On 9/4/2020 10:18 PM, Bruce Kellett wrote:
On Sat, Sep 5, 2020 at 2:42 PM 'Brent Meeker' via Everything List
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On 9/4/2020 7:02 PM, Bruce Kellett wrote:
On Sat, Sep 5, 2020 at 11:29 AM 'Brent Meeker' via Everything
List <everything-list@googlegroups.com
<mailto:everything-list@googlegroups.com>> wrote:
But the theory isn't about the probability of a specific
sequence, it's about the probability of |up> vs |down> in the
sequence without regard for order. So there will, if the
theory is correct, be many more sequences with a frequency of
|up> near some theoretically computed proportion |a|^2 than
sequences not near this proportion.
The theory is about the probabilitiies of observations. The
observation in question here is a sequence of |up> / |down>
results, given that the probability for each individual outcome
is 0.5. If the theory cannot give a probability for the sequence,
It can. But QM only predicts the p=0.5. To have a prediction for
a specific sequence HHTTHHHTTHTHTH... you need extra assumptions
about indenpendence.
Sure. And independence of the sequential observations is clearly
implied by the set up.
And given those assumptions your theory will be contradicted with
near certainty.
Why?
The probability of getting any given entry in the sequence is 1/2, so
the probability of getting the whole sequence right is 1/2^N .
Which is why I say the test of QM is whether p=0.5 is consistent
with the observed sequence in the sense of predicting the relative
frequency of H and T, not in the sense of predicting HHTTHHHTTHTHTH...
I am not attempting to predict a particular sequence.
That's what you seemed to reply when I said QM was only predicting the
relative frequency of H within the sequence. If you now agree with
that, then you will also agree that there will many sequences with a
relative frequency of 0.5 for H and given any epsilon the fraction of
such sequences repetitions with 0.5-epsilon<frequency(H)<0.5+epsilon
goes 1 as N->oo. Which is what we mean by confirming the QM prediction
of 0.5.
All that I have said is that the probability of any such sequence in N
independent trials is 1/2^N. And that is simple probability theory,
which cannot be denied.
then multiply the probabilities for each particular result in
your sequence of measurements. The number of sequences with
particular proportions of up or down results is irrelevant for
this calculation.
Again, you are just attempting to divert attention from the
obvious result that the Born rule calculation gives a different
probability than expected when every outcome occurs for each
measurement. In the Everett case, every possible sequence
necessarily occurs. This does not happen in the genuine
stochastic case, where only one (random) sequence is produced.
In the Everett theory a measurement of spin up for a particle
prepared in spin x results in two outcomes...only one is observed.
If that is enough to dismiss Everett then all the this discussion
of probability and the Born rule is irrelevant.
I have no idea what you are talking about! Nothing like that was ever
suggested. Everett predicts that in such a measurement, both outcomes
obtain -- in separate branches.
As I understand your argument you're saying Everett is falsified
because, no matter what N is, it predicts a branch HHHHHHHHHH...H
which...What? Is wrong? Doesn't occur? Is inconsistent with the Born
rule (it isn't)? Is not observed?
If you just say it predicts something which is not observed; then my
point is that it always predicts outcomes that are not observed unless P=1.
Brent
But the probability of this is one. Repeat N times. N time one is
still just one. There is nothing more to it than that. I think you are
being desperate in your attempts to play 'advocatus diaboli'. The
point is that the Born rule is inconsistent with Everett.
Bruce
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