# Understanding the monogamy of entanglement with respect to GHZ States

Given the GHZ state wrt n = 3: $$\frac{|000⟩ + |111⟩}{\sqrt{2}}$$, I'm trying to understand how the principle of monogamy of entanglement manifests here. I came across this explanation.

Let’s say that Alice, Bob, and Charlie hold random bits, which are either all 0 or all 1 (so, they’re classically correlated). If all three of them get together, they can see that their bits are correlated, and the same is true if only two of them are together.

But now suppose the three players share a GHZ state. With all three of them, they can see that the state is entangled, but what if Charlie is gone? Can Alice and Bob see that they’re entangled with each other? No. To see this, observe that by the No-Communication Theorem, Charlie could’ve measured without Alice and Bob knowing. But if he did, then Alice and Bob would clearly have classical correlation only: either both 0’s (if Charlie got the measurement outcome 0) or both 1 (if Charlie got 1). From this it follows that Alice and Bob have only classical correlation regardless of whether Charlie measured or not.

I've been trying to make sense of this for a while now and I'm at my wit's end. ANY help is appreciated, thanks!

• That paragraph involves several concepts; it's a useful skill to be able to divide such paragraphs into bite-sized chunks. For a start, are you aware of the notion of classical and quantum correlations? I'd start with this. – Sanchayan Dutta Dec 4 '19 at 1:57
• @MarkS Fixed it, thank you! – Apple Meson Dec 4 '19 at 18:12

Let’s say that Alice, Bob, and Charlie hold random bits, which are either all or all (so, they’re classically correlated). If all three of them get together, they can see that their bits are correlated, and the same is true if only two of them are together.

If Alice, Bob, and Charlie all have either the bits $$\{1_A, 1_B, 1_C\}$$ or the bits $$\{0_A, 0_B, 0_C\}$$, upon multiple measurements, Bob and Charlie will notice that their bits are always the same as Alice's bits. This is the classical "correlation" Aaronson mentions.

But now suppose the three players share a GHZ state. With all three of them, they can see that the state is entangled (...)

A GHZ state allows for non-classical (or quantum) correlations. If Alice, Bob, and Charlie are together, then with some measurements they can detect some non-trivial correlations; this concept is well explained here.

but what if Charlie is gone? Can Alice and Bob see that they’re entangled with each other? No. To see this, observe that by the No-Communication Theorem, Charlie could’ve measured without Alice and Bob knowing.

If Charlie measured his qubit then depending on whether he gets $$|0\rangle_C$$ or $$|1\rangle_C$$, the remaining composite state of Alice and Bob's qubits would be either $$|00\rangle_{AB}$$ or $$|11\rangle_{AB}$$ respectively. This is clear from the definition of the GHZ state i.e.,

$$|\mathrm{GHZ}_3\rangle = \frac{|00\rangle_{AB} |0\rangle_C + |11\rangle_{AB}|1\rangle_C}{\sqrt 2}.$$

But if he did, then Alice and Bob would clearly have classical correlation only: either both ’s (if Charlie got the measurement outcome) or both (if Charlie got). From this it follows that Alice and Bob have only classical correlation regardless of whether Charlie measured or not

As the post-measurement composite state of Alice and Bob's qubits would be either $$|0\rangle_A|0\rangle_B$$ or $$|1\rangle_A|1\rangle_B$$, the resulting correlation becomes trivial. If Alice's qubit is measured to be $$|0\rangle_A$$ or $$|1\rangle_A$$ then Bob's qubit would be $$|0\rangle_B$$ or $$|1\rangle_B$$ respectively and vice-versa.

Moreover, even if say Charlie didn't measure his qubit, if Alice measures her qubit to be $$|0\rangle_A$$ (or $$|1\rangle_A$$), Bob's qubit will also be $$|0\rangle_B$$ (or $$|1\rangle_B$$). This is again a trivial classical correlation between Alice and Bob. The fact of the matter is that to observe some non-trivial correlations you need to determine the joint statistics of Alice, Bob and Charlie. Pairwise statistics will only show you trivial classical correlations. The magic of entanglement only manifests in the three-body measurement statistics for the GHZ state!

In the GHZ case, we noticed that the effects of entanglement are only observable in the three-qubit statistics and not in the pairwise statistics. Mathematically, it can be shown that in the GHZ state there is no pairwise entanglement at all (from the reduced density matrix $$\rho_{AB}$$). The "Monogamy of Entanglement" part is a bit of a non-sequitur, indeed. Aaronson states the example of the $$|W_3\rangle$$ where no two qubits are pairwise "maximally entangled" but are entangled to some degree.

The point is that, in a three-qubit state, it is theoretically impossible for more than one pair of qubits to be maximally entangled. The $$|\mathrm{GHZ}_3\rangle$$ and $$|W_3\rangle$$ states are simply examples of this; the interesting fact being that the $$|\mathrm{GHZ}_3\rangle$$ state as a whole is maximally entangled although no two qubits are pairwise maximally entangled. Thus, it is analogous to Borromean rings which consist of three topological circles which are linked but where removing any one ring leaves the other two unconnected.

• IMO the case "if say Charlie didn't measure his qubit" requires better explanation; you can say that Alice-Bob correlations can't depend on whether Charlie measured his qubit or not, but it is better to show it formally: trace out Charlie from the common density matrix and see that the remaining density matrix is not entangled. – kludg Dec 4 '19 at 6:14
• @kludg Well, yes. I was avoiding presuming the knowledge of reduced density operators and partial traces on the OP's part. – Sanchayan Dutta Dec 4 '19 at 7:00
• @SanchayanDutta Thank you so much for the detailed explanation! It definitely helped clear some misconceptions up for me. Regarding this: As the post-measurement composite state of Alice and Bob's qubits would be either ... is again a trivial classical correlation between Alice and Bob. Wouldn't this also be the case for the Bell States? Measuring Alice's qubit would tell you the state of Bob's qubit. But isn't that a maximally entangled state between 2 qubits too? I guess I don't see the difference between the Bell States and this example, when Charlie didn't measure his qubit. – Apple Meson Dec 4 '19 at 17:53
• I'm currently taking an intro course on quantum information, so I do have some knowledge of reduced density operators and partial traces. I do see that the density operator over 2 qubits of the GHZ state by tracing out the third is different from the density matrix for the Bell State. However I'm not sure I know how to draw a strong conclusion from that? – Apple Meson Dec 4 '19 at 17:57
• @AppleMeson Good question. The non-trivial correlation in the Bell state is not evident until you measure in a different basis apart from $\{|0\rangle, |1\rangle\}$ (cf. this). This doesn't occur in the GHZ state and that can be proven from the reduced density matrix for Alice and Bob (using measures like the von Neumann entropy). You will notice that while the density matrix in the Bell state case is maximally entangled, it is not so in the GHZ case. I will try to elaborate on this in my answer tomorrow. – Sanchayan Dutta Dec 4 '19 at 18:23