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

[Bruno Marchal]

On 12 Jul 2017, at 04:59, Bruce Kellett wrote:


On 12/07/2017 12:42 pm, Russell Standish wrote:

On Wed, Jul 12, 2017 at 12:00:40PM +1000, Bruce Kellett wrote:

Well, if that is what it is supposed to imply, then John might well
be right to have problems with it! As I have said before, there is
no analogy between step 3 and quantum many worlds -- the differences
far outweigh any superficial similarities.
I don't think you have properly elucidated what those differences  
are,

other than in passing, maybe. How about concentrating on those
differences in detail - if you can show that the many worlds of FPI
are phenomenally different from the many worlds of quantum mechanics,


I don't know how many differences you need, but in step 3, a person  
is duplicated, not a world. And in that scenario, I could, after  
duplication and finding myself in Moscow, get on a plane and fly to  
Washington and meet up with my duplicate. That sort of interaction  
between duplicates is not possible in MWI (at least in its  
decoherent form).


In that step 3 scenario you are right.

But the step 3 is used only to explain and define the notion of thirs  
and fist person view, and to explain the first person indeterminacy.


Step 4 to 7 is needed to understand that in arithmetic too, nobody can  
talk with its (infinitely many) doppelgangers in "parallel"  
computational history, and why matter is not clonable, and other  
qualitaitive quantum facts.


Then, to get the quantitative aspect, you need to compare quantum  
logic(s) with the arithmetical quantum logics.



Also, in quantum MWI, there is no external observer who can see the  
splitting as there can be external observers of person duplication  
-- the copies do not have to be transported, after all, they might  
both be in the same room.


You get the step 3 point, OK. But you seem to miss the steps 4, 5, 6,  
7 (and 8, which is not necessary, unless you believe in magic and that  
the physical universe is small).


The reversal relies on the fact that with computationalism, we *are*  
at each nano-second (say) multiplied/differentiated into an infinity  
of computations emulated in a (tiny) part of the arithmetical reality.  
I called that the global indeterminacy, where the reconstitution  
domain is simply the structure (N, 0, +, *) (not to confuse with any  
theory on that structure, which plays the role of the observers and  
are seen as numbers *in* that reality).


And the math confirms that we got a quantum logic. Even three of them,  
and the only question is if that quantum logic define, or not, a  
unique measure on the computations "seen from the 1p view, and for  
this we need a mathematical theory of the 1p view. It happens that the  
incompleteness phenomenon rehabilitates the theory of knowledge of  
Theaetetus. Incompleteness makes provability into a rational-belief  
notion, and we can define knowledge by the conjunction of rational- 
belief and truth. On the proposition corresponding to the existence of  
a computational continuations just that give already a quantum logic,  
but we get two other one with adding constraints of consistency.


I was hoping hat such a theory would be quickly refuted, so that we  
could learnj something, but that is not the case. Digital Mechanism is  
the only theory which explains qulaia and quanta and their relations,  
and which does not commit an ontological commitment (in a PRIMARY  
MATTER, or in GOD, or something).


It is the second time that you talk like if step 3 was the last step,  
or like I would have been defending the (ridiculous) idea that the  
quantum superposition was literally a duplication of person's bodies  
or of of worlds. It is more like a notion of infinitely many  
"preparation" in the arithmetical reality (the standard model of  
arithmetic). That is apparent at step 7.


Bruno





Bruce


then you are well on the way to showing a fundamental incompatibility
between computationalism and quantum theory. Then presumably, we can
perform an experiment to show which one is incorrect.  
Computationalism

or QM. Worthy of a Nobel prize, I'd think.

Somehow, I don't know that the task is going to be quite so easy...


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

[Bruce Kellett]

On 12/07/2017 12:42 pm, Russell Standish wrote:

On Wed, Jul 12, 2017 at 12:00:40PM +1000, Bruce Kellett wrote:

Well, if that is what it is supposed to imply, then John might well
be right to have problems with it! As I have said before, there is
no analogy between step 3 and quantum many worlds -- the differences
far outweigh any superficial similarities.

I don't think you have properly elucidated what those differences are,
other than in passing, maybe. How about concentrating on those
differences in detail - if you can show that the many worlds of FPI
are phenomenally different from the many worlds of quantum mechanics,


I don't know how many differences you need, but in step 3, a person is 
duplicated, not a world. And in that scenario, I could, after 
duplication and finding myself in Moscow, get on a plane and fly to 
Washington and meet up with my duplicate. That sort of interaction 
between duplicates is not possible in MWI (at least in its decoherent 
form). Also, in quantum MWI, there is no external observer who can see 
the splitting as there can be external observers of person duplication 
-- the copies do not have to be transported, after all, they might both 
be in the same room.


Bruce


then you are well on the way to showing a fundamental incompatibility
between computationalism and quantum theory. Then presumably, we can
perform an experiment to show which one is incorrect. Computationalism
or QM. Worthy of a Nobel prize, I'd think.

Somehow, I don't know that the task is going to be quite so easy...


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

[Russell Standish]

On Wed, Jul 12, 2017 at 12:00:40PM +1000, Bruce Kellett wrote:
> 
> Well, if that is what it is supposed to imply, then John might well
> be right to have problems with it! As I have said before, there is
> no analogy between step 3 and quantum many worlds -- the differences
> far outweigh any superficial similarities. 

I don't think you have properly elucidated what those differences are,
other than in passing, maybe. How about concentrating on those
differences in detail - if you can show that the many worlds of FPI
are phenomenally different from the many worlds of quantum mechanics,
then you are well on the way to showing a fundamental incompatibility
between computationalism and quantum theory. Then presumably, we can
perform an experiment to show which one is incorrect. Computationalism
or QM. Worthy of a Nobel prize, I'd think.

Somehow, I don't know that the task is going to be quite so easy...


-- 


Dr Russell StandishPhone 0425 253119 (mobile)
Principal, High Performance Coders
Visiting Senior Research Fellowhpco...@hpcoders.com.au
Economics, Kingston University http://www.hpcoders.com.au


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

[Bruce Kellett]

On 12/07/2017 11:25 am, Russell Standish wrote:

On Tue, Jul 11, 2017 at 03:04:13PM +1000, Bruce Kellett wrote:

On 11/07/2017 2:12 pm, Russell Standish wrote:

You're still missing the point. The quantum reality is a 1p thing, it
is the observed phenomenal physics. Substrate independence is a 3p
thing, and may be quantum, classical or whatever, just needing to
support universal computation.

Well, in the Everettian picture, the quantum reality is the whole
structure, multiverse or whatever. The 0p/3p picture of splitting
into separate worlds according to the unitary evolution of the wave
function is the quantum reality. The 1p view of this is just a
projection from the whole according to self-selection. If the
universe were classical, Turing machines could still exist and
support the computations underlying consciousness, but there would
be no splitting into many worlds.

Yes there is. That is the FPI result, which appears at step 3 of the
UDA. (The one that John Clark apparently has problems with, although
he's never clearly enunciated what problems he sees with it(.


Well, if that is what it is supposed to imply, then John might well be 
right to have problems with it! As I have said before, there is no 
analogy between step 3 and quantum many worlds -- the differences far 
outweigh any superficial similarities. It is what I call the "cat = dog" 
argument, though it probably has a more formal name: my dog has four 
legs and my cat has four legs; therefore, my cat is a dog.


Bruce

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

[Russell Standish]

On Tue, Jul 11, 2017 at 03:04:13PM +1000, Bruce Kellett wrote:
> On 11/07/2017 2:12 pm, Russell Standish wrote:

> >You're still missing the point. The quantum reality is a 1p thing, it
> >is the observed phenomenal physics. Substrate independence is a 3p
> >thing, and may be quantum, classical or whatever, just needing to
> >support universal computation.
> 
> Well, in the Everettian picture, the quantum reality is the whole
> structure, multiverse or whatever. The 0p/3p picture of splitting
> into separate worlds according to the unitary evolution of the wave
> function is the quantum reality. The 1p view of this is just a
> projection from the whole according to self-selection. If the
> universe were classical, Turing machines could still exist and
> support the computations underlying consciousness, but there would
> be no splitting into many worlds. 

Yes there is. That is the FPI result, which appears at step 3 of the
UDA. (The one that John Clark apparently has problems with, although
he's never clearly enunciated what problems he sees with it(.


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Visiting Senior Research Fellowhpco...@hpcoders.com.au
Economics, Kingston University http://www.hpcoders.com.au


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

[Bruno Marchal]

On 10 Jul 2017, at 14:29, Bruce Kellett wrote:


On 10/07/2017 9:34 pm, Bruno Marchal wrote:

On 10 Jul 2017, at 03:41, Bruce Kellett wrote:

On 7/07/2017 7:19 pm, Bruno Marchal wrote:

On 07 Jul 2017, at 01:52, Bruce Kellett wrote:


On 7/07/2017 12:50 am, Bruno Marchal wrote:

On 06 Jul 2017, at 14:22, Bruce Kellett wrote:

On 6/07/2017 5:55 pm, Russell Standish wrote:
And assuming conscious classic digital machines, quantum  
phenomenology
appears at the observed level - a result in line with Bruno  
Marchal's

FPI result.


Prove it. Bruno has failed to do so -- his person duplication  
thought experiments do not reproduce quantum behaviour.


Which one? Z1*, X1*, or S4Grz1? If you know about a physical  
facts contradicting those theories, I would be pleased to know.  
The person duplication experience just shows that physics is  
given by a "sum" on all computations, seen from internal points  
of view imposed by incompleteness, and until now, as modest as  
the results can be, the three propositional physics are still  
not refuted. I am not sure you have studied them, because you  
have shown not knowing the basic theories needed to apprehend  
them, so it looks you are just inventing something here.


The point that I was trying to make to Russell was the fact that  
purely classical machines can exhibit consciousness means that  
you cannot derive quantum mechanics from consciousness alone.


That depends on your assumptions. If my consciousness, or my 1p  
experience are invariant for a physical digital substitution, in  
virtue of computing, then there is just no choice in the matter.


Let me spell out the argument more clearly. If consciousness  
implies that the world is quantum mechanical (one can derive  
quantum mechanics from the existence of observer moments), then it  
follows that consciousness is not possible in a non-quantum world  
(modus tollens). But a Turing machine is not a quantum device;


OK. It is an arithmetical entity.



it could exist in a non-quantum world


Indeed, at least seen from outside, in the 0p view. OK.


and exhibit consciousness (given the appropriate computations), so  
something has to give -- either the derivation of the quantum from  
the existence of consciousness, or digital substitution of  
consciousness (substrate independence). Take your pick.


The machine "lives", or "exists" in the arithmetical reality, in  
the eyes of god (in the 3p absolute view, or in the 0p view), but  
from its first personal perspective (1p view) it lives provably in  
a quantum reality. Then we can test if the quantum reality of the  
machine violates or not the quantum that we infer from nature.


You must not identify: "the machine is in arithmetic", with the  
machine's point of view access only a quantum reality (the reality  
of all computations going through its current states, below its  
substitution . We need to always make clear which pov we are  
talking about. The UDA showed that the physical is 1p plural  
statistical. It is not a 3p view.


I don't think it is that simple.


It is not simple at all. Just finding reasonable arithmetical (like  
[]p, or meta-arithmetical like []p & p) definitions is not simple at  
all.




If we have substrate independence, the machine (the conscious  
person) cannot tell what substrate is supporting the computations,  
whether arithmetic, a quantum world, or a classical Newtonian world.


From its first person *singular* point of view.




That would seem to imply that mere consideration of conscious  
observer moments cannot distinguish between these.


Which leads to the global FPI (on all computations going through my  
current (indexical) state/computation. That is why by observation, the  
subject can verify if the physics is, or not, a sum on an infinity of  
computations. QM confirms computationalism, in this way, i.e. that we"  
live in arithmetic" (so to speak), because we "see" the resulting sum  
on histories, obeying formally (until now) to the self-reference  
constraint.





Or else you feel the full force of the conundrum enunciated above:  
if observer moments imply a quantum reality, then the machine can  
indeed determine its substrate, and substrate independence is lost.


It is lost below our substitution level because the appearance of  
substrate is given by a "sum on all computations", and that gives the  
quantum structure. But this is a consequence of the FPI, which is a  
consequence of the computationalist substrate independence.


Bruno







Bruce

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

[Bruce Kellett]

On 11/07/2017 2:12 pm, Russell Standish wrote:

On Mon, Jul 10, 2017 at 10:29:26PM +1000, Bruce Kellett wrote:

I don't think it is that simple. If we have substrate independence,
the machine (the conscious person) cannot tell what substrate is
supporting the computations, whether arithmetic, a quantum world, or
a classical Newtonian world. That would seem to imply that mere
consideration of conscious observer moments cannot distinguish
between these. Or else you feel the full force of the conundrum
enunciated above: if observer moments imply a quantum reality, then
the machine can indeed determine its substrate, and substrate
independence is lost.


You're still missing the point. The quantum reality is a 1p thing, it
is the observed phenomenal physics. Substrate independence is a 3p
thing, and may be quantum, classical or whatever, just needing to
support universal computation.


Well, in the Everettian picture, the quantum reality is the whole 
structure, multiverse or whatever. The 0p/3p picture of splitting into 
separate worlds according to the unitary evolution of the wave function 
is the quantum reality. The 1p view of this is just a projection from 
the whole according to self-selection. If the universe were classical, 
Turing machines could still exist and support the computations 
underlying consciousness, but there would be no splitting into many 
worlds. So if consciousness implies quantum mechanics, then we are able 
to determine the substrate -- if quantum, the substrate cannot be purely 
Newtonian, and the computations are not supported by a classical Turing 
machine. I don't think the 1p-1pp-3p distinctions are helpful here, they 
obscure what is actually going on.


Bruce





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

[Russell Standish]

On Mon, Jul 10, 2017 at 10:29:26PM +1000, Bruce Kellett wrote:
> 
> I don't think it is that simple. If we have substrate independence,
> the machine (the conscious person) cannot tell what substrate is
> supporting the computations, whether arithmetic, a quantum world, or
> a classical Newtonian world. That would seem to imply that mere
> consideration of conscious observer moments cannot distinguish
> between these. Or else you feel the full force of the conundrum
> enunciated above: if observer moments imply a quantum reality, then
> the machine can indeed determine its substrate, and substrate
> independence is lost.
> 

You're still missing the point. The quantum reality is a 1p thing, it
is the observed phenomenal physics. Substrate independence is a 3p
thing, and may be quantum, classical or whatever, just needing to
support universal computation.

-- 


Dr Russell StandishPhone 0425 253119 (mobile)
Principal, High Performance Coders
Visiting Senior Research Fellowhpco...@hpcoders.com.au
Economics, Kingston University http://www.hpcoders.com.au


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

[Bruno Marchal]

On 10 Jul 2017, at 03:41, Bruce Kellett wrote:


On 7/07/2017 7:19 pm, Bruno Marchal wrote:

On 07 Jul 2017, at 01:52, Bruce Kellett wrote:


On 7/07/2017 12:50 am, Bruno Marchal wrote:

On 06 Jul 2017, at 14:22, Bruce Kellett wrote:

On 6/07/2017 5:55 pm, Russell Standish wrote:
And assuming conscious classic digital machines, quantum  
phenomenology
appears at the observed level - a result in line with Bruno  
Marchal's

FPI result.


Prove it. Bruno has failed to do so -- his person duplication  
thought experiments do not reproduce quantum behaviour.


Which one? Z1*, X1*, or S4Grz1? If you know about a physical  
facts contradicting those theories, I would be pleased to know.  
The person duplication experience just shows that physics is  
given by a "sum" on all computations, seen from internal points  
of view imposed by incompleteness, and until now, as modest as  
the results can be, the three propositional physics are still not  
refuted. I am not sure you have studied them, because you have  
shown not knowing the basic theories needed to apprehend them, so  
it looks you are just inventing something here.


The point that I was trying to make to Russell was the fact that  
purely classical machines can exhibit consciousness means that you  
cannot derive quantum mechanics from consciousness alone.


That depends on your assumptions. If my consciousness, or my 1p  
experience are invariant for a physical digital substitution, in  
virtue of computing, then there is just no choice in the matter.


Let me spell out the argument more clearly. If consciousness implies  
that the world is quantum mechanical (one can derive quantum  
mechanics from the existence of observer moments), then it follows  
that consciousness is not possible in a non-quantum world (modus  
tollens). But a Turing machine is not a quantum device;


OK. It is an arithmetical entity.





it could exist in a non-quantum world


Indeed, at least seen from outside, in the 0p view. OK.




and exhibit consciousness (given the appropriate computations), so  
something has to give -- either the derivation of the quantum from  
the existence of consciousness, or digital substitution of  
consciousness (substrate independence). Take your pick.


The machine "lives", or "exists" in the arithmetical reality, in the  
eyes of god (in the 3p absolute view, or in the 0p view), but from its  
first personal perspective (1p view) it lives provably in a quantum  
reality. Then we can test if the quantum reality of the machine  
violates or not the quantum that we infer from nature.


You must not identify: "the machine is in arithmetic", with the  
machine's point of view access only a quantum reality (the reality of  
all computations going through its current states, below its  
substitution . We need to always make clear which pov we are talking  
about. The UDA showed that the physical is 1p plural statistical. It  
is not a 3p view.


Bruno








Bruce

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

[Bruce Kellett]

On 10/07/2017 11:40 am, Russell Standish wrote:

On Fri, Jul 07, 2017 at 10:56:27AM +1000, Bruce Kellett wrote:

On 7/07/2017 10:40 am, Russell Standish wrote:

On Thu, Jul 06, 2017 at 10:22:40PM +1000, Bruce Kellett wrote:


No, position and momentum are dual in the sense I defined. The
observables are not compatible -- position and momentum are not
simultaneously observable.

I know what you mean by "dual", although the conventional term is
"complementary".

Yes, I was avoiding that term because it is sometimes controversial.
But complementarity is a feature of conjugate variables.

I didn't know the terminology was controversial. Why?


Well, the term originated with Bohr, I think, and there are some who 
reject more or less everything Bohr did as part of their rejection of 
the Copenhagen Interpretation. More recently, the term "Black Hole 
Complementarity" has been invented by Susskind and others. They use this 
as a way to escape from the contradictions inherent in their solution of 
the BH information paradox. Their use of the term there has really 
nothing to do with Bohr's notion of complementarity, and I think it 
merely confusing.




Observing S=X+P does not imply simultaneously observing X and P.

As far as I can see, it does. It is not just something you construct
by measuring X then P (or vice versa).

Exactly, the combined error on S will be the sum of errors on X & P,
so there's no point in doing that.



X and P are operators in
different spaces, related by a Fourier transform. Unless you mean
measuring the conjugate variables on different members of an
ensemble of identically prepared states?



Prove that I can't observe S, or provide a reference to someone doing
so. It appears rather crucial to your critique.

Again, I do not accept the reversal of the burden of proof. X and P
are conjugate variables so they are not simultaneously measurable.

Yes - but failure of imagination is not proof of failure. You were
quite clear that X+P could not be measured independently of X or
P. Why? Might there not be something like a squeezed light scenario
that could do it?


Squeezed states are simply states that reach the limits of the spread in 
conjugate variables consistent with the Uncertainly Principle -- you can 
squeeze in one dimension at the expense of an increase in uncertainty in 
the other dimension. I think it is up to you to show how this can be 
used to give meaning to the operator X+P. My argument against this 
possibility relies ultimately on complementarity (or the dual nature of 
position and momentum spaces) -- the spaces are alternatives, not parts 
of the same thing.




On your last point, this is not crucial to my critique of your
theory. Well, not unless you are going to rely on this in your proof
of linearity.

I don't, but clearly if not all linear combinations of observer
moments are observer moments in themselves, then that is an important
critique. If (X+P) is not a physical observable, then (\P_X+\P_P)ψ is
probably not an OM, and it makes it just that much more difficult for
me to demonstrate the emergence of QM. ITOH, if X+P is a physical
observable, then clearly (\P_X+\P_P)ψ is an observer moment.


There is a point about operators on observer moments that I mentioned 
briefly before, but would like to raise in more detail again. Your 
projection operator \P_A produces the set of possible outcomes of the 
observation A, \P{a}*psi = psi_a, the set of possible outcomes of the 
observation. My question was, do you consider it possible for these 
alternative outcomes to interfere, or not? If they can interfere, then 
you have assumed a pure quantum state. If they cannot interfere, then 
you have a mixed state, which is just a range of disjoint possible 
classical outcomes. In which case, the result of the observation is not 
a single vector in some space.


It seems that you take these projection operators to produce pure 
quantum states, since later (in the derivation of the Schrödinger 
equation, for instance) you assume unitarity. It is only pure quantum 
states that preserve unitarity -- mixed states are not reversible, hence 
not the product of unitary evolution from a single predecessor state. 
But pure states are uniquely quantum, so if your operators produce pure 
states, you have already built a large amount of quantum mechanics into 
your derivation.


Bruce

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

[Bruce Kellett]

On 7/07/2017 7:19 pm, Bruno Marchal wrote:

On 07 Jul 2017, at 01:52, Bruce Kellett wrote:


On 7/07/2017 12:50 am, Bruno Marchal wrote:

On 06 Jul 2017, at 14:22, Bruce Kellett wrote:

On 6/07/2017 5:55 pm, Russell Standish wrote:
And assuming conscious classic digital machines, quantum 
phenomenology

appears at the observed level - a result in line with Bruno Marchal's
FPI result.


Prove it. Bruno has failed to do so -- his person duplication 
thought experiments do not reproduce quantum behaviour.


