Part III — The physics

Confinement and tunnelling

Inside a uranium-238 nucleus, an alpha particle has about 4.3 MeV of energy. The electric repulsion that would hurl it away is walled off by a barrier some 28 MeV high, far more than it has. Yet on average, after about four and a half billion years, it leaves. How does a particle get out of a place it lacks the energy to leave?

This essay applies Schrödinger's equation to the two simplest shapes of potential: a box with walls, and a wall with something on the other side. The box shows where discrete energies come from, with nothing assumed beyond the equation and the requirement that the wavefunction make sense. The wall shows that a wave can leak through a region where a classical particle could never be. Together they explain the alpha particle's escape, and why a nucleus whose alpha particles carry about twice the energy decays some $10^{24}$ times faster.

Reading the equation locally

From the wavefunction essay, a state of definite energy $E$ has a shape $\varphi(x)$ obeying

$$-\frac{\hbar^2}{2m}\,\varphi'' = \bigl(E - V(x)\bigr)\,\varphi .$$

Everything in this essay follows from looking at the sign of $E - V$.

Where $E > V$, the region a classical particle could reach, the equation reads $\varphi'' = -k^2\varphi$ with $k = \sqrt{2m(E - V)}/\hbar$. The wavefunction oscillates, with wavelength $2\pi/k$: shorter where the kinetic energy $E - V$ is larger.
Where $E < V$, the region classically forbidden, it reads $\varphi'' = +\kappa^2\varphi$ with $\kappa = \sqrt{2m(V - E)}/\hbar$. The wavefunction does not oscillate; it grows or dies away exponentially, over a distance of about $1/\kappa$.

A particle in a box

Take the simplest confinement: a particle free to move between two walls a distance $L$ apart, with walls so high it can never be found beyond them. Inside, $V = 0$ and the wavefunction oscillates. At the walls it must fall to zero, since the particle is never found outside and $\varphi$ cannot jump. That is the condition on a guitar string, and it has the same consequence: only a whole number of half-wavelengths fits,

$$\varphi_n(x) = \sin\frac{n\pi x}{L},\qquad k_n = \frac{n\pi}{L},$$

for $n = 1, 2, 3, \dots$

Each allowed $k$ gives an allowed energy, $E = \hbar^2k^2/2m$:

$$E_n = \frac{n^2\pi^2\hbar^2}{2mL^2} = n^2 E_1 .$$

Nothing was put in by hand. The energies are discrete because the wave has to fit, which is the idea de Broglie's standing wave hinted at in the Bohr essay, now made exact. Two features are worth noticing. The levels spread out as $n^2$. And the lowest energy is not zero: $n = 0$ would give $\varphi = 0$ everywhere, no particle at all. A confined particle always moves, as the uncertainty essay said it must.

Left: the first three stationary states in a box, each drawn at the height of its energy (to scale, 1 : 4 : 9). Right: the exact stationary state for a wave meeting a barrier twice as high as its energy. Inside the barrier the wave dies away exponentially; a smaller wave, carrying 7 percent of the probability flow, emerges on the far side. Parameters in the script: $E = U_0/2$ and $\kappa a = 2$.

Worked example: an electron in a nanometre box

The energies. For an electron in a box $L = 1$ nm wide, with $\hbar = 1.055\times10^{-34}$ J s and $m = 9.109\times10^{-31}$ kg,

$$\begin{aligned} E_1 &= \frac{\pi^2(1.055\times10^{-34})^2}{2(9.109\times10^{-31})(10^{-9})^2}\ \text{J}\\[4pt] &= 6.02\times10^{-20}\ \text{J} = 0.376\ \text{eV}. \end{aligned}$$

The next levels are $4E_1 = 1.50$ eV and $9E_1 = 3.38$ eV. An electron dropping from $n = 2$ to $n = 1$ emits a photon of $3E_1 = 1.13$ eV, a wavelength of $1240/1.13 = 1100$ nm, in the near infrared.

Squeeze the box. Halve the width to 0.5 nm and every energy rises by four: $E_1 = 1.50$ eV, and the same jump now emits 4.51 eV, in the ultraviolet at 275 nm. Size sets colour. This is how semiconductor quantum dots, crystals a few nanometres across, are tuned: smaller dots glow bluer. (Their electrons move with an effective mass different from the free electron's, so these exact numbers do not carry over, but the $1/L^2$ scaling does.)

