I'm studying the E91 protocol. Suppose Eve has a qubit entangled with Alice's and Bob's qubits.

I understand how Alice and Bob can detect Eve if Eve measures before A&B by testing Bell's inequalities.

I'm confused how Alice and Bob could detect Eve if Eve waits to measure her qubit until after A&B publish their bases.

My current understanding: Alice and Bob would measure collapsing the qubits into some state and destorying the entanglement. Eve could then wait until A&B publish their bases and then perfectly recover the secret key by measuring her qubits. And A&B can do nothing to prevent this.

Could someone help me understand what I'm missing here? Thanks!


1 Answer 1


Let us recall the Ekert-91 (E91) protocol.

  • Initially Alice and Bob each independently receive respective elements of a plurality of Bell pairs (for example, generated by spontaneous parametric down-conversion;

  • Upon deciding to generate a secret key, Alice and Bob each independently measure their pairs in one of three bases, and publicly announce their bases;

  • For the qubit pairs where Alice and Bob measure in the same bases, they keep their measurement results secret, as they will be used to generate the key; and

  • For the qubit pairs where Alice and Bob measured in different bases, they announce publicly their results, so that they can generate statistics inline with the CHSH game to test for interference from the eavesdropper Eve.

If I understand your question, you envision Eve:

  1. Surreptitiously entangling qubits with either Alice or Bob's;

  2. Waiting until Alice and Bob publicly announce each of their bases measurements for each qubit; and then

  3. Measuring each of her entangled qubits in the appropriate bases announced by Alice and Bob.

However, merely the act of entangling Eve's qubits with those of Alice or Bob destroys the previously available maximal entanglement between Alice and Bob.

This is the monogamy of entanglement - qubits can only be maximally entangled with each other, and not with a third party. Eve's effort to entangle her qubits with those of Alice/Bob will enable Alice and Bob to run the Bell tests on their qubits, to detect this corruption.

I also think the principle of deferred measurement may be at play - it should not matter whether Eve measures before or after Alice and Bob contemporaneously announce their bases. If it did, then Eve could send a superluminal message to either Alice or Bob(!) For example, if there is a difference between measuring before or after Alice and Bob communicating, then Eve could turn this around to send a bit to Alice and Bob without actually sending any classical information to either Alice or Bob. This violates the no-signaling theorem.

  • $\begingroup$ Hi, I don't understand why if Eve entangles her qubit with say Alice's qubit the maximal entanglement is lost. If Alice's and Bob's qubits are in the state |00>+|11> for example and Eve uses a CNOT gate on the second qubit with one of her qubits that would give the state |000>+|111>. Isn't that the state of 3 maximally entangled qubits? For the sake of clarity I ignored the probability amplitudes. $\endgroup$
    – Omeglac
    Commented Apr 24 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.