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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!

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  • $\begingroup$ 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? $\endgroup$
    – XXDD
    Oct 29, 2018 at 16:26
  • $\begingroup$ 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. $\endgroup$
    – user4961
    Oct 29, 2018 at 16:30
  • $\begingroup$ Maybe I have a misunderstanding. I will read your question carefully and respond later. $\endgroup$
    – XXDD
    Oct 29, 2018 at 16:35
  • $\begingroup$ 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. $\endgroup$
    – XXDD
    Oct 29, 2018 at 16:50
  • $\begingroup$ 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> $\endgroup$
    – user4961
    Oct 29, 2018 at 16:58

2 Answers 2

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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.

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  • $\begingroup$ 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? $\endgroup$
    – user4961
    Oct 30, 2018 at 8:14
  • $\begingroup$ @QuantaMag Yes. I'll just update my answer to make it clearer... $\endgroup$
    – DaftWullie
    Oct 30, 2018 at 8:14
  • $\begingroup$ 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 $\endgroup$
    – user4961
    Oct 30, 2018 at 8:19
  • $\begingroup$ @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. $\endgroup$
    – DaftWullie
    Oct 30, 2018 at 8:21
  • $\begingroup$ Ive expanded the definitions and come to your result. Only out of interest, is it also possible to show that in reverse? $\endgroup$
    – user4961
    Oct 30, 2018 at 8:59
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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.

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