# Why the quantum entanglement doesn't break quantum cryptography

If the Alice sends to Bob an q-bit $\alpha\left|\mathbf{0}\right> + \beta\left|\mathbf{1}\right>$ in quantum key exchange (such as BB84) it is assumed that Eve cannot copy the state (due to no cloning theorem) and attempts to read the state requires guessing the base which would allow to detect interception.

However if the Eve put $\left|\mathbf{1}\right>$ through C-Not gate than the state would be:

$$\phi_0 = (\alpha\left|\mathbf{0}\right> + \beta\left|\mathbf{1}\right>)\otimes\left|\mathbf{1}\right>$$ $$\phi_1 = \alpha\left|\mathbf{00}\right> + \beta\left|\mathbf{11}\right>$$

Now Eve can store her q-bit somewhere and send the 'original' one to Bob. Since the base disclosure is on public channel she can measure her q-bit in correct base (the state would be collapsed but it shouldn't matter).

Why then quantum entanglement doesn't break quantum cryptography?

Edit: To clarify what I understand (I haven't read the text about density matrices).

Alice sends to Bob two qubits - $\left|\mathbf{0}\right>$ and $\left|\mathbf{+}\right>$. Let's assume they are not discarded then in first case after Eve entangle it's qubit with sent qubit the state in case one:

$$\phi_{\left|\mathbf{0}\right>} = \left(\begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right)\left(\begin{array}{r} 0 \\ 1 \\ 0 \\ 0\end{array}\right) = \left(\begin{array}{r} 1 \\ 0 \\ 0 \\ 0\end{array}\right) = \left|\mathbf{00}\right>$$

Now the only possible value read by Bob is $\mathbf{0}$ assuming he guessed base. Eve can then overhear the disclosure of bases and measure the reduced state on previously stored entangled qubit.

In second case:

$$\phi_{\left|\mathbf{+}\right>} = \left(\begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right)\frac{1}{\sqrt{2}}\left(\begin{array}{r} 0 \\ 1 \\ 0 \\ 1\end{array}\right) = \frac{1}{\sqrt{2}}\left(\begin{array}{r} 1 \\ 0 \\ 0 \\ 1\end{array}\right) = \left|\mathbf{++}\right>$$

Assuming that the Bob guessed the base correctly the Eve would also measured $\mathbf{+}$ in similar way as previously. If Bob didn't guessed it correctly Eve can discard particular qubit.

Now if quantum cryptography works:

• For some reasons the entanglement is forbidden (if yes - what is the difference between sending qubit and processing it inside quantum computer?)
• Reduction of state by Bob does not happen in a way I think (possibly due to not knowing the density matrix).
• I don't understand the question. What exactly are you proposing that Eve do? – Rhys Mar 11 '13 at 11:32
• @Rhys: Use C-Not (controlled Pauli-X) gate to get from state $\phi_0$ to $\phi_1$ (using own q-bit and intercepted q-bit). Then she gets to entangled q-bits and she sends one of them to Bob. Then after Bob communicates bases to Alice she knows how to measure the second one. – Maciej Piechotka Mar 11 '13 at 11:57
• Okay, I get it now. In the meantime, Emilio has given a good answer! I think you are basically trying to violate the no-cloning theorem... – Rhys Mar 11 '13 at 13:17
• @Rhys: Probably - but I'm not sure how the Eve interaction would differ from fanout/fanin resulting in error-correction. I'm not claiming to break the quantum cryptography - I just don't understand what is the difference. – Maciej Piechotka Mar 11 '13 at 14:39
• The basis disclosure happens after Eve has lost her opportunity to clone the qubit. – Peter Shor Mar 11 '13 at 15:02

Since Bob can't receive information from Eve (which would mean Eve giving their game away), Bob effectively receives the qubits in the reduced state $$\rho_\text{Bob}=\text{Tr}_\text{Eve}\left[\left(\alpha|\mathbf{00}\rangle+\beta|\mathbf{11}\rangle\right)\left(\alpha^\ast\langle\mathbf{00}|+\beta^\ast\langle\mathbf{11}|\right)\right] =|\alpha|^2|\mathbf{0}\rangle\langle\mathbf{0}|+|\beta|^2|\mathbf{1}\rangle\langle\mathbf{1}|.$$ Notice that the intervention of Eve has completely killed the off-diagonal elements in this density matrix. For the case of $\alpha=\pm\beta$ this state is the completely mixed state, and Bob can extract no information from it. Thus, even if Alice sends $|+\rangle$, Bob might measure $|-\rangle$, and when they compare results they will find that an unacceptable fractionof their measurements don't match.

• Why does he received the qubits in reduced state? Or rather why what Eve does differs in any way from quantum error correction. All she does is adding a qubit (hidden from Alice & Bob) and putting it through CNOT gate. (And as side note - Bob does communicate with Alice which is intercepted by Eve before her measurement so Eve does not base in which to measure). – Maciej Piechotka Mar 11 '13 at 14:42
• Bob can only implement local measurements, so Bob-Eve entangling operators are out. That means that when tracing over the Bob+Eve spaces to get expectation values, the only relevant part is the reduced state. Maybe these notes will help. – Emilio Pisanty Mar 11 '13 at 17:07
• Ok. I see where I make an error (I took Bell state for $\left|\mathbf{++}\right>$ as pointed out by Peter). – Maciej Piechotka Mar 12 '13 at 0:38

After reading your clarification, I think the problem is that you cannot have a quantum map that takes

$$| 0 \rangle \rightarrow |00 \rangle \qquad \mathrm{and} \qquad | + \rangle \rightarrow |++ \rangle .$$

Isn't this cloning?

The state $|++\rangle$ is $\frac{1}{2}(1,1,1,1)$.

• Doesn't a $CNOT$ gate work for these two special cases? – orome Mar 11 '13 at 23:59
• No. $\text{CNOT}_{1\rightarrow2}|+\rangle|0\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)\neq|+\rangle|+\rangle$. – Emilio Pisanty Mar 12 '13 at 0:42