What two have they cannot give a third

The correlation no instructions produce establishes that two particles can be correlated more strongly than any list of pre-arranged answers allows. The obvious next question is how far that goes — whether a particle can be that strongly correlated with several others at once, and what happens to the correlation when many parties share it.

The answer is a strict prohibition with an exact form. Entanglement is monogamous: the more of it two systems have with each other, the less either can have with anything else, and the trade is quantitative rather than a matter of degree.

The trade-off, and what it rules out #

The classical bound on a CHSH value is 2, the quantum maximum is $2\sqrt2 \approx 2.83$, and the monogamy relation says that the squares of two such values, sharing a party, sum to no more than 8.

The consequence is stronger than a trade-off. Set both values equal and the constraint gives exactly 2 — the classical value. So it is not merely that a party maximally entangled with one other is weakly entangled with a third; it is that no state permits two simultaneous Bell violations sharing a party at all. The figure finds that number by searching along the boundary rather than by rearranging the algebra, and it comes out at 2 to five decimals.

That fact does real work. It is the reason entanglement-based key distribution is secure: if two parties see a Bell violation between them, no third party can be correlated with either strongly enough to know what they will find. The security argument does not depend on any assumption about what the eavesdropper is doing or how clever they are; it depends on the geometry of that quarter circle. In the same way that the state that cannot be copied forbids a passive tap, monogamy forbids a passive correlation.

Where a qubit’s entanglement goes #

The quantitative statement for three qubits is due to Coffman, Kundu and Wootters, and it is an inequality between three numbers.

Take one qubit and ask how entangled it is with the other two taken as a single object. That has a clean answer for a pure state of three: it is $2(1 - \operatorname{tr}\rho^2)$ for the single qubit’s own density matrix, and it measures how mixed that qubit looks on its own. Then ask how entangled it is with each of the other two individually, which is a question about a mixed two-qubit state and is answered by the concurrence.

The theorem is that the whole is at least the sum of the parts: the two pairwise squared concurrences never add up to more than the entanglement with the pair. The difference is a genuinely three-way quantity — the tangle — that no pair holds and no pair can extract.

This is the sense in which entanglement is not a substance. A qubit does not have a quantity of it to distribute; it has a pattern, and the pattern determines both how much any pair can hold and how much is only available to all three at once.

The two extremes #

The two states at the ends of the figure are worth naming, because they are the standard examples and they are as far apart as three qubits can get.

The GHZ state — all three up or all three down, in superposition — has no pairwise entanglement at all. Trace out any one qubit and the remaining two are in a classical mixture: half the time both up, half the time both down, with no coherence between the two possibilities. Its entire tangle is three-way, and the figure gets exactly 1 for it.

The W state — one excitation shared equally among three — has the opposite structure. Every pair is entangled, with concurrence 2/3, and the three-way residue is exactly zero. Trace out a qubit and the other two are still entangled, which is the property GHZ lacks entirely.

Neither can be turned into the other by anything done to individual qubits, however cleverly, however many copies are supplied. They are not two amounts of the same thing; they are two kinds, and the figures compute the numbers that tell them apart.

What the interpolation shows #

The interpolating family makes the trade explicit, and it makes a point the two endpoints alone cannot.

The total — one qubit’s entanglement with the rest — is nearly flat. If entanglement were a quantity, that flatness would say the family is all the same state in different dress. It is not: at one end no pair is entangled and at the other every pair is, and the difference is measurable by an experiment on two qubits that never touches the third.

So the useful reading is that a fixed total is allocated rather than possessed, and the allocation is what an experiment sees. Two parties who can act only on their own qubits care entirely about the pairwise part; a protocol that needs all three to agree at once — and there are such protocols — cares entirely about the residue.

There is a further asymmetry that the curve hints at without showing. The GHZ end is fragile: lose one qubit and the remaining two have nothing. The W end is robust: lose one qubit and the remaining two are still entangled, which is why W-type states are the ones proposed when a party may drop out. Robustness and three-way strength are opposite ends of the same axis, and no state has both.

A control, and what it settles #

An inequality that every state satisfies is worth testing against a state whose answer is known in advance, and the two-pairs arrangement is that test.

Its numbers are dictated by construction. A qubit maximally entangled with one partner and unentangled with anything else must have a total of one, a pairwise concurrence with its partner of one, nothing with the third party, and a residue of zero — and if the machinery produced anything else, every other number in the figure would be suspect.

It also makes the contrast with GHZ visible in one glance. Both states have the same total for the qubit in question, and they distribute it in opposite ways: one puts all of it into a single link, the other into a three-way structure with no links at all. The W state sits between them with three weak links. Three states, one total, three arrangements, and an experiment on any two qubits distinguishes all three.

A conserved-looking total with a free allocation is a common shape in physics and an unusual one here, because the allocation is not a matter of how the state was prepared in some incidental way. It is the state, and no operation short of bringing the parties back together changes it.

Sharing thins it out #

The last figure takes the sharing to its limit, and the falloff is the practical face of monogamy.

A W state of N parties is the most democratic entangled state there is: every party is equivalent, every pair is equivalent, and no party can be separated from the rest. It looks like the right way to distribute entanglement to a network. It is not, because what each pair gets falls as 1/N, and it falls below the threshold for violating a Bell inequality at about six parties.

That is not a defect of the W state; it is monogamy applied N-1 times. Each party is entangled with N-1 others, the squares of those entanglements must fit under a fixed total, and so each is of order 1/N. A network wanting strong pairwise entanglement has to distribute pairs, one link at a time, rather than one big shared state — and that observation is the reason quantum networks are designed as chains of two-party links with entanglement swapping between them rather than as a single global state.

What the yardstick actually measures #

All four figures are drawn in concurrence, and it is worth saying what that number is, because a monogamy statement is only as meaningful as the quantity it constrains.

For a pure two-qubit state, the concurrence is a direct measure of how far the state is from being a product: zero when the two qubits can be described separately, one when neither has any state of its own. That much would be true of several candidate measures. What makes concurrence the right one here is that it extends to mixed states in a way that keeps its meaning — it is the minimum average entanglement over all the ways the mixture could have been prepared, which is exactly the quantity a party can be sure of having.

That “minimum over preparations” is doing something subtle and is the reason GHZ’s pairs come out at zero. Trace one qubit out of GHZ and the remaining two are in a mixture that could have arisen from a classical coin toss deciding whether both are up or both are down. Since there exists a preparation with no entanglement in it, the honest answer is that the pair has none — even though the same mixture has correlations that look striking when written down. The distinction is the same one the answer that was not there before makes between a correlation that a list of instructions could produce and one that could not.

Concurrence also has the practical virtue of being computable. Wootters’ formula turns it into an eigenvalue problem for a four-by-four matrix, which is what the figures solve; almost every other entanglement measure for mixed states requires an optimisation with no closed form. The monogamy inequality was found in concurrence because concurrence is the measure anyone could calculate, and the search for which other measures obey it is still going on.

Why it has to be true #

The theorem has proofs, and it also has a reason that is worth having separately because it connects to something already established.

A qubit strongly entangled with two others would let a measurement be repeated. Measure the first partner and the outcome pins down what the central qubit is; measure the second and it pins it down again, independently. Two independent determinations of the same unknown state amount to a copy of it — which is exactly what the state that cannot be copied forbids, and it is forbidden for the same one-line reason about linearity.

That argument is not a substitute for the inequality, since it gives no numbers, but it says where the prohibition comes from. Monogamy, no-cloning, and the impossibility of signalling faster than light are three faces of one constraint: the correlations quantum mechanics permits are exactly those that cannot be used to learn an unknown state twice.

The classical contrast makes the strangeness sharper. Classical correlation is entirely promiscuous — a hundred parties can all hold copies of the same random bit, each perfectly correlated with all the others, and nothing objects. It is the strength beyond the classical bound that is exclusive, and this is another place where the quantum and classical answers differ by a prohibition rather than by a magnitude.

Monogamy is why the world looks classical #

There is one consequence large enough to be worth stating on its own, and it is not about protocols.

A system in contact with an environment becomes entangled with it — with air molecules, with photons, with the vibrations of whatever it is sitting on. Monogamy then says what that costs: every unit of entanglement the system builds with its surroundings is a unit it cannot have with anything else, including with the other half of a pair it was prepared with.

That is decoherence, stated as a conservation constraint rather than as a process. The entanglement is not destroyed; it is spread into an environment nobody can measure, and it becomes inaccessible for the same reason a W state’s pairwise entanglement becomes negligible at thirty parties. Where the interference goes traces the same disappearance from the other direction, by asking what happened to the coherence rather than where the entanglement went.

It explains a puzzle about scale too. Large objects are hard to entangle not because entanglement has an upper size but because a large object is already thoroughly entangled with its surroundings and has nothing left to offer. Cooling and isolating an experiment is, in this language, a matter of recovering a monopoly.

Where the model stops #

Everything here is qubits. Concurrence has no clean generalisation to larger systems, and there is no single agreed measure of entanglement for three parties of higher dimension. The monogamy statement survives in general form for some measures and fails for others, and which measures are monogamous is an open subject rather than a settled one.

The figures assume real amplitudes. Wootters’ formula holds for any two-qubit state, but the eigenvalue routine used here handles the real case, so the drawn family is restricted to states with real coefficients. That covers GHZ, W and everything between them and is not a restriction on the theorem.

The three-way tangle is defined for pure states. For a mixed state of three qubits it has to be defined by a minimisation over decompositions, is hard to compute, and loses some of the interpretation. The bar charts above are all pure states for that reason.

And monogamy constrains entanglement, not correlation. Two parties can be perfectly correlated with a third in a classical sense — all reading the same bit — while sharing no entanglement at all. Nothing here limits how much information can be shared; it limits how much of a particular non-classical resource can be.

And nothing here says which pairs a state chooses. The inequality bounds the allocation without predicting it, so a state’s pattern of links is an extra fact about the state, not a consequence of monogamy. Deciding that pattern is what a network protocol is for, and it is the same distinction the questions that can be asked together draws between a constraint on which observables are compatible and the separate matter of which ones a particular state makes sharp.

What the pictures cannot show #

The bar charts split a number into parts and cannot show that the parts are different in kind. The three-way residue is not a smaller version of the pairwise entanglement; it is a quantity no two of the parties can access by any operation on their own qubits, and a bar of the same length carries an entirely different practical meaning depending on which fill it is.

The CHSH region is drawn as though a state were a point in it, which is a projection. A state has many CHSH values depending on which measurements are chosen, and what is plotted is the best each pair can do. Two states at the same point in that plane need have nothing else in common.

Where the ladder goes next #

The entanglement ladder began with the correlation no instructions produce and continued with the disagreement that one run settles, where a single trial distinguishes the two accounts rather than a statistical margin. This rung asks how the correlation may be distributed and finds a hard constraint. The rungs after it: entanglement swapping, which moves a link between parties who never met and is exactly the operation a network needs; the monogamy of steering and of contextual correlations, where the same shape appears with different numbers; and the area laws in many-body systems, which are monogamy read across a boundary rather than between three parties.

The habit worth carrying away is that a resource with a conservation-like constraint behaves differently from one without. Entanglement cannot be broadcast, and almost every protocol that uses it — key distribution, teleportation, the design of a network — is shaped by that one fact rather than by how much of it is available.