The motion that cannot be stopped

Cool anything far enough and its motion stops. That is what temperature means, and every thermometer and every refrigerator is built on it. It is also, for a bound particle, false: helium cooled to a thousandth of a degree above absolute zero remains a liquid, flows, and would go on flowing if it were cooled further, because the motion left in it is not thermal and there is nothing to take away.

The argument, in four lines #

The uncertainty relation says a particle confined to a region of size $a$ has a momentum spread of at least $\hbar/2a$. Its kinetic energy is therefore at least $\hbar^2/8ma^2$, and in a spring of stiffness $k$ its potential energy is about $\tfrac12 k a^2$. So

$$E(a) \ge \frac{\hbar^2}{8ma^2} + \frac{1}{2}m\omega^2 a^2.$$

Differentiate and set to zero: $a^4 = \hbar^2/4m^2\omega^2$, so $a = \sqrt{\hbar/2m\omega}$, and substituting back gives

$$E_{\min} = \frac{\hbar\omega}{4} + \frac{\hbar\omega}{4} = \frac{\hbar\omega}{2}.$$

Two things about that deserve attention. The two terms come out equal at the minimum, each contributing a quarter of $\hbar\omega$ — which is the virial theorem for a quadratic potential, arriving without being invoked. And the answer $\hbar\omega/2$ is not an estimate: it is the exact ground-state energy of a quantum harmonic oscillator, obtained here with no wave equation, no boundary conditions and no special functions.

The oscillator’s levels are evenly spaced by $\hbar\omega$ and the lowest sits half a rung above the bottom of the well rather than on it. That half is the whole subject: it is not a small residual motion left over from imperfect cooling, it is the ground state, and there is no state below it to be reached by any means whatever — the microscopic reason absolute zero cannot be reached in a finite number of steps.

Why it is not a small correction #

For a mass on a laboratory spring, $\hbar\omega/2$ is around $10^{-33}$ joules and is beyond any conceivable measurement. For an atom in a solid it is not.

A copper atom in a copper lattice sits in a well with a vibration frequency around $5\times10^{12}$ per second, so its zero-point energy is about 3 millielectronvolts — small against the binding energy of half an electronvolt, but not negligible, and enough to expand the lattice measurably. The effect is directly visible in isotopes: heavier isotopes of the same element have lower zero-point energies, so they are more tightly bound, and lithium-7 melts at a slightly higher temperature than lithium-6 for no other reason.

Near a minimum every well is a parabola, whatever it is a well of — which is what makes the $\hbar\omega/2$ result general rather than a fact about springs. The frequency differs from case to case and the form does not, so a molecule’s bond, an atom in a lattice and a trapped ion all have a zero-point energy given by the same expression with different numbers in it.

The last clause is the one that matters for what follows. When the zero-point motion becomes comparable to the width of the well, the particle is no longer near the bottom of anything.

What was actually being denied #

The history here is short and turns on a single term in an equation.

Planck’s 1900 derivation of the blackbody law gave each oscillator an average energy of $h\nu/(e^{h\nu/kT}-1)$, which goes to zero as the temperature does. In his second theory of 1912 he obtained instead $h\nu/(e^{h\nu/kT}-1) + h\nu/2$, with a residual term that survives at absolute zero, and he was uncomfortable with it. Einstein and Stern used it in 1913 to fit the heat capacity of hydrogen; Einstein later withdrew the argument. The term became respectable only when Heisenberg’s mechanics produced it in 1925 as an unavoidable consequence rather than an adjustable addition.

What kept it doubtful for a decade is exactly what makes it hard to measure: the half-quantum is common to every level, so it cancels in every transition energy. Spectroscopy, which was the instrument of the period and was capable of extraordinary precision, is blind to it by construction. The evidence that settled it came from places where the absolute energy matters rather than a difference — the isotope shifts in molecular vibration frequencies, the lattice spacing of solid neon compared with its heavier isotopes, and above all the plain fact of liquid helium, which had been produced in 1908 and had refused to freeze ever since without anybody knowing why.

The substance that never freezes #

The condition can be made into a single dimensionless number. Compare the de Broglie wavelength an atom has when its kinetic energy equals the depth of the well it sits in, against the range of that well. That ratio is the de Boer parameter,

$$\Lambda^* = \frac{h}{\sigma\sqrt{m\varepsilon}},$$

with $\sigma$ and $\varepsilon$ the length and energy of the interatomic potential.

