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TL;DR     The answer is classical cryptography. Besides a quantum link, secure data communication with Quantum Cryptography (more precisely, Quantum Key Distribution) uses classical links, mathematically provable classical cryptography, and a setup procedure using initially trusted material just as in classical cryptography. To perform the same, classical ...


20

When we say a solution is impractical, we don't mean that it can't work at all; instead, we mean that it has serious deficiencies compared to other ways to solve the same problem. For example, one could use an airplane to go visit your neighbor who lives 100 meters away; we would all agree that there are considerably easier methods, and so using the ...


11

There are a few key distinctions to make Quantum cryptanalysis This is what you hear all the buzzing about. Specifically, there is something called Shor's algorithm, that when used to break modern crypto, can be devastating. If you've encrypted a zip file and told someone the key you're quite safe. But things like PGP and SSL, where you have to agree to a ...


10

“Both 0 and 1 simultaneously” is a lie-to-children: it's a simplification intended to be comprehensible and not fully accurate. It's not a very good one. A better way to present it is that a qubit is partly 0 and partly 1. It's not a single number, it's a function that takes values between 0 and 1. As for why… it's physics. Classical physics says that a ...


10

@fgrieu already wrote a little book, so I'll restrict my answer to a minimum to avoid repetitions. Think of this as an extended comment (which indeed wouldn't have fit the comment size limits). What makes Quantum Cryptography secure? … what makes it more secure than the classical version? In classical crypto, things like three party key distribution ...


9

There are several kinds of quantum key distribution (QKD) protocols as of today. Are you looking for a particular one? The best-known QKD protocol goes by the name BB84 after its inventors Bennett and Brassard and the year in which they presented their work. Searching on the Internet, I found this link http://fredhenle.net/bb84/demo.php with a simulation ...


8

You have to distinguish between "theoretically" and "practically" breakable. Pretty much any cryptosystem can be broken in practice by violating the assumptions that the theoretical security proof uses. For example, with Quantum Key Distribution and a OTP, you might have theoretical perfect secrecy. However, the proof for such a scheme ...


8

Quantum key distribution takes advantage of physics to create a communication channel that can't be cleanly intercepted without corrupting part of the message. This can be used to create a shared secret key for a one-time pad to be used over a classical connection. Particles may have a quantum state which can be thought of as having multiple bases (such as ...


8

Let’s take your questions in order. Note that I’m a physicist working in quantum cryptography, so my opinion on this might be biased 1. What about authentication ? The classical channel between Alice and Bob has to be authenticated in order for the protocol to work. Formally, this is a pre-requisite for quantum key distribution (QKD), and is not part of the ...


7

My turn! In classical cryptography between two peers over a channel such as the internet, an eavesdropper on the channel learns a transcript of information from which secrets could theoretically be derived. For example, the eavesdropper learns Diffie–Hellman public keys $g^a$ and $g^b$, which with unbounded computation could be used to recover the peer's ...


7

What's to guarantee authentication or message integrity (particularly when Alice and Bob are exchanging which filters were correct and so on)? A pre-authenticated classical channel is an essential requirement in addition to the quantum channel on which the quantum key exchange (QKE) is performed. This implies that Alice and Bob must share an initial secret ...


7

It's easier to work with a lot of photons in a stream than to work with a single photon at once. This is explained in the paper introducing continuous-variable quantum key distribution. (With apologies for the paywall. Ms. Greene, tear down this wall! Or read the arXiv preprint if you're not a rebel.) P.S. I originally answered the question as follows, ...


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In addition to the other answers, I want to add that D-Wave makes a type of machine called an Adiabatic Quantum Computer, which is fundamentally different from the general-purpose quantum computers capable of running Shor's algorithm. D-Wave's machines are good at optimization problems, but have zero applicability to cryptography. Moreover, it's not even ...


6

Impracticability of Quantum Key Distribution (QKD), starting with the two showstoppers QKD is network-adverse, because it requires setting up a QDK-compatible link between the two crypto endpoints. AFAIK the only two options are direct line of sight, and laying out an optical fiber. Ordinary fiber transceiver and switches are not usable, thus forget about ...


6

The answer is both yes and no, as I explain below. 1. No, BBM92 is better (or at least, we initially thought so) E91 was the initial idea which led to the more rigorous BBM92. In the E91 paper, there is not actual security proof, ant the attack is not really specified: if the CHSH inequality is violated by only a few percents (say $S=2.05$), how many secret ...


