subject:"STEP 3"



On Tuesday, September 24, 2019 at 2:24:05 PM UTC-6, John Clark wrote:
>
>
> On Tue, Sep 24, 2019 at 4:06 PM Alan Grayson  > wrote:
>
> *> According to Brent, Laplace was aware that due to inherent measurement 
>> inaccuracies we cannot know the exact configuation of the universe at any 
>> time, so in fact we can't predict its past and future with any accuracy 
>> (only for relatively short durations). But this is what Carroll omitted in 
>> his use of Laplace *
>
>
> Alan, for god's sake, your digging yourself into a deeper and deeper hole! 
> Read the man's damn book before you do any more pontificating about Carroll 
> unless you enjoy publicly making a fool of yourself.
>
>   John K Clark
>

It's core claim is crap for the masses, so I won't waste my time. Enjoy 
your fantasy. AG 

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Re: Sean Carroll gets past "Step 3" of the UDA

On Tue, Sep 24, 2019 at 4:06 PM Alan Grayson  wrote:

*> According to Brent, Laplace was aware that due to inherent measurement
> inaccuracies we cannot know the exact configuation of the universe at any
> time, so in fact we can't predict its past and future with any accuracy
> (only for relatively short durations). But this is what Carroll omitted in
> his use of Laplace *


Alan, for god's sake, your digging yourself into a deeper and deeper hole!
Read the man's damn book before you do any more pontificating about Carroll
unless you enjoy publicly making a fool of yourself.

  John K Clark


>

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Re: Sean Carroll gets past "Step 3" of the UDA



On Tuesday, September 24, 2019 at 12:32:13 PM UTC-6, John Clark wrote:
>
>
>
> On Tue, Sep 24, 2019 at 2:27 PM Alan Grayson  > wrote:
>
>>
>>
>> On Tuesday, September 24, 2019 at 12:19:40 PM UTC-6, John Clark wrote:
>>>
>>> On Mon, Sep 23, 2019 at 11:59 AM Alan Grayson  
>>> wrote:
>>>
>>> >>  Sean Carroll:
>
> So Isaac Newton came up with the rules of classical mechanics in the 
> 1600s, but it wasn't until Laplace around the year 1800 that this 
> implication of classical mechanics was realized.
> It's a clockwork universe.  That the way classical mechanics works is 
> if you tell me the state of a system right now at one moment by which in 
> classical mechanics you would mean the position and the velocity of every 
> part, and you knew the laws of physics and you had arbitrarily large 
> computational capacity, 
> Laplace said of vast intelligence okay then to that vast intelligence 
> the past and future would be as determined and known as the present was 
> because that's the clockwork universe is deterministic everything is 
> fixed 
> once you know the present moment.
>
>  
 *> But Laplace was wrong in one very important respect. One can never 
 know the exact position and momentum of any particle, let alone the entire 
 universe. There are no perfect measurements! Further, the situation is 
 further aggravated by the Uncertainty Principle. In sum, using classical 
 mechanics the future is NOT determined by its present, imprecise 
 configuration. Not only is Laplace mistaken, but Carroll as well, who 
 should know better. AG *

>>>
>>> Oh for christ sake! That remark is as stupid as your crap about the 
>>> flying saucer people in New Mexico. Do you really think Sean Carroll, a 
>>> professor of physics at one of the best universities in the world, 
>>> doesn't know that?!  
>>>
>>> John K Clark
>>>
>>
>> *> Before you shoot your mouth off, read what I wrote in response to 
>> Brent. Sean DOES know better, but he deliberately twisted Laplace's view to 
>> fit his foolish agenda. Not very honest. As for flying saucers, they're 
>> really much more probable than* [...]
>>
>
> You sir are an ass.
>
> John K Clark
>

I will also note your dishonesty, or shall we say cowardice, in trucating 
my comment. AG 

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Re: Sean Carroll gets past "Step 3" of the UDA



On Tuesday, September 24, 2019 at 12:32:13 PM UTC-6, John Clark wrote:
>
>
>
> On Tue, Sep 24, 2019 at 2:27 PM Alan Grayson  > wrote:
>
>>
>>
>> On Tuesday, September 24, 2019 at 12:19:40 PM UTC-6, John Clark wrote:
>>>
>>> On Mon, Sep 23, 2019 at 11:59 AM Alan Grayson  
>>> wrote:
>>>
>>> >>  Sean Carroll:
>
> So Isaac Newton came up with the rules of classical mechanics in the 
> 1600s, but it wasn't until Laplace around the year 1800 that this 
> implication of classical mechanics was realized.
> It's a clockwork universe.  That the way classical mechanics works is 
> if you tell me the state of a system right now at one moment by which in 
> classical mechanics you would mean the position and the velocity of every 
> part, and you knew the laws of physics and you had arbitrarily large 
> computational capacity, 
> Laplace said of vast intelligence okay then to that vast intelligence 
> the past and future would be as determined and known as the present was 
> because that's the clockwork universe is deterministic everything is 
> fixed 
> once you know the present moment.
>
>  
 *> But Laplace was wrong in one very important respect. One can never 
 know the exact position and momentum of any particle, let alone the entire 
 universe. There are no perfect measurements! Further, the situation is 
 further aggravated by the Uncertainty Principle. In sum, using classical 
 mechanics the future is NOT determined by its present, imprecise 
 configuration. Not only is Laplace mistaken, but Carroll as well, who 
 should know better. AG *

>>>
>>> Oh for christ sake! That remark is as stupid as your crap about the 
>>> flying saucer people in New Mexico. Do you really think Sean Carroll, a 
>>> professor of physics at one of the best universities in the world, 
>>> doesn't know that?!  
>>>
>>> John K Clark
>>>
>>
>> *> Before you shoot your mouth off, read what I wrote in response to 
>> Brent. Sean DOES know better, but he deliberately twisted Laplace's view to 
>> fit his foolish agenda. Not very honest. As for flying saucers, they're 
>> really much more probable than* [...]
>>
>
> You sir are an ass.
>
> John K Clark
>

According to Brent, Laplace was aware that due to inherent measurement 
inaccuracies we cannot know the exact configuation of the universe at any 
time, so in fact we can't predict its past and future with any accuracy 
(only for relatively short durations). But this is what Carroll omitted in 
his use of Laplace and the alleged predictable universe under CM, which was 
undone by QM.  AG

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Re: Sean Carroll gets past "Step 3" of the UDA

On Tue, Sep 24, 2019 at 2:27 PM Alan Grayson  wrote:

>
>
> On Tuesday, September 24, 2019 at 12:19:40 PM UTC-6, John Clark wrote:
>>
>> On Mon, Sep 23, 2019 at 11:59 AM Alan Grayson 
>> wrote:
>>
>> >>  Sean Carroll:

 So Isaac Newton came up with the rules of classical mechanics in the
 1600s, but it wasn't until Laplace around the year 1800 that this
 implication of classical mechanics was realized.
 It's a clockwork universe.  That the way classical mechanics works is
 if you tell me the state of a system right now at one moment by which in
 classical mechanics you would mean the position and the velocity of every
 part, and you knew the laws of physics and you had arbitrarily large
 computational capacity,
 Laplace said of vast intelligence okay then to that vast intelligence
 the past and future would be as determined and known as the present was
 because that's the clockwork universe is deterministic everything is fixed
 once you know the present moment.

>>> *> But Laplace was wrong in one very important respect. One can never
>>> know the exact position and momentum of any particle, let alone the entire
>>> universe. There are no perfect measurements! Further, the situation is
>>> further aggravated by the Uncertainty Principle. In sum, using classical
>>> mechanics the future is NOT determined by its present, imprecise
>>> configuration. Not only is Laplace mistaken, but Carroll as well, who
>>> should know better. AG *
>>>
>>
>> Oh for christ sake! That remark is as stupid as your crap about the
>> flying saucer people in New Mexico. Do you really think Sean Carroll, a
>> professor of physics at one of the best universities in the world,
>> doesn't know that?!
>>
>> John K Clark
>>
>
> *> Before you shoot your mouth off, read what I wrote in response to
> Brent. Sean DOES know better, but he deliberately twisted Laplace's view to
> fit his foolish agenda. Not very honest. As for flying saucers, they're
> really much more probable than* [...]
>

You sir are an ass.

John K Clark

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Re: Sean Carroll gets past "Step 3" of the UDA



On Tuesday, September 24, 2019 at 12:19:40 PM UTC-6, John Clark wrote:
>
> On Mon, Sep 23, 2019 at 11:59 AM Alan Grayson  > wrote:
>
> >>  Sean Carroll:
>>>
>>> So Isaac Newton came up with the rules of classical mechanics in the 
>>> 1600s, but it wasn't until Laplace around the year 1800 that this 
>>> implication of classical mechanics was realized.
>>> It's a clockwork universe.  That the way classical mechanics works is if 
>>> you tell me the state of a system right now at one moment by which in 
>>> classical mechanics you would mean the position and the velocity of every 
>>> part, and you knew the laws of physics and you had arbitrarily large 
>>> computational capacity, 
>>> Laplace said of vast intelligence okay then to that vast intelligence 
>>> the past and future would be as determined and known as the present was 
>>> because that's the clockwork universe is deterministic everything is fixed 
>>> once you know the present moment.
>>>
>>>  
>> *> But Laplace was wrong in one very important respect. One can never 
>> know the exact position and momentum of any particle, let alone the entire 
>> universe. There are no perfect measurements! Further, the situation is 
>> further aggravated by the Uncertainty Principle. In sum, using classical 
>> mechanics the future is NOT determined by its present, imprecise 
>> configuration. Not only is Laplace mistaken, but Carroll as well, who 
>> should know better. AG *
>>
>
> Oh for christ sake! That remark is as stupid as your crap about the flying 
> saucer people in New Mexico. Do you really think Sean Carroll, a professor 
> of physics at one of the best universities in the world, doesn't know 
> that?!  
>
> John K Clark
>

Before you shoot your mouth off, read what I wrote in response to Brent. 
Sean DOES know better, but he deliberately twisted Laplace's view to fit 
his foolish agenda. Not very honest. As for flying saucers, they're really 
much more probable than believing that some fool who does a double slit 
experiment can create possibly uncountable worlds replete with stars, 
galaxies, and living being; or nothing at all like that. I call it hubris 
on steroids, but to some who are misguided, it seems quite normal. AG 

>
>
>
>  
>
>>
>>> Now quantum mechanics comes along and throws a spanner into the works a 
>>> little bit if you're a many-worlds person Laplace is demon is still 
>>> possible.
>>> So if you know the wave function of the universe exactly and you have 
>>> infinite calculational capacity you could predict the past and the future 
>>> with perfect accuracy.
>>> But! what you're predicting is all of the branches of the wavefunction 
>>> so any individual person inside the wavefunction still experiences 
>>> apparently random events.
>>>
>>> Right, so *you can't predict what will happen to you even if you can 
>>> predict what will happen to the entire universe*.
>>>
>>>
>>> This is the essence of Step 3 of the UDA.  In an experiment involving 
>>> duplication of persons, apparent randomness emerges.  There is no actual 
>>> randomness in the complete system, but individual experiences will have the 
>>> characteristic of randomness, in the sense of not being able to make 
>>> definite predictions concerning their experiences.  Sean Carroll gets 
>>> this.  Max Tegmark gets this.  You got it at least once 6 years ago on this 
>>> list when you agreed that a forking computer process containing AIs could 
>>> not predict which process they would end up in.  This is enough for you to 
>>> proceed to the next step, which adds only a time delay to one of the 
>>> duplicates.  You are almost there.
>>>
>>> Jason
>>>  
>>>
>>  
>>
>>>
>>> Now quantum mechanics comes along and throws a spanner into the works a 
>>> little bit if you're a many-worlds person Laplace is demon is still 
>>> possible.
>>> So if you know the wave function of the universe exactly and you have 
>>> infinite calculational capacity you could predict the past and the future 
>>> with perfect accuracy.
>>> But! what you're predicting is all of the branches of the wavefunction 
>>> so any individual person inside the wavefunction still experiences 
>>> apparently random events.
>>>
>>> Right, so *you can't predict what will happen to you even if you can 
>>> predict what wil

Re: Sean Carroll gets past "Step 3" of the UDA

On Mon, Sep 23, 2019 at 11:59 AM Alan Grayson 
wrote:

>>  Sean Carroll:
>>
>> So Isaac Newton came up with the rules of classical mechanics in the
>> 1600s, but it wasn't until Laplace around the year 1800 that this
>> implication of classical mechanics was realized.
>> It's a clockwork universe.  That the way classical mechanics works is if
>> you tell me the state of a system right now at one moment by which in
>> classical mechanics you would mean the position and the velocity of every
>> part, and you knew the laws of physics and you had arbitrarily large
>> computational capacity,
>> Laplace said of vast intelligence okay then to that vast intelligence the
>> past and future would be as determined and known as the present was because
>> that's the clockwork universe is deterministic everything is fixed once you
>> know the present moment.
>>
>>
> *> But Laplace was wrong in one very important respect. One can never know
> the exact position and momentum of any particle, let alone the entire
> universe. There are no perfect measurements! Further, the situation is
> further aggravated by the Uncertainty Principle. In sum, using classical
> mechanics the future is NOT determined by its present, imprecise
> configuration. Not only is Laplace mistaken, but Carroll as well, who
> should know better. AG *
>

Oh for christ sake! That remark is as stupid as your crap about the flying
saucer people in New Mexico. Do you really think Sean Carroll, a professor
of physics at one of the best universities in the world, doesn't know
that?!

John K Clark





>
>> Now quantum mechanics comes along and throws a spanner into the works a
>> little bit if you're a many-worlds person Laplace is demon is still
>> possible.
>> So if you know the wave function of the universe exactly and you have
>> infinite calculational capacity you could predict the past and the future
>> with perfect accuracy.
>> But! what you're predicting is all of the branches of the wavefunction so
>> any individual person inside the wavefunction still experiences apparently
>> random events.
>>
>> Right, so *you can't predict what will happen to you even if you can
>> predict what will happen to the entire universe*.
>>
>>
>> This is the essence of Step 3 of the UDA.  In an experiment involving
>> duplication of persons, apparent randomness emerges.  There is no actual
>> randomness in the complete system, but individual experiences will have the
>> characteristic of randomness, in the sense of not being able to make
>> definite predictions concerning their experiences.  Sean Carroll gets
>> this.  Max Tegmark gets this.  You got it at least once 6 years ago on this
>> list when you agreed that a forking computer process containing AIs could
>> not predict which process they would end up in.  This is enough for you to
>> proceed to the next step, which adds only a time delay to one of the
>> duplicates.  You are almost there.
>>
>> Jason
>>
>>
>
>
>>
>> Now quantum mechanics comes along and throws a spanner into the works a
>> little bit if you're a many-worlds person Laplace is demon is still
>> possible.
>> So if you know the wave function of the universe exactly and you have
>> infinite calculational capacity you could predict the past and the future
>> with perfect accuracy.
>> But! what you're predicting is all of the branches of the wavefunction so
>> any individual person inside the wavefunction still experiences apparently
>> random events.
>>
>> Right, so *you can't predict what will happen to you even if you can
>> predict what will happen to the entire universe*.
>>
>>
>> This is the essence of Step 3 of the UDA.  In an experiment involving
>> duplication of persons, apparent randomness emerges.  There is no actual
>> randomness in the complete system, but individual experiences will have the
>> characteristic of randomness, in the sense of not being able to make
>> definite predictions concerning their experiences.  Sean Carroll gets
>> this.  Max Tegmark gets this.  You got it at least once 6 years ago on this
>> list when you agreed that a forking computer process containing AIs could
>> not predict which process they would end up in.  This is enough for you to
>> proceed to the next step, which adds only a time delay to one of the
>> duplicates.  You are almost there.
>>
>> Jason
>>
>>
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Re: Sean Carroll gets past "Step 3" of the UDA

On Mon, Sep 23, 2019 at 7:23 PM Jason Resch  wrote:

>> I think you need to indicate how, out of the set of all computations,
>> you can pick the correct ones from the incorrect ones without the help of
>> matter that obeys the laws of physics.
>>
>
> *> How do you suppose the laws of physics pick out the correct physical
> outcomes from among all possibilities? *
>

You don't have to explain why a phenomena works the way it does to prove it
does in fact work that way. I don't need to explain how physical law gained
the ability to tell the difference between things that work and things that
don't because I have concrete (pun intended) proof that it does in fact
have that ability. If physical law says a bridge will not collapse under a
given load then it won't collapse, if it says it will then you'd better not
go on that bridge. That's why bridge engineers study physics and not p-adic
arithmetic.

> > *You presume there is a physical world governed by physical laws.*
>

Yes.

>   > *But you deny an arithmetical world governed by arithmetical laws. *
>

I don't deny that at all, but there are a infinite number of self
consistent arithmetical worlds, including the 3-adic world where 300 is
smaller than 8/45 because in that world 300 is only 1/3 distance units from
zero but 8/45 is 9 units. However out of that infinite number of ways
distance along the number line could be measured one of them is unique, it
stands out for only one reason, it is the only one that is consistent with
physical law, and that is the reason we teach that one and only that one to
our children, and that is the reason first graders say 2+2=4 and the reason
third graders say 300 is larger than 8/45.

> > *Yet, assuming an arithmetical world governed by arithmetical laws, you
> can derive the appearance of a physical universe governed by physical laws.*
>

Baloney! There is no way somebody can start with nothing but arithmetic and
derive the laws of Newton Einstein and Quantum Mechanics without also
deriving a infinite number of other physical laws that do NOT conform with
experimental observation. No way.

>> I'm very surprised that as soon as you mentioned the Planck Time in the
>> above you didn't realize you had left the world of pure dimensionless number
>> s and was talking numbers with physical units associated with them, like
>> measures of time and space and mass and energy and electrical charge.
>>
>
> *> If you think physical laws are computable,*
>

I think physical laws can make computations, and if a clever programer has
access to a physical Turing Machine he can use a few simple physical laws
to predict what will happen when a huge number of those simple laws
interact in astronomically complex ways. That's what a meteorologist does
when he makes a computer model of a hurricane.

> * > then time, space, mass, etc. can all be reduced to computation (and
> computation is the manipulation of pure numbers).*
>

There is no way pure arithmetic can come up with the Planck Time, it can't
find anything special about the number 5.39245 *10^-44 *seconds* because it
is *not* a pure number, there is no way pure arithmetic can know what the
hell a second is, or time in general, or space, or electrical charge, or
angular momentum or...

John K Clark

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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-24 Thread Philip Thrift



On Tuesday, September 24, 2019 at 8:40:57 AM UTC-5, John Clark wrote:
>
> On Mon, Sep 23, 2019 at 4:22 PM Jason Resch  > wrote:
>
> > *This Halloween will mark 6 years since you agreed with Step 3,*
>
>
> *BULLSHIT! *
>
> This is the entire post and even though 6 years has passed I stand by 
> every word and wouldn't change anything:
>
> On Thu, Oct 31, 2013 at 2:12 PM, Jason Resch  > wrote:
>
> *>  A) The test described where the simulation process forks 8 times and 
>> 256 copies are created and they each see a different pattern of the ball 
>> changing color*
>
>
> Duplicating a brain is not enough, the intelligence has NOT forked until 
> there is something different about them, such as one remembering seeing a 
> red ball and the other remember seeing a green ball, only then do they 
> fork. It was the decision made by somebody or something outside the 
> simulation to make sure all 256 saw a difference sequence of colored balls 
> that created 256 distinct minds. And to a simulated physicist a decision 
> made outside the simulation would be indistinguishable from being random, 
> that is to say the simulated laws of physics could not be used to figure 
> out what that decision would be.  
>
> *>  B) A test where the AI is not duplicated but instead a random number 
>> generator (controlled entirely outside the simulation) determines whether 
>> the ball changes to red or blue with 50% probability 8 times Then the AI 
>> (or AIs) could not say whether test A occurred first or test B occurred 
>> first.*
>
>
> Both A and B are identical in that the intelligence doesn't know what it 
> is going to see next; but increasingly convoluted thought experiments are 
> not needed to demonstrate that everyday fact. The only difference is that 
> in A lots of copies are made of the intelligence and in B they are not; but 
> as the intelligence would have no way of knowing if a copy had been made of 
> itself or not nor would it have any way of knowing if it was the original 
> or the copy, subjectively it doesn't matter if A or B is true.
>
> So yes, subjectively the intelligence would have no way of knowing if A 
> was true or B, or to put it another way subjectively it would make no 
> difference.  
>
> *> I reformulated the UDA in a way that does not use any pronouns at all, 
>> and it doesn't matter if you consider the question from one view or from 
>> all the views, the conclusion is the same.*
>
>
> Yes, the conclusion is the same, and that is the not very profound 
> conclusion that you never know what you're going to see next, and Bruno's 
> grand discovery of First Person Indeterminacy is just regular old dull as 
> dishwater indeterminacy first discovered by Og the caveman. After the big 
> buildup it's a bit of a letdown actually.
>
>   John K Clark
>  
>
>> >> important it's crystal clear exactly what the correct prediction would 
>>> have turned out to be.  
>>>
>>
>> > I did a few days ago, but you didn't respond.  I'll post it again:
>>
>
>>
>>
>>
>>
>> *First, consider this experiment:Imagine there is a conscious AI (or 
>> uploaded mind) inside a virtual environment (an open field)Inside that 
>> virtual environment is a ball, which the AI is looking at and next to the 
>> ball is a note which reads:"At noon (when the virtual sun is directly 
>> overhead) the protocol will begin.  In the protocol, the process containing 
>> this simulation will fork (split in two), after the fork, the color of the 
>> ball will change to red for the parent process and it will change to blue 
>> in the child process (forking duplicates a process into two identical 
>> copies, with one called the parent and the other the child). A second after 
>> the color of the ball is set, another fork will happen.  This will happen 8 
>> times leading to 256 processes, after which the simulation will end."Now, 
>> with the understanding of that experiment, consider the following:If the AI 
>> (or all of them) went through two tests, test A, and test B*
>
>  *A) The test described where the simulation process forks 8 times and 
>> 256 copies are created and they each see a different pattern of the ball 
>> changing color*
>
>  *B) A test where the AI is not duplicated but instead a random number 
>> generator (controlled entirely outside the simulation) determines whether 
>> the ball changes to red or blue with 50% probability 8 times*
>
> *Then the AI (or AIs) could not say whether test A occurred first or test 
>> B occurred first.*
>

Re: Sean Carroll gets past "Step 3" of the UDA

On Mon, Sep 23, 2019 at 4:22 PM Jason Resch  wrote:

> *This Halloween will mark 6 years since you agreed with Step 3,*

*BULLSHIT! *

This is the entire post and even though 6 years has passed I stand by every
word and wouldn't change anything:

On Thu, Oct 31, 2013 at 2:12 PM, Jason Resch  wrote:

*>  A) The test described where the simulation process forks 8 times and
> 256 copies are created and they each see a different pattern of the ball
> changing color*

Duplicating a brain is not enough, the intelligence has NOT forked until
there is something different about them, such as one remembering seeing a
red ball and the other remember seeing a green ball, only then do they
fork. It was the decision made by somebody or something outside the
simulation to make sure all 256 saw a difference sequence of colored balls
that created 256 distinct minds. And to a simulated physicist a decision
made outside the simulation would be indistinguishable from being random,
that is to say the simulated laws of physics could not be used to figure
out what that decision would be.

