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Poisson periodization of a normalized Gaussian on a full-rank lattice

Statement

Assume Countable Choice and use the full-rank cell and angular frequencies of Neutral point sources and Fourier data on a three-dimensional torus. Here $S (R^{3})$ denotes the Schwartz space of smooth functions whose derivatives decay faster than every inverse polynomial. For $α > 0$ let $g_{α} (x) = \frac{α^{3}}{π^{3 / 2}} e^{- α^{2} ∣ x ∣^{2}} .$ For every $x \in R^{3}$ the two series below converge absolutely, locally uniformly in $x$ , and $\sum_{λ \in Λ} g_{α} (x + λ) = \frac{1}{V} \sum_{k \in Z^{3}} e^{- ∣ κ_{k} ∣^{2} / (4 α^{2})} e^{i κ_{k} \cdot x} .$ The lattice Poisson theorem is applied only to this Schwartz Gaussian; it is not applied to a compactly supported or finite-smoothness assignment window.

Facts & Assumptions

Given: Countable Choice, an invertible cell matrix $A$ , $V = ∣ \det A ∣$ , $Λ = A Z^{3}$ , $Λ^{*} = A^{- T} Z^{3}$ , $κ_{k} = 2 π A^{- T} k$ , $α > 0$ , and $x \in R^{3}$ .

[A1]

Countable Choice is required by the published lattice Poisson theorem and the Gaussian transform supplier (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For $f \in S (R^{3})$ , both lattice series converge absolutely and the periodization satisfies $\sum_{λ \in Λ} f (x + λ) = V^{- 1} \sum_{ξ \in Λ^{*}} \hat{f} (ξ) e^{2 π i ξ \cdot x}$ under Countable Choice (Poisson summation for a full-rank lattice).

[F2]

With $\hat{f} (ξ) = \int_{R^{3}} f (x) e^{- 2 π i ξ \cdot x} d x$ , $F (e^{- π t ∣ x ∣^{2}}) (ξ) = t^{- 3 / 2} e^{- π ∣ ξ ∣^{2} / t}$ for $t > 0$ under Countable Choice (Euclidean Gaussian transform with the 2π normalization).

[F3]

The normalized Gaussian in the screened-kernel definition is $g_{α} (x) = α^{3} π^{- 3 / 2} e^{- α^{2} ∣ x ∣^{2}}$ (Gaussian screened Coulomb kernel and its self limit).

[F4]

The lattice is $A Z^{3}$ , its dual is $A^{- T} Z^{3}$ , and the physical angular frequency for index $k$ is $κ_{k} = 2 π A^{- T} k$ (Neutral point sources and Fourier data on a three-dimensional torus).

Proof

technique · direct

1.1givenA1F2F3algebra

Put $t = α^{2} / π > 0$ . By [F2], the Fourier transform of $e^{- π t ∣ x ∣^{2}}$ is $t^{- 3 / 2} e^{- π ∣ ξ ∣^{2} / t}$ . Multiplying the input by $α^{3} / π^{3 / 2}$ cancels $t^{- 3 / 2} = (π / α^{2})^{3 / 2}$ , so ${\hat{g}}_{α} (ξ) = e^{- π^{2} ∣ ξ ∣^{2} / α^{2}} .$ Each derivative of $g_{α}$ is a polynomial times $e^{- α^{2} ∣ x ∣^{2}}$ . For every degree $d$ and integer $N$ , choose an integer $m > (d + N) / 2$ ; the inequality $e^{y} \geq y^{m} / m!$ for $y \geq 0$ bounds $∣ x ∣^{d} (1 + ∣ x ∣)^{N} e^{- α^{2} ∣ x ∣^{2}}$ on $∣ x ∣ \geq 1$ , and continuity bounds it on the unit ball. Hence $g_{α} \in S (R^{3})$ , so [F1] applies.

1.2givenF3F4algebra

Fix a compact set $C \subset R^{3}$ and choose $R$ with $∣ x ∣ \leq R$ on $C$ . Since $A$ is invertible, $∣ A n ∣ \geq c ∣ n ∣$ for $c = ∥ A^{- 1} ∥_{2}^{- 1} > 0$ . For $∣ n ∣ \geq 2 R / c$ , $∣ x + A n ∣ \geq c ∣ n ∣ / 2$ uniformly on $C$ , so $g_{α} (x + A n) \leq α^{3} π^{- 3 / 2} e^{- b ∣ n ∣^{2}}$ with $b = α^{2} c^{2} / 4 > 0$ . This majorant is summable: $\sum_{n \in Z^{3}} e^{- b ∣ n ∣^{2}} = (\sum_{j \in Z} e^{- b j^{2}})^{3} < \infty$ , since $e^{- b j^{2}} \leq e^{- b ∣ j ∣}$ for $∣ j ∣ \geq 1$ . The remaining finitely many indices have bounded terms. Bounding every series tail by the corresponding majorant tail proves absolute uniform convergence on $C$ .

2.1A1F1F4step 1.1step 1.2algebra∎

Apply [F1] to $g_{α}$ . Every $ξ \in Λ^{*}$ has the unique form $ξ = A^{- T} k$ with $k \in Z^{3}$ ; by [F4], $2 π ξ = κ_{k}$ . Therefore $π^{2} ∣ ξ ∣^{2} / α^{2} = ∣ κ_{k} ∣^{2} / (4 α^{2})$ and $e^{2 π i ξ \cdot x} = e^{i κ_{k} \cdot x}$ . Substitution in the periodization formula gives exactly the displayed reciprocal series, and [F1] supplies absolute convergence on both sides. At $x = 0$ , absolute convergence of the reciprocal series gives $\sum_{k} e^{- ∣ κ_{k} ∣^{2} / (4 α^{2})} < \infty$ . Since $∣ e^{i κ_{k} \cdot x} ∣ = 1$ , this same sum bounds reciprocal-series tails uniformly in all $x$ . Together with step 1.2 this proves both asserted local uniform convergence claims.

Depends on

Used by

The alpha-independent Ewald potential and renormalized energy Theorem

Dependency tree · two levels

39 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Lior Silberman, Fourier Series and the Poisson Summation Formula (Math 613 lecture notes) (standard reference, not scraped)
Hark Lee and Wei Cai, Ewald Summation for Coulomb Interactions in a Periodic Supercell (2009 lecture notes) (standard reference, not scraped)