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Hermite Ritz exactness

Example

Assume Choice and Countable Choice. Fix $m, ω, ℏ > 0$ and let $H_{o s c}$ be the self-adjoint coefficient-domain realization on $L^{2} (R, d x)$ of $- \frac{ℏ^{2}}{2 m} \frac{d^{2}}{d x^{2}} + \frac{m ω^{2}}{2} x^{2}$ with normalized Hermite eigenfunctions $(ψ_{j})_{j \geq 0}$ , where $H_{o s c} ψ_{j} = E_{j} ψ_{j}$ and $E_{j} = ℏ ω (j + \frac{1}{2})$ . For $n \geq 1$ , let $V_{n} = span {ψ_{0}, \dots, ψ_{n - 1}} \subseteq Q (H_{o s c})$ . The $n$ Ritz values of the associated form on $V_{n}$ are exactly $E_{0}, \dots, E_{n - 1}$ in increasing order, and each continuum residual $(H_{o s c} - E_{j} I) ψ_{j}$ is zero.

Example references

Oscillator scaling and exact realization

Conforming Ritz eigenvalue upper bounds

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

Unbounded Borel functional calculus: domains, products, spectral mapping

Facts & Assumptions

Given: Choice and Countable Choice; $m, ω, ℏ > 0$ ; the oscillator $H_{o s c}$ ; its normalized Hermite eigenfunctions $ψ_{j}$ ; and $V_{n}$ as in the Example.

[A1]

Choice and Countable Choice are the assumptions of the form min–max result and the exact oscillator realization used here (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The oscillator realization is self-adjoint, has the stated coefficient domain, satisfies $H_{o s c} \geq \frac{ℏ ω}{2} I > 0$ , its Hermite functions form a complete orthonormal family, and $H_{o s c} ψ_{j} = ℏ ω (j + \frac{1}{2}) ψ_{j}$ (Oscillator scaling and exact realization).

[F2]

In the spectral-form definition of Conforming Ritz eigenvalue upper bounds, take $A = H_{o s c}$ and $c = 0$ . Since $H_{o s c} \geq 0$ , this gives $Q (H_{o s c}) = D (S)$ and $q (f, g) = ⟨ S f, S g ⟩$ for $S = H_{o s c}^{1 / 2}$ defined by the Borel functional calculus.

[F3]

The spectral/Borel calculus gives $D (H_{o s c}) = {f : \int x^{2} d μ_{f} (x) < \infty}$ and $Q (H_{o s c}) = D (H_{o s c}^{1 / 2}) = {f : \int x d μ_{f} (x) < \infty}$ because $H_{o s c} \geq 0$ ; $x \leq 1 + x^{2}$ on $[0, \infty)$ therefore gives $D (H_{o s c}) \subseteq Q (H_{o s c})$ (Spectral theorem for unbounded self-adjoint operators (PVM form), Unbounded Borel functional calculus: domains, products, spectral mapping).

[F4]

The Ritz values on a finite-dimensional conforming form subspace are the ordered eigenvalues of its exact Gram/form generalized eigenproblem (Conforming Ritz eigenvalue upper bounds).

Proof

1.1A1F1F2F3algebra

By [F1], each $ψ_{j} \in D (H_{o s c})$ , the family is orthonormal, and $H_{o s c} ψ_{j} = E_{j} ψ_{j}$ . By [F3], each $ψ_{j}$ lies in $Q (H_{o s c}) = D (S)$ . The product rule gives $S^{2} = H_{o s c}$ with $D (S^{2}) = D (H_{o s c})$ ; since $S$ is self-adjoint, for $i, j < n$ we have $q (ψ_{j}, ψ_{i}) = ⟨ S ψ_{j}, S ψ_{i} ⟩ = ⟨ S^{2} ψ_{j}, ψ_{i} ⟩ = ⟨ H_{o s c} ψ_{j}, ψ_{i} ⟩ = E_{j} δ_{i j}$ . Thus the Gram matrix on $V_{n}$ is the identity and its form matrix is diagonal with entries $E_{0}, \dots, E_{n - 1}$ .

2.1F1F4step 1.1algebra

Since $ℏ ω > 0$ , $E_{0} < \dots < E_{n - 1}$ . The generalized matrix eigenproblem therefore returns exactly these $n$ ordered Ritz values by [F4], and the oscillator eigenvalue equations in [F1] give $(H_{o s c} - E_{j} I) ψ_{j} = 0$ for each $j < n$ .

3.1step 2.1∎

Hence this conforming Hermite trial space attains the first $n$ oscillator eigenvalues exactly and every corresponding continuum residual vanishes.

Depends on

Dependency tree · 2 levels

Sources

James Binney and David Skinner, The Physics of Quantum Mechanics, Cappella Archive (2008; revised printings 2009–2013)