Re: “Could a Quantum Computer Have Subjective Experience?”

[Bruce Kellett]

On 30/06/2017 1:15 pm, Russell Standish wrote:

On Fri, Jun 30, 2017 at 11:26:50AM +1000, Bruce Kellett wrote:

On 29/06/2017 5:36 pm, Russell Standish wrote:

I haven't established this, because it is not needed to make contact
with the regular set of axioms assumed in quantum theory. But to the
extent it can be shown within regular QM, it will be shown within my theory.

But it is not a derived result in standard QM -- it is assumed as
part of the postulates. The linear spaces for each quantum operator
are disjoint -- spin operators act in a different space from that
for position operators -- so the combined Hilbert space must be a
tensor product of distinct Hilbert spaces. You would have to prove
this from your starting point if you want to make contact with
quantum mechanics.


I have two QM textbooks here: a classic one by Leonard Schiff, who
eschews postulates altogether, and my undergraduate one, by Ramamurti
Shankar. My book follows the 4 postulates given in Shankar's book,
which doesn't include the one you mention above, which I can sumarise
as:

1. Hilbert space
2. Schroedinger equation
3. Born rule
4. Correspondence Principle

For good measure, I consulted that indubitable font of wisdom:
Wikipedia. On Wikipedia's page, the Hilbert space and Born rule are
there, but not the Schrodinger equation one (unless you consider that to be a
specialisation of the Wigner theorem one mentioned there). Finally,
there is the composition postulate you mention.


Skimming through a number of QM texts, most seem to take a historical 
and/or experimental approach. Dirac emphasizes superposition, and von 
Neumann's primary focus is on the formalism of Hilbert space and operators.


The Wikipedia page on the 'Mathematical formulation of QM' lists postulates:
 1. Hilbert space
 2. Hilbert space for composite systems = tensor product of the 
component state spaces

 3. Physical symmetries act unitarily or antunitarily (Wigner's theorem)
 4. Observable are Hermitian matrices on the Hilbert space (operators)
 5. Possible outcomes of measurement are eigenvalues of the 
corresponding operator.


There are clearly various ways of formulating these postulates, but 
linearity (Hilbert space over a complex field) is clearly central to the 
formal development. Linearity, if justified at all, is justified as by 
Dirac, in terms of experimental results such as superposition and 
interference. For Dirac, the state vector is prior to Hilbert space, but 
these are clearly closely related.


The composition principle is essential if one moves beyond measurements 
of just a single variable.


Bruce



Note: my treatment just provides justification for the first 3 of
Shankar's postulates - the correspondence principle is deliberately
left out, with a note that Stenger shows how to get the correspondence
principle from Noether's theorem. In fact, both 2 & 4 can be traced to
Noether's theorem, and can perhaps be rolled into the Wigner theorem
postulate described on Wikipedia.

So maybe the composition postulate can derived from my Nothing
approach, maybe not. Nice open problem. I can't see that it
invalidates my approach, though.



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Re: “Could a Quantum Computer Have Subjective Experience?”

[Russell Standish]

On Fri, Jun 30, 2017 at 11:26:50AM +1000, Bruce Kellett wrote:
> On 29/06/2017 5:36 pm, Russell Standish wrote:
> >On Thu, Jun 29, 2017 at 03:19:40PM +1000, Bruce Kellett wrote:
> >>On 28/06/2017 2:26 pm, Russell Standish wrote:
> >>>On Tue, Jun 27, 2017 at 05:09:49PM +1000, Bruce Kellett wrote:
> On 27/06/2017 10:21 am, Russell Standish wrote:
> >No, you are just dealing with a function from whatever set the ψ and ψ_α
> >are drawn from to that same set. There's never been an assumption that
> >ψ are numbers or functions, and initialy not even vectors, as that
> >later follows by derivation.
> psi(t) is an ensemble, psi_a is an outcome.
> >>>ψ_α is still an ensemble, just a sub-ensemble of the original. Assume
> >>>α is the result (A has the value 2.1 ± 0.15). Then all universes where
> >>>A is greater than 1.95 and 2.25 will be in the ensemble ψ_α.
> >>That might be what you meant, but that is not what the average
> >>reader (such as I am) is going to take from the text. You say the
> >>projections divides the observer moment (set) into a discrete set of
> >>outcomes (indexed by a). You then wish to calculate the probability
> >>that outcome a is observed. Now observers observe one outcome --
> >>even if there is some associated measurement error, there is still
> >>one outcome -- one observer does not see 2.0, 2.1, and 2.2. At
> >>least, that would be a very strange way of talking about an
> >>observation. Other observer might see that, and one observer might
> >>see a range of results on repeated measurements, but that is not
> >>what you appeared to be talking about.
> >>
> >Any measurement on a continuum will have an uncertainty. A measurement
> >of 2.0 ± 0.1 will be compatible with an infinite variety of observer
> >moments, observing the value in the range 1.9 to 2.1.
> 
> I think, in the light of your response to the dichotomous
> measurement below, that this is just a red herring. If the outcome
> psi_a is an ensemble, it has nothing to do with measurement errors.
> Besides, one does not measure errors with a single measurement. A
> measurement of a continuous variable may be subject to error, but a
> single measurement gives a single value: errors can be estimated by
> doing statistics over repeated independent measurements of the same
> variable on similar systems, but that does not seem to be what your
> projections operators are about.
> 