*>  B) A test where the AI is not duplicated but instead a random number
> generator (controlled entirely outside the simulation) determines whether
> the ball changes to red or blue with 50% probability 8 times Then the AI
> (or AIs) could not say whether test A occurred first or test B occurred
> first.*

Both A and B are identical in that the intelligence doesn't know what it is
going to see next; but increasingly convoluted thought experiments are not
needed to demonstrate that everyday fact. The only difference is that in A
lots of copies are made of the intelligence and in B they are not; but as
the intelligence would have no way of knowing if a copy had been made of
itself or not nor would it have any way of knowing if it was the original
or the copy, subjectively it doesn't matter if A or B is true.

So yes, subjectively the intelligence would have no way of knowing if A was
true or B, or to put it another way subjectively it would make no
difference.

*> I reformulated the UDA in a way that does not use any pronouns at all,
> and it doesn't matter if you consider the question from one view or from
> all the views, the conclusion is the same.*

Yes, the conclusion is the same, and that is the not very profound
conclusion that you never know what you're going to see next, and Bruno's
grand discovery of First Person Indeterminacy is just regular old dull as
dishwater indeterminacy first discovered by Og the caveman. After the big
buildup it's a bit of a letdown actually.

  John K Clark

> >> important it's crystal clear exactly what the correct prediction would
>> have turned out to be.
>>
>
> > I did a few days ago, but you didn't respond.  I'll post it again:
>

>
>
>
>
> *First, consider this experiment:Imagine there is a conscious AI (or
> uploaded mind) inside a virtual environment (an open field)Inside that
> virtual environment is a ball, which the AI is looking at and next to the
> ball is a note which reads:"At noon (when the virtual sun is directly
> overhead) the protocol will begin.  In the protocol, the process containing
> this simulation will fork (split in two), after the fork, the color of the
> ball will change to red for the parent process and it will change to blue
> in the child process (forking duplicates a process into two identical
> copies, with one called the parent and the other the child). A second after
> the color of the ball is set, another fork will happen.  This will happen 8
> times leading to 256 processes, after which the simulation will end."Now,
> with the understanding of that experiment, consider the following:If the AI
> (or all of them) went through two tests, test A, and test B*

 *A) The test described where the simulation process forks 8 times and 256
> copies are created and they each see a different pattern of the ball
> changing color*

 *B) A test where the AI is not duplicated but instead a random number
> generator (controlled entirely outside the simulation) determines whether
> the ball changes to red or blue with 50% probability 8 times*

*Then the AI (or AIs) could not say whether test A occurred first or test B
> occurred first.*

*Do you agree that it is impossible for any entity within the simulation to
> determine whether test A was executed first, or whether test B was executed
> first, with higher than a 50% probability?*

Yes of course I agree with that, but that doesn't mean Bruno's "question
isn't gibberish as is his "proof"!  Unlike Bruno's thought experiment you
did not use any personal pronouns and I congratulate you for that, although
why you made it so convoluted is a mystery to me. And unlike Bruno you
didn't demand predicti

Re: Sean Carroll gets past "Step 3" of the UDA

> On 24 Sep 2019, at 03:33, 'Brent Meeker' via Everything List 
>  wrote:
> 
> 
> 
> On 9/23/2019 4:23 PM, Jason Resch wrote:
>> Yet, assuming an arithmetical world governed by arithmetical laws, you can 
>> derive the appearance of a physical universe governed by physical laws.
> 
> If only it were so.  So far it's hand waving aspiration.

Not all. We get first that a physical universe has to be perceive by the 
numbers, that explains why we believe in a physical universe, without 
ontological commitment. Only that is far better than an extrapolation from an 
ostentatious exhibit.

Secondly, the proof is constructive. It says physics is given by those precise 
modes of self-reference. 

Thirdly, we get the full quantum logic, with testable difference (the 
arithmetical quantum logic has been shown richer than most physical quantum 
one).

Fourthly, we get a theory of qualia and consciousness coherent with the 
prediction on pur first person experience, where physicalism, when rigorous, 
has to eliminate or dismiss qualia and consciousness.

Physicalism has never work, except by denying the mind, but with Digital 
Mechanism, we know why, and we know how to improve/correct it.

The real trouble are for those who defend both Mechanism and 
Materialism/Physicalism, be it with one world or many.

Bruno

> 
> Brent
> 
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Re: Sean Carroll gets past "Step 3" of the UDA


> On 24 Sep 2019, at 00:43, John Clark  wrote:
> 
> On Mon, Sep 23, 2019 at 11:41 AM Jason Resch  > wrote:
> 
> Jason thinks I must be suffering from buyer's remorse because I "spent 
> $80,000 when he is already saved by arithmetic" he concludes this because on 
> December 26 2012 at 12:34 PM I said " A better question is do the natural 
> numbers need a reason to exist? I don't know the answer to that but my hunch 
> is no". However in another post on December 26 2012 at 1:26 PM, less than 2 
> hours later I said "it is a fact that thinking of information as something 
> physical has over the last century proven itself to be remarkably fertile and 
> has led to the discovery of new knowledge, while thinking of information as 
> ethereal was found to be sterile and has led to nowhere and nothing".
> 
> The existence of the natural numbers may or may not be a brute fact, but it 
> is certainly NOT a brute fact that we teach our children the particular 
> metric to measure the distance a natural number is from zero that yields 
> results such as 2+2=4 and not one of the infinite number of other self 
> consistent ones that the P-adic metric can provide. It is not a brute fact 
> because there is a reason for it, we teach that one and only that one to 
> children because it is the only one that is consistent with the physical 
> world. And because that one is far more intuitive than any P-adic one. And it 
> is more intuitive precisely because it is consistent with the physical world 
> we see around us and P-adic is not.
>  
> > However he uses the static nature of arithmetical truth to presume that it 
> > cannot represent "real computations". 
> 
> There is a easy way to tell a "real computation" from the other sort. Your 
> computer can make one sort of computation without a battery or a AC power 
> outlet, but for the other sort your computer needs electricity.  And you can 
> *do* something with one sort of calculation, but you can't *do* anything with 
> the other sort of "calculation". 
> 
> > But he has not indicated why fundamental change (which I take to mean 
> > successive creation and destruction of states) should be necessary to 
> > computation,
> 
> Do I really need to indicate why you can't create or destroy something 
> without making a change? I don't think so. But I think you need to indicate 
> how, out of the set of all computations, you can pick the correct ones from 
> the incorrect ones without the help of matter that obeys the laws of physics. 
>  
> 
> I think meaning needs contrast. Michelangelo's David was carved from a single 
> huge block of marble that was a 100 million years old, but it would be silly 
> to say David was 100 million years old and Michelangelo did nothing but 
> unpack it from the marble that was not part of David. And to make a real 
> calculation rather than a pretend toy one you have to differentiate the 
> correct from the incorrect, you not only have to mention the correct answer 
> you have to make it clear that all the other answers, and there are a 
> infinite number of them, are wrong. And for that you need a physical machine.
> 
>  > I think John has also argued against philosophical zombies.
> 
> I have indeed.

But then you accept infinitely many zombie in arithmetic, or deny the theorem 
in arithmetic sating that the computations exist (and *are* computation).



>  
> > John's theory that fundamental change is required leads to an infinity of 
> > philosophical zombies existing within the arithmetical computations,
> 
> My theory is NOTHING exists within arithmetical computations because 
> arithmetical computations don't exist

False with exist taken in the same sense as in “their exists no biggest prime 
number”.



> (existence being defined as stuff that can *do* things),

In metaphysics or theology when done with the scientific attitude, this invoke 
your personal ontological commitment.

That is as funny as the drawing of the guy doing a proof and invoking a miracle.

That is not even religion, but pseudo-religion or pseudo-science.





> but physical computations certainly exist and can *do" all sorts of things.


That is like the priest of the institutionalised religion. You talk like if you 
knew the truth. That is automatically invalid.

Bruno 




>  
> > 1. Can the time evolution of John Clark's brain be described by the 
> > solutions to a particular Diophantine equation? (e.g. an equation with 
> > variables t and s, where t = number of Plank times since start of 
> > emulation, and s = the wave function describing all the particles in your 
> > skull)
> 
> It can unless physics needs Real Numbers and it probably doesn't. Yes  
> Schrodinger's equation uses Real Numbers because it assumes space and time 
> are continuous, but that is probably only approximately true.  And there are 
> a infinite number of equations and mathematically there is absolutely nothing 
> special about Schrodinger's equation,

Re: Sean Carroll gets past "Step 3" of the UDA


> On 23 Sep 2019, at 17:59, Alan Grayson  wrote:
> 
> 
> 
> On Monday, September 23, 2019 at 9:23:29 AM UTC-6, Jason wrote:
> 
> 
> On Mon, Sep 23, 2019 at 5:51 AM John Clark > 
> wrote:
> On Mon, Sep 23, 2019 at 6:31 AM Bruno Marchal  > wrote:
> 
> > if you think that Carroll’s got it right, you do accept step 3, (as Carroll 
> > accept it, according to Jason) 
> 
> If Jason thinks Carroll accepts it then Jason is full of shit. And I've read 
> Carroll's book, you haven't. You two should actually read the book, then 
> we'll talk.
> 
> 
> I guess you never clicked the link I provided at the start of this thread.  
> I'll transcribe it for those who can't access the video:
> 
> Sean Carroll:
> So Isaac Newton came up with the rules of classical mechanics in the 1600s, 
> but it wasn't until Laplace around the year 1800 that this implication of 
> classical mechanics was realized.
> It's a clockwork universe.  That the way classical mechanics works is if you 
> tell me the state of a system right now at one moment by which in classical 
> mechanics you would mean the position and the velocity of every part, and you 
> knew the laws of physics and you had arbitrarily large computational 
> capacity, 
> Laplace said of vast intelligence okay then to that vast intelligence the 
> past and future would be as determined and known as the present was because 
> that's the clockwork universe is deterministic everything is fixed once you 
> know the present moment.
>  
> But Laplace was wrong in one very important respect. One can never know the 
> exact position and momentum of any particle, let alone the entire universe.

Like in arithmetic. We can never know-for-sure which machine we are, nor which 
computations “run” us. 

That does not make arithmetic non deterministic. Likewise, quantum mechanics is 
a purely deterministic theory, independently that we, from inside, cannot use 
it to predict our future 1p (plural) experiments.




> There are no perfect measurements! Further, the situation is further 
> aggravated by the Uncertainty Principle. In sum, using classical mechanics 
> the future is NOT determined by its present, imprecise configuration. Not 
> only is Laplace mistaken, but Carroll as well, who should know better. AG 


Laplace is mistaken with respect to the quantum theory.

Carroll is just incomplete with respect to Mechanism (and with respect to the 
problem of qualia, consciousness, which is not his domain of investigation.

Bruno





> 
> Now quantum mechanics comes along and throws a spanner into the works a 
> little bit if you're a many-worlds person Laplace is demon is still possible.
> So if you know the wave function of the universe exactly and you have 
> infinite calculational capacity you could predict the past and the future 
> with perfect accuracy.
> But! what you're predicting is all of the branches of the wavefunction so any 
> individual person inside the wavefunction still experiences apparently random 
> events.
> 
> Right, so you can't predict what will happen to you even if you can predict 
> what will happen to the entire universe.
> 
> This is the essence of Step 3 of the UDA.  In an experiment involving 
> duplication of persons, apparent randomness emerges.  There is no actual 
> randomness in the complete system, but individual experiences will have the 
> characteristic of randomness, in the sense of not being able to make definite 
> predictions concerning their experiences.  Sean Carroll gets this.  Max 
> Tegmark gets this.  You got it at least once 6 years ago on this list when 
> you agreed that a forking computer process containing AIs could not predict 
> which process they would end up in.  This is enough for you to proceed to the 
> next step, which adds only a time delay to one of the duplicates.  You are 
> almost there.
> 
> Jason
>  
>  
> 
> Now quantum mechanics comes along and throws a spanner into the works a 
> little bit if you're a many-worlds person Laplace is demon is still possible.
> So if you know the wave function of the universe exactly and you have 
> infinite calculational capacity you could predict the past and the future 
> with perfect accuracy.
> But! what you're predicting is all of the branches of the wavefunction so any 
> individual person inside the wavefunction still experiences apparently random 
> events.
> 
> Right, so you can't predict what will happen to you even if you can predict 
> what will happen to the entire universe.
> 
> This is the essence of Step 3 of the UDA.  In an experiment involving 
> duplication of persons, apparent randomness emerges.  There is no actual 
> rando

Re: Sean Carroll gets past "Step 3" of the UDA


> On 23 Sep 2019, at 17:41, Jason Resch  wrote:
> 
> 
> 
> On Mon, Sep 23, 2019 at 5:31 AM Bruno Marchal  <mailto:marc...@ulb.ac.be>> wrote:
> 
>> On 22 Sep 2019, at 11:43, John Clark > <mailto:johnkcl...@gmail.com>> wrote:
>> 
>> On Thu, Sep 19, 2019 at 6:41 AM Jason Resch > <mailto:jasonre...@gmail.com>> wrote:
>> 
>> > Perhaps Carroll's explanation might help others who've struggled to get 
>> > past Step 3.
>> 
>> If Jason Resch reads Carroll's book as John Clark has done then Jason Resch 
>> will find that Carroll goes into considerable detail explaining what the 
>> personal pronoun "you" could mean when there are multiple copies of "you". 
>> And that is something John Clark has done many times on this list, and that 
>> is something Bruno has never done and is what makes step 3 not just wrong 
>> but silly. 
> 
> On the contrary, each time I have used the nuances (1p, 3p, 1-plural-p)  to 
> explain step 3, all your critics have suppress the nuances, usually using 
> mockery and semantic play, without any argument understoodd by any on this 
> list.
> 
> Then, if you think that Carroll’s got it right, you do accept step 3, (as 
> Carroll accept it, according to Jason) and it is even more weird why you have 
> not yet move to step 4.
> 
> Of course, we know that you will have a problem with step 7, as you believe 
> that a computation is ream only off implemented in an assumed physical 
> reality, but this contradict a century of computer science.
> 
> 
> 
> Perhaps it is a manifestation of "buyer's remorse" (he spent $80,000 when he 
> is already saved by arithmetic).

I am not sure. The saving in arithmetic might be close to the Indian Nirvana 
idea. Technological immortality is for those who want save their ego, their 
local memories, and somehow procrastinate the Nirvana, and pursue the Samsara.

Saying “yes” ti the doctor is rather vain, if the goal is only to prolongate 
existence, but it can be sensefull if the goal is being able to see the next 
soccer cup.




> 
> While he might have a problem with step 7, it appears John Clark does support 
> arithmetical realism:

Clark is like my early opponent. They mocked it quite loudly and publicly 
before studying the argument, just because they see word like “consciousness” 
or “reality”, and when they understand there is a reasoning, a theory, means of 
testing it, they do not want to admit they were wrong.

Some people cannot change their mind.

It is sad that people open to the MW shows difficulties for the simpler and 
more obvious (provable) “many-computation” in arithmetic.

Now Clark seems also to have some more genuine  difficulties in mathematical 
logic, as he confused theory and models regularly. To his discharge, 
mathematical logic is poorly taught, when taught.



> 
> John Clark mailto:johnkcl...@gmail.com>> 12/26/12
> to everything-list
> On Sat, Dec 22, 2012 at 11:05 AM, Telmo Menezes  <mailto:te...@telmomenezes.com>> wrote:
> > Why do the natural numbers exist?
> A better question is do the natural numbers need a reason to exist? I don't 
> know the answer to that but my hunch is no.
>  
> However he uses the static nature of arithmetical truth to presume that it 
> cannot represent "real computations".  But he has not indicated why 
> fundamental change (which I take to mean successive creation and destruction 
> of states) should be necessary to computation, while the indexical eternal 
> existence of each successive computational state won't do. John's theory that 
> fundamental change is required leads to an infinity of philosophical zombies 
> existing within the arithmetical computations, but I think John has also 
> argued against philosophical zombies.  I would like him to answer the 
> following questions:
> 
> 1. Can the time evolution of John Clark's brain be described by the solutions 
> to a particular Diophantine equation? (e.g. an equation with variables t and 
> s, where t = number of Plank times since start of emulation, and s = the wave 
> function describing all the particles in your skull)
> 2. Are those brain states found in the collection of solutions to that 
> equation reflective of a philosophical zombie? (e.g. could we build a John 
> Clark robot that behaved exactly as John Clark would by searching for 
> solutions to this equation, which would not be conscious)

I let people guess what John Clark could say, above his taking granted a 
fundamental primary time (like Prigogine) or a fundamental primary physical 
universe (like the Aristotelians).

Bruno 



> 
> Jason
> 
> -- 
> You received this message because you are

Re: Sean Carroll gets past "Step 3" of the UDA


> On 23 Sep 2019, at 12:50, John Clark  wrote:
> 
> On Mon, Sep 23, 2019 at 6:31 AM Bruno Marchal  <mailto:marc...@ulb.ac.be>> wrote:
> 
> > if you think that Carroll’s got it right, you do accept step 3, (as Carroll 
> > accept it, according to Jason) 
> 
> If Jason thinks Carroll accepts it then Jason is full of shit.

Easy, gross and .. not a valid argument.


> And I've read Carroll's book, you haven't. You two should actually read the 
> book, then we'll talk.

More than one people told you since long that QM many-worlds use the first 
person indeterminacy more or less explicitly.

Bruno




> 
> John K Clark
> 
> 
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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-23 Thread 'Brent Meeker' via Everything List





On 9/23/2019 4:23 PM, Jason Resch wrote:
Yet, assuming an arithmetical world governed by arithmetical laws, you 
can derive the appearance of a physical universe governed by physical 
laws.


If only it were so.  So far it's hand waving aspiration.

Brent

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Re: Sean Carroll gets past "Step 3" of the UDA

On Mon, Sep 23, 2019 at 5:44 PM John Clark  wrote:

> On Mon, Sep 23, 2019 at 11:41 AM Jason Resch  wrote:
>
> Jason thinks I must be suffering from buyer's remorse because I "spent
> $80,000 when he is already saved by arithmetic" he concludes this because
> on December 26 2012 at 12:34 PM I said " *A better question is do the
> natural numbers need a reason to exist? I don't know the answer to that but
> my hunch is no*". However in another post on December 26 2012 at 1:26 PM,
> less than 2 hours later I said "*it is a fact that thinking of
> information as something physical has over the last century proven itself
> to be remarkably fertile and has led to the discovery of new knowledge,
> while thinking of information as ethereal was found to be sterile and has
> led to nowhere and nothing*".
>
> The existence of the natural numbers may or may not be a brute fact, but
> it is certainly NOT a brute fact that we teach our children the particular
> metric to measure the distance a natural number is from zero that yields
> results such as 2+2=4 and not one of the infinite number of other self
> consistent ones that the P-adic metric can provide. It is not a brute fact
> because there is a reason for it, we teach that one and only that one to
> children because it is the only one that is consistent with the physical
> world. And because that one is far more intuitive than any P-adic one. And
> it is more intuitive precisely because it is consistent with the physical
> world we see around us and P-adic is not.
>
>
>> > *However he uses the static nature of arithmetical truth to presume
>> that it cannot represent "real computations". *
>>
>
> There is a easy way to tell a "real computation" from the other sort.
> Your computer can make one sort of computation without a battery or a AC
> power outlet, but for the other sort your computer needs electricity.  And you
> can *do* something with one sort of calculation, but you can't *do*
> anything with the other sort of "calculation".
>
> *> But he has not indicated why fundamental change (which I take to mean
>> successive creation and destruction of states) should be necessary to
>> computation,*
>>
>
> Do I really need to indicate why you can't create or destroy something
> without making a change? I don't think so. But I think you need to indicate
> how, out of the set of all computations, you can pick the correct ones from
> the incorrect ones without the help of matter that obeys the laws of
> physics.
>

How do you suppose the laws of physics pick out the correct physical
outcomes from among all possibilities?  You presume there is a physical
world governed by physical laws.  But you deny an arithmetical world
governed by arithmetical laws.  Yet, assuming an arithmetical world
governed by arithmetical laws, you can derive the appearance of a physical
universe governed by physical laws.


>
> I think meaning needs contrast. Michelangelo's David was carved from a
> single huge block of marble that was a 100 million years old, but it would
> be silly to say David was 100 million years old and Michelangelo did
> nothing but unpack it from the marble that was not part of David. And to
> make a real calculation rather than a pretend toy one you have to
> differentiate the correct from the incorrect, you not only have to
> mention the correct answer you have to make it clear that all the other
> answers, and there are a infinite number of them, are wrong. And for that
> you need a physical machine.
>
> * > I think John has also argued against philosophical zombies.*
>
>
> I have indeed.
>
>>
>
>> > *John's theory that fundamental change is required leads to an
>> infinity of philosophical zombies existing within the arithmetical
>> computations,*
>>
>
> My theory is NOTHING exists within arithmetical computations because
> arithmetical computations don't exist (existence being defined as stuff
> that can *do* things), but physical computations certainly exist and can
> *do" all sorts of things.
>

>
>> *> 1. Can the time evolution of John Clark's brain be described by the
>> solutions to a particular Diophantine equation? (e.g. an equation with
>> variables t and s, where t = number of Plank times since start of
>> emulation, and s = the wave function describing all the particles in your
>> skull)*
>>
>
> It can unless physics needs Real Numbers and it probably doesn't. Yes
> Schrodinger's equation uses Real Numbers because it assumes space and time
> are continuous, but that is probably only approximately true.  And there
> are a infinite number of equations and mathematically there is absolutely
> nothing special about Schrodinger's equation, the only thing special about
> that particular equation is it conforms with our observations of how the
> physical world behaves.
>
> And I'm very surprised that as soon as you mentioned the Planck Time in
> the above you didn't realize you had left the world of pure dimensionless
> numbersand was talking numbers with phys

Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-23 Thread John Clark

On Mon, Sep 23, 2019 at 11:41 AM Jason Resch  wrote:

Jason thinks I must be suffering from buyer's remorse because I "spent
$80,000 when he is already saved by arithmetic" he concludes this because
on December 26 2012 at 12:34 PM I said " *A better question is do the
natural numbers need a reason to exist? I don't know the answer to that but
my hunch is no*". However in another post on December 26 2012 at 1:26 PM,
less than 2 hours later I said "*it is a fact that thinking of information
as something physical has over the last century proven itself to be
remarkably fertile and has led to the discovery of new knowledge, while
thinking of information as ethereal was found to be sterile and has led to
nowhere and nothing*".

The existence of the natural numbers may or may not be a brute fact, but it
is certainly NOT a brute fact that we teach our children the particular
metric to measure the distance a natural number is from zero that yields
results such as 2+2=4 and not one of the infinite number of other self
consistent ones that the P-adic metric can provide. It is not a brute fact
because there is a reason for it, we teach that one and only that one to
children because it is the only one that is consistent with the physical
world. And because that one is far more intuitive than any P-adic one. And
it is more intuitive precisely because it is consistent with the physical
world we see around us and P-adic is not.

> > *However he uses the static nature of arithmetical truth to presume
> that it cannot represent "real computations". *
>

There is a easy way to tell a "real computation" from the other sort. Your
computer can make one sort of computation without a battery or a AC power
outlet, but for the other sort your computer needs electricity.  And you
can *do* something with one sort of calculation, but you can't *do*
anything with the other sort of "calculation".

