Assumptions of the Bell theorem

[Lynch101]

That is not EPR, it is your statement. EPR say that if you can predict with 100% certainty the outcome of a measument, then a complete theory must accout for that, the observable must have a value before the measurement. What you are saying is that the system must be somewhere (because it cannot be nowhere), therefore any complete theory must have values for positions at any time.

Yes, EPR did say that, but they also said that was just one possible way, not the only possible way.

It seems to us that this criterion, while far from exhausting all possible ways of recognizing a physical reality, at least provides us with one.

It is a natural assumption that a "system" is alwas there. But it still is an assumption. It is silly to question it in the case of the moon. At least in some sense it is always there. But it is it an irrefutable fact in the case of "objects" like electrons and photons?

H.G. Wells wrote:
"It may be that we exist and cease to exist in alternations, like the minute dots in some form of toned printing or the succession of pictures on a cinema film."
(Science and Ultimate Truth, 1931)

Wells obviously had a more general notion of "system".

If the system ceases to exist, how can it interact with the measurement device? Does the system spontaneously comes into existence at the precise moment we would expect it to interact with the measurement device? Does the system, which is made of 'something' disappear into complete 'nothingness' i.e. go from existing to not-existing? Equally does it physically manifest out of complete nothingness?
If so, this would seem to be more 'spooky action' that needs explaining or describing.

 

[Lynch101]

It is a natural assumption that a "system" is alwas there. But it still is an assumption. It is silly to question it in the case of the moon. At least in some sense it is always there. But it is it an irrefutable fact in the case of "objects" like electrons and photons?

H.G. Wells wrote:
"It may be that we exist and cease to exist in alternations, like the minute dots in some form of toned printing or the succession of pictures on a cinema film."
(Science and Ultimate Truth, 1931)

Wells obviously had a more general notion of "system".

If the system ceases to exist, how can it interact with the measurement device? Does the system spontaneously comes into existence at the precise moment we would expect it to interact with the measurement device? Does it just randomly come into existence at the locations where measurement devices just happen to be located? Does the system, which is made of 'something' disappear into complete 'nothingness' i.e. go from existing to not-existing? Equally does it physically manifest out of complete nothingness?
If so, this would seem to be more 'spooky action' that needs explaining or describing.

 

[Lynch101]

That quantum systems are located somewhere is described by QT in the sense that the probability that the particle is somewhere is 1 (supposed it is sure that it is indeed still there and not annihilated with some antiparticle).

Can it be anywhere?

 

[vanhees71]

It's described by the position-probability distribution, which tells you for any region in space what the probability is to find it there when looking.

 

[Demystifier]

You also cannot conclude from the fact that Q(F)T gives "only" probabilities for the outcomes of measurments, that it must necessarily be an incomplete theory. It may well be that Nature really behaves inherently probabilistic.

You are missing the point. In the first sentence, the emphasis is not on "probabilities". The emphasis is on "measurements". If a theory only predicts outcomes of measurements (irrespective of whether those predictions are probabilistic or deterministic), then the theory is incomplete. A complete theory (according to this philosophy, which you are not obliged to accept) should describe things at the microscopic level without referring to measurements. Quantum theory in its standard minimal form cannot consistently talk about probabilities without talking about measurements. If a probabilistic theory is complete, then this complete theory should talk about probabilities without measurements. Standard minimal QM does not do that.

 

[Demystifier]

It's described by the position-probability distribution, which tells you for any region in space what the probability is to find it there when looking.

Yes, when looking. But according to people who argue that minimal QM is incomplete, the complete theory should say something about probabilities even in the absence of looking.

 

[Lynch101]

It's described by the position-probability distribution, which tells you for any region in space what the probability is to find it there when looking.

This is an explanation of what it means to give the probability of measurement outcomes. What we need is a description of the system prior to measurement.

If we drill down into the statement however, we can try to see if it tells us anything about the system prior to measurement.

We always measure the system to be in a definite location and the probability distribution tells us the likelihood of measuring a definite position for the system, at the given location. Does this mean that the system always has a definite location but due to a lack of information we can only predict the probability of that definite location?

