17

Quantum key distribution requires that you wholesale replace your entire communications infrastructure built out of 5 EUR ethernet cables and 0.50 EUR CPUs by multimillion-euro dedicated fiber links and specialized computers that actually just do classical secret-key cryptography anyway. Plus you have to authenticate the shared secret keys you negotiate ...


11

If it is proven that a given asymmetric encryption protocol relies on a problem which cannot be solved efficiently even by a quantum computer, then quantum cryptography becomes largely irrelevant. The point is that, as of today, no one was able to do this. Indeed, such a result would be a serious breakthrough, as it would prove the existence of $\text{NP}$ ...


5

I emailed Artur Ekert to seek help for this quesiton, and he replied: There are different variants of the E91 protocol that may give you different efficiencies. In my original version the settings used for the keys bits were indeed chosen with the probability 2/9, but others optimised it in all kind of ways. So at least 2/9 is the probability of ...


5

One (obvious) application is the generation as True Random Number Generators, e.g. IDQ, or you can download some here Free True Random Numbers (please do not use these for security relevant application). In order to build such a TRNG, from a quantum circuit perspective, all you need is a single qubit, a Hadamard gate and a measurement. Although there might ...


4

Certainly not exhaustive, but to get the ball rolling... One possible application is blind quantum computation. In this, there is a user who wants to complete a computation, but only has the capability of producing single-qubit (non-entangled) states. These are sent to a server who can (locally) entangle them for the purposes of performing a measurement-...


4

I assume the paper you are reading is referring to information reconciliation. Information reconciliation is a vital part of post-processing in QKD, to limit (or erase in the best-case scenario) the amount of errors/differences between the key of Alice and Bob. In that sense, it is a form of (classical) error-correction, and, broadly speaking, it works like ...


3

1. Existing Products I've not yet come across commercial systems using entanglement based schemes. The commercial products I've seen, like ID Quantique and MagiQ seem to be using BB84/B92 type protocols. It may be relevant to the question to note that I'm only aware of these particular companies because their commercial QKD systems got hacked. The ...


3

I am glad you enjoyed my experiments! :) I'd be happy to talk more about how I ran that project --- dm me at twitter.com/crazy4pi314. To your question, I don't know of any good papers or articles on the setup, but you can get a pretty reasonable demo of polarization-encoded BB84 with a few pretty common components: polarized laser pointer some half wave ...


2

CW from self-answer Reviewing Farhi et al. on quantum money from knots, one can say that the Markov chain applied by the verification algorithm that walks along the Reidemeister graph is far from ergodic, as the graph includes many individual connected components corresponding to separate knots. Each bill corresponds to a uniform superposition over ...


2

Quantum key distribution (QKD) is a secure communication method that enables two parties to produce a shared random secret key known only to them, which can then be used to encrypt and decrypt messages. Quantum key distribution is only used to produce and distribute a key, not to transmit any message data. The algorithm most commonly associated with QKD is ...


2

Post-quantum crypto schemes run on classical computers and are hoped to be secure against quantum attacks. Quantum key distribution such as bb84 or e91 run on quantum hardware (although does not require the full power of a quantum computer) and is provably secure (subject to certain underlying assumptions about lab security etc) against quantum attack.


2

TL;DR: The efficiency is 2/9, not 25%. The Ekert 91 protocol involves many rounds. In each round, Alice and Bob share a Bell pair $$ (|00\rangle+|11\rangle)/\sqrt{2} $$ They both choose randomly which of 3 measurements to make. Alice chooses between the measurement bases $Z$, $(X+Z)/\sqrt{2}$ and $X$. Bob chooses between $(X+Z)/\sqrt{2}$, $X$ and $(X-Z)/\...


2

I'm not sure that I would claim the two situations are entirely equivalent - if Alice and Bob share Bell pairs, there are extra things they could do (e.g. testing Bell inequalities), but in terms of the actual protocol applied, they're equivalent. One way to see this is to think about the protocol (more or less) as described: Alice first sends halves of ...


2

The progress in DIQKD security proofs has been quite rapid in recent years. In particular, the approach of Vazirani and Vidick is no longer what the community uses. The two major approaches that I know of are using the entropy accumulation theorem or quantum probability estimation. These two approaches are quite similar in spirit but the entropy accumulation ...


2

Error correcting codes work the same on entangled qubits as any other qubit. All Alice and Bob have to do is separately encode their qubit into a Shor code. They each run an encoding circuit, apply noise, then run a decoding circuit and apply the corrections it inferred.


2

add the value of the first share to the second (modulo three), and then add the value of the second share to the first means you first perform, on the 2 shares, the 2-qutrit unitary which maps $$|00\rangle\to|00\rangle\\ |01\rangle\to|01\rangle\\ \ldots\\ |22\rangle\to|21\rangle$$ (i.e. adding the first share to the second modulo 3, so $2+2 = 1 \mod 3$, ...


1

I think your misunderstanding is in the wording of the secret key rate; The PLOB-Bound offers an asymptotic, ultimate-upper bound on the secret key rate per use of a lossy bosonic channel. This bound on the secret key rate is computed as a regularisation, where one considers the infinite limit of $ n \rightarrow \infty$ transmissions across the channel in ...


1

I do not know how the crosstalk matrix works. But the way I understand QBER is the following: QBER is the mismatch probability of the signals sent and receieved between Alice and Bob. Let Alice send 100 signals to Bob. Bob makes his own measurement choices and at the end, they compare their results. Let's say, in 60 of these 100 signals, Bob measured in the '...


1

What a Franson interferometer does is to superpose two-photon wavepackets generated at different times (within the coherence time of the pump, supposing we are generating a pair of entangled photons by e.g. SPDC). This superposition is realized by an unbalanced MZI where the time delay between short and long arm should be in principle orders of magnitude ...


1

To see intuitively why this protocol increases the entanglement after each iteration, we can work out an example where our initial state is say $\lvert 00\rangle+\lvert 11\rangle$. Upon passing through a 50:50 Beam Splitter, we get: $$|00\rangle+|11\rangle = |00\rangle+ a_{0}^{\dagger}a_{1}^{\dagger}|00\rangle \hspace{3mm} \text{transforms} \hspace{3mm}|00\...


1

I’m not exactly sure what is meant by “the bits they agree on”, but I would interpret it slightly differently to you. If Alice and Bob don’t agree on the same basis, they discard those bits, so they have a maximum of 6. If everything works perfectly, all 6 bits should be equal. Of course, in the real world, there are experimental imperfections and it could ...


1

In lecture 4 of O'Donnell's series on quantum computing, he introduces the Elitzur-Vaidman bomb tester, which is an interesting application of the quantum Zeno effect. O'Donnell introduces the bomb tester prior to discussing multi-qubit entanglement in lecture 5. In the Elitzur-Vaidman tester, a single qubit in a superposition can be used to probe and ...


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