*> But he has not indicated why fundamental change (which I take to mean
> successive creation and destruction of states) should be necessary to
> computation,*
>

Do I really need to indicate why you can't create or destroy something
without making a change? I don't think so. But I think you need to indicate
how, out of the set of all computations, you can pick the correct ones from
the incorrect ones without the help of matter that obeys the laws of
physics.

I think meaning needs contrast. Michelangelo's David was carved from a
single huge block of marble that was a 100 million years old, but it would
be silly to say David was 100 million years old and Michelangelo did
nothing but unpack it from the marble that was not part of David. And to
make a real calculation rather than a pretend toy one you have to
differentiate the correct from the incorrect, you not only have to mention
the correct answer you have to make it clear that all the other answers,
and there are a infinite number of them, are wrong. And for that you need a
physical machine.

* > I think John has also argued against philosophical zombies.*

I have indeed.

>

> > *John's theory that fundamental change is required leads to an infinity
> of philosophical zombies existing within the arithmetical computations,*
>

My theory is NOTHING exists within arithmetical computations because
arithmetical computations don't exist (existence being defined as stuff
that can *do* things), but physical computations certainly exist and can
*do" all sorts of things.

> *> 1. Can the time evolution of John Clark's brain be described by the
> solutions to a particular Diophantine equation? (e.g. an equation with
> variables t and s, where t = number of Plank times since start of
> emulation, and s = the wave function describing all the particles in your
> skull)*
>

It can unless physics needs Real Numbers and it probably doesn't. Yes
Schrodinger's equation uses Real Numbers because it assumes space and time
are continuous, but that is probably only approximately true.  And there
are a infinite number of equations and mathematically there is absolutely
nothing special about Schrodinger's equation, the only thing special about
that particular equation is it conforms with our observations of how the
physical world behaves.

And I'm very surprised that as soon as you mentioned the Planck Time in the
above you didn't realize you had left the world of pure dimensionless number
sand was talking numbers with physical units associated with them, like
measures of time and space and mass and energy and electrical charge.

*> 2. Are those brain states found in the collection of solutions to that
> equation reflective of a philosophical zombie?*
>

No.

> *could we build a John Clark robot that behaved exactly as John Clark
> would by searching for solutions to this equation, which would not be
> conscious*
>

No. And it would not behave exactly like John Clark, it would not behave at
all because without physics there would be no way to search through
solutions to that equation or to any other.

John K

Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-23 Thread 'Brent Meeker' via Everything List

On 9/23/2019 12:45 PM, Philip Thrift wrote:

On Monday, September 23, 2019 at 2:32:11 PM UTC-5, Brent wrote:

On 9/23/2019 8:59 AM, Alan Grayson wrote:
> But Laplace was wrong in one very important respect. One can never
> know the exact position and momentum of any particle, let alone the
> entire universe. There are no perfect measurements!

Laplace knew that. His point was that the future (and the past) were
completely determined by the present state of the world.  Even
though we
can't measure it perfectly, Laplace assumed that the variables like
position and  momentum had definite values.  That's what is
fundamentally different about quantum mechanics, they don't have
definite values.

> Further, the situation is further aggravated by the Uncertainty
> Principle. In sum, using classical mechanics the future is NOT
> determined by its present, imprecise configuration.

The uncertainty principle is part of QM not CM.  Just because you
can't
measure it precisely, doesn't mean that the present configuration
is not
precise; it means that we are ignorant of the precise values. This
was
Einstein's idea, that QM was incomplete and its randomness was
just an
expression of our ignorance, as in CM.

> Not only is Laplace mistaken, but Carroll as well, who should know
> better. AG

Neither Laplace nor Carroll is mistaken.

Brent

Both were/are superstitious, basically religiously so, in their 
fear/rejection of probabilities.

https://en.wikipedia.org/wiki/Laplace%27s_demon#Quantum_mechanical_irreversibility

Did you tell Carroll that?

Brent

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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-23 Thread Alan Grayson



On Monday, September 23, 2019 at 1:32:11 PM UTC-6, Brent wrote:
>
>
>
> On 9/23/2019 8:59 AM, Alan Grayson wrote: 
> > But Laplace was wrong in one very important respect. One can never 
> > know the exact position and momentum of any particle, let alone the 
> > entire universe. There are no perfect measurements! 
>
> Laplace knew that. His point was that the future (and the past) were 
> completely determined by the present state of the world.  Even though we 
> can't measure it perfectly, Laplace assumed that the variables like 
> position and  momentum had definite values.  That's what is 
> fundamentally different about quantum mechanics, they don't have 
> definite values. 
>
> > Further, the situation is further aggravated by the Uncertainty 
> > Principle. In sum, using classical mechanics the future is NOT 
> > determined by its present, imprecise configuration. 
>
> The uncertainty principle is part of QM not CM.  


Yes, and I didn't indicate otherwise. AG 

Just because you can't 
> measure it precisely, doesn't mean that the present configuration is not 
> precise; it means that we are ignorant of the precise values. This was 
> Einstein's idea, that QM was incomplete and its randomness was just an 
> expression of our ignorance, as in CM. 
>

 That's not the mainstream view today IIUC. It's that position and momentum 
as simultaneous values don't exist, not that we can't measure them 
precisely. AG

>
> > Not only is Laplace mistaken, but Carroll as well, who should know 
> > better. AG 
>
> Neither Laplace nor Carroll is mistaken. 
>

Carroll intentionally misstated Laplace's position in an attempt to make 
his reasoning plausible. So IMO he's not only wrong about MW, but dishonest 
as well. AG 

>
> Brent 
>
>
>

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Re: Sean Carroll gets past "Step 3" of the UDA

On Mon, Sep 23, 2019 at 2:58 PM John Clark  wrote:

> On Mon, Sep 23, 2019 at 11:23 AM Jason Resch  wrote:
>
> *> I guess you never clicked the link I provided at the start of this
>> thread.  *
>
>
> I've done a lot better than click on a link that provides a brief
> synopsis, I've spent hours reading every page in the man's entire book and
> you and Bruno should do the same.
>
> *>You got it at least once 6 years ago on this list when you agreed that a
>> forking computer process containing AIs could not predict which process
>> they would end up in. *
>
>
> I don't know what you're referring to so it's hard to know how to respond, but
> since you can pinpoint the exact time, 6 years ago, you should be able to
> include the exact quote where I said I "got it" and enough context around
> it so it's clear who "they" are that failed to make a prediction, and even
> more important it's crystal clear exactly what the correct prediction would
> have turned out to be.
>

I did a few days ago, but you didn't respond.  I'll post it again:

This Halloween will mark 6 years since you agreed with Step 3, but said it
was a let down (presumably because you thought it so obvious):

https://groups.google.com/d/msg/everything-list/5PR1FXp_CSU/73ltRVEHUtQJ
https://groups.google.com/d/msg/everything-list/5PR1FXp_CSU/PnuTSn_82PwJ

Should we expect another 6 years before you proceed through the next
steps?  There's no rush, since you are freezing yourself this debate could
go on another 10^100 years.



>
> > quoting Carroll: "*Now quantum mechanics comes along and throws a
>> spanner into the works a little bit if you're a many-worlds person Laplace
>> is demon is still possible*".
>
>
> Yes, if Many Worlds is correct then the Schrodinger Wave Equation of the
> Multiverse is all there is, and it is a 100% deterministic equation, so 
> Laplace's
> demon could solve it and in theory *you* could too. And yet the empirical
> fact remains *you*  can NOT predict the future, at least not always and
> not perfectly. If Many Worlds could not explain this obvious glaring
> discrepancy it would be dead dead dead. But Many Worlds can explain it and
> can do so easily; *you* can't answer the question "*What one and only one
> thing will **you** see tomorrow after the universe splits?*" for exactly
> the same reason *you* can't answer Bruno's question "*What one and only
> one thing will **you*  *see tomorrow after **you** are duplicated and *
> *you* *become two and **you** see two different things?*" The  difference
> is in the Many Worlds case, after the universe splits, if I asked *you* today
> what the correct answer *you* should have given yesterday was:
>
> 1) It would be obvious who the question was directed to.
> 2)  It would obvious what would have been the correct answer.
>
> Neither of these things is true for Bruno's "question".
>

What's so special about duplicating universes?  Perhaps you can explain why
one leads to apparent randomness and the other case does not.


>
> Of course Sean Carroll delves into this issue in far greater detail that
> I have here, and you'd know that if you had read the man's book as I have.
>

Is anything I said about Carroll wrong?  What do you hope I will learn from
reading Caroll's book?

Jason

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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-23 Thread John Clark

On Mon, Sep 23, 2019 at 11:23 AM Jason Resch  wrote:

*> I guess you never clicked the link I provided at the start of this
> thread.  *

I've done a lot better than click on a link that provides a brief synopsis,
I've spent hours reading every page in the man's entire book and you and
Bruno should do the same.

*>You got it at least once 6 years ago on this list when you agreed that a
> forking computer process containing AIs could not predict which process
> they would end up in. *

I don't know what you're referring to so it's hard to know how to respond, but
since you can pinpoint the exact time, 6 years ago, you should be able to
include the exact quote where I said I "got it" and enough context around
it so it's clear who "they" are that failed to make a prediction, and even
more important it's crystal clear exactly what the correct prediction would
have turned out to be.

> quoting Carroll: "*Now quantum mechanics comes along and throws a spanner
> into the works a little bit if you're a many-worlds person Laplace is demon
> is still possible*".

Yes, if Many Worlds is correct then the Schrodinger Wave Equation of the
Multiverse is all there is, and it is a 100% deterministic equation,
so Laplace's
demon could solve it and in theory *you* could too. And yet the empirical
fact remains *you*  can NOT predict the future, at least not always and not
perfectly. If Many Worlds could not explain this obvious glaring
discrepancy it would be dead dead dead. But Many Worlds can explain it and
can do so easily; *you* can't answer the question "*What one and only one
thing will **you** see tomorrow after the universe splits?*" for exactly
the same reason *you* can't answer Bruno's question "*What one and only one
thing will **you*  *see tomorrow after **you** are duplicated and
**you* *become
two and **you** see two different things?*" The  difference is in the Many
Worlds case, after the universe splits, if I asked *you* today what the
correct answer *you* should have given yesterday was:

1) It would be obvious who the question was directed to.
2)  It would obvious what would have been the correct answer.

Neither of these things is true for Bruno's "question".

Of course Sean Carroll delves into this issue in far greater detail that I
have here, and you'd know that if you had read the man's book as I have.

John K Clark

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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-23 Thread Philip Thrift



On Monday, September 23, 2019 at 2:32:11 PM UTC-5, Brent wrote:
>
>
>
> On 9/23/2019 8:59 AM, Alan Grayson wrote: 
> > But Laplace was wrong in one very important respect. One can never 
> > know the exact position and momentum of any particle, let alone the 
> > entire universe. There are no perfect measurements! 
>
> Laplace knew that. His point was that the future (and the past) were 
> completely determined by the present state of the world.  Even though we 
> can't measure it perfectly, Laplace assumed that the variables like 
> position and  momentum had definite values.  That's what is 
> fundamentally different about quantum mechanics, they don't have 
> definite values. 
>
> > Further, the situation is further aggravated by the Uncertainty 
> > Principle. In sum, using classical mechanics the future is NOT 
> > determined by its present, imprecise configuration. 
>
> The uncertainty principle is part of QM not CM.  Just because you can't 
> measure it precisely, doesn't mean that the present configuration is not 
> precise; it means that we are ignorant of the precise values. This was 
> Einstein's idea, that QM was incomplete and its randomness was just an 
> expression of our ignorance, as in CM. 
>
>
> > Not only is Laplace mistaken, but Carroll as well, who should know 
> > better. AG 
>
> Neither Laplace nor Carroll is mistaken. 
>
> Brent 
>
>
>
Both were/are superstitious, basically religiously so, in their 
fear/rejection of probabilities.

https://en.wikipedia.org/wiki/Laplace%27s_demon#Quantum_mechanical_irreversibility

@philipthrift

 

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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-23 Thread 'Brent Meeker' via Everything List

On 9/23/2019 8:59 AM, Alan Grayson wrote:
But Laplace was wrong in one very important respect. One can never
know the exact position and momentum of any particle, let alone the
entire universe. There are no perfect measurements!

Laplace knew that. His point was that the future (and the past) were
completely determined by the present state of the world. Even though we
can't measure it perfectly, Laplace assumed that the variables like
position and momentum had definite values. That's what is
fundamentally different about quantum mechanics, they don't have
definite values.

Further, the situation is further aggravated by the Uncertainty
Principle. In sum, using classical mechanics the future is NOT
determined by its present, imprecise configuration.

The uncertainty principle is part of QM not CM. Just because you can't
measure it precisely, doesn't mean that the present configuration is not
precise; it means that we are ignorant of the precise values. This was
Einstein's idea, that QM was incomplete and its randomness was just an
expression of our ignorance, as in CM.

Not only is Laplace mistaken, but Carroll as well, who should know
better. AG

Neither Laplace nor Carroll is mistaken.

Brent

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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-23 Thread Alan Grayson



On Monday, September 23, 2019 at 9:23:29 AM UTC-6, Jason wrote:
>
>
>
> On Mon, Sep 23, 2019 at 5:51 AM John Clark  > wrote:
>
>> On Mon, Sep 23, 2019 at 6:31 AM Bruno Marchal > > wrote:
>>
>> *> if you think that Carroll’s got it right, you do accept step 3, (as 
>>> Carroll accept it, according to Jason) *
>>
>>
>> If Jason thinks Carroll accepts it then Jason is full of shit. And I've 
>> read Carroll's book, you haven't. You two should actually read the book, 
>> then we'll talk.
>>
>>
> I guess you never clicked the link I provided at the start of this 
> thread.  I'll transcribe it for those who can't access the video:
>
> Sean Carroll:
>
> So Isaac Newton came up with the rules of classical mechanics in the 
> 1600s, but it wasn't until Laplace around the year 1800 that this 
> implication of classical mechanics was realized.
> It's a clockwork universe.  That the way classical mechanics works is if 
> you tell me the state of a system right now at one moment by which in 
> classical mechanics you would mean the position and the velocity of every 
> part, and you knew the laws of physics and you had arbitrarily large 
> computational capacity, 
> Laplace said of vast intelligence okay then to that vast intelligence the 
> past and future would be as determined and known as the present was because 
> that's the clockwork universe is deterministic everything is fixed once you 
> know the present moment.
>
>
*(Indentation fixed).*
*But Laplace was wrong in one very important respect. One can never know 
the exact position and momentum of any particle, let alone the entire 
universe. There are no perfect measurements! Further, the situation is 
further aggravated by the Uncertainty Principle. In sum, using classical 
mechanics the future is NOT determined by its present, imprecise 
configuration. Not only is Laplace mistaken, but Carroll as well, who 
should and does know better. AG*
 

>
> Now quantum mechanics comes along and throws a spanner into the works a 
> little bit if you're a many-worlds person Laplace is demon is still 
> possible.
> So if you know the wave function of the universe exactly and you have 
> infinite calculational capacity you could predict the past and the future 
> with perfect accuracy.
> But! what you're predicting is all of the branches of the wavefunction so 
> any individual person inside the wavefunction still experiences apparently 
> random events.
>
> Right, so *you can't predict what will happen to you even if you can 
> predict what will happen to the entire universe*.
>
>
*The premise is wrong. Based on classical or quantum mechanics, one cannot 
predict the future of the universe, in part or in whole. AG *

>
> This is the essence of Step 3 of the UDA.  In an experiment involving 
> duplication of persons, apparent randomness emerges.  There is no actual 
> randomness in the complete system, but individual experiences will have the 
> characteristic of randomness, in the sense of not being able to make 
> definite predictions concerning their experiences.  Sean Carroll gets 
> this.  Max Tegmark gets this.  You got it at least once 6 years ago on this 
> list when you agreed that a forking computer process containing AIs could 
> not predict which process they would end up in.  This is enough for you to 
> proceed to the next step, which adds only a time delay to one of the 
> duplicates.  You are almost there.
>
> Jason
>  
>

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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-23 Thread Alan Grayson



On Monday, September 23, 2019 at 9:23:29 AM UTC-6, Jason wrote:
>
>
>
> On Mon, Sep 23, 2019 at 5:51 AM John Clark  > wrote:
>
>> On Mon, Sep 23, 2019 at 6:31 AM Bruno Marchal > > wrote:
>>
>> *> if you think that Carroll’s got it right, you do accept step 3, (as 
>>> Carroll accept it, according to Jason) *
>>
>>
>> If Jason thinks Carroll accepts it then Jason is full of shit. And I've 
>> read Carroll's book, you haven't. You two should actually read the book, 
>> then we'll talk.
>>
>>
> I guess you never clicked the link I provided at the start of this 
> thread.  I'll transcribe it for those who can't access the video:
>
> Sean Carroll:
>
> So Isaac Newton came up with the rules of classical mechanics in the 
> 1600s, but it wasn't until Laplace around the year 1800 that this 
> implication of classical mechanics was realized.
> It's a clockwork universe.  That the way classical mechanics works is if 
> you tell me the state of a system right now at one moment by which in 
> classical mechanics you would mean the position and the velocity of every 
> part, and you knew the laws of physics and you had arbitrarily large 
> computational capacity, 
> Laplace said of vast intelligence okay then to that vast intelligence the 
> past and future would be as determined and known as the present was because 
> that's the clockwork universe is deterministic everything is fixed once you 
> know the present moment.
>
>  
But Laplace was wrong in one very important respect. One can never know the 
exact position and momentum of any particle, let alone the entire universe. 
There are no perfect measurements! Further, the situation is further 
aggravated by the Uncertainty Principle. In sum, using classical mechanics 
the future is NOT determined by its present, imprecise configuration. Not 
only is Laplace mistaken, but Carroll as well, who should know better. AG 

>
> Now quantum mechanics comes along and throws a spanner into the works a 
> little bit if you're a many-worlds person Laplace is demon is still 
> possible.
> So if you know the wave function of the universe exactly and you have 
> infinite calculational capacity you could predict the past and the future 
> with perfect accuracy.
> But! what you're predicting is all of the branches of the wavefunction so 
> any individual person inside the wavefunction still experiences apparently 
> random events.
>
> Right, so *you can't predict what will happen to you even if you can 
> predict what will happen to the entire universe*.
>
>
> This is the essence of Step 3 of the UDA.  In an experiment involving 
> duplication of persons, apparent randomness emerges.  There is no actual 
> randomness in the complete system, but individual experiences will have the 
> characteristic of randomness, in the sense of not being able to make 
> definite predictions concerning their experiences.  Sean Carroll gets 
> this.  Max Tegmark gets this.  You got it at least once 6 years ago on this 
> list when you agreed that a forking computer process containing AIs could 
> not predict which process they would end up in.  This is enough for you to 
> proceed to the next step, which adds only a time delay to one of the 
> duplicates.  You are almost there.
>
> Jason
>  
>
 

>
> Now quantum mechanics comes along and throws a spanner into the works a 
> little bit if you're a many-worlds person Laplace is demon is still 
> possible.
> So if you know the wave function of the universe exactly and you have 
> infinite calculational capacity you could predict the past and the future 
> with perfect accuracy.
> But! what you're predicting is all of the branches of the wavefunction so 
> any individual person inside the wavefunction still experiences apparently 
> random events.
>
> Right, so *you can't predict what will happen to you even if you can 
> predict what will happen to the entire universe*.
>
>
> This is the essence of Step 3 of the UDA.  In an experiment involving 
> duplication of persons, apparent randomness emerges.  There is no actual 
> randomness in the complete system, but individual experiences will have the 
> characteristic of randomness, in the sense of not being able to make 
> definite predictions concerning their experiences.  Sean Carroll gets 
> this.  Max Tegmark gets this.  You got it at least once 6 years ago on this 
> list when you agreed that a forking computer process containing AIs could 
> not predict which process they would end up in.  This is enough for you to 
> proceed to the next step, which adds only a time delay to one of the 
> duplicates.  You are alm

Re: Sean Carroll gets past "Step 3" of the UDA

On Mon, Sep 23, 2019 at 5:31 AM Bruno Marchal  wrote:

>
> On 22 Sep 2019, at 11:43, John Clark  wrote:
>
> On Thu, Sep 19, 2019 at 6:41 AM Jason Resch  wrote:
>
> *> Perhaps Carroll's explanation might help others who've struggled to get
>> past Step 3.*
>
>
> If Jason Resch reads Carroll's book as John Clark has done then Jason
> Resch will find that Carroll goes into considerable detail explaining
> what the personal pronoun "you" could mean when there are multiple copies
> of "you". And that is something John Clark has done many times on this
> list, and that is something Bruno has never done and is what makes step 3
> not just wrong but silly.
>
>
> On the contrary, each time I have used the nuances (1p, 3p, 1-plural-p)
>  to explain step 3, all your critics have suppress the nuances, usually
> using mockery and semantic play, without any argument understoodd by any on
> this list.
>
> Then, if you think that Carroll’s got it right, you do accept step 3, (as
> Carroll accept it, according to Jason) and it is even more weird why you
> have not yet move to step 4.
>
> Of course, we know that you will have a problem with step 7, as you
> believe that a computation is ream only off implemented in an assumed
> physical reality, but this contradict a century of computer science.
>
>
>
Perhaps it is a manifestation of "buyer's remorse" (he spent $80,000 when
he is already saved by arithmetic).

While he might have a problem with step 7, it appears John Clark does
support arithmetical realism:

John Clark  12/26/12
to everything-list
On Sat, Dec 22, 2012 at 11:05 AM, Telmo Menezes 
wrote:
> Why do the natural numbers exist?
A better question is do the natural numbers need a reason to exist? I don't
know the answer to that but my hunch is no.

However he uses the static nature of arithmetical truth to presume that it
cannot represent "real computations".  But he has not indicated why
fundamental change (which I take to mean successive creation and
destruction of states) should be necessary to computation, while the
indexical eternal existence of each successive computational state won't
do. John's theory that fundamental change is required leads to an infinity
of philosophical zombies existing within the arithmetical computations, but
I think John has also argued against philosophical zombies.  I would like
him to answer the following questions:

1. Can the time evolution of John Clark's brain be described by the
solutions to a particular Diophantine equation? (e.g. an equation with
variables t and s, where t = number of Plank times since start of
emulation, and s = the wave function describing all the particles in your
skull)
2. Are those brain states found in the collection of solutions to that
equation reflective of a philosophical zombie? (e.g. could we build a John
Clark robot that behaved exactly as John Clark would by searching for
solutions to this equation, which would not be conscious)

Jason

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Re: Sean Carroll gets past "Step 3" of the UDA

On Mon, Sep 23, 2019 at 5:51 AM John Clark  wrote:

> On Mon, Sep 23, 2019 at 6:31 AM Bruno Marchal  wrote:
>
> *> if you think that Carroll’s got it right, you do accept step 3, (as
>> Carroll accept it, according to Jason) *
>
>
> If Jason thinks Carroll accepts it then Jason is full of shit. And I've
> read Carroll's book, you haven't. You two should actually read the book,
> then we'll talk.
>
>
I guess you never clicked the link I provided at the start of this thread.
I'll transcribe it for those who can't access the video:

Sean Carroll:

So Isaac Newton came up with the rules of classical mechanics in the 1600s,
but it wasn't until Laplace around the year 1800 that this implication of
classical mechanics was realized.
It's a clockwork universe.  That the way classical mechanics works is if
you tell me the state of a system right now at one moment by which in
classical mechanics you would mean the position and the velocity of every
part, and you knew the laws of physics and you had arbitrarily large
computational capacity,
Laplace said of vast intelligence okay then to that vast intelligence the
past and future would be as determined and known as the present was because
that's the clockwork universe is deterministic everything is fixed once you
know the present moment.

