I happened to learn that encryption by quantum cryptography would be impossible to break as it's state changes the instant of an eavesdropping event(by non-quantum systems) occurs. But I'm not sure if it's still held in the case of the eavesdropper using a quantum computer(not sure how that works either) as well. Couldn't find the answer on Google so felt the community could shed some light on this.
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$\begingroup$ it seems the question I'm asking maybe pointless as quantum computers are computers capable of factoring large numbers efficiently. So I guess quantum cryptography is unbreakable after all...or is it? $\endgroup$– AllenSep 10, 2018 at 16:46
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$\begingroup$ Quantum computers use quantum physics to accelerate certain calculations like factoring. Quantum cryptography uses quantum physics to create a channel to transmit information in such a way that eavesdropping causes the information to change which then can be detected. $\endgroup$– j.p.Sep 11, 2018 at 6:02
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I happened to learn that encryption by quantum cryptography would be impossible to break as it's state changes the instant an eavesdropping event (by non-quantum systems) occurs.
Actually, the QKD system doesn't assume that the eavesdropper is nonquantum; he is allowed to attempt to generate an entangled state with the qubit being transmitted. However, even the act of generating such an entangled state would alter the received bit (with a probability that is bounded away from zero), hence any such eavesdropper is detectable, even if the eavesdropper is allowed a quantum computer.
[Insert standard skepticism about whether QKD is a practical system, compared to the alternatives]
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$\begingroup$ [Also insert standard reference to fgrieu's detailed discussion on the topic] $\endgroup$– SEJPMSep 11, 2018 at 8:50
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$\begingroup$ There is a man in the middle attack for the Quantum key exchange. $\endgroup$– kelalakaSep 12, 2018 at 6:00
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1$\begingroup$ @kelalaka: how is there a MITM attack (assuming that the attacker cannot fake the authenticated parts of the exchange)? Or, are you talking about a side channel attack (where the attacker gets information beyond the actual qubit)? $\endgroup$– ponchoSep 12, 2018 at 10:12