The sharpness a variance cannot see

Sharpness has to be paid for found the uncertainty principle in the mathematics of waves: a wave packet squeezed into a small region of space must contain a wide spread of wavelengths, so a particle described by one cannot have both a sharp position and a sharp momentum. The statement everyone learns is Heisenberg’s inequality, $\sigma_x\sigma_p \geq \hbar/2$, in which $\sigma$ is a standard deviation. It is a fine inequality, and a Gaussian wave packet meets it with equality. But it uses one particular way of measuring spread, and that way of measuring spread has a blind spot large enough for a whole family of states to sit in.

The blind spot is easy to see. A standard deviation measures how far a distribution’s values lie from its mean, on average, squared. That is the natural measure for a single hump. For a distribution with two narrow humps far apart it is a very poor measure of how much is known: the mean lies in the empty space between them, every value is far from the mean, and the standard deviation is about half the separation, however narrow the humps. A particle that is certainly in one of two small boxes a metre apart has a position spread, in Heisenberg’s sense, of half a metre.

Two places at once #

The first figure draws the state. It is a superposition of two narrow wave packets, each well localised, separated by many times their own width. Its position distribution is two sharp peaks. Its momentum distribution is the momentum distribution of one packet — a broad Gaussian envelope, broad because each packet is narrow — modulated by fringes: the two packets interfere in momentum space exactly as the two slits of where the interference goes interfere on a screen, and the fringes are spaced inversely to the separation.

Heisenberg’s product for this state is large. The position spread is dominated by the separation, and the momentum spread by the envelope, which is set by the width of each packet; their product is roughly the separation divided by the width, eight times the minimum for the state drawn, and it grows without limit as the packets are pulled apart. On Heisenberg’s measure the state is hopelessly uncertain. Yet nobody would describe it that way. A measurement of position finds the particle in one of two narrow places. A measurement of momentum finds it within one of a comb of narrow fringes. Both distributions are sharp; they are just not single humps.

Measuring spread by information #

The measure that sees this is the one information theory uses. The Shannon entropy of a distribution, $H = -\int \rho\ln\rho\,dx$, measures how many bits — or here, natural units, nats — it takes on average to say where a sample from the distribution fell, to a given precision. Spreading a distribution evenly over twice the range adds $\ln 2$ to its entropy, one bit. But splitting it into two narrow pieces of the same total width as one also adds only $\ln 2$, however far apart the pieces are: knowing which piece costs exactly one bit, and the empty space between them costs nothing. Entropy is a count found the same quantity in statistical mechanics, the logarithm of the number of ways a system can be arranged; here it is the logarithm of the effective range a distribution occupies.

In 1957 Isidore Hirschman conjectured, and in 1975 Iwo Białynicki-Birula and Jerzy Mycielski, and independently William Beckner, proved, that position and momentum entropies obey

$$H_x + H_p \geq \ln(\pi e\hbar),$$

with equality only for Gaussian packets. It is an uncertainty relation of the same kind as Heisenberg’s: no state can make both distributions sharp. But it measures sharpness by information rather than by spread from the mean.

What each measure says about the split state #

The second figure follows both measures as the two packets are pulled apart. Heisenberg’s product rises steadily, in proportion to the separation. The entropic sum rises a little and stops. The position entropy grows by $\ln 2$ as the peaks separate and stops growing, because once they no longer overlap, making them further apart adds no further uncertainty about where the particle is. The momentum entropy actually falls, by $1 - \ln 2$, because the interference fringes concentrate the momentum distribution into narrow bands within its envelope. The total levels off at $2\ln 2 - 1$, about 0.39 nats above the floor, at any separation.

That is a much truer description of the state. It is not a blurred particle with a spread of metres. It is two sharp alternatives, and the price of not knowing which is a fraction of a coin toss in the entropic account. The variance-based account cannot express that, because a standard deviation has no way to distinguish uncertainty about which of two places from uncertainty spread evenly between them.

The split state is not exotic. Schrödinger’s cat is the famous instance; the states of a matter-wave interferometer, in which an atom travels two paths centimetres apart, are real ones; and the probability that goes below zero found that their quasi-probability distributions carry fringes of negative values between the two peaks, the signature of the coherence between them. Heisenberg’s inequality is true of all of them and informative about none.

The entropic relation contains Heisenberg’s #

The entropic relation is not merely different; it is stronger. Among all distributions with a given standard deviation, the Gaussian has the largest entropy, $\tfrac12\ln(2\pi e\sigma^2)$. So the entropy of any distribution is at most that of a Gaussian with the same spread, and substituting into the entropic bound gives $\ln(2\pi e\sigma_x\sigma_p) \geq \ln(\pi e\hbar)$, which is Heisenberg’s inequality. Any state that satisfies the entropic relation satisfies Heisenberg’s; the reverse fails, as the split state shows.