Now quantum mechanics comes along and throws a spanner into the works a
little bit if you're a many-worlds person Laplace is demon is still
possible.
So if you know the wave function of the universe exactly and you have
infinite calculational capacity you could predict the past and the future
with perfect accuracy.
But! what you're predicting is all of the branches of the wavefunction so
any individual person inside the wavefunction still experiences apparently
random events.

Right, so *you can't predict what will happen to you even if you can
predict what will happen to the entire universe*.

This is the essence of Step 3 of the UDA.  In an experiment involving
duplication of persons, apparent randomness emerges.  There is no actual
randomness in the complete system, but individual experiences will have the
characteristic of randomness, in the sense of not being able to make
definite predictions concerning their experiences.  Sean Carroll gets
this.  Max Tegmark gets this.  You got it at least once 6 years ago on this
list when you agreed that a forking computer process containing AIs could
not predict which process they would end up in.  This is enough for you to
proceed to the next step, which adds only a time delay to one of the
duplicates.  You are almost there.

Jason

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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-23 Thread Philip Thrift



On Monday, September 23, 2019 at 5:23:52 AM UTC-5, Bruno Marchal wrote:
>
>
> On 20 Sep 2019, at 21:32, Philip Thrift > 
> wrote:
>
>
>
> On Friday, September 20, 2019 at 11:22:24 AM UTC-5, Bruno Marchal wrote:
>>
>>
>> On 19 Sep 2019, at 12:41, Jason Resch  wrote:
>>
>> Like Max Tegmark, in "Our Mathematical Universe", who described how 
>> duplication of the person results in an inability to perfectly predict 
>> future outcomes and experiences, in this interview Sean Carroll describes 
>> how even with perfect knowledge of the universe and it's evolution one 
>> could not make future predictions about what one will experience due to 
>> duplication:
>>
>> https://youtube.com/watch?feature=youtu.be&v=TP5W2MG8Jjs&t=1h5m
>>
>> Perhaps Carroll's explanation might help others who've struggled to get 
>> past Step 3.
>>
>>
>> Let us pray ...
>>
>>
>> Bruno
>>
>>
>>
> As Feyerabend saw (and foresaw)  science is now religion.
>
>
> When a scientist proceeds from the mathematics of any theory to any 
> certain ontology of nature, they are being a religious guru, not a 
> scientific one.
>
>
>
> I totally agree with Feyerabend, except I would have said that deducing an 
> ontology, and taking it for granted (dogma) is the pseudo-religious trick.
>
> Science has never been separated from religion, as this is logically 
> impossible, but indeed, science, when used to claim ontology becomes 
> pseudo-religion, and pseudo-science.
>
> The only problem is that Feyerabend seemed to believe in the ontology of 
> matter, in some of his text, and so fall in the trap that he described here.
>
>
> Bruno
>

I wrote the sentence above (n a tweet). :)

I put the word "certain" in "certain ontology" to mean the opposite that it 
is always possibly updatable, and not final.

@philipthrift


>

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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-23 Thread John Clark

On Mon, Sep 23, 2019 at 6:31 AM Bruno Marchal  wrote:

*> if you think that Carroll’s got it right, you do accept step 3, (as
> Carroll accept it, according to Jason) *


If Jason thinks Carroll accepts it then Jason is full of shit. And I've
read Carroll's book, you haven't. You two should actually read the book,
then we'll talk.

John K Clark

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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-23 Thread Bruno Marchal


> On 22 Sep 2019, at 11:43, John Clark  wrote:
> 
> On Thu, Sep 19, 2019 at 6:41 AM Jason Resch  <mailto:jasonre...@gmail.com>> wrote:
> 
> > Perhaps Carroll's explanation might help others who've struggled to get 
> > past Step 3.
> 
> If Jason Resch reads Carroll's book as John Clark has done then Jason Resch 
> will find that Carroll goes into considerable detail explaining what the 
> personal pronoun "you" could mean when there are multiple copies of "you". 
> And that is something John Clark has done many times on this list, and that 
> is something Bruno has never done and is what makes step 3 not just wrong but 
> silly. 

On the contrary, each time I have used the nuances (1p, 3p, 1-plural-p)  to 
explain step 3, all your critics have suppress the nuances, usually using 
mockery and semantic play, without any argument understoodd by any on this list.

Then, if you think that Carroll’s got it right, you do accept step 3, (as 
Carroll accept it, according to Jason) and it is even more weird why you have 
not yet move to step 4.

Of course, we know that you will have a problem with step 7, as you believe 
that a computation is ream only off implemented in an assumed physical reality, 
but this contradict a century of computer science.

Bruno




> 
>  John K Clark
> 
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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-23 Thread Bruno Marchal


> On 20 Sep 2019, at 21:32, Philip Thrift  wrote:
> 
> 
> 
> On Friday, September 20, 2019 at 11:22:24 AM UTC-5, Bruno Marchal wrote:
> 
>> On 19 Sep 2019, at 12:41, Jason Resch > 
>> wrote:
>> 
>> Like Max Tegmark, in "Our Mathematical Universe", who described how 
>> duplication of the person results in an inability to perfectly predict 
>> future outcomes and experiences, in this interview Sean Carroll describes 
>> how even with perfect knowledge of the universe and it's evolution one could 
>> not make future predictions about what one will experience due to 
>> duplication:
>> 
>> https://youtube.com/watch?feature=youtu.be&v=TP5W2MG8Jjs&t=1h5m 
>> <https://youtube.com/watch?feature=youtu.be&v=TP5W2MG8Jjs&t=1h5m>
>> 
>> Perhaps Carroll's explanation might help others who've struggled to get past 
>> Step 3.
> 
> Let us pray ...
> 
> 
> Bruno
> 
> 
> 
> As Feyerabend saw (and foresaw)  science is now religion.
> 
> When a scientist proceeds from the mathematics of any theory to any certain 
> ontology of nature, they are being a religious guru, not a scientific one.


I totally agree with Feyerabend, except I would have said that deducing an 
ontology, and taking it for granted (dogma) is the pseudo-religious trick.

Science has never been separated from religion, as this is logically 
impossible, but indeed, science, when used to claim ontology becomes 
pseudo-religion, and pseudo-science.

The only problem is that Feyerabend seemed to believe in the ontology of 
matter, in some of his text, and so fall in the trap that he described here.


Bruno





> 
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> 
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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-22 Thread John Clark

On Sun, Sep 22, 2019 at 11:21 AM Jason Resch  wrote:

*> Were you being wrong or silly when you accepted it 6 years ago?*
>

I don't know, it depends on what "it" was 6 years ago.

John K Clark

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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-22 Thread Jason Resch

On Sunday, September 22, 2019, John Clark  wrote:

> On Thu, Sep 19, 2019 at 6:41 AM Jason Resch  wrote:
>
> *> Perhaps Carroll's explanation might help others who've struggled to get
>> past Step 3.*
>
>
> If Jason Resch reads Carroll's book as John Clark has done then Jason
> Resch will find that Carroll goes into considerable detail explaining
> what the personal pronoun "you" could mean when there are multiple copies
> of "you". And that is something John Clark has done many times on this
> list, and that is something Bruno has never done and is what makes step 3
> not just wrong but silly.
>
>
Were you being wrong or silly when you accepted it 6 years ago?

Jason

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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-22 Thread John Clark

On Thu, Sep 19, 2019 at 6:41 AM Jason Resch  wrote:

*> Perhaps Carroll's explanation might help others who've struggled to get
> past Step 3.*


If Jason Resch reads Carroll's book as John Clark has done then Jason
Resch will
find that Carroll goes into considerable detail explaining what the
personal pronoun "you" could mean when there are multiple copies of "you".
And that is something John Clark has done many times on this list, and that
is something Bruno has never done and is what makes step 3 not just wrong
but silly.

 John K Clark

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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-20 Thread Philip Thrift



On Friday, September 20, 2019 at 11:22:24 AM UTC-5, Bruno Marchal wrote:
>
>
> On 19 Sep 2019, at 12:41, Jason Resch > 
> wrote:
>
> Like Max Tegmark, in "Our Mathematical Universe", who described how 
> duplication of the person results in an inability to perfectly predict 
> future outcomes and experiences, in this interview Sean Carroll describes 
> how even with perfect knowledge of the universe and it's evolution one 
> could not make future predictions about what one will experience due to 
> duplication:
>
> https://youtube.com/watch?feature=youtu.be&v=TP5W2MG8Jjs&t=1h5m
>
> Perhaps Carroll's explanation might help others who've struggled to get 
> past Step 3.
>
>
> Let us pray ...
>
>
> Bruno
>
>
>
As Feyerabend saw (and foresaw)  science is now religion.

When a scientist proceeds from the mathematics of any theory to any certain 
ontology of nature, they are being a religious guru, not a scientific one.

@philipthrift


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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-20 Thread Bruno Marchal


> On 19 Sep 2019, at 12:41, Jason Resch  wrote:
> 
> Like Max Tegmark, in "Our Mathematical Universe", who described how 
> duplication of the person results in an inability to perfectly predict future 
> outcomes and experiences, in this interview Sean Carroll describes how even 
> with perfect knowledge of the universe and it's evolution one could not make 
> future predictions about what one will experience due to duplication:
> 
> https://youtube.com/watch?feature=youtu.be&v=TP5W2MG8Jjs&t=1h5m 
> <https://youtube.com/watch?feature=youtu.be&v=TP5W2MG8Jjs&t=1h5m>
> 
> Perhaps Carroll's explanation might help others who've struggled to get past 
> Step 3.

Let us pray ...


Bruno



> 
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> 
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Re: Sean Carroll gets past "Step 3" of the UDA

2019-09-19 Thread Philip Thrift



On Thursday, September 19, 2019 at 5:41:44 AM UTC-5, Jason wrote:
>
> Like Max Tegmark, in "Our Mathematical Universe", who described how 
> duplication of the person results in an inability to perfectly predict 
> future outcomes and experiences, in this interview Sean Carroll describes 
> how even with perfect knowledge of the universe and it's evolution one 
> could not make future predictions about what one will experience due to 
> duplication:
>
> https://youtube.com/watch?feature=youtu.be&v=TP5W2MG8Jjs&t=1h5m
>
> Perhaps Carroll's explanation might help others who've struggled to get 
> past Step 3.
>
> Jason
>



There Sean Carroll meets Deepak Chopra.

@philipthrift 

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Sean Carroll gets past "Step 3" of the UDA

2019-09-19 Thread Jason Resch

Like Max Tegmark, in "Our Mathematical Universe", who described how
duplication of the person results in an inability to perfectly predict
future outcomes and experiences, in this interview Sean Carroll describes
how even with perfect knowledge of the universe and it's evolution one
could not make future predictions about what one will experience due to
duplication:

https://youtube.com/watch?feature=youtu.be&v=TP5W2MG8Jjs&t=1h5m

Perhaps Carroll's explanation might help others who've struggled to get
past Step 3.

Jason

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Re: STEP 3

2019-08-22 Thread Bruno Marchal

> On 21 Aug 2019, at 18:04, Jason Resch  wrote:
> 
> 
> 
> On Fri, Aug 16, 2019 at 11:05 AM Bruno Marchal  > wrote:
> 
> 
> The fact that there is a universal Diophantine polynomial is rather 
> extraordinary. It means that all proofs that some machine do something can be 
> verified in less than one hundred operations (of addition and 
> multiplication). From this it can be shown that all (halting) computations 
> can be done in less than one hundred operations. This means that the 
> “dynamical content of a computation” can be limited to 100 operations, and 
> all the rest will belong to statical description of (expectedly) very 
> gigantic numbers and very long multiplications. 
> 
> Maybe this is what I was missing. When you say "very gigantic numbers", how 
> large are they? 

To emulate a part of a long computation, the numbers will be absolutely 
gigantic, a bit like a description of your digital state. Imagine you want to 
simulate only our cluster of galaxies, you will need something like the 
description of the state of that cluster. To emulate the entire multiverse will 
asks for much less information, like a “vacuum state” and the equations. But in 
arithmetic you have both, and with some luck, the least programs are the 
“winner”. This would explain why the physical laws are easy to compute when we 
accept to do some approximations, and very hard to compute when we want all 
details exact.

> Are they on the order of the memory requirements of the computation * the 
> number of computational steps?  Or something along those lines?

They need apparent big amount of contingent information, when we emulate a 
subpart of the whole physical reality. That can can explicitly give the number 
of computational steps, but it use also the content of the special current 
memories of the person, or system, to be emulated.

> 
> If so, then I can more easily see how any computation could be equivalent to 
> the evaluation of a Diophantine equation.  Before it seemed like cheating, a 
> short-circuit to compute (or at least verify) anything far more easily than 
> usual.  Is it the case then, that evaluating the universal Diophantine for 
> some halting program is expected to take as long for a Turing machine to 
> compute as for that same Turing machine to perform the computation?

Yes. You can derive this from the “intensional Church-Turing thesis”. Not only 
all universal machinery compte exactly the same (partial and total) computable 
functions, but they can emulate each other, that is, each universal machine can 
compute each computable function in exactly the same way. You can find a 
diophantine polynomial emulating (simulating exactly) a pattern of of the game 
of life, itself emulating a Fortran programs, etc. When they do that, they take 
the same time, up to some mutiplicative constant (which is not first person 
accessible, and that explain why we can start from any universal machinery).

Implementation is always defined relatively to a universal machinery. It 
happens that the physical reality is Turing complete, so we have a notion of 
physical implementation. That is still a bit of a mystery. Today, we can’t 
claim that Mechanism has evacuated all “white rabbits”, but note that this is 
the case also in quantum field theory, despite the theory of renormalisation 
which solves the problem partially (I think).

>  
> 
> There has been some period where I thought this could refute 
> computationalism. I don’t think it is a threat, even if that is weird. It is 
> weird that you can test x^(y ^(z^(t^…..^r…) = r with 10^1000 nested 
> exponentiations by only 100 additions and multiplications using only much 
> less variable/numbers. Some notorious logicians thought that this would just 
> be impossible, and took some time to discourage the search for a diophantine 
> polynomial computing (just with addition and multiplication) the exponential 
> (Julia Robinson already showed that this would lead to the solution).
> 
> This theorem suggests also that consciousness is mainly in the number 
> relations, not in the operation emulating the computation, but this we 
> already knew: it makes it more striking. 
> 
> 
> Indeed. I think I am now starting to grasp this (at least if the numbers 
> involved, and cost of evaluating the polynomials is as large as I think the 
> might be).

Let me give you a universal system of polynomial Diophantine equations. It is 
“short” as we use many variables, and we allow a degree bigger than 4 (4 is 
enough, but then you need much more variables), and we allow more than one 
equation. The variables range on the non negative integers (= 0 included)
There are 31 unknown variables: A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, 
Q, R, S, T, W, Z, U, Y, Al, Ga, Et, Th, La, Ta, Ph, and two parameters:  Nu and 
X.

The number X is in the recursively enumerable set W_Nu iff   phi_Nu(X) stop if 
and only if there are number A, B, C, …

Re: STEP 3

2019-08-21 Thread Jason Resch

On Fri, Aug 16, 2019 at 11:05 AM Bruno Marchal  wrote:

>
>
> The fact that there is a universal Diophantine polynomial is rather
> extraordinary. It means that all proofs that some machine do something can
> be verified in less than one hundred operations (of addition and
> multiplication). From this it can be shown that all (halting) computations
> can be done in less than one hundred operations. This means that the
> “dynamical content of a computation” can be limited to 100 operations, and
> all the rest will belong to statical description of (expectedly) very
> gigantic numbers and very long multiplications.
>

Maybe this is what I was missing. When you say "very gigantic numbers", how
large are they?  Are they on the order of the memory requirements of the
computation * the number of computational steps?  Or something along those
lines?

If so, then I can more easily see how any computation could be equivalent
to the evaluation of a Diophantine equation.  Before it seemed like
cheating, a short-circuit to compute (or at least verify) anything far more
easily than usual.  Is it the case then, that evaluating the universal
Diophantine for some halting program is expected to take as long for a
Turing machine to compute as for that same Turing machine to perform the
computation?


>
> There has been some period where I thought this could refute
> computationalism. I don’t think it is a threat, even if that is weird. It
> is weird that you can test x^(y ^(z^(t^…..^r…) = r with 10^1000 nested
> exponentiations by only 100 additions and multiplications using only much
> less variable/numbers. Some notorious logicians thought that this would
> just be impossible, and took some time to discourage the search for a
> diophantine polynomial computing (just with addition and multiplication)
> the exponential (Julia Robinson already showed that this would lead to the
> solution).
>
> This theorem suggests also that consciousness is mainly in the number
> relations, not in the operation emulating the computation, but this we
> already knew: it makes it more striking.
>


Indeed. I think I am now starting to grasp this (at least if the numbers
involved, and cost of evaluating the polynomials is as large as I think the
might be).


>
> The theorem is proved in the quite remarkable presentation by Martin
> Davis, in the Scientific American (I think) of the
> Putnam-Davis-Robinson-Matiyazevic's theorem (the universality of
> diophantine polynomials) . It has been reprinted in an appendice in the
> Dover edition of its “Computability and Unsolvability” book.
>

Thanks for the reference, do you happen to know if it is one of these?
https://www.scientificamerican.com/author/martin-davis/
Was it titled "Hilbert's 10th Problem" ?

Jason


> I will, soon or later, make a summary of all the “concrete” universal
> machinery (the phi_i and w_i) that we have encountered (mainly, Boolean
> Graph + Clock/Delay, Elementary Arithmetic, Diophantine Polynomial
> Equation, Turing Machine, Register Machine (coffee-bar), LISP,
> lambda-expression and the combinators). Each have their own phi_i and w_i,
> but all phi_i and w_i obeys the same fundamental “computational” laws,
> largely captured by the combinatory algebras and the Models of Lambda
> Calculus).
>
> But to get the intensional nuances, the simplest way consists in using the
> phi_i and w_i directly.
>
> Basically everything follows from two facts, here below,  about all
> “acceptable” universal machinery (enumeration of the partial computable
> function. Note that a total (everywhere defined) function is a particular
> case of a partial function):
>
> - 1)  it exist a computable function s such that for all number *x*, *y*,
> and *i*,  *phi_i (x,y) = phi_s(i, x) (y)*
>
>(*s* parametrises *x* on *i*)  It is the SMN theorem
> (here the simplest S21 theorem)
>
> - 2)  it exist a universal number *u* such that for all number *x*, *y* 
> *phi_u(x,y)
> = phi_x (y).*
>
> A lot can be deduced from this. I build a self-regeretaig program
> (planaria) using a generalisation by John Case, of the Recursion theorem of
> Kleene, which can be proved in five line from just the SMN theorem.
> The whole logic of self-reference comes from the fact that PA (ZF, …) can
> prove those theorems in arithmetic.
>
>
> That richness has some price, and the universal machine brings a lot of
> mess in in (Arithmetical) Platonia, but that mess is also highly
> structured, which help when deriving a measure on the computational
> histories.
>
> Bruno
>
>
>
>
>
>
>
>
>
>
>
>
> Thank you.
>
> Jason
>
>
>>
>> Note that the parallel worlds are given by perpendicular states. They
>> should be called the perpendicular universes. Once two
>> “universes/histories" are not perpendicular they can interfere
>> “statistically”, and they are inter-reachable “probabilistically” through
>> appropriate measurements/interactions. That imposes also some symmetries.
>>
>>
>> Bruno
>>
>>
>>
> --
> You re

Re: STEP 3

2019-08-19 Thread smitra

On 19-08-2019 11:01, Bruno Marchal wrote:

On 19 Aug 2019, at 06:55, smitra wrote:

On 19-08-2019 03:52, Bruce Kellett wrote:

On Sat, Aug 17, 2019 at 10:28 AM smitra wrote:

On 16-08-2019 09:01, Bruce Kellett wrote:

On Fri, Aug 16, 2019 at 4:43 PM smitra wrote:
I think you need to prove that. In my understanding, A(x) =

to be interpreted as the amplitude for a wave through slit A to

get to

the screen at x. There is nothing 3-dimensional about this. The

'x' is

just the distance from the centre of the screen in the plane of

the

screen. Nothing else is relevant. You do not have to integrate

over

all space because you use a complete set of states in the

x-direction:

int |x>
A quantum state is defined in the position representation by
assigning
an amplitude to all points in space. It can be the case that the
amplitude is zero outside of a narrow volume surrounding the the
screen,
in which case the integration can be approximated as an integral
over
the screen's surface.
The orthogonality can be rigorously proved as follows. If we have a
single particle incident on the two slits described by a time
dependent
wave function psi(x,t) = 1/sqrt(2) [A(x,t) + B(x,t)] such that at
A(x,0)
is nonzero at one slit and B(x,0) at the other slit, then A(x,0)
and
B(x,0) are obviously orthogonal. Since time evolution will preserve
inner products, A(x,t) and B(x,t) will remain orthogonal as a
function
of t. One can then describe the interaction with the screen as an
effective collapse that will happen with the largest probability
when
the peak of the wavefunction has arrived at the screen.
The slits are orthogonal only if they are eigensatates of the
position

operator (in the x direction). That is the case only if you have a
measurement that gives which-way information. Then there is no
interference, as advertised. The amplitudes at the slits aren't
orthogonal just because you say so.

The particle will be in one of the two eigenstates if such a
measurement is made,

From the perspective of the one which has done the measure, but
actually, he just entangle him/herself with the particle position.
“The particle” is a bit ambiguous, as “the observer” is too, in such
self-entanglement.

but if we don't make that measurement then the particle ends up in a
superposition of the two eigenstates.

It means that the observer is still able to get the interference. It
is only his knowing which slit the particle has gone through which
makes him to be unable to get the interference, but only due to
self-entanglement. The measurement just makes some accessible
histories inaccessible (when we don’t assume a collapse).

Indeed, the observer then locates him/herself in a sector of Hilbert
space corresponding the particle moving through either the left or the
right slit.

Saibal

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Re: STEP 3

2019-08-19 Thread smitra

On 19-08-2019 07:56, Bruce Kellett wrote:

On Mon, Aug 19, 2019 at 2:55 PM smitra wrote:

On 19-08-2019 03:52, Bruce Kellett wrote:

On Sat, Aug 17, 2019 at 10:28 AM smitra wrote:

On 16-08-2019 09:01, Bruce Kellett wrote:

On Fri, Aug 16, 2019 at 4:43 PM smitra

wrote:

I think you need to prove that. In my understanding, A(x) =

to be interpreted as the amplitude for a wave through slit A to

get to

the screen at x. There is nothing 3-dimensional about this. The

'x' is

just the distance from the centre of the screen in the plane of

the

screen. Nothing else is relevant. You do not have to integrate

over

all space because you use a complete set of states in the

x-direction:

int |x>

A quantum state is defined in the position representation by
assigning
an amplitude to all points in space. It can be the case that the
amplitude is zero outside of a narrow volume surrounding the the
screen,
in which case the integration can be approximated as an integral
over
the screen's surface.

