Potential Step
V(x) = 2(x > 0)
Open in the app The dials and keys named below are the app's.
What you see
The formula is the potential: x > 0 is 1 to the right of the origin and 0 to the left, so V is a cliff of height 2. There is flat ground on the near side and a raised plateau on the far side. The dashed line across it is the packet's energy, k₀²/2 + 1/4σ² = 2 + 0.51 = 2.51, just above the top of the step.
The packet starts at Packet centre −4 with Momentum k₀ 2, and reaches the step about two seconds in at Speed 1. A ball with more energy than the step would climb it, slow down and carry on. This packet does something else: it divides.
The physics
What matters is not the packet's average energy but each momentum in it separately. A component of momentum k has energy k²/2 and can only pass if k > 2, exactly the packet's central momentum. Half of what it is built from is under the threshold and is turned back whole. The rest crosses and slows to k′ = √(k² − 4), so a component at 2.5 leaves the step at 1.5. The transmitted hump therefore crawls away while the reflected one races back at the full speed 2. Even the components that clear the step are not all transmitted. A step reflects a wave whenever it changes its wavelength abruptly, in the ratio 4kk′/(k + k′)², which vanishes as k′ goes to zero. Adding that up over the packet's momenta gives 0.44.
Where T settles
T overshoots to about 0.63 near t = 3, while the packet is still lying across the origin. It then falls back as the reflected part pulls away. By t = 6 it reads 0.44, the value the momentum sum gives.
Try
- Set Momentum k₀ to 3: the energy is well clear of the step now and T reaches about 0.9.
- Halve the step, 1 · (x > 0): about 0.84 goes on.
- Turn the cliff into a drop, −2 · (x > 0). Nothing is in the way and the particle speeds up, yet about a tenth of it still comes back. Reflection needs a change, not an obstacle.
- Press Side in the deck to read the split against the silhouette of the step.