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

[Bruce Kellett]

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
observer moment. This then gives you a measure, which is then made
to be complex!! The number of observers for each observer moment,
even if there can be more than one, which is not proved, cannot be
complex.

Why? Give me one good reason - other than it doesn't match your
intuition, which is generally not a good reason.


People/observers can be complicated beings, but that does not mean that 
you can have a complex number of them. Since all you are really wanting, 
ISTM, is to give each observer moment a weight. The n umber of observers 
observing tis moment seems a rather arbitrary source for this weight, 
because it is not ever determinable. Why not just give a weight, which 
can be complex if you wish, but it has nothing to do with multiple 
observers.



So your introduction of a measure, or weight for each
superposition really does not make 

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

[Russell Standish]

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.

> 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.

>  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).

> 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
> observer moment. This then gives you a measure, which is then made
> to be complex!! The number of observers for each observer moment,
> even if there can be more than one, which is not proved, cannot be
> complex.

Why? Give me one good reason - other than it doesn't match your
intuition, which is generally not a good reason.

> So your introduction of a measure, or weight for each
> superposition really does not make sense. You then conclude that V,
> the set of all observer moment, is a vector space over the complex
> numbers.
> 

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.

> I remain baffled. You start with an observer moment as a set of
> consistent possibilities. But there is no specification of what
> 'consistent' might mean.

There doesn't need to be a specification. All we need to know is that
some worlds correspond to the one we see, and some don't. We don't
need a constructive procedure for determining which worlds are to be
included, and in all likelihood, no such constructive procedure will
be found anyway.

>  There is also no particular structure
> imposed on this observer moment

It satisfies set axioms, otherwise you cannot apply Kolmogorov's
probability axioms. I have been criticised for this particular
assumption before, however the Nothing (the book, after all is called
"Theory of Nothing") is a set above all.

> , and you conclude, after a number of
> obscure manipulations,  that the set of all observer moments is a
> vector space over the complex numbers. I look in vain for the magic
> that converts an unstructured ensemble into a linear vector space.

The "magic" IMHO is to consider that observers are also drawn from a
distribution according to some measure, rather than just being a
single observer. This is forced onto us by the Multiverse nature of
assumed reality. An observer cannot see a∧b, where a and b are
disjoint, but two different observers in different branches of the
Multiverse can.

> This is surely a non-trivial restriction on the nature of observer
> moments, but you do not restrict the possible generality, you only
> project particular (unstructured) results from this observer 

Re: What lead to free-will denial?

[John Clark]

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. ​



> ​>​
> 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".  ​

> ​> ​
> Then what's to stop me from assuming it?
>
> ​Nothing prevents you except for the embarrassment of speaking gibberish.
​


​John K Clark​







>

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

[Stathis Papaioannou]

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


--
Stathis Papaioannou

> --
Stathis Papaioannou

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Re: What lead to free-will denial?

[Adrian Chira]

John,

I can't point out the contradiction because there is none. And it leads to 
no contradiction because free will can't be assumed.

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? What happens if I 
do it anyway? Per your admission, I won't run into a contradiction. Then 
what's to stop me from assuming it? I point I'll make in an upcoming post 
(mostly in response to Stathis and you) is that people in general, 
including you, make many assumptions that don't make any sense unless you 
make them or unless you make other assumptions which don't make sense 
except in the paradigm that assumes them.

Good heavens! I really didn't think a assumption that gibberish is 
worthless was controversial! ​

You finally feel what I feel! :) I really didn't think that something as 
self-obvious like our experience of free will was controversial! But in 
regards to your concern, you only need to step outside the box and you will 
see useful gibberish all around (again, more in my upcoming post).

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

[John Clark]

​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.

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

 John K Clark

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Re: What lead to free-will denial?

[John Clark]

On Wed, Jun 28, 2017 at 3:19 AM, Bruno Marchal  wrote:

​>> ​
>> Harry Potter doesn't exist, but it would be wrong to say free will
>> doesn't exist. Free will has neither the property of existence nor
>> nonexistence because free will is pure unadulterated gibberish.
>
>
> ​> ​
> Mocking a notion is not more convincing than mocking a person.
>

My apologies to Mr. Harry Potter.