The orthogonality can be rigorously proved as follows. If we

have a

single particle incident on the two slits described by a time
dependent
wave function psi(x,t) = 1/sqrt(2) [A(x,t) + B(x,t)] such that

A(x,0)
is nonzero at one slit and B(x,0) at the other slit, then A(x,0)
and
B(x,0) are obviously orthogonal. Since time evolution will

preserve

inner products, A(x,t) and B(x,t) will remain orthogonal as a
function
of t. One can then describe the interaction with the screen as

effective collapse that will happen with the largest probability
when
the peak of the wavefunction has arrived at the screen.

The slits are orthogonal only if they are eigensatates of the

position

operator (in the x direction). That is the case only if you have

measurement that gives which-way information. Then there is no
interference, as advertised. The amplitudes at the slits aren't
orthogonal just because you say so.

The particle will be in one of the two eigenstates if such a
measurement
is made, but if we don't make that measurement then the particle
ends up
in a superposition of the two eigenstates.

Come on, Saibal. You are just blowing smoke. The particle is not in a
superposition of the two orthogonal eigenstates -- it is in a
superposition of an indefinite number of eigenstates after passing the
slits, and a different superposition at each slit, so there is a
non-zero overlap (It spreads out as a cylindrical wave after each
slit.). Integrating over all 3-space is not going to help you here. In
fact, that is a meaningless operation in the context. Integration over
the direction along the screen is, at most, all that is required for
the complete set of states. And that does not integrate to zero.

If there is no overlap immediately after the wavepacket spreads out of
the slits, then there can't be any overlap later. Immediately after
leaving the slits the amplitudes for left slit is zero at the right slit
and vice versa, so that there is no overlap immediately after the
wavepacket spreads out the slits is trivially true. The overlap is then
zero at all later times even when the wavepackets emanating from the two
slits cross and "overlap" in the colloquial sense, due to unitary time
evolution: If U(t) is the time evolution operator and |psi_j(t)> is the
state the particle would be in, if only slit j were open, then

= =

Therefore, if both slits are open and the initial state is some
superposition of the two states, then it will remain a superposition of
two orthogonal states, even though the two orthogonal components will
"overlap" in the colloquial sense. Orthogonality involves overlap in the
mathematical sense: = 0

And we may evaluate this in the position representation by inserting a
complete set of position eigenstates:

Int d^3x |x>0 = = int d^3x = int d^3x psi_1(x)*
psi_2(x)

So, an integral over all space will yield zero. This means that if an
integral over only the screen would not yield zero, then that
integration doesn't capture all the regions where the wavefunctions have
an overlap in the colloquial sense.

The point raised about there being many different possible states for a
particle when only one slit is open doesn't change the above argument.
What matters is that after some initial wavepacket moves through both
slits, it can be written as a superposition of the two states it would
be in corresponding to only 1 of the two slits being open. And these two
states are orthogonal per the above argument.

Saibal

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Re: STEP 3

2019-08-19 Thread Bruno Marchal



> On 19 Aug 2019, at 06:55, smitra  wrote:
> 
> On 19-08-2019 03:52, Bruce Kellett wrote:
>> On Sat, Aug 17, 2019 at 10:28 AM smitra  wrote:
>>> On 16-08-2019 09:01, Bruce Kellett wrote:
 On Fri, Aug 16, 2019 at 4:43 PM smitra  wrote:
 I think you need to prove that. In my understanding, A(x) = 
>>> is
 to be interpreted as the amplitude for a wave through slit A to
>>> get to
 the screen at x. There is nothing 3-dimensional about this. The
>>> 'x' is
 just the distance from the centre of the screen in the plane of
>>> the
 screen. Nothing else is relevant. You do not have to integrate
>>> over
 all space because you use a complete set of states in the
>>> x-direction:
 int |x>>> A quantum state is defined in the position representation by
>>> assigning
>>> an amplitude to all points in space. It can be the case that the
>>> amplitude is zero outside of a narrow volume surrounding the the
>>> screen,
>>> in which case the integration can be approximated as an integral
>>> over
>>> the screen's surface.
>>> The orthogonality can be rigorously proved as follows. If we have a
>>> single particle incident on the two slits described by a time
>>> dependent
>>> wave function psi(x,t) = 1/sqrt(2) [A(x,t) + B(x,t)] such that at
>>> A(x,0)
>>> is nonzero at one slit and B(x,0) at the other slit, then A(x,0)
>>> and
>>> B(x,0) are obviously orthogonal. Since time evolution will preserve
>>> inner products, A(x,t) and B(x,t) will remain orthogonal as a
>>> function
>>> of t. One can then describe the interaction with the screen as an
>>> effective collapse that will happen with the largest probability
>>> when
>>> the peak of the wavefunction has arrived at the screen.
>> The slits are orthogonal only if they are eigensatates of the position
>> operator (in the x direction). That is the case only if you have a
>> measurement that gives which-way information. Then there is no
>> interference, as advertised. The amplitudes at the slits aren't
>> orthogonal just because you say so.
> 
> The particle will be in one of the two eigenstates if such a measurement is 
> made,

>From the perspective of the one which has done the measure, but actually, he 
>just entangle him/herself with the particle position. “The particle” is a bit 
>ambiguous, as “the observer” is too, in such self-entanglement.



> but if we don't make that measurement then the particle ends up in a 
> superposition of the two eigenstates.

It means that the observer is still able to get the interference. It is only 
his knowing which slit the particle has gone through which makes him to be 
unable to get the interference, but only due to self-entanglement. The 
measurement just makes some accessible histories inaccessible (when we don’t 
assume a collapse).

Bruno


> 
> Saibal
> 
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Re: STEP 3

2019-08-18 Thread Bruce Kellett

On Mon, Aug 19, 2019 at 2:55 PM smitra  wrote:

> On 19-08-2019 03:52, Bruce Kellett wrote:
> > On Sat, Aug 17, 2019 at 10:28 AM smitra  wrote:
> >
> >> On 16-08-2019 09:01, Bruce Kellett wrote:
> >>> On Fri, Aug 16, 2019 at 4:43 PM smitra  wrote:
> >>>
> >>>
> >>> I think you need to prove that. In my understanding, A(x) = 
> >> is
> >>> to be interpreted as the amplitude for a wave through slit A to
> >> get to
> >>> the screen at x. There is nothing 3-dimensional about this. The
> >> 'x' is
> >>> just the distance from the centre of the screen in the plane of
> >> the
> >>> screen. Nothing else is relevant. You do not have to integrate
> >> over
> >>> all space because you use a complete set of states in the
> >> x-direction:
> >>> int |x> >>
> >> A quantum state is defined in the position representation by
> >> assigning
> >> an amplitude to all points in space. It can be the case that the
> >> amplitude is zero outside of a narrow volume surrounding the the
> >> screen,
> >> in which case the integration can be approximated as an integral
> >> over
> >> the screen's surface.
> >>
> >> The orthogonality can be rigorously proved as follows. If we have a
> >>
> >> single particle incident on the two slits described by a time
> >> dependent
> >> wave function psi(x,t) = 1/sqrt(2) [A(x,t) + B(x,t)] such that at
> >> A(x,0)
> >> is nonzero at one slit and B(x,0) at the other slit, then A(x,0)
> >> and
> >> B(x,0) are obviously orthogonal. Since time evolution will preserve
> >>
> >> inner products, A(x,t) and B(x,t) will remain orthogonal as a
> >> function
> >> of t. One can then describe the interaction with the screen as an
> >> effective collapse that will happen with the largest probability
> >> when
> >> the peak of the wavefunction has arrived at the screen.
> >
> > The slits are orthogonal only if they are eigensatates of the position
> > operator (in the x direction). That is the case only if you have a
> > measurement that gives which-way information. Then there is no
> > interference, as advertised. The amplitudes at the slits aren't
> > orthogonal just because you say so.
>
> The particle will be in one of the two eigenstates if such a measurement
> is made, but if we don't make that measurement then the particle ends up
> in a superposition of the two eigenstates.
>

Come on, Saibal. You are just blowing smoke. The particle is not in a
superposition of the two orthogonal eigenstates -- it is in a superposition
of an indefinite number of eigenstates after passing the slits, and a
different superposition at each slit, so there is a non-zero overlap (It
spreads out as a cylindrical wave after each slit.). Integrating over all
3-space is not going to help you here. In fact, that is a meaningless
operation in the context. Integration over the direction along the screen
is, at most, all that is required for the complete set of states. And that
does not integrate to zero.

Bruce

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Re: STEP 3

2019-08-18 Thread smitra

On 19-08-2019 03:52, Bruce Kellett wrote:

On Sat, Aug 17, 2019 at 10:28 AM smitra wrote:

On 16-08-2019 09:01, Bruce Kellett wrote:

On Fri, Aug 16, 2019 at 4:43 PM smitra wrote:

I think you need to prove that. In my understanding, A(x) =

to be interpreted as the amplitude for a wave through slit A to

get to

the screen at x. There is nothing 3-dimensional about this. The

'x' is

just the distance from the centre of the screen in the plane of

the

screen. Nothing else is relevant. You do not have to integrate

over

all space because you use a complete set of states in the

x-direction:

int |x>

The orthogonality can be rigorously proved as follows. If we have a

single particle incident on the two slits described by a time
dependent
wave function psi(x,t) = 1/sqrt(2) [A(x,t) + B(x,t)] such that at
A(x,0)
is nonzero at one slit and B(x,0) at the other slit, then A(x,0)
and
B(x,0) are obviously orthogonal. Since time evolution will preserve

inner products, A(x,t) and B(x,t) will remain orthogonal as a
function
of t. One can then describe the interaction with the screen as an
effective collapse that will happen with the largest probability
when
the peak of the wavefunction has arrived at the screen.

The slits are orthogonal only if they are eigensatates of the position
operator (in the x direction). That is the case only if you have a
measurement that gives which-way information. Then there is no
interference, as advertised. The amplitudes at the slits aren't
orthogonal just because you say so.

The particle will be in one of the two eigenstates if such a measurement
is made, but if we don't make that measurement then the particle ends up
in a superposition of the two eigenstates.

Saibal

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Re: STEP 3

2019-08-18 Thread smitra


On 19-08-2019 04:06, 'Brent Meeker' via Everything List wrote:

On Fri, Aug 16, 2019 at 4:43 PM smitra  wrote:


The orthogonality can be rigorously proved as follows. If we have a
single particle incident on the two slits described by a time
dependent
wave function psi(x,t) = 1/sqrt(2) [A(x,t) + B(x,t)] such that at
A(x,0)
is nonzero at one slit and B(x,0) at the other slit, then A(x,0)
and
B(x,0) are obviously orthogonal.


 Don't you mean "...such that at A(x,0) is _ZERO_ at one slit and
B(x,0) at the other slit, then A(x,0) and
 B(x,0) are obviously orthogonal."  Simply being non-zero at the
alternate slits won't make them orthogonal.
 But of course they are not zero at either .



That's right, the overlap will then be zero.

Saibal

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Re: STEP 3

2019-08-18 Thread 'Brent Meeker' via Everything List




On Fri, Aug 16, 2019 at 4:43 PM smitra > wrote:

The orthogonality can be rigorously proved as follows. If we have a
single particle incident on the two slits described by a time dependent
wave function psi(x,t) = 1/sqrt(2) [A(x,t) + B(x,t)] such that at A(x,0)
is nonzero at one slit and B(x,0) at the other slit, then A(x,0) and
B(x,0) are obviously orthogonal.


Don't you mean "...such that at A(x,0) is /*zero*/ at one slit and 
B(x,0) at the other slit, then A(x,0) and
B(x,0) are obviously orthogonal."  Simply being non-zero at the 
alternate slits won't make them orthogonal.

But of course they are not zero at either .

Brent

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Re: STEP 3

2019-08-18 Thread Bruce Kellett

On Sat, Aug 17, 2019 at 10:28 AM smitra  wrote:

> On 16-08-2019 09:01, Bruce Kellett wrote:
> > On Fri, Aug 16, 2019 at 4:43 PM smitra  wrote:
> >
> >
> > I think you need to prove that. In my understanding, A(x) =  is
> > to be interpreted as the amplitude for a wave through slit A to get to
> > the screen at x. There is nothing 3-dimensional about this. The 'x' is
> > just the distance from the centre of the screen in the plane of the
> > screen. Nothing else is relevant. You do not have to integrate over
> > all space because you use a complete set of states in the x-direction:
> >  int |x>
> A quantum state is defined in the position representation by assigning
> an amplitude to all points in space. It can be the case that the
> amplitude is zero outside of a narrow volume surrounding the the screen,
> in which case the integration can be approximated as an integral over
> the screen's surface.
>
> The orthogonality can be rigorously proved as follows. If we have a
> single particle incident on the two slits described by a time dependent
> wave function psi(x,t) = 1/sqrt(2) [A(x,t) + B(x,t)] such that at A(x,0)
> is nonzero at one slit and B(x,0) at the other slit, then A(x,0) and
> B(x,0) are obviously orthogonal. Since time evolution will preserve
> inner products, A(x,t) and B(x,t) will remain orthogonal as a function
> of t. One can then describe the interaction with the screen as an
> effective collapse that will happen with the largest probability when
> the peak of the wavefunction has arrived at the screen.
>

The slits are orthogonal only if they are eigensatates of the position
operator (in the x direction). That is the case only if you have a
measurement that gives which-way information. Then there is no
interference, as advertised. The amplitudes at the slits aren't orthogonal
just because you say so.

Bruce

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Re: STEP 3

2019-08-16 Thread smitra


On 16-08-2019 09:01, Bruce Kellett wrote:

On Fri, Aug 16, 2019 at 4:43 PM smitra  wrote:


On 16-08-2019 06:31, Bruce Kellett wrote:

On Wed, Aug 14, 2019 at 8:28 AM smitra  wrote:


On 13-08-2019 13:33, Bruce Kellett wrote:


Of course A(x) and B(x) refer to the same point on the screen.

That is

not a collapse, that is just what the notation means.


A(x) and B(x) considered as the representations of |A> and |B>

in

the
position basis, i.e. A(x) =  and B(x) =  are still
orthogonal
states, as they represent the orthogonal states |A> and |B>:

0 =  = Integral over x of d^3x = Integral over x

of


A*(x)B(x) d^3x


I don't think this really works out. You are claiming that the
integral of the interference terms over the whole screen

vanishes. If

we look at the usual derivation of the interference from two

slits, we

get something like

Intensity I = 2 A^2 (sin^2(beta)/beta^2) (1 + cos(delta))

where the term involving the angle beta is the superposed

diffraction

pattern from the finite width of the slits. The cos (delta) term

is

the interference, but it has this form only in a small angle
approximation, and the phase difference delta is, of course,

limited

by the separation of the slits. So, although the cos(delta) term

may

integrate to zero over small angles, the presence of the

diffraction

envelope, and the limitations of the small angle approximation,

mean

that is almost certainly will not vanish when integrated over the
whole screen.


Yes, this is in the small angle approximation, if you go beyond
that
then the itnegral over the screen won't vanish.



So  will not vanish in general.

No, because  is the integral over all space and this is
exactly
zero. What happens is that when the small angle approximation
becomes
invalid and the integral over only the screen becomes nonzero, the
integrals over surfaces parallel to the screen will have a values
that
differ by a phase factor that depends on the distance in the
direction
orthogonal to the screen. This then causes the integral over all
space
to vanish.


I think you need to prove that. In my understanding, A(x) =  is
to be interpreted as the amplitude for a wave through slit A to get to
the screen at x. There is nothing 3-dimensional about this. The 'x' is
just the distance from the centre of the screen in the plane of the
screen. Nothing else is relevant. You do not have to integrate over
all space because you use a complete set of states in the x-direction:
 int |x>

A quantum state is defined in the position representation by assigning 
an amplitude to all points in space. It can be the case that the 
amplitude is zero outside of a narrow volume surrounding the the screen, 
in which case the integration can be approximated as an integral over 
the screen's surface.


The orthogonality can be rigorously proved as follows. If we have a 
single particle incident on the two slits described by a time dependent 
wave function psi(x,t) = 1/sqrt(2) [A(x,t) + B(x,t)] such that at A(x,0) 
is nonzero at one slit and B(x,0) at the other slit, then A(x,0) and 
B(x,0) are obviously orthogonal. Since time evolution will preserve 
inner products, A(x,t) and B(x,t) will remain orthogonal as a function 
of t. One can then describe the interaction with the screen as an 
effective collapse that will happen with the largest probability when 
the peak of the wavefunction has arrived at the screen.


Saibal

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Re: STEP 3

2019-08-16 Thread Bruno Marchal


> On 15 Aug 2019, at 23:22, Jason Resch  wrote:
> 
> 
> 
> On Sun, Aug 11, 2019 at 12:24 PM Bruno Marchal  > wrote:
> 
>> On 10 Aug 2019, at 20:34, Jason Resch > > wrote:
>> 
>> 
>> 
>> On Fri, Aug 9, 2019 at 10:20 AM Bruno Marchal > > wrote:
>> 
>>> On 9 Aug 2019, at 13:09, Jason Resch >> > wrote:
>>> 
>>> 
 
 Bruno,
 
 Forgive me if I have asked this before, but can you elaborate on the 
 how/why the math suggests negative interference?
 
 I currently have no intuition for why this should be.
 
 I recall reading something on continuous probability as being more natural 
 and leading to something much like the probability formulas in quantum 
 mechanics. Is that related?
>>> 
>>> 
>>> It is not intuitive at all. With the UDA, we can have have the intuition 
>>> coming from the first person indeterminacy on all all computational 
>>> continuation in arithmetic, but in the AUDA (the Arithmetical UDA), the 
>>> probabilities are constrained by the logic of self-reference G and G*. So 
>>> the reason why we can hope for negative amplitude of probability comes from 
>>> the fact that modal variant of the first person on the (halting) 
>>> computations, which is given by the arithmetical interpretation of:
>>> 
>>> []p & p
>>> 
>>>  or
>>> 
>>> []p & <>t
>>> 
>>> or
>>> 
>>> []p & <>t & p
>>> 
>>>  With, as usual, [] = Beweisbar, and p is an arbitrary sigma_1 sentences 
>>> (partial computable formula).
>>> 
>>> They all give a quantum logic enough close to Dalla Chiara’s presentation 
>>> of them, to have the quantum features like complimentary observable, and 
>>> what I have called a sort of abstract linear evolution build on a highly 
>>> symmetrical core (than to LASE: the little Schroeder equation: p -> []<>p, 
>>> which provides a quantisation of the sigma_1 arithmetical reality.
>>> 
>>> It is mainly the presence of this quantisation which justify that the 
>>> probabilities behave in a quantum non boolean way, but this is hard to 
>>> verify because the nesting of boxes in the G* translation makes those 
>>> formula … well, probably in need of a quantum computer to be evaluated. But 
>>> normally, if mechanism (and QM) are correct this should work.
>>> 
>>> This is explained with more detail in “Conscience et Mécanisme”.
>>> 
>>> Bruno
>>> 
>>> 
>>> Thank you Bruno for your explanation and references. 
>> 
>> Y’re welcome.
>> 
>> 
>>> Regarding “Conscience et Mécanisme”, is there a web/html or English version 
>>> available?  Unfortunately my browser cannot do translations of PDFs but can 
>>> translate web pages.  If not don't worry, I can copy and paste into a 
>>> translator.
>> 
>> Yes, There is no HTML page for the long text. But you can consult also my 
>> paper:
>> 
>> Marchal B. The Universal Numbers. From Biology to Physics, Progress in 
>> Biophysics and Molecular Biology, 2015, Vol. 119, Issue 3, 368-381.
>> https://www.ncbi.nlm.nih.gov/pubmed/26140993 
>> 
>> 
>> You will still need some background in quantum logic, like  the paper by 
>> Goldblatt which makes the link between minimal quantum logic and the B modal 
>> logic. 
>> 
>> There is also a paper by Rawling and Selesnick which shows how to build a 
>> quantum NOT gate, from the Kripke semantics of the B logic. It is not 
>> entirely clear if this can be used in arithmetic, because we loss the 
>> necessitation rule in “our” B logic. Open problem. A positive solution on 
>> this would be a great step toward an explanation that the universal machine 
>> has necessarily a quantum structure and can exploit the “parallel 
>> computations in arithmetic” in the limit of the 1p indeterminacy..
>> 
>> Rawling JP and Selesnick SA, 2000, Orthologic and Quantum Logic: Models and 
>> Computational Elements, Journal of the ACM, Vol. 47, n° 4, pp. 721-T51.
>> 
>> Ask question, online or here. It *is* rather technical at some point.
>> 
>> Bruno
>> 
>> 
>> 
>> I've been reading those references, and have found a few more which might be 
>> related and of interest.  Effectively, they provide arguments for the 
>> quantum probability theory based on the requirement for continuous 
>> reversible operations, or the juxtaposition between finite information-carry 
>> capacity and smoothness.
>> 
>> 
>> Lucien Hardy's "Quantum Theory From Five Reasonable Axioms" 
>> https://arxiv.org/abs/quant-ph/0101012 
>> 
>> 
>> The usual formulation of quantum theory is based on rather obscure axioms 
>> (employing complex Hilbert spaces, Hermitean operators, and the trace rule 
>> for calculating probabilities). In this paper it is shown that quantum 
>> theory can be derived from five very reasonable axioms. The first four of 
>> these are obviously consistent with both quantum theory and classical 
>> probability theory. Axiom

Re: STEP 3

2019-08-16 Thread Bruce Kellett

On Fri, Aug 16, 2019 at 4:43 PM smitra  wrote:

> On 16-08-2019 06:31, Bruce Kellett wrote:
> > On Wed, Aug 14, 2019 at 8:28 AM smitra  wrote:
> >
> >> On 13-08-2019 13:33, Bruce Kellett wrote:
> >>>
> >>> Of course A(x) and B(x) refer to the same point on the screen.
> >> That is
> >>> not a collapse, that is just what the notation means.
> >>
> >> A(x) and B(x) considered as the representations of |A> and |B> in
> >> the
> >> position basis, i.e. A(x) =  and B(x) =  are still
> >> orthogonal
> >> states, as they represent the orthogonal states |A> and |B>:
> >>
> >> 0 =  = Integral over x of d^3x = Integral over x of
> >>
> >> A*(x)B(x) d^3x
> >
> > I don't think this really works out. You are claiming that the
> > integral of the interference terms over the whole screen vanishes. If
> > we look at the usual derivation of the interference from two slits, we
> > get something like
> >
> >  Intensity I = 2 A^2 (sin^2(beta)/beta^2) (1 + cos(delta))
> >
> > where the term involving the angle beta is the superposed diffraction
> > pattern from the finite width of the slits. The cos (delta) term is
> > the interference, but it has this form only in a small angle
> > approximation, and the phase difference delta is, of course, limited
> > by the separation of the slits. So, although the cos(delta) term may
> > integrate to zero over small angles, the presence of the diffraction
> > envelope, and the limitations of the small angle approximation, mean
> > that is almost certainly will not vanish when integrated over the
> > whole screen.
>
> Yes, this is in the small angle approximation, if you go beyond that
> then the itnegral over the screen won't vanish.
>
> >
> > So  will not vanish in general.
> No, because  is the integral over all space and this is exactly
> zero. What happens is that when the small angle approximation becomes
> invalid and the integral over only the screen becomes nonzero, the
> integrals over surfaces parallel to the screen will have a values that
> differ by a phase factor that depends on the distance in the direction
> orthogonal to the screen. This then causes the integral over all space
> to vanish.
>

I think you need to prove that. In my understanding, A(x) =  is to be
interpreted as the amplitude for a wave through slit A to get to the screen
at x. There is nothing 3-dimensional about this. The 'x' is just the
distance from the centre of the screen in the plane of the screen. Nothing
else is relevant. You do not have to integrate over all space because you
use a complete set of states in the x-direction:  \int |x> > Which is what I would have
> > thought because the paths through the separate slits are not
> > independent -- each particle essentially has to see both slits (go
> > through both slits) in order to maintain coherence. So they cannot be
> > orthogonal (independent).
>
> Coherence and orthogonality have nothing to do with each other.
>
> >
> > In practice, to see the interference pattern you need coherent
> > illumination over both slits. This is easy these days with lasers, but
> > in older books, coherence was ensured by having a preparatory single
> > slit followed by suitable condenser lenses. If the slits could be
> > treated as independent entities, this would not have been necessary.
>
> This has nothing to do with orthogonality of the states. What matters is
> that the interference pattern shouldn't get washed out due to each
> wavefunction of each particle near the screen having its peaks and
> fringes at different places. This can be prevented by using an
> approximate monochromatic light source and making sure that the light
> passes through a collimator. Without a collimator, the interference
> pattern due to the light from one part of the source will be shifted
> w.r.t. to the other part causing the pattern to get washed out.
>
> Saibal
>
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Re: STEP 3

2019-08-15 Thread smitra

On 16-08-2019 06:31, Bruce Kellett wrote:

On Wed, Aug 14, 2019 at 8:28 AM smitra wrote:

On 13-08-2019 13:33, Bruce Kellett wrote:

Of course A(x) and B(x) refer to the same point on the screen.

