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Negative-subspace completeness and degeneracy

Statement

Assume the Axiom of Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice). Let $γ > 0$ and let $C = - Δ - γ / ∣ \cdot ∣$ on $D (C) = H^{2} (R^{3}; C)$ be the self-adjoint operator of Coulomb infinitesimal kinetic bound with essential spectrum $[0, \infty)$ . For $n \geq 1$ , $0 \leq l \leq n - 1$ and $- l \leq m \leq l$ let $u_{n l}$ be the regular solution of Kummer recurrence and termination at $k = γ / (2 n)$ (the polynomial-exponential solution, unique up to a nonzero scalar), and put $ψ_{n l m} (x) : = \frac{u_{n l} (r)}{r} Y_{l m} (Ω) (x = r Ω) .$ Then:

(a) each $ψ_{n l m}$ lies in $D (C)$ and $C ψ_{n l m} = - \frac{γ^{2}}{4 n^{2}} ψ_{n l m}$ ; the vectors $ψ_{n l m}$ with distinct labels are pairwise orthogonal, and for fixed $n$ the eigenspace of $C$ at $- γ^{2} / (4 n^{2})$ is exactly their span; it has dimension $\sum_{l = 0}^{n - 1} (2 l + 1) = n^{2}$ ;

(b) after normalization, the family $(ψ_{n l m} / ∥ u_{n l} ∥_{2})_{n \geq 1, 0 \leq l \leq n - 1, - l \leq m \leq l}$ is a complete orthonormal family of the negative spectral subspace $E_{C} ((- \infty, 0)) L^{2} (R^{3})$ ;

(c) the family is not complete in $L^{2} (R^{3})$ : the nonnegative spectral subspace $E_{C} ([0, \infty)) L^{2} (R^{3})$ is nonzero (indeed infinite dimensional), so the bound states alone do not span $L^{2}$ .

All claims concern the mathematical spectral decomposition of $C$ ; no physical interpretation is asserted.

Facts & Assumptions

Given: $γ > 0$ , the operator $C$ with essential spectrum $[0, \infty)$ and spectral measure $E$ ; the labels $n, l, m$ ; the functions $ψ_{n l m}$ .

[F1]

The negative eigenvalues of $C$ are exactly $- γ^{2} / (4 n^{2})$ ( $n \geq 1$ ), and every eigenfunction of $C$ with negative eigenvalue $E = - γ^{2} / (4 n^{2})$ has, in each block $(l, m)$ with $l \leq n - 1$ , a scalar multiple of $u_{n l}$ , while every block with $l \geq n$ vanishes (because $- γ^{2} / (4 n^{2})$ is not a negative eigenvalue of $A_{l}$ when $l \geq n$ ) (Laguerre bound energy exhaustion).

[F2]

$U$ is unitary and $∥ ψ ∥_{L^{2}}^{2} = \sum_{l, m} ∥ u_{l m} ∥_{2}^{2}$ for $U ψ = (u_{l m})$ ; in particular the inverse image of a vector supported in one block $(l, m)$ with entry $u$ has norm $∥ u ∥_{2}$ (Angular radial unitary reduction (a), (b)). If $u \in D (A_{l})$ and $A_{l} u = E u$ , that one-block vector belongs to the direct-sum operator domain since $∥ u ∥_{2}^{2} + ∥ E u ∥_{2}^{2} < \infty$ , so its inverse image $ψ (x) = u (r) Y_{l m} (Ω) / r$ lies in $D (C)$ and $C ψ = E ψ$ for every allowed $m$ (Transported radial endpoint realization (d)).

[F3]

The negative spectral subspace $E_{C} ((- \infty, 0)) L^{2}$ is the closed linear span of the eigenspaces of $C$ for its negative eigenvalues (Coulomb essential spectrum (c)).

[F4]

For the self-adjoint (possibly unbounded) $C$ , if $C u = E u$ and $C v = E^{'} v$ with $u, v \in D (C)$ and distinct real $E, E^{'}$ , then $E ⟨ u, v ⟩ = ⟨ C u, v ⟩ = ⟨ u, C v ⟩ = E^{'} ⟨ u, v ⟩$ , hence $⟨ u, v ⟩ = 0$ by symmetry on its domain; $E ({λ})$ is the orthogonal projection onto $\ker (C - λ)$ ; and $λ \in σ_{e s s} (C)$ iff $rank E ((λ - ε, λ + ε)) = \infty$ for every $ε > 0$ (Symmetric, self-adjoint and essentially self-adjoint operators, Discrete and essential spectrum of a self-adjoint operator, Projection valued measure).

