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Degenerate Feshbach error

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be an arbitrary complex Hilbert space (Hilbert space) and let $H_{0}$ be self-adjoint on its exact dense domain $D (H_{0})$ (Symmetric, self-adjoint and essentially self-adjoint operators). Use $σ (H_{0})$ for the resolvent spectrum (Resolvent and spectrum of an unbounded operator). Let $E \in σ (H_{0})$ be isolated, with $\dim \ker (H_{0} - E) = r \geq 1$ and $dist (E, σ (H_{0}) ∖ {E}) \geq g > 0$ . Write $P = E_{H_{0}} ({E})$ for its orthogonal spectral projection and $Q = I - P$ , using the CC arbitrary-H calculus of CC spectral calculus on an arbitrary complex Hilbert space by compatible vector-local Cayley subspaces. Let $V = V^{*} \in B (H)$ with $v = ∥ V ∥ > 0$ , and let $λ \in R$ satisfy $∣ λ ∣ v < g / 4$ (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum). Put $H_{λ} = H_{0} + λ V$ on $D (H_{0})$ and assume it is self-adjoint there. Then:

(a) For every real $z$ with $∣ z - E ∣ + ∣ λ ∣ v \leq g / 2$ , the operator $T_{λ} (z) = Q (H_{λ} - z) Q$ on $Q H$ , with domain $Q D (H_{0}) = D (H_{0}) \cap Q H$ , has a bounded two-sided inverse $B (z) : Q H \to Q D (H_{0})$ with $∥ B (z) ∥ \leq 4 / g$ . This includes $z = E$ and the case $Q H = {0}$ , with its unique zero-space inverse.

(b) If such $z$ is an eigenvalue of $H_{λ}$ with eigenvector $ψ \neq 0$ , then $φ = P ψ \neq 0$ , $χ = Q ψ = - λ B (z) Q V P φ$ , and

$[E P + λ P V P - λ^{2} P V Q B (z) Q V P] φ = z φ .$

$dist (z, E + λ {spec}_{P H} (P V P)) \leq \frac{4 λ^{2} v^{2}}{g},$

where ${spec}_{P H} (P V P)$ is the spectrum of the Hermitian operator $(P V P) ∣_{P H}$ on the finite-dimensional space $P H$ . The statement retains all finite multiplicities $r$ and does not identify any eigenvector of $P V P$ with an exact eigenvector of $V$ .

Facts & Assumptions

Given: CC, arbitrary complex $H$ with first-variable-linear inner product, $H_{0}, E, g, r, P, Q, V, v, λ, H_{λ}$ as above, and real $z$ with $∣ z - E ∣ + ∣ λ ∣ v \leq g / 2$ .

[F1]

CC spectral calculus on an arbitrary complex Hilbert space by compatible vector-local Cayley subspaces proves under CC alone on arbitrary complex Hilbert spaces the unique representing PVM, concentration on the spectrum, the nonzero-neighborhood characterization of spectral points, bounded Borel product and adjoint laws, and the exact unbounded square-integrability domain. In particular bounded Borel $b$ has $∥ b (H_{0}) ∥ \leq \sup ∣ b ∣$ , and if $t b (t)$ is bounded then $b (H_{0}) H \subset D (H_{0})$ with the corresponding multiplier identity. Projection valued measure specifies the projections and strong countable additivity.

[F2]

Resolvent and spectrum of an unbounded operator, A bounded linear operator between normed spaces and The operator norm as the least bound and as the unit-sphere or unit-ball supremum specify the inverse domains and operator norm products. For a bounded operator $K$ with $∥ K ∥ < 1$ , the norm-convergent series $\sum_{j \geq 0} (- K)^{j}$ is the inverse of $I + K$ : multiplying its finite sums by $I + K$ leaves a remainder of norm at most $∥ K ∥^{N + 1}$ , which tends to zero. Its norm is at most $(1 - ∥ K ∥)^{- 1}$ .

[F3]

Every finite-dimensional real or complex inner product space has an orthonormal basis supplies an orthonormal basis in each finite-dimensional real or complex inner product space, by finite Gram–Schmidt; no arbitrary-index choice is involved. Finite sums in such a basis satisfy the Pythagorean norm identity. The spectral decomposition of the finite Hermitian matrix used below is derived in the proof from [F1], rather than imported from an AC or Weyl theorem.

