I Would studying MWI be a waste of time?

[zonde]

I don't understand this argument, because you assume that something is non-local to argue that something is non-local.

I took example that is non-local in sense of (1) but is local in sense of (2). So it is counterexample to your statement that (2) implies (1).

The linked-cluster principle (1) follows from (2) as shown in Weinberg's book. Whether (2) is also necessary for (1) is, as far as I know, not known yet.

I don't have Weinberg's book so if you could give short definition of linked-cluster principle (or give some link) it could help discussion.

 

[stevendaryl]

Let me say a little more about why I consider the minimal interpretation to be NOT actually minimal.

As I said already, it strikes me as odd to have the Born rule as a fundamental law, because it is written in terms of concepts that are not primitives of the theory--namely, measurements. If you try to unravel what a "measurement" means, you'll be forced into having to face the exact problems (in collapse interpretations, or in MWI, or in Bohm) that you think you're solving by being "minimal". The minimal interpretation in my opinion just amounts to pretending that the problems don't exist by impoverishing the language that you would use to analyze those problems. In other words, it's a cop-out.

I actually don't mind the cop-out. If that's the best that we can do, then that's what we have to deal with. But I would prefer that people be honest about what they're doing (as they are in the "shut up and calculate" interpretation).

Here's the big issue that I have with the Born rule. First let me state the rule in a particularly simplified form:

Suppose that we have some observable corresponding to Hermitian operator \hat{A}. This operator has a complete set of eigenstates |\psi_j\rangle with corresponding eigenvalues a_j (for simplicity, let's assume that there is no degeneracy--different eigenstates correspond to different eigenvalues). Then a measurement of the observable on a state |\psi\rangle = \sum_j C_j |\psi_j\rangle will result in a_j with probability |C_j|^2.

But what does it mean to say that you're measuring some observable \hat{A}? Here's a stab at an answer.

Classically, you would describe it this way: You have a device that is meta-stable. That means that it has a "neutral" state that it can remain in for long periods of time if it is left unmolested, but a very tiny interaction can nudge it into one of a number of "pointer states" that are stable against small perturbations. For illustration, a coin can with care be balanced on its edge, but that's a precarious state. A small nudge will flip it into one of two stable states: "heads" or "tails". So a measurement device for some property \hat{A} of some small subsystem is a metastable device that interacts with the subsystem so that when it is initially in the "neutral" state, and the subsystem is in the state |\psi_j\rangle, the device will tend to make the transition to a corresponding pointer state S_j, where for i \neq j, S_i and S_j are distinguishable. (What does "distinguishable" mean? Yeah, that's a good question. In practice, we know what it means, but it's hard to define it rigorously.)

The above description is a mish-mash of classical and quantum concepts. We're treating the measuring device classically, and we're treating the system to be measured quantum-mechanically. For the "shut-up and calculate" interpretation, I think that's fine. But if we really believe that QM is the correct theory of matter and fields, then there should be a quantum-mechanical description of the measuring device, as well. The fact that something is a measuring device for property \hat{A} should in theory be deducible from QM. Here's a tentative way to formalize it (which is actually not correct, for reasons that I will get to later):

Suppose we formalize the measuring device as a quantum system with states |\Phi_j\rangle plus a special, neutral state |\Phi_{\emptyset}\rangle. Then we assume that the usual rules for state evolution (the Schrodinger equation) results in the following transitions:

|\psi_j\rangle \otimes |\Phi_{\emptyset}\rangle \Longrightarrow |\Phi_j\rangle

(I haven't written the right-hand side as a product state, because many measurements are destructive, such as detecting a photon, so the final state no longer has a component corresponding to the subsystem being measured.)

Then the linearity of QM would imply that:

(\sum_j C_j |\psi_j\rangle) \otimes |\Phi_{\emptyset}\rangle \Longrightarrow \sum_j C_j |\Phi_j\rangle

In terms of this model of measurement (which is incorrect, as I said), the Born rule says the following:

If you have a macroscopic system that can be written as a superposition \sum_j C_j |\Phi_j\rangle of macroscopically distinguishable pointer states |\Phi_j\rangle, then this superposition is to be interpreted as the system actually being in one of the pointer states |\Phi_j\rangle with probability |C_j|^2.

