Quantum entangled-photon Chinese satellite.

[jim bell]

 From: Zenaan Harkness <zen at freedbms.net>
On Fri, Aug 05, 2016 at 09:10:42AM -0600, Mirimir wrote:
...
> Here, from <http://www.scottaaronson.com/blog/?p=2464>:
> > The violation of the Bell inequality has a schizophrenic status in
> > physics.  To many of the physicists I know, Nature’s violating the
> > Bell inequality is so trivial and obvious that it’s barely even
> > worth doing the experiment: if people had just understood and
> > believed Bohr and Heisenberg back in 1925, there would’ve been no
> > need for this whole tiresome discussion.

>Seriously, I am none the wiser and cannot yet make sense of what they
>are saying.
>China apparently is putting this experiment in space - are they winning
>a game on prediction of one particular bit with > 75% probability, and
>if so, can they run that game numerous times to get that probability
>close to 100%, and if so, can the random inputs to each side be made not
>random so that the result of the game is transmission of information?
>I cannot begin to answer any of these questions sorry...

I will explain what I think they are doing in the fiber-optic version of the experiment,at least so nobody is permanently misled by my previous analogy.
Imagine a central location on earth, let's call it Location B.  20,500 meters west of that is Location A, and 20,500 meters to the east of "B" is Location C.  There's anoptical fiber going from "A" to "B", and another optical fiber going from "B" to "C".Two entangled photons are produced at Location B, then one is launched into fiber going to "A", and the other photon is launched from "B" into the fiber going to "C".
After about 100 microseconds later (since the speed of light in that fiber is about 'c'/1.4584, where 1.4584 is the index of refraction of infrared in silica, thus 205.5 meters /microsecond), those photons emerge from their respective ends.  Notnecessarily at the same time, because the length of the fibers may not be quiteidentical.  They do the detection at Location A, and through prior arrangement theyschedule the detection at "C" a few nanoseconds later, possibly adjusting the physical length of the fiber to get the timing close to being correct..  Good synchronization could be achieved by GPS-controlled clocks, or perhaps a third fiber being used to synchronize local clocks at "A" and "C".
They first detect at "A", and then detect at "C".  And they might reverse the order, forcompleteness.  But that's not the end:To determine that there has been more than a 50% correlation of the measured spins, they haveto transmit the type (angle) of measurement they make by ordinary optical fiber.  (Although,it wouldn't have to be on an optical fiber:  It could be a USB memory stick glued to theshell of a fast snail, I suppose.  the important thing is that the information eventually getsto the other side, not how fast it takes to get there.)
  The information eventually gets to the other end, and they do the calculations andverify that SOMEHOW, the fact that a measurement at "A" somehow affected themeasurement at "C".   If they "schmoo plot" (meaning carefully adjust, then plot on a graph)  https://en.wikipedia.org/wiki/Shmoo   , they can determine how fast some affecting particle or signal would have to travel to affect the receiver at the other end.  Since the delay from Location "A" to "Location "C" would be (100+100) = 200 microseconds in a fiber, it would be 200/1.4584 = 137.136 microseconds from location A to location C,by air.  (or in a vacuum, or by radio, etc.)
That figure I found from an article a few years ago, that said it would have to travel at least 10,000x that of 'c' to affect the measurement, would require that the delay is measured by:137.136 microseconds/10,000 = 13.7126 nanoseconds.   If the measurement at "A" occurred only 13.7136 nanoseconds before the measurement at "C", and yet there wasstill correlation, this shows that a velocity of at least 10,000 'c' to affect the outcome at "C".
Therefore, I conclude that it would be easy to measure the minimum effective speed of the 
hypothetical interfering particle or wave.  That particle or wave would have to travel 41,000meters in less than 13.7 nanoseconds, to achieve that interference.  Time measurement to 1 nanosecond is easy, to 1 picosecond is doable, and in fact measurement of time valuesfar less than 1 picosecond can be accomplished.
             Jim Bell


  
-------------- next part --------------
A non-text attachment was scrubbed...
Name: not available
Type: text/html
Size: 11272 bytes
Desc: not available
URL: <http://lists.cpunks.org/pipermail/cypherpunks/attachments/20160806/9d1b7ae0/attachment-0002.txt>