That is

not a collapse, that is just what the notation means.

A(x) and B(x) considered as the representations of |A> and |B> in
the
position basis, i.e. A(x) = and B(x) = are still
orthogonal
states, as they represent the orthogonal states |A> and |B>:

0 = = Integral over x of d^3x = Integral over x of

A*(x)B(x) d^3x

I don't think this really works out. You are claiming that the
integral of the interference terms over the whole screen vanishes. If
we look at the usual derivation of the interference from two slits, we
get something like

Intensity I = 2 A^2 (sin^2(beta)/beta^2) (1 + cos(delta))

where the term involving the angle beta is the superposed diffraction
pattern from the finite width of the slits. The cos (delta) term is
the interference, but it has this form only in a small angle
approximation, and the phase difference delta is, of course, limited
by the separation of the slits. So, although the cos(delta) term may
integrate to zero over small angles, the presence of the diffraction
envelope, and the limitations of the small angle approximation, mean
that is almost certainly will not vanish when integrated over the
whole screen.

Yes, this is in the small angle approximation, if you go beyond that
then the itnegral over the screen won't vanish.

So will not vanish in general.
No, because is the integral over all space and this is exactly
zero. What happens is that when the small angle approximation becomes
invalid and the integral over only the screen becomes nonzero, the
integrals over surfaces parallel to the screen will have a values that
differ by a phase factor that depends on the distance in the direction
orthogonal to the screen. This then causes the integral over all space
to vanish.

Which is what I would have
thought because the paths through the separate slits are not
independent -- each particle essentially has to see both slits (go
through both slits) in order to maintain coherence. So they cannot be
orthogonal (independent).

Coherence and orthogonality have nothing to do with each other.

In practice, to see the interference pattern you need coherent
illumination over both slits. This is easy these days with lasers, but
in older books, coherence was ensured by having a preparatory single
slit followed by suitable condenser lenses. If the slits could be
treated as independent entities, this would not have been necessary.

This has nothing to do with orthogonality of the states. What matters is
that the interference pattern shouldn't get washed out due to each
wavefunction of each particle near the screen having its peaks and
fringes at different places. This can be prevented by using an
approximate monochromatic light source and making sure that the light
passes through a collimator. Without a collimator, the interference
pattern due to the light from one part of the source will be shifted
w.r.t. to the other part causing the pattern to get washed out.

Saibal

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Re: STEP 3

2019-08-15 Thread Bruce Kellett

On Wed, Aug 14, 2019 at 8:28 AM smitra  wrote:

> On 13-08-2019 13:33, Bruce Kellett wrote:
> >
> > Of course A(x) and B(x) refer to the same point on the screen. That is
> > not a collapse, that is just what the notation means.
>
> A(x) and B(x) considered as the representations of |A> and |B> in the
> position basis, i.e.  A(x) =  and B(x) =  are still orthogonal
> states, as they represent the orthogonal states |A> and |B>:
>
> 0 =  = Integral over x of d^3x =  Integral over x of
> A*(x)B(x) d^3x
>

I don't think this really works out. You are claiming that the integral of
the interference terms over the whole screen vanishes. If we look at the
usual derivation of the interference from two slits, we get something like

 Intensity I = 2 A^2 (sin^2(beta)/beta^2) (1 + cos(delta))

where the term involving the angle beta is the superposed diffraction
pattern from the finite width of the slits. The cos (delta) term is the
interference, but it has this form only in a small angle approximation, and
the phase difference delta is, of course, limited by the separation of the
slits. So, although the cos(delta) term may integrate to zero over small
angles, the presence of the diffraction envelope, and the limitations of
the small angle approximation, mean that is almost certainly will not
vanish when integrated over the whole screen.

So  will not vanish in general. Which is what I would have thought
because the paths through the separate slits are not independent -- each
particle essentially has to see both slits (go through both slits) in order
to maintain coherence. So they cannot be orthogonal (independent).

In practice, to see the interference pattern you need coherent illumination
over both slits. This is easy these days with lasers, but in older books,
coherence was ensured by having a preparatory single slit followed by
suitable condenser lenses. If the slits could be treated as independent
entities, this would not have been necessary.

But you interpret this as the total counts of particles on the entire
> screen not changing which you call an absence of interference. However
> interference is what we detect locally on each point on the screen. You
> can't say that for each point x0 on the screen,  A(x0) and B(x0) are
> particle states. These values are not the quantum states of the particle
> before it hits the screen,

No, they are the amplitudes of the wave function at each point on the
screen. This is what give the probability of the particle being detected
(by the screen) at this point.

Bruce

unless you would have done a measurement
> localizing the particle near x0.
>
> So, your argument only makes sense if you invoke collapse via a position
> measurement.
>

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Re: STEP 3

2019-08-15 Thread Jason Resch

On Sun, Aug 11, 2019 at 12:24 PM Bruno Marchal  wrote:

>
> On 10 Aug 2019, at 20:34, Jason Resch  wrote:
>
>
>
> On Fri, Aug 9, 2019 at 10:20 AM Bruno Marchal  wrote:
>
>>
>> On 9 Aug 2019, at 13:09, Jason Resch  wrote:
>>
>> 
>>
>>>
>>> Bruno,
>>>
>>> Forgive me if I have asked this before, but can you elaborate on the
>>> how/why the math suggests negative interference?
>>>
>>> I currently have no intuition for why this should be.
>>>
>>> I recall reading something on continuous probability as being more
>>> natural and leading to something much like the probability formulas in
>>> quantum mechanics. Is that related?
>>>
>>>
>>>
>>> It is not intuitive at all. With the UDA, we can have have the intuition
>>> coming from the first person indeterminacy on all all computational
>>> continuation in arithmetic, but in the AUDA (the Arithmetical UDA), the
>>> probabilities are constrained by the logic of self-reference G and G*. So
>>> the reason why we can hope for negative amplitude of probability comes from
>>> the fact that modal variant of the first person on the (halting)
>>> computations, which is given by the arithmetical interpretation of:
>>>
>>> []p & p
>>>
>>>  or
>>>
>>> []p & <>t
>>>
>>> or
>>>
>>> []p & <>t & p
>>>
>>>  With, as usual, [] = Beweisbar, and p is an arbitrary sigma_1 sentences
>>> (partial computable formula).
>>>
>>> They all give a quantum logic enough close to Dalla Chiara’s
>>> presentation of them, to have the quantum features like complimentary
>>> observable, and what I have called a sort of abstract linear evolution
>>> build on a highly symmetrical core (than to LASE: the little Schroeder
>>> equation: p -> []<>p, which provides a quantisation of the sigma_1
>>> arithmetical reality.
>>>
>>> It is mainly the presence of this quantisation which justify that the
>>> probabilities behave in a quantum non boolean way, but this is hard to
>>> verify because the nesting of boxes in the G* translation makes those
>>> formula … well, probably in need of a quantum computer to be evaluated. But
>>> normally, if mechanism (and QM) are correct this should work.
>>>
>>> This is explained with more detail in “Conscience et Mécanisme”.
>>>
>>> Bruno
>>>
>>>
>> Thank you Bruno for your explanation and references.
>>
>>
>> Y’re welcome.
>>
>>
>> Regarding “Conscience et Mécanisme”, is there a web/html or English
>> version available?  Unfortunately my browser cannot do translations of PDFs
>> but can translate web pages.  If not don't worry, I can copy and paste into
>> a translator.
>>
>>
>> Yes, There is no HTML page for the long text. But you can consult also my
>> paper:
>>
>> Marchal B. The Universal Numbers. From Biology to Physics, Progress in
>> Biophysics and Molecular Biology, 2015, Vol. 119, Issue 3, 368-381.
>> https://www.ncbi.nlm.nih.gov/pubmed/26140993
>>
>> You will still need some background in quantum logic, like  the paper by
>> Goldblatt which makes the link between minimal quantum logic and the B
>> modal logic.
>>
>> There is also a paper by Rawling and Selesnick which shows how to build a
>> quantum NOT gate, from the Kripke semantics of the B logic. It is not
>> entirely clear if this can be used in arithmetic, because we loss the
>> necessitation rule in “our” B logic. Open problem. A positive solution on
>> this would be a great step toward an explanation that the universal machine
>> has necessarily a quantum structure and can exploit the “parallel
>> computations in arithmetic” in the limit of the 1p indeterminacy..
>>
>> Rawling JP and Selesnick SA, 2000, Orthologic and Quantum Logic: Models
>> and Computational Elements, Journal of the ACM, Vol. 47, n° 4, pp. 721-T51.
>>
>> Ask question, online or here. It *is* rather technical at some point.
>>
>> Bruno
>>
>>
>>
> I've been reading those references, and have found a few more which might
> be related and of interest.  Effectively, they provide arguments for the
> quantum probability theory based on the requirement for continuous
> reversible operations, or the juxtaposition between finite
> information-carry capacity and smoothness.
>
>
> Lucien Hardy's "Quantum Theory From Five Reasonable Axioms"
> https://arxiv.org/abs/quant-ph/0101012
>
> The usual formulation of quantum theory is based on rather obscure axioms
> (employing complex Hilbert spaces, Hermitean operators, and the trace rule
> for calculating probabilities). In this paper it is shown that quantum
> theory can be derived from five very reasonable axioms. The first four of
> these are obviously consistent with both quantum theory and classical
> probability theory. Axiom 5 (which requires that there exists continuous
> reversible transformations between pure states) rules out classical
> probability theory. If Axiom 5 (or even just the word "continuous" from
> Axiom 5) is dropped then we obtain classical probability theory instead.
> This work provides some insight into the reasons quantum theory is the way
> it is. For example, it explains the n

Re: STEP 3

> On 15 Aug 2019, at 13:28, Bruce Kellett  wrote:
> 
> On Thu, Aug 15, 2019 at 7:49 PM Bruno Marchal  <mailto:marc...@ulb.ac.be>> wrote:
> On 15 Aug 2019, at 02:54, Bruce Kellett  <mailto:bhkellet...@gmail.com>> wrote:
>> On Wed, Aug 14, 2019 at 10:10 PM Bruno Marchal > <mailto:marc...@ulb.ac.be>> wrote:
>> On 12 Aug 2019, at 14:42, Bruce Kellett > <mailto:bhkellet...@gmail.com>> wrote:
>>> 
>>> That is simply incorrect. I refer you again to Zurek, who works in a 
>>> basically Everettian framework, but he stresses the importance of 
>>> environmental induced superselection (einselection) in producing the 
>>> preferred pointer basis. This then breaks things, in the sense that no 
>>> other basis is stable against decoherence, and other sets of basis vectors 
>>> rapidly (in times of the order of femtoseconds) collapse on to the 
>>> preferred pointer states. This is the basis of the emergence of the 
>>> classical world from the quantum substrate. And this occurs in Everett's 
>>> relative state approach just as much as in a Copenhagen-like collapse 
>>> models.
>> 
>> That explains why the many histories will look classical. But if I observe a 
>> cat in the dead+alive state,
>> 
>> The point of the existence of a preferred basis is that you will never 
>> observe a cat in a "dead+alive" state.
> 
> If that is what you mean, we both agree that Everett + Zurek solves that 
> problem. But the point is that If I observe the cat with an apparatus 
> deciding between alive and dead, I will put myself in the corresponding 
> superposition, unless some physical collapse occurs, but then we are no more 
> in Everett’s QM (QM without collapse).
> 
> No, you do not see any superposition.

Yes. We agree on this. The point is that I will not see it, because I am part 
of it, nor because it would have disappeared in some way.

> The cat itself is never in a superposition because decoherence brings about a 
> definite live state or dead state.

So we do disagree, at least if you claim that the superposition state disappear 
and QM applies to the cat. 

If there is no wave collapse, saying that some unknown particle interacting 
with the cat demolished the superposition state means only that the particle 
has been entangled with the state of the cat, and that being unable to track 
that particle makes me unable to handle the superposition anymore, but this 
means that the cat remains superposed, the particle get also superposed, and my 
whole environment get superposed. By the double QM linearity, the superposition 
never disappeared at all. The cat + the particle + me + the universe evolved 
through a unitary rotation where I am locally described in the two terms of the 
wave: seeing the cat alive and seeing the cat dead.

Decoherence in Everett (entanglement), and in Zurek (at least the perhaps old 
paper I read) does not make the “other branches vanishing”, it makes just my 
alternate history inaccessible.

> By the time you open the box, you have also split according to the classical 
> basis,

Ah, but that was my point. 

> so your probability of seeing a live or dead cat is just the classical 
> ignorance probability.

Yes, like the H guy in Moscow and in Washington, already duplicated, but before 
they open the door of the reconstitution box (in the 2 cities). The H-guy is 
ignorant on which branch of the computation he/she belongs.

> Opening the box does not collapse anything.

Thanks for reassuring me. That was my point.

> The point is that even if there is some superposition, it lasts no more than 
> a few nanoseconds.

That is misleading, as some people will interpret this by a collapse. I am 
happy you reject that interpretation. I agree that, even we’ll before I open 
the box, I have already been multiplied, unless some engineer trick to isolate 
the cat completely, which is just impossible today.

> After that time, your position is one of classical ignorance -- you are 
> either in the branch with the live cat, or the branch with the dead cat.

Yes, like after the WM-duplication, I am either in Moscow or in Washington. 
Pure classical ignorance.

> Opening the box does not change your relative state, or collapse anything.

Exactly my point. 

> There is no "you" that is in a superposition of these branches (just as there 
> is no "you" that is in both Washington and Moscow in step 3).

Exactly. In the first person perspective. Obviously, the 3p observer which 
looks at the entire setup can see me in both cities.
The only difference with the superposition, is that the 3p-view is technically 
lost from the experimental local (first person plural) view.

> So it is just like someo

Re: STEP 3

2019-08-15 Thread Bruce Kellett

On Thu, Aug 15, 2019 at 7:49 PM Bruno Marchal  wrote:

> On 15 Aug 2019, at 02:54, Bruce Kellett  wrote:
>
> On Wed, Aug 14, 2019 at 10:10 PM Bruno Marchal  wrote:
>
>> On 12 Aug 2019, at 14:42, Bruce Kellett  wrote:
>>
>>
>> That is simply incorrect. I refer you again to Zurek, who works in a
>> basically Everettian framework, but he stresses the importance of
>> environmental induced superselection (einselection) in producing the
>> preferred pointer basis. This then breaks things, in the sense that no
>> other basis is stable against decoherence, and other sets of basis vectors
>> rapidly (in times of the order of femtoseconds) collapse on to the
>> preferred pointer states. This is the basis of the emergence of the
>> classical world from the quantum substrate. And this occurs in Everett's
>> relative state approach just as much as in a Copenhagen-like collapse
>> models.
>>
>>
>> That explains why the many histories will look classical. But if I
>> observe a cat in the dead+alive state,
>>
>
> The point of the existence of a preferred basis is that you will never
> observe a cat in a "dead+alive" state.
>
>
> If that is what you mean, we both agree that Everett + Zurek solves that
> problem. But the point is that If I observe the cat with an apparatus
> deciding between alive and dead, I will put myself in the corresponding
> superposition, unless some physical collapse occurs, but then we are no
> more in Everett’s QM (QM without collapse).
>

No, you do not see any superposition. The cat itself is never in a
superposition because decoherence brings about a definite live state or
dead state. By the time you open the box, you have also split according to
the classical basis, so your probability of seeing a live or dead cat is
just the classical ignorance probability. Opening the box does not collapse
anything. The point is that even if there is some superposition, it lasts
no more than a few nanoseconds. After that time, your position is one of
classical ignorance -- you are either in the branch with the live cat, or
the branch with the dead cat. Opening the box does not change your relative
state, or collapse anything. There is no "you" that is in a superposition
of these branches (just as there is no "you" that is in both Washington and
Moscow in step 3). So it is just like someone tossing a classical coin that
you can't see.

Bruce

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Re: STEP 3


> On 15 Aug 2019, at 02:54, Bruce Kellett  wrote:
> 
> On Wed, Aug 14, 2019 at 10:10 PM Bruno Marchal  > wrote:
> On 12 Aug 2019, at 14:42, Bruce Kellett  > wrote:
>> 
>> That is simply incorrect. I refer you again to Zurek, who works in a 
>> basically Everettian framework, but he stresses the importance of 
>> environmental induced superselection (einselection) in producing the 
>> preferred pointer basis. This then breaks things, in the sense that no other 
>> basis is stable against decoherence, and other sets of basis vectors rapidly 
>> (in times of the order of femtoseconds) collapse on to the preferred pointer 
>> states. This is the basis of the emergence of the classical world from the 
>> quantum substrate. And this occurs in Everett's relative state approach just 
>> as much as in a Copenhagen-like collapse models.
> 
> That explains why the many histories will look classical. But if I observe a 
> cat in the dead+alive state,
> 
> The point of the existence of a preferred basis is that you will never 
> observe a cat in a "dead+alive" state.

If that is what you mean, we both agree that Everett + Zurek solves that 
problem. But the point is that If I observe the cat with an apparatus deciding 
between alive and dead, I will put myself in the corresponding superposition, 
unless some physical collapse occurs, but then we are no more in Everett’s QM 
(QM without collapse).

Bruno




> 
> Bruce
> 
> -- 
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Re: STEP 3


> On 13 Aug 2019, at 18:37, Jason Resch  wrote:
> 
> 
> 
> On Tuesday, August 13, 2019, John Clark  > wrote:
> On Wed, Aug 7, 2019 at 9:40 AM Bruno Marchal  > wrote:
> 
> On 7 Aug 2019, at 15:08, Bruce Kellett  > wrote:
>  
> >> What exactly is the difference between something that it is impossible in 
> >> principle to detect and something that does not exist?
>  
> > It is like the difference between the human existence and the human non 
> > existence, for an alien situated in a very far away galaxy. The fact that 
> > this alien cannot detect us does not make the human disappearing. 
> 
> You're dodging the question. We can certainly detect ourselves, and it is not 
> impossible in principle for a alien in a distant galaxy to detect us, and 
> that is very different from your silly phantom computations that pure numbers 
> are suposed to be able to perform. 
> 
> > It is like the other side of the moon before we built rocket. 
> 
> Nobody ever said there was a philosophical problem in observing the far side 
> of the moon, it was always just a matter of engineering, but no amount of 
> engineering can make your ridiculous phantom calculations real. If they 
> existed it would be possible in principle to count the number of angels that 
> were sitting on the head of a pin, but your non-material Turing Machine is 
> hopeless. 
> 
> 
> The conversation was about other branches of the wavefunction.  (Which I 
> think you believe in despite not being able to see, nor them being able to 
> make a CPU useful to you).

Well seen. 

Bruno



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> 
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Re: STEP 3

> On 13 Aug 2019, at 16:53, John Clark  wrote:
> 
> Nobody ever said there was a philosophical problem in observing the far side 
> of the moon, it was always just a matter of engineering, but no amount of 
> engineering can make your ridiculous phantom calculations real.

There is a philosophical problem even on milk in the fridge when the fridge is 
close. We can use Einstein’s reality principle, but as you know, this leads to 
other problem, and people different on the existence or not of a collapse. Yes, 
those are not problem if we adopt a FAPP-philosophy.

> If they existed it would be possible in principle to count the number of 
> angels that were sitting on the head of a pin,

With mechanism, that is the same as the problem of how much bits or qubits we 
can process in a volume similar to the head of a pin. It makes sense.

> but your non-material Turing Machine is hopeless. 

Then Church, Turing, Post, Kleene, Gödel and many more are all hopeless, in 
fact elementary arithmetic becomes hopeless.

Bruno

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Re: STEP 3

2019-08-14 Thread smitra


On 15-08-2019 06:25, Bruce Kellett wrote:

On Thu, Aug 15, 2019 at 7:23 AM smitra  wrote:


On 14-08-2019 00:44, Bruce Kellett wrote:


You over-elaborate a simple schematic. My A(x) and B(x) are

simply the

amplitudes of the wave function at point x on the screen from the

two

slits. To get the intensity at x, you add the amplitudes and take

the

modulus squared -- you do not add the intensities from each slit
separately.


What matters is that you can get interference of the orthogonal
components in the superposition. Whether or not you do actually get

interference is not orthogonality before measurement, but after
measurement. If the photons moving through the slits interact with
another system initially in some state |C> such that if the photon
moves
through slit A the state |C> changes to |D> while it changes to |E>
if
the photon moves through slit B, then the interference pattern will
be
the function Re[A(x)*B(x)], this will become zero if |C> and
|D>
are orthogonal, which means that you still have a superposition of
two
orthogonal terms in the MWI sector where a photon lands on some
specific
spot on the screen. The system then has perfect which-way
information.


I don't think this quite works either. Look at it this way. Take |A>
as the wave function for going through slit A, and |B> the
corresponding wave function for going through slit B. The
superposition (|A> + |B>) is the wave function that propagates to the
screen. At the screen, the intensity is the square of this:
   (A| + |B>)() =  |A|^2 + |B|^2 +  + ,
and there is interference. This is because you have added together the
paths through each slit, but you don't know which slit was traversed,
so A and B are coherent.

If you now add a detector at the slits that registers L (for Left) if
A was traversed, and R (for Right) if B was traversed, the original
superposition becomes
   (|A>|L> + |B>|R>)
because the detector becomes entangled with the slit through which the
photon went. If we now go to the intensity at the screen, we obviously
get
   |A|^2 + |B|^2 +  + 
The detector states recording L or R for which slit the photon
traversed are orthogonal,  =  = 0, so  detecting which slit
the photon went through destroys the interference. If there is no
detection at the slits, the amplitudes are coherent, they are not
orthogonal, and there is interference.



