Smooth Step
V(x) = 1 + erf(x/2)
Open in the app The dials and keys named below are the app's.
What you see
The formula is the potential. erf runs from −1 far to the left to +1 far to the right, so 1 + erf(x/2) climbs from 0 to 2. The ground is flat on both sides and the whole rise happens in the middle.
The 2 inside sets how long that rise takes. At x = 2 the potential has reached 1 + erf(1) = 1.84, and at x = −2 it is only 0.16, so the step is essentially complete across four units of x. Its half height falls at x = 0.
The physics
Momentum k₀ 2 and Packet width 0.7 give an energy of k₀²/2 + 1/(4σ²) = 2 + 0.51 = 2.51. The step tops out at 2. The packet is therefore travelling above the step, and a classical particle would cross every time, merely slowing down as it climbed.
A quantum one does not. Part of it turns back from a rise it has more than enough energy to climb, which is called quantum reflection, and it depends on how abruptly the potential changes rather than on how high it goes. A step that turns on gently over several wavelengths is almost invisible to the packet. A step that turns on within one is a wall it can partly bounce off.
What T and R read
By t ≈ 5 at Speed 1 the readouts hold near T = 0.76. About a quarter of the packet comes back from a step it should have cleared. Read the numbers before t ≈ 7, after which the wall at x = −8 returns the reflected part and mixes the two.
Try
- Make the step abrupt with 2 · (x > 0). Same height, no gradient at all, and T drops to about 0.48. The edge alone accounts for the difference.
- Make it gentler with 1 + erf(x/4). T rises to about 0.83, closer to the classical answer of everything through.
- Make it steeper with 1 + erf(x/0.5). T falls to about 0.54, nearly the sharp step again.
- Raise Momentum k₀ to 3. T reaches about 0.95: a faster packet has a shorter wavelength, so the same rise looks gentler to it.