Which one? Z1*, X1*, or S4Grz1? If you know about a physical facts 
contradicting those theories, I would be pleased to know. The person 
duplication experience just shows that physics is given by a "sum" 
on all computations, seen from internal points of view imposed by 
incompleteness, and until now, as modest as the results can be, the 
three propositional physics are still not refuted. I am not sure you 
have studied them, because you have shown not knowing the basic 
theories needed to apprehend them, so it looks you are just 
inventing something here.


The point that I was trying to make to Russell was the fact that 
purely classical machines can exhibit consciousness means that you 
cannot derive quantum mechanics from consciousness alone.


That depends on your assumptions. If my consciousness, or my 1p 
experience are invariant for a physical digital substitution, in 
virtue of computing, then there is just no choice in the matter.


Let me spell out the argument more clearly. If consciousness implies 
that the world is quantum mechanical (one can derive quantum mechanics 
from the existence of observer moments), then it follows that 
consciousness is not possible in a non-quantum world (modus tollens). 
But a Turing machine is not a quantum device; it could exist in a 
non-quantum world and exhibit consciousness (given the appropriate 
computations), so something has to give -- either the derivation of the 
quantum from the existence of consciousness, or digital substitution of 
consciousness (substrate independence). Take your pick.


Bruce

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

[Russell Standish]

On Fri, Jul 07, 2017 at 10:56:27AM +1000, Bruce Kellett wrote:
> On 7/07/2017 10:40 am, Russell Standish wrote:
> >On Thu, Jul 06, 2017 at 10:22:40PM +1000, Bruce Kellett wrote:
> >
> >>No, position and momentum are dual in the sense I defined. The
> >>observables are not compatible -- position and momentum are not
> >>simultaneously observable.
> >I know what you mean by "dual", although the conventional term is
> >"complementary".
> 
> Yes, I was avoiding that term because it is sometimes controversial.
> But complementarity is a feature of conjugate variables.

I didn't know the terminology was controversial. Why?

> 
> 
> >Observing S=X+P does not imply simultaneously observing X and P.
> 
> As far as I can see, it does. It is not just something you construct
> by measuring X then P (or vice versa). 

Exactly, the combined error on S will be the sum of errors on X & P,
so there's no point in doing that.



> X and P are operators in
> different spaces, related by a Fourier transform. Unless you mean
> measuring the conjugate variables on different members of an
> ensemble of identically prepared states?
> 



> >Prove that I can't observe S, or provide a reference to someone doing
> >so. It appears rather crucial to your critique.
> 
> Again, I do not accept the reversal of the burden of proof. X and P
> are conjugate variables so they are not simultaneously measurable.

Yes - but failure of imagination is not proof of failure. You were
quite clear that X+P could not be measured independently of X or
P. Why? Might there not be something like a squeezed light scenario
that could do it?

> 
> On your last point, this is not crucial to my critique of your
> theory. Well, not unless you are going to rely on this in your proof
> of linearity. 

I don't, but clearly if not all linear combinations of observer
moments are observer moments in themselves, then that is an important
critique. If (X+P) is not a physical observable, then (\P_X+\P_P)ψ is
probably not an OM, and it makes it just that much more difficult for
me to demonstrate the emergence of QM. ITOH, if X+P is a physical
observable, then clearly (\P_X+\P_P)ψ is an observer moment.

> But as I pointed out a while ago, additivity of
> operators does not imply that the state is a vector in a linear
> space.
> 

That's irrelevant, since its additivity of states that counts.


-- 


Dr Russell StandishPhone 0425 253119 (mobile)
Principal, High Performance Coders
Visiting Senior Research Fellowhpco...@hpcoders.com.au
Economics, Kingston University http://www.hpcoders.com.au


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

[Bruno Marchal]

On 07 Jul 2017, at 01:52, Bruce Kellett wrote:


On 7/07/2017 12:50 am, Bruno Marchal wrote:

On 06 Jul 2017, at 14:22, Bruce Kellett wrote:

On 6/07/2017 5:55 pm, Russell Standish wrote:
And assuming conscious classic digital machines, quantum  
phenomenology
appears at the observed level - a result in line with Bruno  
Marchal's

FPI result.


Prove it. Bruno has failed to do so -- his person duplication  
thought experiments do not reproduce quantum behaviour.


Which one? Z1*, X1*, or S4Grz1? If you know about a physical facts  
contradicting those theories, I would be pleased to know. The  
person duplication experience just shows that physics is given by a  
"sum" on all computations, seen from internal points of view  
imposed by incompleteness, and until now, as modest as the results  
can be, the three propositional physics are still not refuted. I am  
not sure you have studied them, because you have shown not knowing  
the basic theories needed to apprehend them, so it looks you are  
just inventing something here.


The point that I was trying to make to Russell was that since purely  
classical machines can exhibit consciousness means that you cannot  
derive quantum mechanics from consciousness alone.


That depends on your assumptions. If my consciousness, or my 1p  
experience are invariant for a physical digital substitution, in  
virtue of computing, then there is just no choice in the matter.  
Physics (the science of observable prediction) has to be retrieved  
from arithmetical self-reference. Precisely, the logic of "measure  
one" (the yes-no experiments) must be retrieved from two things:


1) the modal arithmetical nuances (brought by incompleteness) of  
Gödel's arithmetical provability predicate ([]p): precisely either []p  
& p, or []p & <>t, or []p & <>t & p.  (by the UDA reasoning).


2) the restriction of the arithmetical realization to the sigma_1  
sentence.(by the UD itself).





And neither does the fact that you might have found a couple of  
objects that do not commute mean that you have derived QM.


I found more than that (a quantum logic) but the points is that this  
is found at the place where UDA justifies that we must find the logic  
of the observable.





Else the person who first noted that rotations in 3 dimensions do  
not commute could be said to have discovered QM!


Which would be ridiculous indeed, but has nothing to do with what I  
have explained we need to do to solve the mind-body problem, or more  
generally the 1p/3p relation problem.


If you want to save both Digital Mechanism (in cognitive science, not  
in physics) and physicalism, you need to explain what a physical  
universe is, and how it can select some subset of the set of all  
computations emulated in (a tiny part) of Arithmetic (the model, not  
any theory). If not, the physicalist misuse the (metaphysical) noyion  
of physical universe in the same manner a creationist use God to  
criticize Evolution.


Bruno





Bruce

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



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

['scerir' via Everything List]

Just 2 (old) references

On quantum-mechanical automata
-David Z. Albert
Physics Letters A
Volume 98, Issues 5–6, 24 October 1983, Pages 249-252
Abstract
An automaton whose states are solutions of quantum-mechanical equations of 
motion is described, and the capacities of such an automaton to “measure” and 
to “know” and to “predict” certain physical properties of the world are 
considered. It is inquired what sort of empirical description such an automaton 
would produce of itself. It turns out that this description would be a very 
novel one, such as was never imagined in conventional theories of measurement.

On quantum-mechanical automata
-Asher Peres
Physics Letters A
Volume 101, Issues 5–6, 2 April 1984, Pages 249-250
Abstract
An automaton which can “measure” or “know” or “predict” the values of physical 
quantities cannot be described by quantum mechanics.

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

[Bruce Kellett]

On 7/07/2017 10:40 am, Russell Standish wrote:

On Thu, Jul 06, 2017 at 10:22:40PM +1000, Bruce Kellett wrote:


No, position and momentum are dual in the sense I defined. The
observables are not compatible -- position and momentum are not
simultaneously observable.

I know what you mean by "dual", although the conventional term is
"complementary".


Yes, I was avoiding that term because it is sometimes controversial. But 
complementarity is a feature of conjugate variables.




Observing S=X+P does not imply simultaneously observing X and P.


As far as I can see, it does. It is not just something you construct by 
measuring X then P (or vice versa). X and P are operators in different 
spaces, related by a Fourier transform. Unless you mean measuring the 
conjugate variables on different members of an ensemble of identically 
prepared states?



Prove that I can't observe S, or provide a reference to someone doing
so. It appears rather crucial to your critique.


Again, I do not accept the reversal of the burden of proof. X and P are 
conjugate variables so they are not simultaneously measurable.


On your last point, this is not crucial to my critique of your theory. 
Well, not unless you are going to rely on this in your proof of 
linearity. But as I pointed out a while ago, additivity of operators 
does not imply that the state is a vector in a linear space.


Bruce

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

[Russell Standish]

On Thu, Jul 06, 2017 at 10:22:40PM +1000, Bruce Kellett wrote:

> No, position and momentum are dual in the sense I defined. The
> observables are not compatible -- position and momentum are not
> simultaneously observable.

I know what you mean by "dual", although the conventional term is
"complementary". 

Observing S=X+P does not imply simultaneously observing X and P.

Prove that I can't observe S, or provide a reference to someone doing
so. It appears rather crucial to your critique.


-- 


Dr Russell StandishPhone 0425 253119 (mobile)
Principal, High Performance Coders
Visiting Senior Research Fellowhpco...@hpcoders.com.au
Economics, Kingston University http://www.hpcoders.com.au


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

[Bruce Kellett]

On 7/07/2017 12:50 am, Bruno Marchal wrote:

On 06 Jul 2017, at 14:22, Bruce Kellett wrote:

On 6/07/2017 5:55 pm, Russell Standish wrote:

And assuming conscious classic digital machines, quantum phenomenology
appears at the observed level - a result in line with Bruno Marchal's
FPI result.


Prove it. Bruno has failed to do so -- his person duplication thought 
experiments do not reproduce quantum behaviour.


Which one? Z1*, X1*, or S4Grz1? If you know about a physical facts 
contradicting those theories, I would be pleased to know. The person 
duplication experience just shows that physics is given by a "sum" on 
all computations, seen from internal points of view imposed by 
incompleteness, and until now, as modest as the results can be, the 
three propositional physics are still not refuted. I am not sure you 
have studied them, because you have shown not knowing the basic 
theories needed to apprehend them, so it looks you are just inventing 
something here.


The point that I was trying to make to Russell was that since purely 
classical machines can exhibit consciousness means that you cannot 
derive quantum mechanics from consciousness alone. And neither does the 
fact that you might have found a couple of objects that do not commute 
mean that you have derived QM. Else the person who first noted that 
rotations in 3 dimensions do not commute could be said to have 
discovered QM!


Bruce

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

[Bruno Marchal]

On 06 Jul 2017, at 14:22, Bruce Kellett wrote:


On 6/07/2017 5:55 pm, Russell Standish wrote:

On Thu, Jul 06, 2017 at 04:18:49PM +1000, Bruce Kellett wrote:

On 6/07/2017 2:33 pm, Russell Standish wrote:

Establishing linearity is key.

Yes, and you haven't made progress with that.

All I ask is to give me some more time on this. I have some further
ideas in this regard, but need some dedicated time to think about it.


Fair enough.


Establishing the resultant vector space
is a Hilbert space does follow fairly easily from Kolmogorov's  
axioms

(although its possible you have beef with those :).

I think the issue I would have here is that you assume that a
projection can give a range of results. I don't think that is
necessarily true even in a multiverse. Since projections are as
undefined as everything else here, it could be the case that a
projection gave a single value -- the same value in every world of
the multiverse, so that you had a version of classical mechanics.
Assuming a range of different values, assigning probabilities to
individual outcomes makes some sort of sense, but that is assuming a
lot of quantum mechanics at the start. Hence my worries about
circularity.


If there is only a single outcome, then the projection concerned
will be the certain event projection. With probability 1. That case  
is covered.


A more serious consideration occurred to me -- the projection might  
have a range of possible outcomes, but these might be merely an  
illustration of classical probability arising from ignorance. In  
quantum terms, you have to distinguish between pure and mixed  
states. You appear to assume that your projections are from a pure  
state, but that is circular: you can only have pure states in that  
sense in a quantum theory. Pure states can exhibit interference,  
mixed states cannot.



S=X+P is definitely an observable, and corresponds to measuring  
the sum

of position and momentum crisply. You can work out the equivalent
Heisenberg uncertainty relations too
ΔsΔx ≥ ℏ and ΔsΔp ≥ ℏ
So if you measure with S, you will have uncertainty in both  
position

and momentum.

So it is not a quantum operator or quantum observable. These are
defined to have precise values: eigenvalues corresponding to the
eigenfuntions of the Hermitian operator. And the eigenfunctions span
the corresponding Hilbert space.

I don't get your point. The operator S=X+P satisfies all of those.


No, position and momentum are dual in the sense I defined. The  
observables are not compatible -- position and momentum are not  
simultaneously observable.



My point here is that position
space, spanned by the eigenfunctions of the position operator, is
'dual' to momentum space, spanned by the eigenfunctions of the
momentum operator. These spaces are related by the Fourier transform
and each serves to give a distinct complete representation of the
underlying quantum state. (That is the sense in which I refer to
these operators and spaces as 'dual' -- it makes no quantum sense to
add them,)

Operators operate on the Hilbert state space, not position or  
momentum

space, which is more of classical mechanics concept. And yes, one can
add them (modulo an assumed unit conversion constant) mathematically,
and the sum X+P is meaningful physically.


Classically perhaps, but not in quantum mechanics.


In general, any bounded hermitian operator is an observable of some
sort (one can create a machine, albeit weirdly wonderful in a Heath
Robisonesque way, that will measure that particular quantity).

Certainly. If one has the Hilbert space for some quantum operator,
linear combinations of the eigenvector basis vectors that also span
the space will be the eigenvalues of some other hermitian operator
in the space. As you say, not all of the resultant operators
correspond to natural observables.

I don't say that. You're saying that, and I fail to see your point.


It is really the (dead-cat +- live-cat basis) issue. A superposition  
of polarization states becomes entangled with a cat in Schrödinger's  
thought experiment. We don't observe the superposition because the  
(dead +- alive) basis does not correspond to a natural observable --  
decoherence sees to that. Similarly for most other quantum  
variables. The normal position basis, delta(x-x_i), can be  
superposed to give some alternative basis, and there will be a  
corresponding operator on the Hilbert space, but superpositions of  
position results are not given by any achievable measurement --  
again, decoherence leads to the preferred stable basis: all  other  
bases do not give results that are stable under decoherence. Not all  
possible operators are possible measurements (ie, not all operators  
give stable results).


The sum of two bounded hermitian operators is also a bounded  
hermition

operator. In fact a linear combination with real coefficients will
too, but not necessarily complex coeficients (since the hermitian
property may not be preserved).

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

[Bruce Kellett]

On 6/07/2017 5:55 pm, Russell Standish wrote:

On Thu, Jul 06, 2017 at 04:18:49PM +1000, Bruce Kellett wrote:

On 6/07/2017 2:33 pm, Russell Standish wrote:

Establishing linearity is key.

Yes, and you haven't made progress with that.

All I ask is to give me some more time on this. I have some further
ideas in this regard, but need some dedicated time to think about it.


Fair enough.


Establishing the resultant vector space
is a Hilbert space does follow fairly easily from Kolmogorov's axioms
(although its possible you have beef with those :).

I think the issue I would have here is that you assume that a
projection can give a range of results. I don't think that is
necessarily true even in a multiverse. Since projections are as
undefined as everything else here, it could be the case that a
projection gave a single value -- the same value in every world of
the multiverse, so that you had a version of classical mechanics.
Assuming a range of different values, assigning probabilities to
individual outcomes makes some sort of sense, but that is assuming a
lot of quantum mechanics at the start. Hence my worries about
circularity.


If there is only a single outcome, then the projection concerned
will be the certain event projection. With probability 1. That case is covered.


A more serious consideration occurred to me -- the projection might have 
a range of possible outcomes, but these might be merely an illustration 
of classical probability arising from ignorance. In quantum terms, you 
have to distinguish between pure and mixed states. You appear to assume 
that your projections are from a pure state, but that is circular: you 
can only have pure states in that sense in a quantum theory. Pure states 
can exhibit interference, mixed states cannot.




S=X+P is definitely an observable, and corresponds to measuring the sum
of position and momentum crisply. You can work out the equivalent
Heisenberg uncertainty relations too
ΔsΔx ≥ ℏ and ΔsΔp ≥ ℏ
So if you measure with S, you will have uncertainty in both position
and momentum.

So it is not a quantum operator or quantum observable. These are
defined to have precise values: eigenvalues corresponding to the
eigenfuntions of the Hermitian operator. And the eigenfunctions span
the corresponding Hilbert space.

I don't get your point. The operator S=X+P satisfies all of those.


No, position and momentum are dual in the sense I defined. The 
observables are not compatible -- position and momentum are not 
simultaneously observable.



My point here is that position
space, spanned by the eigenfunctions of the position operator, is
'dual' to momentum space, spanned by the eigenfunctions of the
momentum operator. These spaces are related by the Fourier transform
and each serves to give a distinct complete representation of the
underlying quantum state. (That is the sense in which I refer to
these operators and spaces as 'dual' -- it makes no quantum sense to
add them,)


Operators operate on the Hilbert state space, not position or momentum
space, which is more of classical mechanics concept. And yes, one can
add them (modulo an assumed unit conversion constant) mathematically,
and the sum X+P is meaningful physically.


Classically perhaps, but not in quantum mechanics.


In general, any bounded hermitian operator is an observable of some
sort (one can create a machine, albeit weirdly wonderful in a Heath
Robisonesque way, that will measure that particular quantity).

Certainly. If one has the Hilbert space for some quantum operator,
linear combinations of the eigenvector basis vectors that also span
the space will be the eigenvalues of some other hermitian operator
in the space. As you say, not all of the resultant operators
correspond to natural observables.

I don't say that. You're saying that, and I fail to see your point.


It is really the (dead-cat +- live-cat basis) issue. A superposition of 
polarization states becomes entangled with a cat in Schrödinger's 
thought experiment. We don't observe the superposition because the (dead 
+- alive) basis does not correspond to a natural observable -- 
decoherence sees to that. Similarly for most other quantum variables. 
The normal position basis, delta(x-x_i), can be superposed to give some 
alternative basis, and there will be a corresponding operator on the 
Hilbert space, but superpositions of position results are not given by 
any achievable measurement -- again, decoherence leads to the preferred 
stable basis: all  other bases do not give results that are stable under 
decoherence. Not all possible operators are possible measurements (ie, 
not all operators give stable results).



The sum of two bounded hermitian operators is also a bounded hermition
operator. In fact a linear combination with real coefficients will
too, but not necessarily complex coeficients (since the hermitian
property may not be preserved).

Yes, but my concern here is that this generality does not extend to
tensor products of 

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

[Russell Standish]

On Thu, Jul 06, 2017 at 04:18:49PM +1000, Bruce Kellett wrote:
> On 6/07/2017 2:33 pm, Russell Standish wrote:
> >Establishing linearity is key.
> 
> Yes, and you haven't made progress with that.

All I ask is to give me some more time on this. I have some further
ideas in this regard, but need some dedicated time to think about it.

> 
> >Establishing the resultant vector space
> >is a Hilbert space does follow fairly easily from Kolmogorov's axioms
> >(although its possible you have beef with those :).
> 
> I think the issue I would have here is that you assume that a
> projection can give a range of results. I don't think that is
> necessarily true even in a multiverse. Since projections are as
> undefined as everything else here, it could be the case that a
> projection gave a single value -- the same value in every world of
> the multiverse, so that you had a version of classical mechanics.
> Assuming a range of different values, assigning probabilities to
> individual outcomes makes some sort of sense, but that is assuming a
> lot of quantum mechanics at the start. Hence my worries about
> circularity.
> 

If there is only a single outcome, then the projection concerned
will be the certain event projection. With probability 1. That case is covered.

> >
> >S=X+P is definitely an observable, and corresponds to measuring the sum
> >of position and momentum crisply. You can work out the equivalent
> >Heisenberg uncertainty relations too
> >ΔsΔx ≥ ℏ and ΔsΔp ≥ ℏ
> >So if you measure with S, you will have uncertainty in both position
> >and momentum.
> 
> So it is not a quantum operator or quantum observable. These are
> defined to have precise values: eigenvalues corresponding to the
> eigenfuntions of the Hermitian operator. And the eigenfunctions span
> the corresponding Hilbert space.

I don't get your point. The operator S=X+P satisfies all of those.

> My point here is that position
> space, spanned by the eigenfunctions of the position operator, is
> 'dual' to momentum space, spanned by the eigenfunctions of the
> momentum operator. These spaces are related by the Fourier transform
> and each serves to give a distinct complete representation of the
> underlying quantum state. (That is the sense in which I refer to
> these operators and spaces as 'dual' -- it makes no quantum sense to
> add them,)
> 

Operators operate on the Hilbert state space, not position or momentum
space, which is more of classical mechanics concept. And yes, one can
add them (modulo an assumed unit conversion constant) mathematically,
and the sum X+P is meaningful physically.

> >In general, any bounded hermitian operator is an observable of some
> >sort (one can create a machine, albeit weirdly wonderful in a Heath
> >Robisonesque way, that will measure that particular quantity).
> 
> Certainly. If one has the Hilbert space for some quantum operator,
> linear combinations of the eigenvector basis vectors that also span
> the space will be the eigenvalues of some other hermitian operator
> in the space. As you say, not all of the resultant operators
> correspond to natural observables.

I don't say that. You're saying that, and I fail to see your point.

> 
> >The sum of two bounded hermitian operators is also a bounded hermition
> >operator. In fact a linear combination with real coefficients will
> >too, but not necessarily complex coeficients (since the hermitian
> >property may not be preserved).
> 
> Yes, but my concern here is that this generality does not extend to
> tensor products of Hilbert spaces, and the product space is what one
> has in quantum mechanics -- projections in one space are not
> projections in another of the component spaces. You have also to
> distinguish carefully between compatible and incompatible
> observations -- commuting and non-commuting observables
> (projections). I don't see how you could establish this distinction
> in your approach without just assuming that it exists.
> 

Can you give an example of a tensor product where the summation
failed? The way I see it, if observables A and B operated in separate
subspaces, AA and BB (think say X and Y positions), then one can
extend the operators A and B to the product space AA⊗BB by making A a
constant along all axes of BB and vice-versa for B. In matrix
terminology, we create A' by copying A all down the diagonal,
similarly with B:

  /a11 a12 a13\
  |a21 a22 a23|
  |a31 a32 a33|
  |a11 a12 a13|
A' =  |a21 a22 a23|
  |a31 a32 a33|
  |a11 a12 a13|
  |a21 a22 a23|
  \a21 a32 a33/


  /b11 b12 b13\
  |b11 b12 b13|
  |b11 b12 b13|
  |b21 b22 b23|
B' =  |b21 b22 b23|
  

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

[Bruce Kellett]

On 6/07/2017 2:33 pm, Russell Standish wrote:

Sorry for having gone dark, although maybe you relished the
respite. I've been travelling, and its not been all that convenient to
check and respond to emails.