>> Unicorns don't exist,
>
>

>In which theory?


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

John K Clark

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Re: What lead to free-will denial?

[John Clark]

On Tue, Jun 27, 2017 at 11:28 PM, Adrian Chira  wrote:

​>> ​
>> free will makes no sense in any context
>
>

​> ​
> Except the context which assumes free will. If the assumption of free will
> leads to a logical contradiction please point that out.
>

​I can't point out the contradiction because there is none. And it leads to
no contradiction because free will can't be assumed. And free will can't be
assumed because there is nothing there to assume. In other words "free
will" is gibberish. Assuming there is a largest prime number leads to a
logical contradiction, but assuming free will exists is like assuming
Klogknee exists.


> ​>
> an implicit and, as of yet, unsupported assumption that you make is that
> anything that doesn't make sense is worthless.
>
> ​Good heavens! I really didn't think a assumption that gibberish is
worthless was controversial! ​

> ​> ​
> It would be great if you could share why you think that's the case.
>
> ​Ok this is my response as to why I think that is the case:
​"Phlobneegob". Do you think my response has worth or is it worthless?

John K Clark







>
>

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Re: What lead to free-will denial?

[Bruno Marchal]

On 27 Jun 2017, at 16:14, John Clark wrote:

On Tue, Jun 27, 2017 at 4:16 AM, Bruno Marchal   
wrote:


​> ​As Brent just mentioned, it makes sense in a court law,

​No,​ ​free will makes no sense in any context​;​ and  
that's exactly why the court system in most countries is such a  
ridiculous chaotic self contradictory mess.


So free-will exists, because there is certainly one country having a  
better court system than another, and indeed, just compare a court  
system of a democracy and a totalitarian state.






​> ​and indeed, I am not sure a lot of human value (justice,  
freedom, responsibility,


​Justice and responsibility are intrinsically linked with the  
concept of punishment, and the only logical reason to punish anybody  
for anything is to discourage similar acts in the future; if it  
can't do that then there is no point to the punishment. It's true  
that the reptilian part of our brain can also get enjoyment from  
making somebody we don't like suffer just for the sake of suffering,  
but I'm not proud of that part of my brain and so the more recently  
evolved parts of that organ have decided to refuse to defend such a  
feeling.


As for freedom that just means the ability to do what you want to  
do, and sometimes you can but usually you can not because something  
restrains you, if by nothing else the laws of physics.


The ability to do what you want to do is freedom, OK, but that notion  
needs some free-will to make sense, and then some amount of degrees of  
freedom to enact the free-will.






 ​

​> ​To decide that free-will does not exist​ [...]​

​Unicorns don't exist,


In which theory?




the largest prime number doesn't exist,


The existence of a largest prime number is consistent with RA. I guess  
you assume PA or second order arithmetic. "existence" is a notion  
depending of the theory, and its interpretation.






Harry Potter doesn't exist, but it would be wrong to say free will  
doesn't exist. Free will has neither the property of existence nor  
nonexistence because free will is pure unadulterated gibberish.


Mocking a notion is not more convincing than mocking a person.







​> ​The program e imagine itself doing two tasks and choose to  
actually proceed on one of them by comparing mentally the  
consequences.


​Sure, the program decided to do X rather than Y for a reason, just  
as a cog in a cuckoo clock decided to turn left rather than right  
for a reason.


You are using the incompatibilist notion of free-will, but we have  
already discuss this, and we both agree that incompatibilist free-will  
does not make sense.


You do the same thing with all terms in theology. You choose an absurd  
theory, and then condemn the concept by mocking the theory, instead of  
criticizing the theory and searching a better one.
You could say that Earth does not exist given that a flat earth does  
not make sense.








​> ​Free-will denial is first person denial.

​Please please I'm begging you, lets not wade back into that  
cesspool of pointless peepee! ​ ​


Mocking a notion is not more convincing than mocking a person. Here  
you do both. I guess it is a rhetorical trick to hide an absence of  
arguments.


Bruno





 John K Clark​



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http://iridia.ulb.ac.be/~marchal/



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