Quantum Tunneling LabBeta

Watch a wave packet reflect and tunnel through barriers in real time

This tool is a beta version still being stabilized. Please let us know if a result looks off.

Throw a ball at a wall and it always bounces back. But a tiny particle like an electron is a wave, so part of it leaks through a wall even when its energy is lower than the wall. That's quantum tunneling.

🏁 Goal Keep the energy below the wall and make the transmission 45–55% — build a quantum coin toss.

Setting up the quantum lab…

Press the canvas to fire from there · drag the white handles to change the barrier's height, width and position

How to read the picture
  • Height of the glowing curve = probability |ψ|² of finding the particle there. Color = the wave's phase (a rainbow sweep means it's moving forward).
  • Gray landscape = potential V(x). Dashed line = particle energy E; the pale band is the packet's energy spread.
  • Red hatching = where a classical particle can never go for lack of energy. The wave seeps in, decaying exponentially.
  • Boxes at both ends = detectors. Arriving probability piles up there; a measurement makes just one of them click.
  • Units are natural units with ħ = m = 1. Shapes and ratios match the real phenomena; the numbers are illustrative.

About Quantum Tunneling Lab

This tool solves the one-dimensional time-dependent Schrödinger equation in units where ħ = m = 1. A box of length 260 is split into 2,600 grid points (spacing about 0.1), and each step of Δt = 0.04 advances the wave function with the Crank–Nicolson method. In these units a particle with momentum k has energy E = k²/2 and its packet moves at speed k, so the default E = 2 means k = 2.

The last 20 units at each end of the box carry a complex absorbing potential, so waves that reach the edge fade out instead of bouncing back. The tool tallies probability absorbed at the left edge as the reflection detector and probability absorbed at the right edge as the transmission detector. The R and T on screen are therefore the probabilities that actually left on each side once the packet has finished interacting with the barrier.

Why Crank–Nicolson

Push ψ forward with the simplest scheme, forward Euler, and the norm of the wave function grows a little every step until the simulation blows up. Apply the method John Crank and Phyllis Nicolson published in 1947 for the heat equation, and the update becomes (1 + iHΔt/2)ψ(t+Δt) = (1 − iHΔt/2)ψ(t). When H is Hermitian this map is unitary: total probability is conserved to rounding error, and larger time steps lose accuracy but never diverge.

The price is a linear system to solve every step. In one dimension H is tridiagonal, so the Thomas algorithm solves it in time proportional to the number of grid points. The tool factorizes the matrix only when the potential changes and then does one forward and one backward sweep per step. Ever since Goldberg, Schey and Schwartz produced computer-generated films of Gaussian packets splitting at barriers in 1967 for teaching, simulations like this have been a standard part of quantum mechanics courses.

The accuracy limits can be put in numbers. At grid spacing 0.1, a k = 2 wave has a lattice kinetic energy about 0.3% below the continuum value, and the phase error per time step is around 0.05%. Even at the top of the energy slider, E = 5 (k ≈ 3.2), the grid error stays under 1%.

The transmission formula for a rectangular barrier

For a plane wave of energy E < V hitting a barrier of height V and width a, the transmission probability is T = 1 / [1 + V² sinh²(κa) / (4E(V − E))], with κ = √(2(V − E)) when ħ = m = 1. For a thick barrier the sinh grows like an exponential and T ≈ 16E(V − E)/V² · e^(−2κa). The essence of tunneling is right there: doubling the width does not halve T, it multiplies it by another factor of e^(−2κa).

Above the barrier, E > V, T is still not 1. The sinh turns into a sine, T = 1 / [1 + V² sin²(qa) / (4E(E − V))] with q = √(2(E − V)), and part of the wave reflects even though a classical particle would sail over. The step scenario is built around exactly this reflection.

The theory curve on screen is not the bare plane-wave formula. A packet of width σ carries a spread of momenta of roughly σk = 1/(2σ), so the tool computes T(k) with transfer matrices and averages it over that momentum distribution. Narrow the packet and its energy spread widens, which blunts sharp features such as resonance peaks.