[F5]

For a complete orthonormal family $(e_{i})$ of a closed subspace $S$ of a Hilbert space one has $x = \sum_{i} ⟨ x, e_{i} ⟩ e_{i}$ and $∥ x ∥^{2} = \sum_{i} ∣ ⟨ x, e_{i} ⟩ ∣^{2}$ for $x \in S$ (Parseval equivalences for an orthonormal family, Orthonormal families, complete orthonormal systems and Hilbert bases, Hilbert space).

Proof

1.1F1F2F4givenalgebra

[Eigenvectors, exact norms and orthogonality] For each label put $E_{n} = - γ^{2} / (4 n^{2})$ . The Kummer termination and Laguerre exhaustion suppliers give a nonzero $u_{n l} \in D (A_{l})$ with $A_{l} u_{n l} = E_{n} u_{n l}$ ; [F2] transports the vector supported in block $(l, m)$ to $ψ_{n l m} \in D (C)$ with $C ψ_{n l m} = E_{n} ψ_{n l m}$ and $∥ ψ_{n l m} ∥_{2} = ∥ u_{n l} ∥_{2} > 0$ . For different $n$ , the numbers $E_{n}$ are distinct and self-adjointness gives orthogonality by [F4]. For equal $n$ and different $(l, m)$ , the direct-sum vectors have disjoint supports, hence their inner product is zero; unitarity transfers that orthogonality to $L^{2} (R^{3})$ . Thus $e_{n l m} : = ψ_{n l m} / ∥ u_{n l} ∥_{2}$ is an orthonormal family.

2.1step 1.1F1F2algebra

[The entire eigenspace and its dimension] Let $C ψ = E_{n} ψ$ . The block-domain identity in [F2] and exhaustion in [F1] give $U ψ = (a_{l m} u_{n l})_{0 \leq l \leq n - 1, - l \leq m \leq l}$ with all other blocks zero. There are only $\sum_{l = 0}^{n - 1} (2 l + 1)$ such blocks, so inversion gives the finite sum $ψ = \sum_{l = 0}^{n - 1} \sum_{m = - l}^{l} a_{l m} ψ_{n l m}$ ; conversely each such sum belongs to the eigenspace by step 1.1. Its squared norm is $\sum_{l = 0}^{n - 1} \sum_{m = - l}^{l} ∣ a_{l m} ∣^{2} ∥ u_{n l} ∥_{2}^{2}$ by [F2]. If the sum is zero, every coefficient vanishes because each radial norm is positive; thus these vectors form a basis. Their number is $2 \sum_{l = 0}^{n - 1} l + n = n (n - 1) + n = n^{2}$ , proving (a), including full $C$ degeneracy rather than only radial simplicity.

3.1step 1.1step 2.1F1F3F5algebra

[The bound orthonormal basis of the negative subspace] Let $S = E_{C} ((- \infty, 0)) L^{2}$ . By [F3], $S$ is the closed span of all negative eigenspaces, and by [F1] and step 2.1 these are exactly the finite-dimensional spaces spanned by $ψ_{n l m}$ for $n \geq 1$ . Each normalization scalar in step 1.1 is finite and nonzero, so the normalized and unnormalized families have exactly the same finite linear span, and therefore the same closed span $S$ . The orthonormal family $(e_{n l m})$ is consequently complete in $S$ . For $f \in S$ , [F5] gives $f = \sum_{n, l, m} ⟨ f, e_{n l m} ⟩ e_{n l m}$ in $L^{2}$ and $∥ f ∥_{2}^{2} = \sum_{n, l, m} ∣ ⟨ f, e_{n l m} ⟩ ∣^{2}$ , with the declared first-entry-linear convention. This proves (b).

4.1step 3.1F4givenalgebra

[An infinite-dimensional complementary subspace] Since $1 / 4 \in σ_{e s s} (C) = [0, \infty)$ , [F4] with radius $1 / 4$ gives infinite rank for $E_{C} ((0, 1 / 2))$ . Its range is contained in $E_{C} ([0, \infty)) L^{2}$ , because $(0, 1 / 2) \subseteq [0, \infty)$ and nested spectral projections satisfy $E_{C} ([0, \infty)) E_{C} ((0, 1 / 2)) = E_{C} ((0, 1 / 2))$ . Thus the nonnegative subspace is infinite dimensional. Spectral projections for disjoint sets satisfy $E_{C} ((- \infty, 0)) E_{C} ([0, \infty)) = 0$ , so this nonzero subspace is orthogonal to $S$ . Step 3.1 identifies the closed span of the bound states with $S$ , proving that family cannot be complete in all of $L^{2}$ and establishing (c).

5.1step 1.1step 2.1step 3.1step 4.1∎

[Conclusion] Step 1.1 proves domain membership and orthogonality; step 2.1 proves the full eigenspace and $n^{2}$ degeneracy in (a); step 3.1 proves (b); step 4.1 proves (c).

Negative-subspace completeness and degeneracy

Statement

Facts & Assumptions

Proof

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