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Finite Hermite expansion and coefficient recurrences for a Cartesian Gaussian pair

Statement

In one dimension, let gi(xA,a)=xAie−axA2 and gj(xB,b)=xBje−bxB2, with xA=x−A, xB=x−B, p=a+b, P=(aA+bB)/p, and K=e−ab(A−B)2/p. There are unique coefficients Eijt such that gi(xA,a)gj(xB,b)=∑t=0i+jEijtΛt(x;p,P). With E000=K and Eijt=0 outside 0≤t≤i+j, they satisfy Ei+1,jt=Eijt−12p+(P−A)Eijt+(t+1)Eijt+1, Ei,j+1t=Eijt−12p+(P−B)Eijt+(t+1)Eijt+1. In three dimensions the pair expansion is the product of the three one-dimensional expansions, so it is a finite sum of Λtuv with 0≤t≤α1+β1, 0≤u≤α2+β2, and 0≤v≤α3+β3.

Facts & Assumptions

Given: Primitive Gaussian factors and indices as above, the product identity of Gaussian product theorem and finite polynomial shifts, and Λt from Hermite Gaussian and Boys Coulomb auxiliary functions.

[F1]

Gaussian product theorem and finite polynomial shifts: the product exponential equals Ke−p(x−P)2 and the remaining polynomial has degree at most i+j.

[F2]

Hermite Gaussian and Boys Coulomb auxiliary functions: Λt=∂Pte−p(x−P)2.

[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

1.1givenF2algebrainductionL9

Differentiating [F2] once gives Λt+1=2p ∂Pt((x−P)e−p(x−P)2). The product rule applied t times gives ∂Pt((x−P)h)=(x−P)∂Pth−t∂Pt−1h, hence (x−P)Λt=Λt+12p+tΛt−1, with the last term zero for t=0.

2.1givenF1F2step 1.1algebrainduction

The polynomial factors xAixBj have degree at most i+j. By [F1] they multiply a common Gaussian. The polynomial part of Λt has degree t and leading coefficient (2p)t, as follows inductively from Λt+1=2p(x−P)Λt−2ptΛt−1. Thus these polynomials form a triangular basis of the polynomials of degree at most i+j, which gives existence and uniqueness of the finite coefficients Eijt.

3.1step 1.1step 2.1algebra

Multiplying the expansion for gigj by xA=(x−P)+(P−A) gives the expansion for gi+1gj. Substitute the identity in step 1.1 and collect the coefficient of each Λt; it is Eijt−1/(2p)+(P−A)Eijt+(t+1)Eijt+1. Replacing xA by xB=(x−P)+(P−B) proves the second recurrence. The zero-outside-range convention handles both endpoints.

4.1givenF1step 2.1algebra∎

In three dimensions the primitive product factors into the product of its three one-dimensional coordinate factors. Multiplying their finite expansions gives the stated rectangular index range and a finite expansion in Λtuv.

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