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The article cited in the question (Eleni Diamanti, Hoi-Kwong Lo, Bing Qi, and Zhiliang Yuan: Practical challenges in quantum key distribution, in npj Quantum Information 2016, a Nature Partner Journal) gives a list of challenges facing Quantum Key Distribution, but forgets several important ones: Not fulfilling the market demand for simplified key ...


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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, ...


5

How Alice and Bob agree on which of the two beam splitters bases (Horizontal-Vertical and Right-Left) correspond to which of the classical bits (1 and 0) ? How do Alice and Bob agree on when actual transmission starts, when it is time to send/receive a photon taking note of event ? Alice and Bob operate according to an agreed-upon protocol, which can define ...


5

There is some confusion regarding QKD. The confusion revolves around the underlying principle and it's nuts & bolts implementation in the physical world. The two are conflated, which I believe is unfair as QKD is an emerging technology so shouldn't be so harshly compared with centuries year old cipher principles. So in this quantum version, what makes ...


5

Since the question was labelled unclear, I’ll first clarify my understanding of the question, and then give various motivations (academics, then practical). This answer turned out to be quite long Full disclosure: I am an academic who has worked on continuous variable quantum cryptography since my PhD thesis (defended in 2003), so I am biased towards this ...


4

Actually recently I found out about a complete QKD simulation toolkit that has become available, accessible online via this link, QKD simulator. It is a parameter-based simulator, so different scenarios (qubit numbers, Eve's influence, etc.) can be set up and simulated.


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Bob and Alice then take the correct configurations, and they convert those to a prearranged code, which could be, as an example, that a correctly guessed diagonal detector is 0, and a correctly guessed rectilinear detector is 1. This way an encryption key will be composed. I believe that's where you got it a bit off. A correctly guessed rectilinear ...


3

The difference between cryptographic algorithms and quantum key distribution (QKD) is that the algorithms operate purely in the mathematical realm of information theory. QKD is a physical process. Thus, it can do things that algorithms cannot do. Let's say Alice wants to send a message to Bob without Eve getting the information. If Alice uses a ...


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That question is essentially A.1 of this lengthy post, with some elements of answer in B.1 For your convenience, here’s quoting those specific sections of his linked answer: A. Issues prevent direct use of sifted bits … Errors creep in the sifted bits The many imperfections of the model (heavily simplified compared to the actual ...


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Which QKD schemes don’t utilize an authenticated classical channel? None that can be trusted. All secure Key Distribution systems, quantum or not, require an authenticated channel at setup, before the distributed key can be trusted. This is typically a classical channel, like a person carrying (before the Key Distribution occurs) a paper with some secret ...


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Has anyone implemented the BB84 protocol and put it in production? There are a number of commercial QKD devices on the market, including ones by IDQuantique, QuintessenceLabs, MagicQ and SeQureNet. In addition, there are a number of other companies that are working on it (and may be actively marketing them as well). Some of these companies use an internal ...


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I do not immediately see how Zero Knowledge Proof could be used for secure transmission of information, and thus compete with Quantum Key Distribution (when trivially complemented with the One Time Pad for confidentiality and a Carter-Wegman hash for integrity). Much less do I see how traditional ZKP could work absent some unproved mathematical assumption (...


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Do the bits generated password using QKD protocols need to be converted to another base before being used as a key for the one time pad? No, if the OTP encrypts symbols that are bits, as does the most usual OTP. Yes, in all the other cases. One simple method is rejection sampling: if symbols can take $n$ values, we group input QKD bits by blocks of $b=\left\...


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As you know, most computers talk in binary " (with only 1s and 0s)". So $E_K(P) = C$ in any base as long as encryption function $E$ is a bijection. Therefore $E'_{K}(C) = P$ reveals the original message. As long as $E$ is deterministic and bijective, the numeric base is kinda irrelevant, but computers do love their octets. [This is not the quantum ...


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The density operator of a pure quantum state $|\psi\rangle$ is simply its outer product with itself: $|\psi\rangle\langle\psi|$. For each of the four BB84 states $|0\rangle$, $|1\rangle$, $|+\rangle$, $|-\rangle$ you should be able to write down the matrix representation of the corresponding operator in the standard computational basis.


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