Note 1: There is no need to postulate that the measurement results in an eigenvalue of the observable that is being measured. That's true by the definition of what it means to be a measuring device, together with the linearity of quantum mechanics.

Note 2: This way of stating the Born rule says that a superposition of a particular type simply means a probability of being one of the elements of the superposition. The usual quantum mechanical phrase is that the amplitude squared gives the probability of measuring the system to be in such-and-such a state. But if you interpret superpositions of the measuring device that way, then you're plunging into an infinite regress. You need another measuring device to measure the state of the first device? Then another to measure the state of the second? Etc, etc.

To me, trying to unravel the meaning of the Born rule shows how ad hoc it is. You have a rule that applies to a particular collection of orthonormal states (that the coefficients represent probabilities, without the need for measurements or observations) that does not apply to other (microscopic) collections. You're assuming that superpositions of pointer states don't happen--that there is always a definite value for a pointer variable. That isn't true for microscopic properties such as the z-component of spin of an electron.

Let me return to the question of why the proposed transition for a measurement isn't actually correct. I wrote:

|\psi_j\rangle \otimes |\Phi_{\emptyset}\rangle \Longrightarrow |\Phi_j\rangle

But quantum-mechanically, the evolution equations are reversible. If it's possible for state |A\rangle to evolve into state |B\rangle, then it's possible for |B\rangle to evolve into |A\rangle. That doesn't seem to correctly describe the measurement process. Measurements are irreversible. How can we describe that, quantum-mechanically?

The part that is left out is the environment (the electromagnetic field, the walls and floors of the laboratory, the air, the researchers, etc.). The macroscopic pointer states will in general be "entangled" with the rest of the universe, so it would not be possible to write the device as being in a superposition. So what's really going on in measurements is a lot more complicated. But I think that what I've said (incomplete or even wrong as it is) is enough to give a feeling for why I reject the Born rule as a fundamental law of physics. At best, it has to be a rule of thumb.

 

[Demystifier]

Only, because the value of an observable is indetermined it doesn't mean that it doesn't exist. That's all I'm saying.

Consider a massive spin-1 particle prepared in the state $|z+\rangle$, that is in the state with spin $+1$ in the $z$-direction. Clearly, the spin in the $x$-direction is indetermined. Does the spin in the $x$-direction exist?

 

[stevendaryl]

Couldn't one say that it became true after Bob's measurement? Just because Bob's and Alice's measurements are simultaneous, and technically the statement became true after Alice's measurement, it doesn't mean that Alice's act has anything to do with it.

There is no requirement in EPR that the measurements have to be simultaneous. Experimenters try to make them simultaneous to rule out possible hidden-variable explanations for EPR results. But Bob can take his sweet time to measure his particle's spin, and it makes no difference. So I'm talking about a case in which Alice performs her measurement well before Bob performs his. In that case, during the time between the measurements, Alice knows something about Bob's measurement result before he performs that measurement.

 

[vanhees71]

Consider a massive spin-1 particle prepared in the state $|z+\rangle$, that is in the state with spin $+1$ in the $z$-direction. Clearly, the spin in the $x$-direction is indetermined. Does the spin in the $x$-direction exist?

Of course, why not?

 

[martinbn]

There is no requirement in EPR that the measurements have to be simultaneous. Experimenters try to make them simultaneous to rule out possible hidden-variable explanations for EPR results. But Bob can take his sweet time to measure his particle's spin, and it makes no difference. So I'm talking about a case in which Alice performs her measurement well before Bob performs his. In that case, during the time between the measurements, Alice knows something about Bob's measurement result before he performs that measurement.

But they are space-like, so there is a frame in which they are simultaneous (or in which Bob measures first).