The third figure tests four ordinary single-humped shapes against both. Every one lies above both bounds, and only the Gaussian touches either. They do not even rank the shapes alike: a packet shaped like a parabolic cap comes closer to Heisenberg’s bound than one with exponential tails, and further from the entropic one, because the cap’s sharp edges cost information in momentum without costing much variance. Which shape is “least uncertain” depends on what uncertainty is taken to mean.

A pair of questions the variance form cannot constrain #

The weakness of the variance form is worse for observables with discrete outcomes. The questions that can be asked together found Robertson’s general inequality, $\Delta A\,\Delta B \geq \tfrac12|\langle[A,B]\rangle|$, which bounds the product of two uncertainties by the expectation value of their commutator. The trouble is that the expectation value depends on the state, and for many states it is zero. For a qubit, the commutator of its $x$ and $z$ spin components is proportional to its $y$ component, so for any state with $\langle Y\rangle = 0$ — every state pointing somewhere in the $x$–$z$ plane — Robertson’s bound is zero and forbids nothing.

The fourth figure walks round that circle. The variance product does fall to zero, at the four points where the qubit points exactly along $z$ or $x$ and one of the two questions has a certain answer — and there the bound says nothing more than that a product of two non-negative numbers is non-negative. But the two questions plainly cannot both be answered: when the $z$ answer is certain, the $x$ answer is a fair coin. The entropic relation of Hans Maassen and Jos Uffink, from 1988, says so: for two observables whose bases have largest overlap $c$, $H(A) + H(B) \geq -2\ln c$, and for a qubit’s $x$ and $z$, $c^2 = 1/2$, so the sum of the two entropies is never less than one bit. It is reached exactly at the four points where Robertson’s product vanishes. The entropic bound depends only on the observables, never on the state, so it cannot become trivial.

An angle, where the variance form breaks entirely #

The angle the uncertainty principle cannot be written for found the case where a variance-based relation fails outright. An angle wraps round: a distribution spread evenly round the circle has a finite variance, at most that of the uniform distribution, while its partner, the angular momentum, is then perfectly sharp. A naive $\Delta\phi\,\Delta L \geq \hbar/2$ would be violated by the simplest state there is, a rotor with definite angular momentum, and the repairs that essay discussed all involve choosing a window or a circular measure of spread. It named the entropic relation as the alternative that needs no repair, and the fifth figure computes it.

For a rotor the entropic bound is $H(\phi) + H(L) \geq \ln 2\pi$, where the angle’s entropy is that of a continuous distribution on the circle and the angular momentum’s is that of a discrete one on the integers. A state of definite angular momentum spreads its angle evenly round the circle: its angle entropy is $\ln 2\pi$, its angular-momentum entropy is zero, and it sits on the bound. As the angle is aimed more sharply, the angular momentum spreads over more values and the sum rises, approaching $\ln(\pi e)$, the position–momentum bound, because a sharply aimed rotor occupies a short arc that might as well be a straight line. At no point does the relation need adjusting for the wrap-round. Entropy does not care whether a distribution lives on a line or a circle, because it never refers to a mean.

A family of relations, and the guessing game #

Shannon’s is not the only entropy, and the choice matters less than it might seem. Alfréd Rényi defined a family of entropies, indexed by a number $\alpha$, that weight a distribution’s peaks and tails differently: $\alpha = 1$ is Shannon’s, $\alpha = 2$ is the logarithm of the probability that two independent samples land in the same place, and $\alpha \to \infty$ depends only on the largest single probability. Białynicki-Birula showed in 2006 that position and momentum obey a whole family of entropic relations, one for each pair of Rényi indices satisfying $1/\alpha + 1/\beta = 2$, and the Shannon relation is the symmetric member of the family.

The limiting member has the most direct meaning of all. The min-entropy of a distribution measures how well its outcome can be guessed in one try: it is minus the logarithm of the probability of the single most likely outcome. An uncertainty relation in min-entropies is therefore a statement about a game. A referee prepares a state and asks either for its position or for its momentum, to some precision; the player knows the state but not which question will come. The relation bounds the player’s chance of being right on both, whatever state is prepared. For a qubit’s $x$ and $z$, asked at random, the best possible preparation is answered correctly with probability $\tfrac12 + 1/2\sqrt2 \approx 0.85$, never 1, and that number, not any standard deviation, is what a cryptographer needs to know about an eavesdropper.

