On Mon, Dec 02 2013, Jim Bell wrote:
My fiber optic invention has a certain relevance here. A typical modern germania-doped-core (GeO2) silica optical fiber has a loss of about 0.19 decibels/kilometer (db/km). Over 50 km, the loss is (50 km x 0.19 db/km) = 9.5 db, ignoring splice losses. (A good splice has a loss of about 0.10 db.) So, the quote above is indicating that above a loss of about 10 db, a quantum system is hard to maintain. I have suggested in my patent application that isotope-modified fiber (where the Si-29 level is brought from nature's 4.67% (atom/atom) to 0.10 %, the loss might decrease by a factor of 10 to 20. This means that the ultimate distance limit might increase to 50 x 10 = 500 km, to 50 x 20 = 1000 km. That would be a major improvement if it works. The reason that this new fiber would be necessary is this: Ever since the invention of the EDFA (Erbium-doped fiber amplifier http:/ /en.wikipedia.org/wiki/Optical_amplifier ) in 1986, it has been used to amplify IR signals in the 1510-1560 nanometer band. Using it and ordinary signals (not quantum signals) it is possible to go about 125 kilometers between amplifiers. (In other words, that usage tolerates about 25 db of optical loss before an EDFA is necessary.) However, apparently an EDFA cannot be used to amplify a quantum system. http: //www.nict.go.jp/en/press/2010/02/08-1.html Or, at least, not directly.
While reducing loss will certainly help, the NICT paper you link to will
not. Quantum cryptography relies on only having a single entangled
particle going to each end, so that if anyone intercepts either particle
and attempts to measure whatever property you're using to derive the key
(polarization generally), the keys will not match because the quantum
state will be destroyed.
The paper you link to talks about creating large numbers of entangled
particles. While this is useful for sharing quantum computations over
long distances, it is not at all useful for quantum cryptography,
because one could intercept a small number of these particles, measure
them on each of the possible axes used for the cryptosystem, and figure
out the shared key.
--
Sean Richard Lynch