 

[stevendaryl]

But they are space-like, so there is a frame in which they are simultaneous (or in which Bob measures first).

I'm talking about the case in which Alice measures her spin well before Bob measures his. As I said, Bob can wait years before measuring his particle's spin, and it makes no difference. It doesn't make any difference whether the interaction is spacelike or not. The purpose of choosing a spacelike interval is to rule out a slower-than-light hidden-variables explanation.

 

[Demystifier]

Of course, why not?

Because of the Kochen-Specker theorem. More precisely, if you believe that the particle has both z-spin value and x-spin value, and that it is the value that would be determined by an appropriately oriented SG apparatus, then the Kochen-Specker theorem proves that you are wrong.

The Kochen-Specker theorem is an important step towards the correct understanding of the Bell theorem, as explained e.g. in the Ballentine's book Sec. 20.6. Apparently you miss this important step, which can explain why you don't understand why many people think that QM is either non-local or non-ontological. You think that QM is both local and ontological, and now it seems that it can be boiled down to your lack of understanding of the Kochen-Specker theorem.

 

[vanhees71]

I did NOT say that I believe that the particle has both determined spin-z and spin-x value. This is in clear contradiction to standard QT! You said you've prepared the particle's spin in the state $|\sigma_z=1 \rangle \langle \sigma_z=1|$. This implies that the value of the spin-z component is 1, while the value of the spin-x component is indetermined. Of course all the three spin components always exist according to QT, because spin is an observable (in non-relativistic QT). This does not imply that all their values are always determined. As you example shows that depends on the state, and your choice implies that $\sigma_z=1$ is determined, while the other two spin components are not determined, but their measurement leads a random result with probabilities predicted by QT (I'm to lazy to evaluate them know):
$$P_x(\sigma_x)=|\langle \sigma_x|\sigma_z=1 \rangle|^2, \quad \sigma_x \in \{-1,0,1 \}.$$

 

[zonde]

Suppose we formalize the measuring device as a quantum system with states |\Phi_j\rangle plus a special, neutral state |\Phi_{\emptyset}\rangle. Then we assume that the usual rules for state evolution (the Schrodinger equation) results in the following transitions:

|\psi_j\rangle \otimes |\Phi_{\emptyset}\rangle \Longrightarrow |\Phi_j\rangle

(I haven't written the right-hand side as a product state, because many measurements are destructive, such as detecting a photon, so the final state no longer has a component corresponding to the subsystem being measured.)

Something makes me feel uneasy about this representation of measurement. Maybe you have some example on mind for this type of measurement device? Because the only examples that come to my mind do not really fit with this description.
Say you measure photon polarization. You make a setup where H polarized photons end up in one place and place a detector there (and place another detector where V polarized photons end up). So you do not detect polarization of photon but rather simply a photon and from experimental setup you infer that it was H polarized. But photon itself is not represented by ray in Hilbert space (in NRQM). So detector does not measure eigenvalue. It just counts photons.

 

[PeterDonis]

We can consider a toy model.

In your example, the measurement events are not spacelike separated.

(if they somehow draw the balls non-localy).

They can't because both draws come from the same box, and the box's worldline must be timelike.

 

[stevendaryl]

Something makes me feel uneasy about this representation of measurement.

You should feel uneasy, because it's not actually true. What's really going on is a lot more complicated. Basically, because in order to be metastable, the measuring device has to make irreversible changes. You can't represent irreversible changes using simple state vectors.

Maybe you have some example on mind for this type of measurement device? Because the only examples that come to my mind do not really fit with this description.
Say you measure photon polarization. You make a setup where H polarized photons end up in one place and place a detector there (and place another detector where V polarized photons end up). So you do not detect polarization of photon but rather simply a photon and from experimental setup you infer that it was H polarized. But photon itself is not represented by ray in Hilbert space (in NRQM). So detector does not measure eigenvalue. It just counts photons.