Measured, not only derived #

Entropic relations have been tested in the laboratory directly, by preparing states, measuring the two incompatible properties on many copies, building the two distributions and computing their entropies. Experiments with the polarisations of single photons in 2011 found the sums lying on or above the predicted bounds across the families of states prepared, and touching them where theory says they should; experiments with the spins of neutrons in a polarimeter, in 2015, tested the related entropic relations for the error a measurement makes and the disturbance it causes. The photon experiments also tested the version with a quantum memory: when the photon being measured was entangled with a second photon held by the experimenter, the uncertainty about the first, conditioned on measurements of the second, fell below the ordinary bound by exactly the amount the entanglement allowed.

Those experiments make the point that an uncertainty relation is a property of the predictions of quantum mechanics that can be checked like any other, not a statement about the limitations of instruments. The neutron experiment was about the second kind of statement, which is easy to confuse with the first: the relations drawn here constrain how sharply a state can be prepared, and the error–disturbance relations constrain what a measurement does to it, and the two have different bounds.

What the stronger relation is used for #

Entropic uncertainty relations are not only a more honest description; they are the form in which the uncertainty principle is put to work. Quantum key distribution rests on the fact that an eavesdropper who learns about one of two incompatible properties of a signal must disturb the other, and a proof of security has to say how much. Entropy is exactly the currency that question is asked in — how many bits an eavesdropper can know — and since 2010 the security proofs of practical systems have used entropic relations extended to the case where the eavesdropper holds quantum memory entangled with the signal. In that form the relation’s floor can be lowered by entanglement, and how far it can be lowered is a measure of the entanglement.

The same relations sharpen the statements of the state that cannot be copied: a copy good enough to reveal both of two incompatible properties would let the copier beat the entropic bound, and it is the entropic form that gives the strongest limit on how good an imperfect copy can be. And they link the uncertainty principle to the thermodynamic cost of information that the bit that has to be paid for measured: uncertainty in bits is a quantity with a price in energy.

Where each relation is the right tool #

None of this makes Heisenberg’s inequality wrong or useless. For single-humped distributions, which is what wave packets in most physical situations are, the two relations rank states almost alike and Heisenberg’s is far easier to compute with: a standard deviation is a second moment, and second moments obey simple equations of motion, so the motion that cannot be stopped could compute the zero-point energy of an oscillator directly from it. The entropic relation needs the whole distribution and its logarithm, and entropies of continuous distributions depend on the units in which they are measured, which is why the bound contains $\hbar$ and why only the sum of the two entropies, not either alone, has a unit-independent meaning.

The differential entropy also has a technical weakness of its own. It can be negative, and for a distribution concentrated on a very small set it is a large negative number; it measures relative, not absolute, spread. That is harmless in the sum, where the two units cancel, but it means a single position entropy is not a count of anything in the way a discrete entropy is.

Distributions, not measurements #

The figures compute distributions and their entropies; they do not show measurements. The entropic relation, like Heisenberg’s, is a statement about the distributions a state predicts for two different measurements made on separate copies of it, not about measuring both properties of one particle at once, and nothing in the figures shows the disturbance a measurement causes. The qubit figure draws only states on one great circle of the Bloch sphere; off that circle Robertson’s bound becomes nonzero but remains state-dependent, while the entropic one stays fixed. And the angle figure uses a single family of aimed states; other families give other curves, all above the same line.

Still open: the tightest relation for more than two questions #

For two observables the entropic bound is known exactly in many cases, and for position and momentum it is reached by Gaussians. For three or more mutually incompatible observables — the three spin components of a particle, or the many sets of measurements used in modern quantum cryptography — the tightest possible entropic bound is known only in special cases, and in general finding it is a hard optimisation problem. Whether the bounds can be computed efficiently, and whether entanglement with a quantum memory can lower them in every case as it does for two observables, are open questions with direct consequences for how much key a quantum cryptosystem can safely produce.

The habit worth carrying away is to ask what a measure of spread is blind to. A particle split between two narrow packets has a Heisenberg product that grows without limit with their separation, eight times the minimum for packets 8 units apart, while its entropic sum $H_x + H_p$ stays within $2\ln 2 - 1$ of the floor $\ln(\pi e\hbar)$; the entropic relation implies Heisenberg’s, still holds at one bit for a qubit’s x and z where Robertson’s bound vanishes, and needs no repair for an angle. Uncertainty is missing information, and a standard deviation counts empty space as information missing.