If not, then what does it mean, in our universe, to not have a definitive position but to still be located somewhere in the universe?
Does it mean:
- the system is located in more than one place at a time?
- the system pops in an out of existence?
- [insert other possible questions].If the answer is that it only allows us to make probabilistic predictions as to the definite location of the system, after it interacts with a measurement device, then, by definition, it is an incomplete description of physical reality.

 

[facenian]

I believe that trying to "prove" the incompleteness issue as a mathematical theorem is not the correct approach. It could lead to endless unproductive discussions. You either postulate its truth or reject it. It is an epistemological possion.
EPR tried to prove it but Einstein did not like it. That is a very important point. We should listen to Einstein, not to EPR. In my opinion, EPR produced much confusion introducing unnecessary metaphysical elements.
Einstein "proved" incompleteness by assuming locality (through his separation principle). That means that completeness and locality cannot go together when we assume a given notion of locality.
One of the important things that the Bell theorem tells us when it is correctly interpreted is that to keep locality we must change its usual meaning and replace "Local Causality" with "No-signaling". That is one way, that locality and completeness can get along together.
I do not agree with those who cheerfully declare that quantum theory is local by construction. That explains nothing.
Not to mention the absurd claim that QM is local because the Bell inequality is a classical result.

 

[Demystifier]

I do not agree with those who cheerfully declare that quantum theory is local by construction. That explains nothing.

Indeed. If one looks carefully at the QFT axioms, one can see that some axioms are local in a certain sense by construction, but some are not. In particular, the Born rule axiom is not local by construction (here I'm not saying that it is nonlocal by construction). So one must look carefully at consequences of all axioms together to see whether the theory as a whole is local or not. Depending on how exactly one defines locality of theory as a whole, one obtains that the theory is local in one sense and nonlocal in another. Which definition of locality is more relevant or makes more physical sense is a matter of philosophy.

 

[WernerQH]

Does the system, which is made of 'something' disappear into complete 'nothingness' i.e. go from existing to not-existing? Equally does it physically manifest out of complete nothingness?

Yes. Why not? (It was obviously conceivable for Wells.)

Your addition "which is made of 'something'" indicates a metaphysical belief that the universe is made from some "stuff". Feel free to seek a description of Nature that conforms more closely to your intuition.

 

[Lynch101]

Yes. Why not? (It was obviously conceivable for Wells.)

I would question how conceivable it actually is. How does 'something' cease to exist?

EDIT: or the more difficult age old question, how do you get 'something' from 'absolutely nothing'?

Your addition "which is made of 'something'" indicates a metaphysical belief that the universe is made from some "stuff". Feel free to seek a description of Nature that conforms more closely to your intuition.

'Something' is simply the bluntest descriptor we can use for existence. We know that there is existence because there is experience. Whatever it is that exists we apply the label 'something' to it. Whatever exists is made of 'something' - you might call it 'stuff'. It can't be made of 'nothing' because 'nothing', by definition, does not exist.

It's not possible for 'something' i.e. existence to come from 'nothing' i.e. non-existence - at least not without some 'spooky action'.

Equally, how 'something' could suddenly cease to exist would require similar 'spooky action'.

 

[vanhees71]

You are missing the point. In the first sentence, the emphasis is not on "probabilities". The emphasis is on "measurements". If a theory only predicts outcomes of measurements (irrespective of whether those predictions are probabilistic or deterministic), then the theory is incomplete. A complete theory (according to this philosophy, which you are not obliged to accept) should describe things at the microscopic level without referring to measurements. Quantum theory in its standard minimal form cannot consistently talk about probabilities without talking about measurements. If a probabilistic theory is complete, then this complete theory should talk about probabilities without measurements. Standard minimal QM does not do that.

Then no physics is realistic or complete, because all of physics describes or predicts observations and thus, in quantified form, the outcome of measurements. How do you want to describe anything if not by referring to the phenomena we can observe about it?

 

[vanhees71]

Yes, when looking. But according to people who argue that minimal QM is incomplete, the complete theory should say something about probabilities even in the absence of looking.

If I never look, I can't check any predicted probabilities.

 

[vanhees71]

This is an explanation of what it means to give the probability of measurement outcomes. What we need is a description of the system prior to measurement.

