On 5/14/2022 4:35 AM, Bruce Kellett wrote:
On Sat, May 14, 2022 at 9:19 PM John Clark <johnkcl...@gmail.com> wrote:
On Fri, May 13, 2022 at 10:41 PM Bruce Kellett
<bhkellet...@gmail.com> wrote:
>> After my body has been duplicated but before I have open
the door of the duplicating chamber to see where I was I
won't know if Iwill be the John Clark who has seen Moscow
or the John Clark who has seen Helsinki, and indeed the
distinction between the two would be meaningless because
the two would be identical until the door is opened and
they differentiate because then one has the memory of
seeing Moscow but the other has the memory of seeing
Helsinki. So if both decided to place a bet on what they
would see after the door was opened (and if one decided to
place a bet then the other certainly would too because
they're identical) then, provided they were logical,and I
think I am at least most of the time, they would both put
the odds at 50-50.
/> So how do you accommodate a situation in which there is a
90% chance of seeing Moscow and a 10% chance of seeing Helsinki?/
*You've asked that exact same question several times before so
I'll answer it the exact same way I did before because you never
made an argument against what I said, you just keep asking the
same question again. If I know the duplicating machine has made 10
copies of me and that 9 of them are in Helsinki and 1 is in Moscow
then then 9 John Clark's will remember seeing Helsinki but only 1
will remember seeing Moscow; so if they place odds after the
duplication but before the door was opened and they observe where
they are they would all say there was a 90% chance they were in
Helsinki, and 90% of them would turn out to be correct and would
win their bet. *
The trouble is that the duplicating machine makes only one copy, so
there is one for Moscow and one for Helsinki. There are no
multiple copies in the original scenario. Changing the nature of the
question is not an answer.
The reason I repeat this is that the Schrodinger equation gives one
branch for each component of the superposition -- one branch for each
dimension of the Hilbert space. So I ask again, how do you accommodate
a situation in which there is a 90% chance of being on one branch and
a 10% chance of being on the other branch, as per the Born rule?
Changing the number of branches (or duplicates) is fine in a
general theory, but not in QM. The SE gives only one branch for each
outcome. What you are really saying is that the SE is inconsistent
with the Born rule -- a point I have been making all along.
Even Sean Carroll who is a proponent of MWI says that it's necessary to
associate "weights" or "amplitudes" with branches. I think it's
possible to do it with branch counting if you assume some sufficiently
large number are available to split...but that's not much different than
assigning amplitudes.
Brent
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