# Is quantum teleportation fully secure (in principle)?

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?
• What makes you think anyone capable of producing, developing or come to that, even describing quantum teleportation would not be equally capable of cracking whatever security it had to offer? Oct 1 at 21:57

• 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.