For those of you wondering, 2.05 = 7 ^ exp(-1)
where 7 comes from the "seven symbols [encoded] onto a combination of the orbital angular momentum (“twisted” light) and their angular position".
Orbital angular momentum – in photonics short-handed as “twisted light” – can help expand the capacity of comms systems using quantum communications. Since quantum key distribution (QKD) is already a commercial, if pricey, market segment, quantum communication science is of a more practical application today than quantum …
Doesn't look like that's where the 2.05 bits/photon comes from though.
Eq.(5) gives the bits per photon: 1/log(2) * (log(d) + F*log(F) + (1-F) * log((1-F)/(d-1)) where d is the number of modes (symbols) and F is the probability of correct detection.
They're using d = 7 and an experimentally measured F = 0.895 which gives 2.0513 bits/photon. They point out that the ideal value (F ->1) is 2.8, and that if they didn't have any dark counts (which contribute 4% towards the error) they'd get 2.29 bits per photon (F = 0.935). That their measured value is close to d^exp(-1) (= 2.0460) appears to be coincidental, but then you never quite know with e.
In other news, the 4.17 bits/photon relates to another paper (they'd need 25 symbols and F = 0.955 here) and 4 kHz refers to the burst operation of the encoding mirror; the actual secure data rate (after including uploading to the mirror and the like) is down around 6.5 bps. I'm also available for children's parties and Bar Mitzvahs.
They point out that the ideal value (F ->1) is 2.8
... because if you have 7 symbols with perfect reliability, you have log2 7 = ~2.8 bits - just to clarify that point.
Thanks for reading and citing the paper, by the way. I was puzzled about the derivation of the OP's formulation. As you say, you never know where e will rear its head, but the association in this case wasn't clear. I'm not even sure what a real-world interpretation of 71/e would be, so I also suspect it's accidental. But IANAM.
I think this is the first reg article I've read where there are obviously lots of English words, forming grammatically correct sentences, but I have absolutely no comprehension of their meaning as a whole...
In other words, I didn't understand any of it..but isn't that the point of anything quantum? :D
Executive summary: Photons can do various things as they move along. If you can pick which of those various things they do when you emit them, and figure out which they're doing when you receive them, then you can use that to encode messages. This team has shown a way to use more of the things photons can do.
After WW2, all a physicist needed to get funding was to propose work on something "nuclear" as possibly having a military application. Currently, the magic word is "quantum".
@themoononastick
"I'm also available for children's parties and Bar Mitzvahs."
Presumably, very slow but extremely secure children's parties and bar mitzvahs.
No bat mitzvahs? Looks like quantum encoding is just as male dominated as all the other branches of physics.
Yes, can someone tell me the point of this, or is it just plain research learning etc, when we already can get information to travel at bagigahertz down a fibre using light, it just seems this is a bit late.
I get that quantum stuff may mean higher security, but I could hide 6 bits in 60 trillion bits over fibre and you'd have a hard time finding them
Yes, can someone tell me the point of this ... we already can get information to travel at bagigahertz down a fibre using light ... I could hide 6 bits in 60 trillion bits over fibre and you'd have a hard time finding them
There are several different aspects to this. First, being able to increase bandwidth over existing infrastructure is worth a good deal. Just because we currently have achieved "lots" does not mean that it is not desirable to achieve "lots more." Quite the contrary, in fact.
Second, you are confounding two different security measures: steganography and encryption. Yes, you can hide a needle in a haystack, but then you have to send the entire haystack along in order to move the needle. While steganography is useful, it generally should be used in conjunction with encryption and not by itself. There are plenty of mass surveillance tools in use these days that are likely to catch all the "needles" no matter how many stacks accompany them.
Finally, the way data is encoded here in part uses the polarization of photons which makes intercept detection a built in feature - desirable for obvious reasons.
There have been a number of articles concerning this here on El Reg. Here too is a primer on the subject:
http://resources.infosecinstitute.com/quantum-cryptography/
Yes, you can hide a needle in a haystack, but then you have to send the entire haystack along in order to move the needle.
And the recipient has to know how to find the needle. If you have a secure way to tell the recipient that, why do you need the haystack in the first place?
That is the Key Distribution Problem. Asymmetric cryptography is one way of addressing it; quantum key distribution is another.
The objections raised in the first couple of posts in this thread miss the point entirely. "What do I need a telephone for? I've been able to send an entire novel by post for years! That's a lot better than reading it to someone over the phone." Well, yes. And driving a car is a better way to travel a thousand miles than walking; but it's not really suitable for going from the bedroom to the bathroom.