About this experiment
A ball rolling at a hill too high for its energy always rolls back. A quantum particle does not: its wavefunction leaks into the barrier, decays exponentially, and if the barrier is thin enough some of it comes out the other side. This lab solves the Schrödinger equation exactly for one or two rectangular barriers. Change the energy, height and width, swap the electron for a heavier muon or proton, and see why tunnelling matters for electrons in chips and microscopes but almost never for everyday objects.
Learning goals
- Write the solutions of the time-independent Schrödinger equation inside and outside a rectangular barrier
- Explain how matching ψ and dψ/dx at each edge fixes the transmission and reflection
- Show that the transmission falls roughly as e^(−2κa) for a thick, high barrier
- Explain why transmission depends so strongly on the particle mass
- Describe transmission resonances above a barrier and resonant tunnelling through two barriers
How to use it
- 1With the default electron, read T. Double the barrier width. By roughly what factor does T fall? Compare with the estimate e^(−2κa).
- 2Switch on the log scale and watch how log T changes as you widen the barrier. Why is it almost a straight line?
- 3Replace the electron with a muon, then a proton, keeping E, V₀ and a fixed. Explain the change using κ.
- 4Raise E above V₀. Find energies where T = 1 exactly. Show that at these energies the barrier width holds a whole number of half wavelengths.
- 5Choose two barriers and drag E slowly on the lower plot. Find the sharp peaks where T reaches 1 even though one barrier alone lets through only a few percent.
- 6At a resonance, look at |ψ|² between the barriers. What does the particle do there?
The model
Assumptions
- One dimension, a time-independent potential made of flat steps, and a steady beam of particles arriving from the left.
- The particle has a definite energy, so the wavefunction is a stationary state. A real particle is a wave packet that spreads over a range of energies.
- The animated curve is the real part of ψ e^(−iEt/ħ). The phase turns far faster in reality; it is slowed down so the motion of the crests can be seen.
- ψ and dψ/dx are continuous at every edge. The coefficients are found exactly with 2 × 2 transfer matrices, with no numerical integration.
- Very thick barriers for heavy particles give transmissions below 10^−300. Then only the estimate e^(−2κa) is shown.
Try this in class
- Q1An electron with 1 eV meets a 3 eV barrier. Find κ and the distance over which ψ falls by a factor e.
- Q2Use the thick-barrier formula to estimate T for the same electron and a barrier 1 nm wide.
- Q3Why is T not zero when E is a little above V₀ but exactly 1 at certain higher energies?
- Q4A scanning tunnelling microscope current falls by about a factor 10 when the tip moves 0.1 nm further away. Use e^(−2κd) to estimate κ.
- Q5Explain why alpha decay half-lives range from microseconds to billions of years for similar alpha particle energies.
Frequently asked questions
Does the particle lose energy inside the barrier?
No. The stationary state has one energy everywhere. Inside the barrier the kinetic energy term is formally negative, which is why the solution is a decaying exponential instead of a wave. A transmitted particle has the same energy as the incident one.
Where is tunnelling used?
In the scanning tunnelling microscope, flash memory (electrons tunnel through a thin oxide to charge the floating gate), tunnel diodes and resonant tunnelling diodes, Josephson junctions, alpha decay and nuclear fusion in the Sun, where protons tunnel through their electric repulsion.
Why does T reach 1 at sharp energies with two barriers?
The region between the barriers acts like a box with quasi-bound energy levels. When E matches one of them, the waves reflected back and forth interfere so that all reflected waves cancel, as in a Fabry-Perot interferometer for light. This is resonant tunnelling.