Why does this not break the security of quantum teleportation

Question

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?

Answer 1

I am going to assume in this answer that you're concerned about the security of quantum cryptography not teleportation.

You say that Eve gives Alice and Bob the following state: $$|\Psi_{ABE}^-\rangle=\frac1{\sqrt{2}}\left[|0_A1_B1_E\rangle-|1_A0_B0_E\rangle\right]$$ instead of $$|\Psi_{AB}^-\rangle=\frac1{\sqrt{2}}\left[|0_A1_B\rangle-|1_A0_B\rangle\right].$$

The state $|\Psi_{AB}^-\rangle$ is a pure entangled state Alice and Bob and they can check that it is pure an entangled by performing measurements on repeated instances of the same state, which are required to send a message anyway.

The reduced density matrix for Alice and Bob in the state $|\Psi_{ABE}^-\rangle$ is $$\rho_{AB}=\tfrac{1}{2}(|0_A1_B\rangle\langle 0_A1_B|+|1_A0_B\rangle\langle 1_A0_B|),$$ which isn't a pure state and can be distinguished from the state $|\Psi_{AB}^-\rangle$ by repeated measurements of the same state. Also, the state for Eve is given by $$\rho_E = \tfrac{1}{2}(|0_E\rangle\langle 0_E|+|1_E\rangle\langle 1_E|)$$ and measuring this state gives the holder no information about Alice's or Bob's state. This might prevent Alice and Bob from sending information, but it doesn't allow Eve to intercept messages.

Entanglement doesn't magically give you information about systems you haven't interacted with. Rather, it allows you conceal and unlock quantum information in a controlled manner:

https://arxiv.org/abs/quant-ph/9906007

Answer 2

I would like to show how your text is correct in saying that the teleportation protocol is secure. However, it needs a bunch of Bell pairs for this -- not just one.

Let's say Alice and Bob have decided to use $n$ Bell pairs for each bit of information they want to communicate. They (try to) get $n$ Bell pairs distributed between themselves using the quantum channel. However, having access to the quantum channel, Eve sneakily entangles each Bell pair to a third qubit that she keeps with herself and thus Alice and Bob end up having two parts of a three-way-entangled three-qubit system rather than a Bell pair. I have sofar only described your attack proposal in my own words. Here comes the security plan of Alice and Bob:

0. Alice and Bob keep track of the order in which the qubits arrive to them and index them accordingly (say, by putting the $i^{\rm th}$ qubit in a box labeled $i$). So, ideally, the $i^{\rm th}$ qubit of Alice and the $i^{\rm th}$ qubit of Bob belong to the same Bell pair for each $i$.
1. Alice generates $k$ distinct random integers between $1$ and $n$ and broadcasts them over a public channel. Alice also picks a random direction $\hat{n}$ and broadcasts this over a public channel. For each random number $i$ communicated over the public channel, Alice and Bob measure their $i^{\rm th}$ qubit along $\hat{n}$ (I am adopting a language in which the qubit is a spin$-1/2$ particle and thus the basis of measurement is in one-to-one correspondence with a direction in $3d$ space).
2. They repeat the previous step multiple times wherein they measure randomly selected qubits (among the unmeasured ones) in a randomly selected direction -- and broadcast their results.
3. After having repeated this procedure a sufficient number of times, they can construct the density matrix of the two-qubit systems that were distributed to them over the quantum channel. If this is a pure-state density matrix, they decide to communicate. If it is not, they terminate their plans.

The important thing to notice here is that there is nothing that Eve can do here that can prevent Alice and Bob from finding out that their Bell pairs had been messed with because Eve would have to affect the density matrix of the two-qubit systems distributed b/w Alice and Bob and this is something that cannot be done by Eve. You cannot impact remote density matrices even if you have some entanglement with them! So, if Eve messes with the Bell pair distribution, Alice and Bob will find out.

The security protocol that I describe here is quite inefficient and it can be improved a lot by deciding to do tomography more smartly. However, I hope the central idea is conveyed.

PS: I'd like to point out that in my understanding, the purpose of quantum teleportation is not quantum cryptography. Rather, it's just transportation. You want to, in effect, transport qubits over classical channels because high-quality quantum channels are very hard to maintain. I assume there is some expectation of security for all systems but I have not really seen a discussion of security in the context of teleportation.

Answer 3

It's Alice who generates all the 3 qubits and entangles them pair-wise. Alice only sends qubit $B$ to Bob. So, the attack scenarios do not work.