All I'm trying to say here is that an infinitely precise measurement
is a fiction. Before and after are ensembles of infinite length
strings, and the set of strings compatible with a finitely precise
measurement will still be infinite, and have non-zero measure.

> >The average reader should remember that, at least if they're scientist.
> >
> The projection produces
> the outcome from the observer moment psi. It maps the ensemble to
> one member of that ensemble.
> >>>No. See above. It always maps to an ensemble with an infinite number
> >>>of members.
> >>Not necessarily. A photon polarization measurement is dichotomous --
> >>you see a photon downstream of the polarizer, or you do not. No
> >>uncertainty involved.
> >>
> >A discrete partitioning of the "Nothing" will still involve infinite
> >sized ensembles. Just in this case, there is not way of disciminating
> >them - with this observable at least.
> 
> I think this is revealing. What you are really saying is that a
> measurement does not project out subsets from the observer moment,
> it produces a complete set of new observer moments, \P_{a} maps the
> ensemble of observer moments onto itself, with one new observer
> moment for each possible measurement outcome a.

Not quite. \P_{a} maps the ensemble of observer moments to that
compatible with the observation a. \P_S does what you say above.

> That could have been
> made clearer. But it also needs justification because it is not
> implicit in the notion of an observation, unless you assume QM and
> MWI from the outset.
> 

The justification is summed up in what I call the projection postulate,
that observers will observe a unique outcome out of a disjoint set of
outcomes, with certain probability measure. There is a paragraph
writeup on page 119, but essentially the justification is in terms of
Lewontin's second principle of evolution (aka Selection).

> 
> That is where I must object. 3p, or 0p as I would prefer to refer to
> the bird view, is not an observer moment because there is no such
> observer.
> 
> If we take a simple example, where observer moments are, say, the
> sets {1,2,3}, {4,5,6}, and {7,8,9}, then the set of all observer
> moments, a set of sets, each being one of these sets, does not
> contain the union of the first two observer moments, which is the
> set {1,2,3,4,5,6}. So the sum or union of two observer moments is
> not necessarily another observer moment, in fact, I doubt that it
> ever could be; 3p is not an observer moment.
> 

Yes - this is a fair point. \P_{a}ψ+\P_{b}ψ obviously refers 

Re: “Could a Quantum Computer Have Subjective Experience?”

[Bruce Kellett]

On 29/06/2017 5:36 pm, Russell Standish wrote:

On Thu, Jun 29, 2017 at 03:19:40PM +1000, Bruce Kellett wrote:

On 28/06/2017 2:26 pm, Russell Standish wrote:

On Tue, Jun 27, 2017 at 05:09:49PM +1000, Bruce Kellett wrote:

On 27/06/2017 10:21 am, Russell Standish wrote:

No, you are just dealing with a function from whatever set the ψ and ψ_α
are drawn from to that same set. There's never been an assumption that
ψ are numbers or functions, and initialy not even vectors, as that
later follows by derivation.

psi(t) is an ensemble, psi_a is an outcome.

ψ_α is still an ensemble, just a sub-ensemble of the original. Assume
α is the result (A has the value 2.1 ± 0.15). Then all universes where
A is greater than 1.95 and 2.25 will be in the ensemble ψ_α.

That might be what you meant, but that is not what the average
reader (such as I am) is going to take from the text. You say the
projections divides the observer moment (set) into a discrete set of
outcomes (indexed by a). You then wish to calculate the probability
that outcome a is observed. Now observers observe one outcome --
even if there is some associated measurement error, there is still
one outcome -- one observer does not see 2.0, 2.1, and 2.2. At
least, that would be a very strange way of talking about an
observation. Other observer might see that, and one observer might
see a range of results on repeated measurements, but that is not
what you appeared to be talking about.


