On 3/22/2022 11:05 PM, Bruce Kellett wrote:
On Wed, Mar 23, 2022 at 4:01 PM Brent Meeker <meekerbr...@gmail.com>
wrote:
On 3/22/2022 7:55 PM, Bruce Kellett wrote:
Actually, that is where I started. I assumed that Alice and Bob
were both able to collect results from N trials before they met.
Then there are 2^N copies of each experimenter, and a potential
(2^N)^(2^N) pairs when they meet. The trouble to be explained is
that there are actually only 2^N pairs in a real experiment, each
with inequality-violating correlations. What has happened to all
the extra pairings that MWI must produce? (Most of which have
correlations violating the quantum predictions.)
Don't you mean (2^N)x(2^N).
Probably. Both Alice and Bob split into 2^N copies for binary results.
So each of the 2^N Alice's splits into 2^N further copies, one for
each copy of Bob.
Why would Alice split again for each copy of Bob. There's no QM
involved. Each Alice and each Bob has seen N events and has recorded
them as a binary number of N bits in their notebook. They are
quasi-classical observers and they're going to meet in the future. If
there are not (2^N)x(2^N) meetings there must be some quantum magic that
it is eliminating or otherwise weighting them.
Brent
That seems to give (2^N)*(2^N) pairings. That is still a lot of
redundant copies when only one copy of Bob meets each copy of Alice in
the actual experiment.
Bruce.
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