Well, photons can't really be treated using QM without going to QFT, so let's use an electron instead. So in that case, you send an electron through a Stern-Gerlach device, and a spin-up electrons is deflected to the left, where it hits one detector and a spin-down electrons is deflected right, where it hits a different detector.

Then the whole measuring device setup might be described by a state such as |S_L, S_R, ...\rangle, where S_L describes the state of the left detector, either "on" or "off" ("on" means "detected something" and "off" means "detected nothing so far"), and where S_R describes the state of the right detector, and ... represents other irrelevant degrees of freedom of the measuring device.

So in my terms, the state |\Phi_\emptyset\rangle = |off, off, ...\rangle, |\Phi_u\rangle = |on, off, ...\rangle and |\Phi_d\rangle = |off, on, ...\rangle

Then the assumption is that
|u\rangle \otimes |\Phi_\emptyset\rangle \Longrightarrow |\Phi_u\rangle
|d\rangle \otimes |\Phi_\emptyset\rangle \Longrightarrow |\Phi_d\rangle

 

[PeterDonis]

Measurements are irreversible.

I think this is actually interpretation-dependent. More precisely, it is interpretation-dependent whether measurements are irreversible in principle (in a collapse interpretation), or only irreversible FAPP (in a no collapse interpretation like MWI).

You can't represent irreversible changes using simple state vectors.

Per the above, I think this is only an issue for collapse interpretations, where measurements are irreversible in principle. In the MWI, you can in principle represent everything using a state vector; it's just that there are so many degrees of freedom in the environment that you can't actually keep track of them all, so the measurement looks irreversible because you don't know which exact state vector represents the system.

 

[Demystifier]

I did NOT say that I believe that the particle has both determined spin-z and spin-x value.

And I didn't say that you said that. You said that both values exist. And I say that their existence is in contradiction with Kochen-Specker theorem.

Of course all the three spin components always exist according to QT, because spin is an observable (in non-relativistic QT).

No, according to the Kochen-Specker theorem, which is derived from QT, all three components do not exist. The fact that they are observables does not imply that they exist before observation. Kochen-Specker theorem plus the fact they are observables imply that the values are not merely revealed by observation. Instead, they must be somehow created by observation.

 

[stevendaryl]

I think this is actually interpretation-dependent. More precisely, it is interpretation-dependent whether measurements are irreversible in principle (in a collapse interpretation), or only irreversible FAPP (in a no collapse interpretation like MWI).

Okay, but for something to actually count as a "measurement", it must at least be irreversible FAPP.

At least that's true with an old-fashioned notion of measurement. You've measured something if you can write down the result and compile statistics and publish in a journal. I have seen articles that use a much milder notion of "measurement", under which definition, an interaction between a photon and an electron could be considered a measurement of the electron's position. With that milder notion, measurements can be reversible.

Per the above, I think this is only an issue for collapse interpretations, where measurements are irreversible in principle. In the MWI, you can in principle represent everything using a state vector; it's just that there are so many degrees of freedom in the environment that you can't actually keep track of them all, so the measurement looks irreversible because you don't know which exact state vector represents the system.

Yes, that's the reason why the notation

|\psi_j\rangle \otimes |\Phi_\emptyset\rangle \Longrightarrow |\Phi_j\rangle

isn't quite right. To be a measurement (in the old-fashioned sense), there must be many more degrees of freedom. So there isn't just a single state |\Phi_j\rangle, there are many, many different states corresponding to the same value of j.

 

[vanhees71]

This is nonsense: In your example $\sigma_z$ was even determined to be 1. The other two perpendicular components are then necessarily indetermined, but this doesn't imply that they don't exist. Of course they do, because you can measure them, as you measured $\sigma_z$. It's this nebulous formulation about "existence" that causes the confusion, not QT itself! I admit that it's difficult to accept that observables' values can be (in fact in most of the cases are) indetermined, because we are used to a classical world view, but it's wrong to say that something doesn't exist, only because it hasn't a definite value. If you'd claim this to be true, then anything you don't know wouldn't exist, i.e., almost the entire universe wouldn't exist. That's a bit solipsistic, isn't it?

