# Why does entanglement of 3 qubits break this?

Another user asked about the Ekert Quantum Key Distribution protocol: Alice and Bob each randomly choose one of three bases separated by $$120^{\circ}$$ in which to measure, and then later compare the results both when choosing the same base and when choosing a different base. For fun, I decided to implement this in Quirk. Quirk: Alice and Bob. Picture below.

[The gate "0/1/2" maps $$|00\rangle$$ to $$(|00\rangle + |01\rangle + |10\rangle)/\sqrt3$$ so that each of the three basis is chosen with equal probability.]

As you can see, Alice and Bob always get the same result when they use the same basis, and agree $$\frac14$$ of the time when they use a different basis. Exactly what theory demands.

I was uncertain how to implement Eve. To start, I gave Eve a copy of the entangled qubit by adding an X gate on the empty line between the two spacers. Quirk: Eve. Much to my surprise, that was all that was needed. The probabilities immediately changed to $$\frac34$$ and $$\frac38$$, as predicted by theory.

I have two questions:

1. Why does the existence of the third entangled qubit destroy the algorithm? I could understand if we measured the qubit or used it to control something, then we'd get another algorithm. But somehow the very fact that Alice and Bob's pair is actually a triplet is all that is necessary. I don't quite grok why that is sufficient.

2. If Eve decides not to e[a]vesdrop, is there anything that she can do to her qubit so that Alice and Bob can exchange keys in peace? Or is the very fact that the third entangled qubit was created mean the algorithm won't work?

Update: Adding response to Craig Gidney's answer.

You didn't quite answer my question, but you got me awfully close.

You gave me the insight I needed to realize that I could simplify my query to the following circuit. If you only look at two qubits, things look a lot alike until suddenly they don't. I need to look a bit more with the linear algebra to see why the third qubit makes such a big difference.

## 2 Answers

Why wouldn't it break it? Involving Eve requires, for example, putting a CNOT gate from one of the distributed qubits to some outside qubit. That CNOT operation doesn't commute with the controlled Y rotations. So by default it's going to mess everything up to add that CNOT gate to the circuit.

• Comments added as update to my question. Nov 24, 2022 at 1:18

Remember that, from the perspective of Alice and Bob, Eve's qubit(s) are an external system that they cannot control. The best description that they have of their system involves tracing out Eve's qubit. Mathematically, the tracing out is what does the business of reducing the fidelity.

One of the points of the reduced density matrix of a system is that all of Alice and Bob's measurements have the correct statistics no matter what Eve does to her system (provided they don't learn what she did and what outcomes she got, in which case they would be able to update their description). So the average across all Eve's possible (single-qubit) measurements and results is exactly that given by the reduced density matrix. So from Alice and Bob's perspective, it does not matter what Eve does beyond the initial act of entangling/measuring.