If we drill down into the statement however, we can try to see if it tells us anything about the system prior to measurement.

We always measure the system to be in a definite location and the probability distribution tells us the likelihood of measuring a definite position for the system, at the given location. Does this mean that the system always has a definite location but due to a lack of information we can only predict the probability of that definite location?

If not, then what does it mean, in our universe, to not have a definitive position but to still be located somewhere in the universe?
Does it mean:
- the system is located in more than one place at a time?
- the system pops in an out of existence?
- [insert other possible questions].If the answer is that it only allows us to make probabilistic predictions as to the definite location of the system, after it interacts with a measurement device, then, by definition, it is an incomplete description of physical reality.

As I repeatedly said, the description of the system prior to measurement is given by the quantum state (statistical operator) at the initial time (after the "preparation" is finished). In classical mechanics it's given by the point in phase space at the initial time.

What does it mean not to have a definite position is clear: You cannot predict with certainty to find the particle in a given region in space but you can give only the probability to find the particle in any given region. That's all we know given the quantum state of the particle, and according to all observations today that's also all we can know in principle.

 

[vanhees71]

Indeed. If one looks carefully at the QFT axioms, one can see that some axioms are local in a certain sense by construction, but some are not. In particular, the Born rule axiom is not local by construction (here I'm not saying that it is nonlocal by construction). So one must look carefully at consequences of all axioms together to see whether the theory as a whole is local or not. Depending on how exactly one defines locality of theory as a whole, one obtains that the theory is local in one sense and nonlocal in another. Which definition of locality is more relevant or makes more physical sense is a matter of philosophy.

Can you specify more clearly, what you mean by the Born rule were not local? I have no clue, how the Born rule may be considered as local or non-local at all.

 

[Lynch101]

you can give only the probability to find the particle in any given region.

What does this tell us about the system?

We can only predict, with probability, that we will find the particle in any given region. Yet, when we measure the particle, we always measure it with a definite location. Does this mean that the particle always has a definite location but, due to missing information, we can only predict [with probability] where this definite location will be?

That's all we know given the quantum state of the particle, and according to all observations today that's also all we can know in principle.

There is a distinction to be made, as you have here, between all that we can know in principle and all there is to know. The latter would constitute a complete description of physical reality. It's possible that no more complete theory is possible but that would not mean that the given theory is a complete description of the univervse.

 

[PeterDonis]

The system must be located somewhere in the universe prior to measurement. If it were not, then it couldn't interact with the measurement device in the first place.

QM has no such requirement. In QM it is perfectly possible for systems that do not have definite positions to interact. In fact, since in QM states with definite positions (eigenstates of the position operator) are not even normalizable, they are not realizable physically, so strictly speaking, nothing in QM ever has a definite position. Yet things interact in QM just fine.

 

[PeterDonis]

I'm not suggesting the theory must have definite values, but it must have some description of its location - whatever form that may take.

A theory/interpretation which only gives the probability of a location after interacting with a measurement device does not, by definition, describe the location prior to measurement. Thus making it, by definition, an incomplete description of physical reality.

Again, QM has no such requirement. See my previous post just now.

 

[PeterDonis]

We always measure the system to be in a definite location

No, we don't. All measurements of position have finite error bars; it is impossible to measure position with infinite precision.

 

[vanhees71]

QM has no such requirement. In QM it is perfectly possible for systems that do not have definite positions to interact. In fact, since in QM states with definite positions (eigenstates of the position operator) are not even normalizable, they are not realizable physically, so strictly speaking, nothing in QM ever has a definite position. Yet things interact in QM just fine.

Yes, and indeed that implies that positions are always to a certain extent "uncertain". A $\delta$-distribution is indeed not a square-integrable function!

 

[PeterDonis]

Yet, when we measure the particle, we always measure it with a definite location.

No, we don't. See my previous post just now.

Does this mean that the particle always has a definite location but, due to missing information, we can only predict [with probability] where this definite location will be?

Your question here is meaningless since it is based on a false premise. See above.

 

[Lynch101]

QM has no such requirement. In QM it is perfectly possible for systems that do not have definite positions to interact. In fact, since in QM states with definite positions (eigenstates of the position operator) are not even normalizable, they are not realizable physically, so strictly speaking, nothing in QM ever has a definite position. Yet things interact in QM just fine.

