measure (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

[F6]

Since Lebesgue measure on $R^{3}$ is the completion of its Borel restriction, every Lebesgue-measurable function agrees almost everywhere with a Borel-measurable function; apply this to $∣ u ∣^{2}$ ( $L (R^{n})$ is exactly the completion of the restriction of $λ_{n}$ 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 $H_{0}$ puts $z$ in $ρ (H_{0})$ ; self-adjointness of $H_{λ}$ in [A1] puts it in $ρ (H_{λ})$ . Also $d (z) \neq 0$ : if $d (z) = 0$ , then $g \neq 0$ , $u = R_{0} (z) g \neq 0$ , and $(H_{λ} - z I) u = g + λ ⟨ R_{0} (z) g, g ⟩ g = 0$ , contradicting $z \in ρ (H_{λ})$ . 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 $⟨ R_{λ} (z) f, h ⟩ = ⟨ R_{0} (z) f, h ⟩ - \frac{λ a_{f} (z) b_{h} (z)}{d (z)}$ . 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, $d_{c o n t} (z) = (z - z_{0})^{k} q (z)$ near $z_{0}$ , where $q$ is holomorphic and $q (z_{0}) \neq 0$ . The product $a_{f} b_{h}$ is holomorphic and nonzero at $z_{0}$ , so its quotient by $d_{c o n t}$ has a pole of order exactly $k$ . The first term $⟨ R_{0} (z) f, h ⟩_{c o n t}$ is holomorphic and cannot cancel that pole. Thus the continued $H_{λ}$ matrix element has the asserted resonance pole. If $k = 1$ , its residue is $- λ a_{f} (z_{0}) b_{h} (z_{0}) / d_{c o n t}^{'} (z_{0}) = - a_{f} (z_{0}) b_{h} (z_{0}) / m_{c o n t}^{'} (z_{0})$ .

3.1F2F4F5F6step 2.1algebra

In the Gaussian model take exactly the positive center $E_{⋆}$ , disk excluding zero, and normalized holomorphic branch $k_{D}$ 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 $u \in L^{2} (R^{3})$ , $μ_{u} (B) = \int_{{∣ p ∣^{2} \in B}} ∣ u (p) ∣^{2} d p = \int_{B \cap (0, \infty)} \frac{\sqrt{E}}{2} \int_{S^{2}} ∣ u (\sqrt{E} ω) ∣^{2} d σ (ω) d E$ . The substitution $E = r^{2}$ gives $r^{2} d r = (\sqrt{E} / 2) d E$ . Thus $E_{H_{0}} (N) = 0$ for every Lebesgue-null $N \subset (0, \infty)$ , so any open $I \subset (0, \infty)$ meets the Statement's absolute-continuity condition. For $f = h = g$ , one has $a_{g} = b_{g} = m_{c o n t}$ . At a zero of $1 + λ m_{c o n t}$ , $m_{c o n t} (z_{0}) = - 1 / λ \neq 0$ . 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 $u \in D (H_{λ})$ satisfied $H_{λ} u = z_{0} u$ with $Im z_{0} < 0$ , self-adjointness in [A1] would make $⟨ H_{λ} u, u ⟩$ real, but it equals $z_{0} ∥ u ∥^{2}$ , which is nonreal. This contradiction proves that the continued pole is not a nonreal eigenvalue.

Resonance poles in an analytic rank-one model

Statement

Statement references

Facts & Assumptions

Proof

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