Proof

1.1F1givenalgebra

[The atom and domain projections] By [F1], the projection on ${E}$ has range $\ker (H_{0} - E)$ : a vector on that atom has finite second moment and operator value $E$ times itself, while the identity $∥ (H_{0} - E) u ∥^{2} = \int ∣ t - E ∣^{2} d ⟨ E_{H_{0}} (t) u, u ⟩$ forces every eigenvector's scalar measure onto the atom. Thus $P H \subset D (H_{0})$ , $H_{0} P = E P$ , and $P$ and $Q$ preserve $D (H_{0})$ and commute with $H_{0}$ there by the multiplier product and exact-domain identities. In particular $Q D (H_{0}) = D (H_{0}) \cap Q H$ . The gap and spectral concentration show $E_{H_{0}} ((E - g, E + g)) = P$ , so every vector in $Q H$ has its scalar measure supported where $∣ t - E ∣ \geq g$ .

1.2F1F3givenconstructalgebra

[Finite Hermitian diagonalization under the actual CC supplier] Apply [F1] to the bounded Hermitian operator $(P V P) ∣_{P H}$ on the $r$ -dimensional Hilbert space $P H$ . Its spectrum has at most $r$ points: if there were $r + 1$ distinct spectral points, choose disjoint bounded neighborhoods around this finite list. Each neighborhood projection is nonzero by the supplied spectral characterization; choosing and normalizing one vector in each range gives $r + 1$ mutually orthogonal nonzero vectors in $P H$ , impossible. These are finitely many choices. The spectrum is nonempty, since otherwise concentration on the spectrum would contradict the total projection $I_{P H}$ with $r \geq 1$ . Concentration and finite additivity now give $I_{P H}$ as the finite sum of its atom projections. By the exact-domain residual formula each atom range is the eigenspace for that spectral value. Choose an orthonormal basis in each of these finitely many orthogonal eigenspaces using [F3], and combine them into an orthonormal eigenbasis $(e_{j})_{j = 1}^{r}$ , with eigenvalues $μ_{j}$ . This proves precisely the finite diagonalization needed here using CC, without an AC spectral theorem or a Weyl perturbation inequality.

2.1step 1.1F1F2algebra

[The actual reduced resolvent, including the unperturbed eigenvalue] Put $A_{Q} = H_{0} ∣_{D (H_{0}) \cap Q H}$ on $Q H$ and $T_{0} (z) = A_{Q} - z$ . Define $b_{z} (t) = (t - z)^{- 1}$ when $∣ t - E ∣ \geq g$ and $b_{z} (t) = 0$ otherwise. Since $∣ z - E ∣ \leq g / 2$ , $∣ t - z ∣ \geq g - ∣ z - E ∣ \geq g / 2$ on the first region. Thus $b_{z}$ is bounded by $2 / g$ , and $t b_{z} (t)$ is bounded. By [F1], $R_{Q} (z) = b_{z} (H_{0}) ∣_{Q H}$ maps $Q H$ into $Q D (H_{0})$ , with norm at most $2 / g$ . The multiplier products equal one on the $Q$ -spectral support, giving $T_{0} (z) R_{Q} (z) = I_{Q H}$ and $R_{Q} (z) T_{0} (z) = I$ on $Q D (H_{0})$ . Hence this is the genuine reduced inverse, also at $z = E$ ; no full inverse of $H_{0} - E$ is used. If $Q H = 0$ , all these identities concern the unique map of the zero space, whose identity equals its zero map.

3.1step 2.1F2givenalgebra

[The correctly ordered Neumann factors] Let $W = (Q V Q) ∣_{Q H}$ , a bounded Hermitian operator with $∥ W ∥ \leq v$ . On exactly $Q D (H_{0})$ one has $T_{λ} (z) = T_{0} (z) + λ W = (I_{Q H} + λ W R_{Q} (z)) T_{0} (z)$ , because $R_{Q} (z) T_{0} (z)$ is the domain identity. The rightmost unbounded factor acts first. Set $S_{z} = I_{Q H} + λ W R_{Q} (z)$ ; then $∥ λ W R_{Q} (z) ∥ \leq 2 ∣ λ ∣ v / g < 1 / 2$ , so [F2] gives its norm-convergent inverse. Define $B (z) = R_{Q} (z) S_{z}^{- 1}$ . This maps into $Q D (H_{0})$ , and $T_{λ} (z) B (z) = S_{z} T_{0} (z) R_{Q} (z) S_{z}^{- 1} = I_{Q H}$ ; on the domain, $B (z) T_{λ} (z) = R_{Q} (z) S_{z}^{- 1} S_{z} T_{0} (z) = I$ . Its norm is at most $(2 / g) / (1 - 2 ∣ λ ∣ v / g) \leq 4 / g$ . This proves (a) with both inverse equations and the exact restricted domain.