On Sat, Jul 01, 2017 at 02:56:58PM +1000, Bruce Kellett wrote:

On 1/07/2017 11:18 am, Russell Standish wrote:
I'm not asking for your sympathy. Actually, it is helpful if you 
aren't sympathetic, as you're more likely to find a critical flaw. 
Obviously, you need to be motivated enough to poke around in it, though. 

Maybe you are leaning on the fact that if A and B are projections,
then aA+bB is not in general a projection, since idempotency is not preserved?

That could be an issue, but it is not the main issue here. I think
the concentration on projections and events/observations/outcomes is
probably a mistake. You start with the concept of an observer moment
as a set of possibilities consistent with what is known at time t.
The elements of this set are infinite strings of bits encoding the
information and possible continuations. This is fair enough, I
suppose. But if you want to make contact with ordinary QM, you have
to see this psi(t) essentially as a wave function (or equivalent).
So it is this OM that is to be interpreted as a vector or ray in
some space, and you have to establish that this is a linear space,
with an defined inner product, so it is a Hilbert space.


Establishing linearity is key.


Yes, and you haven't made progress with that.


Establishing the resultant vector space
is a Hilbert space does follow fairly easily from Kolmogorov's axioms
(although its possible you have beef with those :).


I think the issue I would have here is that you assume that a projection 
can give a range of results. I don't think that is necessarily true even 
in a multiverse. Since projections are as undefined as everything else 
here, it could be the case that a projection gave a single value -- the 
same value in every world of the multiverse, so that you had a version 
of classical mechanics. Assuming a range of different values, assigning 
probabilities to individual outcomes makes some sort of sense, but that 
is assuming a lot of quantum mechanics at the start. Hence my worries 
about circularity.





In this endeavour, concentration on projections as measurements is
not actually helpful. These projections would correspond to
operators on the vector space of OMs, and the existence and/or
actions of operators is not actually going to help with establishing
linearity.

Observations are projections (in the general sense, not necessarily
linear space sense). This is why I focus on the projection operator
form of QM.


The point here is that the action of an operator on a
wave function does not actually change the wave function, it just
gives a set of eigenfunctions in terms of which the original wave
function can be expanded.

That is the case of a traditional observable - a full rank Hermitian
operator.

Going from a projection operator formalism to the traditional observer
formulation is mathematically quite trivial, however.


However, I don't think that's what I'm relying on. Given a starting
vector ψ, the vector (αA+βB)ψ is going the the result of some
projection C, where Cψ=(αA+βB)ψ, modulo an arbitrary complex
factor, and so (αA+βB)ψ is still an observer moment.

Well, that might be the case for some linear operators, but it is
not generally the case. One of the significant ways in which QM
differs from classical mechanics is that some observations are
mutually exclusive - the corresponding operators do not commute, so
adding them does not result in another possible operator: (x + p) is
not an operator in either x-space or p-space. In formal developments
of QM, such as Landau and Lifschitz, or von Neumann, a lot of care
is taken to distinguish between compatible and incompatible
observations (commuting and non-commuting operators). I do not see
how this could be incorporated in your approach -- just like the
question of tensor product Hilbert spaces for different (commuting)
operators.


S=X+P is definitely an observable, and corresponds to measuring the sum
of position and momentum crisply. You can work out the equivalent
Heisenberg uncertainty relations too
ΔsΔx ≥ ℏ and ΔsΔp ≥ ℏ
So if you measure with S, you will have uncertainty in both position
and momentum.


So it is not a quantum operator or quantum observable. These are defined 
to have precise values: eigenvalues corresponding to the eigenfuntions 
of the Hermitian operator. And the eigenfunctions span the corresponding 
Hilbert space. My point here is that position space, spanned by the 
eigenfunctions of the position operator, is 'dual' to momentum space, 
spanned by the eigenfunctions of the momentum operator. These spaces are 
related by the Fourier transform and each serves to give a distinct 
complete representation of the underlying quantum state. (That is the 
sense in which I refer to these operators and spaces as 'dual' 

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

[Russell Standish]

Sorry for having gone dark, although maybe you relished the
respite. I've been travelling, and its not been all that convenient to
check and respond to emails.

On Sat, Jul 01, 2017 at 02:56:58PM +1000, Bruce Kellett wrote:
> On 1/07/2017 11:18 am, Russell Standish wrote:
> >To summarise, you are happy with everything up to (D.6), but think the
> >move to the linear superposition (D.7) is not justified, because you
> >say it cannot be an observer moment.
> 
> I would not say that I am happy with everything up to (D.7)! I find
> the notation confusing, and some of the manipulations do not seem to
> make operational sense. Your comments through these exchanges have
> certainly helped me to see what is going on but, as you know, I am
> fairly out of sympathy with the general approach anyway

I'm not asking for your sympathy. Actually, it is helpful if you
aren't sympathetic, as you're more likely to find a critical
flaw. Obviously, you need to be motivated enough to poke around in it, though.

> 
> >Maybe you are leaning on the fact that if A and B are projections,
> >then aA+bB is not in general a projection, since idempotency is not 
> >preserved?
> 
> That could be an issue, but it is not the main issue here. I think
> the concentration on projections and events/observations/outcomes is
> probably a mistake. You start with the concept of an observer moment
> as a set of possibilities consistent with what is known at time t.
> The elements of this set are infinite strings of bits encoding the
> information and possible continuations. This is fair enough, I
> suppose. But if you want to make contact with ordinary QM, you have
> to see this psi(t) essentially as a wave function (or equivalent).
> So it is this OM that is to be interpreted as a vector or ray in
> some space, and you have to establish that this is a linear space,
> with an defined inner product, so it is a Hilbert space.
> 

Establishing linearity is key. Establishing the resultant vector space
is a Hilbert space does follow fairly easily from Kolmogorov's axioms
(although its possible you have beef with those :).

> In this endeavour, concentration on projections as measurements is
> not actually helpful. These projections would correspond to
> operators on the vector space of OMs, and the existence and/or
> actions of operators is not actually going to help with establishing
> linearity. 

Observations are projections (in the general sense, not necessarily
linear space sense). This is why I focus on the projection operator
form of QM.

> The point here is that the action of an operator on a
> wave function does not actually change the wave function, it just
> gives a set of eigenfunctions in terms of which the original wave
> function can be expanded.

That is the case of a traditional observable - a full rank Hermitian
operator.

Going from a projection operator formalism to the traditional observer
formulation is mathematically quite trivial, however.

> 
> >However, I don't think that's what I'm relying on. Given a starting
> >vector ψ, the vector (αA+βB)ψ is going the the result of some
> >projection C, where Cψ=(αA+βB)ψ, modulo an arbitrary complex
> >factor, and so (αA+βB)ψ is still an observer moment.
> 
> Well, that might be the case for some linear operators, but it is
> not generally the case. One of the significant ways in which QM
> differs from classical mechanics is that some observations are
> mutually exclusive - the corresponding operators do not commute, so
> adding them does not result in another possible operator: (x + p) is
> not an operator in either x-space or p-space. In formal developments
> of QM, such as Landau and Lifschitz, or von Neumann, a lot of care
> is taken to distinguish between compatible and incompatible
> observations (commuting and non-commuting operators). I do not see
> how this could be incorporated in your approach -- just like the
> question of tensor product Hilbert spaces for different (commuting)
> operators.


S=X+P is definitely an observable, and corresponds to measuring the sum
of position and momentum crisply. You can work out the equivalent
Heisenberg uncertainty relations too
ΔsΔx ≥ ℏ and ΔsΔp ≥ ℏ
So if you measure with S, you will have uncertainty in both position
and momentum.

In general, any bounded hermitian operator is an observable of some
sort (one can create a machine, albeit weirdly wonderful in a Heath
Robisonesque way, that will measure that particular quantity).

The sum of two bounded hermitian operators is also a bounded hermition
operator. In fact a linear combination with real coefficients will
too, but not necessarily complex coeficients (since the hermitian
property may not be preserved).

All bets are off with unbounded operators, of course, but my attitude
is that unbounded operators are stictly unphysical, albeit sometimes
convenient for computational purposes.

> 
> 
> >Thinking along those lines some more, I'm incorrect to say that the
> >vector space V is 

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

[Bruce Kellett]

On 1/07/2017 11:18 am, Russell Standish wrote:

On Fri, Jun 30, 2017 at 04:57:00PM +1000, Bruce Kellett wrote:

In other words, observer moments are not vectors (or rays) in a
linear vector space because a linear superposition of observer
moments is not another observer moment:  your derivation of Hilbert
space fails.

Bruce


To summarise, you are happy with everything up to (D.6), but think the
move to the linear superposition (D.7) is not justified, because you
say it cannot be an observer moment.


I would not say that I am happy with everything up to (D.7)! I find the 
notation confusing, and some of the manipulations do not seem to make 
operational sense. Your comments through these exchanges have certainly 
helped me to see what is going on but, as you know, I am fairly out of 
sympathy with the general approach anyway



Maybe you are leaning on the fact that if A and B are projections,
then aA+bB is not in general a projection, since idempotency is not preserved?


That could be an issue, but it is not the main issue here. I think the 
concentration on projections and events/observations/outcomes is 
probably a mistake. You start with the concept of an observer moment as 
a set of possibilities consistent with what is known at time t. The 
elements of this set are infinite strings of bits encoding the 
information and possible continuations. This is fair enough, I suppose. 
But if you want to make contact with ordinary QM, you have to see this 
psi(t) essentially as a wave function (or equivalent). So it is this OM 
that is to be interpreted as a vector or ray in some space, and you have 
to establish that this is a linear space, with an defined inner product, 
so it is a Hilbert space.


In this endeavour, concentration on projections as measurements is not 
actually helpful. These projections would correspond to operators on the 
vector space of OMs, and the existence and/or actions of operators is 
not actually going to help with establishing linearity. The point here 
is that the action of an operator on a wave function does not actually 
change the wave function, it just gives a set of eigenfunctions in terms 
of which the original wave function can be expanded. Particular results 
can then be projected out of the expansion by saying that an observation 
corresponds to a particular eigenfunction, and the measurement result is 
the corresponding eigenvalue. In your notation, the action of the 
operator seems to be \P_A*psi = \Sigma_a \P_{a}*psi, where \P_{a} is the 
projection of a particular result. Since applying the operator is 
\P_A*psi, the result is still an observer moment -- in fact, the 
original observer moment in some sense. The projections of individual 
results, \P_{a}*psi, take one branch of the total wave function, but 
this is still a wave function, and also still a possible observer 
moment. So taking projections of OMs leads to further OMs, and it is the 
linearity of these OMs that is the issue. If they are vectors in a 
linear space, the the operators or projections act in a linear space, 
and their linearity is a separate issue.


In order to establish that OMs are vectors in a linear space, you have 
to show that linear combinations of OMs are just further OMs.



However, I don't think that's what I'm relying on. Given a starting
vector ψ, the vector (αA+βB)ψ is going the the result of some
projection C, where Cψ=(αA+βB)ψ, modulo an arbitrary complex
factor, and so (αA+βB)ψ is still an observer moment.


Well, that might be the case for some linear operators, but it is not 
generally the case. One of the significant ways in which QM differs from 
classical mechanics is that some observations are mutually exclusive - 
the corresponding operators do not commute, so adding them does not 
result in another possible operator: (x + p) is not an operator in 
either x-space or p-space. In formal developments of QM, such as Landau 
and Lifschitz, or von Neumann, a lot of care is taken to distinguish 
between compatible and incompatible observations (commuting and 
non-commuting operators). I do not see how this could be incorporated in 
your approach -- just like the question of tensor product Hilbert spaces 
for different (commuting) operators.




Thinking along those lines some more, I'm incorrect to say that the
vector space V is the set of all observer moments. It must be the set
of all successor observer moments to ψ, or all continuations as I
think you put it earlier. The clue lies in the linear span (D.8). OMs
that can't be reached from ψ just simply cannot be put in the linear
span of outcomes from an observation on ψ.


Any observer moment is the successor of some previous observer moment, 
so concentrating on successor OMs of psi achieves nothing.




I need to think some more about what a linear superposition of
observers actually means in terms of selecting out subsets of infinite
strings, which is the original model I proposed. It may help resolve
the 

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

[Russell Standish]

On Fri, Jun 30, 2017 at 04:57:00PM +1000, Bruce Kellett wrote:
> 
> In other words, observer moments are not vectors (or rays) in a
> linear vector space because a linear superposition of observer
> moments is not another observer moment:  your derivation of Hilbert
> space fails.
> 
> Bruce
> 

To summarise, you are happy with everything up to (D.6), but think the
move to the linear superposition (D.7) is not justified, because you
say it cannot be an observer moment.

Maybe you are leaning on the fact that if A and B are projections,
then aA+bB is not in general a projection, since idempotency is not preserved?

However, I don't think that's what I'm relying on. Given a starting
vector ψ, the vector (αA+βB)ψ is going the the result of some
projection C, where Cψ=(αA+βB)ψ, modulo an arbitrary complex
factor, and so (αA+βB)ψ is still an observer moment.

Thinking along those lines some more, I'm incorrect to say that the
vector space V is the set of all observer moments. It must be the set
of all successor observer moments to ψ, or all continuations as I
think you put it earlier. The clue lies in the linear span (D.8). OMs
that can't be reached from ψ just simply cannot be put in the linear
span of outcomes from an observation on ψ.

I need to think some more about what a linear superposition of
observers actually means in terms of selecting out subsets of infinite
strings, which is the original model I proposed. It may help resolve
the observer measure issue.

Would it be fair to say that the remainder of the appendix follows
through once linearity is demonstrated? It seems fairly
uncontroversial to me (aside from the assumption of continuous time,
which for me is a necessary ad-hoc assumption to make contact with
regular QM).

-- 


Dr Russell StandishPhone 0425 253119 (mobile)
Principal, High Performance Coders
Visiting Senior Research Fellowhpco...@hpcoders.com.au
Economics, Kingston University http://www.hpcoders.com.au


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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:

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

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 to the
observer moment prior to observation, ie indicating an ignorant OM,
but a\P_{A}ψ+b\P_{B}ψ in equation (D.7) cannot refer to a single
observer moment, but rather an aggregate OM. It seems to me to
describe a observers in OM ψ choosing to measure on partition A,¬A and
b observers choosing to measure on partition B,¬B.


In other words, observer moments are not vectors (or rays) in a linear 
vector space because a linear superposition of observer moments is not 
another observer moment:  your derivation of Hilbert space fails.


Bruce

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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: “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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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 <bhkell...@optusnet.com.au>
To: everything-list <everything-list@googlegroups.com>
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 o

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: “Could a Quantum Computer Have Subjective Experience?”

[Russell Standish]

On Tue, Jun 27, 2017 at 08:52:15AM +1000, Bruce Kellett wrote:
> On 26/06/2017 3:57 pm, Russell Standish wrote:
> >On Mon, Jun 26, 2017 at 11:50:45AM +1000, Bruce Kellett wrote:
> >>That is not what is normally meant by the '+' symbol. You have
> >>simply defined a conjunction to be a disjunction!
> >We are constructively defining +. I would not be so cruel as to use +
> >if the end point were not the usual group operation.
> 
> Yes, the endpoint is that the '+' is simple addition. It seems to me
> that if you actually wrote
> 
> psi_a v psi_b,
> 
> where 'v' stands for disjunction, or 'or', you would not have got
> very far with your derivation. By writing the sum
> 
>psi_a + psi_b = psi_{ab}
> 
> you have, in fact, simply assumed linearity.

Not at all. The latter equation is identical to the first with the
symbol v replaced by +.

> A significant property
> of linear systems is that if you have two solutions, the sum is also
> a solution. If you are dealing with sets, the the operation is the
> union of sets, which is different. But you specifically state that
> your projection operator acting on the ensemble produces a single
> outcome psi_a = \P_{a}*psi, so you are dealing with addition of
> numbers or functions, not the union of sets.
> 

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.

> 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. 


-- 


Dr Russell StandishPhone 0425 253119 (mobile)
Principal, High Performance Coders
Visiting Senior Research Fellowhpco...@hpcoders.com.au
Economics, Kingston University http://www.hpcoders.com.au


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

[spudboy100 via Everything List]

For me, what needs to be established, is can we ever 'transact' with sister 
universes, in a true way? Being nicely pessimistic for the moment, let us shout 
"Hell No!" Thus, the discussion becomes wonderful for readers of fantasy and 
scifi, but unimportant as in boring, for people who want to see science and the 
philosophy of science, mathematics, improve human life. Physicist David Deutsch 
at Oxford, has long proposed that quantum computing actions would flip and flop 
back and forth between 1+ more universes. Better, would be trade with other 
universes with information and goods. I wish would could do this in the Galaxy, 
but unless ya like Tabby's Star, things appear dead as a door nail. 



-Original Message-
From: John Clark <johnkcl...@gmail.com>
To: everything-list <everything-list@googlegroups.com>
Sent: Mon, Jun 26, 2017 1:23 pm
Subject: Re: “Could a Quantum Computer Have Subjective Experience?”



On Mon, Jun 26, 2017 at 1:57 AM, Russell Standish <li...@hpcoders.com.au> 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. 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.   ​


 John K Clark




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

[Bruce Kellett]

On 26/06/2017 3:57 pm, Russell Standish wrote:

On Mon, Jun 26, 2017 at 11:50:45AM +1000, Bruce Kellett wrote:

That is not what is normally meant by the '+' symbol. You have
simply defined a conjunction to be a disjunction!

We are constructively defining +. I would not be so cruel as to use +
if the end point were not the usual group operation.


Yes, the endpoint is that the '+' is simple addition. It seems to me 
that if you actually wrote


psi_a v psi_b,

where 'v' stands for disjunction, or 'or', you would not have got very 
far with your derivation. By writing the sum


   psi_a + psi_b = psi_{ab}

you have, in fact, simply assumed linearity. A significant property of 
linear systems is that if you have two solutions, the sum is also a 
solution. If you are dealing with sets, the the operation is the union 
of sets, which is different. But you specifically state that your 
projection operator acting on the ensemble produces a single outcome 
psi_a = \P_{a}*psi, so you are dealing with addition of numbers or 
functions, not the union of sets.


Thus, for the sum to make sense you must assume linearity. 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.


Bruce

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

[Bruno Marchal]

On 26 Jun 2017, at 03:50, Bruce Kellett wrote:


On 26/06/2017 2:14 am, Bruno Marchal wrote:


Keep in mind that to refute Mechanism (in cognitive science), it is
not enough to show that a piece of matter is not Turing emulable. You
need mainly to show that its behavior is not retrievable from the
statistics of the first person indeterminacy on all computations.


I do not accept this reversal of the burden of proof. I do not have  
to prove a universal negative: you have to prove the positive by  
actually deriving physics from the statistics of the first person  
indeterminacy on all computations.


I was just saying that to refute computationalism by invoking infinite  
matters does not work, as computationalism predicts infinities in the  
material domain. You need to show that there are *special* infinities,  
not recovered by the global First Person Indeterminacy. It is a not a  
reversal of the burden of the proof, it is just a question of validity.


If not that would beg the question (as shown by the UDA). You give a  
criteria of verification, which I accept and indeed, if nature violate  
Z1*, we will know that the classical computationalist theory of mind  
and matter of the computationalist universal (Löbian) machine is  
wrong. But until today it is rather confirmed (even in the startling  
aspects), and it is to my knowledge the only clear account of the 1p  
relation with the 3p relations (measurable or not).


The reversal of the burden of the proof is the main basic first  
result. It shows that using an universal extrapolation of the physical  
laws from a finite number of number measurements cannot be invoked to  
prevent the need to address the infinite renormalization procedure  
that the arithmetical reality provides when seen from the (true,  
consistent, provable, and conjunctions) possible points of view of the  
universal numbers.


I do not criticize the theory which would assume a primitive physical  
reality. I criticized its misuse in the "philosophy of mind",  
especially when both a (primitive) matter and mechanism are assumed.


With mechanism, we need to make *infinite sum* on the histories  
(computations 1p-filtrated), the reason why we get "negative  
probabilities" might be related to the fact that 1+2+3+4+5+... "="   
-1/12.


It is just because I am not enough competent in algebra that I am  
unable to make clear the general "Galois theory" of (mechanical)  
consciousness. Consciousness is on the side of truth, and semantics,  
and meanings, and thus of models. Like in number equation, the more  
you have equation, resp. axioms, the less you have solutions, resp.  
models (in the mathematical sense of the logicians).
So I still don't know if (assuming mechanism) consciousness increase  
or decrease with the number of neurons, the 1p spectrum grow, and  
consciousness is related to relative "spectrum anticipation".


Around 2000, I read the book by William Seager, which encourages me in  
the belief that in the Theaetetus "[]p & p" (I believe p & it is the  
case that p), the consciousness/knowledge still relied in the key role  
of the machine's body/representation/3p encapsulated in the box []p,  
or Bp, (Gödel's provability predicate). But today, I think  
consciousness is more on the side of "p" or "Dt". It is not unrelated  
with Brent's insistence to call up an environment. The machine  
"understand" this already (arguably, in some precise technical sense  
using the many modal logics of "self" reference). My current theories,  
written in G, is that consciousness is Dt v t. D is the diamond of G  
(consistency) and Dp v p is the diamond of S4Grz(1). A very good book,  
btw, that William Seager (even if physicalist by default).


