Finite Wells and Quantum Tunnelling

Introduction

The infinite box is an elegant first model, but its perfectly impenetrable walls are idealisations. A real electron often sees a potential that rises by a finite amount. Its wavefunction can then extend into a region where classical mechanics would predict no motion. If that region has finite width, there can be a nonzero chance of detecting the electron beyond it. This is quantum tunnelling, an effect of the wave equation and its boundary conditions rather than a temporary violation of energy conservation.

Core explanation

For a particle of energy E in a region of constant potential V with V > E, the time-independent Schrödinger equation is −(ℏ²/2m)d²ψ/dx² + Vψ = Eψ. Rearrangement gives d²ψ/dx² = κ²ψ, where κ = √[2m(V − E)]/ℏ. Its solutions are exponentials, not the oscillating sine waves of a classically allowed region. For a bound state outside a finite well, the physically acceptable tail decreases away from the well, such as ψ ∝ exp(−κx) on the right. Because probability density is ψ ², it decays as exp(−2κx). The tail length scale is 1/κ.

The wavefunction and its first derivative must normally join continuously at a finite step in potential. An infinite wall instead forces a node at the boundary. Consequently, a finite well's bound-state energies differ from those of an infinite box, and a finite well of given depth and width supports only a limited number of bound states. A bound state has energy below the outside potential, but its probability density outside the nominal well is not zero.

Now place a barrier of finite width a between two classically allowed regions. A wave entering from the left produces a decaying component inside the barrier and generally a transmitted component on the right. For a sufficiently thick rectangular barrier, transmission has the approximate exponential dependence T ∝ exp(−2κa), although its exact prefactor depends on energy, barrier height and boundary matching. Increasing a or V − E suppresses transmission strongly; reducing mass also changes κ. During transmission through a static barrier, the particle's total energy remains E. The model describes probabilities from an incident wave, not a particle secretly acquiring extra energy to climb over.

Step-by-step reasoning

Sketch V against position and mark the particle's total energy E. Identify intervals where E > V as oscillatory and intervals where E < V as exponential. In each interval write the appropriate Schrödinger solution. Reject any bound-state solution that diverges at infinity, match ψ and its derivative at finite boundaries, and then interpret ψ ². For a barrier question, use κ and width a to compare relative transmission trends, distinguishing an approximation from an exact numerical probability.

Visual explanation

Draw a low central well bounded by two finite steps. A wave-like curve oscillates inside, crosses each boundary smoothly and tapers exponentially outside. On a second sketch, draw a raised rectangular barrier of width a. Show an incoming wave on the left, a short exponential region in the barrier and a smaller outgoing wave on the right.

Real-world analogy

A sound wave can become weak across a wall yet remain detectable on the far side; a thicker wall usually weakens it more. Tunnelling also concerns a wave amplitude through an obstacle. The analogy is limited because a quantum probability amplitude is not ordinary sound and the transmitted electron does not lose the missing energy as sound attenuates.

Real-world example

In a scanning tunnelling microscope, electrons tunnel across a small gap between a conductive tip and a surface. The tunnelling current responds strongly to the gap width, allowing the instrument to map surface structure at very small scales. The apparatus needs a suitable bias and electronic states; the elementary rectangular-barrier model explains the sensitivity without reproducing every detail of the measured current.

Why?

An exponentially decaying wavefunction cannot generally stop abruptly at a finite potential step while meeting the equation's boundary conditions. Its nonzero amplitude reaches into the forbidden region. A finite barrier allows matching to an outgoing wave beyond it, which is why the transmission probability can be nonzero. The word forbidden describes classical motion at E < V, not a rule that the quantum wavefunction must vanish there.

Common misconception

Tunnelling is sometimes described as an electron borrowing energy for a short time. That language suggests an energy change that is unnecessary and misleading for a stationary barrier. The incident and transmitted waves have the same total energy in this model. Another error is to say that a forbidden region has negative probability; probability density remains ψ ² ≥ 0, while the wavefunction changes its mathematical form.

Worked example

Suppose two barriers have the same V − E and electron mass, but barrier B is 0.20 nm wider than barrier A. If κ = 10 nm⁻¹, the thick-barrier approximation predicts T B/T A ≈ exp[−2κ(0.20 nm)] = exp(−4) ≈ 0.018. Thus B transmits roughly 1.8% as much as A under this approximation. The ratio uses the same prefactor for comparison; without the complete potential and boundary conditions it is not an exact absolute transmission probability.

Quick check

1. What happens to the decay length 1/κ if the barrier height rises while E stays fixed? Answer: V − E grows, so κ = √[2m(V − E)]/ℏ increases. Its inverse falls, and the wavefunction penetrates a shorter distance into the barrier.

Exam focus

Label E and V before calling a region allowed or forbidden. For E < V, use an exponential and square its magnitude to discuss probability. Quote exp(−2κa) as an approximate thick-barrier trend, not a universal equality. For finite wells, do not impose the zero-at-wall condition inherited from the infinite-box model.

Advanced insight

Barrier transmission depends on coherent matching of wave amplitude and slope at both interfaces. A resonant structure with more than one barrier can transmit strongly at selected energies even when a simple single-barrier exponential would suggest suppression. This is why the shape of the potential, not just its maximum height, is important in molecular and solid-state electronic transport.

Summary

Finite walls allow bound-state wavefunctions to leak into classically forbidden regions as exponential tails. For V > E, κ = √[2m(V − E)]/ℏ sets their decay scale. A finite-width barrier can transmit a particle with nonzero probability, commonly falling roughly as exp(−2κa) for a thick rectangular barrier. Energy remains conserved in a static potential.

Practice questions

1. A bound electron has a wavefunction tail ψ ∝ exp(−κx). How does its probability density vary with distance? Answer: Probability density is ψ ², so it varies as exp(−2κx). At a distance 1/κ the density has fallen by a factor e⁻² relative to its value at the boundary. 2. Barrier X and barrier Y have equal widths, but Y has a larger V − E. Which has the smaller tunnelling transmission in the elementary model? Answer: Y has a larger κ and hence a stronger exponential decrease across the same width. Its transmission is smaller, assuming otherwise comparable interfaces and no special resonance. 3. Does a particle detected on the far side of a static barrier necessarily emerge with less energy than it had before entering? Answer: No. The time-independent barrier does not supply or remove total energy. A transmitted component has the same energy E as the incoming stationary state; only the probabilities of reflection and transmission differ.