Questions: Quantum Harmonic Oscillator and Molecular Vibrations

Question 1

The ground-state energy of a quantum harmonic oscillator is E_0 = ℏω/2, not zero. Which principle most directly requires this non-zero ground state?

• A. The Pauli exclusion principle, which prevents two particles from sharing the same quantum state.
• B. The Heisenberg uncertainty principle, which forbids a particle from simultaneously having zero position uncertainty (at equilibrium) and zero momentum.
• C. The Born-Oppenheimer approximation, which separates nuclear and electronic motion.
• D. Conservation of energy, which requires vibrational energy to be stored somewhere at all temperatures.

Answer: (B)

If the oscillator had zero energy, it would sit motionless at the equilibrium position — its position and momentum would both be exactly zero, violating the uncertainty relation ΔxΔp ≥ ℏ/2. The zero-point energy ℏω/2 is the minimum energy consistent with this uncertainty. This is not a measurement artifact; it has physical consequences: even at 0 K, molecular bonds vibrate, which affects isotope effects, tunnel rates, and thermodynamic properties.

Question 2

The quantum harmonic oscillator has equally spaced energy levels. This equal spacing means a real diatomic molecule's vibrational transitions should most appear at exactly the same frequency in an IR spectrum.

• T. True
• F. False

Answer: False

Equal spacing holds only for the ideal harmonic potential. Real bonds follow an anharmonic potential (approximated by the Morse potential), where the restoring force weakens as the bond stretches toward dissociation. This causes the energy level spacing to decrease with increasing v. As a result, the fundamental (v=0→1) and overtones (v=0→2, 0→3) appear at slightly different frequencies, and at high enough energy the levels converge and the bond dissociates.

Question 3

A quantum harmonic oscillator wavefunction has finite amplitude in regions where the total energy is less than the potential energy — classically forbidden zones. What is this phenomenon called, and what observable spectroscopic consequence does it have?

Model answer: Quantum mechanical tunneling. Because wavefunctions decay exponentially rather than abruptly into classically forbidden regions, there is a non-zero probability of finding the bond stretched beyond its classical turning point. In spectroscopy, tunneling enables hydrogen-atom transfer reactions and contributes to the intensities of combination bands. It also underlies the fact that zero-point energy is non-zero: the particle cannot be localized at the potential minimum without kinetic energy.

In classical mechanics, a particle with energy E cannot penetrate a region where V > E. In quantum mechanics the wavefunction decays exponentially in such regions (evanescent behavior) but does not vanish, allowing tunneling through thin barriers. This is directly observable: reaction rates of proton-transfer reactions are much faster than predicted classically at low temperatures, due to tunneling through the activation barrier.