Harmonic Well
V(x) = x²/2
Open in the app The dials and keys named below are the app's.
What you see
The formula is the potential, V = x²/2: a parabola, the same bowl that holds a mass on a spring. Written as ½ω²x² it has ω = 1. Its silhouette on the plane looks shallow in the middle, because with the walls at ±8 its far edges reach 32.
The packet is released at Packet centre 3 with Momentum k₀ 0, at rest and off centre, with Packet width 0.7. It slides down, through the middle, up the far side to x = −3, and back, over and over.
Why it never spreads
A free packet flattens as it goes; this one does not. The bowl's own ground state has width 1/√(2ω) ≈ 0.7071, and 0.7 is that width to within a per cent. What is drawn is therefore a coherent state: the same shape carried along a classical path. Its width holds at 0.70 through swing after swing, and the round trip takes 2π/ω ≈ 6.28 seconds at Speed 1. That period is the same whatever the amplitude, as for any harmonic oscillator. Ehrenfest's theorem already guarantees that the mean position obeys the classical equation when V is quadratic. The coherent state is the case where the whole shape follows it too.
T as a clock
Here the readouts count out the swing. T is 1 at the start, 0.5 as the packet crosses the middle at t ≈ 1.57, and 0 at the far turning point at t ≈ 3.14. It is 1 again at 6.28. The energy readout is the packet's kinetic part alone, 1/4σ² = 0.51, and the dashed line draws that, not the potential energy it also has at x = 3. That is why the line sits near the floor of the bowl.
Try
- Squeeze it: Packet width 0.35, half the natural width. The swing is unchanged, but the shape now breathes out to 1.43 as it crosses the middle and back to 0.35 at each turning point, twice a period.
- Push it as you release it: Momentum k₀ 2 gives 3 cos t + 2 sin t, an amplitude of √13 ≈ 3.61 and the same 6.28 second period.
- Steepen the bowl to 2x²: now ω = 2 and the period halves to π ≈ 3.14. The natural width drops to 0.5, so the packet breathes again.
- Pull Span of x in to 4 so the parabola fills more of the plate.