The scattering quantum mechanics returns unchanged

Where the quantum picture hands back the old one found that the classical answer usually returns from quantum mechanics by blurring: the quantum density oscillates at every quantum number, and the classical density is what is left when the oscillations are too fine for anything to resolve. The return is approximate, and it improves as the quantum numbers grow. This essay is about the one famous case where the return is not approximate at all — where a calculation made with classical orbits is exactly right, at every energy, in a regime where quantum mechanics ought to dominate — and about how a single change, making the two colliding particles identical, makes it wrong in a way that only quantum mechanics can repair.

Hyperbolas and a count #

Ernest Rutherford’s model of the atom rested on a calculation he made in 1911 to explain why a few alpha particles fired at gold foil bounced straight back. Treat the alpha particle as a point charge on a hyperbolic orbit around the nucleus, repelled by Coulomb’s force. An alpha particle aimed at distance $b$ from the nucleus, the impact parameter, is deflected through an angle $\theta$ given by

$$\tan\frac{\theta}{2} = \frac{d}{2b},$$

where $d = Z_1Z_2e^2/4\pi\varepsilon_0E$ is the closest approach in a head-on collision, the distance at which all the projectile’s kinetic energy $E$ has been turned into electrostatic energy.

A beam is a uniform rain of impact parameters, and counting how many particles land in each range of angles turns the deflection law into a cross-section — the effective target area per unit solid angle:

$$\frac{d\sigma}{d\Omega} = \left(\frac{d}{4}\right)^2\frac{1}{\sin^4(\theta/2)}.$$

The orbits also cast a shadow. Every path is turned aside before reaching a paraboloid with the nucleus at its focus, and nothing in the classical picture ever enters it; the tip of the shadow is the closest approach $d$. Hans Geiger and Ernest Marsden measured the angular distribution over four orders of magnitude in rate and found the $\sin^{-4}$ law exactly. Collisions are easier than forces followed how conservation laws alone turned that measurement into the discovery of the nucleus.

A calculation that had no right to work #

The trouble, looked at with hindsight, is that a five-mega-electronvolt alpha particle has a de Broglie wavelength of about six femtometres, comparable to the distance of closest approach to a gold nucleus, about forty-five. Everything has a wavelength made the point that a classical trajectory is a good description only when the wavelength is much shorter than every length the motion explores. Here the two are within a factor of a few, and for lighter nuclei or slower projectiles the wavelength is the longer of the two. A planetary-orbit calculation in that regime ought to be badly wrong.

When quantum mechanics arrived, the scattering of a wave by a Coulomb field was one of the first problems solved, by Gregor Wentzel and by J. Robert Oppenheimer in the lowest-order approximation in 1926 and 1927, and exactly by Walter Gordon and Nevill Mott in 1928. All of them got Rutherford’s formula. The exact quantum amplitude for scattering through $\theta$ is

$$f(\theta) = -\frac{\eta}{2k\sin^2(\theta/2)}\exp\!\left(-i\eta\ln\sin^2\frac{\theta}{2} + 2i\sigma_0\right),$$

where $k$ is the wavenumber and $\eta = Z_1Z_2e^2/4\pi\varepsilon_0\hbar v$, the Sommerfeld parameter, is half the closest approach measured in units of the wavelength over $2\pi$. The rate is the square of the magnitude. The magnitude contains $\eta/k = d/2$ — a purely classical length, with no $\hbar$ in it — and the angle; the whole of the quantum mechanics is in the phase, and the phase disappears from the square.

The reason it works is that the Coulomb potential has no length of its own. A cross-section has the dimensions of an area, and the only lengths in a collision between two point charges are the closest approach $d$, which is classical, and the wavelength, which is quantum. The cross-section could have depended on their ratio, $\eta$, in any way at all; for a general force it does. For a screened charge — a nucleus surrounded by its electron cloud, which gives the potential a range — the quantum cross-section departs from Rutherford’s at small angles, where the projectile passes far enough away to feel the screening, and the departure depends on the wavelength. For the pure inverse-square force, the dependence on $\eta$ happens to cancel out of the magnitude exactly.

“Happens to” hides a structure. The arrow that says which way the orbit points found that the inverse-square force has a conserved quantity no other central force has, the Runge–Lenz vector, which fixes the orientation of every orbit. In quantum mechanics the same hidden symmetry makes the hydrogen atom’s energy levels more degenerate than rotation alone requires and lets the Schrödinger equation for the Coulomb field be solved in parabolic coordinates, where scattering separates as neatly as the bound states do. The exact classical cross-section and the extra degeneracy of hydrogen are the same accident, seen in scattering and in binding.