Any measurement on a continuum will have an uncertainty. A measurement
of 2.0 ± 0.1 will be compatible with an infinite variety of observer
moments, observing the value in the range 1.9 to 2.1.


I think, in the light of your response to the dichotomous measurement 
below, that this is just a red herring. If the outcome psi_a is an 
ensemble, it has nothing to do with measurement errors. Besides, one 
does not measure errors with a single measurement. A measurement of a 
continuous variable may be subject to error, but a single measurement 
gives a single value: errors can be estimated by doing statistics over 
repeated independent measurements of the same variable on similar 
systems, but that does not seem to be what your projections operators 
are about.



The average reader should remember that, at least if they're scientist.


The projection produces
the outcome from the observer moment psi. It maps the ensemble to
one member of that ensemble.

No. See above. It always maps to an ensemble with an infinite number
of members.

Not necessarily. A photon polarization measurement is dichotomous --
you see a photon downstream of the polarizer, or you do not. No
uncertainty involved.


A discrete partitioning of the "Nothing" will still involve infinite
sized ensembles. Just in this case, there is not way of disciminating
them - with this observable at least.


I think this is revealing. What you are really saying is that a 
measurement does not project out subsets from the observer moment, it 
produces a complete set of new observer moments, \P_{a} maps the 
ensemble of observer moments onto itself, with one new observer moment 
for each possible measurement outcome a. That could have been made 
clearer. But it also needs justification because it is not implicit in 
the notion of an observation, unless you assume QM and MWI from the outset.





That's right. If a linear combination of observer moments is also and
observer moment, then the set of all observer moments is a vector
space. This is linear algebra 101.

Again, that is not what your text says. You say that linearity comes
from the additive property of measure, and that is really what (D.7)
appears to be about. Except that it is not the additivity of measure
that is doing the work there, it is the additivity of probabilities
for disjoint observations (when the probability measure is
normalized to unity).

I think you have to do more that just asserting that a linear
combination of observer moments is also an observer moment. The
notion of an observer moment has become opaque. An observer moment
is the set of possibilities consistent with what is known at that
point in time. So it is complete in itself -- how can you add two
observer moments? You clearly cannot add them for a single observer,
because adding two moments in time is not a defined operation -- it
would not be an observer moment since no observer observes two
moments in time simultaneously. Other observers at that time? Again,
if you add observer moments for different observers, you have no
guarantee that there is another observer who has just this
combination of possibilities consistent with what they know at that
point in time. That would have to be proved, rather than just
asserted. In fact, ISTM that such a result would require that every
observer knows everything at every time, and that everything that is
ever possible is part of the set of things consistent with what is
known by that observer at that point in time -- and the notion of
observer moments 

Re: is there any acceptable definition of free-will?

[John Clark]

On Thu, Jun 29, 2017 at 12:48 PM, Bruno Marchal  wrote:

​
>> ​>> ​
>> I can think of 2 definitions of "free will" that are not gibberish,
>> although neither is useful:​
>
>
>> 1) It is the inability to know what you will decide to do next before you
>> decide to do it.
>
>
> ​> ​
> Not to bad.
>

Thanks, of course a
​ ​
cuckoo clock
​ has free will too, but never mind.​


> ​> ​
> Useful?
>

​No, but at least it's not gibberish. ​



> ​> ​
> If not try a perhaps better definition.
>

​I seen lots of definitions of free will but all the others are either
circular, ​self contradictory, or so vague it could mean anything. If
philosophers never typed the letters "free will" again their field would be
much improved.

​>>​
>> 2) It is a sequence of letters that lots of people on the internet like
>> to type.
>
>
> ​> ​
> Unfortunately, there are many sequence of letters that people like to
> type. So this definition is much to large,
>

​I don't see why you say it's too large, one random sequence of letters is
much like another,
there is nothing special about the sequence "free will"​

​that sets it apart from all the other gibberish.


> ​> ​
> and besides free-will, if it exist or not, like the unicorn,
>

​
Free will is not like a unicorn. A unicorn either exists or it doesn't, but
that's not the case with "fluxnotwadelskrunk". Saying fluxnotwadelskrunk
​ ​
is untrue makes no more sense than saying
fluxnotwadelskrunk
​ ​
is true, and it's the same with "free will", a phrase that is so bad it's
not even wrong.
​

John K Clark​

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What lead to free-will denial?

[John Clark]

On Thu, Jun 29, 2017 at 10:58 AM, Bruno Marchal  wrote:


​> ​
> you just said to Adrian Chira: "Assuming there is a largest prime number
> leads to a logical contradiction, but assuming free
> will exists is like assuming Klogknee exists."


