# Tunnelling Barrier

Quantum · ψ under V(x)

`V(x) = 3(abs(x) < 0.4)`

[Open in the app](https://www.wavelace.com/app#p=69) · [This page](https://www.wavelace.com/presets/tunnelling-barrier)

### What you see

The formula is the potential. `abs(x) < 0.4` is 1 where it holds and 0 where it does not, so `V` is a rectangular barrier 0.8 wide and 3 tall sitting on the origin. The dashed line drawn across it is the packet's energy, `k₀²/2 + 1/4σ² = 2 + 0.51 = 2.51`, and it passes below the top.

The packet starts at `Packet centre` `−4` with `Momentum k₀` 2, so it takes about two seconds at `Speed` 1 to reach the barrier. A ball with that energy would bounce off. Watch what the packet does instead: it splits, and part of it appears on the far side.

### The physics

Inside a barrier it cannot climb, the wave stops oscillating and decays, at the rate `κ = √(2(V − E)) = 0.99`. A plane wave at the packet's central energy therefore comes out the far side with about `e^(−2κ · 0.8) ≈ 0.21` of its probability. Thinner or lower barriers pass far more, and the dependence on width is exponential. The packet is not one plane wave, though. Its width of 0.7 gives it a momentum spread of `1/2σ ≈ 0.71`, and a quarter of it carries more than the 3 the barrier asks for and simply goes over the top. Weighting the exact barrier formula across those momenta gives 0.36.

### Where T settles

The collision is over by `t ≈ 4`, and there `T` reads 0.375 against `R` 0.625, close to the 0.36 the momentum sum predicts. That is the number to read early. The reflected half returns from the wall at `x = −8` around `t = 7`, and after that the two mix again.

### History

In 1928 George Gamow, and independently Ronald Gurney and Edward Condon, used this to explain alpha decay. An alpha particle inside a nucleus is held by a barrier it has no energy to climb, so it leaks out instead. Because the leak is exponential in the barrier, a narrow range of decay energies produces half-lives spread over twenty orders of magnitude.

### Try

- Raise the barrier, `6 · (abs(x) < 0.4)`: `T` falls to about 0.08.
- Double its width, `3 · (abs(x) < 0.8)`: about 0.20 gets through, so width bites harder than height.
- Set `Momentum k₀` to 3. The energy is now 5, well above the top, and still only about 0.71 goes on: a wave is partly reflected by any sudden change, even a favourable one.
- Use `Step` to walk the collision frame by frame.

### Read more

- [Quantum tunnelling](https://en.wikipedia.org/wiki/Quantum_tunnelling)
- [Rectangular potential barrier](https://en.wikipedia.org/wiki/Rectangular_potential_barrier)
- [Alpha decay](https://en.wikipedia.org/wiki/Alpha_decay)
- [Wave packet](https://en.wikipedia.org/wiki/Wave_packet)