The reason is exactly the competition in the first figure. To build a crystal, each atom must be localised at a lattice site — squeezed into a region much smaller than the spacing — and the kinetic price of that squeezing is the zero-point energy. For argon the price is a few per cent of the binding energy and the crystal forms happily. For helium the price exceeds the binding energy: the lattice would cost more to hold together than it saves, so the atoms decline to be localised, and what remains is a liquid at absolute zero.

Applying pressure changes the arithmetic by making the well narrower and deeper relative to the spacing, which is why helium can be frozen by squeezing and by nothing else.

Below 2.17 K helium-4 becomes a superfluid rather than a solid, and the fraction in the ground state rises from nothing to nearly all. That is what the liquid does instead of freezing, and it happens because the zero-point motion is comparable with the binding — so the atoms cannot be localised enough to form a lattice, and the alternative they find is a condensate.

Helium-3, twice as quantum by this measure, is a fermion and cannot share the ground state; it becomes superfluid three orders of magnitude colder, by pairing, in a way that is a much closer analogue of superconductivity than of helium-4’s transition.

The same floor, in an atom #

The hydrogen atom is the other place where the argument does all the work.

A classical electron orbiting a proton would radiate and spiral in, reaching the nucleus in about sixteen picoseconds. What stops it is the same competition: localising the electron near the nucleus costs kinetic energy as $1/a^2$ while the Coulomb attraction only repays as $1/a$.

$$E(a) \approx \frac{\hbar^2}{2ma^2} - \frac{e^2}{4\pi\epsilon_0 a}$$

has a minimum at the Bohr radius and a value of −13.6 eV. Both numbers are exactly right, from the same two-line argument.

The ground state’s radial distribution peaks at the Bohr radius and has no weight at the origin at all. That is the same floor in an atom: the electron cannot sit on the nucleus, because localising it there would cost more kinetic energy than the Coulomb attraction returns — and the Bohr radius is precisely where those two costs balance.

That is the answer to why matter has volume. Not that electrons are objects of a certain size, and not that they repel: a hydrogen atom is a tenth of a nanometre across because that is the width at which the kinetic cost of confinement stops being worth the potential saving, and every atomic radius in the periodic table is the same calculation with more electrons in it. Run it on matter dense enough that the electrons share one box and the same term becomes a pressure with no temperature in it, holding a white dwarf up.

A thermal helium atom’s wavelength is comparable with the spacing between atoms in the liquid, and that comparison is the whole criterion. Where the wavelength is small compared with the spacing, the substance behaves classically; where it is not, the zero-point motion dominates — and helium is the only element for which it is not, at any pressure a laboratory reaches.

The uncertainty relation is not the whole reason #

It is worth separating two arguments that often get merged, because only one of them is exact.

The uncertainty relation gives a bound, and a bound is not an equality. Minimising an energy subject to $\Delta x \Delta p \ge \hbar/2$ gives a number that the true ground state must exceed — except when the true ground state happens to saturate the relation, which is the case only for a Gaussian wavefunction, which is the case only for a quadratic potential. That the harmonic oscillator’s answer comes out exactly right is a coincidence of the harmonic oscillator.

The general version is the variational principle: the expectation of the energy in any trial state is at least the ground-state energy, with equality only for the ground state itself. Choosing a family of trial states and minimising over the family gives an upper bound that improves as the family is widened. That argument is exact, it is what the first figure is really doing, and it works for any potential — including ones with no closed-form solution, where it is the practical method rather than a heuristic.

A wavefunction localised in space is a superposition of many wavenumbers, and momentum is Planck’s constant times wavenumber — so a narrow particle is necessarily a fast one. That is the uncertainty relation, and it is worth being clear that it is not the cause of the zero-point energy: it is a statement about Fourier transforms, and the energy follows from it only once a Hamiltonian is supplied.

The estimate, made honest #

Two things are worth checking about the four-line argument, because both are the kind of step that can be right by luck.

The first is the choice of $\Delta x = a$ and $\Delta p = \hbar/2a$. Writing the kinetic energy as $\langle p^2\rangle/2m$ and noting that $\langle p\rangle = 0$ for a bound state gives $\langle p^2 \rangle = (\Delta p)^2$ exactly, so the kinetic term is honest. The potential term is not quite: $\langle x^2 \rangle = (\Delta x)^2$ likewise, so $\langle V \rangle = \tfrac12 m\omega^2 (\Delta x)^2$ is exact too — for the harmonic oscillator. For any other potential the expectation of $V$ is not a function of $\Delta x$ alone, and the argument becomes a scaling estimate rather than a calculation.