 

[stevendaryl]

No, according to the Kochen-Specker theorem, which is derived from QT, all three components do not exist. The fact that they are observables does not imply that they exist before observation. Kochen-Specker theorem plus the fact they are observables imply that the values are not merely revealed by observation. Instead, they must be somehow created by observation.

This might be just a quibble over what it means for an observable to exist. @vanhees71 might be saying that the observable always exists (it's represented as an operator), although the value may be indeterminate.

 

[vanhees71]

Okay, but for something to actually count as a "measurement", it must at least be irreversible FAPP.

At least that's true with an old-fashioned notion of measurement. You've measured something if you can write down the result and compile statistics and publish in a journal. I have seen articles that use a much milder notion of "measurement", under which definition, an interaction between a photon and an electron could be considered a measurement of the electron's position. With that milder notion, measurements can be reversible.
Yes, that's the reason why the notation

|\psi_j\rangle \otimes |\Phi_\emptyset\rangle \Longrightarrow |\Phi_j\rangle

isn't quite right. To be a measurement (in the old-fashioned sense), there must be many more degrees of freedom. So there isn't just a single state |\Phi_j\rangle, there are many, many different states corresponding to the same value of j.

Yes, and this solves all apparent "measurement problems". Measurement apparti are classical (FAPP ;-)), because they consist of very many particles and a pointer reading is a very coarse-grained notion, i.e., it's from a microscopic point of view, the average over very many microstates. That makes it behave classically (keyword decoherence).

 

[vanhees71]

This might be just a quibble over what it means for an observable to exist. @vanhees71 might be saying that the observable always exists (it's represented as an operator), although the value may be indeterminate.

Again: An observable is something measurable (or just observable) in the real world. The operators are representatives of such defined observables in the formalism. Also in classical physics, an electric field is also not a triple of three real numbers defined at any place and any time, but it's defined as something measurable via the forces on test charges. The three real numbers (it's components wrt. to a Cartesian coordinate system) are representatives within the formalism (aka Maxwell euations).

 

[Demystifier]

This is nonsense:

So how do you interpret the Kochen-Specker theorem?

In your example $\sigma_z$ was even determined to be 1. The other two perpendicular components are then necessarily indetermined, but this doesn't imply that they don't exist. Of course they do, because you can measure them, as you measured $\sigma_z$.

Isn't it possible that the measurement itself creates them, so that they don't exist before measurement?

It's this nebulous formulation about "existence" that causes the confusion, not QT itself!

Are you saying that Kochen-Specker theorem is nebulous?

If you'd claim this to be true, then anything you don't know wouldn't exist, i.e., almost the entire universe wouldn't exist. That's a bit solipsistic, isn't it?

Yes, a sort of solipsism is a logical possibility
https://arxiv.org/abs/1112.2034
compatible with the Bell and Kochen-Specker theorems.

 

[stevendaryl]

Yes, and this solves all apparent "measurement problems".

Not true at all, in my opinion.

 

[Demystifier]

This might be just a quibble over what it means for an observable to exist. @vanhees71 might be saying that the observable always exists (it's represented as an operator), although the value may be indeterminate.

I think he made quite clear that for him "indeterminate" means unknown. Very much like the head or tail dilemma before you look at the coin. And he refuses to understand that Kochen-Specker disproves him.

 

[martinbn]

I'm talking about the case in which Alice measures her spin well before Bob measures his. As I said, Bob can wait years before measuring his particle's spin, and it makes no difference. It doesn't make any difference whether the interaction is spacelike or not. The purpose of choosing a spacelike interval is to rule out a slower-than-light hidden-variables explanation.

Yes, but "before" is not an absolution notion, as you well know it is frame dependent. So pick a frame in which the two events (Alice measures and Bob measures) are simultaneous.

 

[stevendaryl]

Yes, but "before" is not an absolution notion, as you well know it is frame dependent. So pick a frame in which the two events (Alice measures and Bob measures) are simultaneous.

