Great Idea...
...lets bring another single point of failure (charlie) into the mix!
Over recent years, the gap between theoretical security of quantum crytography and practical implementation has provided plenty of fun for super-geniuses the world over. Yes, quantum cryptography is supposed to be unbreakable. After all, if anybody even observes the state of a qubit that Alice has prepared, entangled with …
They're all derived by Bruce Schneier I think.
WP says the names Alice and Bob were first used by Ron Rivest (the 'R' of "RSA") in 1978, and that other names have been added as need arose. The popularity of the naming scheme may have increased as a result of John Gordon's oft-quoted after-dinner speech in 1984. Bruce scheier has certainly used them, not least in Applied Cryptography (1994).
http://en.wikipedia.org/wiki/Alice_and_Bob
Beer, because there's no port icon and a good after-dinner speech deserves a full glass.
"They're all derived by Bruce Schneier I think."
This sentence could be:
"They're all devised by Bruce Schneier I think." or:
"They're all derived from Bruce Schneier I think."
Which is it?
Personally I think that the first makes more sense though both are reasonably correct, grammatically.
Because when Charlie sends his yes or no to Bob, then Bob will know the polarisation state of the photon he has sent to Alice. Therefore a measurement has taken place and the quantum state of the other photon
(the one Alice gets) will collapse to the opposite state. So instead of needing to intercept the signal at Bob's end you only need to intercept it at 'Charlie's place' to intercept the key.
You are not supposing for a second these guys haven't thought about this, do you?
We are not in classical space here.
I will have to read the paper later but I suppose that in this case you "measure" whether two qubits are equal or not without measuring the qbits...
So, your original state is one of |00>, |01>, |10>, |11>
Then you ask the question: is the state one of (|01>, |10>) or one of (|11>, |00>)
This gives you the classical Yes/No bit, at which point your photon pair is still either
(|01>, |10>)
or
(|11>, |00>)
But you don't know which, and A and B will still have to measure to see whether the 0 or the 1 is on their end. Charlie doesn't know anything helpful about that.
We need the "are you serious, commenter" icon ASAP-
And Charlie can't possibly have malicious intent, instead of comparing Alice' polarization with Bobs polarization to his own polarization for each, thus create false keys and enable a suitably equipped third party, lets call him Mallory, to be the man in the middle with his new keys, which are viable to transceive signals to Alice and Bob?