BB84 attack with entangled qubits example

Question

BB84 attack with entangled qubits example

Hi, I'am interested in an attack for BB84 protocol with entangled quibits. Lets say Alice sends a qubit $x$ in state $\left|1\right>_x$ to Bob and Eve takes the CNOT gate to entangle the states. Therefore, Eve uses a qubit $e$ in state $\left|0\right>_e$. Using CNOT gate the result of this operation is $$\left|1\right>_x\left|0\right>_e\rightarrow \left|1\right>\left|1\right>.$$ Let's say now Bob measures base B in 90° and 0° orientation (the $|0\rangle/|1\rangle basis). Alice and Bob communicate their choice of basis over the classical channel. Eve now knows the orientation and therefore measures her entangled qubit in the right orientation. That means Eve knows now the bit value of the key, lets say 1.

But what would be the case if Alice sends now a qubit in the state $$\frac{1}{\sqrt{2}}(\left|0\right>_x-\left|1\right>_x)?$$ Eve would create the entangled state $$\frac{1}{\sqrt{2}}(\left|00\right>-\left|11\right>)$$ There are two different cases depending on Bob's choice of basis:

• case 1: Alice sends the qubit to Bob and Bob measures in the wrong base B 90° and 0° orentation, therefore nothing happens, because Alice and Bob have different bases.

• case 2: But what if Bob measures in diagonal base 45° and -45° ($|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt{2}$). Someone said that BB84 protocol covers this in 50% of cases. But why is it like that?

A measurement in 45° and -45° basis is equal to use the Hadamard transform and a measurement in base B (90°,0°). So it results in something like this (Bob measures the first bit): $$H(\left|x\right>)\left|e\right>=\frac{1}{\sqrt{2}}((H\left|0\right>)\left|0\right> - (H\left|1\right>)\left|1\right>)$$ this comes to $$\frac{1}{2}(\left|00\right>-\left|01\right>+\left|10\right>+\left|11\right>)$$

But why does this result does not agree with Alice's bit? Why does the BB84 protocol expose 50% of cases (in my example)?

I hope I made understandable what I wanted to ask. I know that it is complicated. I would be very happy to receive an answer. Thank you!

Answer 1

Firstly, it's not entirely clear that your described eavesdropping strategy is the best there could be. But it is useful for trying to work through what's going on. As you say, If Alice sends the state $|-\rangle$ to Bob, then by Eve performing the controlled-not, they are left with $$ |-\rangle_x|0\rangle_e\rightarrow\frac{1}{\sqrt{2}}(|0\rangle_x|0\rangle_e-|1\rangle_x|1\rangle_e) $$ We can rewrite this as $$ \frac{1}{\sqrt{2}}(|+\rangle_x|-\rangle_e+|-\rangle_x|+\rangle_e). $$ Remember that these are the qubits that Bob and Eve hold at this point. Alice is expecting Bob to get measurement result $-\rangle$. But you can see from the way that I've written the state that Bob gets answer $|+\rangle$ 50% of the time, and $|-\rangle$ 50% of the time. So, half the time, his answer does not match Alice's expectation. However, Eve always knows exactly what answer Bob got by using the same measurement basis (once it's been announced) because her answers are always the opposite of what Bob gets.

Answer 2

Your strategy only works if Alice and Bob encode their message on the computational basis, for in that case CNOT copies the qubit's state with perfect fidelity. In general, however, the fidelity of the CNOT as a copy machine depends on the initial state. In particular, copying eigenstates of Pauli $X$ produces the lowest fidelity possible: $50%$.

When you individualize each qubit in a Bell pair, you find that they are described by an ensemble that has uniform distribution over all possible qubit states. In other words, you cannot possibly obtain any information about the qubits by just measuring one of them. A better attack would use a phase covariant copying machine.