I think I said this about three times: I'm talking about the case in which Alice's measurement and Bob's are timelike separated. Alice's measurement occurs before Bob's in every frame. That's not the usual case, but that's the case I'm talking about.

 

[LeandroMdO]

No, according to the Kochen-Specker theorem, which is derived from QT, all three components do not exist.

It's not even clear that "existence" (of the value of an observable prior to measurement) is a concept that can be defined in a sufficiently precise manner that would allow one to prove theorems about it.

 

[martinbn]

I think I said this about three times: I'm talking about the case in which Alice's measurement and Bob's are timelike separated. Alice's measurement occurs before Bob's in every frame. That's not the usual case, but that's the case I'm talking about.

I missed that. You did say though that it didn't matter whether they are space-like or not. And I agree, for the point that you are making your reasoning should work in the space-like case as well. Also for a pair of photons, the two events cannot be time-like no matter how far Bob is. Anyway this is probably way off topic, so I should stop. There is something about this that I don't understand and has always bothered me, I'll probably start a new thread about it.

 

[Denis]

It's not even clear that "existence" (of the value of an observable prior to measurement) is a concept that can be defined in a sufficiently precise manner that would allow one to prove theorems about it.

It can be, and has been. There are realistic and causal interpretations of quantum theory - in particular dBB theory. Once you can prove theorems in dBB theory, and given that it is a realistic interpretation, this already allows you to prove various theorems, like that there exists a realistic interpretation.

Then, there is the notion of realism used in various proofs of Bell's theorem. Bell's theorem is, essentially, the theorem which starts with some precised definition of realism, together with Einstein causality and derives Bell inequalities.

That one can also use imprecise philosophical nonsense to define "realism" so that it is impossible to derive anything from it is irrelevant once precise definitions, sufficient to prove theorems, exist.

 

[Denis]

It is obvious (even classically) that a measurement result doesn't exist as long as no measurement is taken. Nevertheless the object to be measured must exist even before the measurement; otherwise it cannot be measured at all! This doesn't change in the quantum realm.

I disagree. If something does not exist, it cannot be measured.

There are, of course, a lot of things we observe in classical physics, which have not existed before. But in this case, it is not appropriate to name this observation a measurement. The name "measurement" is simply misleading and confusing, because it suggests that what is measured already existed before the measurement. See Bell's paper "Against measurement".

I'm thinking about a word which would give the correct associations. My actual favorite is "coordination", because it has some typical properties which are present in quantum theory too:

1.) Symmetry: A coordinates with B is the same as B coordinates with A. There is a similar symmetry in a quantum interaction too.
2.) Repetition gives the same result: Simply repeating the same coordination procedure will give the same result. This property that repetition gives the same result is what is shared with a measurement, and probably an explanation why this word has been used.
3.) There may be incompatible coordinations: Trump coordinates something with Turkey. After this, he coordinates something with the Kurds. It makes sense to assume that these two actions are incompatible with each other, and that the former coordination with Turkey has no longer any value. If, after this, Trump coordinates the same question with Turkey again, it is quite probable that the result is not the same as that of the first coordination. The coordination with the Kurds has destroyed the value of the former coordination with Turkey.

Any proposals for other words which give similar correct associations?

 

[LeandroMdO]

It can be, and has been. There are realistic and causal interpretations of quantum theory - in particular dBB theory.

It's not clear that the informal phrase "existence of values of observables prior to measurement" corresponds to a hidden variable model.

 

[Denis]

It's not clear that the informal phrase "existence of values of observables prior to measurement" corresponds to a hidden variable model.

This informal phrase has, of course, also another sufficiently certain meaning, which is used in the Kochen-Specker theorem.

In other words, the imprecise "existence" has several well-define precise implementations, some of them are compatible with quantum theory, others already falsified.

 

[Dadface]

Still here trying to get the gist of what's going on in these discussions.

