Double Well
V(x) = (x² − 4)²/8
Open in the app The dials and keys named below are the app's.
What you see
The formula is the potential, V = (x² − 4)²/8. It is zero at x = ±2 and rises to V(0) = 2 between them: two wells with a hill in the middle, the classic shape for a thing that can be in one of two places. With the walls at ±6, where V reaches 128, the wells are drawn almost flat. Pull Span of x in and they show.
The packet starts in the left well, at Packet centre −2, at rest, with Packet width 0.5.
Why it leaks
Near the bottom of a well the potential is a parabola. At x = −2 it is V ≈ 2(x + 2)², which is ½ω²u² with ω = 2. That well's own ground state has width 1/√(2ω) = 0.5, exactly the width the packet is given. So it settles in almost at rest and only breathes a little, out to about 1.0 and back, because the true well is not quite parabolic.
Its energy is the readout's 1/4σ² = 1 of motion plus about 0.52 of potential where it sits, so roughly 1.52. That is below the 2 at the top of the hill, and it should stay where it is forever. Instead T creeps up: 0.07 by t = 3, 0.10 by 12, 0.21 by 20 at Speed 1. The particle is tunnelling through the middle, and given long enough the two wells share it.
Where it turns up
Ammonia is built like this. The nitrogen atom of NH₃ sits above or below the plane of its three hydrogens, two wells with a barrier between, and it tunnels through at 23.8 GHz. That line is what the first maser amplified, built by Gordon, Zeiger and Townes in 1954.
Try
- Lower the hill, (x² − 4)²/16: the top drops to 1, no longer above the packet's energy, so it spills over instead of tunnelling and T reaches 0.83 by t = 20.
- Push the wells apart, (x² − 9)²/8, with Packet centre −3: the barrier is now 10 tall and six wide, and almost nothing crosses.
- Start on top of the hill: Packet centre 0. It splits evenly and stays even, half in each well.
- Pull Span of x in to 4 to see the shape of the two wells.