LESSON NOTES · 03

Harmonic oscillator and zero-point motion

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.

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

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

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.

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.

References and further reading

• MIT 8.04: Harmonic oscillator

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

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