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Quantum computing allows us to encrypt information in a different way compared to what we use today, but quantum computers are much more powerful than today's computers. So if we manage to build quantum computers (hence use quantum cryptography), will the so-called "hackers" have more or fewer chances of "hacking" into the systems? Or is it impossible to determine it?

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If you are talking specifically about quantum key distribution (quantum cryptography being an umbrella term that could apply to lots of stuff), then once we have a quantum key distribution scheme, this is theoretically perfectly secure. Rather than computational security that much of current cryptography is based on, quantum key distribution is perfectly secure.

That said, it is only perfectly secure subject to certain assumptions relating mainly to lab security. These same assumptions are essentially present in the classical case as well, just that because the quantum experiments are a lot more fiddly, it might be harder to be completely on top of all the possible attacks. Realistically, these are already the directions in which cryptography is attacked, rather than trying to brute force a crack. For example, exploits relying on a bad implementation of a protocol (rather than the protocol itself being flawed).

What quantum crypto, or post quantum crypto, is aiming to do is sidestep the loss of computational security implied by a quantum computer. It will never avoid these implementation issues.

In a completely shameless plug, you might be interested in my introduction to quantum cryptography video. It talks a little about this computational vs perfect security question (although doesn't really talk about possible hacks of QKD).

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  • $\begingroup$ I just had a short presentation about BB84 which is a quantum key distribution scheme. So my understanding is that this is "safer" (even stronger it is "provably secure") than any classical cryptography. So I am just wondering what you mean by "because the quantum experiments are a lot more fiddly, it might be harder to be completely on top of all the possible attacks". $\endgroup$
    – JanVdA
    Jun 15, 2018 at 17:49
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    $\begingroup$ @JanVdA BB84 and other schemes might be "provably secure", but it's provably secure subject to assumptions. Particularly, that both parties have secure labs. For instance, if someone else can see your monitor, they can read your decrypted message. It doesn't matter how good your cryptography is. But in the quantum case, there are some very subtle (hardware dependent) things that can happen. that yield something similar. For instance, in BB84, if it takes some time to change the basis in which you're sending photons, time analysis can reveal what basis is being used. $\endgroup$
    – DaftWullie
    Jun 15, 2018 at 17:56
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Most attacks now on classical computers don't actually break the encryption, they trick the systems / communication protocols into using it in a weak way, or into exposing information via side channels or directly (via exploits like buffer overflows).

Or they trick humans into doing something (social engineering).

I.e. currently you don't attack the crypto itself (because things like AES or RSA are very well tested), you attack the system built around it and the people using it.


All of these avenues of attack will sill be present when computers communicate via quantum encryption. However, with quantum encryption theoretically giving perfect security instead of just computational security, tricking a system into weakening its encryption (by using weak keys or wrong keys or keys you already know) shouldn't be a problem.

Possibly there will be weaknesses that systems need to avoid in quantum crypto, especially practical implementations that work over imperfect channels.


TL:DR: When quantum computers can practically attack RSA and the world switches over to quantum crypto for communication without a pre-shared secret, we'll be back in the same situation we are now: the crypto itself is not the weakest link.

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    $\begingroup$ Welcome to quantum computing! It's worth adding/mentioning that in theory, there is the potential possibility of a quantum computer someday cracking classical encryption entirely. However, I agree that this answer explains that actual issues in cryptography in that the encryption method is no longer (currently) the weakest link $\endgroup$
    – Mithrandir24601
    Jun 14, 2018 at 23:05
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    $\begingroup$ @Mithrandir24601: I meant currently you don't attack the crypto, thanks for pointing out the ambiguity in the way I put it. Updated to clarify. $\endgroup$ Jun 14, 2018 at 23:23
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There is one big difference between quantum cryptography and quantum computing. Yes, both belong together but when talking about quantum computing and cryptography most people mean implementing a quantum algorithm to attack currently used cryptographic schemes (RSA, ECC, AES) using factoring, calculating discrete logarithms or implementing a fast search algorithm which searches in O(sqrt(N)). Quantum cryptography uses entanglement to create correlations in a state and uses this as a resource for 1) detecting whether there is an attacker on the channel (QKD), 2) secret sharing, anonymous transmission etc in quantum networks and 3) verifying that the network state distributed is really the state (or sufficiently close to the state) which is desired and necessary to implement a certain protocol. Yes, the results can also be used for theoretical foundations in quantum information/computation but still both are different. I think there is one quantum key exchange protocol which does not rely on entanglement and only uses the No-cloning theorem for its security proof but I am not entirely sure about that. The security of QKD relies on detecting when the key is intercepted and throwing out that part of a potential key. The security after the key exchange relies on the one time pad which is mathematically secure when using a truly random key and using it only once. The randomness is again guaranteed by creating entropy for a quantum source in the first place. Using it once is upon the user.

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