Monogamy of entanglement

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

For the sake of illustration, consider the three-qubit state ${\displaystyle |\psi \rangle \in (\mathbb {C} ^{2})^{\otimes 3}}$ consisting of qubits A, B, and C. Suppose that A and B form a (maximally entangled) EPR pair. We will show that:

${\displaystyle |\psi \rangle =|{\text{EPR}}\rangle _{AB}\otimes |\phi \rangle _{C}}$

for some valid quantum state ${\displaystyle |\phi \rangle _{C}}$. By the definition of entanglement, this implies that C must be completely disentangled from A and B.

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

${\displaystyle |\psi \rangle =|00\rangle \otimes (\alpha _{0}|0\rangle +\alpha _{1}|1\rangle )+|11\rangle \otimes (\beta _{0}|0\rangle +\beta _{1}|1\rangle )}$

for some ${\displaystyle \alpha _{0},\alpha _{1},\beta _{0},\beta _{1}\in \mathbb {C} }$ such that ${\displaystyle |\alpha _{0}|^{2}+|\alpha _{1}|^{2}=|\beta _{0}|^{2}+|\beta _{1}|^{2}={\frac {1}{2}}}$. We can rewrite the states of A and B in terms of diagonal basis vectors ${\displaystyle |+\rangle }$ and ${\displaystyle |-\rangle }$:

${\displaystyle |\psi \rangle ={\frac {1}{2}}(|++\rangle +|+-\rangle +|-+\rangle +|--\rangle )\otimes (\alpha _{0}|0\rangle +\alpha _{1}|1\rangle )+{\frac {1}{2}}(|++\rangle -|+-\rangle -|-+\rangle +|--\rangle )\otimes (\beta _{0}|0\rangle +\beta _{1}|1\rangle )}$

${\displaystyle ={\frac {1}{2}}(|++\rangle +|--\rangle )\otimes ((\alpha _{0}+\beta _{0})|0\rangle +(\alpha _{1}+\beta _{1})|1\rangle )+{\frac {1}{2}}(|+-\rangle +|-+\rangle )\otimes ((\alpha _{0}-\beta _{0})|0\rangle +(\alpha _{1}-\beta _{1})|1\rangle )}$

Being maximally entangled, A and B collapse to one of the two states ${\displaystyle |++\rangle }$ or ${\displaystyle |--\rangle }$ when measured in the diagonal basis. The probability of observing outcomes ${\displaystyle |+-\rangle }$ or ${\displaystyle |-+\rangle }$ 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 ${\displaystyle |\psi \rangle }$ accordingly:

${\displaystyle |\psi \rangle =(|++\rangle +|--\rangle )\otimes (\alpha _{0}|0\rangle +\alpha _{1}|1\rangle )}$

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

${\displaystyle =|{\text{EPR}}\rangle _{AB}\otimes |\phi \rangle _{C}}$

This shows that the original state can be written as a product of a pure state in AB and a pure state in C, which means that the EPR state in qubits A and B is not entangled with the qubit C.