“Know how to solve every problem that has been solved.” “What I cannot create, I do not understand.” — Richard Feynman

Gaussian Integrals by Hand — Exercises

Exercises

Quantum Chemistry course › Unit 4 — Integrals

The technique: evaluate molecular integrals over s-type Gaussians by hand, using the Gaussian product theorem to collapse every two-center product into a single Gaussian. Done a few times with real exponents, the closed forms stop being formulas and start being objects you can estimate at a glance.

Worked example: one overlap, start to finish

Problem. Two normalized s-type primitive Gaussians, both with exponent $α = β = 0.5$ , sit at $A = (0, 0, 0)$ and $B = (0, 0, 1)$ bohr. Compute their overlap $S = ⟨ g_{A} ∣ g_{B} ⟩$ to six decimal places.

Diagnosis. A two-center integral, so the move is the product theorem: rewrite $g_{A} g_{B}$ as one Gaussian times the prefactor $K = exp (- \frac{α β}{p} ∣ A - B ∣^{2})$ with $p = α + β$ , then integrate a single Gaussian — which is just a normalization bookkeeping exercise.

Predict before reading on: Both Gaussians are normalized and have equal exponents. Before computing anything: do you expect the normalization factors to survive into the final answer, or cancel?

Solution. Assemble the pieces of the closed form:

S = N_{α} N_{β} (\frac{π}{p})^{3/2} exp (- \frac{α β}{p} ∣ A - B ∣^{2})

Step 1 — the exponential. $p = 1.0$ , $α β / p = 0.25$ , $∣ A - B ∣^{2} = 1$ , so $K = e^{- 0.25} = 0.778801$ .

Step 2 — the prefactor. With $N = (2 α / π)^{3/4}$ and equal exponents, $N_{α} N_{β} (π / p)^{3/2} = (1/ π)^{3/2} \cdot π^{3/2} = 1$ exactly: for equal exponents the normalization and volume factors cancel to one.

Predict before reading on: Given that cancellation, what is the overlap of these two Gaussians at any separation R, in one symbol?

Step 3 — assemble: $S = 1 \times K = 0.778801$ . For equal-exponent normalized s-Gaussians, the overlap is the product-theorem prefactor.

Verification. Limits: at $R = 0$ , $K = 1$ — self-overlap of a normalized function ✓. As $R \to \infty$ , $S \to 0$ ✓. And the project-2 code agrees: prim_overlap(0.5, A, 0.5, B) returns 0.778801.

Articulate: State the technique in one sentence — what does the product theorem buy you, operationally, in every integral of this family?

Practice

P.1 geometry — weighted averages

The tightest STO-3G primitive (

α = 3.42525

) sits at the origin; the most diffuse one (

β = 0.16886

) sits at

(0, 0, 1.4)

. Where is the product center P? Compute its z-coordinate.

Find the analogue: The worked example's P sat at the midpoint. Which inequality between the exponents broke that symmetry here, and toward which center must P move?

show answer

P_{z} = \frac{β \cdot 1.4}{α + β} = \frac{0.16886 \times 1.4}{3.59411} = 0.0658

bohr — P hugs the tight Gaussian. The exponent-weighted average is dominated by the larger exponent.

P.2 screening — why big molecules are sparse

For the worked example's pair (

α = β = 0.5

), compute the prefactor K at separations R = 1.4 and R = 2.8 bohr. By what factor does doubling the distance shrink it?

Find the analogue: Same Step 1 as the worked example, run twice — only $∣ A - B ∣^{2}$ changes. What does its appearance squared in the exponent do to the decay?

show answer

K (1.4) = e^{- 0.25 \times 1.96} = 0.6126

K (2.8) = e^{- 0.25 \times 7.84} = 0.1409

— a factor of 4.3 from doubling R, and it accelerates (Gaussian, not exponential, decay). This is the suppression Schwarz screening exploits to skip distant integrals.

P.3 basis design — who carries the bond?

Compute the overlap of the two tightest STO-3G primitives (

α = β = 3.42525

), one on each H atom of H₂ at R = 1.4 bohr. Compare with the full contracted overlap S₁₂ = 0.659.