 = 0 due to orthogonality (and this is preserved as the screen is 
approached as the state evolves in a unitary way), what we measure at 
the screen is  +  as a function of the position x on 
the screen. If we add up all the counts on the screens from all 
positions then the interference effect averages out to zero due to 
orthogonality, but that doesn't matter. We can still see fringes and 
peaks on the screen.


Saibal

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Re: STEP 3

2019-08-14 Thread Bruce Kellett

On Thu, Aug 15, 2019 at 7:23 AM smitra  wrote:

> On 14-08-2019 00:44, Bruce Kellett wrote:
> >
> > You over-elaborate a simple schematic. My A(x) and B(x) are simply the
> > amplitudes of the wave function at point x on the screen from the two
> > slits. To get the intensity at x, you add the amplitudes and take the
> > modulus squared -- you do not add the intensities from each slit
> > separately.
>
>
> What matters is that you can get interference of the orthogonal
> components in the superposition. Whether or not you do actually get
> interference is not orthogonality before measurement, but after
> measurement. If the photons moving through the slits interact with
> another system initially in some state |C> such that if the photon moves
> through slit A the state |C> changes to |D> while it changes to |E> if
> the photon moves through slit B, then the interference pattern will be
> the function Re[A(x)*B(x)], this will become zero if |C> and |D>
> are orthogonal, which means that you still have a superposition of two
> orthogonal terms in the MWI sector where a photon lands on some specific
> spot on the screen. The system then has perfect which-way information.
>

I don't think this quite works either. Look at it this way. Take |A> as the
wave function for going through slit A, and |B> the corresponding wave
function for going through slit B. The superposition (|A> + |B>) is the
wave function that propagates to the screen. At the screen, the intensity
is the square of this:
   (A| + |B>)() =  |A|^2 + |B|^2 +  + ,
and there is interference. This is because you have added together the
paths through each slit, but you don't know which slit was traversed, so A
and B are coherent.

If you now add a detector at the slits that registers L (for Left) if A was
traversed, and R (for Right) if B was traversed, the original superposition
becomes
   (|A>|L> + |B>|R>)
because the detector becomes entangled with the slit through which the
photon went. If we now go to the intensity at the screen, we obviously get
   |A|^2 + |B|^2 +  + 
The detector states recording L or R for which slit the photon traversed
are orthogonal,  =  = 0, so  detecting which slit the photon went
through destroys the interference. If there is no detection at the slits,
the amplitudes are coherent, they are not orthogonal, and there is
interference.

Bruce

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Re: STEP 3

2019-08-14 Thread Bruce Kellett

On Wed, Aug 14, 2019 at 10:10 PM Bruno Marchal  wrote:

> On 12 Aug 2019, at 14:42, Bruce Kellett  wrote:
>
>
> That is simply incorrect. I refer you again to Zurek, who works in a
> basically Everettian framework, but he stresses the importance of
> environmental induced superselection (einselection) in producing the
> preferred pointer basis. This then breaks things, in the sense that no
> other basis is stable against decoherence, and other sets of basis vectors
> rapidly (in times of the order of femtoseconds) collapse on to the
> preferred pointer states. This is the basis of the emergence of the
> classical world from the quantum substrate. And this occurs in Everett's
> relative state approach just as much as in a Copenhagen-like collapse
> models.
>
>
> That explains why the many histories will look classical. But if I observe
> a cat in the dead+alive state,
>

The point of the existence of a preferred basis is that you will never
observe a cat in a "dead+alive" state.

Bruce

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Re: STEP 3

2019-08-14 Thread smitra


On 14-08-2019 00:44, Bruce Kellett wrote:

On Wed, Aug 14, 2019 at 8:28 AM smitra  wrote:


On 13-08-2019 13:33, Bruce Kellett wrote:


Of course A(x) and B(x) refer to the same point on the screen.

That is

not a collapse, that is just what the notation means.


A(x) and B(x) considered as the representations of |A> and |B> in
the
position basis, i.e. A(x) =  and B(x) =  are still
orthogonal
states, as they represent the orthogonal states |A> and |B>:

0 =  = Integral over x of d^3x = Integral over x of

A*(x)B(x) d^3x

But you interpret this as the total counts of particles on the
entire
screen not changing which you call an absence of interference.
However
interference is what we detect locally on each point on the screen.
You
can't say that for each point x0 on the screen, A(x0) and B(x0)
are
particle states. These values are not the quantum states of the
particle
before it hits the screen, unless you would have done a measurement

localizing the particle near x0.


The spot on the screen may well be such an experiment.


So, your argument only makes sense if you invoke collapse via a
position
measurement.


You over-elaborate a simple schematic. My A(x) and B(x) are simply the
amplitudes of the wave function at point x on the screen from the two
slits. To get the intensity at x, you add the amplitudes and take the
modulus squared -- you do not add the intensities from each slit
separately.

Perhaps the problem stems from my using the ket vector representation
for single complex numbers. But complex numbers are never orthogonal,
so using the ket generalises this interference result to general state
vectors. Integrating over the screen is not relevant to the intensity
at each point.


What matters is that you can get interference of the orthogonal 
components in the superposition. Whether or not you do actually get 
interference is not orthogonality before measurement, but after 
measurement. If the photons moving through the slits interact with 
another system initially in some state |C> such that if the photon moves 
through slit A the state |C> changes to |D> while it changes to |E> if 
the photon moves through slit B, then the interference pattern will be 
the function Re[A(x)*B(x)], this will become zero if |C> and |D> 
are orthogonal, which means that you still have a superposition of two 
orthogonal terms in the MWI sector where a photon lands on some specific 
spot on the screen. The system then has perfect which-way information.


Saibal

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Re: STEP 3

2019-08-14 Thread Bruno Marchal


> On 12 Aug 2019, at 14:42, Bruce Kellett  wrote:
> 
> On Mon, Aug 12, 2019 at 7:36 PM Bruno Marchal  > wrote:
> On 12 Aug 2019, at 04:06, Bruce Kellett  > wrote:
>> 
>> If you do not measure which slit the photon went through, then the 
>> superposition of slits is not broken by decoherence.
> Decoherence break things only if there is a collapse.
> 
> That is simply incorrect. I refer you again to Zurek, who works in a 
> basically Everettian framework, but he stresses the importance of 
> environmental induced superselection (einselection) in producing the 
> preferred pointer basis. This then breaks things, in the sense that no other 
> basis is stable against decoherence, and other sets of basis vectors rapidly 
> (in times of the order of femtoseconds) collapse on to the preferred pointer 
> states. This is the basis of the emergence of the classical world from the 
> quantum substrate. And this occurs in Everett's relative state approach just 
> as much as in a Copenhagen-like collapse models.

That explains why the many histories will look classical. But if I observe a 
cat in the dead+alive state, some people will still live the two alternative 
quasi-classical histories. 



> 
> Without collapse, even if I measure which slit the photon went through, the 
> two terms of the superposition continue to exist, describing me seeing both 
> outcomes, and both me feel like if there has been a collapse,
> 
> No, that is not really true. Complementarity plays a role. Measurement of 
> which slit the photon went through is incompatible with the effective 
> momentum measurement that gives interference at the screen. That is 
> essentially why the slit measurement destroys the possibility of 
> interference. There is an incompatibility between the measurements -- 
> basically related to the non-commutation of position and momentum operators 
> and the Heisenberg uncertainty principle.

I agree with this, but fail to see why it makes my statement above false.



> 
> and that decoherence is physical real, but that illusion is explained by the 
> formalism, in a manner similar to the WM duplication: it is just first person 
> indeterminacy, not in a self-duplication, but in a self entanglement.
> 
> Not really true, either. Decoherence, the formation of a stable pointer 
> basis, and the like, are all essential for the emergence of the classical 
> from the quantum, and the formation of objective physical states. This is the 
> operation of Zurek's Quantum Darwinism -- the environment acting as witness. 
> So this is third person objective physics -- it is not first person 
> indeterminacy.


Why? The environment, in a world with brains, acts as a witness of all 
quasi-classical histories. Decoherence explains why those histories look very 
classical, but it does not select one classical history (cat alive, say), it 
put itself in a superposition of the (two, here) classical histories. 

Bruno



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Re: STEP 3

2019-08-14 Thread Bruno Marchal


> On 12 Aug 2019, at 14:28, Bruce Kellett  wrote:
> 
> On Mon, Aug 12, 2019 at 7:30 PM Bruno Marchal  > wrote:
> On 11 Aug 2019, at 14:09, Bruce Kellett  > wrote:
> 
>> It is not a matter of the difference between collapse or no-collapse models 
>> -- it is a matter of the basic interpretation of what Everett's "relative 
>> states" actually are, and why the basis problem is so important for Everett.
> 
> Everett makes clear that the choice of the base os irrelevant for the global 
> description,
> 
> Maybe for the global description, but not for recovering the results of 
> experience. Everett got this latter problem wrong (he probably wasn't even 
> aware that there was a problem.).

That is not my feeling, but I will take a look back to what Everett said in his 
long text.




> Zurek says:
> "The basis ambiguity is not limited to pointers of measuring devices. One can 
> show that also very large systems (such as satellites or planets) can evolve 
> into very non-classical superpositions.

I agree (with Zurek).



> In reality, this does not seem to happen.

OK.



> So there is something that picks out certain preferred quantum states, and 
> makes them effectively classical.

… relatively to the observer. The observer is itself the result of many 
interaction/measurement which picked the base defining its computational 
properties.




> Before there is collapse, a set of preferred states, one of which is selected 
> by the collapse must somehow be chosen.

OK, with “apparent collapse” instead of “collapse”. The context should be given 
if this is what Zurek means. Some texts by Zurek explicitly rejects the idea 
that the collapse is a physical event (like in Everett).




> There is nothing in the writings of Everett that would hint at such a 
> criterion for such preferred states, and nothing to hint that he was aware of 
> this question The preferred basis problem was settled by environment 
> induced superselection (einselection), usually discussed along with 
> decoherence.


OK. And this explains the appearance of quasi classical histories.



> When environmental monitoring is focussed on a specific observable of the 
> system, its eigenstates form a pointer basis:


Yes. This is how the “brain” has been able to evolve, and why the position 
observable plays a key role for macroscopic entities.




> They entangle least with the environment (and, therefore, are least perturbed 
> by it.) This resolves the basic ambiguity." (Zurek, arxiv:0707.2832 July 2007)

OK.



> 
> Zurek goes on to develop this at some length, but the basic point is 
> stability under environmental interactions, so the permanent, objective 
> records of the result can be formed in the environment. This is the basis of 
> intersubjective agreement about the result,

Yes, it provides a notion of first person plural sharable observation.



> namely, objectivity. Quantum Darwinism means that the results that we observe 
> are classically stable.
> 
> and the choice of the base is done by the observer for the local description,
> 
> The observer does not chose the basis -- he might choose what to measure, but 
> it is the environment that ultimately decides what bases is stable, and which 
> other bases rapidly collapse on to states in the pointer basis.

OK. My way to express this was perhaps unclear. It is evolution which has 
“chosen” the base, somehow.




>  
> exactly like with digital mechanism, where the notion of universal numbers 
> play the role of the base. The whole theology and physics does not depend on 
> the choice of the universal machinery we posit at the start, but from the 
> points if view of each relative universal numbers, such difference play a key 
> role in the local happenings and predictions.
> 
> This is no longer quantum mechanics, nor has it anything to do with physics.

Only because you work in the materialist and non mechanist frame, I guess. With 
mechanism, those becomes key element in the explanation of why and how a 
physical reality appears and get sharable by many independent universal 
machines. 

Bruno



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Re: STEP 3

2019-08-13 Thread Bruce Kellett

On Wed, Aug 14, 2019 at 8:28 AM smitra  wrote:

> On 13-08-2019 13:33, Bruce Kellett wrote:
> >
> > Of course A(x) and B(x) refer to the same point on the screen. That is
> > not a collapse, that is just what the notation means.
>
> A(x) and B(x) considered as the representations of |A> and |B> in the
> position basis, i.e.  A(x) =  and B(x) =  are still orthogonal
> states, as they represent the orthogonal states |A> and |B>:
>
> 0 =  = Integral over x of d^3x =  Integral over x of
> A*(x)B(x) d^3x
>
> But you interpret this as the total counts of particles on the entire
> screen not changing which you call an absence of interference. However
> interference is what we detect locally on each point on the screen. You
> can't say that for each point x0 on the screen,  A(x0) and B(x0) are
> particle states. These values are not the quantum states of the particle
> before it hits the screen, unless you would have done a measurement
> localizing the particle near x0.
>

The spot on the screen may well be such an experiment.

> So, your argument only makes sense if you invoke collapse via a position
> measurement.
>

You over-elaborate a simple schematic. My A(x) and B(x) are simply the
amplitudes of the wave function at point x on the screen from the two
slits. To get the intensity at x, you add the amplitudes and take the
modulus squared -- you do not add the intensities from each slit separately.

Perhaps the problem stems from my using the ket vector representation for
single complex numbers. But complex numbers are never orthogonal, so using
the ket generalises this interference result to general state vectors.
Integrating over the screen is not relevant to the intensity at each point.

Bruce

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Re: STEP 3

2019-08-13 Thread smitra


On 13-08-2019 13:33, Bruce Kellett wrote:

On Tue, Aug 13, 2019 at 7:41 PM smitra  wrote:


On 13-08-2019 05:14, Bruce Kellett wrote:

On Tue, Aug 13, 2019 at 12:16 PM smitra  wrote:


On 13-08-2019 02:49, Bruce Kellett wrote:

On Tue, Aug 13, 2019 at 10:43 AM smitra 

wrote:



On 13-08-2019 01:41, Bruce Kellett wrote:

On Tue, Aug 13, 2019 at 9:09 AM smitra 

wrote:



On 12-08-2019 08:29, Bruce Kellett wrote:


Look at this another way. It is just an illustration of
complementarity. Measuring which slit the photon went

through

is

a

position measurement at the slits. Measuring the

interference

pattern

at the screen is equivalent to a momentum measurement at

the

slits.

Such measurement operators do not commute -- the

measurements

are

complementary and cannot be performed simultaneously.



It doesn't matter for orthogonality of the states whether or

not

they
are measured.


Of course it does. The slits are not orthogonal states unless

they are

measured position eigenstates. If they are not measured, they

are

individually superpositions of many position eigenstates

(including

eigenstates that overlap both slits), so the slits themselves

are

no

longer orthogonal. Orthogonal states cannot interfere, that

is

why a

position measurement at the slits makes the interference

pattern

on

the screen disappear.

The fact remains, that orthogonal states cannot interfere:

( + |B>) =  +  + 2 

and the interference term  vanishes if |A> and |B> are
orthogonal. You can't get away from this basic fact about

quantum

mechanics.



 is zero in the two slit experiment, if you integrate the
interference term over the screen you get zero.

Thing is that the interference we can observe at some position

x

on

the
screen is Re[], which for general x is nonzero

despite

the
fact that  = 0.


So you agree that if the overlap vanishes you do not get

interference.

You go to some lengths to try and avoid this fact. Saying that

the

integral over the screen vanishes is beside the point.


 is the integral over all space of  and this is
zero, but
the interference we observe at some point x on the screen is
Re[], and that's in general nonzero even for

orthogonal

|A>
and |B>.


I don't know what you are trying to prove, but that is not what

my

formula meant. What I intended was that |A> represents the wave

at the

screen coming from slit A, and |B> is the wave at the screen from

slit

B. There is already and implicit dependence on the position along

the

screen A = A(x), etc. So inserting a complete set of position

states,

int dx |x>
the

point.


The states |A> and |B> have components psi_A(x) = , and
psi_B(X) =
, these components depend on x, not the states |A> and |B>. We
can
work with these components (wavefunctions) psi_A(x), and psi_B(x)
instead of |A> and |B> describing the particle just before it
interacts
with the screen, but |A> and |B> are still orthogonal, the
wavefunctions
psi_A(x), and psi_B(x) are obviously also orthogonal. You can argue
that
when restricting the two wavefunctions in some small region near x
= x0
must yield two functions that are not orthogonal, otherwise there
cannot
be interference at x = x0. However, what you are then doing is
letting
the states collapse by letting them get localized near x0. So,
you're
not talking about |A> and |B>, you are instead talking about two
states
that are proportional to |x0>


If we take a general state |A>, which depends on x, the overlap

of two

such is , and this cannot vanish if there is to be

interference.

Classically, the intensity at x from slit A is |A|^2, and

similarly

for slit B. The fact that quantum mechanically we add amplitudes,

not

intensities, means that there is the overlap term . If this

does

not vanish, [ |A> and 
whether

the slit states are orthogonal or not. In general, they are not.


As explained above, what you call |A> and |B> are in fact both
proportional to some arbitrarily chosen |x0>. You have effectively
replaced |A> by  |x0> and |B> by  |x0>, so you are
letting
the two states collapse at some point x0 on the screen first which
yield
states that depend on x0 and then you argue on the basis that the
squared norm of what you now have should have an interference term,

which then precludes the two states being orthogonal. But you made
them
co-linear by letting them collapse at the same position first.


Of course A(x) and B(x) refer to the same point on the screen. That is
not a collapse, that is just what the notation means.


A(x) and B(x) considered as the representations of |A> and |B> in the 
position basis, i.e.  A(x) =  and B(x) =  are still orthogonal 
states, as they represent the orthogonal states |A> and |B>:


0 =  = Integral over x of d^3x =  Integral over x of 
A*(x)B(x) d^3x


But you interpret this as the total counts of particles on the entire 
screen not changing which you call an absence of interference. However 
interference is what we detect locally on each point on the scr

Re: STEP 3

2019-08-13 Thread Jason Resch

On Tuesday, August 13, 2019, John Clark  wrote:

> On Wed, Aug 7, 2019 at 9:40 AM Bruno Marchal  wrote:
>
> On 7 Aug 2019, at 15:08, Bruce Kellett  wrote:
>
>>
>> >> What exactly is the difference between something that it is
>> impossible in principle to detect and something that does not exist?
>>
>
>
> *> It is like the difference between the human existence and the human non
>> existence, for an alien situated in a very far away galaxy. The fact that
>> this alien cannot detect us does not make the human disappearing. *
>>
>
> You're dodging the question. We can certainly detect ourselves, and it is
> not impossible in principle for a alien in a distant galaxy to detect us,
> and that is very different from your silly phantom computations that pure
> numbers are suposed to be able to perform.
>
> > *It is like the other side of the moon before we built rocket. *
>>
>
> Nobody ever said there was a philosophical problem in observing the far
> side of the moon, it was always just a matter of engineering, but no amount
> of engineering can make your ridiculous phantom calculations real. If they
> existed it would be possible in principle to count the number of angels
> that were sitting on the head of a pin, but your non-material Turing
> Machine is hopeless.
>
>
The conversation was about other branches of the wavefunction.  (Which I
think you believe in despite not being able to see, nor them being able to
make a CPU useful to you).

Jason



> John K Clark
>
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> 
> .
>

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Re: STEP 3

2019-08-13 Thread John Clark

On Wed, Aug 7, 2019 at 9:40 AM Bruno Marchal  wrote:

On 7 Aug 2019, at 15:08, Bruce Kellett  wrote:

>
> >> What exactly is the difference between something that it is impossible
> in principle to detect and something that does not exist?
>

*> It is like the difference between the human existence and the human non
> existence, for an alien situated in a very far away galaxy. The fact that
> this alien cannot detect us does not make the human disappearing. *
>

You're dodging the question. We can certainly detect ourselves, and it is
not impossible in principle for a alien in a distant galaxy to detect us,
and that is very different from your silly phantom computations that pure
numbers are suposed to be able to perform.

> *It is like the other side of the moon before we built rocket. *
>

Nobody ever said there was a philosophical problem in observing the far
side of the moon, it was always just a matter of engineering, but no amount
of engineering can make your ridiculous phantom calculations real. If they
existed it would be possible in principle to count the number of angels
that were sitting on the head of a pin, but your non-material Turing
Machine is hopeless.

John K Clark

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Re: STEP 3

2019-08-13 Thread Bruce Kellett

On Tue, Aug 13, 2019 at 7:41 PM smitra  wrote:

> On 13-08-2019 05:14, Bruce Kellett wrote:
> > On Tue, Aug 13, 2019 at 12:16 PM smitra  wrote:
> >
> >> On 13-08-2019 02:49, Bruce Kellett wrote:
> >>> On Tue, Aug 13, 2019 at 10:43 AM smitra  wrote:
> >>>
>  On 13-08-2019 01:41, Bruce Kellett wrote:
> > On Tue, Aug 13, 2019 at 9:09 AM smitra 
> >> wrote:
> >
> >> On 12-08-2019 08:29, Bruce Kellett wrote:
> >>>
> >>> Look at this another way. It is just an illustration of
> >>> complementarity. Measuring which slit the photon went through
>  is
> >> a
> >>> position measurement at the slits. Measuring the interference
> >> pattern
> >>> at the screen is equivalent to a momentum measurement at the
> >> slits.
> >>> Such measurement operators do not commute -- the measurements
>  are
> >>> complementary and cannot be performed simultaneously.
> >>>
> >>
> >> It doesn't matter for orthogonality of the states whether or
> >> not
> >> they
> >> are measured.
> >
> > Of course it does. The slits are not orthogonal states unless
>  they are
> > measured position eigenstates. If they are not measured, they
> >> are
> > individually superpositions of many position eigenstates
>  (including
> > eigenstates that overlap both slits), so the slits themselves
> >> are
>  no
> > longer orthogonal. Orthogonal states cannot interfere, that is
>  why a
> > position measurement at the slits makes the interference
> >> pattern
>  on
> > the screen disappear.
> >
> > The fact remains, that orthogonal states cannot interfere:
> >
> > ( + |B>) =  +  + 2 
> >
> > and the interference term  vanishes if |A> and |B> are
> > orthogonal. You can't get away from this basic fact about
> >> quantum
> > mechanics.
> >
> 
>   is zero in the two slit experiment, if you integrate the
>  interference term over the screen you get zero.
> 
>  Thing is that the interference we can observe at some position x
> >> on
>  the
>  screen is Re[], which for general x is nonzero despite
>  the
>  fact that  = 0.
> >>>
> >>> So you agree that if the overlap vanishes you do not get
> >> interference.
> >>> You go to some lengths to try and avoid this fact. Saying that
> >> the
> >>> integral over the screen vanishes is beside the point.
> >>>
> >>  is the integral over all space of  and this is
> >> zero, but
> >> the interference we observe at some point x on the screen is
> >> Re[], and that's in general nonzero even for orthogonal
> >> |A>
> >> and |B>.
> >
> > I don't know what you are trying to prove, but that is not what my
> > formula meant. What I intended was that |A> represents the wave at the
> > screen coming from slit A, and |B> is the wave at the screen from slit
> > B. There is already and implicit dependence on the position along the
> > screen A = A(x), etc. So inserting a complete set of position states,
> > int dx |x> > point.
>
> The states |A> and |B> have components psi_A(x) = , and psi_B(X) =
> , these components depend on x, not the states |A> and |B>. We can
> work with these components (wavefunctions) psi_A(x), and psi_B(x)
> instead of |A> and |B> describing the particle just before it interacts
> with the screen, but |A> and |B> are still orthogonal, the wavefunctions
> psi_A(x), and psi_B(x) are obviously also orthogonal. You can argue that
> when restricting the two wavefunctions in some small region near x = x0
> must yield two functions that are not orthogonal, otherwise there cannot
> be interference at x = x0. However, what you are then doing is letting
> the states collapse by letting them get localized near x0. So, you're
> not talking about |A> and |B>, you are instead talking about two states
> that are proportional to |x0>
>
>
> > If we take a general state |A>, which depends on x, the overlap of two
> > such is , and this cannot vanish if there is to be interference.
> > Classically, the intensity at x from slit A is |A|^2, and similarly
> > for slit B. The fact that quantum mechanically we add amplitudes, not
> > intensities, means that there is the overlap term . If this does
> > not vanish, [ |A> and  > interference. There is no more to it than that. The waves at the
> > screen are not orthogonal! It really has nothing to do with whether
> > the slit states are orthogonal or not. In general, they are not.
>
> As explained above, what you call |A> and |B> are in fact both
> proportional to some arbitrarily chosen |x0>. You have effectively
> replaced |A> by  |x0> and |B> by  |x0>, so you are letting
> the two states collapse at some point x0 on the screen first which yield
> states that depend on x0 and then you argue on the basis that the
> squared norm of what you now have should have an interference term,
> which then precludes the two states being orthogonal. But you made them
> co-linear by letting them col

Re: STEP 3

2019-08-13 Thread smitra

On 13-08-2019 10:21, Philip Thrift wrote:

On Monday, August 12, 2019 at 7:43:51 PM UTC-5, smitra wrote:

Thing is that the interference we can observe at some position x on
the
screen is Re[], which for general x is nonzero despite the

fact that = 0.

