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Cartesian Gaussian raising, lowering, and kinetic-energy recurrences

Statement

For $g_{α, a, A}$ from Cartesian Gaussian primitives, contractions, and integral conventions, let $e_{j}$ be the $j$ th coordinate vector. With terms having negative indices interpreted as zero, $\partial_{x_{j}} g_{α, a, A} = α_{j} g_{α - e_{j}, a, A} - 2 a g_{α + e_{j}, a, A}, \partial_{A_{j}} g_{α, a, A} = - α_{j} g_{α - e_{j}, a, A} + 2 a g_{α + e_{j}, a, A} .$ The second spatial derivative and kinetic operator are $\partial_{x_{j}}^{2} g_{α, a, A} = α_{j} (α_{j} - 1) g_{α - 2 e_{j}, a, A} - 2 a (2 α_{j} + 1) g_{α, a, A} + 4 a^{2} g_{α + 2 e_{j}, a, A},$ $- \frac{1}{2} Δ g_{α, a, A} = \sum_{j = 1}^{3} (- \frac{α_{j} (α_{j} - 1)}{2} g_{α - 2 e_{j}, a, A} + a (2 α_{j} + 1) g_{α, a, A} - 2 a^{2} g_{α + 2 e_{j}, a, A}) .$ Consequently every kinetic integral of finite contractions is a finite linear combination of overlap integrals with these shifted multi-indices.

Facts & Assumptions

Given: A primitive $g_{α, a, A}$ with $a > 0$ as in Cartesian Gaussian primitives, contractions, and integral conventions, and the integral definitions there.

[F1]

Cartesian Gaussian primitives, contractions, and integral conventions: $g_{α, a, A} (x) = (x - A)^{α} e^{- a ∥ x - A ∥^{2}}$ and $T (ϕ, ψ) = - \frac{1}{2} \int \overline{ϕ} Δ ψ$ .

[F2]

Absolute convergence and local parameter domination for Gaussian basis integrals: each shifted primitive and the corresponding kinetic integrand are absolutely integrable.

[L9]

The exponential function is smooth and $(\exp)^{'} = \exp$ and Sum, product, quotient, and chain rules from difference quotients: the exponential is smooth with derivative itself; finite sums, products and polynomial compositions obey the product, power and chain rules. Applying these rules one coordinate or parameter at a time justifies every displayed finite Gaussian derivative.

Proof

1.1givenF1algebraL9

Differentiate the product $(x_{j} - A_{j})^{α_{j}} e^{- a (x_{j} - A_{j})^{2}}$ . The polynomial derivative contributes $α_{j} g_{α - e_{j}}$ and the exponential derivative contributes $- 2 a g_{α + e_{j}}$ , proving the spatial formula. Differentiating with respect to $A_{j}$ reverses the polynomial derivative sign and changes the exponential derivative to $+ 2 a (x_{j} - A_{j})$ , proving the center formula.

2.1step 1.1algebraL9

Differentiate the spatial formula once more. The derivative of the lower term is $α_{j} (α_{j} - 1) g_{α - 2 e_{j}} - 2 a α_{j} g_{α}$ and the derivative of the raised term is $- 2 a (α_{j} + 1) g_{α} + 4 a^{2} g_{α + 2 e_{j}}$ . Combining the two middle coefficients gives $- 2 a (2 α_{j} + 1)$ and proves the second-derivative formula.

3.1givenF1F2step 2.1algebraL9∎

Multiplying the second-derivative identity by $- 1 / 2$ and summing over $j$ gives the displayed Laplacian formula pointwise. By [F2], all shifted terms are integrable, so substituting this finite sum into the kinetic integral and using finite linearity expresses it as the asserted finite combination of overlaps. Conjugation and finite contraction sums preserve the identity.

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Sources

Trygve Helgaker, Molecular Integral Evaluation, 10th Sostrup Summer School lecture notes (2008) (standard reference, not scraped)
L. E. McMurchie and E. R. Davidson, One and Two Electron Integrals over Cartesian Gaussian Functions, UCRL-13791 (1976) (standard reference, not scraped)