On 2/21/2020 4:57 PM, Bruce Kellett wrote:
On Sat, Feb 22, 2020 at 11:35 AM 'Brent Meeker' via Everything List
<[email protected]
<mailto:[email protected]>> wrote:
On 2/21/2020 4:19 PM, Bruce Kellett wrote:
On Sat, Feb 22, 2020 at 11:11 AM smitra <[email protected]
<mailto:[email protected]>> wrote:
On 16-02-2020 06:34, Bruce Kellett wrote:
>
> The probabilistic interpretation of QM arose in a single-world,
> collapse, model. Attempting to graft probability on to
many-worlds is
> a failure, as my arguments against Everett show. If the
data for any
> sequence of trials are independent of the amplitudes, then
some ad hoc
> probability interpretation of the amplitudes is not going
to affect
> the data. But the data is what we use to infer that Born rule
> probabilities are what we observe. This is a single-world
result.
>
> Bruce
It's only a failure when attempting to describe the state in the
multiverse in that particular way where you can argue that the
amplitudes don't matter. It's not an argument against the
multiverse
idea, as whether or not you have a large number of copies
should not
affect the statistical outcome observed in any experiment.
No. And it has not been suggested that it will. But that is just
to take a single-world approach: statistics work fine in each
branch of the multiverse (each world), but the concepts of
probability and the Born rule break down when every result is
obtained in experiments because the data obtained in each world
are independent of the amplitudes of the original state.
This objection seems to rest on the idea that the set of sequences
(across all the worlds) cannot be regarded as an ensemble from
which one is picked by self-location.
I am not sure what this means. The set of all sequences (all branches
or worlds) is just the set of all 2^N possible binary strings, so any
set of results that an individual might obtain is a selection from
this set. If you want to regard this as self-location, then fine, but
I don't know what that gains you.
If you regard his self-location as equi-probable over the ensemble then
his expectation values and other statistics are the same as for a
N-string in one world.
Otherwise applying probability theory to the ensemble recovers the
one-world statistics. Right?
In the two-outcome case, any branch of the ensemble is just a
one-world set of Bernoulli trials, so the statistics of that branch
correspond to the binomial distribution with a probability
characteristic of the branch. Different branches will, in general,
correspond to different p-values. The argument against Everett is that
this set of binary strings is independent of the amplitudes of the
initial quantum state so there is no room for the Born rule.
Yes, that's true. But all the advocates of MWI I know of assume that
there is some kind of weighting that follows from the state preparation,
so that if you are measuring a state with is in a superposition
a|up>+b|dwn> in the instrument up/dwn basis, then the branches are in
proportion to a and b, rather than just one each. I don't know whether
you can justify this by a detailed interaction term in the Hamiltonian,
but it is commonly assumed as "that's the way it must be".
Brent
Of course you can say there really is no ensemble (that is
accessible to anyone), but that's true of the hypothetical
ensembles invoked in statistics all the time.
I can't really grasp what your objection to the case I have made is.
Is it anything more than that the result goes against your basic
intuitions?
Bruce
--
You received this message because you are subscribed to the Google
Groups "Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send
an email to [email protected]
<mailto:[email protected]>.
To view this discussion on the web visit
https://groups.google.com/d/msgid/everything-list/CAFxXSLQBd9%3DwkC7aw6GUxhLFg_%2BCcVz-toqLP5O1Ghs%2BmwZtLg%40mail.gmail.com
<https://groups.google.com/d/msgid/everything-list/CAFxXSLQBd9%3DwkC7aw6GUxhLFg_%2BCcVz-toqLP5O1Ghs%2BmwZtLg%40mail.gmail.com?utm_medium=email&utm_source=footer>.
--
You received this message because you are subscribed to the Google Groups
"Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to [email protected].
To view this discussion on the web visit
https://groups.google.com/d/msgid/everything-list/37ff5fe2-cc0e-f818-c4bb-7330890e426b%40verizon.net.