Saibal

To a probability (or measure) theorist, one making the conversion from
classical summing rules to quantum summing rules, this should be
easier:

https://arxiv.org/pdf/gr-qc/9401003.pdf

Where quantum theory differs from classical mechanics (in this view)
is in its dynamics, which of course is stochastic rather than
deterministic. As such, the theory functions by furnishing
probabilities for sets of histories. More formally, it associates to a
set A of histories a non-negative real number |A|, which I will call
its quantum measure |A|; and it is this measure that enters into the
sum-rules we will be concerned with.
...

NOTIONS SUCH AS STATE-VECTORS AND OBSERVABLES NEVER APPEAR, EXCEPT FOR
THE SAKE OF COMPUTATIONAL CONVENIENCE.

...

In the two-slit experiment, for example, the probability that a
particular detector will register the arrival of the electron is
(proportional to) the measure |C| of the set C of all electron world
lines which in fact pass close enough to that detector to trigger it.
When we contemplate also blocking off one or the other slit, there are
(for a fixed detector) three sets of histories to consider: the set A
of histories which arrive at the detector after traversing the
“first” slit, the corresponding set B for the “second” slit,
and the original set C = A ∐ B, the disjoint † union of A and B.

It is of course characteristic of quantum probability that the
interference term

I(A, B) := |A ∐ B| − |A| − |B|

between the slits is not zero. The surprising thing (once one has
gotten used to the fact of interference itself) is that this violation
of the classical probability sum-rules is in a certain sense so mild,
since the corresponding sum-rule for three alternatives
remains valid.

In any case, the important thing from the standpoint of interpretation
is that the electron follows one and only one path, not somehow two at
once. If probabilities are involved, it is only because the path is
not determined in advance, just as it is
initially undetermined in a classical stochastic process.

Given the failure of the sum rule I(A, B) = 0, it is clear that
quantum probabilities cannot be interpreted in the same manner that
classical ones are wont to be interpreted, in terms of (actual or
fictitious) ensemble frequencies.

Thanks, I'll read that article.

Saibal

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Re: STEP 3

2019-08-13 Thread smitra


On 13-08-2019 05:14, Bruce Kellett wrote:

On Tue, Aug 13, 2019 at 12:16 PM smitra  wrote:


On 13-08-2019 02:49, Bruce Kellett wrote:

On Tue, Aug 13, 2019 at 10:43 AM smitra  wrote:


On 13-08-2019 01:41, Bruce Kellett wrote:

On Tue, Aug 13, 2019 at 9:09 AM smitra 

wrote:



On 12-08-2019 08:29, Bruce Kellett wrote:


Look at this another way. It is just an illustration of
complementarity. Measuring which slit the photon went through

is

a

position measurement at the slits. Measuring the interference

pattern

at the screen is equivalent to a momentum measurement at the

slits.

Such measurement operators do not commute -- the measurements

are

complementary and cannot be performed simultaneously.



It doesn't matter for orthogonality of the states whether or

not

they
are measured.


Of course it does. The slits are not orthogonal states unless

they are

measured position eigenstates. If they are not measured, they

are

individually superpositions of many position eigenstates

(including

eigenstates that overlap both slits), so the slits themselves

are

no

longer orthogonal. Orthogonal states cannot interfere, that is

why a

position measurement at the slits makes the interference

pattern

on

the screen disappear.

The fact remains, that orthogonal states cannot interfere:

( + |B>) =  +  + 2 

and the interference term  vanishes if |A> and |B> are
orthogonal. You can't get away from this basic fact about

quantum

mechanics.



 is zero in the two slit experiment, if you integrate the
interference term over the screen you get zero.

Thing is that the interference we can observe at some position x

on

the
screen is Re[], which for general x is nonzero despite
the
fact that  = 0.


So you agree that if the overlap vanishes you do not get

interference.

You go to some lengths to try and avoid this fact. Saying that

the

integral over the screen vanishes is beside the point.


 is the integral over all space of  and this is
zero, but
the interference we observe at some point x on the screen is
Re[], and that's in general nonzero even for orthogonal
|A>
and |B>.


I don't know what you are trying to prove, but that is not what my
formula meant. What I intended was that |A> represents the wave at the
screen coming from slit A, and |B> is the wave at the screen from slit
B. There is already and implicit dependence on the position along the
screen A = A(x), etc. So inserting a complete set of position states,
int dx |x>

The states |A> and |B> have components psi_A(x) = , and psi_B(X) = 
, these components depend on x, not the states |A> and |B>. We can 
work with these components (wavefunctions) psi_A(x), and psi_B(x) 
instead of |A> and |B> describing the particle just before it interacts 
with the screen, but |A> and |B> are still orthogonal, the wavefunctions 
psi_A(x), and psi_B(x) are obviously also orthogonal. You can argue that 
when restricting the two wavefunctions in some small region near x = x0 
must yield two functions that are not orthogonal, otherwise there cannot 
be interference at x = x0. However, what you are then doing is letting 
the states collapse by letting them get localized near x0. So, you're 
not talking about |A> and |B>, you are instead talking about two states 
that are proportional to |x0>




If we take a general state |A>, which depends on x, the overlap of two
such is , and this cannot vanish if there is to be interference.
Classically, the intensity at x from slit A is |A|^2, and similarly
for slit B. The fact that quantum mechanically we add amplitudes, not
intensities, means that there is the overlap term . If this does
not vanish, [ |A> and 

As explained above, what you call |A> and |B> are in fact both 
proportional to some arbitrarily chosen |x0>. You have effectively 
replaced |A> by  |x0> and |B> by  |x0>, so you are letting 
the two states collapse at some point x0 on the screen first which yield 
states that depend on x0 and then you argue on the basis that the 
squared norm of what you now have should have an interference term, 
which then precludes the two states being orthogonal. But you made them 
co-linear by letting them collapse at the same position first.


Before collapse you can write |A> in the position  representation as 
integral over |x> d^3x and similarly  B = integral over |y> 
d^3y, and we not only have that |A> and |B> are orthogonal, but also the 
simple fact that |x> and |y> are orthogonal for x not equal to y.


Saibal

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Re: STEP 3

2019-08-13 Thread Philip Thrift

On Monday, August 12, 2019 at 7:43:51 PM UTC-5, smitra wrote:
>
>
> Thing is that the interference we can observe at some position x on the 
> screen is Re[], which for general x is nonzero despite the 
> fact that  = 0. 
>
> Saibal 
>

To a probability (or measure) theorist, one making the conversion from 
classical summing rules to quantum summing rules, this should be easier:

https://arxiv.org/pdf/gr-qc/9401003.pdf

Where quantum theory differs from classical mechanics (in this view) is in 
its dynamics, which of course is stochastic rather than deterministic. As 
such, the theory functions by furnishing probabilities for sets of 
histories. More formally, it associates to a set A of histories a 
non-negative real number |A|, which I will call its quantum measure |A|; 
and it is this measure that enters into the sum-rules we will be concerned 
with.
...
*Notions such as state-vectors and observables never appear, except for the 
sake of computational convenience.*

...

In the two-slit experiment, for example, the probability that a particular 
detector will register the arrival of the electron is (proportional to) the 
measure |C| of the set C of all electron world lines which in fact pass 
close enough to that detector to trigger it. When we contemplate also 
blocking off one or the other slit, there are (for a fixed detector) three 
sets of histories to consider: the set A of histories which arrive at the 
detector after traversing the “first” slit, the corresponding set B for the 
“second” slit, and the original set C = A ∐ B, the disjoint † union of A 
and B.

It is of course characteristic of quantum probability that the interference 
term

I(A, B) := |A ∐ B| − |A| − |B|

between the slits is not zero. The surprising thing (once one has gotten 
used to the fact of interference itself) is that this violation of the 
classical probability sum-rules is in a certain sense so mild, since the 
corresponding sum-rule for three alternatives
remains valid.

In any case, the important thing from the standpoint of interpretation is 
that the electron follows one and only one path, not somehow two at once. 
If probabilities are involved, it is only because the path is not 
determined in advance, just as it is
initially undetermined in a classical stochastic process.

Given the failure of the sum rule I(A, B) = 0, it is clear that quantum 
probabilities cannot be interpreted in the same manner that classical ones 
are wont to be interpreted, in terms of (actual or fictitious) ensemble 
frequencies.

...

@philipthrift 

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Re: STEP 3

On Tue, Aug 13, 2019 at 12:16 PM smitra  wrote:

> On 13-08-2019 02:49, Bruce Kellett wrote:
> > On Tue, Aug 13, 2019 at 10:43 AM smitra  wrote:
> >
> >> On 13-08-2019 01:41, Bruce Kellett wrote:
> >>> On Tue, Aug 13, 2019 at 9:09 AM smitra  wrote:
> >>>
>  On 12-08-2019 08:29, Bruce Kellett wrote:
> >
> > Look at this another way. It is just an illustration of
> > complementarity. Measuring which slit the photon went through
> >> is
>  a
> > position measurement at the slits. Measuring the interference
>  pattern
> > at the screen is equivalent to a momentum measurement at the
>  slits.
> > Such measurement operators do not commute -- the measurements
> >> are
> > complementary and cannot be performed simultaneously.
> >
> 
>  It doesn't matter for orthogonality of the states whether or not
>  they
>  are measured.
> >>>
> >>> Of course it does. The slits are not orthogonal states unless
> >> they are
> >>> measured position eigenstates. If they are not measured, they are
> >>> individually superpositions of many position eigenstates
> >> (including
> >>> eigenstates that overlap both slits), so the slits themselves are
> >> no
> >>> longer orthogonal. Orthogonal states cannot interfere, that is
> >> why a
> >>> position measurement at the slits makes the interference pattern
> >> on
> >>> the screen disappear.
> >>>
> >>> The fact remains, that orthogonal states cannot interfere:
> >>>
> >>> ( + |B>) =  +  + 2 
> >>>
> >>> and the interference term  vanishes if |A> and |B> are
> >>> orthogonal. You can't get away from this basic fact about quantum
> >>> mechanics.
> >>>
> >>
> >>  is zero in the two slit experiment, if you integrate the
> >> interference term over the screen you get zero.
> >>
> >> Thing is that the interference we can observe at some position x on
> >> the
> >> screen is Re[], which for general x is nonzero despite
> >> the
> >> fact that  = 0.
> >
> > So you agree that if the overlap vanishes you do not get interference.
> > You go to some lengths to try and avoid this fact. Saying that the
> > integral over the screen vanishes is beside the point.
> >
>  is the integral over all space of  and this is zero, but
> the interference we observe at some point x on the screen is
> Re[], and that's in general nonzero even for orthogonal |A>
> and |B>.
>

I don't know what you are trying to prove, but that is not what my formula
meant. What I intended was that |A> represents the wave at the screen
coming from slit A, and |B> is the wave at the screen from slit B. There is
already and implicit dependence on the position along the screen A = A(x),
etc. So inserting a complete set of position states, \int dx |x>, which depends on x, the overlap of two such
is , and this cannot vanish if there is to be interference.
Classically, the intensity at x from slit A is |A|^2, and similarly for
slit B. The fact that quantum mechanically we add amplitudes, not
intensities, means that there is the overlap term . If this does not
vanish, [ |A> and https://groups.google.com/d/msgid/everything-list/CAFxXSLScH0pCEA1%2BdQm9%3DT6BwujQKz7vCVp2QQF-SNpcnSv0DQ%40mail.gmail.com.

Re: STEP 3

2019-08-12 Thread smitra

On 13-08-2019 02:49, Bruce Kellett wrote:

On Tue, Aug 13, 2019 at 10:43 AM smitra wrote:

On 13-08-2019 01:41, Bruce Kellett wrote:

On Tue, Aug 13, 2019 at 9:09 AM smitra wrote:

On 12-08-2019 08:29, Bruce Kellett wrote:

Look at this another way. It is just an illustration of
complementarity. Measuring which slit the photon went through

position measurement at the slits. Measuring the interference

pattern

at the screen is equivalent to a momentum measurement at the

slits.

Such measurement operators do not commute -- the measurements

are

complementary and cannot be performed simultaneously.

It doesn't matter for orthogonality of the states whether or not
they
are measured.

Of course it does. The slits are not orthogonal states unless

they are

measured position eigenstates. If they are not measured, they are
individually superpositions of many position eigenstates

(including

eigenstates that overlap both slits), so the slits themselves are

longer orthogonal. Orthogonal states cannot interfere, that is

why a

position measurement at the slits makes the interference pattern

the screen disappear.

The fact remains, that orthogonal states cannot interfere:

( + |B>) = + + 2

and the interference term vanishes if |A> and |B> are
orthogonal. You can't get away from this basic fact about quantum
mechanics.

is zero in the two slit experiment, if you integrate the
interference term over the screen you get zero.

Thing is that the interference we can observe at some position x on
the
screen is Re[], which for general x is nonzero despite
the
fact that = 0.

So you agree that if the overlap vanishes you do not get interference.
You go to some lengths to try and avoid this fact. Saying that the
integral over the screen vanishes is beside the point.

is the integral over all space of and this is zero, but
the interference we observe at some point x on the screen is
Re[], and that's in general nonzero even for orthogonal |A>
and |B>.

Saibal

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Re: STEP 3

On Tue, Aug 13, 2019 at 10:43 AM smitra  wrote:

> On 13-08-2019 01:41, Bruce Kellett wrote:
> > On Tue, Aug 13, 2019 at 9:09 AM smitra  wrote:
> >
> >> On 12-08-2019 08:29, Bruce Kellett wrote:
> >>>
> >>> Look at this another way. It is just an illustration of
> >>> complementarity. Measuring which slit the photon went through is
> >> a
> >>> position measurement at the slits. Measuring the interference
> >> pattern
> >>> at the screen is equivalent to a momentum measurement at the
> >> slits.
> >>> Such measurement operators do not commute -- the measurements are
> >>> complementary and cannot be performed simultaneously.
> >>>
> >>
> >> It doesn't matter for orthogonality of the states whether or not
> >> they
> >> are measured.
> >
> > Of course it does. The slits are not orthogonal states unless they are
> > measured position eigenstates. If they are not measured, they are
> > individually superpositions of many position eigenstates (including
> > eigenstates that overlap both slits), so the slits themselves are no
> > longer orthogonal. Orthogonal states cannot interfere, that is why a
> > position measurement at the slits makes the interference pattern on
> > the screen disappear.
> >
> > The fact remains, that orthogonal states cannot interfere:
> >
> > ( + |B>) =  +  + 2 
> >
> > and the interference term  vanishes if |A> and |B> are
> > orthogonal. You can't get away from this basic fact about quantum
> > mechanics.
> >
>
>  is zero in the two slit experiment, if you integrate the
> interference term over the screen you get zero.
>
> Thing is that the interference we can observe at some position x on the
> screen is Re[], which for general x is nonzero despite the
> fact that  = 0.
>

So you agree that if the overlap vanishes you do not get interference. You
go to some lengths to try and avoid this fact. Saying that the integral
over the screen vanishes is beside the point.

Bruce

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Re: STEP 3

2019-08-12 Thread smitra

On 13-08-2019 01:41, Bruce Kellett wrote:

On Tue, Aug 13, 2019 at 9:09 AM smitra wrote:

On 12-08-2019 08:29, Bruce Kellett wrote:

Look at this another way. It is just an illustration of
complementarity. Measuring which slit the photon went through is

position measurement at the slits. Measuring the interference

pattern

at the screen is equivalent to a momentum measurement at the

slits.

Such measurement operators do not commute -- the measurements are
complementary and cannot be performed simultaneously.

It doesn't matter for orthogonality of the states whether or not
they
are measured.

Of course it does. The slits are not orthogonal states unless they are
measured position eigenstates. If they are not measured, they are
individually superpositions of many position eigenstates (including
eigenstates that overlap both slits), so the slits themselves are no
longer orthogonal. Orthogonal states cannot interfere, that is why a
position measurement at the slits makes the interference pattern on
the screen disappear.

The fact remains, that orthogonal states cannot interfere:

( + |B>) = + + 2

and the interference term vanishes if |A> and |B> are
orthogonal. You can't get away from this basic fact about quantum
mechanics.

is zero in the two slit experiment, if you integrate the
interference term over the screen you get zero.

Thing is that the interference we can observe at some position x on the
screen is Re[], which for general x is nonzero despite the
fact that = 0.

Saibal

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Re: STEP 3

On Tue, Aug 13, 2019 at 9:09 AM smitra  wrote:

> On 12-08-2019 08:29, Bruce Kellett wrote:
> >
> > Look at this another way. It is just an illustration of
> > complementarity. Measuring which slit the photon went through is a
> > position measurement at the slits. Measuring the interference pattern
> > at the screen is equivalent to a momentum measurement at the slits.
> > Such measurement operators do not commute -- the measurements are
> > complementary and cannot be performed simultaneously.
> >
>
> It doesn't matter for orthogonality of the states whether or not they
> are measured.


Of course it does. The slits are not orthogonal states unless they are
measured position eigenstates. If they are not measured, they are
individually superpositions of many position eigenstates (including
eigenstates that overlap both slits), so the slits themselves are no longer
orthogonal. Orthogonal states cannot interfere, that is why a position
measurement at the slits makes the interference pattern on the screen
disappear.

The fact remains, that orthogonal states cannot interfere:

( + |B>) =  +  + 2 

and the interference term  vanishes if |A> and |B> are orthogonal. You
can't get away from this basic fact about quantum mechanics.

Bruce

It's of course true that if we measure the position at the
> screen we're measuring something else than the which way information and
> the position eigenstates are not the original  superposition. But it
> remain a fact that the state of which we're measuring the position
> operator on the screen is a superposition of two orthogonal states and
> we're then seeing an interference effect defined as the difference in
> the number of dots in the screen of the actual counts and the sum of
> what would be seen if slit 1 were cloes and slit 2 open and vice versa.
>
> Saibal
>

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Re: STEP 3

2019-08-12 Thread smitra

On 12-08-2019 08:29, Bruce Kellett wrote:

On Mon, Aug 12, 2019 at 4:09 PM Bruce Kellett
wrote:

On Mon, Aug 12, 2019 at 3:56 PM smitra wrote:

On 12-08-2019 04:06, Bruce Kellett wrote:

On Mon, Aug 12, 2019 at 3:48 AM Bruno Marchal

wrote:

In the sense you mention I am OK, but we have a slight

vocabulary

problem. Not important, if you agree that measurement are
self-entanglement, so that the superposition of the orthogonal

state

SlitA and SlitB, say some oblique (with sqrt(2) = 1) SlitA +

SlitB is

inherited by the observer “looking” which is which.

If you do not measure which slit the photon went through, then

the

superposition of slits is not broken by decoherence. But the
interference at the screen depends only on things like the

wavelength

of the light, the separation of the slits, and the distance

between

the slits and the screen. If you refine this calculation by

taking the

finite width of the slits into account, you convolute the

interference

pattern with the diffraction pattern due to finite slit width.

This is

an elementary calculation in physical optics, not even

requiring

quantum mechanics. But the waves at the screen cannot be

orthogonal,

or else they would not interfere.

The states at the screen are orthogonal because they were at the
start
and inner product is conserved under the unitary time evolution.

The sits are orthogonal if you measure which slit the photon went
through, in which case the interference pattern disappears, as
required by orthogonality. But they are not orthogonal if they are
not measured, else there would be no interference. Orthogonal states
cannot interfere.

Look at this another way. It is just an illustration of
complementarity. Measuring which slit the photon went through is a
position measurement at the slits. Measuring the interference pattern
at the screen is equivalent to a momentum measurement at the slits.
Such measurement operators do not commute -- the measurements are
complementary and cannot be performed simultaneously.

It doesn't matter for orthogonality of the states whether or not they
are measured. It's of course true that if we measure the position at the
screen we're measuring something else than the which way information and
the position eigenstates are not the original superposition. But it
remain a fact that the state of which we're measuring the position
operator on the screen is a superposition of two orthogonal states and
we're then seeing an interference effect defined as the difference in
the number of dots in the screen of the actual counts and the sum of
what would be seen if slit 1 were cloes and slit 2 open and vice versa.

Saibal

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Re: STEP 3

On Mon, Aug 12, 2019 at 7:36 PM Bruno Marchal  wrote:

> On 12 Aug 2019, at 04:06, Bruce Kellett  wrote:
>
>
> If you do not measure which slit the photon went through, then the
> superposition of slits is not broken by decoherence.
>
> Decoherence break things only if there is a collapse.
>

That is simply incorrect. I refer you again to Zurek, who works in a
basically Everettian framework, but he stresses the importance of
environmental induced superselection (einselection) in producing the
preferred pointer basis. This then breaks things, in the sense that no
other basis is stable against decoherence, and other sets of basis vectors
rapidly (in times of the order of femtoseconds) collapse on to the
preferred pointer states. This is the basis of the emergence of the
classical world from the quantum substrate. And this occurs in Everett's
relative state approach just as much as in a Copenhagen-like collapse
models.

Without collapse, even if I measure which slit the photon went through, the
> two terms of the superposition continue to exist, describing me seeing both
> outcomes, and both me feel like if there has been a collapse,
>

No, that is not really true. Complementarity plays a role. Measurement of
which slit the photon went through is incompatible with the effective
momentum measurement that gives interference at the screen. That is
essentially why the slit measurement destroys the possibility of
interference. There is an incompatibility between the measurements --
basically related to the non-commutation of position and momentum operators
and the Heisenberg uncertainty principle.

and that decoherence is physical real, but that illusion is explained by
> the formalism, in a manner similar to the WM duplication: it is just first
> person indeterminacy, not in a self-duplication, but in a self entanglement.
>

Not really true, either. Decoherence, the formation of a stable pointer
basis, and the like, are all essential for the emergence of the classical
from the quantum, and the formation of objective physical states. This is
the operation of Zurek's Quantum Darwinism -- the environment acting as
witness. So this is third person objective physics -- it is not first
person indeterminacy.

Bruce

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Re: STEP 3