Re: Schmidhuber II implies FTL communications

From: Wei Dai <>
Date: Fri, 6 Sep 2002 14:49:25 -0700

On Thu, Sep 05, 2002 at 07:32:49PM -0700, Hal Finney wrote:
> This was an interesting paper but unfortunately the key point seemed
> to pass by without proof. On page 5, the proposal is to use entangled
> particles to try to send a signal by measuring at one end in a sequence
> of different bases which are chosen by an algorithmically incompressible
> mechanism. The assumption is that this will result in an algorithmically
> incompressible set of results at both ends, in contrast to the state
> where stable measurements are done, which we assume for the purpose of
> the paper produces algorithmically compressible results.
> The author writes: "This process of scrambling with the random template T
> guarantees that Bob's modified N-bit long string of quantum measurements
> is almost surely p-incompressible..., and that Alice's corresponding
> string (which is now different from Bob's) is also (almost surely)
> p-incompressible"
> It's not clear to me that this follows. Why couldn't Bob's measurement
> results, when using a randomly chosen set of bases, still have a
> compressible structure? And why couldn't Alice's?

I think you're right, the author skipped some steps in his reasoning here.
I'll try to fill in with my guesses.

When Bob randomizes his measurement directions, it could have one of three
effects. It could predictably change Alice's measurement outcomes, it
could unpredictably randomize Alice's measurement outcomes, or it could
not affect Alice's measurement outcomes at all. The first two can both be
used for FTL communications. The last one is ruled out by experiments
relating to Bell's inequality.

To see why, let Alice and Bob agree on 3 coplanar directions each
seperated by 120 degrees, and label them a, b, and c. They're going to
independently randomly choose one of these directions and measure the spin
of their particles along it. They do this a bunch of times and then
compare notes.

Suppose Alice can predict her outcomes as a function of her choice of
direction independent of Bob's choice of direction. She can say something
like if I choose a, outcome will be 0, if I choose b, outcome will be 1,
and if I choose c, outcome will be 1. These predictions must apply to Bob
just as well. So it's like there is a local hidden variable in the
particle that determines what the outcome is in all three cases. Bell's
inequality says if you look at the times when Alice and Bob choose
different directions, their measurements should agree at least 1/3 of the
time. Take the <0,1,1> case above. There are 6 ways that Alice and Bob can
choose different directions, and at least 2 of them (b,c and c,b) result
in agreement. This applies to every possible prediction, hence the 1/3
figure. However, QM predicts and actual experiment shows they agree only
1/4 of the time, therefore Alice must not be able to predict her outcomes
without knowing Bob's choice of direction.
Received on Fri Sep 06 2002 - 14:49:55 PDT

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