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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!

Is the CNOT operation dependent on the basis used by the qubit sent from Alice to Bob? If Eve does not know the basis, how can he carry out a CNOT? – XXDD – 2018-10-29T16:26:43.583

I'm not sure what you mean exactly. I just assumed some conditions, for example that Eve uses the CNOT operation on Alice's qubit x to create an entangled state. – None – 2018-10-29T16:30:12.770

Maybe I have a misunderstanding. I will read your question carefully and respond later. – XXDD – 2018-10-29T16:35:34.923

Are you trying to clone the qubit from A to B by CNOT? The no-cloning rule says you can not succeed with 100 percent. – XXDD – 2018-10-29T16:50:01.770

Hey, I used cnot gate to entagle |x> in state |1> (alice) and Eves bit |e>. Eve has prepared a bit |e> in state |0>. Entanglement on |x>|e>-->|1>|0> works likes this "negate the second bit if first bit was 1" that means |1>|0> --> |1>|1> is the entangled state.

But in my second example I used another state for |x> but the same |e> – None – 2018-10-29T16:58:25.933

Yes, cnot can be regarded as a kind of copy operator. But it only works well when x is 0 or 1. if x is 0+1 or 0-1, then it's not a perfect copy machine, this is exactly what you find that the copy machine does not work perfectly. – XXDD – 2018-10-29T17:03:16.940

Suppose Alice qubit is in the state $ \left|x\right>=\frac{1}{\sqrt{2}}(\left|0\right>-\left|1\right>) $ and the cnot gate transfers this to the state $ \left|x\right>\left|e\right>=\frac{1}{\sqrt{2}}(\left|00\right>-\left|11\right>) $, what comes out if Bob measures the first Bit of the entangled state |x> in base (45°,-45°)? – None – 2018-10-29T17:07:59.393

0 or 1 with both p=1/2 – XXDD – 2018-10-29T17:31:47.207

Okay, maybe that's just the point where I'm stuck. Can you explain exactly how you come to your conclusion? The two qubit system confuses me a bit. Did you measured the first Bit of the entangled state |x> ? – None – 2018-10-29T17:33:37.040

just as you mentioned in your question, this can be understood as to first apply a H on the first qubit and then measure it in the 0/1 basis, the result is then 0 or 1 with p=1/2 – XXDD – 2018-10-29T17:40:02.150

so if I understand it right. Measuring the first qubit means: $ \frac{1}{2}(\left|00\right>-\left|01\right>+\left|10\right>+\left|11\right>) $, the propability for state 0 ist then |1/2|^2 + |1/2|^2. First qubit is 00 and 01 right? – None – 2018-10-29T17:46:49.337