I'm not saying that it must have a definite location in order to interact with the measurement device. I said it 'must be located somewhere in the universe'. If it is not in the universe it cannot interact with the measurement device, which is.

Also, 'somewhere in the universe' is also not a definite location.

 

[Lynch101]

No, we don't. All measurements of position have finite error bars; it is impossible to measure position with infinite precision.

The reasoning still applies just with the caveat, 'within finite error bars' appended.

 

[Lynch101]

No, we don't. See my previous post just now.Your question here is meaningless since it is based on a false premise. See above.

The question can be rephrased as:

does this mean that the particle always has a definite location, within the given error bars, but, due to missing information, we can only predict [with probability] where this definite location, within the given error bars, will be?

 

[vanhees71]

No. Take a wave function of the form
$$\psi(\vec{x})=N \{\exp[-(\vec{x}-\vec{a})^2/(4 \sigma_1^2)] + \exp[-(\vec{x}-\vec{b})^2/(4 \sigma_2^2)] \},$$
where $|\vec{a}-\vec{b}| \gg \text{max}(\sigma_1,\sigma_2)$.

 

[PeterDonis]

I'm not saying that it must have a definite location in order to interact with the measurement device. I said it 'must be located somewhere in the universe'.

Okay, so what does "located somewhere in the universe" mean? What does it correspond to in the math of QM?

I strongly suspect that you will be unable to give a valid answer to this question that supports the claims you are making--i.e., that is inconsistent with the things other people are saying that you have been objecting to.

does this mean that the particle always has a definite location, within the given error bars, but, due to missing information, we can only predict [with probability] where this definite location, within the given error bars, will be?

If "has a definite location" means "is located at some single point", then the answer to this is obviously no, since, as I have already stated, states with definite positions--where a particle is located at a single point--are not normalizable and hence cannot be realized physically.

If "has a definite location" means something else, then you need to explain what you are using that term to mean.

 

[vanhees71]

I think the tautology that a "particle is located somewhere in the universe" is the very weak assumption that, given that there is a particle of a certain kind and that it has a position observable, then
$$\int_{\mathbb{R}^3} \mathrm{d}^3 x \rho(t,\vec{x},\vec{x})=1.$$
This is indeed already in the very foundations of quantum theory, because it merely says that a quantum state is described by a statistical operator (self-adjoint positive semidefinite operator of trace 1).

 

[Lynch101]

No. Take a wave function of the form
$$\psi(\vec{x})=N \{\exp[-(\vec{x}-\vec{a})^2/(4 \sigma_1^2)] + \exp[-(\vec{x}-\vec{b})^2/(4 \sigma_2^2)] \},$$
where $|\vec{a}-\vec{b}| \gg \text{max}(\sigma_1,\sigma_2)$.

OK, so it doesn't have a definite location (within the given error bars) prior to measurement. But, when we measure the system we measure it at a single location (within the given error bars), as opposed to at multiple locations.

Where was it prior to measurement then? It must have had a location within the universe, somewhere. If it wasn't located somewhere within the universe it couldn't interact with a measurement device which is located in the universe. This 'somewhere' doesn't have to be a single, pre-determined location but it does require a description in order to give a complete description of physical reality, as per EPR.

If we have probabilistic predictions for measuring the system at a given location, does this mean that the system is in multiple locations at once?

 

[PeterDonis]

It must have had a location within the universe, somewhere.

Unless and until you can give this statement a definite meaning in the math of QM, it is meaningless noise. I have asked you once already to do that. Can you?

 

[vanhees71]

Yes indeed, for a single realization of this experiment you cannot know before the measurement in which spot (around $\vec{a}$ or around $\vec{b}$) you will find the particle. It could even by found (with correspondingly smaller probability of course) somewhere else far away from both $\vec{a}$ and $\vec{b}$. All you know and (as far as quantum theory is complete) all you can know are the probabilities given by the probability distribution $|\psi(\vec{x})|^2$. That it is "somewhere" is clear, because by definition
$$\int_{\mathbb{R}^3} \mathrm{d}^3 x |\psi(\vec{r})|^2=1.$$

 

[Lynch101]

Okay, so what does "located somewhere in the universe" mean? What does it correspond to in the math of QM?

I strongly suspect that you will be unable to give a valid answer to this question that supports the claims you are making--i.e., that is inconsistent with the things other people are saying that you have been objecting to.

'Located somewhere in the universe' simply means it is in the universe, such that it will be included in any full description of the universe aka 'physical reality'.

That is, however, the crux of the issue. It's that, by definition, interpretations which say that the mathematics only give us probabilistic predictions for the outcomes of experiments do not have anything in the mathematics which correspond to "located somewhere in the universe".

Since the quantum system must be located somewhere in the universe, at all times, a theory which only gives probabilistic predictions for when the system interacts with a measurement device does not, by definition, describe the system prior to measurement.

If "has a definite location" means "is located at some single point", then the answer to this is obviously no, since, as I have already stated, states with definite positions--where a particle is located at a single point--are not normalizable and hence cannot be realized physically.

If "has a definite location" means something else, then you need to explain what you are using that term to mean.

If the system is not "located at some single point" then the probability distribution isn't telling us that the system is always located at some single point and it's just a lack of information on our part that gives us the probabilistic predictions.

Given that the system must be located somewhere, what does the probability distribution tell us? Does it tell us the system is distributed across a broader area than a single location, with 'more' of it (or a greater density of it) in one location than another?

If the probability distribution only tells us the probability of the system interacting with a measurement device at a given location, then it doesn't describe the system prior to that interaction - as a matter of definition.

 

[Lynch101]

Unless and until you can give this statement a definite meaning in the math of QM, it is meaningless noise. I have asked you once already to do that. Can you?

The issue is that some claim that the math of QM does not have such a thing, when, in order to be considered a complete description of physical reality, it should.

 

[EPR]

That QT appears to violate realism doesn't necessarily mean that it's incomplete. It could be that your notion of physical reality could be wrong.

 

[PeterDonis]

'Located somewhere in the universe' simply means it is in the universe, such that it will be included in any full description of the universe aka 'physical reality'.

In terms of QM, this would mean that it has a quantum state, or at least is included in the quantum state of the universe, correct?

If so, @vanhees71 has correctly pointed out in post #657 that this statement, while true, is useless, since it just tells us something we already know anyway.

by definition, interpretations which say that the mathematics only give us probabilistic predictions for the outcomes of experiments do not have anything in the mathematics which correspond to "located somewhere in the universe".

If "located somewhere in the universe" means "is part of the quantum state of the universe", as above, then your claim here is simply false.

Given that the system must be located somewhere, what does the probability distribution tell us?

If "must be located somewhere" means "is part of the quantum state of the universe" as above, the all the probability distribution tells us is what any probability distribution tells us. Namely, the probabilities of possible measurement results.

The issue is that some claim that the math of QM does not have such a thing,

We are not talking about what "some" claim. We are talking about what you claim. You are making very strong claims that, so far, you have failed to back up with a definition of "located somewhere in the universe" that supports them. The only definition you have come up with so far does not tell us anything useful, as I have shown above. So you either need to come up with a different definition or drop this line of argument since you cannot support it.

 

[Lynch101]

Yes indeed, for a single realization of this experiment you cannot know before the measurement in which spot (around $\vec{a}$ or around $\vec{b}$) you will find the particle. It could even by found (with correspondingly smaller probability of course) somewhere else far away from both $\vec{a}$ and $\vec{b}$. All you know and (as far as quantum theory is complete) all you can know are the probabilities given by the probability distribution $|\psi(\vec{x})|^2$. That it is "somewhere" is clear, because by definition
$$\int_{\mathbb{R}^3} \mathrm{d}^3 x |\psi(\vec{r})|^2=1.$$

To find it around $\vec{a}$ or around $\vec{b}$, or somewhere else far away,
It must be:
- either located at $\vec{a}$ or around $\vec{b}$, or somewhere else far away
- located at all $\vec{a}$ and around $\vec{b}$, and somewhere else far away

If it is not located at any of these, how could it possibly interact with a measurement device that is located there?