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5

No, weak measurement and quantum tomography don't break BB84. I recommend that you create an explicit quantum circuit that implements the weak measurement or the quantum tomography, and check for yourself that it actually fails. The basic problem comes down to the fact that there is a trade-off between how much information you get and how likely you are to ...


3

Tomography generally speaking uses a collection of measurements to reproduce an underlying state. So you experimentally reproduce the same situation over and over, collect statistics and find the most likely estimate for that state. In QKD, information is sent once and doesn't repeat. So for each unit of information, you're never ever able to collect enough ...


3

In schemes like E91, the idea behind using an entangled state is that: in a particular measurement basis (for both parties), the measurement outcomes are perfectly correlated but completely random (50:50 outcomes). you can perform a Bell test on the state to verify its nature. Using a maximally entangled state gives you the property of the 50:50 outcomes (...


3

It's the dimension of the Hilbert spaces. In DV-QKD you have a finite dimensional Hilbert space (like a qubit). Thus your measurement outcomes come from a finite set. On the other hand a CV-QKD protocol uses infinite dimensional systems and therefore you can have a continuum of measurement outcomes. If there's no specification as to whether a QKD scheme is ...


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

Indeed, the PLOB bound is an ultimate upper bound for repeaterless quantum communications, and is thus derived by averaging over $n\rightarrow \infty$ uses of the communications channel - Hence why the capacity is given as bits per channel use. If you wish to more closely compare the PLOB bound to other well known protocols, look at Supplementary Note 6 in ...


1

Quantum computing is so far away from anything that would ever have to worry about security risks. It's very much debated how far away we are from having a quantum computer, but I personally think it'll be decades before "security risks" are even worth speculating about. There is an effort to build a "quantum internet" which connects ...


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

As far as I know, the error threshold depends on the security proof. Meaning, BB84 has different security proofs, each with different assumptions, resulting in different security threshold. The 'optimal' bound is the one that has the best 'error-rate'. Meaning, what is the highest noise(error) that can be tolerated, and still, the protocol can be proved ...


1

They did use them in the Vienna loophole-free Bell test, which is a pre-condition for implementing DIQKD. Their benefit is that you can do a loophole-free violation of the CHSH inequality efficiency down to $2/3$, whereas if you use maximally entangled states the required efficiency is much higher, $2\sqrt2-2$. Therefore I fully expect them to be used for ...


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