The answer to this question leads us to an important fundamental property of entanglement called monogamy: if two objects are maximally entangled to each other, then neither object is entangled to a third one. More generally, it says that the stronger the entanglement between two objects is, the weaker their entanglements are with other objects.

To really understand what monogamy means, we first need to describe how entanglement between pairs of objects are measured. That way, saying 2 objects are maximally entangled has a precise meaning.

Measuring entanglement in general is a not so easy thing: it turns out you can get different numbers depending on context. However, to just understand the basic idea, it will be sufficient to consider 2 qubits.

Suppose we have qubits A and B that are in a pure state which we can write as

*|A,B) = a |0,0) + b |0,1) + c |1,0) + d|1,1),*

where |0) and |1) represent a pair of distinguishable states for a qubit, and the notation |0,1) means that qubit A is in the state |0) and qubit B is in the state |1), and so on.

In 1997, Hill and Wootters described a general formula for measuring the entanglement of 2 qubits called concurrence. For the state |A,B), the concurrence

*C(A,B)*is given by*C(A,B) = 2 sqrt{ det(M) }*

where

Here M is the matrix for the reduced state of A, that is, it describes the state of qubit A alone given that the state of the qubits A and B is |A,B). det(M) refers to the determinant of the matrix M. For our purposes, it is sufficient to just know the formula for the concurrence, in terms of the numbers a,b,c, and d.

The only other thing we need to know about concurrence is the following. If

*C(A,B) = 0*, this means qubits A and B have no entanglement between them. If*C(A,B) = 1*, qubits A and B are maximally entangled.
For example, when we consider the state

*|Bell) = 1/sqrt(2) |0,0) + 1/sqrt(2) |1,1)*

then

Intuitively, this is nice because for |Bell), whenever we measure the two qubits in the same way, we find them in the same state, e.g., they are both in |0) or |1), and this is kind of what you expect for a state that supposedly has the maximum amount of entanglement.

*det(M) = 1/4*and so*C(A,B) = 2 sqrt(1/4) = 1*. This means we have a maximally entangled state for 2 qubits.Intuitively, this is nice because for |Bell), whenever we measure the two qubits in the same way, we find them in the same state, e.g., they are both in |0) or |1), and this is kind of what you expect for a state that supposedly has the maximum amount of entanglement.

Now that we know how to measure entanglement between pairs of qubits, we can ask what happens if we have three qubits A,B,C in a pure state and how the concurrence between different pairs of qubits are related.

Again, the situation is quite complicated to analyze in general so we consider a simple example. Suppose we have the 3 qubits in the state

*|A,B,C) = x |1,0,0) + y|0,1,0) + z |0,0,1).*

We can calculate the concurrences

*C(A,B)*and*C(A,C)*and find that*C(A,B) = 2|xy|,*

*C(A,C) = 2|yz|.*

If we compute the reduced state of A to get the matrix

*M*in this case, we find that
Comparing with the concurrences, we find that

*C(A,B)^2 + C(A,C)^2 = 4 det(M).*

This is a quantitative statement of the monogamy of entanglement. If we treat qubits B and C as one object, the right-hand side of the formula above is equal to the concurrence

*C(A,BC)^2*.
Roughly speaking, the formula says that the entanglement between qubits A and B and the entanglement between A and C is bounded by the entanglement between A and the joint system formed by qubits B and C.

Of course, the above formula was obtained for a special example. Coffman, Kondu, and Wootters showed that if we consider a general state for 3 qubits, the formula becomes an inequality:

*C(A,B)^2 + C(A,C)^2 <= 4 det(M).*

**References:**

Scott Hill and William K. Wootters, "Entanglement of a Pair of Qubits," Phys. Rev. Lett.

**78**, 5022 (1997).

Valerie Coffman, Joydip Kundu, and William K. Wootters, "Distributed Entanglement," Phys. Rev. A,

**61**, 052306 (2000).

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