The second is whether the minimum is a minimum. The kinetic term diverges as $a \to 0$ and the potential as $a \to \infty$, and both are convex, so the sum has exactly one stationary point and it is a minimum. That much survives for any potential rising at infinity, which is why the existence of a ground-state energy above the bottom of the well is far more robust than its value.

Applying the same method to a potential proportional to $|x|$ — a particle bouncing under gravity — gives $E \approx 1.36(\hbar^2 g^2 m)^{1/3}$ against the exact 1.86, which is the honest performance of the argument when it is not being helped by a Gaussian. Wrong by a quarter, right in every exponent.

Where the model stops #

The harmonic approximation. The parabola describes the bottom of a well and nothing else. In helium the zero-point motion is a third of the interatomic spacing, so the atom explores the anharmonic part of its well thoroughly and $\hbar\omega/2$ is a rough guide rather than a value. The de Boer parameter is used precisely because it does not assume the motion is small.

One particle. Everything here confines a single particle in a fixed well. In a solid the wells are made by the neighbours, which are themselves moving, so the correct treatment is a coupled problem in all the coordinates at once — and its zero-point energy is a sum over normal modes rather than a single $\hbar\omega/2$.

Statistics ignored. For identical particles the ground state is constrained by symmetry as well as by energy. Fermions cannot share a state, so the zero-point energy of a collection of them is enormously larger than the one-particle answer times the number of particles — which is what holds up a white dwarf, and is a zero-point energy in exactly this sense.

As a substance is cooled, each degree of freedom stops absorbing heat when $kT$ falls below its own quantum — so the heat capacity falls away and the third law is satisfied. That is the one thermal consequence worth naming, and it is why a zero-point energy is compatible with a vanishing heat capacity: the energy is there and nothing can be added to it or taken from it in small amounts.

What the pictures cannot show #

The first figure plots energy against a width, which is a parameter of a trial wavefunction rather than an observable. No experiment measures the width of a particle’s wavefunction directly; what is measured is a spread of results over many repetitions, and the identification of the two is a piece of interpretation the figure quietly assumes.

Nor does anything here show motion. “Zero-point motion” is a phrase, and the ground state is stationary: its probability distribution does not change with time, nothing oscillates, and a photograph of it at two moments would be identical. What is non-zero is the expectation of $p^2$, and there is no figure of a still thing having momentum.

And the de Boer chart puts seven substances on one axis as though they were a series. They are not: each is a separate measurement of two constants, and the smooth ordering is a consequence of the periodic table rather than of any underlying continuous variable.

The bond that is harder to break when it is heavier #

The lithium melting-point difference was mentioned in a clause. The same effect in chemistry is large, routinely exploited, and is the most direct evidence that the half-quantum is really there.

Two isotopes of an element have identical electron clouds, so the potential well holding a bond together is the same for both to a very good approximation. What differs is the mass in the oscillator, and since the frequency goes as one over the square root of the reduced mass, the zero-point energy differs.

A carbon–hydrogen stretch vibrates at about 3,000 reciprocal centimetres, which puts its zero-point energy near 18 kilojoules per mole. Substituting deuterium roughly doubles the reduced mass, lowers the frequency by a factor of about 1.36, and lowers the zero-point energy to about 13. The well is the same depth; the deuterated bond simply starts five kilojoules per mole further down in it.

So breaking a C–D bond costs five kilojoules per mole more than breaking a C–H one. If that bond-breaking is the slow step of a reaction, the rate carries an exponential of the barrier over $kT$, and five kilojoules per mole at room temperature is a factor of about seven and a half.

Measured primary kinetic isotope effects are six to eight. The agreement is close enough that the ratio is used as a diagnostic: a large effect on deuterating a position means that position’s bond is broken in the rate-determining step, and a small one means it is not. It is one of the standard ways of establishing a reaction mechanism, and the whole of it rests on a half-quantum of vibrational energy.

The same arithmetic explains why heavy water is toxic. Nothing about deuterium is chemically strange; every enzymatic step involving a proton transfer simply runs several times slower, and an organism whose body water is more than about half deuterated cannot keep its metabolism coordinated.

