# Why does this not break the security of quantum teleportation

Reposted from my post on the physics stack exchange, apparently this is a better place.

I am taking an course on quantum mechanics, and we have just encountered the quantum teleportation protocol that allows for the transfer of one qubit from Alice to Bob. I think I have a way for Eve to break the security of this protocol. For completeness, the protocol is as follows.

$$\textbf{Protocol:}$$ Alice and Bob, beforehand share the entangled state $$|\Psi_{AB}^-\rangle=\frac1{\sqrt{2}}\left[|0_A1_B\rangle-|1_A0_B\rangle\right]$$ which is known to both of them. Then Alice's quantum computer spits out a state $$|\chi_{A^{\prime}}\rangle=c|0_{A^{\prime}}\rangle+d|1_{A^{\prime}}\rangle$$ ($$c$$ and $$d$$ potentially unknown to her) which she wishes to send to Bob far away. To do this, she considers the entangled state $$|\Phi_{A^{\prime}AB}\rangle~=~c|0_{A^{\prime}}\rangle|\Psi_{AB}^{-}\rangle+d|1_{A^{\prime}}\rangle|\Psi_{AB}^{-}\rangle$$ which may be written in the Bell basis as $$|\Phi_{A^{\prime}AB}\rangle = \frac12[|\Psi_{A^{\prime}A}^-\rangle(-c|0_B\rangle-d|1_B\rangle)+|\Psi_{A^{\prime}A}^+\rangle(-c|0_B\rangle+d|1_B\rangle)+|\Phi_{A^{\prime}A}^-\rangle(c|1_B\rangle+d|0_B\rangle)+|\Phi_{A^{\prime}A}^+\rangle(c|1_B\rangle-d|0_B\rangle)]$$. Finally, Alice makes a Bell measurement, yielding one of the Bell states ($$|\Psi_{A^{\prime}A}^{-}\rangle,|\Psi_{A^{\prime}A}^{+}\rangle,|\Phi_{A^{\prime}A}^{-}\rangle,|\Phi_{A^{\prime}A}^{+}\rangle$$) and she then knows exactly which of the four possible options Bob's state is now in, sends this information along a (potentially unsecure) classical channel to Bob, who then either has the state Alice had to start, or knows what unitary transformation he must apply to his state to get Alice's original state $$|\chi_{B}\rangle=c|0_{B}\rangle+d|1_{B}\rangle$$.

Our text says that this secure in two specific ways.

$$\textbf{1}.$$ If the "quantum channel" (the channel by which Alice and Bob recieve their entangled state) is secure, then even if the classical communication is intercepted by Eve, she does not have the entangled state, and so cannot know what state Bob was sent.

$$\textbf{2}$$ Quoting our text "Suppose when the entangled particles are being sent to Alice and Bob, a third party, Eve, intercepted the particles. She could secretly record their states and along with the information from the classical channel could then read the message being sent. Eve could then send particles to Alice and Bob in hopes they do not recognise she has been eavesdropping. In doing this Eve would send particles to Alice and Bob that would have had their quantum states altered, and the entanglement between them would have been lost. Thus if Alice and Bob did tests on their entangled particles, they could check if the correlation between them is lost, thereby know that a third party is intefereing with their communication. As such this means of communication can be completely secure."

I believe there exists a way for Eve to evesdrop on the communication in such a way as she receives all information Bob receives, and it is impossible for Alice and Bob to discover her meddling. We shall assume that Eve has access to all quantum and classical communication, as in the worst case scenario.