Three calculations, one answer #

The lowest-order quantum approximation, named after Max Born, agrees as well, and for a reason that makes the absence of a length even plainer. In that approximation the scattering amplitude is the Fourier transform of the potential, evaluated at the momentum transferred to the particle, $q = 2k\sin(\theta/2)$. The Fourier transform of $1/r$ is $1/q^2$, with no other factor: an inverse-square law in momentum, as scale-free as the inverse-distance potential it came from. Its square is $1/q^4$, which is Rutherford’s $1/\sin^4(\theta/2)$ with the right coefficient. For any potential with a range, the transform has the range in it — a screened charge gives $1/(q^2 + a^{-2})$ — and the lowest-order answer differs both from the exact one and from the classical one. Coulomb’s force is the only one for which the classical count, the first quantum approximation and the exact quantum result coincide.

The coincidence also explains why Rutherford’s own experiment could not have told him the difference. For a five-mega-electronvolt alpha particle on gold, $\eta$ is about twenty, and the motion is classical by any standard; the cross-section would have been close to classical whatever the force. The exactness matters when $\eta$ is small — fast electrons on light nuclei, where the wavelength is far longer than the closest approach and no orbit could be a fair description of anything — and there the measured rate still follows Rutherford’s law, corrected only for the electron’s spin and its relativistic speed, until the electron comes close enough to see the size of the nucleus. The departures from that law are how the charge distributions of nuclei were measured, a method the length no experiment can resolve traced to its limit.

A cross-section that is classical and a wave that is not #

The agreement is in the cross-section, the rate per unit solid angle far from the nucleus. Close to the nucleus the wave and the orbits look quite different. In the first figure, every point outside the shadow lies on two classical paths, one with a smaller impact parameter that has been deflected more and one with a larger impact parameter deflected less, crossing there. A wave reaching the same point by both routes interferes with itself, and the quantum density near the nucleus is striped with fringes that no classical count contains. Along the shadow’s edge the two paths merge, the classical density piles up without limit, and the wave replaces the pile-up with a smooth peak and a set of decaying ripples running into the shadow, which the wave enters a little way where no orbit can. The edge of the shadow is a caustic, the same structure that makes the bright edge of a rainbow.

Far away, those near-field fringes leave no trace in the angular distribution, because each direction of travel at large distance corresponds to exactly one impact parameter: the deflection falls steadily as $b$ grows, and two paths never leave in the same direction. Interference between two classical routes needs two routes ending at the same place, and for a single Coulomb centre the only place where that happens is close in. The average that obeys Newton found the mean of a wavepacket following a classical path exactly for forces at most quadratic in position; Coulomb’s force is not one of them, and its packet does not follow an orbit in any detail. What survives exactly is a count — how much of the beam ends up in each direction — and not the motion.

When the two nuclei are the same #

Now fire alpha particles not at gold but at helium, whose nuclei are alpha particles too. Classically nothing new happens. In the centre-of-mass frame, if the projectile is deflected by $\theta$, the target recoils at $180° - \theta$, and a detector at angle $\theta$ counts both projectiles scattered by $\theta$ and targets knocked into that direction. The classical rate is the sum of two Rutherford terms, one for $\theta$ and one for $180° - \theta$.

Quantum mechanics does not let the two be told apart. An alpha particle arriving at the detector is not labelled as the one that came from the accelerator; the two outcomes — projectile at $\theta$ and target at $180° - \theta$, or the other way round — lead to the same final state, and the outcomes identical photons refuse found what that means: the amplitudes for the two routes add before anything is squared. For spinless nuclei, the sum is $f(\theta) + f(180° - \theta)$, and the square has a cross term.

The cross term carries the phases of the two amplitudes, and those phases contain $\hbar$ through $\eta$. They differ by $\eta\ln\tan^2(\theta/2)$, so the interference oscillates in angle, with fringes whose spacing shrinks as $\eta$ grows. At exactly 90° the two routes are mirror images, their phases are equal, and the amplitudes add in step: the rate there is exactly twice the classical one, whatever the energy. Mott worked out the formula in 1930 and James Chadwick confirmed it the same year with alpha particles scattered in helium at low energy; thirty years later, heavy ions of carbon on carbon and oxygen on oxygen, at energies low enough to stay below the nuclear force’s range, showed oscillating angular distributions that followed Mott’s formula fringe for fringe.

How the classical count comes back #

The Coulomb cross-section for distinguishable particles is classical at every $\eta$. For identical particles, the classical sum is recovered only in the limit, and the way it is recovered is the general mechanism of the correspondence principle in miniature.

The fringes never get weaker; they get closer together. Their number between any two angles grows in proportion to $\eta$, and $\eta$ is inversely proportional to $\hbar$ — making a collision more classical, by raising the charges or slowing the particles so the closest approach spans more wavelengths, packs more fringes into each degree. A detector has a finite angular width, and once several fringes fit inside it, it averages them to nothing and records the classical sum. A sharper detector needs a larger $\eta$ before it stops seeing the fringes. The classical answer is not what the particles do; it is what an instrument that cannot resolve the fringes reports — exactly the conclusion the particle in a box of where the quantum picture hands back the old one led to, here appearing in a collision with no box at all.