Check: the classical limit. The spacing between neighbouring levels is $(2n+1)E_1$, and $E_1$ falls as $1/mL^2$. For a box a centimetre wide the spacing is $10^{-14}$ times smaller than for our nanometre box, and for a heavy object smaller still. The levels merge into a continuum, and any energy looks allowed, as classically it is.

Optional: walls that are not infinitely high

Real walls have a finite height $U_0$. Then the wavefunction need not vanish at a wall. Inside the wall, where $E < U_0$, it dies away as $e^{-\kappa x}$, and the rule is that $\varphi$ and its slope must join smoothly at the edge.

Two things change. The wave now has a little more room, since its tails extend into the walls, so its wavelength is slightly longer and every energy slightly lower than for the infinite box. And there are only finitely many bound states: once $E$ reaches $U_0$ the particle is no longer held. Solving the matching conditions requires a short numerical calculation, but the picture is the one in the figure's left panel with the sine waves' ends leaking a little way into the walls.

A wave meeting a wall

Now turn the situation round. A particle with energy $E$ approaches a barrier of height $U_0 > E$ and width $a$. Classically it bounces back, every time. Quantum mechanically, the wave arriving at the barrier does not stop dead at its edge. It continues inside as a decaying exponential, $e^{-\kappa x}$. If the barrier is thin enough that something is left at the far edge, the wave carries on beyond it as an ordinary travelling wave, smaller than the one that arrived. The particle has some probability of being found on the far side: it has tunnelled.

The probability of getting through, the transmission coefficient $T$, is fixed mostly by how far the amplitude decays across the barrier. The amplitude falls by $e^{-\kappa a}$, the probability by its square, and so for a barrier more than about one decay length thick,

$$T \approx e^{-2\kappa a},\qquad \kappa = \frac{\sqrt{2m(U_0 - E)}}{\hbar},$$

up to a factor of order one in front that depends on $E/U_0$. The exponent is what matters. It makes tunnelling extraordinarily sensitive to the barrier: an electron facing a barrier 4 eV above its energy has $\kappa = 10.2$ nm⁻¹, and every extra 0.1 nm of barrier cuts $T$ by a factor of about 8. The scanning tunnelling microscope turns that sensitivity into images of single atoms, by measuring the electron current across the gap between a sharp tip and a surface.

Two checks. For a very thick or very high barrier, $\kappa a \to \infty$ and $T \to 0$: the classical answer, complete reflection. And for a particle with more energy than the barrier, classical physics says it always passes; the wave says it is sometimes reflected even then, about 19 percent of the time in one example the script computes. Neither behaviour has a classical counterpart; both follow from the same equation.

Optional: the exact transmission through a rectangular barrier

Write the wavefunction as $e^{ikx} + r\,e^{-ikx}$ to the left (arriving plus reflected), $A e^{\kappa x} + B e^{-\kappa x}$ inside, and $t\,e^{ikx}$ to the right (transmitted). Requiring $\varphi$ and $\varphi'$ to be continuous at both edges gives four equations for the four unknowns $r, A, B, t$. Solving them,

$$T = |t|^2 = \left[\,1 + \frac{U_0^2\,\sinh^2(\kappa a)}{4E\,(U_0 - E)}\,\right]^{-1}.$$

For $\kappa a \gg 1$, $\sinh\kappa a \approx \tfrac12 e^{\kappa a}$ and this becomes $T \approx 16\,\tfrac{E}{U_0}\bigl(1 - \tfrac{E}{U_0}\bigr)\,e^{-2\kappa a}$, the exponential with a prefactor between 0 and 4. For the figure's barrier, $E = U_0/2$ and $\kappa a = 2$, the exact value is $T = 0.071$, against $4e^{-4} = 0.073$ from the approximation. The reflected probability is $1 - T$: nothing is lost.

How an alpha particle escapes

In 1928 George Gamow, and independently Ronald Gurney and Edward Condon, applied this to radioactive decay. Picture an alpha particle, two protons and two neutrons, held inside a nucleus by the short-range nuclear force. Just outside the nuclear surface the nuclear force stops and electric repulsion takes over. The potential energy there forms a hill, highest at the nuclear surface and falling off as $1/r$ further out. For uranium the top of the hill is about 28 MeV. The alpha particle has 4.3 MeV, so classically it is trapped forever.