You might need to read/study at least Smullyan Forever undecided, and  
well, also "To mock a Mocking Bird" to study G, G*, S4Grz1, X1*, etc.  
Or better, Mendelson and/or Boolos' 1979 book.


Bruno





Bruce

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



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

[John Clark]

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. 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.   ​

 John K Clark

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

[Russell Standish]

On Mon, Jun 26, 2017 at 11:50:45AM +1000, Bruce Kellett wrote:
> 
> You mean your statement about the variation upon which anthropic
> selection acts? Does this mean that the continuations that are
> anthropically allowed are those the permit the observer's continued
> existence? Or is something more implied?

At this point in the argument, there is no discussion of
continuations, interesting though that discussion is. We're just
talking about an observation that takes place in the context of an
Observer Moment. So we should just park that for now.

> 
> That is not what is normally meant by the '+' symbol. You have
> simply defined a conjunction to be a disjunction!

We are constructively defining +. I would not be so cruel as to use +
if the end point were not the usual group operation.

> 
> Bur that does not work for the equations immediately following,
> where you simply sum over all the possible outcomes of the operator
> 'A'.

All the outcomes of A are disjoint. A is a partition, by definition.

>  Why use the symbol \Sigma if you mean a disjunction? And you
> then go on to say that if the outcome is a continuous set, you
> replace the sum by an integral with uniform measure. It is difficult
> to avoid the conclusion that you do actually mean that '+' implies
> standard addition.

Not yet at this point.

> 
> Since psi_a is in general just a set of continuations selected by
> the operation 'A', it is by no means clear that such a summation has
> any meaning in general.

A is any partition of the observer moment. + (D.1) is perfectly well
defined over any such partition.

>  Of course, there may be some sets that can
> sensibly be summed, but that does not seem a reasonable proposition
> for sets of possible continuations such as the one I have given in
> terms of cats and dogs, walking, and talking.
> 

Again, why not? The sum simply corresponds to the observation that
either of the distinct observation hold. Assuming, for the sake of the
argument that walking the dog is mutually exlusive with stroking the cat.

> In order for the development you outline to make sense, you would
> have to specify the operator 'A' in a lot more detail, so that it
> only selected things over which summation was meaningful. IOW, you
> actually want 'A' to be a measurement of a quantum state. And you
> specifically want a quantum state, because you want there to be more
> than one possible result for the measurement 'A'. If 'A' is a
> classical measurement of position, for example, then there is only
> one possible outcome, and your further development of the situation
> becomes trivial, not giving you quantum mechanics at all.
> 

Not so fast. We're situated in a Multiverse, not a class Universe, so
talking about an two identical copies seeing different disjoint
outcomes is a perfectly reasonable thing to want to describe.

> It seems, therefore, that in order to get quantum mechanics out, you
> have to essentially assume quantum mechanics right from the start.
> You have, at least, to assume that there is variability in the
> results of operation 'A', and that this variability can sensibly be
> superposed.
> 

Of course. That is reasonable supposition for a Multiverse. The things
not assumed are linearity, Hilbert spaces, Born's probability rule,
and unitarity, all of which are normally assumed axiomatically in QM,
but is derived in this treatment.

> >>It is clear that your are trying to introduce the concept of a
> >>quantum superposition by the back door, without doing any work, and
> >>relying on the inherent ambiguity in the '+' operation.
> >Why is this ambiguous?
> 
> Se the above. Does '+' mean 'or', or not?

No it does not. In particular it will differ markedly from it when non
disjoint events are considered.

> 
> >>  If you have
> >>nothing but classical outcomes from your observer moment psi(t),
> >>then you cannot simply add these outcomes as if they were separate
> >>eigenfunctions of a quantum operator. There are no such things as
> >>superpositions in classical physics.
> >There are, it's just that they're not particularly interesting from a
> >physical point of view.
> >
> >A superposition of a blue ball and a red ball just means that we don't
> >know what colour the ball is.
> 
> That is a matter of ignorance, not a superposition of different colours.
> 

In the Multiverse, it is also a superposition. Before a measurement is
made, I am both the person who observes a blue ball, and the one who
observes a red ball. When we finally get around to the full vector
space version, the relevant state vector is a linear superposition of
both states. Since it is headed that way, then why not call the state
a superposition, even before linearity is proved. Just like it is
reasonable to use the symbol +, even though I have only defined (at
that time) on a subset of its full range.

> >>Sorry, but the whole procedure is nonsense on stilts. It does not
> >>get any better from then on in, but I refrain from 

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

[Bruce Kellett]

On 25/06/2017 9:25 pm, Russell Standish wrote:

On Sun, Jun 25, 2017 at 04:25:07PM +1000, Bruce Kellett wrote:

On 24/06/2017 8:36 pm, Russell Standish wrote:

On Sat, Jun 24, 2017 at 06:29:54PM +1000, Bruce Kellett wrote:

On 24/06/2017 5:23 pm, Russell Standish wrote:

OK, it was possibly the case that you gave arguments earlier in the
book. But I was going on the basis of the Appendix "Derivation of
Quantum postulates".

But the problems only begin with the assumption of a probabilistic
model. Psi(t) is the set of possibilities consistent with what is
known at time t. But how do you limit this set? At the moment, I
could go to the pub for a drink, could open a bottle of wine at
home, stroke the cat, turn on the telly, talk to my wife, etc,
etc,. The possibilities consistent with what is known at this
time is not a well defined set, or limited in any way.

The everything is the set of all infinite length strings, each of
which describes a universe to infinite detail. Some of these strings
will describe universes compatible with our current observer moment -
an infinite number even, as the information content of our OM is
finite. Others will not. It is a well defined subset of the everything.

What does "compatible" mean? Is this linked to our current moment by
law-like behaviour, or just any any string which happens, by chance
to contain our present moment? If the latter (and also in compete
generality, given your definitions), then my characterization of the
list of possibilities is correct, and the string could contain any
future indeed. If the former, how do you know that there is not just
a single string that contains the only law-like continuation of your
present state?

There is an interpreter of the strings. In Solomonoff-Levin's theory,
the interpreter is some given universal machine U, but in this case,
the only possible interpreter is an observer of the system.

Compatible just means that the string is interpreted as giving the
current observer moment by the observer.

At this point, there is no theory about what likely successor strings
might be. The most obvious model (ie just read some more bits of a
given string), doesn't work.

What I propose in the appendix D work is to apply an evolutionary
process to it.


You mean your statement about the variation upon which anthropic 
selection acts? Does this mean that the continuations that are 
anthropically allowed are those the permit the observer's continued 
existence? Or is something more implied?




Because you then go on to define projection operators in terms of a
sum over the members of this set of possible outcomes. That is
meaningless unless you are already assuming the the outcomes are
just possible results for a well-defined measurement, and that this
measurement process can be defined in a linear vector space.


Summation of the projection operators is defined in equation D.1 for
disjoint observations a and b (ie where it is impossible to observe a
and b simultaneously). Linearity is not assumed at this point.

That is where you have an enormous problem, following on from the
previous point that you have not limited the possible continuations
in any way. You define a projection as occurring when the observer
applies an operator A (again undefined and unlimited) to the
observer moment, which operator divides it into a discrete set of
outcomes, psi_a. Note that 'a' is only an index, not an eigenvalue
or any such. Note that you explicitly state that \P_{a} is not
assumed linear at this point. (I use \P as a notation for your
script P.) You then *define* addition for two distinct outcomes a
=/= b as:

 \P_{a} psi  + \P_{b} psi = \P_{a,b} psi.

This is, of course meaningless if psi_a is 'taking the dog for a
walk,' and psi_b is 'stroking the cat'. You can define a '+' sign as
anything you like, but such a definition does not ensure that the
result has any meaning -- as in my example, which follows completely
logically from the definitions that you make.


Note that {a,b} is just {a} ∪ {b}, so that \P_{a,b}ψ is is just the
observation "taking the dog for a walk _or_ stroking the cat". That is
a perfectly acceptable classical observation.


That is not what is normally meant by the '+' symbol. You have simply 
defined a conjunction to be a disjunction!


Bur that does not work for the equations immediately following, where 
you simply sum over all the possible outcomes of the operator 'A'. Why 
use the symbol \Sigma if you mean a disjunction? And you then go on to 
say that if the outcome is a continuous set, you replace the sum by an 
integral with uniform measure. It is difficult to avoid the conclusion 
that you do actually mean that '+' implies standard addition.


Since psi_a is in general just a set of continuations selected by the 
operation 'A', it is by no means clear that such a summation has any 
meaning in general. Of course, there may be some sets that can sensibly 
be summed, but that does not seem a reasonable 

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

[Bruce Kellett]

On 26/06/2017 2:14 am, Bruno Marchal wrote:


Keep in mind that to refute Mechanism (in cognitive science), it is
not enough to show that a piece of matter is not Turing emulable. You
need mainly to show that its behavior is not retrievable from the
statistics of the first person indeterminacy on all computations.


I do not accept this reversal of the burden of proof. I do not have to 
prove a universal negative: you have to prove the positive by actually 
deriving physics from the statistics of the first person indeterminacy 
on all computations.


Bruce

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

[Bruno Marchal]

On 23 Jun 2017, at 03:20, Bruce Kellett wrote:


On 22/06/2017 7:22 pm, Bruno Marchal wrote:

On 22 Jun 2017, at 01:31, Bruce Kellett wrote:

On 22/06/2017 1:44 am, Bruno Marchal wrote:

On 21 Jun 2017, at 08:21, Bruce Kellett wrote:

On 21/06/2017 4:03 pm, Russell Standish wrote:

On Mon, Jun 19, 2017 at 12:15:31PM +1000, Bruce Kellett wrote:

On 19/06/2017 10:23 am, Russell Standish wrote:
I know Scott wouldn't go as far as me. For me, all such  
irreversible

processes are related to conscious entities in some way. Whilst
agreeing that Geiger counters are unlikely to be conscious, I  
would
say that the output of Geiger counter is not actually  
discrete until

observed by a conscious experimenter.

That sounds remarkably like the "many minds" interpretation of
quantum mechanics. This is disfavoured by most scientists  
because it
leaves the physics of the billions of years before the  
emergence of
the first "conscious" creature unresolved -- the first  
consciousness

would cause an almighty collapse on the many minds reading.


Each consciousness causes "an almighty collapse" in er own mind
independently of any other. It's a pure 1p phenomena.


It is actually a 3p phenomenon because there is inter-subjective  
agreement about the fact that measurements give definite results.


Inter-subjectivity does not imply 3p, as it can be "only" 1p  
plural. Let me illustrate this with a variant of the WM  
duplication.


Imagine that Bruce and John are undergoing the WM-duplication  
*together*.


By this I mean they both enter the scanning-annihilating box, and  
are both reconstituted in Washington and in Moscow.


And let us assume they do it repetitively, which means they come  
back to Helsinki, and do it again together.


Obviously, the line-life past that each copies describes in its  
personal diaries grows like H followed by a sequence of W and M.  
The number of copies grows exponentially (2^n). After ten  
iterations, we have 2^10 = 1024 individuals, who share an  
indeterminate experiences. With minor exceptions, they all agree  
that the experience has always given each times a precise  
outcome, always belonging to {W, M}. Importantly  the duplicated  
couples agreed (which was the Washington or Moscow outcome) in  
all duplication. They mostly all agreed they did not found any  
obvious algorithm to predict the sequence (the exception might  
concerned the guys in nameable stories, like:


WW

MM

Or the development of some remarkable real number in binary, like  
the binary expansion of PI, sqrt(2), sqr(n), etc. In this case,  
the computable is made rare (and more and more negligible when n  
grows, those histories are "white rabbits histories").


That is what I mean by first person plural. It concerns  
population of machine sharing self-multiplication. it is  
interesting to compare the quantum linear self-superposition with  
the purely arithmetical one.


Sure, that would seem to be reasonably described as 1p-plural.  
except that there is no need to have two people enter the  
duplicating machine


? Then it is just 1p singular. We need two (or more) people  
entering the duplication device so that we get the intersubjective  
agreement.

and undergo different teleportations afterwards.


? They undergo the same teleportations. They are both reconstituted  
in the two different locations, and, obviously (we assume Digital  
Mechanism) they agree that the outcome is well determined from  
their common first person view, and that this the 1p plural.


Surely it is sufficient to consider one person doing a series of  
polarization measurements on a sequence of photons from an  
unpolarized source.


You need two persons. With one person, you can't distinguish 1p  
from 1pp (1p plural).


I think I understand it now. The problems, I think, have arisen  
because you are using the same terminology for both the classical  
duplication of persons and the quantum branching of worlds. I think  
that conflating these two situations is a mistake because they are  
intrinsically different.



Perhaps. That is why I propose a precise testing. The logic of the 1p  
in the classical duplication is given by the Z1* logic. This gives a  
quantum logic, so it looks currently that it might be similar. I am  
not sure why you think there are or have to be different.





In classical person duplication, there is no entanglement.



There is. Both formally and intuitively. The duplication of population  
is only a way to get the intuitive view of the classical mechanist  
entanglement, which is seen as a sort of duplication-contagion, like  
when two people enter the duplication box, or like with the  
schroedinger cat in QM.







Even if you duplicate two or more persons simultaneously, and  
subject them to the same teleportations, there is no real  
entanglement, just a simulation that mimics some features of  
entanglement.


Yes, but we know that in the real case, it involves sum on 

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

[Russell Standish]

On Sun, Jun 25, 2017 at 10:12:20AM +0100, David Nyman wrote:
> 
> 
> ​I always wondered about this aspect of your theory Russell. I assume that
> you mean "describes under some interpretation". If so, the only available
> interpretation (whether explicitly computationalist, a la UDA, or
> otherwise) would have to be intrinsic to the strings themselves. Hence some
> interpretative scheme or other must also be implicitly assumed. Isn't that
> so?
> 

Interpreted by the observer. Yes, I do think the observer is reflected
in the string, which is a manifestation of physical supervenience. My
argument for that goes by the name of the Occam catastrophe.

-- 


Dr Russell StandishPhone 0425 253119 (mobile)
Principal, High Performance Coders
Visiting Senior Research Fellowhpco...@hpcoders.com.au
Economics, Kingston University http://www.hpcoders.com.au


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

[Russell Standish]

On Sat, Jun 24, 2017 at 11:45:23PM -0700, Brent Meeker wrote:
> 
> There's a paper (actually several) by Sorkin which develops QM from
> a measure on histories point of view that starts by the contrary of
> the above equation;  http://arXiv.org/abs/gr-qc/9401003v2
> 

Sounds interesting. I've downloaded it, and added it to my stack.

-- 


Dr Russell StandishPhone 0425 253119 (mobile)
Principal, High Performance Coders
Visiting Senior Research Fellowhpco...@hpcoders.com.au
Economics, Kingston University http://www.hpcoders.com.au


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

[Russell Standish]

On Sun, Jun 25, 2017 at 04:25:07PM +1000, Bruce Kellett wrote:
> On 24/06/2017 8:36 pm, Russell Standish wrote:
> >On Sat, Jun 24, 2017 at 06:29:54PM +1000, Bruce Kellett wrote:
> >>On 24/06/2017 5:23 pm, Russell Standish wrote:
> >>
> >>OK, it was possibly the case that you gave arguments earlier in the
> >>book. But I was going on the basis of the Appendix "Derivation of
> >>Quantum postulates".
> >>
> >>But the problems only begin with the assumption of a probabilistic
> >>model. Psi(t) is the set of possibilities consistent with what is
> >>known at time t. But how do you limit this set? At the moment, I
> >>could go to the pub for a drink, could open a bottle of wine at
> >>home, stroke the cat, turn on the telly, talk to my wife, etc,
> >>etc,. The possibilities consistent with what is known at this
> >>time is not a well defined set, or limited in any way.
> >The everything is the set of all infinite length strings, each of
> >which describes a universe to infinite detail. Some of these strings
> >will describe universes compatible with our current observer moment -
> >an infinite number even, as the information content of our OM is
> >finite. Others will not. It is a well defined subset of the everything.
> 
> What does "compatible" mean? Is this linked to our current moment by
> law-like behaviour, or just any any string which happens, by chance
> to contain our present moment? If the latter (and also in compete
> generality, given your definitions), then my characterization of the
> list of possibilities is correct, and the string could contain any
> future indeed. If the former, how do you know that there is not just
> a single string that contains the only law-like continuation of your
> present state?

There is an interpreter of the strings. In Solomonoff-Levin's theory,
the interpreter is some given universal machine U, but in this case,
the only possible interpreter is an observer of the system.

Compatible just means that the string is interpreted as giving the
current observer moment by the observer.

At this point, there is no theory about what likely successor strings
might be. The most obvious model (ie just read some more bits of a
given string), doesn't work.

What I propose in the appendix D work is to apply an evolutionary
process to it.

> 
> >>Because you then go on to define projection operators in terms of a
> >>sum over the members of this set of possible outcomes. That is
> >>meaningless unless you are already assuming the the outcomes are
> >>just possible results for a well-defined measurement, and that this
> >>measurement process can be defined in a linear vector space.
> >>
> >Summation of the projection operators is defined in equation D.1 for
> >disjoint observations a and b (ie where it is impossible to observe a
> >and b simultaneously). Linearity is not assumed at this point.
> 
> That is where you have an enormous problem, following on from the
> previous point that you have not limited the possible continuations
> in any way. You define a projection as occurring when the observer
> applies an operator A (again undefined and unlimited) to the
> observer moment, which operator divides it into a discrete set of
> outcomes, psi_a. Note that 'a' is only an index, not an eigenvalue
> or any such. Note that you explicitly state that \P_{a} is not
> assumed linear at this point. (I use \P as a notation for your
> script P.) You then *define* addition for two distinct outcomes a
> =/= b as:
> 
> \P_{a} psi  + \P_{b} psi = \P_{a,b} psi.
> 
> This is, of course meaningless if psi_a is 'taking the dog for a
> walk,' and psi_b is 'stroking the cat'. You can define a '+' sign as
> anything you like, but such a definition does not ensure that the
> result has any meaning -- as in my example, which follows completely
> logically from the definitions that you make.
> 

Note that {a,b} is just {a} ∪ {b}, so that \P_{a,b}ψ is is just the
observation "taking the dog for a walk _or_ stroking the cat". That is
a perfectly acceptable classical observation.


> It is clear that your are trying to introduce the concept of a
> quantum superposition by the back door, without doing any work, and
> relying on the inherent ambiguity in the '+' operation.

Why is this ambiguous?

>  If you have
> nothing but classical outcomes from your observer moment psi(t),
> then you cannot simply add these outcomes as if they were separate
> eigenfunctions of a quantum operator. There are no such things as
> superpositions in classical physics.

There are, it's just that they're not particularly interesting from a
physical point of view. 

A superposition of a blue ball and a red ball just means that we don't
know what colour the ball is. 

> 
> Sorry, but the whole procedure is nonsense on stilts. It does not
> get any better from then on in, but I refrain from analysing further
> -- my blood pressure will not stand it!
> 

I think what you're reacting to is the transition from D.6 which
describes 

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

[David Nyman]

On 24 June 2017 at 11:36, Russell Standish  wrote:

> On Sat, Jun 24, 2017 at 06:29:54PM +1000, Bruce Kellett wrote:
> > On 24/06/2017 5:23 pm, Russell Standish wrote:
> >
> > OK, it was possibly the case that you gave arguments earlier in the
> > book. But I was going on the basis of the Appendix "Derivation of
> > Quantum postulates".
> >
> > But the problems only begin with the assumption of a probabilistic
> > model. Psi(t) is the set of possibilities consistent with what is
> > known at time t. But how do you limit this set? At the moment, I
> > could go to the pub for a drink, could open a bottle of wine at
> > home, stroke the cat, turn on the telly, talk to my wife, etc,
> > etc,. The possibilities consistent with what is known at this
> > time is not a well defined set, or limited in any way.
>
> The everything is the set of all infinite length strings, each of
> which describes a universe to infinite detail.


​I always wondered about this aspect of your theory Russell. I assume that
you mean "describes under some interpretation". If so, the only available
interpretation (whether explicitly computationalist, a la UDA, or
otherwise) would have to be intrinsic to the strings themselves. Hence some
interpretative scheme or other must also be implicitly assumed. Isn't that
so?