The same figure says something about the phase that cancelled out of the single-particle cross-section. It was never physically meaningless, only invisible to a measurement that squares one amplitude. Two amplitudes with different phases make it visible, and the Coulomb phase $\eta\ln\sin^2(\theta/2)$ — a logarithm, the trace of the force’s infinite range, which keeps distorting the wave however far it travels — is measured by the positions of the fringes.

Bosons and fermions at right angles #

The sign with which the two amplitudes add depends on what the nuclei are. For spinless nuclei, bosons like helium-4, carbon-12 and oxygen-16, the amplitudes add. For identical spin-½ particles, fermions like protons, the total state must change sign when the two are exchanged, and how that constrains the spatial part depends on the spins. A pair in the spin singlet is antisymmetric in spin and so symmetric in space, and its amplitudes add; a pair in one of the three triplet states is symmetric in spin and so antisymmetric in space, and its amplitudes subtract. In an unpolarised beam a quarter of the pairs are singlets and three quarters triplets, and the cross term is weighted by $\tfrac14 - \tfrac34 = -\tfrac12$.

At 90°, where the two routes are indistinguishable even in principle, the result is a pure statement about exchange: twice the classical rate for spinless bosons, half for unpolarised spin-½ fermions. For triplet pairs alone it would be zero — two identical fermions in the same spin state never emerge at right angles to each other in the centre-of-mass frame, which is the exclusion principle appearing as a gap in an angular distribution. The general rule for identical nuclei of spin $I$, in unpolarised beams, weights the cross term by $(-1)^{2I}/(2I+1)$: plus one for spin zero, minus a half for spin one-half, plus a third for spin one, and so on, shrinking as the spin grows because more of the pairs are in spin states that do not interfere at all. So the depth of the fringes at 90° measures the spin of the nucleus and their sign says whether it is a boson or a fermion, and before nuclear spins could be read from spectra, identical-particle scattering was one of the ways to read them. Carbon-12 on carbon-12 gives the full boson pattern and carbon-13, with spin one-half, the fermion pattern, inverted and at half the depth. Proton–proton scattering at low energy shows the fermion pattern, modified by the nuclear force once the protons come close enough to feel it, and the modification is how the strength of that force was first measured.

What the pictures leave out #

The figures treat point charges interacting only by Coulomb’s law. Real nuclei have a size and a short-range nuclear force, and once a collision brings them within a few femtometres of each other — at energies above the Coulomb barrier — the cross-section departs from both Rutherford’s and Mott’s, which is how nuclear radii were measured. Real targets are atoms, and their electrons screen the nucleus at small angles, as the second figure shows. The centre-of-mass angles of the identical-particle figures must be converted to laboratory angles for any real experiment, and for equal masses the laboratory angle is half the centre-of-mass one, so 90° in the figures is 45° in the laboratory.

What the pictures cannot show is the most interesting thing about the exact Coulomb result: that a wave scattered by a long-range force never becomes a free wave. Even far from the nucleus, the incoming and outgoing waves carry logarithmic phase distortions, and the very idea of an asymptotic plane wave, on which the usual theory of scattering rests, has to be modified for the Coulomb case. The cross-section is classical; the wave is thoroughly quantum. The domain of the drawings is point charges below the barrier, in the centre-of-mass frame, at Sommerfeld parameters from one to a few thousand.

Still open: how classical the classical limit of exchange is #

That identical-particle interference vanishes from any measurement of finite resolution as $\eta$ grows is not in doubt. What is less settled is what it means for large systems. Two identical molecules, or two identical clusters of atoms, scattering off each other have enormous $\eta$ and enormous numbers of internal states, and in practice they never interfere because their internal states almost never match; but the exchange symmetry is still there in principle, and experiments that collide carefully prepared identical molecules at very low energy are beginning to see its effects on how they react. Where in the growth of a system from nucleus to molecule to dust grain exchange stops mattering for any measurement that could be made, and whether the answer is a matter of resolution alone, is the question the classical limit has raised since the particle in a box, posed here for the one interaction where every other quantum correction cancels.

The calculation underneath is short. For a pure Coulomb force the quantum amplitude has Rutherford’s magnitude at every angle and energy, because the only length the problem offers is the closest approach and ħ enters only the phase, $e^{-i\eta \ln \sin^2(\theta/2)}$; identical nuclei let that phase interfere, doubling the rate at 90° for bosons and halving it for spin-½ fermions, with fringes whose number grows as η — and a detector ±2° wide returns the classical sum to 0.6 per cent by η = 3000 by being too coarse to see them. Rutherford was exactly right about one nucleus and exactly wrong about two.