Module 03 ended with an exponential tail: in the classically forbidden region the wavefunction is not zero, merely decaying.
Now make the wall thin. The tail has not finished decaying by the time it reaches the far side, so an outgoing wave appears beyond the wall. The particle has gone through.
Tunnelling: a wave packet hits a wall it cannot climb
A classical ball with E < V₀ always bounces back. Watch the wave packet hit the barrier instead: part reflects, part decays exponentially inside the wall, and part genuinely turns up on the far side.
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Barrier
Incident wave packet
A wider σ means a sharper momentum, and a transmission closer to the plane-wave formula
E < V₀: classically forbidden
Evolution
The vertical scale is pinned to the initial packet, so the transmitted blob really is that small. Magnifying only helps you see its shape.
Transmission T(E)
The blue line is the current energy, the grey line the barrier height. Note that T does not equal 1 once E > V₀: it oscillates towards 1, because the waves reflected at the two faces interfere (resonant transmission).
- ψ (complex tube — the twist is the phase)
- |ψ|²
- barrier V(x)
- Classical particle
What to look for
- With the default settings E < V₀. Press play and keep your eyes on the right of the barrier: a small blob of probability density emerges from the wall and keeps travelling. Its area is the transmission T.
- Pause the moment the packet is pressed against the barrier and switch to the barrier close-up. Inside the wall |ψ|² is not zero but decays as e^{−2κx}. The wider the wall, the less gets through — that is where T ≈ e^{−2κa} comes from.
- Take the width from 1.6 to 3.2 — merely doubling it — and see how many orders of magnitude T loses. That exponential sensitivity is why a scanning tunnelling microscope can resolve a single atomic step.
- Push E just above V₀. Classically everything should pass, yet some of the packet still bounces back. "Quantum reflection" is the other face of wave behaviour.
- Switch on the classical particle: the ball rebounds along its own path and never appears on the right. Run the two side by side once and the difference speaks for itself.
✕ "The particle borrows energy from the wall and climbs over"
Nothing is borrowed. In a stationary state the energy is definite, and the wavefunction is simply non-zero in the forbidden region — a decaying real exponential, not a climb. Treating "borrowed energy" as a real mechanism makes calculations go wrong immediately.
✕ "Inside the barrier the particle has imaginary velocity"
There is no well-defined classical velocity inside the barrier. The wave number k = iκ is purely imaginary, which means decay, not propagation; talking about "velocity" there has already left the domain of the theory.
✕ "Transmission falls off linearly with width"
It falls exponentially. Double the width and T is squared. That distinction is the basis of every application of tunnelling — STM, α decay, stellar fusion.
Think it through
- A larger σ (sharper momentum) brings the numerical T closer to the plane-wave formula; a small σ pulls them apart. Why? (Hint: a packet is a superposition of many k, each with its own T.)
- Half-lives for α decay run from 10⁻⁷ seconds to 10¹⁷ years — more than thirty orders of magnitude — while the energies differ by only a few MeV. Explain that ferocious sensitivity with T ≈ e^{−2κa}.
- Is T exactly zero for an infinitely high barrier? Push V₀ to its maximum, then reconsider where the boundary condition of the infinite square well comes from.
This scene really is solving the equation
The earlier scenes had either closed-form solutions or eigenvalue problems. This one is different: it integrates the time-dependent Schrödinger equation
frame by frame in your browser, using the split-operator method: the evolution operator is split symmetrically,
with the kinetic part diagonal in momentum space, so an FFT there and back does the work. The scheme is exactly unitary: run it for ten thousand steps and the norm does not drift. That is what the “remaining norm” readout is watching (the absorbing layer at the edges does show up honestly as a decrease).
The algorithm is documented in appendix C.
Transmission: exponentially sensitive
For a plane wave on a rectangular barrier with the exact result is
When , and
Exponential in the width and in . That single fact underlies three apparently unrelated things:
| Phenomenon | Mechanism |
|---|---|
| An STM resolving single atoms | 0.1 nm of tip height changes the tunnelling current by an order of magnitude |
| α-decay half-lives spanning 30 orders of magnitude | A few MeV of energy, exponentially amplified |
| The Sun burning at all | Protons only get through the Coulomb barrier by tunnelling |
Above the barrier: quantum reflection and resonances
Push the energy above the barrier height. Classically everything passes, yet part of the packet still comes back — because an abrupt change in potential reflects waves, exactly as light partially reflects entering glass.
Keep raising the energy and does not climb monotonically to 1: it oscillates towards it, hitting exactly 1 whenever . That is destructive interference between the waves reflected at the two faces — resonant transmission (the Ramsauer–Townsend effect is its three-dimensional cousin).
The curve in the control panel draws all of this, with the current working point marked.
Why the packet's T disagrees with the plane-wave formulaadvanced
The above is for a plane wave at a single energy. The scene launches a Gaussian packet, which is a superposition of many :
Each component has its own , so the measured transmission is a weighted average
Since rises exponentially with energy for , the high-energy tail of the distribution is amplified out of proportion, and generally comes out above . Increase in the scene (sharper momentum) and the two converge. The discrepancy is quantitative and checkable — it is not numerical error.
Key formulas
Decay constant
The rate of decay in the forbidden region
Thin-barrier limit
Double the width and T is squared
Exact transmission
Rectangular barrier, E < V₀
Resonance condition
T = 1 above the barrier
Think it through
- Take the barrier width from 1.6 to 3.2 and read off how much falls. Predict the ratio from first, then compare.
- As the barrier height goes to infinity, . How does that connect to the infinite square well’s boundary condition that ψ vanishes at the wall?
- Reflection and transmission should sum to one, yet the readout often falls short. Look at the middle segment of the bar labelled “still near the barrier” and explain the deficit.