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LESSON NOTES · 05

5. Harmonic oscillator and zero-point motion

Position in the course: Lesson 5 of 10. Complete the preceding derivation and use the explained exercises to check understanding.

Prerequisites: Canonical commutator; differential equations

1. Approximation near a stable minimum

Shift position and energy origins. A smooth potential is approximately \(m\omega^2x^2/2\) near a stable minimum. Cubic and higher terms are neglected, so this is best for small amplitude. Ladder operators turn the differential problem into algebra. Positivity of \(a^\dagger a\) supplies the spectrum’s lower end.

\[ \hat H=\frac{\hat p^2}{2m}+\frac12m\omega^2\hat x^2,\quad a=\sqrt{\frac{m\omega}{2\hbar}}\hat x+\frac{i\hat p}{\sqrt{2m\hbar\omega}},\quad[a,a^\dagger]=1. \]

2. Construct equally spaced energies

Substitute the definitions into \(H\) using \([x,p]=i\hbar\). Commuting \(H\) with the ladder operators shifts energy by \(\hbar\omega\). Repeated lowering cannot produce negative norm; the lowest state is annihilated by \(a\). Repeated raising gives normalized higher states.

\[ \begin{aligned}H&=\hbar\omega(a^\dagger a+\tfrac12),\\{}[H,a^\dagger]&=\hbar\omega a^\dagger,\quad[H,a]=-\hbar\omega a,\\E_n&=\hbar\omega(n+\tfrac12),\quad |n\rangle=\frac{(a^\dagger)^n}{\sqrt{n!}}|0\rangle.\end{aligned} \]

3. Solve the annihilation equation

Use \(p=-i\hbar\partial_x\) in \(a\psi_0=0\). The first-order equation integrates to a Gaussian; its squared integral fixes normalization. The wavefunction has nonzero tails beyond the classical turning points, not negative probabilities.

\[ \frac{d\psi_0}{dx}=-\frac{m\omega}{\hbar}x\psi_0,\quad\psi_0=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-m\omega x^2/(2\hbar)}. \]

4. Zero-point fluctuations

The symmetric Gaussian has zero mean position and momentum but nonzero spreads. Evaluate the two second moments to see that the ground state saturates the uncertainty bound. Zero-point energy is not thermal energy.

\[ \langle x^2\rangle_0=\frac{\hbar}{2m\omega},\quad\langle p^2\rangle_0=\frac{m\hbar\omega}{2},\quad\Delta x\Delta p=\hbar/2. \]

Harmonic oscillator and zero-point motion — schematic under the stated model assumptions

Teaching schematic drawn from the equations or algorithm steps in this lesson; not measured data.

5. Worked isotope scaling

For two atoms use reduced mass \(\mu=m_1m_2/(m_1+m_2)\) and force constant \(k\), giving \(\omega=\sqrt{k/\mu}\). Increasing \(\mu\) fourfold at fixed \(k\) halves spacing and zero-point energy. Ground-state position width scales as \(\mu^{-1/4}\) and falls by \(\sqrt2\). This is a harmonic analytical scaling, not a molecular spectrum prediction.

6. Check yourself and limits

Can \(x\) and \(p\) both be sharply zero?

Answer and reasoning

No, the canonical uncertainty relation forbids it.

Why do high molecular vibration levels deviate from equal spacing?

Answer and reasoning

Anharmonic terms and eventual dissociation invalidate the quadratic approximation.

7. Matrix elements, selection rules and thermal populations

Invert the ladder definitions to obtain \(x=\sqrt{\hbar/(2m\omega)}(a+a^\dagger)\). Since \(a|n\rangle=\sqrt n|n-1\rangle\), the position matrix element is nonzero only between neighboring levels:

\[ \langle m|x|n\rangle=\sqrt{\frac{\hbar}{2m\omega}}\left(\sqrt n\,\delta_{m,n-1}+\sqrt{n+1}\,\delta_{m,n+1}\right). \]

A dipole interaction linear in the normal coordinate therefore has the harmonic selection rule \(\Delta n=\pm1\). This is conditional: the dipole must change with that coordinate for the vibration to be infrared active. Anharmonic potentials and nonlinear dipole terms permit additional transitions. An energy-level diagram alone cannot determine activity or intensity.

For a canonical oscillator, summing its geometric-series partition function gives \(Z=e^{-\beta\hbar\omega/2}/(1-e^{-\beta\hbar\omega})\). Differentiating \(-\partial_\beta\ln Z\) gives mean energy \(\hbar\omega/2+\hbar\omega/(e^{\beta\hbar\omega}-1)\). At low temperature the thermal part vanishes but zero-point energy remains. At high temperature the expression approaches \(k_BT\) to leading order. The quantum heat capacity therefore vanishes at low temperature, unlike a classical mode with equipartition at every temperature.

Additional exercise: Evaluate \(\langle1|x|0\rangle\). Only the raising term survives, giving \(\sqrt{\hbar/(2m\omega)}\). Evaluate \(\langle2|x|0\rangle\) and obtain zero. Neither result says the first excited state is more spatially localized; transition amplitudes and position widths are different quantities. These analytical relations connect basic quantum mechanics to vibrational spectroscopy and to the quantum corrections missing from classical MD.

Further conceptual check

A coordinate displacement can change the origin without changing the spacing. A frequency change alters both spacing and wavefunction width. In an actual normal-mode calculation, the mass-weighted Hessian supplies frequencies, while a negative curvature indicates an unstable direction rather than a real bound oscillator.

8. References and further reading

Derivations and numerical examples here are original teaching constructions, not copied passages or reported research data.

Molecular methods · Quantum Monte Carlo · Gaussian · VASP

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