​Yes, gibberish in ​
gibberish
​ out.


> But Robinson Arithmetic, a weak but already sigma_1 complete theory (and
> thus Turing universal) is consistent with "there is a
>  biggest prime number".


​If Robinson says there is a largest prime number then ​Robinson is an ass.


​> ​
> You were probably assuming Peano Arithmetic (which is Robinson arithmetic
> + the induction axiom),


​No I was not making that assumption, and neither was Euclid when he showed
2500 years before Peano or Robinson were born that there is no largest
prime number.

​> ​
>  the here-and-now 1p experience


​To hell with peepee!!​

​

John K Clark​

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Re: is there any acceptable definition of free-will?

[Bruno Marchal]

On 28 Jun 2017, at 21:30, Stathis Papaioannou wrote:



On Wed, 28 Jun 2017 at 3:29 pm, Adrian Chira   
wrote:
I suggest there is and that we don’t need to throw the notion of  
free will in the garbage bin of meaninglessness.


https://adrianmchira.wordpress.com/2017/06/28/is-there-any-acceptable-definition-of-free-will/

It's OK not to have an explicit definition, but even the intuitive  
notion of free will leads either to triviality (free will is when I  
do what I want to do)



Is it really trivial? Free-will and consciousness are "obvious" from  
the 1p pov, yet irreductible to 3p, except with the poor lawyer  
defense "my client just obeyed to the physical laws".


Today, if you build a machine, and if the machine kills a human, you  
will be judged responsible, but for how long?


Higher order structure exist, and indeed even electron and proton will  
themselves be "reduced" has sort of collective hallucination (higher  
order structures)., without changing anything in physics, but a lot in  
metaphysics, theology and theotechnologies, etc.


It is not because we can represent something in something else that  
the "something" will disappears.


"Free" is easy to define, even if hard to obtain, the mystery is more  
in the will, even if we can intellectually understand that it has to  
be mysterious so as to proceed and survive.






or nonsense (free will is not compatible with my behaviour being  
either determined or random).



Or something in between. Something we can be confronted with in our  
everyday "higher order" life. We are not "just" colony of bacterias  
and amoebas.


Bruno


We are not human beings having spiritual experiences from times to  
times. We are spiritual beings having human experiences from times to  
times. (Chardin).





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Re: is there any acceptable definition of free-will?

[Bruno Marchal]

On 28 Jun 2017, at 16:12, John Clark wrote:

​I can think of 2 definitions of "free will" that are not  
gibberish, although neither is useful:​


1) It is the inability to know what you will decide to do next  
before you decide to do it.



Not to bad. Useful? If not try a perhaps better definition.






2) It is a sequence of letters that lots of people on the internet  
like to type.



Unfortunately, there are many sequence of letters that people like to  
type. So this definition is much to large, and besides free-will, if  
it exist or not, like the unicorn, is anything but a sequence of  
letter. The sequence of letters is only a pointer.


Once I saw the definition, for unicorn,  "horse with a frontal horn".  
In that case it exists, even if rare. Usually the term "mythical" is  
added. But according to google, a unicorn is also a start-up company  
valued at more than a billion dollars, typically in the software or  
technology sector. Which I guess settles the problem of its existence :)



Google:

unicorn
ˈjuːnɪkɔːn/
noun
1.
a mythical animal typically represented as a horse with a single  
straight horn projecting from its forehead.


2.
a start-up company valued at more than a billion dollars, typically in  
the software or technology sector.







 John K Clark


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Re: “Could a Quantum Computer Have Subjective Experience?”

[Bruno Marchal]

On 26 Jun 2017, at 19:23, John Clark wrote:

On Mon, Jun 26, 2017 at 1:57 AM, Russell Standish  wrote:


​> ​I've started with a different set of metaphysical  
assumptions,​ ​namely that we live in a Multiverse,


​Do you assume the number of universes are ​denumerable​? ​

​> ​and that observer moments are​ ​drawn from a much more  
general measure than classical probability​ ​theory allows.


​Infinite sets can cause problems with probability even if you can  
count the elements, and if you can't it certainly doesn't help.



On the infinite enumerable set, you might need to relinquish the  
additivity axiom, or to use local relations (like with the prime  
number theory).


But on the non enumerable sets, (Like R, C) probabilities and measure  
have nice theories, like the theories of Riemann, Lebesgue, or the  
Haar measure on the Lie groups.







