Can you explain this statement? I understand that algorithms such as RSA might require particular padding or what have you to be secure in practice, but is quantum-resistant crypto much different?
What I'm talking about is "quantum cryptography", in which qubits are actually used in the protocol. Quantum cryptography is also known to be secure against quantum computers (indeed it's secure against arbitrary amounts of computing resources, unless our theories of physics are wrong).
It's also simpler than RSA or elliptic curves, so I hope that (after the kinks are worked out) it will also be less susceptible to bad implementations and side-channel attacks.
>Not to mention a small initial secret
This is interesting, what are you referring to here?
As soon as there's any device between you and the recipient that breaks the entanglement, all the guarantees of quantum encryption go out of the window, and it's possible to attack the comms at that retransmission point.
No, the retransmitters can preserve the entanglement (and A and B can verify that this has been done).
In fact (providing qubits can be stored) the transmission can be done indirectly and in advance.
The telecom company produces lots of entangled qubits and gives a bunch to each customer, keeping half of each pair for itself. Then when Alice wants to communicate securely with Bob they ask the telecom company to take the corresponding qubits and perform a joint measurement. This entangles Alice's qubits with Bob's (like making a connection at a telephone exchange). Then Alice and Bob can measure their qubits (in various bases) to create a one-time-pad.
The clever thing is that Alice and Bob can (by checking that on a portion of the qubits their results always matched when they used the same basis) verify that they did indeed have maximally entangled qubits and therefore no one was listening-in.
Does it change your thoughts on the utility of quantum cryptography?