Quantum Tunneling
A particle with too little energy to climb the wall comes out the other side anyway — and nothing was borrowed to do it.
Before this: Superposition & Measurement
In this lesson you’ll learn
- State the classical turning-point argument and identify exactly which step quantum mechanics breaks
- Explain why the Schrödinger equation gives exponentials rather than oscillations when E < V
- Use T ≈ 16(E/V)(1 − E/V)e^{−2κw} to predict how transmission responds to a thicker barrier
- Recognise tunnelling at work in STMs, flash memory, alpha decay and stellar fusion
The classical impossibility
Roll a ball up a hill. Kinetic energy converts into potential energy as it climbs, and energy conservation holds the whole way: ½mv² + V(x) = E, where E is the total energy it started with and never changes. Solve for the speed and you get v(x) = √(2(E − V(x))/m).
Now let the hill be taller than the ball can manage. At the height where V(x) = E the speed is exactly zero. That point is called the turning point, and it is a wall in the most literal sense the theory allows: beyond it V(x) > E, so E − V(x) is negative, and v(x) would be the square root of a negative number. There is no such speed. The ball stops dead at the turning point, reverses, and rolls back. The region beyond is not merely hard to reach — it is classically forbidden, and the probability of ever finding the ball there is exactly zero, not approximately zero.
This is worth stating carefully, because it is the precise claim quantum mechanics is about to violate, and it has nothing to do with friction or with the ball being too slow. It is two lines of algebra applied to energy conservation: anything with a definite position and a definite momentum is barred from every region where V exceeds E.
Pose the same problem as a rectangular barrier — height V₀, width w, a particle of energy E < V₀ arriving from the left — and the classical answer is stark. Reflection: 100%. Transmission: 0%, for every width, for every height above E, forever. Hold onto that number, because the experiment gives something else.
Predict before you read on
Quantum mechanics replaces the trajectory with a wavefunction ψ(x), and inside the barrier — where E < V — the Schrödinger equation reads d²ψ/dx² = (2m/ħ²)(V − E)ψ, with a coefficient that is now positive. What do its solutions look like in there?
Commit to an answer — the next section is the payoff.