The Finite Well and Quantum Tunnelling

In the infinite square well we built a comfortable little fiction: walls so high that the particle is trapped perfectly, the wavefunction slammed to exactly zero at each edge, and an endless ladder of allowed energies climbing forever. It is a beautiful first model — but no real box has infinitely high walls. Drop the walls to a finite height V_0 and two strange, deeply physical things happen at once. First, the ladder of bound states becomes finite: a shallow well holds only a handful of trapped states, and a shallow enough one holds just a single state. Second — and this is the astonishing part — the wavefunction refuses to vanish at the wall. It seeps a little way into the wall, into a region where a classical particle simply could not be, because it does not have enough energy to be there.

That leakage has a name and a spectacular consequence. Where the wave leaks into a forbidden region, it decays exponentially rather than stopping dead; and if the forbidden region is a barrier of finite width — a wall with another allowed region on the far side — a sliver of the wave makes it all the way through and comes out the other side. The particle can appear on a side it could never classically reach. This is quantum tunnelling, and it is not a laboratory curiosity: it is why the Sun shines, why radioactive nuclei decay, and how we photograph individual atoms. This page is about one idea — a particle can be found where classical physics forbids it, because its wavefunction decays rather than dies at a finite wall — and everything that follows from it.

A finite well holds only a few states

Put a particle of mass m in a well of depth V_0 and width L. Inside, where V = 0, the wavefunction still oscillates like a sine, exactly as in the infinite well. But now look at what happens outside, where the potential is V_0 and the particle's energy E is less than V_0. The time-independent Schrödinger equation there reads

-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V_0\,\psi = E\,\psi \quad\Longrightarrow\quad \frac{d^2\psi}{dx^2} = \frac{2m(V_0 - E)}{\hbar^2}\,\psi = \kappa^2\,\psi.

Because V_0 - E > 0, the right-hand side has the same sign as \psi — and a function equal to a positive constant times itself is not a sine but an exponential. The physically sensible solution (the one that does not blow up at infinity) is a decaying exponential:

\psi(x) \propto e^{-\kappa x}, \qquad \kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar}.

So the wave does not stop at the wall; it bends over and dies away smoothly. To knit the inside sine onto the outside exponential, both \psi and its slope \psi' must match at each wall. That matching condition can only be satisfied at certain energies — and, crucially, for a finite well there are only finitely many of them. Every finite well, however shallow, holds at least one bound state; a deeper or wider well holds more. Contrast that with the infinite well's endless ladder: finiteness of the walls buys finiteness of the states.

The penetration depth: how far the wave leaks

The decay constant \kappa is the whole story of the leak. Its reciprocal,

\delta = \frac{1}{\kappa} = \frac{\hbar}{\sqrt{2m(V_0 - E)}},

is the penetration depth: the distance over which the wavefunction falls to 1/e of its value at the wall. Read the formula and it tells you exactly what you would hope. The deeper the particle is buried below the top of the wall (the larger V_0 - E), the larger \kappa, the smaller \delta — the wave dies away fast and barely penetrates. A particle only just below the top of the wall (V_0 - E small) has a small \kappa and leaks a long way in. A heavier particle (larger m) also penetrates less. The chart below lets you feel this: slide the barrier's height and width and watch the exponential tail steepen or stretch.

Notice what the transmitted wave on the right does not do: it does not vanish. It is small — often fantastically small — but it is there. A particle that hits the barrier has a genuine, nonzero chance of being found on the far side. That chance is what we call the transmission probability, and it is the engine of tunnelling.

Tunnelling: through the wall, not over it

Take the leak to its logical end. Instead of a single wall, make a barrier of finite width a — a region of height V_0 with allowed space on both sides, and a particle of energy E < V_0 coming in from the left. Classically it must bounce off; it hasn't the energy to be inside the barrier, let alone cross it. Quantum-mechanically, the incoming wave decays across the barrier as e^{-\kappa x}, and if the barrier is not too wide there is still some amplitude left when it reaches the far edge. That surviving amplitude launches a small transmitted wave beyond. The particle has tunnelled.