David

Some of these strings
> will describe universes compatible with our current observer moment -
> an infinite number even, as the information content of our OM is
> finite. Others will not. It is a well defined subset of the everything.
>
> >
> > Because you then go on to define projection operators in terms of a
> > sum over the members of this set of possible outcomes. That is
> > meaningless unless you are already assuming the the outcomes are
> > just possible results for a well-defined measurement, and that this
> > measurement process can be defined in a linear vector space.
> >
>
> Summation of the projection operators is defined in equation D.1 for
> disjoint observations a and b (ie where it is impossible to observe a
> and b simultaneously). Linearity is not assumed at this point.
>
> > Another problem occurs further down when you seem to have complex
> > numbers of observers observing an observer moment. Why you should
> > have more than one observer for any observer moment is a mystery yet
> > to be solved.
>
> It's more a measure over observer moments. In a branching multiverse,
> not all observer moments are equally likely, but one would expect
> across a branching point, measure should be conserved.
>
> Why the measure is complex, not real is more tricky. With the
> everything, subsets naturally induce a real valued measure. But we do
> know that complex measures are more general, and we need a good reason
> not to choose the most general. But complex measures are not the most
> general. I do say "more general division algebras cannot support
> equations of the form (D.7)", but I confess, I'm still not completely
> happy with that line.
>
> > But then you go on, in eq. B8 to define the inner
> > product in terms of the probability function. But you have merely
> > multiplied together two expansions in terms of projections over
> > possible outcomes -- assuming that there is a linear span over the
> > space in the process. This gives the Born rule, sure, because you
> > have built it into your derivation of the inner product.
> >
>
> By the time we get to equation D.8, we have proved that the set of
> observer moments is a vector space, so yes, this construction is
> allowed. We are entitled to define any real-valued bilinear operator on
> that space and call it an inner product.
>
> By using that particular inner product, you get the Born rule in the
> usual form. If we'd chosen another, we'd have a different expression
> that is equivalent to the Born rule.
>
> >
> > >>So you know about QM from the start, and devise a strategy to get
> > >>you there. One of the problems that many-worlders face in their
> > >>attempts to derive the Born rule from within MWI is that they cannot
> > >>independently justify a probabilistic model.
> > >Yes, but I don't start with the MWI (namely, I don't start with a
> > >Hilbert space and unitary equation of motion - ie Schroedinger's
> > >equation). I start with evolution in a generic multiverse.
> >
> > Why a multiverse? You no doubt argue for it elsewhere, but that is
> > not apparent in your quantum derivation.
> >
>
> Yes - of course. The whole book is premised on it.
>
> > And I do not understand why the most general equation for computing
> > psi as a function  of time is a first order differential equation.
> > The equation could clearly be non-linear in psi -- such things have
> > been postulated after all, as in general relativity and GRW for
> > instance.
> >
>
> A first order differential equation needn't be linear. Linearity comes
> from assuming that the laws of physics don't change every time you
> observe something, more specifically 

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

[Brent Meeker]

On 6/24/2017 11:25 PM, Bruce Kellett wrote:

On 24/06/2017 8:36 pm, Russell Standish wrote:

On Sat, Jun 24, 2017 at 06:29:54PM +1000, Bruce Kellett wrote:

On 24/06/2017 5:23 pm, Russell Standish wrote:

OK, it was possibly the case that you gave arguments earlier in the
book. But I was going on the basis of the Appendix "Derivation of
Quantum postulates".

But the problems only begin with the assumption of a probabilistic
model. Psi(t) is the set of possibilities consistent with what is
known at time t. But how do you limit this set? At the moment, I
could go to the pub for a drink, could open a bottle of wine at
home, stroke the cat, turn on the telly, talk to my wife, etc,
etc,. The possibilities consistent with what is known at this
time is not a well defined set, or limited in any way.

The everything is the set of all infinite length strings, each of
which describes a universe to infinite detail. Some of these strings
will describe universes compatible with our current observer moment -
an infinite number even, as the information content of our OM is
finite. Others will not. It is a well defined subset of the everything.


What does "compatible" mean? Is this linked to our current moment by 
law-like behaviour, or just any any string which happens, by chance to 
contain our present moment? If the latter (and also in compete 
generality, given your definitions), then my characterization of the 
list of possibilities is correct, and the string could contain any 
future indeed. If the former, how do you know that there is not just a 
single string that contains the only law-like continuation of your 
present state?



Because you then go on to define projection operators in terms of a
sum over the members of this set of possible outcomes. That is
meaningless unless you are already assuming the the outcomes are
just possible results for a well-defined measurement, and that this
measurement process can be defined in a linear vector space.


Summation of the projection operators is defined in equation D.1 for
disjoint observations a and b (ie where it is impossible to observe a
and b simultaneously). Linearity is not assumed at this point.


That is where you have an enormous problem, following on from the 
previous point that you have not limited the possible continuations in 
any way. You define a projection as occurring when the observer 
applies an operator A (again undefined and unlimited) to the observer 
moment, which operator divides it into a discrete set of outcomes, 
psi_a. Note that 'a' is only an index, not an eigenvalue or any such. 
Note that you explicitly state that \P_{a} is not assumed linear at 
this point. (I use \P as a notation for your script P.) You then 
*define* addition for two distinct outcomes a =/= b as:


\P_{a} psi  + \P_{b} psi = \P_{a,b} psi.



There's a paper (actually several) by Sorkin which develops QM from a 
measure on histories point of view that starts by the contrary of the 
above equation;  http://arXiv.org/abs/gr-qc/9401003v2


/The additivity of classical probabilities is only the first in a//
//hierarchy of possible sum-rules, each of which implies its succes-//
//sor. The first and most restrictive sum-rule of the hierarchy yields//
//measure-theory in the Kolmogorov sense, which physically is appro-//
//priate for the description of stochastic processes such as Brownian//
//motion. The next weaker sum-rule defines a generalized measure//
//theory which includes quantum mechanics as a special case. The//
//fact that quantum probabilities can be expressed “as the squares//
//of quantum amplitudes” is thus derived in a natural manner, and//
//a series of natural generalizations of the quantum formalism is de-//
//lineated. Conversely, the mathematical sense in which classical//
//physics is a special case of quantum physics is clarified. The present//
//paper presents these relationships in the context of a “realistic” in-//
//terpretation of quantum mechanics.

/Brent/
/


This is, of course meaningless if psi_a is 'taking the dog for a 
walk,' and psi_b is 'stroking the cat'. You can define a '+' sign as 
anything you like, but such a definition does not ensure that the 
result has any meaning -- as in my example, which follows completely 
logically from the definitions that you make.


It is clear that your are trying to introduce the concept of a quantum 
superposition by the back door, without doing any work, and relying on 
the inherent ambiguity in the '+' operation. If you have nothing but 
classical outcomes from your observer moment psi(t), then you cannot 
simply add these outcomes as if they were separate eigenfunctions of a 
quantum operator. There are no such things as superpositions in 
classical physics.


Sorry, but the whole procedure is nonsense on stilts. It does not get 
any better from then on in, but I refrain from analysing further -- my 
blood pressure will not stand it!


Bruce





Another problem occurs further down 

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

[Bruce Kellett]

On 24/06/2017 8:36 pm, Russell Standish wrote:

On Sat, Jun 24, 2017 at 06:29:54PM +1000, Bruce Kellett wrote:

On 24/06/2017 5:23 pm, Russell Standish wrote:

OK, it was possibly the case that you gave arguments earlier in the
book. But I was going on the basis of the Appendix "Derivation of
Quantum postulates".

But the problems only begin with the assumption of a probabilistic
model. Psi(t) is the set of possibilities consistent with what is
known at time t. But how do you limit this set? At the moment, I
could go to the pub for a drink, could open a bottle of wine at
home, stroke the cat, turn on the telly, talk to my wife, etc,
etc,. The possibilities consistent with what is known at this
time is not a well defined set, or limited in any way.

The everything is the set of all infinite length strings, each of
which describes a universe to infinite detail. Some of these strings
will describe universes compatible with our current observer moment -
an infinite number even, as the information content of our OM is
finite. Others will not. It is a well defined subset of the everything.


What does "compatible" mean? Is this linked to our current moment by 
law-like behaviour, or just any any string which happens, by chance to 
contain our present moment? If the latter (and also in compete 
generality, given your definitions), then my characterization of the 
list of possibilities is correct, and the string could contain any 
future indeed. If the former, how do you know that there is not just a 
single string that contains the only law-like continuation of your 
present state?



Because you then go on to define projection operators in terms of a
sum over the members of this set of possible outcomes. That is
meaningless unless you are already assuming the the outcomes are
just possible results for a well-defined measurement, and that this
measurement process can be defined in a linear vector space.


Summation of the projection operators is defined in equation D.1 for
disjoint observations a and b (ie where it is impossible to observe a
and b simultaneously). Linearity is not assumed at this point.


That is where you have an enormous problem, following on from the 
previous point that you have not limited the possible continuations in 
any way. You define a projection as occurring when the observer applies 
an operator A (again undefined and unlimited) to the observer moment, 
which operator divides it into a discrete set of outcomes, psi_a. Note 
that 'a' is only an index, not an eigenvalue or any such. Note that you 
explicitly state that \P_{a} is not assumed linear at this point. (I use 
\P as a notation for your script P.) You then *define* addition for two 
distinct outcomes a =/= b as:


\P_{a} psi  + \P_{b} psi = \P_{a,b} psi.

This is, of course meaningless if psi_a is 'taking the dog for a walk,' 
and psi_b is 'stroking the cat'. You can define a '+' sign as anything 
you like, but such a definition does not ensure that the result has any 
meaning -- as in my example, which follows completely logically from the 
definitions that you make.


It is clear that your are trying to introduce the concept of a quantum 
superposition by the back door, without doing any work, and relying on 
the inherent ambiguity in the '+' operation. If you have nothing but 
classical outcomes from your observer moment psi(t), then you cannot 
simply add these outcomes as if they were separate eigenfunctions of a 
quantum operator. There are no such things as superpositions in 
classical physics.


Sorry, but the whole procedure is nonsense on stilts. It does not get 
any better from then on in, but I refrain from analysing further -- my 
blood pressure will not stand it!


Bruce





Another problem occurs further down when you seem to have complex
numbers of observers observing an observer moment. Why you should
have more than one observer for any observer moment is a mystery yet
to be solved.

It's more a measure over observer moments. In a branching multiverse,
not all observer moments are equally likely, but one would expect
across a branching point, measure should be conserved.

Why the measure is complex, not real is more tricky. With the
everything, subsets naturally induce a real valued measure. But we do
know that complex measures are more general, and we need a good reason
not to choose the most general. But complex measures are not the most
general. I do say "more general division algebras cannot support
equations of the form (D.7)", but I confess, I'm still not completely
happy with that line.


But then you go on, in eq. B8 to define the inner
product in terms of the probability function. But you have merely
multiplied together two expansions in terms of projections over
possible outcomes -- assuming that there is a linear span over the
space in the process. This gives the Born rule, sure, because you
have built it into your derivation of the inner product.


By the time we get to equation D.8, we 

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

[Russell Standish]

On Sat, Jun 24, 2017 at 06:29:54PM +1000, Bruce Kellett wrote:
> On 24/06/2017 5:23 pm, Russell Standish wrote:
> 
> OK, it was possibly the case that you gave arguments earlier in the
> book. But I was going on the basis of the Appendix "Derivation of
> Quantum postulates".
> 
> But the problems only begin with the assumption of a probabilistic
> model. Psi(t) is the set of possibilities consistent with what is
> known at time t. But how do you limit this set? At the moment, I
> could go to the pub for a drink, could open a bottle of wine at
> home, stroke the cat, turn on the telly, talk to my wife, etc,
> etc,. The possibilities consistent with what is known at this
> time is not a well defined set, or limited in any way.

The everything is the set of all infinite length strings, each of
which describes a universe to infinite detail. Some of these strings
will describe universes compatible with our current observer moment -
an infinite number even, as the information content of our OM is
finite. Others will not. It is a well defined subset of the everything.

> 
> Because you then go on to define projection operators in terms of a
> sum over the members of this set of possible outcomes. That is
> meaningless unless you are already assuming the the outcomes are
> just possible results for a well-defined measurement, and that this
> measurement process can be defined in a linear vector space.
> 

Summation of the projection operators is defined in equation D.1 for
disjoint observations a and b (ie where it is impossible to observe a
and b simultaneously). Linearity is not assumed at this point. 

> Another problem occurs further down when you seem to have complex
> numbers of observers observing an observer moment. Why you should
> have more than one observer for any observer moment is a mystery yet
> to be solved. 

It's more a measure over observer moments. In a branching multiverse,
not all observer moments are equally likely, but one would expect
across a branching point, measure should be conserved.

Why the measure is complex, not real is more tricky. With the
everything, subsets naturally induce a real valued measure. But we do
know that complex measures are more general, and we need a good reason
not to choose the most general. But complex measures are not the most
general. I do say "more general division algebras cannot support
equations of the form (D.7)", but I confess, I'm still not completely
happy with that line.

> But then you go on, in eq. B8 to define the inner
> product in terms of the probability function. But you have merely
> multiplied together two expansions in terms of projections over
> possible outcomes -- assuming that there is a linear span over the
> space in the process. This gives the Born rule, sure, because you
> have built it into your derivation of the inner product.
> 

By the time we get to equation D.8, we have proved that the set of
observer moments is a vector space, so yes, this construction is
allowed. We are entitled to define any real-valued bilinear operator on
that space and call it an inner product.

By using that particular inner product, you get the Born rule in the
usual form. If we'd chosen another, we'd have a different expression
that is equivalent to the Born rule.

> 
> >>So you know about QM from the start, and devise a strategy to get
> >>you there. One of the problems that many-worlders face in their
> >>attempts to derive the Born rule from within MWI is that they cannot
> >>independently justify a probabilistic model.
> >Yes, but I don't start with the MWI (namely, I don't start with a
> >Hilbert space and unitary equation of motion - ie Schroedinger's
> >equation). I start with evolution in a generic multiverse.
> 
> Why a multiverse? You no doubt argue for it elsewhere, but that is
> not apparent in your quantum derivation.
> 

Yes - of course. The whole book is premised on it.

> And I do not understand why the most general equation for computing
> psi as a function  of time is a first order differential equation.
> The equation could clearly be non-linear in psi -- such things have
> been postulated after all, as in general relativity and GRW for
> instance.
> 

A first order differential equation needn't be linear. Linearity comes
from assuming that the laws of physics don't change every time you
observe something, more specifically the solutions ψ_α are also solutions.

A higher order equation can be transformed into a first order equation
by adding new variables - a trick commonly done in dynamical systems theory.

Perhaps there's an implicit assumption that the evolution should be
Markovian. I think one could make a convincing case that it should be,
but perhaps that assumption needs to be made explicit.

> Besides, you do not show that the operator H is the Hamiltonian and
> the energy operator. You do not derive the basic commutation
> relations between position and momentum operators -- a relation that
> is central to the whole 

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

[Bruce Kellett]

On 24/06/2017 5:23 pm, Russell Standish wrote:

On Sat, Jun 24, 2017 at 03:59:56PM +1000, Bruce Kellett wrote:

Well, I have just taken a quick look. What strikes me is that the
first paragraph of Appendix D defines "Observer moments psi(t) are
sets of possibilities consistent with what is known at that point in
time, providing variation upon which anthropic selection acts. ...
We wish to determine the probability of outcome a being observed."
So you assume a probabilistic model from the outcome. Why would you
do that? Why not a deterministic model?

I do spend over a hundred pages prior to the chapter on QM going into
the reasons! But to try to signpost this, and maybe save you the
effort of reading my book, the basic reason is that our 1p view must
be the result of evolution - not biological evolution, per se, but
anthropic evolution - the result of variation of possible futures, and
anthropic selection from those possible futures to the actual result
seen. Along with heredity (which in QM gives rise to unitarity), we
have the three pillars of evolution as espoused by Lewontin.

Consequently, the probabilistic model needs to be there right from the
start to provide the variation on which anthropic selection acts.


OK, it was possibly the case that you gave arguments earlier in the 
book. But I was going on the basis of the Appendix "Derivation of 
Quantum postulates".


But the problems only begin with the assumption of a probabilistic 
model. Psi(t) is the set of possibilities consistent with what is known 
at time t. But how do you limit this set? At the moment, I could go to 
the pub for a drink, could open a bottle of wine at home, stroke the 
cat, turn on the telly, talk to my wife, etc, etc,. The 
possibilities consistent with what is known at this time is not a well 
defined set, or limited in any way.


Because you then go on to define projection operators in terms of a sum 
over the members of this set of possible outcomes. That is meaningless 
unless you are already assuming the the outcomes are just possible 
results for a well-defined measurement, and that this measurement 
process can be defined in a linear vector space.


Another problem occurs further down when you seem to have complex 
numbers of observers observing an observer moment. Why you should have 
more than one observer for any observer moment is a mystery yet to be 
solved. But then you go on, in eq. B8 to define the inner product in 
terms of the probability function. But you have merely multiplied 
together two expansions in terms of projections over possible outcomes 
-- assuming that there is a linear span over the space in the process. 
This gives the Born rule, sure, because you have built it into your 
derivation of the inner product.




So you know about QM from the start, and devise a strategy to get
you there. One of the problems that many-worlders face in their
attempts to derive the Born rule from within MWI is that they cannot
independently justify a probabilistic model.

Yes, but I don't start with the MWI (namely, I don't start with a
Hilbert space and unitary equation of motion - ie Schroedinger's
equation). I start with evolution in a generic multiverse.


Why a multiverse? You no doubt argue for it elsewhere, but that is not 
apparent in your quantum derivation.


And I do not understand why the most general equation for computing psi 
as a function  of time is a first order differential equation. The 
equation could clearly be non-linear in psi -- such things have been 
postulated after all, as in general relativity and GRW for instance.


Besides, you do not show that the operator H is the Hamiltonian and the 
energy operator. You do not derive the basic commutation relations 
between position and momentum operators -- a relation that is central to 
the whole of QM.




If you have a
probabilistic model in 3 or more dimensions, Gleason's theorem tells
you that the Born rule is the only consistent model for
probabilities.

My arguments go through in fewer than 3 dimensions as well, AFAIK,
although that would a relatively uninteresting world - very black and
white :). Which is why I suspect it is independent of Gleason.


But you have to say why you want a probabilistic
interpretation in the first place. Deutsch's attempts founder on the
fact that he has to assume that small amplitudes have small
probabilities, even to get started, so his argument is manifestly
circular.


Yes - I think the problem with those approaches is that they start
with a Hilbert space and unitary equation of motion (ie a classic
MWI), and then fail to generate the Born rule because there is no
observer in their mechanics.


As I said, you build a probabilistic model in at the start, so
Gleason's theorem is going to get you the Born rule automatically.
Or if you don't assume Gleason, you have an equivalent result by
another route. Assuming a probabilistic model is a very powerful
starting point..

Sure - but it is necessary. 

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

[Russell Standish]

On Sat, Jun 24, 2017 at 03:59:56PM +1000, Bruce Kellett wrote:
> 
> Well, I have just taken a quick look. What strikes me is that the
> first paragraph of Appendix D defines "Observer moments psi(t) are
> sets of possibilities consistent with what is known at that point in
> time, providing variation upon which anthropic selection acts. ...
> We wish to determine the probability of outcome a being observed."
> So you assume a probabilistic model from the outcome. Why would you
> do that? Why not a deterministic model?

I do spend over a hundred pages prior to the chapter on QM going into
the reasons! But to try to signpost this, and maybe save you the
effort of reading my book, the basic reason is that our 1p view must
be the result of evolution - not biological evolution, per se, but
anthropic evolution - the result of variation of possible futures, and
anthropic selection from those possible futures to the actual result
seen. Along with heredity (which in QM gives rise to unitarity), we
have the three pillars of evolution as espoused by Lewontin.

Consequently, the probabilistic model needs to be there right from the
start to provide the variation on which anthropic selection acts.

> 
> So you know about QM from the start, and devise a strategy to get
> you there. One of the problems that many-worlders face in their
> attempts to derive the Born rule from within MWI is that they cannot
> independently justify a probabilistic model.

Yes, but I don't start with the MWI (namely, I don't start with a
Hilbert space and unitary equation of motion - ie Schroedinger's
equation). I start with evolution in a generic multiverse.

> If you have a
> probabilistic model in 3 or more dimensions, Gleason's theorem tells
> you that the Born rule is the only consistent model for
> probabilities. 

My arguments go through in fewer than 3 dimensions as well, AFAIK,
although that would a relatively uninteresting world - very black and
white :). Which is why I suspect it is independent of Gleason.

> But you have to say why you want a probabilistic
> interpretation in the first place. Deutsch's attempts founder on the
> fact that he has to assume that small amplitudes have small
> probabilities, even to get started, so his argument is manifestly
> circular.
> 

Yes - I think the problem with those approaches is that they start
with a Hilbert space and unitary equation of motion (ie a classic
MWI), and then fail to generate the Born rule because there is no
observer in their mechanics.

> 
> As I said, you build a probabilistic model in at the start, so
> Gleason's theorem is going to get you the Born rule automatically.
> Or if you don't assume Gleason, you have an equivalent result by
> another route. Assuming a probabilistic model is a very powerful
> starting point..

Sure - but it is necessary. If evolution did not work the way it did,
we could only ever be Boltzmann brains, isolated observers existing
fleetingly, barely having time to consider what to have for lunch, let
alone figuring out the meaning of the universe. Fortunately for us,
evolution does work to generate complex worlds from simple beginnings,
meaning an evolved world is overwhelming more likely to occur in the
Multiverse of Everything than Boltzmann brain existences.


-- 


Dr Russell StandishPhone 0425 253119 (mobile)
Principal, High Performance Coders
Visiting Senior Research Fellowhpco...@hpcoders.com.au
Economics, Kingston University http://www.hpcoders.com.au


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

[Bruce Kellett]

On 24/06/2017 3:02 pm, Russell Standish wrote:

On Sat, Jun 24, 2017 at 01:09:41PM +1000, Bruce Kellett wrote:

On 24/06/2017 11:20 am, Russell Standish wrote:

The 3p is what is left after removing all personal baggage of each 1p
view point. It is literally the view from nowhere (since location is
just such a baggage), and cannot be conscious in itself (for exactly
the reason you outline below)

It is this characterization of 3p that I find misleading. If all
personal baggage has been removed, how come you still talk as
thought this were a third person view? I think this terminology is
an unfortunate carry-over from the classical person-duplication
thought experiments. The bird view would more properly be called a
0p view, since there are no 'persons' or 'person' who has this view.


I have no problem with you arguing for a change in terminology, but to
be clear, the term 3p has been used consistently on this list to mean
roughly what I describe above for almost 2 decades.

Until this discussion, it never even ocurred to me that it might be
confusing - I never thought of 3p as a person.


Maybe you guys should get out more?


