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Hermite orthogonality

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $L^{2} = L^{2} (R, d y)$ be the complex Lebesgue space with inner product linear in the first argument, and let $g_{0}, P, a, a^{†}, g_{n}$ and $H_{n}$ be as in Hermite recurrence. Then, for all $m, n \in N$ :

(a) $⟨ g_{n}, g_{m} ⟩ = δ_{n m}$ ; in particular $(g_{n})_{n \in N}$ is an orthonormal family in $L^{2}$ (Orthonormal families, complete orthonormal systems and Hilbert bases);

(b) the eigenvalue equation $(- \partial_{y}^{2} + Y^{2}) g_{n} = (2 n + 1) g_{n}$ holds as an identity of $L^{2}$ classes; more precisely $Y^{2} - \partial_{y}^{2} = 2 a^{†} a + I$ as operators on $P$ , so that on $P$

$- φ^{''} + y^{2} φ = 2 a^{†} a φ + φ (φ \in P);$

No completeness of the family $(g_{n})$ is asserted here.

Facts & Assumptions

Given: the family $(g_{n})$ of Hermite recurrence with $g_{0} (y) = π^{- 1 / 4} e^{- y^{2} / 2}$ , the ladder operators $a, a^{†}$ on $P$ , and the identities

[F1]

$∥ g_{0} ∥ = 1$ ; $g_{n} = (a^{†})^{n} g_{0} / \sqrt{n!}$ ; $a g_{n} = \sqrt{n} g_{n - 1}$ for $n \geq 1$ with $a g_{0} = 0$ ; $a^{†} g_{n} = \sqrt{n + 1} g_{n + 1}$ ; $a^{†} a g_{n} = n g_{n}$ ; $a a^{†} g_{n} = (n + 1) g_{n}$ ; $a^{†} a - a a^{†} = - I$ on $P$ (Hermite recurrence(a),(d),(e)).

[F2]

$⟨ a^{†} φ, ψ ⟩ = ⟨ φ, a ψ ⟩$ for all $φ, ψ \in P$ (Hermite recurrence(c)).

[F3]

$⟨ \cdot, \cdot ⟩$ is linear in the first argument and conjugate-linear in the second, and $∥ v ∥^{2} = ⟨ v, v ⟩ \geq 0$ with equality only for $v = 0$ (The space $L^{p} (μ)$ as the quotient by null functions, Complex Lp classes and Euclidean test-function conventions, Cauchy-Schwarz inequality for $L^{2}$ ).

[F4]

$Y^{2} - \partial_{y}^{2} = 2 a^{†} a + I$ on $P$ , where $Y$ is multiplication by $y$ and $\partial_{y}$ is the derivative on smooth representatives; this is the computation $(Y - \partial_{y}) (Y + \partial_{y}) = Y^{2} - \partial_{y}^{2} - [\partial_{y}, Y] = Y^{2} - \partial_{y}^{2} - I$ combined with $a^{†} a = \frac{1}{2} (Y - \partial_{y}) (Y + \partial_{y})$ (Hermite recurrence(b),(d)).

Proof

1.1givenF1F2algebra

[Orthogonality] Let $m, n \in N$ with $n \geq 1$ . By [F2] with $φ = a g_{n}$ and $ψ = g_{m}$ , $⟨ a^{†} a g_{n}, g_{m} ⟩ = ⟨ a g_{n}, a g_{m} ⟩$ ; by [F2] with $φ = a g_{m}$ and $ψ = g_{n}$ and then conjugating, $⟨ g_{n}, a^{†} a g_{m} ⟩ = ⟨ a g_{n}, a g_{m} ⟩$ as well. Using $a^{†} a g_{k} = k g_{k}$ from [F1] on both sides, $n ⟨ g_{n}, g_{m} ⟩ = m ⟨ g_{n}, g_{m} ⟩$ . If $n = m$ this gives no information; if $n \neq m$ and $n, m \geq 1$ it gives $⟨ g_{n}, g_{m} ⟩ = 0$ . If $m = 0$ use $a g_{0} = 0$ : then $⟨ a g_{n}, a g_{0} ⟩ = 0$ , and the first identity reads $n ⟨ g_{n}, g_{0} ⟩ = 0$ , so $⟨ g_{n}, g_{0} ⟩ = 0$ for $n \geq 1$ ; conjugate symmetry gives $⟨ g_{0}, g_{n} ⟩ = 0$ as well.

1.2givenF1F4algebra

[Eigenvalue equation] By [F4], $Y^{2} - \partial_{y}^{2} = 2 a^{†} a + I$ on $P$ . Evaluating at $g_{n}$ and using [F1] gives $(- \partial_{y}^{2} + Y^{2}) g_{n} = 2 n g_{n} + g_{n} = (2 n + 1) g_{n}$ as classes in $L^{2}$ ; the left side is computed on the smooth representative $g_{n}$ , and the equality of classes holds because the two sides are continuous functions agreeing in $L^{2}$ . The displayed form for $φ \in P$ with $Y^{2} φ = y^{2} φ$ is the same identity.

2.1step 1.1F1F2F3algebra

[Unit norms] $∥ g_{0} ∥ = 1$ by [F1]. For $n \geq 0$ , applying [F2] with $φ = g_{n}$ , $ψ = a^{†} g_{n}$ gives $∥ a^{†} g_{n} ∥^{2} = ⟨ a^{†} g_{n}, a^{†} g_{n} ⟩ = ⟨ g_{n}, a a^{†} g_{n} ⟩ = (n + 1) ∥ g_{n} ∥^{2}$ by $a a^{†} g_{n} = (n + 1) g_{n}$ from [F1]. Since $g_{n + 1} = a^{†} g_{n} / \sqrt{n + 1}$ by [F1], $∥ g_{n + 1} ∥^{2} = ∥ a^{†} g_{n} ∥^{2} / (n + 1) = ∥ g_{n} ∥^{2}$ . Induction gives $∥ g_{n} ∥ = 1$ for every $n$ . Together with step 1.1, $⟨ g_{n}, g_{m} ⟩ = δ_{n m}$ .

3.1step 2.1F1algebra∎

[Ladder pairings] For $n \neq m$ , $⟨ a g_{n}, a g_{m} ⟩ = \sqrt{n m} ⟨ g_{n - 1}, g_{m - 1} ⟩ = 0$ by step 2.1 when $n, m \geq 1$ , and $⟨ a g_{0}, \cdot ⟩ = 0$ ; for $n = m$ it equals $n ∥ g_{n - 1} ∥^{2} = n$ . Likewise $⟨ a^{†} g_{n}, a^{†} g_{m} ⟩ = \sqrt{(n + 1) (m + 1)} ⟨ g_{n + 1}, g_{m + 1} ⟩ = (n + 1) δ_{n m}$ .

Hermite orthogonality

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources