Inquantum physics,monogamy is the property ofquantum entanglement that restricts entanglement from being freely shared between arbitrarily many parties.

In order for two qubitsA andB to bemaximally entangled, they must not be entangled with any third qubitC whatsoever. Even ifA andB are not maximally entangled, the degree of entanglement between them constrains the degree to which either can be entangled withC. In full generality, for${\displaystyle n\ge 3}$ qubits${\displaystyle A_1,\dots ,A_n}$, monogamy is characterized by theCoffman–Kundu–Wootters (CKW)inequality, which states that

${\displaystyle \sum _{k=2}^n\tau ({\rho }_{A_1{A}_k})\le \tau ({\rho }_{A_1(A_2\dots A_n)})}$

where${\displaystyle {\rho }_{A_1{A}_k}}$ is thedensity matrix of the substate consisting of qubits${\displaystyle A_1}$ and${\displaystyle A_k}$ and${\displaystyle \tau }$ is the "tangle", a quantification of bipartite entanglement equal to the square of theconcurrence.^[1][2]

Monogamy, which is closely related to theno-cloning property,^[3][4] is purely a feature of quantum correlations, and has no classical analogue. Supposing that two classical random variablesX andY are correlated, we can copy, or "clone",X to create arbitrarily many random variables that all share precisely the same correlation withY. If we letX andY be entangled quantum states instead, thenX cannot be cloned, and this sort of "polygamous" outcome is impossible.

The monogamy of entanglement has broad implications for applications ofquantum mechanics ranging fromblack hole physics toquantum cryptography, where it plays a pivotal role in the security ofquantum key distribution.^[5]

The monogamy of bipartite entanglement was established for tripartite systems in terms of concurrence by Valerie Coffman, Joydip Kundu, andWilliam Wootters in 2000.^[1] In 2006, Tobias Osborne andFrank Verstraete extended this result to the multipartite case, proving the CKW inequality.^[2][6]

For the sake of illustration, consider the three-qubit state${\displaystyle |\psi ⟩\in ({\mathbb{C}}^2{)}^{\otimes 3}}$ consisting of qubitsA,B, andC. Suppose thatA andB form a (maximally entangled)EPR pair, e.g.

${\displaystyle |\text{EPR}{⟩}_{AB}=\frac{(|00⟩+|11⟩)}{\sqrt{2}}}$

We will show that:

${\displaystyle |\psi ⟩=|\text{EPR}{⟩}_{AB}\otimes |\varphi {⟩}_C}$

for some valid quantum state${\displaystyle |\varphi {⟩}_C}$. By the definition of entanglement, this implies thatC must be completely disentangled fromA andB.

When measured in the standard basis,A andB collapse to the states${\displaystyle |00⟩}$ and${\displaystyle |11⟩}$ with probability${\displaystyle \frac{1}{2}}$ each. It follows that:

${\displaystyle |\psi ⟩=|00⟩\otimes ({\alpha }_0|0⟩+{\alpha }_1|1⟩)+|11⟩\otimes ({\beta }_0|0⟩+{\beta }_1|1⟩)}$

for some${\displaystyle {\alpha }_0,{\alpha }_1,{\beta }_0,{\beta }_1\in \mathbb{C}}$ such that${\displaystyle |{\alpha }_0{|}^2+|{\alpha }_1{|}^2=|{\beta }_0{|}^2+|{\beta }_1{|}^2=\frac{1}{2}}$.

By defining diagonal basis vectors${\displaystyle |+⟩}$ and${\displaystyle |-⟩}$ such that:

${\displaystyle |0⟩=\frac{1}{\sqrt{2}}(|+⟩+|-⟩)}$

${\displaystyle |1⟩=\frac{1}{\sqrt{2}}(|+⟩-|-⟩)}$

we can rewrite the states ofA andB in terms of${\displaystyle |++⟩}$,${\displaystyle |+-⟩}$,${\displaystyle |-+⟩}$, and${\displaystyle |--⟩}$ as:

${\displaystyle |\psi ⟩=\frac{1}{2}(|++⟩+|+-⟩+|-+⟩+|--⟩)\otimes ({\alpha }_0|0⟩+{\alpha }_1|1⟩)+\frac{1}{2}(|++⟩-|+-⟩-|-+⟩+|--⟩)\otimes ({\beta }_0|0⟩+{\beta }_1|1⟩)}$

${\displaystyle =\frac{1}{2}(|++⟩+|--⟩)\otimes (({\alpha }_0+{\beta }_0)|0⟩+({\alpha }_1+{\beta }_1)|1⟩)+\frac{1}{2}(|+-⟩+|-+⟩)\otimes (({\alpha }_0-{\beta }_0)|0⟩+({\alpha }_1-{\beta }_1)|1⟩)}$

Being maximally entangled,A andB collapse to one of the two states${\displaystyle |++⟩}$ or${\displaystyle |--⟩}$ when measured in the diagonal basis. The probability of observing outcomes${\displaystyle |+-⟩}$ or${\displaystyle |-+⟩}$ is zero. Therefore, according to the equation above, it must be the case that${\displaystyle {\alpha }_0-{\beta }_0=0}$ and${\displaystyle {\alpha }_1-{\beta }_1=0}$. It follows immediately that${\displaystyle {\alpha }_0={\beta }_0}$ and${\displaystyle {\alpha }_1={\beta }_1}$. We can rewrite our expression for${\displaystyle |\psi ⟩}$ accordingly:

${\displaystyle |\psi ⟩=(|++⟩+|--⟩)\otimes ({\alpha }_0|0⟩+{\alpha }_1|1⟩)}$

${\displaystyle =|\text{EPR}{⟩}_{AB}\otimes (\sqrt{2}{\alpha }_0|0⟩+\sqrt{2}{\alpha }_1|1⟩)}$

${\displaystyle =|\text{EPR}{⟩}_{AB}\otimes |\varphi {⟩}_C}$

This shows that the original state can be written as a product of a pure state inAB and a pure state inC, which means that the EPR state in qubitsA andB is not entangled with the qubitC.

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