4.1step 3.1givenalgebra

[Self-adjointness of the inverse] For real $z, λ$ , $T_{λ} (z)$ is symmetric on its restricted domain, since $H_{0}, V$ are symmetric and $Q$ is orthogonal. For arbitrary $f, h \in Q H$ put $u = B (z) f$ , $w = B (z) h$ , which belong to that domain. Then $⟨ B (z) f, h ⟩ = ⟨ u, T_{λ} (z) w ⟩ = ⟨ T_{λ} (z) u, w ⟩ = ⟨ f, B (z) h ⟩$ . Thus the bounded everywhere-defined inverse is self-adjoint. The same assertion is valid on the zero complement.

4.2step 1.1step 3.1givenalgebra

[The exact eigenvector reduction] Let $ψ \in D (H_{0})$ be a nonzero eigenvector at the specified $z$ , and set $φ = P ψ$ , $χ = Q ψ$ . Step 1.1 places $φ \in P H \subset D (H_{0})$ and $χ \in Q D (H_{0})$ . Applying $Q$ to $H_{λ} ψ = z ψ$ gives $T_{λ} (z) χ = - λ Q V P φ$ , hence $χ = - λ B (z) Q V P φ$ by step 3.1. If $φ = 0$ , this forces $χ = 0$ and contradicts $ψ \neq 0$ . Applying $P$ gives $E P φ + λ P V P φ + λ P V Q χ = z φ$ . Substitution proves the matrix equation in (b); each use of $B (z)$ has an input in $Q H$ .

5.1step 3.1step 4.1step 4.2F2algebra

[The Hermitian correction with its retained constant] On $P H$ define $M_{0} = E I_{P H} + λ (P V P) ∣_{P H}$ and $K_{z} = - λ^{2} (P V Q) B (z) (Q V P) ∣_{P H}$ . The adjoint identity $(Q V P)^{*} = P V Q$ , step 4.1 and real $λ$ make both $M_{0}$ and $K_{z}$ Hermitian. Operator norm products give $∥ K_{z} ∥ \leq λ^{2} v^{2} ∥ B (z) ∥ \leq 4 λ^{2} v^{2} / g$ . Step 4.2 says $(M_{0} + K_{z}) φ = z φ$ with $φ \neq 0$ .

6.1step 2.1step 3.1step 4.2step 5.1step 1.2F3algebra∎

[The residual estimate and conclusion] Write $φ = \sum_{j = 1}^{r} ⟨ φ, e_{j} ⟩ e_{j}$ in the eigenbasis of step 1.2. Then $M_{0} e_{j} = (E + λ μ_{j}) e_{j}$ , so the finite Pythagorean identity gives $∥ (M_{0} - z) φ ∥^{2} = \sum_{j = 1}^{r} ∣ E + λ μ_{j} - z ∣^{2} ∣ ⟨ φ, e_{j} ⟩ ∣^{2} \geq dist (z, E + λ {spec}_{P H} (P V P))^{2} ∥ φ ∥^{2}$ . Step 5.1 gives $(M_{0} - z) φ = - K_{z} φ$ and bounds its norm by $4 λ^{2} v^{2} ∥ φ ∥ / g$ . Since $φ \neq 0$ , division proves (c). Together with steps 2.1 and 3.1 this proves all parts for arbitrary $H$ , every finite $r \geq 1$ , including $λ = 0$ , $z = E$ and $Q H = 0$ .

Remarks

The real-coupling condition follows already from self-adjointness of $H_{0} + λ V$ with $V = V^{*}$ and $v > 0$ : symmetry yields $(λ - \overline{λ}) V = 0$ on the dense domain and hence on $H$ . Thus $λ$ is real. The spectral-distance conclusion is conditional on an actual interior eigenvalue and does not assert an ordered eigenvalue correspondence.

Degenerate Feshbach error

Statement

Facts & Assumptions

Proof

Remarks

Depends on

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