Is quantum teleportation fully secure (in principle)?

Question

Suppose two parties Alice and Bob have access to as many EPR pairs as needed, with each one having a qubit of each pair. Now Alice can create a sequence of qubit states $|i_{1}\rangle, |i_{2}\rangle, |i_{3}\rangle, \ldots\in\{|0\rangle, |1\rangle\}$ encoding a message. Then Alice groups each qubit of their message with an EPR qubit and performs repeated Bell measurements on them. To complete the teleportation, Alice has to send classical bits to Bob to tell them how to adjust their share of qubits. Lastly, all Bob has to do is measure their qubits with respect to the computational basis $\{|0\rangle, |1\rangle\}$ (let's say both parties decided ahead to time to encode/decode through the computational basis) and the message $i_{1}i_{2}i_{3}\cdots$ is recreated for Bob.

Now in the intermediate step, Alice has to send classical bits to Bob, and this is where a spy can intercept the info. But as far as I can naively tell, there is no correlation between what bits you need to send to Bob with what the state that you are trying to teleport. Is this correct? If so, doesn't this mean that no spy can deduce the message being teleported by intercepting the classical bits?

For some reason, I've not been able to find any sources that affirm or deny this explicitly. So I am wondering:

• Can a spy obtain the message in any way, shape, or form?
• Does the removal of idealization assumptions in the scenario change the answer?
• Are there any references (papers, textbooks, etc.) that comment about this question?

Answer 1

Adding to Mateus Araújo's answer:

An easy way to see that the classical bits are uncorrelated with the message in the idealized protocol is that any correlation would amount to a measurement of the teleported state, which would mean the teleportation didn't work properly.

If you just want to send a classical message, there is an easier way: measure one Bell pair per bit along some agreed-upon axis and use the outcomes as a one-time pad. This only requires half as many bits to be sent.

Quantum teleportation, like the classical one-time pad, is also subject to a man-in-the-middle attack if the attacker can mess with the Bell state generation process. This gives the attacker full access to the quantum channel as though it was an ordinary physical channel. They are still limited by the no-cloning theorem, but they could read and alter your classical message.

Answer 2

• Yes, in the ideal case it is perfectly secure, the probabilities of each outcome in the measurement in the Bell basis are 1/4 independently of the input state.
• Indeed it does. For example, suppose that the initial entangled state Alice and Bob share is not the maximally entangled state, but the partially entangled $\cos \theta |00\rangle + \sin \theta |11\rangle$. Then the probabilities of obtaining $\phi^+, \phi^-, \psi^+, \psi^-$ in the Bell basis measurement are, respectively, $\cos^2 \theta, \cos^2\theta, \sin^2\theta,\sin^2\theta$ if Alice wants to send $|0\rangle$, and $\sin^2\theta,\sin^2\theta, \cos^2 \theta, \cos^2\theta$ if Alice wants to send $|1\rangle$. Now the classical message is correlated with the bit Alice wants to send. A similar problem appears if the Bell state measurement is not ideal.
• I'm not aware of any reference about it, but I bet one exists just because teleportation is such an old and well-known protocol.