For a barrier that is reasonably "thick" (\kappa a \gg 1) the transmission probability — the fraction of particles that get through — takes a beautifully simple approximate form:

That factor of two in the exponent is just the two-way trip of the probability (|\psi|^2 carries the square of the amplitude, and amplitude decays by e^{-\kappa a} across the width). The headline is the exponential: double the barrier width and you don't halve the transmission, you square the smallness. It is precisely this savage sensitivity that makes tunnelling both rare enough to keep atoms stable and controllable enough to build instruments on.

Where tunnelling runs the world

Because T is so violently sensitive to width and height, tunnelling is either negligible or decisive — and where it is decisive it tends to be running something enormous.

Worked examples

Example 1 — the decay constant and penetration depth. An electron (m = 9.11\times10^{-31}\ \text{kg}) sits V_0 - E = 4.0\ \text{eV} = 6.4\times10^{-19}\ \text{J} below the top of a wall. Then

\kappa = \frac{\sqrt{2m(V_0-E)}}{\hbar} = \frac{\sqrt{2(9.11\times10^{-31})(6.4\times10^{-19})}}{1.05\times10^{-34}} \approx 1.0\times10^{10}\ \text{m}^{-1}.

The penetration depth is \delta = 1/\kappa \approx 1.0\times10^{-10}\ \text{m} — about one atomic diameter. That is exactly the scale on which an STM works: shift the tip by an ångström and the tunnelling current changes by a factor of several.

Example 2 — widen the barrier, watch T collapse. Keep \kappa = 1.0\times10^{10}\ \text{m}^{-1} and compare two barrier widths, a_1 = 0.10\ \text{nm} and a_2 = 0.20\ \text{nm}. The exponents are

2\kappa a_1 = 2(10^{10})(1.0\times10^{-10}) = 2.0, \qquad 2\kappa a_2 = 2(10^{10})(2.0\times10^{-10}) = 4.0.

So T_1 \approx e^{-2.0} \approx 0.14 but T_2 \approx e^{-4.0} \approx 0.018. Doubling the width dropped the transmission by nearly a factor of eight, from about 14\% to under 2\% — and this is only a modest, "thin" barrier. For a thick barrier the same doubling can turn one-in-a-thousand into one-in-a-million.

Example 3 — how many bound states? A useful shorthand for a finite well of width L and depth V_0 defines the dimensionless strength

z_0 = \frac{L}{2\hbar}\sqrt{2mV_0}, \qquad N = \left\lfloor \frac{z_0}{\pi/2} \right\rfloor + 1,

where N is the number of bound states. If z_0 = 5, then z_0/(\pi/2) \approx 3.18, so N = \lfloor 3.18 \rfloor + 1 = 4: this well traps four states, no more. Make the well shallower until z_0 < \pi/2 and only the single ground state survives — but it never disappears entirely. A finite well always keeps at least one.

The classic trap. It is tempting to imagine the tunnelling particle briefly "borrowing" energy to leap over the top of the wall, or somehow speeding up to smash through and emerging drained on the far side. Both pictures are wrong. The particle that tunnels comes out with exactly the same energy E it went in with — energy is conserved throughout, and at no point does the particle have more than E. It never goes over the barrier; it goes through the classically forbidden region, where its wavefunction is a decaying exponential rather than an oscillation.

The right way to hold it: inside the barrier the particle does not have a well-defined "kinetic energy that has gone negative". It has a wavefunction that leaks. Ask "where is it?" and there is a small but real probability the answer is "on the far side". Nothing was borrowed, nothing was repaid, and no energy-time uncertainty loan is being called in — the transmitted particle simply has the same E as the incident one. Tunnelling is a statement about the shape of \psi, not about a particle secretly acquiring the energy to jump.

The whole game is in \kappa a, and it hides an enormous mass in the denominator's cousin. For an everyday object m is colossal, so \kappa = \sqrt{2m(V_0-E)}/\hbar is astronomically large and the exponent 2\kappa a is a number with dozens of digits. The transmission e^{-2\kappa a} is then so unthinkably tiny that a football thrown at a wall every second since the Big Bang would not yet have tunnelled through once. For an electron, with its featherweight mass and a barrier only a nanometre wide, 2\kappa a is a number like 2 or 20 — small enough that tunnelling happens constantly. Same equation, wildly different worlds, and the switch between them is just how big the mass and the width are.