Quantum mechanically it rattles back and forth inside, striking the wall about $10^{21}$ times a second, and at each strike it has a tiny probability of tunnelling out. The barrier is not rectangular, so the exponent becomes a sum of $\kappa$ over its width, but the logic is the same. And the energy enters twice. A lower-energy alpha particle faces a higher barrier above it, so a larger $\kappa$, and a thicker barrier, because the Coulomb hill only falls to its energy further out: about 61 fm for uranium-238, against 26 fm for polonium-212, whose alpha decay releases 9.0 MeV. (These are the energies released in each decay, the ones the barrier calculation needs; the alpha particle itself leaves with about 2 percent less, the recoiling nucleus taking the rest.)

The result is dramatic. The tunnelling probability per strike is about $10^{-38}$ for uranium-238 and $10^{-14}$ for polonium-212. Multiplied by the rate of strikes, that gives half-lives of about $10^{17}$ seconds, some billions of years, and about $10^{-7}$ seconds. The measured values are 4.5 billion years and 0.3 microseconds: lifetimes $10^{24}$ times apart, reproduced by this simple model to within a factor of a few. Hans Geiger and John Nuttall had found in 1911 that alpha emitters with higher energies live shorter lives, following a steep empirical rule nobody could explain. Barrier penetration explained it seventeen years later.

It is worth seeing what that agreement rests on, because the model is crude. It assumes the alpha particle already exists inside the nucleus, ready to leave, when in reality it has to form from the protons and neutrons there, with a probability that can be well below one. It takes the barrier to begin at a nuclear radius estimated from a simple formula, and the answer is sensitive to that choice: making the uranium radius 10 percent smaller lengthens the predicted half-life about fifty-fold. And the strike rate, $10^{21}$ per second, is a classical estimate of a particle bouncing inside a box. Each of these could shift an answer by a factor of ten or more, so the near-exact match for uranium owes something to luck. What the model gets robustly right is the enormous spread and where it comes from.

Most of that spread is the energy. The two nuclei also differ, polonium's daughter having charge 82 against thorium's 90, and a lower charge means a lower barrier. Separating the two effects in the script: raising uranium's alpha energy alone from 4.3 to 9.0 MeV speeds the tunnelling by a factor of about $10^{22}$; lowering the charge from 90 to 82 adds a factor of a few hundred more.

Two questions

1. Widening the gap. In a tunnelling microscope the gap between tip and surface grows from 0.5 nm to 0.7 nm, with an effective barrier 4 eV above the electrons' energy. By roughly what factor does the tunnelling current change?

Worked answer

The current is proportional to $T \approx e^{-2\kappa a}$ with $\kappa = 10.2$ nm⁻¹. Widening by 0.2 nm multiplies $T$ by $e^{-2(10.2)(0.2)} = e^{-4.1} \approx 1/60$. The current falls about sixty-fold.

The easy slip is to reason proportionally: 40 percent more gap, 40 percent less current. In an exponential, equal additions to the width give equal factors, here about 8 for each 0.1 nm, and that is why the microscope can feel the height of single atoms.

2. Why not zero? A classical ball in a box can sit still at the bottom with zero kinetic energy. Why can't the quantum particle in a box have zero energy? What is wrong with taking $n = 0$?

Worked answer

The formula $\sin(n\pi x/L)$ with $n = 0$ gives zero everywhere. That is not a state of lowest energy; it is no particle at all, since $|\varphi|^2 = 0$ means there is nowhere to find it. The lowest real state is $n = 1$, with energy $E_1 > 0$.

The physical reason is the one from the uncertainty essay: a particle confined within a width $L$ must have a momentum spread of at least about $\hbar/L$, and a spread of momentum means kinetic energy. The tempting wrong turn is to treat $n$ as a count of something that can be zero, like a number of quanta. Here it counts half-wavelengths, and a wave needs at least one to exist in the box.

Where this leaves us

The alpha particle leaves because its wavefunction does not stop at the edge of the region it is classically allowed in. It dies away exponentially through the barrier, and what is left on the far side is a small but real probability of being outside. That probability depends exponentially on the barrier's height and thickness, which is why about twice the energy, with a slightly lower barrier, becomes a factor of $10^{24}$ in lifetime. The same equation, applied to a box, gives discrete energies with no rule added: waves that must fit can only have certain wavelengths.

A box has hard walls and levels that spread out as $n^2$. Most real systems are held more gently: an atom in a molecule, pulled back towards its place by a force that grows the further it strays. That is the harmonic oscillator, the most widely used model in physics, and its energy levels turn out to be evenly spaced, as Planck had assumed in 1900 without knowing why. It is the subject of the next essay.

Quantum Mechanics, Explained