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Two-Level Systems by Hand — Bonding and antibonding are not symmetric partners

Exercises

Introductory Quantum Mechanics › Unit 3 · Two-level systems & spin-1/2 › Two-Level Systems by Hand › all problems

Practice Problem 4 of 9

Bonding and antibonding are not symmetric partners

Problem

Two identical atomic orbitals have

H_{11} = H_{22} = - 0.60

, coupling

H_{12} = - 0.35

, and overlap

S_{12} = 0.4

(hartree units). Solve

det (H - ES) = 0

for both roots. How much is the bonding level lowered, how much is the antibonding level raised, and why are the two numbers different?

Solution

Predict before reading on. The new ingredient relative to the worked example is the overlap matrix — the basis is not orthonormal. By symmetry the eigenvectors are still the $\pm$ combinations; where does $S_{12}$ enter the eigenvalues?

E_{\pm} = \frac{H _{11} \pm H _{12}}{1 \pm S _{12}} : E_{+} = \frac{- 0.95}{1.4} = - 0.679, E_{-} = \frac{- 0.25}{0.6} = - 0.417

Bonding drops by 0.079; antibonding rises by 0.183 — more than twice as much. The $1 \pm S_{12}$ denominators do it: overlap renormalizes the two combinations differently. This asymmetry is why filling both levels (4 electrons in 2 orbitals) is net anti-bonding — the reason two helium atoms do not form He $_{2}$ .

Result

E_{+} = - 0.679, E_{-} = - 0.417 E_{h}; antibonding destabilized more than bonding stabilizes