Free Packet
V(x) = 0
Open in the app The dials and keys named below are the app's.
What you see
The formula is the potential, and here it is 0. Nothing pushes the particle about, so there is no silhouette standing on the plane and no energy line across it, only the packet.
It starts as a Gaussian at Packet centre −4, of width σ = 0.5, carrying Momentum k₀ 3: ψ₀ = exp(−(x + 4)²/4σ²) · exp(3ix). Momentum is also speed here, so at Speed 1 the hump slides right at 3 units a second. It passes the origin at t ≈ 1.33, and meets the wall at x = 8 near t = 4, where it bounces.
Why it spreads
A hump that narrow has to be built from a range of momenta, Δk = 1/2σ = 1 either side of 3. Each of them travels at its own speed, so the packet smears as it goes: σ(t) = σ · √(1 + (t/2σ²)²) = 0.5 · √(1 + 4t²). It is twice as wide by t ≈ 0.87, and four times by t = 2. Nothing is lost while it flattens. The area under the curve is fixed, and the norm readout holds at 1.0000, because the solver is exactly unitary.
Where T settles
Nothing turns this packet back, so T climbs through 0.19 at t = 1 and 0.63 at t = 1.5. It then levels near 0.96 rather than 1, because by that time the packet is wide enough to leave a tail behind the origin. Once the wall has returned it, the two readouts settle near 0.5 apiece.
Try
- Raise Packet width to 2: less momentum spread, so it holds together, only half as wide again by t = 4.
- Set Momentum k₀ to 0: the packet stays where it is and spreads both ways at once.
- Put something in its path: 3 · (abs(x) < 0.4) is a barrier, and most of it comes back.
- Press Top in the deck: the trailing past of the curve reads as a wake, widening as it goes.