If you stab the number line at random with an infinitely sharp  
needle your chances of hitting a rational number, or even a  
computable number, are zero even though there are a infinite number  
of them.   ​


That is why we define the measure space by forbidding the infinite  
intersection of interval or open sets, which can lead to points. We  
allow infinite unions. In the iterated duplication, or any Bernouilli  
experience, the probabilities of "successes" are given by the binomial  
coefficients, but with big numbers the continuous Gaussian e^(- x2),  
conveniently renormalized, simplifies the life of the statistician.


From a logician perspective, and assuming computationalism, the  
infinities are sort of oversimplifications made by the finite things/ 
minds trying to understand the finite things/minds. Anyway, with  
Mechanism, it is easier to classified them as epistemological, and  
that is why the ontology is given by Robinson Arithmetic, and the  
phenomenology by what will concern the "observer" PA (Peano  
Arithmetic) which is emulated by Robinson Arithmetic (and thus  
emulated in all models, even those with a biggest prime!).


Bruno



 John K Clark


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Belief in free will predicts criminal punishment support

[Philip Benjamin]

Philip Benjamin  Sent: Thursday, June 29, 2017 3:35 PM
PhysicalSciences PhysicalSciences Subject: PhysOrg [News]: Belief in free will 
predicts criminal punishment support

[Philip Benjamin]

This seems to be a flawed "academedia" type paper (defined below). Free will is 
not a universally held belief. It is practically confined to the Christian 
population [mostly Arminianism named after Prof. James Arminius (1560-1609), 
Theologian and leader of the opposition of Calvinism  in Amsterdam and Leiden]

http://www.adherents.com/images/rel_pie.gif

[http://www.adherents.com/images/rel_pie.gif]

 According to Pew "research the religious profile of the world is rapidly 
changing, driven primarily by differences in fertility rates and the size of 
youth populations among the world’s major religions, as well as by people 
switching faiths. Over the next four decades, Christians will remain the 
largest religious group, but Islam will grow faster than any other major 
religion. If current trends continue, by 2050 …"

2015 Statistics in billions: 
http://www.pewforum.org/2015/04/02/religious-projections-2010-2050/

Christians: 2.17.  Muslims: 1.69.  Hindus: 1.03 . Buddhists:  0.49.  Jews: 0.01 
Unaffiliated: 1.13. Folk religions: o.4. Others: 0.06.
The question of free will is primarily a Christin theological issue. 
Practically, there is no place for "free will" in other religions including 
atheism (an irrational belief system), where nothing happens unless it is 
predetermined by a deity or fate/kismet. Belief in free will is essentially 
limited to the Christian population. So, "belief in free will predicts criminal 
punishment support" seems to be contrary to the fact that Christian populations 
are increasingly becoming "criminal friendly". Death penalty is practically 
banned in most  predominantly Christian countries. "Criminal illegal 
immigrants" have sanctuary cities almost everywhere in the USA which is 
supposed to be predominantly "Christian". "Abortions" and homosexuality used to 
be criminal acts, but no more for most of the mainline Christian populations 
who are very strong on "free will" and "human rights". (The conservative 
Calvinist and Lutheran contentions for "bound will" refer only to the issues 
related to "Rebirth" or "Regeneration" through the Word only and by the Spirit 
only, but not to the routines of living).
Evidentialist
Philip Benjamin, DLT
(Dark/Light Twins)

Note: Academedia (acade-media): The monstrous double headed hybrid of a small 
minority of all academics including seminarians and a large majority of all 
media including the Hollywood, with no-question-asked Marxist-like 
authoritarianism as their modus operandi. Based on the works of Rabbi Daniel 
Lapin, Ben Stein, Victor Mordecai, ex-Marxist David Horwitz

Upon decoupling, the unenergized (unregenerated), non-entropic bio dark-matter 
bodies co-created at the moment of conception will be lost in their abodes of 
the dark-matter realms (black holes), by their own willful choice. Adapted from 
"Ten Implications of Bio Dark-Matter Chemistry" and "Spiritual Body or Physical 
Spirit" by Philip Benjamin PhD MSc MA


Nathan D. Martin, Davide Rigoni, and Kathleen D. Vohs  . June 26, 2017 Journal 
PNAS (2017) 
https://phys.org/news/2017-06-belief-free-criminal-unethical-actions.html. 
www.pnas.org/cgi/doi/10.1073/pnas.1702119114
   Belief in free will predicts criminal punishment support, disapproval of 
unethical actions .they used data from the World Values Survey to analyze 
more than 65,000 residents in 46 countries.



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Re: What lead to free-will denial?

[Bruno Marchal]

On 28 Jun 2017, at 21:59, John Clark wrote:

On Wed, Jun 28, 2017 at 1:59 PM, Adrian Chira   
wrote:


​> ​You assume that it can't be assumed but you bring no support  
for your universal claim.​ How do you know that it can't be done?


​I know because there is nothing there to assume or not to assume.  
With free will there is no there there. ​



Only because you allude to the incompatibilist theory of free will,  
which is indeed arguably senseless.


But formally, you can retrieve 95% of that theory by replacing  
"indeterminacy" by self-indeterminacy. Then theoretical computer  
science and mathematical logic provides models of free-will and  
provides reason for its existence and role. It is related to  
consciousness and the 1p (the person from the person pov).








 ​>​What happens if I do it anyway?

​What will happen if you assume "free will" is exactly precisely  
the same thing that will happen if you assume "phlobnegob".  ​



No, if you assume free-will you ease the work of the judge when  
invoking its conviction in hard cases. If you assume "phlobnegob", you  
are at risk, given that up to now it might mean anything. It can be  
used as a variable or as a constant in a logical theory.






​> ​Then what's to stop me from assuming it?

​Nothing prevents you except for the embarrassment of speaking  
gibberish.  ​



Or maybe you don't understand what we are talking about. We have  
already discussed this, and at some point you have proposed a  
definition, adding it is was not interesting. Since then, we always  
talk on something around your definition, and insisting that you don't  
find it interesting does not add anything interesting to it.


Free-will is (arguably) a higher order logical notion, in relation  
with other notion used everyday, like responsibility, jurisprudence,  
etc. Basically large part of the human science, and larger and larger  
part of mathematical logic and computer science and "artificial"  
intelligence. ("artificial" is an indexical, and is artificial, it is  
a distinction that "we" do).


It is important, I think. There are technic to demolish free-will, and  
make you into slaves, if not into bomb.


Call it "will", if you prefer. Free-will is basically "will", and is  
more a contextual attitude than a mean. Some people can abandon it for  
a penny, other people keeps it under torture.


Bruno





​John K Clark​









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Re: What lead to free-will denial?

[David Nyman]

On 29 Jun 2017 15:58, "Bruno Marchal"  wrote:


On 28 Jun 2017, at 16:03, John Clark wrote:

snip

>> Unicorns don't exist,
>
>

>In which theory?


Oh for christ's sake! This is really getting silly, now existence depends
on theories.



Of course not.


But *asserting* an existence depends on a theory or belief of some subject
doing the assertion.

 For example, you just said to Adrian Chira: "Assuming there is a largest
prime number leads to a logical contradiction, but assuming free will
exists is like assuming Klogknee exists."

But Robinson Arithmetic, a weak but already sigma_1 complete theory (and
thus Turing universal) is consistent with "there is a biggest prime
number". You can add that axiom to RA without getting any contradiction.
You were probably assuming Peano Arithmetic (which is Robinson arithmetic +
the induction axiom), or you were assuming the standard model of arithmetic
(which is a much stronger assumption than the axioms of PA).

Except for the here-and-now 1p experience (consciousness), all existence
assertions presupposes some belief, that is, an unconscious or a conscious
theory.

And the unicorn certainly exists. It is the national animal of Scotland!


On the contrary, as a proud Scot, I can assure you that the national animal
of Scotland is the haggis.

David


Bruno







John K Clark


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Re: What lead to free-will denial?

[Bruno Marchal]

On 28 Jun 2017, at 16:03, John Clark wrote:


snip

>> Unicorns don't exist,

>In which theory?

Oh for christ's sake! This is really getting silly, now existence  
depends on theories.



Of course not.


But *asserting* an existence depends on a theory or belief of some  
subject doing the assertion.


 For example, you just said to Adrian Chira: "Assuming there is a  
largest prime number leads to a logical contradiction, but assuming  
free will exists is like assuming Klogknee exists."


But Robinson Arithmetic, a weak but already sigma_1 complete theory  
(and thus Turing universal) is consistent with "there is a biggest  
prime number". You can add that axiom to RA without getting any  
contradiction.
You were probably assuming Peano Arithmetic (which is Robinson  
arithmetic + the induction axiom), or you were assuming the standard  
model of arithmetic (which is a much stronger assumption than the  
axioms of PA).


Except for the here-and-now 1p experience (consciousness), all  
existence assertions presupposes some belief, that is, an unconscious  
or a conscious theory.


And the unicorn certainly exists. It is the national animal of Scotland!

Bruno








John K Clark


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Re: “Could a Quantum Computer Have Subjective Experience?”

[Russell Standish]

On Thu, Jun 29, 2017 at 03:19:40PM +1000, Bruce Kellett wrote:
> On 28/06/2017 2:26 pm, Russell Standish wrote:
> >On Tue, Jun 27, 2017 at 05:09:49PM +1000, Bruce Kellett wrote:
> >>On 27/06/2017 10:21 am, Russell Standish wrote:
> >>>No, you are just dealing with a function from whatever set the ψ and ψ_α
> >>>are drawn from to that same set. There's never been an assumption that
> >>>ψ are numbers or functions, and initialy not even vectors, as that
> >>>later follows by derivation.
> >>psi(t) is an ensemble, psi_a is an outcome.
> >ψ_α is still an ensemble, just a sub-ensemble of the original. Assume
> >α is the result (A has the value 2.1 ± 0.15). Then all universes where
> >A is greater than 1.95 and 2.25 will be in the ensemble ψ_α.
> 
> That might be what you meant, but that is not what the average
> reader (such as I am) is going to take from the text. You say the
> projections divides the observer moment (set) into a discrete set of
> outcomes (indexed by a). You then wish to calculate the probability
> that outcome a is observed. Now observers observe one outcome --
> even if there is some associated measurement error, there is still
> one outcome -- one observer does not see 2.0, 2.1, and 2.2. At
> least, that would be a very strange way of talking about an
> observation. Other observer might see that, and one observer might
> see a range of results on repeated measurements, but that is not
> what you appeared to be talking about.
> 

Any measurement on a continuum will have an uncertainty. A measurement
of 2.0 ± 0.1 will be compatible with an infinite variety of observer
moments, observing the value in the range 1.9 to 2.1.

The average reader should remember that, at least if they're scientist.

> 
> >>The projection produces
> >>the outcome from the observer moment psi. It maps the ensemble to
> >>one member of that ensemble.
> >No. See above. It always maps to an ensemble with an infinite number
> >of members.
> 
> Not necessarily. A photon polarization measurement is dichotomous --
> you see a photon downstream of the polarizer, or you do not. No
> uncertainty involved.
> 

A discrete partitioning of the "Nothing" will still involve infinite
sized ensembles. Just in this case, there is not way of disciminating
them - with this observable at least.

> >>  Certainly, psi is not a number or a
> >>function, it is a set of possible outcomes: the psi_a are single
> >>outcomes, be they numbers, functions or vectors, but the are not
> >>just further ensembles.
> >>
> Thus, for the sum to make sense you must assume linearity.
> >>>If you are objecting that the use of the symbol '+' implies linearity
> >>>where no such thing is assumed, then feel free to replace it with the
> >>>symbol of your choice. Then once linearity is established, feel free
> >>>to replace it back again to + so that the formulae following D.8 have
> >>>a more usual notation. Fine - that is a presentational quibble. My
> >>>taste is that it is unnecessarily cumbersome, but if you find it helps
> >>>prevent confusion in your mind, please do so.
> >>>
>   Now
> linearity is at the bottom of most distinctive quantum behaviour
> such as superposition, interference, and entanglement. It is not
> surprising, therefore, that if you assume linearity at the start,
> you can get QM with minimal further effort.
> 
> >>>Except that I don't assume linearity from the outset.
> >>There seems to be some confusion between outcomes of observations
> >>and sets of possible outcomes. The \P_A*psi is actually defined as a
> >>superposition in (D.2), ad you then seek to determine the
> >>probability of this superposition? You define the probability of a
> >>set of outcomes by P_psi(\P_A*psi), which is P_psi(\Sigma psi_a). I
> >>find it hard to interpret what this might mean -- the probability of
> >>a superposition of measurement outcomes (with equal weights, what is
> >>more)?
> >>
> >The weights aren't equal. They're denoted P(ψ_α).
> 
> Again, that is not what the text says; (D.2) is a sum over outcomes
> with equal weights.

Yes - at D.2, there is no possibility of weighting - we're just
aggregating disjoint outcomes.

> 
> I find the notation confusing again. You have A contained in S, with
> probability P_psi(\P_A*psi). A is original defined as an observable,
> which divides the observer moment into a set of discrete outcomes.
> But S is the set of possible outcomes: a is a member of S, so A
> contained in S seems to be a different A -- the operator is not a
> subset of the outcomes. To make something out of this, I took the
> latter use of A to be the set of possible outcomes a -- not every
> operator has the same set of possible outcomes.
> 

Yes - you are right, it is confusing. A is used for two unrelated
concepts. Initially it is the observable in paragraph 1, then it is
related to the notion of an event coming in from the Kolmogorov
axioms. After the first paragraph, A never refers to an observable
again - so 

Re: “Could a Quantum Computer Have Subjective Experience?”