There is still another datum. Because of the reasoning used in my
derivation of QM (appendix D of my book), I equate the 3p with the
quantum multiverse. Of course, my derivation may well be faulty - to
my knowledge, only a handful of people have dug into and critiqued
the argument in its 17 years of existence, without finding any fatal
flaw - however assuming its validity, then we can equate the 3p with
the bird view of Tegmark's level 3 multiverse.

I have not had the time or energy to delve deeply into your
derivation. My experience of other attempts to 'derive' quantum
mechanics is that basic quantum concepts are introduced by
sleight-of-hand -- in other words, they usually beg the question.


That is still quite possible in my case, of course, but I have tried
my utmost to make the assumptions explicit, and give reasonable
justifications for them in terms of observer properties. Brent found 1
(or maybe 2, memory's a little hazy) hidden assumptions in my first
version of the argument 15 years ago, which I have since corrected.


Well, I have just taken a quick look. What strikes me is that the first 
paragraph of Appendix D defines "Observer moments psi(t) are sets of 
possibilities consistent with what is known at that point in time, 
providing variation upon which anthropic selection acts. ... We wish to 
determine the probability of outcome a being observed." So you assume a 
probabilistic model from the outcome. Why would you do that? Why not a 
deterministic model?


So you know about QM from the start, and devise a strategy to get you 
there. One of the problems that many-worlders face in their attempts to 
derive the Born rule from within MWI is that they cannot independently 
justify a probabilistic model. If you have a probabilistic model in 3 or 
more dimensions, Gleason's theorem tells you that the Born rule is the 
only consistent model for probabilities. But you have to say why you 
want a probabilistic interpretation in the first place. Deutsch's 
attempts founder on the fact that he has to assume that small amplitudes 
have small probabilities, even to get started, so his argument is 
manifestly circular.




The work has passed peer review, but as you well know, that's only a
minimal hurdle. It is not enough for the work to be taken seriously by
the field, nor (obviously) for to actually be right.

The most worrying aspect of my derivation is the requirement that the
complex field is the most general measure applicable for sets of
observers. Complex numbers are not the most general measure (Banach
spaces are), and if we must restrict the type of measure for any
reason, then why not restrict all the way to real measures (which kind
of seems natural).

Another open problem is what is the relationship with the Gleason
theorem? The Born rule naturally falls out of my construction, so the
question is whether my derivation is independent of Gleason's theorem,
or just incorporates it in disguise.


As I said, you build a probabilistic model in at the start, so Gleason's 
theorem is going to get you the Born rule automatically. Or if you don't 
assume Gleason, you have an equivalent result by another route. Assuming 
a probabilistic model is a very powerful starting point..


Bruce

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

[Russell Standish]

On Sat, Jun 24, 2017 at 01:09:41PM +1000, Bruce Kellett wrote:
> On 24/06/2017 11:20 am, Russell Standish wrote:
> >The 3p is what is left after removing all personal baggage of each 1p
> >view point. It is literally the view from nowhere (since location is
> >just such a baggage), and cannot be conscious in itself (for exactly
> >the reason you outline below)
> 
> It is this characterization of 3p that I find misleading. If all
> personal baggage has been removed, how come you still talk as
> thought this were a third person view? I think this terminology is
> an unfortunate carry-over from the classical person-duplication
> thought experiments. The bird view would more properly be called a
> 0p view, since there are no 'persons' or 'person' who has this view.
> 

I have no problem with you arguing for a change in terminology, but to
be clear, the term 3p has been used consistently on this list to mean
roughly what I describe above for almost 2 decades.

Until this discussion, it never even ocurred to me that it might be
confusing - I never thought of 3p as a person.

> 
> >There is still another datum. Because of the reasoning used in my
> >derivation of QM (appendix D of my book), I equate the 3p with the
> >quantum multiverse. Of course, my derivation may well be faulty - to
> >my knowledge, only a handful of people have dug into and critiqued
> >the argument in its 17 years of existence, without finding any fatal
> >flaw - however assuming its validity, then we can equate the 3p with
> >the bird view of Tegmark's level 3 multiverse.
> 
> I have not had the time or energy to delve deeply into your
> derivation. My experience of other attempts to 'derive' quantum
> mechanics is that basic quantum concepts are introduced by
> sleight-of-hand -- in other words, they usually beg the question.
> 

That is still quite possible in my case, of course, but I have tried
my utmost to make the assumptions explicit, and give reasonable
justifications for them in terms of observer properties. Brent found 1
(or maybe 2, memory's a little hazy) hidden assumptions in my first
version of the argument 15 years ago, which I have since corrected.

The work has passed peer review, but as you well know, that's only a
minimal hurdle. It is not enough for the work to be taken seriously by
the field, nor (obviously) for to actually be right.

The most worrying aspect of my derivation is the requirement that the
complex field is the most general measure applicable for sets of
observers. Complex numbers are not the most general measure (Banach
spaces are), and if we must restrict the type of measure for any
reason, then why not restrict all the way to real measures (which kind
of seems natural).

Another open problem is what is the relationship with the Gleason
theorem? The Born rule naturally falls out of my construction, so the
question is whether my derivation is independent of Gleason's theorem,
or just incorporates it in disguise.

Cheers
-- 


Dr Russell StandishPhone 0425 253119 (mobile)
Principal, High Performance Coders
Visiting Senior Research Fellowhpco...@hpcoders.com.au
Economics, Kingston University http://www.hpcoders.com.au


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

[Bruce Kellett]

On 24/06/2017 11:20 am, Russell Standish wrote:

On Wed, Jun 21, 2017 at 04:21:09PM +1000, Bruce Kellett wrote:

On 21/06/2017 4:03 pm, Russell Standish wrote:

On Mon, Jun 19, 2017 at 12:15:31PM +1000, Bruce Kellett wrote:

On 19/06/2017 10:23 am, Russell Standish wrote:

I know Scott wouldn't go as far as me. For me, all such irreversible
processes are related to conscious entities in some way. Whilst
agreeing that Geiger counters are unlikely to be conscious, I would
say that the output of Geiger counter is not actually discrete until
observed by a conscious experimenter.

That sounds remarkably like the "many minds" interpretation of
quantum mechanics. This is disfavoured by most scientists because it
leaves the physics of the billions of years before the emergence of
the first "conscious" creature unresolved -- the first consciousness
would cause an almighty collapse on the many minds reading.


Each consciousness causes "an almighty collapse" in er own mind
independently of any other. It's a pure 1p phenomena.

It is actually a 3p phenomenon because there is inter-subjective
agreement about the fact that measurements give definite results.


There is no collapse at all at the 3p level, nor even decoherence as such.

Decoherence is a well-understood physical phenomenon that has been
widely observed. I do not know what you mean by saying "nor even
decoherence as such."

Thinking more on this, I think we may be arguing at cross purposes. In
the above, I was using decoherence to refer to idea of einselection in
a Multiverse, promoted by Zurek et al. In the multiverse, this
strictly speaking can't happen, because of unitarity. Einselection is
only a FAPP process, that ultimately manifests in the 1p of a
scientist.

It's dawned on me that when you refer to decoherence as a physical
phenomenon, you're really talking about this latter stuff, like how
difficult it is to maintain a coherent state in a quantum computer,
for instance.


Yes, in this instance I was referring just to the ubiquity of 
environmental interactions, leading to the loss of coherence in quantum 
states. The question of selecting a stable basis, or einselection in 
Zurek's terminology, is clearly related to decoherence, though it does 
take the idea a bit further. The physical problem that we face in 
quantum mechanics is that we observe things in particular bases -- the 
live/dead basis, not the (live+dead)/(liver-dead) basis, in the 
Schrödinger cat thought experiment for example. The basis in which we 
observe things is stable against further decoherence by environmental 
interactions. It is argued (see Schlosshauer, for example) that the 
stable einselected basis is the one in which the corresponding operators 
commute with the interaction Hamiltonian. This is certainly a necessary 
condition, but my feeling has been that this relationship is rather too 
circular to really firmly ground the einselection argument.



You're not implying that you have physical evidence for
einselection in a multiverse (which is how I first read you, and if
true, deserving of a Nobel, I'd think). In any case, this latter stuff
is pure 1p. (1p plural, as Bruno Marchal might insist).


I think there is very good observational evidence for einselction -- the 
basis in which we observe the world (or bases for different operators) 
are very stable. It may, in the final analysis turn out to be FAPP, but 
the fact that decoherence generally removes important coherence 
information at the speed of light, the recombination time for natural 
recovery of coherence tends to infinity in an expanding universe -- 
there is no Poincare recurrence time in de Sitter space! Such 
considerations tend to suggest that decoherence is  permanent feature of 
the multiverse, not something that can ever be reversed. As I have said 
elsewhere, unitarity is only a theoretical construct: a property of the 
equations rather than a proven property of the universe.




Also, you seem to be confusing the
inter-subjective 3p view with Tegmark's bird view.

The 3p is what is left after removing all personal baggage of each 1p
view point. It is literally the view from nowhere (since location is
just such a baggage), and cannot be conscious in itself (for exactly
the reason you outline below)


It is this characterization of 3p that I find misleading. If all 
personal baggage has been removed, how come you still talk as thought 
this were a third person view? I think this terminology is an 
unfortunate carry-over from the classical person-duplication thought 
experiments. The bird view would more properly be called a 0p view, 
since there are no 'persons' or 'person' who has this view.




Tegmark's "bird view" is somewhat poetic, but also ambiguous term,
in that it depends on what level of Multiverse you're talking about.


Sure, but we are talking about the level III quantum many worlds. I 
don't know if one could give any meaning to a 'bird' view of Tegmark's 
level IV multiverse. 

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

[Russell Standish]

On Wed, Jun 21, 2017 at 04:21:09PM +1000, Bruce Kellett wrote:
> On 21/06/2017 4:03 pm, Russell Standish wrote:
> >On Mon, Jun 19, 2017 at 12:15:31PM +1000, Bruce Kellett wrote:
> >>On 19/06/2017 10:23 am, Russell Standish wrote:
> >>>I know Scott wouldn't go as far as me. For me, all such irreversible
> >>>processes are related to conscious entities in some way. Whilst
> >>>agreeing that Geiger counters are unlikely to be conscious, I would
> >>>say that the output of Geiger counter is not actually discrete until
> >>>observed by a conscious experimenter.
> >>That sounds remarkably like the "many minds" interpretation of
> >>quantum mechanics. This is disfavoured by most scientists because it
> >>leaves the physics of the billions of years before the emergence of
> >>the first "conscious" creature unresolved -- the first consciousness
> >>would cause an almighty collapse on the many minds reading.
> >>
> >Each consciousness causes "an almighty collapse" in er own mind
> >independently of any other. It's a pure 1p phenomena.
> 
> It is actually a 3p phenomenon because there is inter-subjective
> agreement about the fact that measurements give definite results.
> 
> >There is no collapse at all at the 3p level, nor even decoherence as such.
> 
> Decoherence is a well-understood physical phenomenon that has been
> widely observed. I do not know what you mean by saying "nor even
> decoherence as such." 

Thinking more on this, I think we may be arguing at cross purposes. In
the above, I was using decoherence to refer to idea of einselection in
a Multiverse, promoted by Zurek et al. In the multiverse, this
strictly speaking can't happen, because of unitarity. Einselection is
only a FAPP process, that ultimately manifests in the 1p of a
scientist.

It's dawned on me that when you refer to decoherence as a physical
phenomenon, you're really talking about this latter stuff, like how
difficult it is to maintain a coherent state in a quantum computer,
for instance. You're not implying that you have physical evidence for
einselection in a multiverse (which is how I first read you, and if
true, deserving of a Nobel, I'd think). In any case, this latter stuff
is pure 1p. (1p plural, as Bruno Marchal might insist).

> Also, you seem to be confusing the
> inter-subjective 3p view with Tegmark's bird view.

The 3p is what is left after removing all personal baggage of each 1p
view point. It is literally the view from nowhere (since location is
just such a baggage), and cannot be conscious in itself (for exactly
the reason you outline below)

Tegmark's "bird view" is somewhat poetic, but also ambiguous term,
in that it depends on what level of Multiverse you're talking about.

If we propose the "antirealist" ontology, then all things must be
grounded in observerhood. I scare-quoted antirealist, because even
though we agreed on this definition in an ongoing conversation with a
colleague, it still causes confusion. But regardless of whether the
terminology is good, it is the idea that the 3p is the maximum
ontology, nothing else exists. In such a case, the 3p can be
identified with Tegmark's bird view.

There is still another datum. Because of the reasoning used in my
derivation of QM (appendix D of my book), I equate the 3p with the
quantum multiverse. Of course, my derivation may well be faulty - to
my knowledge, only a handful of people have dug into and critiqued
the argument in its 17 years of existence, without finding any fatal
flaw - however assuming its validity, then we can equate the 3p with
the bird view of Tegmark's level 3 multiverse.

>  There is no
> person, body, or consciousness that ever has the bird view -- the
> bird is a purely formal construct and has nothing to do with mind or
> consciousness. Even though everything might remain unitary at that
> level, no one can ever experience the consequences of that unitary
> evolution.
> 

Yes, of course. And the same applies to the 3p.


-- 


Dr Russell StandishPhone 0425 253119 (mobile)
Principal, High Performance Coders
Visiting Senior Research Fellowhpco...@hpcoders.com.au
Economics, Kingston University http://www.hpcoders.com.au


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

[Bruce Kellett]

On 22/06/2017 7:22 pm, Bruno Marchal wrote:

On 22 Jun 2017, at 01:31, Bruce Kellett wrote:

On 22/06/2017 1:44 am, Bruno Marchal wrote:

On 21 Jun 2017, at 08:21, Bruce Kellett wrote:

On 21/06/2017 4:03 pm, Russell Standish wrote:

On Mon, Jun 19, 2017 at 12:15:31PM +1000, Bruce Kellett wrote:

On 19/06/2017 10:23 am, Russell Standish wrote:
I know Scott wouldn't go as far as me. For me, all such 
irreversible

processes are related to conscious entities in some way. Whilst
agreeing that Geiger counters are unlikely to be conscious, I would
say that the output of Geiger counter is not actually discrete 
until

observed by a conscious experimenter.

That sounds remarkably like the "many minds" interpretation of
quantum mechanics. This is disfavoured by most scientists because it
leaves the physics of the billions of years before the emergence of
the first "conscious" creature unresolved -- the first consciousness
would cause an almighty collapse on the many minds reading.


Each consciousness causes "an almighty collapse" in er own mind
independently of any other. It's a pure 1p phenomena.


It is actually a 3p phenomenon because there is inter-subjective 
agreement about the fact that measurements give definite results.


Inter-subjectivity does not imply 3p, as it can be "only" 1p plural. 
Let me illustrate this with a variant of the WM duplication.


Imagine that Bruce and John are undergoing the WM-duplication 
*together*.


By this I mean they both enter the scanning-annihilating box, and 
are both reconstituted in Washington and in Moscow.


And let us assume they do it repetitively, which means they come 
back to Helsinki, and do it again together.


Obviously, the line-life past that each copies describes in its 
personal diaries grows like H followed by a sequence of W and M. The 
number of copies grows exponentially (2^n). After ten iterations, we 
have 2^10 = 1024 individuals, who share an indeterminate 
experiences. With minor exceptions, they all agree that the 
experience has always given each times a precise outcome, always 
belonging to {W, M}. Importantly  the duplicated couples agreed 
(which was the Washington or Moscow outcome) in all duplication. 
They mostly all agreed they did not found any obvious algorithm to 
predict the sequence (the exception might concerned the guys in 
nameable stories, like:


WW

MM

Or the development of some remarkable real number in binary, like 
the binary expansion of PI, sqrt(2), sqr(n), etc. In this case, the 
computable is made rare (and more and more negligible when n grows, 
those histories are "white rabbits histories").


That is what I mean by first person plural. It concerns population 
of machine sharing self-multiplication. it is interesting to compare 
the quantum linear self-superposition with the purely arithmetical one.


Sure, that would seem to be reasonably described as 1p-plural. except 
that there is no need to have two people enter the duplicating machine


? Then it is just 1p singular. We need two (or more) people entering 
the duplication device so that we get the intersubjective agreement.

and undergo different teleportations afterwards.


? They undergo the same teleportations. They are both reconstituted in 
the two different locations, and, obviously (we assume Digital 
Mechanism) they agree that the outcome is well determined from their 
common first person view, and that this the 1p plural.


Surely it is sufficient to consider one person doing a series of 
polarization measurements on a sequence of photons from an 
unpolarized source.


You need two persons. With one person, you can't distinguish 1p from 
1pp (1p plural).


I think I understand it now. The problems, I think, have arisen because 
you are using the same terminology for both the classical duplication of 
persons and the quantum branching of worlds. I think that conflating 
these two situations is a mistake because they are intrinsically 
different. In classical person duplication, there is no entanglement. 
Even if you duplicate two or more persons simultaneously, and subject 
them to the same teleportations, there is no real entanglement, just a 
simulation that mimics some features of entanglement. The lesson of Bell 
is that classical simulations of entanglement can never reproduce the 
quantum results.


In classical person duplication, it is only the person that is 
duplicated, not the whole world, so the 1p experience becomes central. 
Because the world is not duplicated, you can have an external observer 
who can see both ends of the duplication -- the 3p view. However, in the 
quantum case, a quantum event, when magnified to macro significance by 
decoherence, results in a branching of the whole world into disjoint 
copies. The role of the observer is diminished here, to the extent that 
an observer is not even required: if it is a quantum measurement, the 
experimenter is entangled with the result simply as part of the wider 

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

[Bruno Marchal]

On 22 Jun 2017, at 03:46, Bruce Kellett wrote:


On 22/06/2017 10:32 am, David Nyman wrote:
On 22 Jun 2017 00:31, "Bruce Kellett"   
wrote:

On 22/06/2017 1:44 am, Bruno Marchal wrote:
On 21 Jun 2017, at 08:21, Bruce Kellett wrote:
On 21/06/2017 4:03 pm, Russell Standish wrote:
On Mon, Jun 19, 2017 at 12:15:31PM +1000, Bruce Kellett wrote:
On 19/06/2017 10:23 am, Russell Standish wrote:
I know Scott wouldn't go as far as me. For me, all such irreversible
processes are related to conscious entities in some way. Whilst
agreeing that Geiger counters are unlikely to be conscious, I would
say that the output of Geiger counter is not actually discrete until
observed by a conscious experimenter.
That sounds remarkably like the "many minds" interpretation of
quantum mechanics. This is disfavoured by most scientists because it
leaves the physics of the billions of years before the emergence of
the first "conscious" creature unresolved -- the first consciousness
would cause an almighty collapse on the many minds reading.

Each consciousness causes "an almighty collapse" in er own mind
independently of any other. It's a pure 1p phenomena.

It is actually a 3p phenomenon because there is inter-subjective  
agreement about the fact that measurements give definite results.


Inter-subjectivity does not imply 3p, as it can be "only" 1p  
plural. Let me illustrate this with a variant of the WM duplication.


Imagine that Bruce and John are undergoing the WM-duplication  
*together*.


By this I mean they both enter the scanning-annihilating box, and  
are both reconstituted in Washington and in Moscow.


And let us assume they do it repetitively, which means they come  
back to Helsinki, and do it again together.


Obviously, the line-life past that each copies describes in its  
personal diaries grows like H followed by a sequence of W and M.  
The number of copies grows exponentially (2^n). After ten  
iterations, we have 2^10 = 1024 individuals, who share an  
indeterminate experiences. With minor exceptions, they all agree  
that the experience has always given each times a precise outcome,  
always belonging to {W, M}. Importantly  the duplicated couples  
agreed (which was the Washington or Moscow outcome) in all  
duplication. They mostly all agreed they did not found any obvious  
algorithm to predict the sequence (the exception might concerned  
the guys in nameable stories, like:


WW

MM

Or the development of some remarkable real number in binary, like  
the binary expansion of PI, sqrt(2), sqr(n), etc. In this case, the  
computable is maderare (and more and more  
negligible when n grows, those histories are "white rabbits  
histories").


That is what I mean by first person plural. It concerns population  
of machine sharing self-multiplication. it is interesting to  
compare the quantum linear self-superposition with the purely  
arithmetical one.


Sure, that would seem to be reasonably described as 1p-plural.  
except that there is no need to have two people enter the  
duplicating machine and undergo different teleportations  
afterwards. Surely it is sufficient to consider one person doing a  
series of polarization measurements on a sequence of photons from  
an unpolarized source. That person will record some sequence of '+'  
and '-' results. If the experiment is repeated N times, there will  
be 2^N sequences, one in each of the generated worlds.


But that has nothing to do with inter-subjective agreement between  
different observers. To see that, consider just one polarization  
measurement: In order for it to be said that the measurement gave a  
result, there has to be decoherence and the formation of  
irreversible records. I think it is Zurek who talks about multiple  
copies of the result entangled with the environment. So many  
different individuals can observe the result of this single  
experiment, and they will all agree that the result was what the  
experimenter wrote in her lab book. That is inter-subjective  
agreement. It clearly has nothing to do with 1p, or 1p-plural  
pictures.


I think there may be a terminological confusion here. IIUC, 1p- 
plural denotes, amongst other things, just such inter-subjective  
agreement between mutually entangled observers.


That seems a remarkably confusing terminology. The example Bruno  
gave to illustrate 1p-plural was not an example of inter-subjective  
agreement -- there were just repeated measurements by the one person.


?

By TWO persons. You do miss the point, it seems.

