# Harmonic Well

Quantum · ψ under V(x)

`V(x) = x²/2`

[Open in the app](https://www.wavelace.com/app#p=71) · [This page](https://www.wavelace.com/presets/harmonic-well)

### 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.

### Read more

- [Quantum harmonic oscillator](https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator)
- [Coherent state](https://en.wikipedia.org/wiki/Coherent_state)
- [Ehrenfest theorem](https://en.wikipedia.org/wiki/Ehrenfest_theorem)
- [Schrödinger equation](https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation)