Does the Observer have anything to do with conscious choice

[rkastner]

Under TI, disentanglement occurs very quickly due to absorption, and resulting collapse. At its root, this is a relativistic process (via field coupling), not taken into account in the usual attempts to interpret QM. This is explained in my book and this paper: http://arxiv.org/abs/1204.5227 We like to think that nonrelativistic QM is a 'stand-alone' theory, but it isn't. There are relativistic processes going on all the time.

 

[Maui]

With regards to decoherence, if an apparatus is in a lab - that is as much as possible isolated from the environment - does the apparatus still experience decoherence due to the 'universal entanglement' (term from http://www.decoherence.de/). Therefore, the apparatus is not isolated from the environment, and never can be if we take Quantum Mechanics seriously?

This explanation assumes that there is an apriori classical environment(made of bullet-like or ball-like particles?) that causes quantum systems to collapse during interaction. It seems decoherence can not go without a theory of measurements(or something like the MWI) and at this point the MWI becomes the only natural sounding option.

 

[rkastner]

This explanation assumes that there is an apriori classical environment(made of bullet-like or ball-like particles?) that causes quantum systems to collapse during interaction. It seems decoherence can not go without a theory of measurements(or something like the MWI) and at this point the MWI becomes the only natural sounding option.

It's true that decoherence cannot account for collapse. However MWI is certainly not the only natural option (for one thing, it has big trouble accounting for the Born Rule in natural physical terms). Have you considered PTI? Intro material here:

rekastner.wordpress.com

 

[bhobba]

It seems decoherence can not go without a theory of measurements(or something like the MWI) and at this point the MWI becomes the only natural sounding option.

It does not seem that way way at all - there are many different ways of dealing with it and even the option the I take of simply saying, basically, it don't matter.

The key issue is the difference between a proper and an improper mixture. MWI assumes nothing happens - we simply partition the universal wave-function into the states of the mixture. The basic axiom of my approach to QM has all systems of the same information carrying capacity equivalent so there is no issue. The TI has another take and still others like GRW, and Consistent Histories exist.

Thanks
Bill

 

[Maui]

It does not seem that way way at all - there are many different ways of dealing with it and even the option the I take of simply saying, basically, it don't matter.

The key issue is the difference between a proper and an improper mixture. MWI assumes nothing happens - we simply partition the universal wave-function into the states of the mixture. The basic axiom of my approach to QM has all systems of the same information carrying capacity equivalent so there is no issue. The TI has another take and still others like GRW, and Consistent Histories exist.

Thanks
Bill

A system's information capacity has what to do with its probability for a given measurement outcome? I don't see a connection. If it's another no-interpretation interpretation, because supposedly that is what you observe and when you shut up and calcule you get the probabilities - ok, but the role of an interpretation is to provide a framework why we observe what we do.

 

[meBigGuy]

I am not sure you are up on exactly what the relational interpretation is saying. Wikipedia gives a nice account:
http://en.wikipedia.org/wiki/Relational_quantum_mechanics

For me its a bit redundant - my view of QM includes it anyway, its just not stated explicitly - its more or less understood its like that. And its only of any concern prior to decoherence which generally occurs very very quickly meaning all observers agree on it after that. You undoubtedly can cook up situations where it's not like that but they in general are not what we see in the everyday world around us.
l

I have read that paragraph, but maybe I'm not understanding it. If you are saying "its more or less understood that it's like that" (which is what I think), why does it have a separate heading? And the description doesn't seem to say it is "like that". Or, is that just your personal view?

The thing I keep getting back to is the interaction of two particles, like an electron and a proton. That interaction gives rise to the probabilities associated with a subsequent observation. But the two particles are interacting, so must be observing each other. The proton "knows" more about the electron than we do since it is interacting and affecting it. We know about the pair as a probability distribution until we actually observe it. That, to my ignorant mind, seems to drive one to a relational interpretation.

 

[bhobba]

I have read that paragraph, but maybe I'm not understanding it. If you are saying "its more or less understood that it's like that" (which is what I think), why does it have a separate heading? And the description doesn't seem to say it is "like that". Or, is that just your personal view?

Different interpretations emphasize different things. Just because one interpretation chooses not to worry about it doesn't meant it doesn't agree with it. It's just not important IMHO because decoherence occurs very very quickly and all observers agree on the resultant mixed state.

The thing I keep getting back to is the interaction of two particles, like an electron and a proton. That interaction gives rise to the probabilities associated with a subsequent observation. But the two particles are interacting, so must be observing each other.

For some reason you don't seem to get an observation is something that leaves a mark here in the macro world.

Thanks
Bill

 

[bhobba]

A system's information capacity has what to do with its probability for a given measurement outcome?

Measurement outcomes convey information. The actual values of an outcome are arbitrary - its the information it conveys that is important. Think of the Stern-Gerlach experiment. It has spin up or down. That coveys a bit of information - what we call the bit on and off is irrelevant. But to be clearer its equivalent to saying systems that are observationally equivalent are equivalent. Information seems to be the buzzword in physics these days. What it means here is since an improper mixed state is observationally the same as a proper one then by the axiom they are equivalent - measurement problem solved.

Thanks
Bill

 

[arkajad]

But to be clearer its equivalent to saying systems that are observationally equivalent are equivalent.

It would be even more clear if we could specify what we mean by a "system". Different axiomatizations may mean different things. Do we mean a "single object" or an "ensemble of objects"?
And under what conditions? People using the same word "system" but with different meaning or undefined at all will argue without realizing that they are talking about different things.

So, one should define "system", "observationally equivalent", "equivalent".

 

[bhobba]

It would be even more clear if we could specify what we mean by a "system". Different axiomatizations may mean different things. Do we mean a "single object" or an "ensemble of objects"? And under what conditions? People using the same word "system" but with different meaning or undefined at all will argue without realizing that they are talking about different things. So, one should define "system", "observationally equivalent", "equivalent".

See the paper I linked to before where the 3 axioms came from. That paper is a bit sloppy but when read with Hardy's seminal paper its a lot clearer:
http://arxiv.org/pdf/quant-ph/0101012v4.pdf

But the terms you mention seem pretty clear.

Can I ask you a question? Is your background math/physics or philosophy? In my experience people that get hung up on what at least seems to me pretty obvious semantics are more likely into philosophy. Its the same sort of stuff you see when people think observation implies an observer. The better books like Dirac and Ballentine are rather careful to use the term measurement instead of observer but once you understand what's meant its not confusing.

Thanks
Bill

 

[arkajad]

See the paper I linked to before where the 3 axioms came from. That paper is a bit sloppy but when read with Hardy's seminal paper its a lot clearer

Axiom 5 (in Hardy, 3 in yours) involves the term "pure state" that has not been defined.

Can I ask you a question? Is your background math/physics or philosophy?

To get my math/physics degree I had to pass an exam in philosophy as well.

 

[bhobba]

Axiom 5 involves the term "pure state" that has not been defined.

It's a standard term in QM and probability theory.

It means a state that can't be expressed as the convex sum of other states. In probability theory that would be the possible outcomes and in QM a state (here state is the general state defined by Von Neumann being a positive operator of unit trace) with only one eigenvector.

My background is applied math so I am not that sensitive to what others think of as semantic imprecision.

Thanks
Bill

 

[arkajad]

It means a state that can't be expressed as the convex sum of other states.

Where is the proof that the set of states is a convex set? It should precede the use of the term "pure state".

 

[bhobba]

Where is the proof that the set of states is a convex set? It should precede the use of the term "pure state".

Its a definition - no proof required.

These are standard terms.

Thanks
Bill

 

[micromass]

Where is the proof that the set of states is a convex set? It should precede the use of the term "pure state".

Where do you need that the set of state is convex in order to define pure states?

 

[bhobba]

Where do you need that the set of state is convex in order to define pure states?

Its in Ballentine page 50 Section 2.3 - its a very simple theorem from the definition of a pure state proven on p52. A state is pure iff it can't be expressed as the convex sum of other states.

Scratching my head why standard stuff like this needs to be explained.

Thanks
Bill

 

[arkajad]

Where do you need that the set of state is convex in order to define pure states?

This is what boobba wrote:

"It means a state that can't be expressed as the convex sum of other states."

"Its a definition - no proof required."

But I do not see the term "convex" in the definition.

 

[micromass]

Its in Ballentine page 50 Section 2.3 - its a very simple theorem form the definition of a pure state proven on p52. A state is pure iff it can't be expressed as the convex sum of other states.

Scratching my head why standard stuff like this needs to be explained.

Thanks
Bill

Sure, I know perfectly what a pure state is. But I fail to see why you need to prove that the states form a convex set in order to define a pure state. A pure state is a state that can't be expressed as the nontrivial convex combination of other states. Nowhere is needed that the states actually do form a convex set.

Sorry if I'm being pedantic.

 

[arkajad]

Its in Ballentine page 50 Section 2.3

But I am not asking about Belinfante's set of axioms. I am asking about about sets of axioms in the two mentioned papers.

 

[arkajad]

Sorry if I'm being pedantic.

It's OK. One should first define "convex combination of states" starting with the axioms and only with them. Otherwise why would we need axioms?

 

[micromass]

It's OK. One should first define "convex combination of states" starting with the axioms and only with them. Otherwise why would we need axioms?

OK, but don't you think that the concept of "convex combination" is well known to everybody reading this paper. I mean, convex combinations don't even have anything to do with QM, they are rather a mathematical concept.

 

[bhobba]

This is what boobba wrote:

"It means a state that can't be expressed as the convex sum of other states."

"Its a definition - no proof required."

But I do not see the term "convex" in the definition.

Its dead simple. Look the outcomes with dead certainty are unit vectors. These are the pure states of probability theory. A vector with positive entries that adds up to one is a probability vector. Only the vectors with unit entries can't be expressed as a convex sum of other vectors. I suppose no proof required might be a little off the mark - but not by much - it's rather trivial.

Both those papers are a bit like that - they are somewhat of a slog to go through in that if you haven't come across these terms before it may take a bit of getting used to.

Thanks
Bill

 

[bhobba]

OK, but don't you think that the concept of "convex combination" is well known to everybody reading this paper. I mean, convex combinations don't even have anything to do with QM, they are rather a mathematical concept.

My bad. Like I say my background is in math so I guess stuff that's common in math I take more or less as given.

Both those papers are of the mathematical physics variety so that sort of terminology tends to be rampant.

Thanks
Bill

 

[arkajad]

OK, but don't you think that the concept of "convex combination" is well known to everybody reading this paper.

Convex combination of "what"? Of two real numbers? I know. Of two trees? I do not know, because I do not know how to add trees and multiply them by numbers to get other trees. The same with "states"? State has been introduced in the axioms. But no operations on state have been discussed. Neither the meaning of these operations.

 

[bhobba]

Convex combination of "what"? Of two real numbers? I know. Of two trees?

Scratching head - of probability vectors.

Thanks
Bil

 

[micromass]

Convex combination of "what"? Of two real numbers? I know. Of two trees? I do not know, because I do not know how to add trees and multiply them by numbers to get other trees. The same with "states"? State has been introduced in the axioms. But no operations on state have been discussed. Neither the meaning of these operations.

Well, since a state is defined as a certain operator, I think it's not difficult to infer that we take the sum and scalar multiplication of operators as defined in mathematics. If somebody doesn't know how to add two operators, then I don't think he should be reading this paper.

I mean, it's clear this paper takes for granted the required math such as linear algebra/functional analysis. So we should read the paper with that in mind. Otherwise, the paper would be far too long, as they would also need to explain what hermitian is and what the trace is.

 

[arkajad]

Well, since a state is defined as a certain operator.

Definition of "State" from Hardy's paper:

"The state associated with a particular preparation is defined to be (that thing represented by) any
mathematical object that can be used to determine the probability associated with the outcomes of any measurement that may be performed on a system prepared by the given preparation."

So it is not defined as an operator.

And then we have

"Axiom 5 Continuity. There exists a continuous reversible transformation on a system between any
two pure states of that system.

The first four axioms are consistent with classical probability theory but the fifth is not (unless the
word “continuous” is dropped). "

Nowhere in the axioms we have operators.

 

[bhobba]

Nowhere in the axioms we have operators.

Its a definitional thing. Given a measurement setup with outcomes yi like they have in the paper one defines the operator associated with the setup as Ʃ yi |bi><bi| for some |bi> not yet determined (they are the states that are the possible outcomes). This is the first axiom of the treatment you find in Ballentine. The second axiom in Ballentine, Born's rule, is what the paper is concerned about - it really is the key axiom. It's a very interesting thing that QM basically depends on just these two axioms. And also a very interesting thing that you can start with the first axiom as a definition (of course you need to justify why you would want such a cockeyed definition in the first place - but that's another story) and derive the second via Gleason's theorem with the assumption of non-contextuality. When people ask me what's the simplest way of viewing QM so it doesn't look like its pulled out of a hat its what I tell them. At its foundations, when viewed the right way, QM is quite natural - but only when viewed the right way.

Although I haven't seen it, and need to check them out when I get a bit of time, I am sure a lot of this stuff is in the video lectures on a course in Quantum Foundations the Perimeter Institute have archived:
http://www.pirsa.org/C12002

Because its something that really interests me I sometimes forget it not as well known as perhaps it should be. This is one of the reasons I am such a big fan of Ballentine - he develops QM this way from the start and examines closely exactly what it means.

Thanks
Bill

 

[meBigGuy]

For some reason you don't seem to get an observation is something that leaves a mark here in the macro world.

I totally get what you are saying, but it is a relative thing (relational). My "here in the macro world" is different than a particle's "here in the macro world". When two particles interact, what happens? Did one particle leave its mark on the other, or not? Yet to us they are still represented by wave functions.

Actually, I have a problem with the phrase "here in the macro world" as if there is some other world. Where is "there in the non-macro world"? Seems there is one world that things exist in, and our knowledge about them is limited, so we express what we know as a probability: the probability of what we might measure when we really pin it down. Relational interpretations are saying that different entities can see different things about an objects state or potential measurable state. One might see it as a superposition, and the other might not. ("for example, to one observer at a given point in time, a system may be in a single, "collapsed" eigenstate, while to another observer at the same time, it may be in a superposition of two or more states."

And this isn't limited to the micro world:
"However, it is held by relational quantum mechanics that this applies to all physical objects, whether or not they are conscious or macroscopic."

I just want to understand two interacting particles that I have not observed but know things about as a pair. How does each particle view the other particle. How is that view different than how I view the pair. They obviously have to be consistant, but not neccessarily the same.

Sometimes you seem to take a pure instrumentalist view, that these things are beyond QM, but then other times you acknowledge there is something more.

 

[bhobba]

When two particles interact, what happens? Did one particle leave its mark on the other, or not?

A mark is something visible here in the macro world. Obviously particles interacting do not do that. When they interact you consider their tensor product and via their interaction Hamiltonian they become entangled - that's the very essence of entanglement.

Basically for particles to leave a mark they need to interact with a macro object. It becomes entangled with the macro object and leaves a mark. QM is a theory about such marks. What its doing or any other visualizations like particles leaving marks on other particles are not part of the theory. Sure it can become entangled with it but that is entirely different to being measured, leaving a mark, observed or the other words bandied about.

Can I suggest you study a book like Ballentine so you know the detail of this sort of stuff?

Or at least have a look at Susskinds lectures on it:
http://theoreticalminimum.com/courses/quantum-entanglement/2006/fall

Thanks
Bill