Bruno



If you conflate 1p-plural with inter-subjective, what on earth is  
3p? The notation suggested to me 'third person', or the view of an  
outsider watching the experiment. This outsider certainly gets  
entangled with the experimenter and his result, but the many copies  
give rise to the inter-subjective agreement about what that result  
was. Bruno has certainly used 3p in this way many times 

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

[Bruno Marchal]

On 22 Jun 2017, at 01:31, Bruce Kellett wrote:


On 22/06/2017 1:44 am, Bruno Marchal wrote:

On 21 Jun 2017, at 08:21, Bruce Kellett wrote:

On 21/06/2017 4:03 pm, Russell Standish wrote:

On Mon, Jun 19, 2017 at 12:15:31PM +1000, Bruce Kellett wrote:

On 19/06/2017 10:23 am, Russell Standish wrote:
I know Scott wouldn't go as far as me. For me, all such  
irreversible

processes are related to conscious entities in some way. Whilst
agreeing that Geiger counters are unlikely to be conscious, I  
would
say that the output of Geiger counter is not actually discrete  
until

observed by a conscious experimenter.

That sounds remarkably like the "many minds" interpretation of
quantum mechanics. This is disfavoured by most scientists  
because it
leaves the physics of the billions of years before the emergence  
of
the first "conscious" creature unresolved -- the first  
consciousness

would cause an almighty collapse on the many minds reading.


Each consciousness causes "an almighty collapse" in er own mind
independently of any other. It's a pure 1p phenomena.


It is actually a 3p phenomenon because there is inter-subjective  
agreement about the fact that measurements give definite results.


Inter-subjectivity does not imply 3p, as it can be "only" 1p  
plural. Let me illustrate this with a variant of the WM duplication.


Imagine that Bruce and John are undergoing the WM-duplication  
*together*.


By this I mean they both enter the scanning-annihilating box, and  
are both reconstituted in Washington and in Moscow.


And let us assume they do it repetitively, which means they come  
back to Helsinki, and do it again together.


Obviously, the line-life past that each copies describes in its  
personal diaries grows like H followed by a sequence of W and M.  
The number of copies grows exponentially (2^n). After ten  
iterations, we have 2^10 = 1024 individuals, who share an  
indeterminate experiences. With minor exceptions, they all agree  
that the experience has always given each times a precise outcome,  
always belonging to {W, M}. Importantly  the duplicated couples  
agreed (which was the Washington or Moscow outcome) in all  
duplication. They mostly all agreed they did not found any obvious  
algorithm to predict the sequence (the exception might concerned  
the guys in nameable stories, like:


WW

MM

Or the development of some remarkable real number in binary, like  
the binary expansion of PI, sqrt(2), sqr(n), etc. In this case, the  
computable is made rare (and more and more negligible when n grows,  
those histories are "white rabbits histories").


That is what I mean by first person plural. It concerns population  
of machine sharing self-multiplication. it is interesting to  
compare the quantum linear self-superposition with the purely  
arithmetical one.


Sure, that would seem to be reasonably described as 1p-plural.  
except that there is no need to have two people enter the  
duplicating machine


? Then it is just 1p singular. We need two (or more) people entering  
the duplication device so that we get the intersubjective agreement.







and undergo different teleportations afterwards.


? They undergo the same teleportations. They are both reconstituted in  
the two different locations, and, obviously (we assume Digital  
Mechanism) they agree that the outcome is well determined from their  
common first person view, and that this the 1p plural.




Surely it is sufficient to consider one person doing a series of  
polarization measurements on a sequence of photons from an  
unpolarized source.


You need two persons. With one person, you can't distinguish 1p from  
1pp (1p plural).




That person will record some sequence of '+' and '-' results. If the  
experiment is repeated N times, there will be 2^N sequences, one in  
each of the generated worlds.


Glad to hear that. You might try to explain this to John. But you need  
two person doing the experiment, or discussing it at least, in which  
case the secondf person will entangle with the first, already  
entangled with the observed particle (say).


O2 O1 (up + down) ==>O2 (O1 up + O1 down) ==> O2 O1 up + O2 O1 down =>  
O2 O1[up] up + O2 O1[down] down => etc. There is subjective agreement  
between O1 and O2, because the superposition and measurement (in the  
same base, here) propogate from O1 to O2.








But that has nothing to do with inter-subjective agreement between  
different observers.


? Because you withdrew the second person. I think me or you miss  
something.





To see that, consider just one polarization measurement: In order  
for it to be said that the measurement gave a result, there has to  
be decoherence and the formation of irreversible records. I think it  
is Zurek who talks about multiple copies of the result entangled  
with the environment. So many different individuals can observe the  
result of this single experiment, and they will all agree that the  
result was 

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

[David Nyman]

On 22 Jun 2017 2:46 a.m., "Bruce Kellett"  wrote:

On 22/06/2017 10:32 am, David Nyman wrote:

On 22 Jun 2017 00:31, "Bruce Kellett" < 
bhkell...@optusnet.com.au> wrote:

On 22/06/2017 1:44 am, Bruno Marchal wrote:

> On 21 Jun 2017, at 08:21, Bruce Kellett wrote:
>
>> On 21/06/2017 4:03 pm, Russell Standish wrote:
>>
>>> On Mon, Jun 19, 2017 at 12:15:31PM +1000, Bruce Kellett wrote:
>>>
 On 19/06/2017 10:23 am, Russell Standish wrote:

> I know Scott wouldn't go as far as me. For me, all such irreversible
> processes are related to conscious entities in some way. Whilst
> agreeing that Geiger counters are unlikely to be conscious, I would
> say that the output of Geiger counter is not actually discrete until
> observed by a conscious experimenter.
>
 That sounds remarkably like the "many minds" interpretation of
 quantum mechanics. This is disfavoured by most scientists because it
 leaves the physics of the billions of years before the emergence of
 the first "conscious" creature unresolved -- the first consciousness
 would cause an almighty collapse on the many minds reading.

 Each consciousness causes "an almighty collapse" in er own mind
>>> independently of any other. It's a pure 1p phenomena.
>>>
>>
>> It is actually a 3p phenomenon because there is inter-subjective
>> agreement about the fact that measurements give definite results.
>>
>
> Inter-subjectivity does not imply 3p, as it can be "only" 1p plural. Let
> me illustrate this with a variant of the WM duplication.
>
> Imagine that Bruce and John are undergoing the WM-duplication *together*.
>
> By this I mean they both enter the scanning-annihilating box, and are both
> reconstituted in Washington and in Moscow.
>
> And let us assume they do it repetitively, which means they come back to
> Helsinki, and do it again together.
>
> Obviously, the line-life past that each copies describes in its personal
> diaries grows like H followed by a sequence of W and M. The number of
> copies grows exponentially (2^n). After ten iterations, we have 2^10 = 1024
> individuals, who share an indeterminate experiences. With minor exceptions,
> they all agree that the experience has always given each times a precise
> outcome, always belonging to {W, M}. Importantly  the duplicated couples
> agreed (which was the Washington or Moscow outcome) in all duplication.
> They mostly all agreed they did not found any obvious algorithm to predict
> the sequence (the exception might concerned the guys in nameable stories,
> like:
>
> WW
>
> MM
>
> Or the development of some remarkable real number in binary, like the
> binary expansion of PI, sqrt(2), sqr(n), etc. In this case, the computable
> is made rare (and more and more negligible when n grows, those histories
> are "white rabbits histories").
>
> That is what I mean by first person plural. It concerns population of
> machine sharing self-multiplication. it is interesting to compare the
> quantum linear self-superposition with the purely arithmetical one.
>

Sure, that would seem to be reasonably described as 1p-plural. except that
there is no need to have two people enter the duplicating machine and
undergo different teleportations afterwards. Surely it is sufficient to
consider one person doing a series of polarization measurements on a
sequence of photons from an unpolarized source. That person will record
some sequence of '+' and '-' results. If the experiment is repeated N
times, there will be 2^N sequences, one in each of the generated worlds.

But that has nothing to do with inter-subjective agreement between
different observers. To see that, consider just one polarization
measurement: In order for it to be said that the measurement gave a result,
there has to be decoherence and the formation of irreversible records. I
think it is Zurek who talks about multiple copies of the result entangled
with the environment. So many different individuals can observe the result
of this single experiment, and they will all agree that the result was what
the experimenter wrote in her lab book. That is inter-subjective agreement.
It clearly has nothing to do with 1p, or 1p-plural pictures.


I think there may be a terminological confusion here. IIUC, 1p-plural
denotes, amongst other things, just such inter-subjective agreement between
mutually entangled observers.


That seems a remarkably confusing terminology. The example Bruno gave to
illustrate 1p-plural was not an example of inter-subjective agreement --
there were just repeated measurements by the one person.


That's why I said "amongst other things".

If you conflate 1p-plural with inter-subjective, what on earth is 3p?


3p is a (sometimes imaginary) perspective on some state of affairs at one
remove from the 1p views of any of the supposed participants of interest.


The notation suggested to me 'third person', or the view of an outsider

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

[Bruce Kellett]

On 22/06/2017 10:32 am, David Nyman wrote:
On 22 Jun 2017 00:31, "Bruce Kellett" > wrote:


On 22/06/2017 1:44 am, Bruno Marchal wrote:

On 21 Jun 2017, at 08:21, Bruce Kellett wrote:

On 21/06/2017 4:03 pm, Russell Standish wrote:

On Mon, Jun 19, 2017 at 12:15:31PM +1000, Bruce
Kellett wrote:

On 19/06/2017 10:23 am, Russell Standish wrote:

I know Scott wouldn't go as far as me. For me,
all such irreversible
processes are related to conscious entities in
some way. Whilst
agreeing that Geiger counters are unlikely to
be conscious, I would
say that the output of Geiger counter is not
actually discrete until
observed by a conscious experimenter.

That sounds remarkably like the "many minds"
interpretation of
quantum mechanics. This is disfavoured by most
scientists because it
leaves the physics of the billions of years before
the emergence of
the first "conscious" creature unresolved -- the
first consciousness
would cause an almighty collapse on the many minds
reading.

Each consciousness causes "an almighty collapse" in er
own mind
independently of any other. It's a pure 1p phenomena.


It is actually a 3p phenomenon because there is
inter-subjective agreement about the fact that
measurements give definite results.


Inter-subjectivity does not imply 3p, as it can be "only" 1p
plural. Let me illustrate this with a variant of the WM
duplication.

Imagine that Bruce and John are undergoing the WM-duplication
*together*.

By this I mean they both enter the scanning-annihilating box,
and are both reconstituted in Washington and in Moscow.

And let us assume they do it repetitively, which means they
come back to Helsinki, and do it again together.

Obviously, the line-life past that each copies describes in
its personal diaries grows like H followed by a sequence of W
and M. The number of copies grows exponentially (2^n). After
ten iterations, we have 2^10 = 1024 individuals, who share an
indeterminate experiences. With minor exceptions, they all
agree that the experience has always given each times a
precise outcome, always belonging to {W, M}. Importantly  the
duplicated couples agreed (which was the Washington or Moscow
outcome) in all duplication. They mostly all agreed they did
not found any obvious algorithm to predict the sequence (the
exception might concerned the guys in nameable stories, like:

WW

MM

Or the development of some remarkable real number in binary,
like the binary expansion of PI, sqrt(2), sqr(n), etc. In this
case, the computable is made rare (and more and more
negligible when n grows, those histories are "white rabbits
histories").

That is what I mean by first person plural. It concerns
population of machine sharing self-multiplication. it is
interesting to compare the quantum linear self-superposition
with the purely arithmetical one.


Sure, that would seem to be reasonably described as 1p-plural.
except that there is no need to have two people enter the
duplicating machine and undergo different teleportations
afterwards. Surely it is sufficient to consider one person doing a
series of polarization measurements on a sequence of photons from
an unpolarized source. That person will record some sequence of
'+' and '-' results. If the experiment is repeated N times, there
will be 2^N sequences, one in each of the generated worlds.

But that has nothing to do with inter-subjective agreement between
different observers. To see that, consider just one polarization
measurement: In order for it to be said that the measurement gave
a result, there has to be decoherence and the formation of
irreversible records. I think it is Zurek who talks about multiple
copies of the result entangled with the environment. So many
different individuals can observe the result of this single
experiment, and they will all agree that the result was what the
experimenter wrote in her lab book. That is inter-subjective
agreement. It clearly has nothing to do with 1p, or 1p-plural
pictures. 



I think there may be a terminological confusion here. IIUC, 

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

[spudboy100 via Everything List]

Unless someone expects "consciousness" to leak in from the cosmic foam, I am 
guessing there is nothing intrinsic in the quantum (aside from wave-function) 
that possesses a 'consciousness,' possibility. Moreover the other two 
possibilities for hypercomputing, can come from two other technologies. One id 
the Stanford developed photonic computing, and the other is various work being 
done with chunks of dna. I am guessing complexity brings consciousness. Would a 
boltzmann brain be a conscious observer? Only if it was complex enough. 



-Original Message-
From: Bruce Kellett <bhkell...@optusnet.com.au>
To: everything-list <everything-list@googlegroups.com>
Sent: Wed, Jun 21, 2017 7:31 pm
Subject: Re: “Could a Quantum Computer Have Subjective Experience?”

On 22/06/2017 1:44 am, Bruno Marchal wrote:
> On 21 Jun 2017, at 08:21, Bruce Kellett wrote:
>> On 21/06/2017 4:03 pm, Russell Standish wrote:
>>> On Mon, Jun 19, 2017 at 12:15:31PM +1000, Bruce Kellett wrote:
>>>> On 19/06/2017 10:23 am, Russell Standish wrote:
>>>>> I know Scott wouldn't go as far as me. For me, all such irreversible
>>>>> processes are related to conscious entities in some way. Whilst
>>>>> agreeing that Geiger counters are unlikely to be conscious, I would
>>>>> say that the output of Geiger counter is not actually discrete until
>>>>> observed by a conscious experimenter.
>>>> That sounds remarkably like the "many minds" interpretation of
>>>> quantum mechanics. This is disfavoured by most scientists because it
>>>> leaves the physics of the billions of years before the emergence of
>>>> the first "conscious" creature unresolved -- the first consciousness
>>>> would cause an almighty collapse on the many minds reading.
>>>>
>>> Each consciousness causes "an almighty collapse" in er own mind
>>> independently of any other. It's a pure 1p phenomena.
>>
>> It is actually a 3p phenomenon because there is inter-subjective 
>> agreement about the fact that measurements give definite results.
>
> Inter-subjectivity does not imply 3p, as it can be "only" 1p plural. 
> Let me illustrate this with a variant of the WM duplication.
>
> Imagine that Bruce and John are undergoing the WM-duplication *together*.
>
> By this I mean they both enter the scanning-annihilating box, and are 
> both reconstituted in Washington and in Moscow.
>
> And let us assume they do it repetitively, which means they come back 
> to Helsinki, and do it again together.
>
> Obviously, the line-life past that each copies describes in its 
> personal diaries grows like H followed by a sequence of W and M. The 
> number of copies grows exponentially (2^n). After ten iterations, we 
> have 2^10 = 1024 individuals, who share an indeterminate experiences. 
> With minor exceptions, they all agree that the experience has always 
> given each times a precise outcome, always belonging to {W, M}. 
> Importantly  the duplicated couples agreed (which was the Washington 
> or Moscow outcome) in all duplication. They mostly all agreed they did 
> not found any obvious algorithm to predict the sequence (the exception 
> might concerned the guys in nameable stories, like:
>
> WW
>
> MM
>
> Or the development of some remarkable real number in binary, like the 
> binary expansion of PI, sqrt(2), sqr(n), etc. In this case, the 
> computable is made rare (and more and more negligible when n grows, 
> those histories are "white rabbits histories").
>
> That is what I mean by first person plural. It concerns population of 
> machine sharing self-multiplication. it is interesting to compare the 
> quantum linear self-superposition with the purely arithmetical one.

Sure, that would seem to be reasonably described as 1p-plural. except 
that there is no need to have two people enter the duplicating machine 
and undergo different teleportations afterwards. Surely it is sufficient 
to consider one person doing a series of polarization measurements on a 
sequence of photons from an unpolarized source. That person will record 
some sequence of '+' and '-' results. If the experiment is repeated N 
times, there will be 2^N sequences, one in each of the generated worlds.

But that has nothing to do with inter-subjective agreement between 
different observers. To see that, consider just one polarization 
measurement: In order for it to be said that the measurement gave a 
result, there has to be decoherence and the formation of irreversible 
records. I think it is Zurek who talks about multiple copies of the 
result entangled with the environment. So many different individuals can 
observ

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

[David Nyman]

On 22 Jun 2017 00:31, "Bruce Kellett"  wrote:

On 22/06/2017 1:44 am, Bruno Marchal wrote:

> On 21 Jun 2017, at 08:21, Bruce Kellett wrote:
>
>> On 21/06/2017 4:03 pm, Russell Standish wrote:
>>
>>> On Mon, Jun 19, 2017 at 12:15:31PM +1000, Bruce Kellett wrote:
>>>
 On 19/06/2017 10:23 am, Russell Standish wrote:

> I know Scott wouldn't go as far as me. For me, all such irreversible
> processes are related to conscious entities in some way. Whilst
> agreeing that Geiger counters are unlikely to be conscious, I would
> say that the output of Geiger counter is not actually discrete until
> observed by a conscious experimenter.
>
 That sounds remarkably like the "many minds" interpretation of
 quantum mechanics. This is disfavoured by most scientists because it
 leaves the physics of the billions of years before the emergence of
 the first "conscious" creature unresolved -- the first consciousness
 would cause an almighty collapse on the many minds reading.

 Each consciousness causes "an almighty collapse" in er own mind
>>> independently of any other. It's a pure 1p phenomena.
>>>
>>
>> It is actually a 3p phenomenon because there is inter-subjective
>> agreement about the fact that measurements give definite results.
>>
>
> Inter-subjectivity does not imply 3p, as it can be "only" 1p plural. Let
> me illustrate this with a variant of the WM duplication.
>
> Imagine that Bruce and John are undergoing the WM-duplication *together*.
>
> By this I mean they both enter the scanning-annihilating box, and are both
> reconstituted in Washington and in Moscow.
>
> And let us assume they do it repetitively, which means they come back to
> Helsinki, and do it again together.
>
> Obviously, the line-life past that each copies describes in its personal
> diaries grows like H followed by a sequence of W and M. The number of
> copies grows exponentially (2^n). After ten iterations, we have 2^10 = 1024
> individuals, who share an indeterminate experiences. With minor exceptions,
> they all agree that the experience has always given each times a precise
> outcome, always belonging to {W, M}. Importantly  the duplicated couples
> agreed (which was the Washington or Moscow outcome) in all duplication.
> They mostly all agreed they did not found any obvious algorithm to predict
> the sequence (the exception might concerned the guys in nameable stories,
> like:
>
> WW
>
> MM
>
> Or the development of some remarkable real number in binary, like the
> binary expansion of PI, sqrt(2), sqr(n), etc. In this case, the computable
> is made rare (and more and more negligible when n grows, those histories
> are "white rabbits histories").
>
> That is what I mean by first person plural. It concerns population of
> machine sharing self-multiplication. it is interesting to compare the
> quantum linear self-superposition with the purely arithmetical one.
>

Sure, that would seem to be reasonably described as 1p-plural. except that
there is no need to have two people enter the duplicating machine and
undergo different teleportations afterwards. Surely it is sufficient to
consider one person doing a series of polarization measurements on a
sequence of photons from an unpolarized source. That person will record
some sequence of '+' and '-' results. If the experiment is repeated N
times, there will be 2^N sequences, one in each of the generated worlds.

But that has nothing to do with inter-subjective agreement between
different observers. To see that, consider just one polarization
measurement: In order for it to be said that the measurement gave a result,
there has to be decoherence and the formation of irreversible records. I
think it is Zurek who talks about multiple copies of the result entangled
with the environment. So many different individuals can observe the result
of this single experiment, and they will all agree that the result was what
the experimenter wrote in her lab book. That is inter-subjective agreement.
It clearly has nothing to do with 1p, or 1p-plural pictures.


I think there may be a terminological confusion here. IIUC, 1p-plural
denotes, amongst other things, just such inter-subjective agreement between
mutually entangled observers. Physics, in this usage, is considered as
1p-plural at least in terms of its phenomenology, because those phenomena
essentially reduce to the sum of all possible measurements of this sort.

David

But it is precisely that inter-subjective agreement that is essential for
physics -- people have to agree that experiments have definite results, and
they have to agree what those results are. Inter-subjective agreement
occurs in just one world -- although there may be similar agreements
between copies of those people entangled be decoherence with the other
possible experimental results.  Each world is then characterized by
inter-subjective agreement about the result obtained in that world.

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

[Bruce Kellett]

On 22/06/2017 1:44 am, Bruno Marchal wrote:

On 21 Jun 2017, at 08:21, Bruce Kellett wrote:

On 21/06/2017 4:03 pm, Russell Standish wrote:

On Mon, Jun 19, 2017 at 12:15:31PM +1000, Bruce Kellett wrote:

On 19/06/2017 10:23 am, Russell Standish wrote:

I know Scott wouldn't go as far as me. For me, all such irreversible
processes are related to conscious entities in some way. Whilst
agreeing that Geiger counters are unlikely to be conscious, I would
say that the output of Geiger counter is not actually discrete until
observed by a conscious experimenter.

That sounds remarkably like the "many minds" interpretation of
quantum mechanics. This is disfavoured by most scientists because it
leaves the physics of the billions of years before the emergence of
the first "conscious" creature unresolved -- the first consciousness
would cause an almighty collapse on the many minds reading.


Each consciousness causes "an almighty collapse" in er own mind
independently of any other. It's a pure 1p phenomena.


It is actually a 3p phenomenon because there is inter-subjective 
agreement about the fact that measurements give definite results.


Inter-subjectivity does not imply 3p, as it can be "only" 1p plural. 
Let me illustrate this with a variant of the WM duplication.


Imagine that Bruce and John are undergoing the WM-duplication *together*.

By this I mean they both enter the scanning-annihilating box, and are 
both reconstituted in Washington and in Moscow.


