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!