BB84 attack with entangled qubits example

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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!

user4961

Posted 2018-10-29T16:21:40.073

Reputation:

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

Answers

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.

DaftWullie

Posted 2018-10-29T16:21:40.073

Reputation: 35 722

First, thank you for improving my question! My problem is actually to see Bobs qubits in the equation. After that Ive read so far Bobs qubits are these (blue) $ \frac{1}{\sqrt{2}}(\left|\color{blue}{+}-\right>+\left|\color{blue}{-}+\right>) $, is this correct? – None – 2018-10-30T08:14:15.790

@QuantaMag Yes. I'll just update my answer to make it clearer... – DaftWullie – 2018-10-30T08:14:55.883

Thanks, so than the 50% percent value is clear so far, because $ |\frac{1}{\sqrt{2}}|^2 = 1/2 $ that the first qubit is in state $ \left|+\right> $. Ok, the only thing I have not quite understood yet is your transformation to equation 2 – None – 2018-10-30T08:19:57.237

@QuantaMag Yes. You can at least verify my transformation by expanding the definitions of $|\pm\rangle$. Equivalently, take the original entangled state and apply Hadamard to both qubits. – DaftWullie – 2018-10-30T08:21:54.750

Ive expanded the definitions and come to your result. Only out of interest, is it also possible to show that in reverse? – None – 2018-10-30T08:59:32.747

@QuantaMag Yes, absolutely. It's exactly as you were trying to do - if you know you're going to measure in the $\pm$ basis, that's equivalent to applying Hadamards and measuring in the 0/1 basis. So apply Hadamards to both qubits. – DaftWullie – 2018-10-30T09:03:21.027

If I may ask again. To execute Hadamard on two qubits looks like this? $$ \frac{1}{\sqrt{2}}(\left|00\right>-\left|11\right>) $$ apply Hadamard transform on two qubits $$ \left|00\right>\rightarrow H \rightarrow \frac{1}{2}((\left|0\right>+\left|1\right>)(\left|0\right>+\left|1\right>))=\frac{1}{2}(\left|00\right>+\left|10\right>+\left|01\right>+\left|11\right>)$$ $$ \left|11\right>\rightarrow H \rightarrow \frac{1}{2}((\left|0\right>-\left|1\right>)(\left|0\right>-\left|1\right>))=\frac{1}{2}(\left|00\right>-\left|10\right>-\left|01\right>+\left|11\right>)$$ – None – 2018-10-30T10:23:04.460

all in all $$\frac{1}{\sqrt{2}}(\frac{1}{2}[(\left|00\right>+\left|10\right>+\left|01\right>+\left|11\right>) -(\left|00\right>-\left|10\right>-\left|01\right>+\left|11\right>)]) $$ $$\frac{1}{\sqrt{2}}[\left|10\right>+\left|01\right>] $$ – None – 2018-10-30T10:23:15.210

@QuantaMag Yes! So if you now measured in the 0/1 basis, you'd get either answer with 50:50 probability, and Bob and Eve always get opposite answers (from which Eve can infer Bob's outcome) – DaftWullie – 2018-10-30T10:52:05.817

Excellent! Thanks again for the extraordinary help! Now Ive understand this so far :) – None – 2018-10-30T11:13:20.303