And let us assume they do it repetitively, which means they come back 
to Helsinki, and do it again together.


Obviously, the line-life past that each copies describes in its 
personal diaries grows like H followed by a sequence of W and M. The 
number of copies grows exponentially (2^n). After ten iterations, we 
have 2^10 = 1024 individuals, who share an indeterminate experiences. 
With minor exceptions, they all agree that the experience has always 
given each times a precise outcome, always belonging to {W, M}. 
Importantly  the duplicated couples agreed (which was the Washington 
or Moscow outcome) in all duplication. They mostly all agreed they did 
not found any obvious algorithm to predict the sequence (the exception 
might concerned the guys in nameable stories, like:


WW

MM

Or the development of some remarkable real number in binary, like the 
binary expansion of PI, sqrt(2), sqr(n), etc. In this case, the 
computable is made rare (and more and more negligible when n grows, 
those histories are "white rabbits histories").


That is what I mean by first person plural. It concerns population of 
machine sharing self-multiplication. it is interesting to compare the 
quantum linear self-superposition with the purely arithmetical one.


Sure, that would seem to be reasonably described as 1p-plural. except 
that there is no need to have two people enter the duplicating machine 
and undergo different teleportations afterwards. Surely it is sufficient 
to consider one person doing a series of polarization measurements on a 
sequence of photons from an unpolarized source. That person will record 
some sequence of '+' and '-' results. If the experiment is repeated N 
times, there will be 2^N sequences, one in each of the generated worlds.


But that has nothing to do with inter-subjective agreement between 
different observers. To see that, consider just one polarization 
measurement: In order for it to be said that the measurement gave a 
result, there has to be decoherence and the formation of irreversible 
records. I think it is Zurek who talks about multiple copies of the 
result entangled with the environment. So many different individuals can 
observe the result of this single experiment, and they will all agree 
that the result was what the experimenter wrote in her lab book. That is 
inter-subjective agreement. It clearly has nothing to do with 1p, or 
1p-plural pictures. But it is precisely that inter-subjective agreement 
that is essential for physics -- people have to agree that experiments 
have definite results, and they have to agree what those results are. 
Inter-subjective agreement occurs in just one world -- although there 
may be similar agreements between copies of those people entangled be 
decoherence with the other possible experimental results.  Each world is 
then characterized by inter-subjective agreement about the result 
obtained in that world.


Again, this bears no relation to Tegmark's 'bird' view. You might well 
call the bird view the 0p view, because there is no person or 
consciousness that can ever experience that view.




There is no collapse at all at the 3p level, nor even decoherence as 
such.


Decoherence is a well-understood physical phenomenon that has been 
widely observed.


I can't agree more. It might be, and should be when assuming digital 
mechanism, a first person plurality phenomenon. In the (quantum) MW, 
is the fission/differentiation of histories brought by measurement, 
and the 

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

[Bruno Marchal]

On 21 Jun 2017, at 08:21, Bruce Kellett wrote:


On 21/06/2017 4:03 pm, Russell Standish wrote:

On Mon, Jun 19, 2017 at 12:15:31PM +1000, Bruce Kellett wrote:

On 19/06/2017 10:23 am, Russell Standish wrote:
I know Scott wouldn't go as far as me. For me, all such  
irreversible

processes are related to conscious entities in some way. Whilst
agreeing that Geiger counters are unlikely to be conscious, I would
say that the output of Geiger counter is not actually discrete  
until

observed by a conscious experimenter.

That sounds remarkably like the "many minds" interpretation of
quantum mechanics. This is disfavoured by most scientists because it
leaves the physics of the billions of years before the emergence of
the first "conscious" creature unresolved -- the first consciousness
would cause an almighty collapse on the many minds reading.


Each consciousness causes "an almighty collapse" in er own mind
independently of any other. It's a pure 1p phenomena.


It is actually a 3p phenomenon because there is inter-subjective  
agreement about the fact that measurements give definite results.


Inter-subjectivity does not imply 3p, as it can be "only" 1p plural.  
Let me illustrate this with a variant of the WM duplication.


Imagine that Bruce and John are undergoing the WM-duplication  
*together*.


By this I mean they both enter the scanning-annihilating box, and are  
both reconstituted in Washington and in Moscow.


And let us assume they do it repetitively, which means they come back  
to Helsinki, and do it again together.


Obviously, the line-life past that each copies describes in its  
personal diaries grows like H followed by a sequence of W and M. The  
number of copies grows exponentially (2^n). After ten iterations, we  
have 2^10 = 1024 individuals, who share an indeterminate experiences.  
With minor exceptions, they all agree that the experience has always  
given each times a precise outcome, always belonging to {W, M}.  
Importantly  the duplicated couples agreed (which was the Washington  
or Moscow outcome) in all duplication. They mostly all agreed they did  
not found any obvious algorithm to predict the sequence (the exception  
might concerned the guys in nameable stories, like:


WW

MM

Or the development of some remarkable real number in binary, like the  
binary expansion of PI, sqrt(2), sqr(n), etc. In this case, the  
computable is made rare (and more and more negligible when n grows,  
those histories are "white rabbits histories").


That is what I mean by first person plural. It concerns population of  
machine sharing self-multiplication. it is interesting to compare the  
quantum linear self-superposition with the purely arithmetical one.









There is no collapse at all at the 3p level, nor even decoherence  
as such.


Decoherence is a well-understood physical phenomenon that has been  
widely observed.


I can't agree more. It might be, and should be when assuming digital  
mechanism, a first person plurality phenomenon. In the (quantum) MW,  
is the fission/differentiation of histories brought by measurement,  
and the measurement itself is part of the histories.







I do not know what you mean by saying "nor even decoherence as such."


Maybe Russell meant in the (3-1) view of the (assumed by Everett)  
Universal wave. Plausible. The universal wave describes a change of  
base. It is God's vision (in this still physicalist view).


Everett, that is QM without the collapse axiom, looks already like a  
solution of the computationalist mind-body problem. But it works only  
if Everett QM is itself derivable from (intensional) arithmetic.




Also, you seem to be confusing the inter-subjective 3p view with  
Tegmark's bird view. There is no person, body, or consciousness that  
ever has the bird view -- the bird is a purely formal construct and  
has nothing to do with mind or consciousness.



That is an interesting remark, but it is a highly debatable question.  
See my conversation with David Nyman, about the "the nature" of the 0p  
view: is it more 1p or 3p? Is it more like a thing or a person? Well,  
I don't know. Is the arithmetical reality conceivable as a person? You  
can see it has an infinite (and highly non mechanical) body of  
(arithmetical) knowledge, but this would be a poetical acknowledgment  
of our ignorance.




Even though everything might remain unitary at that level, no one  
can ever experience the consequences of that unitary evolution.


Hmm... You speculate that there is no global 1p for the global unitary  
evolution, which is an open problem to me. Hard to know.


Nevertheless, assuming QM, you do *experience* the *consequences* of  
the unitary evolution, right here and right now, directly, and  
indirectly, as you are using a machine whose miniaturization has been  
made possible by the QM laws + human inference of the QM laws.


With mechanism, the QM laws have to be derived from the first person  
views 

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

[Telmo Menezes]

On Wed, Jun 21, 2017 at 8:21 AM, Bruce Kellett
 wrote:
> On 21/06/2017 4:03 pm, Russell Standish wrote:
>>
>> On Mon, Jun 19, 2017 at 12:15:31PM +1000, Bruce Kellett wrote:
>>>
>>> On 19/06/2017 10:23 am, Russell Standish wrote:

 I know Scott wouldn't go as far as me. For me, all such irreversible
 processes are related to conscious entities in some way. Whilst
 agreeing that Geiger counters are unlikely to be conscious, I would
 say that the output of Geiger counter is not actually discrete until
 observed by a conscious experimenter.
>>>
>>> That sounds remarkably like the "many minds" interpretation of
>>> quantum mechanics. This is disfavoured by most scientists because it
>>> leaves the physics of the billions of years before the emergence of
>>> the first "conscious" creature unresolved -- the first consciousness
>>> would cause an almighty collapse on the many minds reading.
>>>
>> Each consciousness causes "an almighty collapse" in er own mind
>> independently of any other. It's a pure 1p phenomena.
>
>
> It is actually a 3p phenomenon because there is inter-subjective agreement
> about the fact that measurements give definite results.

It is actually not, if many minds is true.

>> There is no collapse at all at the 3p level, nor even decoherence as such.
>
>
> Decoherence is a well-understood physical phenomenon that has been widely
> observed. I do not know what you mean by saying "nor even decoherence as
> such." Also, you seem to be confusing the inter-subjective 3p view with
> Tegmark's bird view.

Are you capable of arguing your position without talking down to people?

Telmo.

> There is no person, body, or consciousness that ever
> has the bird view -- the bird is a purely formal construct and has nothing
> to do with mind or consciousness. Even though everything might remain
> unitary at that level, no one can ever experience the consequences of that
> unitary evolution.
>
> Bruce
>
>
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Re: “Could a Quantum Computer Have Subjective Experience?”

[Bruce Kellett]

On 21/06/2017 4:03 pm, Russell Standish wrote:

On Mon, Jun 19, 2017 at 12:15:31PM +1000, Bruce Kellett wrote:

On 19/06/2017 10:23 am, Russell Standish wrote:

I know Scott wouldn't go as far as me. For me, all such irreversible
processes are related to conscious entities in some way. Whilst
agreeing that Geiger counters are unlikely to be conscious, I would
say that the output of Geiger counter is not actually discrete until
observed by a conscious experimenter.

That sounds remarkably like the "many minds" interpretation of
quantum mechanics. This is disfavoured by most scientists because it
leaves the physics of the billions of years before the emergence of
the first "conscious" creature unresolved -- the first consciousness
would cause an almighty collapse on the many minds reading.


Each consciousness causes "an almighty collapse" in er own mind
independently of any other. It's a pure 1p phenomena.


It is actually a 3p phenomenon because there is inter-subjective 
agreement about the fact that measurements give definite results.



There is no collapse at all at the 3p level, nor even decoherence as such.


Decoherence is a well-understood physical phenomenon that has been 
widely observed. I do not know what you mean by saying "nor even 
decoherence as such." Also, you seem to be confusing the 
inter-subjective 3p view with Tegmark's bird view. There is no person, 
body, or consciousness that ever has the bird view -- the bird is a 
purely formal construct and has nothing to do with mind or 
consciousness. Even though everything might remain unitary at that 
level, no one can ever experience the consequences of that unitary 
evolution.


Bruce

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

[Russell Standish]

On Mon, Jun 19, 2017 at 12:15:31PM +1000, Bruce Kellett wrote:
> On 19/06/2017 10:23 am, Russell Standish wrote:
> >I know Scott wouldn't go as far as me. For me, all such irreversible
> >processes are related to conscious entities in some way. Whilst
> >agreeing that Geiger counters are unlikely to be conscious, I would
> >say that the output of Geiger counter is not actually discrete until
> >observed by a conscious experimenter.
> 
> That sounds remarkably like the "many minds" interpretation of
> quantum mechanics. This is disfavoured by most scientists because it
> leaves the physics of the billions of years before the emergence of
> the first "conscious" creature unresolved -- the first consciousness
> would cause an almighty collapse on the many minds reading.
> 

Each consciousness causes "an almighty collapse" in er own mind
independently of any other. It's a pure 1p phenomena.

There is no collapse at all at the 3p level, nor even decoherence as such.

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Visiting Senior Research Fellowhpco...@hpcoders.com.au
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Re: “Could a Quantum Computer Have Subjective Experience?”

[Bruce Kellett]

On 19/06/2017 1:17 pm, 'cdemorse...@yahoo.com' via Everything List wrote:


On Sun, Jun 18, 2017 at 7:35 PM, Kim Jones
 wrote:


> Jun 2017, at 12:15 pm, Bruce Kellett > wrote:
>
> That sounds remarkably like the "many minds" interpretation of
quantum mechanics. This is disfavoured by most scientists because
it leaves the physics of the billions of years before the
emergence of the first "conscious" creature unresolved -- the
first consciousness would cause an almighty collapse on the many
minds reading.
>
> Bruce


How do these scientists know for sure that conscious creatures
didn't evolve elsewhere before the Earth? How can we be sure that
conscious observers have been around since "the beginning"...

Kim

I would further point out: How do we know that the primary mover
of everything does not itself depend upon the emergence of primal
consciousness within its own dawning awareness... that is drawing
infinite histories out of nothing.



We know that anything like the above is a load of baloney! :-)

Bruce

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

[Bruce Kellett]

On 19/06/2017 12:35 pm, Kim Jones wrote:

Jun 2017, at 12:15 pm, Bruce Kellett  wrote:

That sounds remarkably like the "many minds" interpretation of quantum mechanics. This is 
disfavoured by most scientists because it leaves the physics of the billions of years before the 
emergence of the first "conscious" creature unresolved -- the first consciousness would 
cause an almighty collapse on the many minds reading.

Bruce

How do these scientists know for sure that conscious creatures didn't evolve elsewhere 
before the Earth? How can we be sure that conscious observers have been around since 
"the beginning"...


I presume you mean "haven't been around since the beginning"?

Well, assuming that consciousness requires a modicum of structure and 
stability, then we can be quite sure that there was no consciousness 
before the cosmological evolution of some stable structure following the 
big bang -- of the order of several million (or billion) years. I am not 
so terracentric as to require that the only possible consciousness 
evolved on earth.


The aim of physics in general, and quantum theory in particular, is to 
understand the evolution of the universe in terms that do not require 
the presence of mind or consciousness. Mind and consciousness are merely 
epiphenomena on the vastness of the physical universe: they are not central.


Bruce

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

['cdemorse...@yahoo.com' via Everything List]

Sent from Yahoo Mail on Android 
 
  On Sun, Jun 18, 2017 at 7:35 PM, Kim Jones wrote:   
> Jun 2017, at 12:15 pm, Bruce Kellett  wrote:
> 
> That sounds remarkably like the "many minds" interpretation of quantum 
> mechanics. This is disfavoured by most scientists because it leaves the 
> physics of the billions of years before the emergence of the first 
> "conscious" creature unresolved -- the first consciousness would cause an 
> almighty collapse on the many minds reading.
> 
> Bruce


How do these scientists know for sure that conscious creatures didn't evolve 
elsewhere before the Earth? How can we be sure that conscious observers have 
been around since "the beginning"...

Kim
I would further point out: How do we know that the primary mover of everything 
does not itself depend upon the emergence of primal consciousness within its 
own dawning awareness... that is drawing infinite histories out of nothing.

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

[Kim Jones]

> Jun 2017, at 12:15 pm, Bruce Kellett  wrote:
> 
> That sounds remarkably like the "many minds" interpretation of quantum 
> mechanics. This is disfavoured by most scientists because it leaves the 
> physics of the billions of years before the emergence of the first 
> "conscious" creature unresolved -- the first consciousness would cause an 
> almighty collapse on the many minds reading.
> 
> Bruce


How do these scientists know for sure that conscious creatures didn't evolve 
elsewhere before the Earth? How can we be sure that conscious observers have 
been around since "the beginning"...

Kim

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

[Bruce Kellett]

On 19/06/2017 10:23 am, Russell Standish wrote:

On Mon, Jun 05, 2017 at 12:48:51PM -0700, Brent Meeker wrote:

Here Scott Aaronson addresses the "pretty-hard problem of consciousness"

http://www.scottaaronson.com/blog/?p=1951

His idea of "participation in the Arrow of Time" is a narrower and
more technical version of my idea that consciousness only exists in
the context of an environment in which it can both perceive and act.

I would say that my "PROJECTION postulate" is a more specific version
of this, particularly in light of my observation that in the
Multiverse, all irreversible processes are Darwinian evolutionary
processes, with selection given by this PROJECTION postulate. I even
give a mechanism for how this might occur, in the form of applied
chaos theory as outline on page 105/6 of my book.

I know Scott wouldn't go as far as me. For me, all such irreversible
processes are related to conscious entities in some way. Whilst
agreeing that Geiger counters are unlikely to be conscious, I would
say that the output of Geiger counter is not actually discrete until
observed by a conscious experimenter.


That sounds remarkably like the "many minds" interpretation of quantum 
mechanics. This is disfavoured by most scientists because it leaves the 
physics of the billions of years before the emergence of the first 
"conscious" creature unresolved -- the first consciousness would cause 
an almighty collapse on the many minds reading.


Bruce



But then Scott is also a lapsed MWIist, so that could be the source of
our disagreement :).



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

[Russell Standish]

On Mon, Jun 05, 2017 at 12:48:51PM -0700, Brent Meeker wrote:
> Here Scott Aaronson addresses the "pretty-hard problem of consciousness"
> 
> http://www.scottaaronson.com/blog/?p=1951
> 
> His idea of "participation in the Arrow of Time" is a narrower and
> more technical version of my idea that consciousness only exists in
> the context of an environment in which it can both perceive and act.

I would say that my "PROJECTION postulate" is a more specific version
of this, particularly in light of my observation that in the
Multiverse, all irreversible processes are Darwinian evolutionary
processes, with selection given by this PROJECTION postulate. I even
give a mechanism for how this might occur, in the form of applied
chaos theory as outline on page 105/6 of my book.

I know Scott wouldn't go as far as me. For me, all such irreversible
processes are related to conscious entities in some way. Whilst
agreeing that Geiger counters are unlikely to be conscious, I would
say that the output of Geiger counter is not actually discrete until
observed by a conscious experimenter.

But then Scott is also a lapsed MWIist, so that could be the source of
our disagreement :).


-- 


Dr Russell StandishPhone 0425 253119 (mobile)
Principal, High Performance Coders
Visiting Senior Research Fellowhpco...@hpcoders.com.au
Economics, Kingston University http://www.hpcoders.com.au


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

[Bruno Marchal]

On 05 Jun 2017, at 21:48, Brent Meeker wrote:

Here Scott Aaronson addresses the "pretty-hard problem of  
consciousness"


http://www.scottaaronson.com/blog/?p=1951



Not much time to read it all, but very interesting. But they miss the  
point. They still don't listen to the machine.


I would say they (still) miss the "third incompleteness theorem of  
Gödel-1931", like Penrose (and this despite I think Aaronson is aware  
of the main mistake made by Penrose).


The "third theorem" of Gödel is at the end of the 1931 paper where he  
explains that the proof of the second theorem (no consistent theory/ 
machine can prove its own consistency) can be carried out in (I prefer  
to say by) the theory/machine itself.


If the machine/theory is PA, or a PA theorem prover, the second  
theorem says that if PA is consistent then consistent(PA) is not  
provable by PA.


The "third theorem" says that PA already knew it. It says that PA  
proves (consistent(PA) -> Non(provable(consistent(PA)).


Gödel was quite quick on this, and that will be proved with all the  
precision required by Hilbert and Bernays in 1939, and generalized and  
embellished in an utter strange way by Löb, in 1955.


That changes everything. Consciousness becomes almost easy, but matter  
needs revision.




His idea of "participation in the Arrow of Time"


I will have to look at that.



is a narrower and more technical version of my idea that  
consciousness only exists in the context of an environment in which  
it can both perceive and act.


Absolutely. We have already discuss this. But you don't to reify it.

I have to explain you that for the (Löbian) machine or number Gödel's  
COMpleteness theorem (1930), a machine/theory/number is consistent if  
and only if there is a reality satisfying its beliefs. Logicians uses  
"model" for reality, but physicists uses model for theory. So I will  
use reality. A reality is "modelled" by a structured collection. The  
(standard) model of PA is the structure (N, 0, +, *) with their usual  
intepretation.


You ask for an environment. Translated in arithmetic, this is asking  
for a reality, and thus (by completeness) for consistency, which ~Bf =  
Dt (D = ~B~).


When you ask, for consciousness, that we add the reality to the  
machine, so that things are contextualized, your argument, in a  
language that PA can understand, is to replace the simple *belief of  
p* by *belief of p and consistency (of p)*. You motivate for the  
passage of []p to the passage of []p & <>t.

(Note that we have Bp & Dp equivalent with Bp & Dt)

Gödel makes <>p unavailable by logic, so that indeed it makes sense,  
and changes the logic, to add such a requirement. <>p means (for us,  
the machine can miss this, or misapplies this, ...) the existence of a  
structure (environment) which satisfies p. It is also the necessary  
requirement to make sense of a probability or any measure on  
uncertainty.


We get it by the passage from []p to []p & p, or Bp to Bp & p. The  
idea of Theaetetus. This entails Dp. Truth and correctness implies  
consistency, but consistency

does not imply correctness/truth.

You are right we must take the environment, but as we cannot justify  
it, that requirement can be used to define that (type) of consciousness.


Consciousness is (first person self)-knowledge, provided more aptly by  
Bp & p (than the mere representational belief Bp), so your requirement  
for consciousness is more given by Bp & p & Dt. Amazingly perhaps,  
incompleteness differentiates again the logics, and it corresponds  
more to immediate perception, sensibility. Bp & p & Dt is less  
solipsistic than the "pure" Bp & p.


The universal machine gives us already a "theory of consciousness"  
which is S4Grz, and X, X*, X1, X1*, and all nuances imposed by self- 
referential correctness. It is also empirically testable, as physics  
should be obtained with S4Grz1, Z1*, X1*.


They should listen to the machine, or to those listening to the machine.

The physicists have the good motivations, but the bad strategy. The  
logicians have the good strategy, but the bad motivation.


The book by Cohen(*) shed a lot of light on the recent origin of this  
situation, and why logicians are anxious with the possibility that  
logic could be applied in philosophy, theology, biology (and when I  
was young, even in computer science!).


Thanks for the link, I will surely come back to it, and comment.

Bruno

(*) Cohen J. Daniel, 2007. Equations from God, Pure Mathematics and  
Victorian Faith, John Hopkins Press, Baltimore.






A lot of good comments too.









Brent

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