# Barrier or Well

Quantum · ψ under V(x)

`V(x) = V₀ · (abs(x) < abs(w))`

[Open in the app](https://www.wavelace.com/app#p=144) · [This page](https://www.wavelace.com/presets/barrier-or-well)

### What you see

The formula is the potential. `abs(x) < abs(w)` is 1 where it holds and 0 where it does not, so `V` is a rectangle of half width `abs(w)` standing on the origin, and `V₀` is its height. A negative `w` draws the same rectangle as a positive one. It opens at `w = 1` and `V₀ = −6`, so the rectangle hangs below the ground instead of above it: a hole 2 wide and 6 deep.

The packet starts at `Packet centre` `−8` with `Momentum k₀` 2, and the dashed line across the plate is its energy, `k₀²/2 + 1/4σ² = 2 + 0.10 = 2.10`.

### The two sides of zero

Drag `V₀` up and the rectangle becomes a wall, and the number that matters is the energy 2.10. `T` reads 0.88 at `V₀ = 1`, 0.33 at 2 and 0.05 at 3: a factor of nineteen over two units of the slider, as the wall rises past the packet's energy.

Drag it down instead and the rectangle is a hole, which a classical particle would cross without noticing. This one does not: `T` falls to 0.66 at `V₀ = −6`, so a third of the packet turns round at a drop in the ground.

### Why a hole reflects, and then stops

Any abrupt change in the potential reflects part of a wave, and the well has two edges. Inside it the wavenumber is `√(2(E − V₀))`, and when a whole number of half wavelengths fits across the full width `2w`, the reflections from the two edges cancel and everything goes through. For this energy and width that falls at `V₀ ≈ −2.8` and `≈ −9.0`, and `T` agrees: 0.97 near `−3` and 0.94 near `−9`, with the dip at `−6` between them. Deepening the well lets less through, then more.

### History

Carl Ramsauer and John Sealy Townsend found the same thing in 1921, independently, firing slow electrons through noble gases: at one particular electron energy the gas was almost transparent. Nothing in classical collision theory allows that. Treating the atom as a well of finite depth does, and the minimum is this same cancellation, found by moving the energy rather than the depth.

### Try

- Drag `V₀` to 3. The rectangle stands above the packet's energy now, and only about a twentieth of it tunnels through.
- Set `V₀` to `−3`, then `−6`, then `−9`, and watch `T`: 0.97, then 0.66, then 0.94 again.
- Set `Packet width` to 0.7. A narrower packet carries a wider spread of momenta, and the cancellation smears out: `T` is 0.86 at `V₀ = −3` against 0.64 at `−6`.
- With `V₀` at 3, raise `Momentum k₀` to 3. The energy is 4.6 and the wall is well below it, yet `T` is only 0.81: a wave is partly turned back by any sudden step, even one it clears.

### Read more

- [Ramsauer–Townsend effect](https://en.wikipedia.org/wiki/Ramsauer%E2%80%93Townsend_effect)
- [Finite potential well](https://en.wikipedia.org/wiki/Finite_potential_well)
- [Rectangular potential barrier](https://en.wikipedia.org/wiki/Rectangular_potential_barrier)
- [Wave packet](https://en.wikipedia.org/wiki/Wave_packet)