[spudboy100 via Everything List]

With the question, yes, I like quantum stuff as well, but a better question, is 
can we make a computer of any type complex enough? It is complexity that 
determines conscious, self-referencing awareness. Can we make a computer 
possessing subjective experience out of carbon + water? There's been some 
researchers at Stanford who have putzed around with DNA based computing, so 
maybe because of this, and because we, ourselves,  have subjective experiences, 
I'd say yes. 



-Original Message-
From: Bruce Kellett 
To: everything-list 
Sent: Thu, Jun 29, 2017 1:19 am
Subject: Re: “Could a Quantum Computer Have Subjective Experience?”

On 28/06/2017 2:26 pm, Russell Standish wrote:
> On Tue, Jun 27, 2017 at 05:09:49PM +1000, Bruce Kellett wrote:
>> On 27/06/2017 10:21 am, Russell Standish wrote:
>>> No, you are just dealing with a function from whatever set the ψ and ψ_α
>>> are drawn from to that same set. There's never been an assumption that
>>> ψ are numbers or functions, and initialy not even vectors, as that
>>> later follows by derivation.
>> psi(t) is an ensemble, psi_a is an outcome.
> ψ_α is still an ensemble, just a sub-ensemble of the original. Assume
> α is the result (A has the value 2.1 ± 0.15). Then all universes where
> A is greater than 1.95 and 2.25 will be in the ensemble ψ_α.

That might be what you meant, but that is not what the average reader 
(such as I am) is going to take from the text. You say the projections 
divides the observer moment (set) into a discrete set of outcomes 
(indexed by a). You then wish to calculate the probability that outcome 
a is observed. Now observers observe one outcome -- even if there is 
some associated measurement error, there is still one outcome -- one 
observer does not see 2.0, 2.1, and 2.2. At least, that would be a very 
strange way of talking about an observation. Other observer might see 
that, and one observer might see a range of results on repeated 
measurements, but that is not what you appeared to be talking about.


>> The projection produces
>> the outcome from the observer moment psi. It maps the ensemble to
>> one member of that ensemble.
> No. See above. It always maps to an ensemble with an infinite number
> of members.

Not necessarily. A photon polarization measurement is dichotomous -- you 
see a photon downstream of the polarizer, or you do not. No uncertainty 
involved.

>>   Certainly, psi is not a number or a
>> function, it is a set of possible outcomes: the psi_a are single
>> outcomes, be they numbers, functions or vectors, but the are not
>> just further ensembles.
>>
 Thus, for the sum to make sense you must assume linearity.
>>> If you are objecting that the use of the symbol '+' implies linearity
>>> where no such thing is assumed, then feel free to replace it with the
>>> symbol of your choice. Then once linearity is established, feel free
>>> to replace it back again to + so that the formulae following D.8 have
>>> a more usual notation. Fine - that is a presentational quibble. My
>>> taste is that it is unnecessarily cumbersome, but if you find it helps
>>> prevent confusion in your mind, please do so.
>>>
   Now
 linearity is at the bottom of most distinctive quantum behaviour
 such as superposition, interference, and entanglement. It is not
 surprising, therefore, that if you assume linearity at the start,
 you can get QM with minimal further effort.

>>> Except that I don't assume linearity from the outset.
>> There seems to be some confusion between outcomes of observations
>> and sets of possible outcomes. The \P_A*psi is actually defined as a
>> superposition in (D.2), ad you then seek to determine the
>> probability of this superposition? You define the probability of a
>> set of outcomes by P_psi(\P_A*psi), which is P_psi(\Sigma psi_a). I
>> find it hard to interpret what this might mean -- the probability of
>> a superposition of measurement outcomes (with equal weights, what is
>> more)?
>>
> The weights aren't equal. They're denoted P(ψ_α).

Again, that is not what the text says; (D.2) is a sum over outcomes with 
equal weights.

I find the notation confusing again. You have A contained in S, with 
probability P_psi(\P_A*psi). A is original defined as an observable, 
which divides the observer moment into a set of discrete outcomes. But S 
is the set of possible outcomes: a is a member of S, so A contained in S 
seems to be a different A -- the operator is not a subset of the 
outcomes. To make something out of this, I took the latter use of A to 
be the set of possible outcomes a -- not every operator has the same set 
of possible outcomes.

>> You then talk about this as though you were still partitioning sets,
>> but the probability is not defined on a set, only on a
>> superposition. If it is a set, then (D.2) makes no sense.
>>
>> You then introduce, quite arbitrarily, multiple observers for each
>>