Kronig–Penney
V(x) = Σ_(k=0)⁶ 4 · exp(−30(x − 2k)²)
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What you see
The formula is the potential, and it is a sum. Σ 4 · exp(−30(x − 2k)²) over k = 0…6 puts seven gaussian barriers on the axis, centred at x = 0, 2, 4 and so on to 12. Each stands 4 tall and falls to 1/e of that within 1/√30 ≈ 0.18 of its centre.
Halfway between two of them the potential has effectively gone: at x = 1 it is 4 · exp(−30) ≈ 4 · 10⁻⁸. The packet therefore crosses flat ground seven times and meets seven walls, all identical and all equally spaced. That regularity is the whole point.
The physics
Momentum k₀ 1.9 and Packet width 1.6 give an energy of k₀²/2 + 1/(4σ²) = 1.81 + 0.10 = 1.90, well under the barrier tops. Every crossing is tunnelling, so a single barrier would already turn much of the packet back.
Seven do something a single one cannot. Each barrier reflects a little, and those reflections interfere with each other. At some momenta they cancel and the lattice is nearly clear; at others they reinforce and it is nearly closed. The closed intervals are band gaps, and a solid conducts or insulates for this reason. A weak lattice of spacing a opens its first gap near k = π/a = 1.57; here the barriers also lift the average potential by 0.65, which pushes the measured gap a little higher.
What T and R read
By t ≈ 12 at Speed 1 the numbers settle near T = 0.29 and R = 0.71. This momentum sits inside a gap, so most of the packet comes back.
Try
- Raise Momentum k₀ to 2.6. T climbs to about 0.83: the same lattice, now transparent, because this momentum lies in a pass band.
- Lower Momentum k₀ to 1.3. T is about 0.50, between the two, on the shoulder of the gap.
- Keep one barrier: 4 · exp(−30x²). It passes about 0.64. Seven acting independently would pass 0.64⁷ ≈ 0.04, and the real seven pass 0.29, so they are not independent at all.
- Widen the spacing to 3 with sum(4 · exp(−30(x − 3k)²), k, 0, 4). T rises to about 0.55: a different spacing puts this momentum outside the gap.