The physics behind the scenarios

The alpha-decay scenario is the problem George Gamow solved in 1928, with Gurney and Condon reaching the same idea independently at about the same time. An alpha particle escapes the nucleus even though its energy is below the Coulomb barrier, and a small change in energy changes the barrier thickness so much that half-lives span dozens of orders of magnitude. In the tool you can see it as T swinging wildly on the log scale when you nudge the E slider.

The STM scenario is the principle of the scanning tunneling microscope that Binnig and Rohrer built in 1982. For an electron crossing a vacuum gap between metals with a work function around 4 eV, κ is about 1.0 Å⁻¹, so widening the gap by 1 Å multiplies T by e^(−2) ≈ 0.13, roughly one order of magnitude. That is why a one-atom step shows up so clearly in the current. The resonant tunneling diode scenario follows the double-barrier resonance observed by Chang, Esaki and Tsu in 1974: when the particle energy matches a quasi-bound level in the well between two barriers, total transmission approaches 1 even though each barrier on its own transmits very little.

Worked examples

Working out the basic scenario by hand

Input
E = 2, V = 2.4, width a = 1.6 (ħ = m = 1)
Output
κ = √0.8 ≈ 0.894, κa ≈ 1.431, sinh(κa) ≈ 1.972, so T = 1 / (1 + 5.76 × 3.889 / 3.2) ≈ 0.125

Double the width to 3.2 and T ≈ 0.0073, about 17 times smaller. Halve it to 0.8 and T ≈ 0.48. The goal of the basic scenario (T between 45% and 55% with E < V) is reached by narrowing the barrier to around 0.8. The on-screen value differs slightly from this plane-wave number because the packet has a spread of momenta.

Reflection from a barrier you have enough energy to clear

Input
E = 3, V = 2.4, a = 1.6
Output
q = √1.2 ≈ 1.095, sin(qa) ≈ 0.983, so T = 1 / (1 + 5.76 × 0.966 / 7.2) ≈ 0.56

A classical ball would clear the barrier every time, but 44% of the wave comes back. At energies where qa is a whole multiple of π the sine vanishes and T = 1, a transmission resonance.

How to use it

  1. Pick a scenario: basic barrier, alpha decay, STM, flash memory, resonant tunneling diode, step, or free experiment.
  2. Set particle energy, barrier height and width, and packet width with the sliders or the handles on the graph.
  3. Fire the packet and compare the reflection and transmission detector readings with the theory curve.
  4. Use the measurement button to repeat the experiment many times and watch each particle land in only one detector.

Frequently asked questions

Does the particle have negative kinetic energy inside the barrier?

Inside the barrier the wave function decays exponentially instead of oscillating. But if you measure position and find the particle inside, that measurement disturbs its momentum so much that its energy is no longer pinned at E. You never observe a particle with negative kinetic energy.

Why is transmission a fraction instead of all or nothing?

The wave function is a probability amplitude. On any single shot the particle is caught by one detector only; repeat the shot many times and the fraction caught on the right approaches T. The measurement statistics in the tool show exactly that.

If ħ = m = 1, what are the real-world sizes?

Lengths and energies are in arbitrary units. To map them onto an electron, pick a length unit and set the energy unit to ħ²/(m·length²). With 1 Å as the length unit, the energy unit is about 7.6 eV.

Do the absorbing edges affect the result?

The absorbing potential switches on smoothly with depth (W ∝ s²) to keep reflections small. The tool locks in its result once T and R stop changing, and at that moment the wave that got past the barrier has not reached the right-hand absorber yet. Running several step and barrier cases at E = 0.2, the locked-in R and T matched the analytic values to within 0.2 percentage points. When the transmitted wave is very slow, as just above a step (wavenumber 0.45 or less), the right absorber does send part of it back, so if the simulation kept running after the verdict R would creep up by a few points (about 3.7 points at E = 0.25 over a 0.2 step). The verdict and the repeated shots use the locked-in values, so that late echo never enters them.

Sources

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