Classical vs Quantum Anharmonic Oscillator

A classical particle and a quantum wave packet are released together at the same point in the same well: a spring with a quartic stiffening term added. Following both shows how closely the quantum average tracks the classical motion, and how the quartic term changes that.

What's on screen

• Potential V(x) (slate): the well, with its formula in the header.
• Classical particle (pink dot): rides on the curve at its own x.
• Quantum |ψ(x)|² (blue fill): the probability density, rescaled to fit the frame.
• E_cl (gold dashed): the classical energy V(A), labelled with its value.
• ⟨E⟩_q (green dashed): the packet's mean energy, labelled with its value.
• Statistics: classical x, ⟨x⟩ quantum, Δx (spread) and the clock. The Guided Inquiry reveals Δx and ⟨x⟩ step by step.

Press ⌖ on a row to see where that control sits in the sim.

Fixed: ħ = 1, so energies and times are in the sim's own units. The wave function lives on a periodic grid x ∈ [−12, 12]. A red warning banner means the numerics have degraded and the sim has paused. It is not physics.

The whole sim is one potential: a spring plus a quartic term. ε sets how much stiffer the well gets away from the centre.

Both travellers start at x = +A. The dot is at rest. The packet is a Gaussian centred on +A with zero mean momentum, with the width a pure parabola would keep unchanged:

They do not start with the same energy. The dot has E_cl = V(A). The packet's mean energy ⟨E⟩_q is higher, by ħω/2 (the zero-point energy) plus a share from the quartic term acting on the packet's width:

At the defaults ⟨E⟩_q = 10.30 and E_cl = 9.23, a gap of 1.07, of which ħω/2 = 0.65. Both lines then stay put for the whole run.

The classical force is the spring pull plus a cubic term:

At the turning point x = A the cubic term is 4εA²/(mω²) times the spring term: 0.43 at A = 3 and 1.70 at A = 6 (defaults). A larger swing is pulled harder than a spring would pull it, and that more than makes up for the longer path. A pure spring is different: its pull is proportional to the stretch, so any swing takes the same time.

At the defaults one round trip takes 4.21 time units from A = 3 and 3.22 from A = 6. A spring alone would take 2π/ω = 4.83 at every amplitude.

The round trip has an exact form. The substitution x = A sin θ removes the singularity at the turning point:

Dividing by ω shows that A enters only through λ = 2εA²/(mω²), which is half the force ratio of section 04. At ε = 0, λ = 0 and T = 2π/ω whatever the amplitude. For ε > 0, λ grows as A², so a wider swing is quicker. Changing m or ω does not remove this. It changes λ, and so how large the effect is. At ω = 4 (default m and ε), A = 3 and A = 6 take 1.55 and 1.48 against 2π/ω = 1.57.

The packet obeys the Schrödinger equation in the same well, with ħ = 1:

In a pure parabola the Gaussian start is a coherent state: a packet the parabola carries without distortion. It swings along the classical path with its width fixed, so Δx stays at 0.62 (defaults) for ever. With ε > 0 the energy levels are no longer evenly spaced, so the parts of the packet that sit in different levels drift out of step. The shape then changes: at the defaults Δx rises and falls over the first few round trips, and each peak is higher than the one before (about 1.0, 1.3, 1.5, 1.7).

Ehrenfest's equations are exact for any packet:

d⟨x⟩/dt = ⟨p⟩/m     d⟨p⟩/dt = −mω²⟨x⟩ − 4ε⟨x³⟩

The dot obeys the same pair with ⟨x³⟩ replaced by ⟨x⟩³. They differ because ⟨x³⟩ = ⟨x⟩³ + 3⟨x⟩(Δx)² + ⟨(x−⟨x⟩)³⟩. For ε = 0 the ⟨x³⟩ term is absent and ⟨x⟩ follows the dot exactly. For ε > 0 the extra terms depend on the packet's width and skewness, so ⟨x⟩ drifts away from the dot.

At ε = 0.10, A = 3 and the default m and ω, Δx passes 2 by t ≈ 5. Then ⟨x⟩ shrinks towards 0 (it stays below 0.5 after t ≈ 7.5) while the dot still reaches ±3. ⟨x⟩ is an average over |ψ|². It is not the place where the packet is.

• Molecular vibration. A bond is a well that is nearly a parabola near its floor. The Morse form fitted to HCl puts its first two rungs 2885 and 2780 cm⁻¹ apart: the levels close up, which a pure parabola cannot do.
• Superconducting qubits. A transmon is a slightly anharmonic oscillator. Its 1→2 transition sits a few hundred MHz below 0→1, so a microwave drive can address the lowest two levels alone.
• Any real oscillator. Near the bottom every smooth well is a spring, and the corrections are what make real periods depend on amplitude.

Key points

• V(x) = ½mω²x² + εx⁴. The packet's mean energy starts at E_cl + ħω/2 + 3εA²/(mω) + 3ε/(4m²ω²).
• The round trip depends on A only through λ = 2εA²/(mω²). At ε = 0 it is 2π/ω.
• At ε = 0, Δx is constant and ⟨x⟩ equals the classical x exactly.
• At ε > 0, Δx changes, and ⟨x⟩ departs from the dot because ⟨x³⟩ ≠ ⟨x⟩³.

Common mix-ups

• "Changing m or ω makes the period independent of amplitude." They rescale the time and change λ, but only ε = 0 removes the dependence.
• "⟨x⟩ is where the particle is." It is an average over |ψ|², and it can sit where |ψ|² is small.
• "The quantum packet should start at E_cl." It starts higher, by the zero-point energy and the quartic share.

Things to try on your own

• At the defaults, find the amplitude, in steps of 0.5, whose round trip is closest to 4.0 time units.
• Set ε = 0.10 and A = 3. Read the time after which |⟨x⟩| stays below 0.5.
• Set ω = 4 and compare the round trips at A = 3 and A = 6 with those at the defaults.