The effect is also visible directly in dissociation energies, where it has to be corrected for rather than exploited. The depth of the potential well of molecular hydrogen is 4.75 electronvolts; the energy actually required to pull the molecule apart from its lowest state is 4.48, and the difference of 0.27 is exactly the zero-point energy. For deuterium, whose zero-point energy is lower, the measured dissociation energy is 0.076 electronvolts larger — a molecule made of heavier atoms held by an identical bond, and measurably harder to break.

And the same difference decides a piece of low-temperature technology. Helium-3 has a larger zero-point energy in the liquid than helium-4, so it is less strongly bound and correspondingly more volatile: at one kelvin its vapour pressure is around a hundred times higher. That difference is what makes it possible to pump on liquid helium-3 to reach a few hundred millikelvin, and it is the basis of the circulation in a dilution refrigerator.

The amplitude that has been measured #

The section on what the pictures cannot show says that no experiment measures the width of a wavefunction. That is right about the wavefunction and understates what is available, because a particular moment of the distribution is measured routinely, and its value at absolute zero is the zero-point amplitude.

When X-rays or neutrons diffract from a crystal, the intensity of each Bragg reflection is reduced by a factor depending on how much the atoms are displaced from their ideal sites — the more they move, the more the scattered waves get out of step, and the reduction grows with the square of the scattering vector. That is the Debye–Waller factor, and inverting it gives the mean square displacement of the atoms directly.

Measure it as a function of temperature and extrapolate to zero. The mean square displacement does not go to zero. What is left is the zero-point motion, and it is a number in ångströms: about four hundredths of an ångström for carbon in diamond, and around nine hundredths for lead, whose atoms are heavier but whose lattice is far softer.

For solid helium — which exists only under pressure, for the reason this essay is about — the root-mean-square zero-point displacement is a quarter to a third of the distance between neighbouring atoms. An atom in that crystal is smeared over a substantial fraction of the space between its neighbours, which is why it is called a quantum solid and why almost none of the ordinary theory of crystals applies to it.

There is a second technique that measures the momentum side instead. Scattering neutrons at high energy from a light nucleus is a Compton-like collision, and the width of the recoil peak reports the momentum distribution of the nucleus before it was struck. Applied to a proton in ice it returns a zero-point kinetic energy of around 150 millielectronvolts, directly.

So the honest position is narrower than the essay’s caution. The wavefunction is not measured; the mean square position and the mean square momentum are, in separate experiments, and both are nonzero at absolute zero by amounts the argument on this page predicts.

What it is not #

Two things travel under the same name and are worth separating from this one.

The first is the zero-point energy of a field, which is the sum of $\hbar\omega/2$ over every mode of the electromagnetic field in a region. That sum diverges, and the divergence is real rather than a slip: there are infinitely many modes and each contributes. What is finite, and measurable, is the difference between the sums for two arrangements — two conducting plates admit fewer modes between them than empty space does, and the difference produces the Casimir attraction, measured to a few per cent. The energy in that calculation is not the energy of any particle in any well, and using the same phrase for both invites the conclusion that one has been measured because the other has.

The second is the claim that this energy can be extracted. It cannot, and the reason is definitional rather than technological: the zero-point energy is the energy of the ground state, and extracting energy means leaving the system in a lower state, of which there is none. A device that repeatedly took energy out of the ground state of anything would be a perpetual motion machine of the first kind. What Casimir measurements extract is work done by a change of configuration — moving the plates — and the work is bounded by the change in the difference, which is finite and small and available only once per approach.

The honest statement is that zero-point energy is unavailable in exactly the way the ground state of any system is unavailable, and interesting for exactly the reason a ground state is interesting: it sets the floor everything else is measured from.

Where the ladder goes next #

The uncertainty relation began as a statement about what can be known at once and has become a statement about what has to be paid. The next rungs are the ones this essay leaned on without deriving: the variational principle proper, which turns the argument into a method; the zero-point energy of a field rather than a particle, where summing $\hbar\omega/2$ over every mode gives an infinite answer whose differences are nevertheless measurable as the Casimir force; and the pressure a gas of fermions exerts at absolute zero.

The habit worth carrying away is the shape of the argument rather than its answer. Two terms, one falling and one rising, with a minimum in between, and the answer read off the minimum — it settles the size of an atom, the energy of an oscillator, the stability of a lattice and the radius of a white dwarf, and in each case the physics is entirely in the two exponents. When a quantity is set by a competition, what decides the answer is which power of the variable each side carries, and the numbers in front rarely matter at all.