On 5/12/2020 10:08 PM, Bruce Kellett wrote:
On Wed, May 13, 2020 at 2:06 PM 'Brent Meeker' via Everything List
<everything-list@googlegroups.com
<mailto:everything-list@googlegroups.com>> wrote:
On 5/12/2020 7:12 PM, Bruce Kellett wrote:
> If we now turn our attention to the quantum case, we have a
> measurement (or sequence of measurements) on a binary quantum state
>
> |psi> = a|0> + b|1>,
>
> where |0> is to be counted as a "success", |1> represents anything
> else or a "fail", and a^2 + b^2 = 1. In a single measurement, we
can
> get either |0> or 1>, (or we get both on separate branches in the
> Everettian case). Over a sequence of N similar trials, we get a
set of
> 2^N sequences of all possible bit strings of length N. (These all
> exist in separate "worlds" for the Everettian, or simply represent
> different "possible worlds" (or possible sequences of results)
in the
> single-world case.) This set of bit strings is independent of the
> coefficients 'a' and 'b' from the original state |psi>, but if we
> carry the amplitudes of the original superposition through the
> sequence of results, we find that for every zero in a bit string we
> get a factor of 'a', and for every one, we get a factor of 'b'.
This is what you previously argued was not part of the Schroedinger
equation and was a cheat to slip the Born rule in. It's what I
said was
Carroll's "weight" or splitting of many pre-existing worlds.
This is not what Carroll does. He looks at a single measurement, and
boosts the number of components of the wave function so that all have
the same amplitude. That, I argue, is a mistake.
>
> Consequently, the amplitude multiplying any sequence of M zeros and
> (N-M) ones, is a^M b^(N-M). Again, differentiating with respect
to 'a'
> to find the turning point (and the value of 'a' that maximizes this
> amplitude), we find
>
> |a|^2 = M/N,
Maximizing this amplitude, instead of simply counting the number of
sequences with M zeroes as a fraction of all sequences (which is
independent of a) is effectively assuming |a|^2 is a probability
weight. The "most likely" number of zeroes, the number that
occurs most
often in the 2^N sequences, is is N/2.
I agree that if you simply look for the most likely number of zeros,
ignoring the amplitudes, then that is N/2. But I do not see that
maximising the amplitude for any particular value of M is to
effectively assume that it is a probability.
I think it is. How would you justify ".. the amplitude multiplying any
sequence of M zeros and (N-M) ones, is a^M b^(N-M)..." except by saying
a is a probability, so a^M is the probability of M zeroes. If it's not a
probability why should it be multiplied into and expression to be maximized?
In any case though, I don't see the form of the Born rule as something
problematic. It's getting from counting branches to probabilities.
Once you assume there is a probability measure, you're pretty much
forced to the Born rule as the only consistent probability measure.
Brent
When you do this, you can see by analogy that it is a probability, but
one did not assume this at the start.
Bruce
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