# Entanglement-based attack on E91 protocol

My first question is if the bell pair that Alice and Bob share for the E91 protocol matters since I've seen different resources use different pairs for the explanation. My second question is say they each get one qubit of the bell state |00> + |11> (for the sake of clarity I'll ignore normalization constants) and while one of those qubits is on the way to Alice, Eve intercepts it and entangles it with one of hers creating the overall state |000>+|111>. Now Eve waits for Alice and Bob to declare in which qubits they have measured with the same basis and does the same. Why wouldn't this work as an attack? I've heard people say that if Eve entangles her own qubit maximal entanglement is lost and the protocol doesn't work anymore but isn't the state |000>+|111> maximally entangled?

If for example the bell pair |01>-|10> is used instead Eve could entangle her qubit with the first to create the state |001>-|110> which I believe is also a maximally entangled state.

If Eve manages to create the overall state $$|GHZ\rangle=|000\rangle+|111\rangle$$, then it is certainly true that if both Alice and Bob choose to measure in the $$Z$$ basis, they will still get perfectly correlated results, and Eve will know that result.
However, what happens if Alice and Bob use some different basis? Let's imagine they both measure in the $$X$$ basis, which is equivalent to applying Hadamard to the qubits and measuring in the standard basis. $$(H\otimes H\otimes H)(|000\rangle+|111\rangle)=(|000\rangle+|011\rangle+|110\rangle+|101\rangle).$$ We see from this that Alice and Bob do not always get the same result. So, effectively, part of the protocol is that Alice and Bob reveal some of their measurement results and check whether they've got the same answers. If not, it's assumed to be because of some sort of attack like this. I said "effectively" because that's an easy way to think of it. Specifically for the E91 protocol, they don't actually reveal measurement results from cases when key could have been generated. Instead, they reveal the measurement results corresponding to when the CHSH inequality is being tested. So, the real way to see what's going on is to evaluate the effect of replacing the Bell state with the GHZ state on the value of the CHSH inequality. The usual CHSH is equivalent to evaluating $$\text{Tr}(|\psi\rangle\langle\psi|\sqrt{2}(XX+ZZ))=2\sqrt{2}.$$ where $$|\psi\rangle$$ is the $$+1$$ eigenstate of both $$XX$$ and $$ZZ$$ (which is the Bell state you chose).
Meanwhile, because the eavesdropper is not returning any measurement values, if they created this GHZ state, we have to trace out their qubit: $$\rho=\text{Tr}_E|GHZ\rangle\langle GHZ|=\frac{I}{4}.$$ So, in this case, if Alice and Bob evaluate the CHSH inequality by revealing some of their measurement results, they get the value $$\text{Tr}(\rho\sqrt{2}(XX+ZZ))=0.$$ There is a very clear signal that bad things have happened, and Alice and Bob should not proceed.
• IIRC the only state with perfect (anti) correlation for all bases, between the first two parties, is the state $(|01\rangle-|10\rangle)_{AB}\otimes|\phi\rangle_E$. Commented Apr 26 at 10:45