$$\textbf{Eve's Interception Plan:}$$ Let us suppose that Alice and Bob always use the entangled state $$|\Psi_{AB}^-\rangle=\frac1{\sqrt{2}}\left[|0_A1_B\rangle-|1_A0_B\rangle\right]$$ for their communication (if they used a different state each time, they would need to communicate that information classsically, and Eve would know, and all is the same). When the entangled state $$|\Psi_{AB}^-\rangle=\frac1{\sqrt{2}}\left[|0_A1_B\rangle-|1_A0_B\rangle\right]$$ is sent out to Alice and Bob, Eve intercepts it, and throws it in the bin. She then creates a new three way entangled state $$|\Psi_{ABE}^-\rangle=\frac1{\sqrt{2}}\left[|0_A1_B1_E\rangle-|1_A0_B0_E\rangle\right]$$ and sends off a particle to Alice and Bob, who each think all is as before. Alice does all of her things as before, makes a Bell measurement on particle $$A$$ and $$A'$$ as before on what she thinks is the state $$|\Phi_{A'AB} \rangle$$ but is actually the state $$|\Phi_{A'ABE} \rangle$$. Alice sends off the result of her measurement to Bob (and Eve intercepts it) and after both Bob and Eve perform the appropriate unitary transformations on their qubits, Bob and Eve both share the entangled state $$|\chi\rangle=c|0_B 0_E\rangle+d|1_B 1_E\rangle$$, when bob thinks he has the state $$c|0_B\rangle+d|1_B\rangle$$.

I claim there is no way for Alice and Bob to detect Eve given this plan. In particular, Bob and Eve cannot perform EPR tests to distinguish between $$c|0_B 0_E\rangle+d|1_B 1_E\rangle$$ and $$c|0_B\rangle+d|1_B\rangle$$ and so the communication is not completely secure.

The only resolutions I have are as follows:

$$\textbf{a)}$$ Maybe, in order to be secure, Alice and Bob $$\textbf{need}$$ a secure classical channel to communicate which entangled state they share, so that Eve dosn't know. If they have this, then they're fine, but then they might as well send their secret message through this classical channel. (If they have a secure channel, they could build up a secure bank of entangled states and then use them when their channel becomes insecure- this should work). Is this true?

$$\textbf{b)}$$ If Eve get's many qubit from Alice in this way, they are only useful to her after she has performed some measurements on them. If she, for example, receives many many qubits of the form $$c|0_B 0_E\rangle+d|1_B 1_E\rangle$$, and performs a series of experiments on them to extract their information, before Bob does, this will collapse Bob's states (though he doesn't know). When Bob performs experiments on his qubit, the outcomes of the experiments (depending on which experiments he chooses!) will already be pre-determined. I feel like if this is the case, then maybe Bob could perform some Bell inequality tests to get a probibalistic idea of if Eve has intercepted the communication. However, just because Eve has put Bob's qubits into known states, that doesn't mean that the outcome of ALL experiments are predetermined on Bob's qubit, only some of them, and so without knowing exactly what experiments Eve performed, this plan may not work. Will this work?

• Can you update the posting with a circuit of what you propose? I think you are violating the monogamy of entanglement when you have Eve create $|\Phi_{A^{\prime}AB}\rangle~=~c|0_{A^{\prime}}\rangle|\Psi_{AB}^{-}\rangle+d|1_{A^{\prime}}\rangle|\Psi_{AB}^{-}\rangle$. Mar 7 at 20:04
• I don't see where she creates A'AB, in the story she creates $|\Psi_{ABE}^-\rangle=\frac1{\sqrt{2}}\left[|0_A1_B1_E\rangle-|1_A0_B0_E\rangle\right]$. And that's easy to make with a CNOT gate controlled by B and a blank as input for E. Actually that's simplest without throwing the received state in the bin, but instead using it: just combine with a blank $|0_E\rangle$ and apply CNOT on (B,E). Mar 7 at 21:40

It is possible for Eve to do this, but then you don't achieve teleportation. Bob does not recieve the pure state $$c|0_B\rangle+d|1_B\rangle$$, which was the goal.
Bob can measure things that only depend on the amplitude ratio of $$|c|$$ and $$|d|$$ but nothing that depends on the phase relation, because that is lost in the entanglement with Eve. Instead of two real numbers of information, as a pure qubit state would have, he has only one, in the density matrix of his mixed state: $$\begin{bmatrix} |c|^2 & 0 \\ 0 & |d|^2 \end{bmatrix} = \begin{bmatrix} |c|^2 & 0 \\ 0 & 1-|c|^2 \end{bmatrix}$$ and actually the same is true for Eve. Both have half the information (the same half, in fact). And if Eve and Bob would team up (it might happen...) then Eve's state would purify Bob's and they would again have the complete information.