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Sandbox Physics
Molecular electrons and the breaking of a bond

L1 · Chapter 1

Choosing an orbital for one hydrogen electron

Sandbox Physics ·

A trial orbital gives a useful molecular answer when it allows the electrons enough freedom to lower their energy. Hydrogen lets us see what that statement means before a second electron complicates the picture.

An orbital predicts where an electron may be found

Repeated position measurements on identically prepared hydrogen atoms do not draw a tiny planetary orbit. They build a spatial probability distribution. A one-electron orbital is a function whose squared magnitude gives that distribution.

Write the orbital as ϕ(r)\phi(\mathbf r). The position vector r\mathbf r tells us where a small volume lies relative to the nucleus. Normalization means the probability, added over all space, is one. The orbital itself can have a positive or negative sign; probabilities cannot. That sign matters when two amplitudes are added.

∫∣ϕ(r)∣2 d3r=1\int |\phi(\mathbf r)|^2\,d^3r=1
The integral adds the probability density over small volume elements throughout space.

Reading: A. E. DePrince III

Tightening the cloud has a cost

Hold one proton at the origin. Use atomic units, in which electron mass, elementary charge and the reduced Planck constant have numerical value one. Distances are in bohr and energies in hartree. Try a spherical orbital that decays as e−ζre^{-\zeta r}, where positive ζ\zeta controls how tightly it is concentrated.

A tighter orbital puts more probability close to the attractive nucleus, which lowers potential energy. It also changes more rapidly in space, which raises kinetic energy. For this normalized exponential, both contributions can be integrated exactly.

The slope of its energy is dE/dζ=ζ−1dE/d\zeta=\zeta-1. It vanishes at ζ=1\zeta=1, where the energy is −1/2-1/2 hartree. Increasing the exponent to two makes the total energy zero: squeezing the cloud twice as hard is a poor bargain.

ϕζ(r)=(ζ3π)1/2e−ζr\phi_\zeta(\mathbf r)=\left(\frac{\zeta^3}{\pi}\right)^{1/2}e^{-\zeta r}
The prefactor normalizes the orbital; the exponent sets its radial scale.
E(ζ)=⟨T⟩+⟨V⟩=ζ22−ζE(\zeta)=\langle T\rangle+\langle V\rangle=\frac{\zeta^2}{2}-\zeta
Kinetic energy grows quadratically with the exponent, whereas the attractive contribution grows linearly.
0120−0.5minimum
Horizontal: exponential decay parameterVertical: energy in hartree
E(ζ)=ζ22−ζE(\zeta)=\frac{\zeta^2}{2}-\zeta
Analytic trial energy of a normalized exponential orbital around one fixed hydrogen nucleus, in atomic units. Tightening the orbital increases kinetic energy faster than it improves attraction.

Reading: A. E. DePrince III

Why a trial energy is an upper bound

Imagine expressing a normalized trial state as a sum of exact energy states Ψ=∑kckΨk\Psi=\sum_k c_k\Psi_k. The squared coefficients are nonnegative weights and add to one. The trial energy is their weighted average. It cannot lie below the lowest exact energy E0E_0.

This is the variational principle. The Hamiltonian is the energy operator: averaging it in a state gives that state’s energy. The useful freedom lies in choosing which trial states are allowed. Enlarging the allowed set can lower the best energy. An optimization that stops early can leave it unnecessarily high. Neither a larger set nor a longer optimization changes the Hamiltonian whose energy is being averaged.

For hydrogen the best exponential happens to be the exact nonrelativistic ground state. For a molecule, a short list of trial functions usually cannot represent the exact state. Its best energy still gives an upper bound, provided the same Hamiltonian and a normalized admissible wavefunction are used.

E[Ψ]=∑k∣ck∣2Ek≥E0,∑k∣ck∣2=1E[\Psi]=\sum_k |c_k|^2 E_k\geq E_0,\qquad \sum_k|c_k|^2=1
A weighted average of exact energies cannot fall below their minimum. This argument applies to wavefunction energies, not automatically to an approximate density functional.

Reading: A. E. DePrince III

Two atomic functions can share a sign or oppose it

Place identical normalized atomic functions χA\chi_A and χB\chi_B on two nuclei. Their overlap S=∫χAχB d3rS=\int\chi_A\chi_B\,d^3r measures how much the functions occupy the same region. The symmetric sum reinforces amplitude between the nuclei. The antisymmetric difference has a node halfway between them.

Do not divide both combinations by 2\sqrt2 unless the overlap is zero. Squaring the sum produces two cross terms, which add 2S2S to its norm. Squaring the difference subtracts the same amount. The denominators below follow directly.

The lower-energy combination is commonly called bonding, and the other antibonding. The labels refer to the resulting molecular orbitals and their energetic role. A visible cloud does not yet tell us how two interacting electrons should occupy them; that is the next problem.

ϕg=χA+χB2(1+S),ϕu=χA−χB2(1−S)\phi_g=\frac{\chi_A+\chi_B}{\sqrt{2(1+S)}},\qquad \phi_u=\frac{\chi_A-\chi_B}{\sqrt{2(1-S)}}
These normalized combinations apply to identical real atomic functions in a symmetric diatomic molecule.

Reading: A. E. DePrince III · Basis Set Exchange

Try the prediction

Look for the shared density between the nuclei

  1. Start near the usual molecular bond length and identify the two nuclei and the density between them.
  2. Increase the separation. Predict how the overlap changes before looking at the updated cloud.
  3. Explain what extra information a joint two-electron distribution supplies beyond a one-electron cloud.

Exercises and explanations

  1. For the hydrogen trial orbital, calculate the energy at ζ=1/2\zeta=1/2 and compare it with the optimum.

    Read the explanation

    E=1/8−1/2=−3/8E=1/8-1/2=-3/8 hartree. The cloud is too diffuse. It gains less attraction than the optimum orbital; its smaller kinetic energy does not compensate. The result lies above −1/2-1/2 hartree, as the variational bound requires.

  2. Two atomic functions have overlap S=0.2S=0.2. What is the squared norm of their unnormalized sum?

    Read the explanation

    It is 1+1+2S=2.41+1+2S=2.4. Divide the sum by 2.4\sqrt{2.4}. Using 2\sqrt2 would leave an incorrectly normalized state and spoil any energy expectation computed without its norm denominator.

  3. A larger set of trial orbitals gives a higher optimized energy. What should you inspect first?

    Read the explanation

    Check that the Hamiltonian and basis definitions really match and that the smaller trial space is contained in the larger one. Then inspect convergence and compare stationary solutions to find the lowest state available in the trial space.

Sources and calculation scope

  1. A. E. DePrince III, The Hartree–Fock Self-Consistent Field Procedure

    Author tutorial: electronic Hamiltonian, variational principle, density matrices, SCF, basis convergence and spin-symmetry breaking. The browser calculation specializes these equations to two hydrogen STO-3G functions.

  2. Basis Set Exchange, STO-3G basis data

    Hydrogen STO-3G contraction: one contracted s function per atom, made from three Gaussian primitives. Basis coefficients and normalization are fixed in the calculation manifest.

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