Find the analogue: Equal exponents again — which one-symbol shortcut from the worked example applies verbatim?

show answer

S = K = exp (- \frac{3.42525}{2} \times 1.96) = 0.0348

. The tight primitives barely see each other; the 0.659 contracted overlap is carried almost entirely by the diffuse primitives. Cusp-patchers don't bond — tails do.

P.4 operator limits — kinetic energy at one center

Take the general kinetic formula

T = \frac{α β}{p} (3 - 2 \frac{α β}{p} ∣ A - B ∣^{2}) S

to the same-center limit

A = B

for equal exponents

α = β = 1

(normalized, so S = 1). What is T?

Find the analogue: The worked example evaluated a general formula at specific values. Here the "value" is a limit — which term dies when $∣ A - B ∣^{2} \to 0$ ?

show answer

The

∣ A - B ∣^{2}

term vanishes, leaving

T = \frac{α β}{p} \cdot 3 \cdot S

. With

α β / p = 1/2

and

S = 1

T = 1.5

hartree. In general the same-center kinetic energy is

\frac{3 α}{2}

— tighter Gaussians cost kinetic energy linearly in α, the confinement half of the ζ tug-of-war.

P.5 normalization — the (2α/π)^(3/4) factor

Compute the normalization constant

N = (2 α / π)^{3/4}

for

α = 1

, and explain in one sentence why N grows with α.

Find the analogue: In the worked example the N's cancelled. Here you compute one alone — what role did each N play before the cancellation?

show answer

N = (2/ π)^{3/4} = 0.7127

. A tighter Gaussian concentrates its probability in a smaller volume, so its amplitude must rise to keep

\int ∣ g ∣^{2} = 1

P.6 special functions — the Boys series

Evaluate

F_{0} (0.1)

using the small-argument series

1 - T /3 + T^{2} /10

, and compare with the closed form

\frac{1}{2} π / T erf (T) = 0.967643

. How many digits does the 3-term series deliver?

Find the analogue: Like the worked example's verification step: a second, independent route to the same number. What plays the role of the "limit check" here?

show answer

Series:

1 - 0.03333 + 0.001 = 0.967667

. Error

2.4 \times 1 0^{- 5}

— four good digits from three terms, because the next term is

- T^{3} /42 \sim 2 \times 1 0^{- 5}

. The series isn't a fallback; near T = 0 it's the better method.

Check problems

Check 1 derivation

Derive, from the general overlap formula, the statement used twice above: for two normalized, equal-exponent s-Gaussians at any separation,

S = K = exp (- \frac{α}{2} ∣ A - B ∣^{2})

. Where exactly does each factor of the normalization cancel?

show solution sketch

With

α = β

p = 2 α

, so

N^{2} (π / p)^{3/2} = (2 α / π)^{3/2} (π /2 α)^{3/2} = 1

— the normalization squared is exactly the inverse of the volume factor of the product Gaussian. What survives is the prefactor with

α β / p = α /2

, giving

S = e^{- α ∣ A - B ∣^{2} /2}

. The cancellation breaks for unequal exponents, which is why the general case keeps all three factors.

Check 2 articulation

The Boys function appears in the nuclear-attraction and electron-repulsion integrals but never in overlap or kinetic energy. Explain, in 100–300 words, what property of the operator decides whether F₀ shows up — and why the product theorem alone can't finish the Coulomb integrals.

show solution sketch

Overlap and kinetic energy involve operators (1 and

- \frac{1}{2} \nabla^{2}

) that keep the integrand a polynomial times a Gaussian, so after the product theorem the integral is a standard Gaussian moment with a closed elementary form. The Coulomb operator

1/ r

is different in kind: it is singular at a point and couples the collapsed Gaussian to a fixed center (a nucleus, or the other electron's product Gaussian). Writing

1/ r

via its integral representation introduces one leftover one-dimensional integral over an auxiliary variable — and that integral is the Boys function. So the product theorem reduces the geometry (many centers → one), while F₀ absorbs the operator's singularity; you need both, and only Coulomb-type operators require the second.

Hands cramping? Good. Now make the machine do it: Project 2 — Overlap turns this technique into code.