On a serious note I'm thinking of putting a tenner on Shantou Flyer to win in the Grand National today. I might lose my money but I know that other versions of me will win in other universes. By the way I think it was another version of me who popped over and wrote this comment.

 

[vanhees71]

I think he made quite clear that for him "indeterminate" means unknown. Very much like the head or tail dilemma before you look at the coin. And he refuses to understand that Kochen-Specker disproves him.

You misunderstand me. The Kochen-Specker theorem follows from QT, so it's part of QT, and I don't claim that the indeterminate values of observables of a system in a state, where these observables are indeterminate, in fact have a certain value which is just unknown, but I say that they are "really" (sic!) indeterminate. That's the point. You just have to accept that some observables for a system in a state have indeterminate values, and all quibbles resolve.

For me, at the moment, all hidden-variable attempts are ruled out by all the Bell experiments (and Kochen-Specker is very closely related with Bell's inequality).

 

[Denis]

For me, at the moment, all hidden-variable attempts are ruled out by all the Bell experiments (and Kochen-Specker is very closely related with Bell's inequality).

Clearly wrong. What is ruled out are only those hidden-variable attempts which want to assign hidden values to all observables. Hidden variable attempts which assign hidden variables only to a subset of commuting observables, with the configuration space observables as the straightforward choice, have no problem at all, neither with Bell nor with Kochen-Specker.

 

[stevendaryl]

I missed that. You did say though that it didn't matter whether they are space-like or not. And I agree, for the point that you are making your reasoning should work in the space-like case as well.

It's not so much that I'm making a point--I'm just asking questions.

Also for a pair of photons, the two events cannot be time-like no matter how far Bob is.

That's not true; you can delay the arrival of a photon by bouncing it around using mirrors.

 

[LeandroMdO]

This informal phrase has, of course, also another sufficiently certain meaning, which is used in the Kochen-Specker theorem.

In other words, the imprecise "existence" has several well-define precise implementations, some of them are compatible with quantum theory, others already falsified.

Not really. It's just an informal word. You can't prove any theorems at all unless you furnish a precise definition, in which case the theorem is about that precise definition and not the original informal one.

 

[Denis]

So what? The "informal word" is used in the EPR/Bell discussions in a quite precise meaning, namely in those precise ways used in Bell's theorem and similar theorems, where one needs, together with realism, also Einstein causality to prove the Bell or similar inequalities.

For what is ruled out by Kochen-Specker the phrase which is usually used instead of your "existence of values of observables prior to measurement" is conterfactual definiteness.

So, the standard use of the word "realism" as used in the EPR/Bell discussions refers to a precise meaning. There may be different variants of this precise meaning in different variants of proofs of Bell inequalities. But this does not change the basic point that 1.) a precise meaning exists. 2.) realism together with Einstein causality gives the Bell inequalities, so that this combination is dead. So realists have to accept a hidden preferred frame. But if one is ready to accept it, everything else is fine and unproblematic, as shown explicitly by dBB theory.

 

[LeandroMdO]

So what? The "informal word" is used in the EPR/Bell discussions in a quite precise meaning

One can in principle make up whatever precise definition they like for an informal word, but they must not mistake their definition for the original thing lest they put words in somebody else's mouth.

Besides, even in contexts where more precision could be expected you'll find that different people think of different things when it comes to words such as "realism". The KS theorem, or Bell's inequality, or GHZ theorem, etc say nothing whatsoever about "existence" in the vague informal sense. In particular, if one adopts a many-worlds ontology it becomes quite fair to say that all possible results exist prior to measurement, just like a classical wave packet "contains" all frequencies in the support of its momentum space representation. Clearly, what matters are the precise statements, not whatever words one chooses to use when informally communicating these results.

But if one is ready to accept it, everything else is fine and unproblematic, as shown explicitly by dBB theory.

I wouldn't call pilot wave interpretations either "fine" or "unproblematic", but that is neither here nor there.

 

[Denis]

The KS theorem, or Bell's inequality, or GHZ theorem, etc say nothing whatsoever about "existence" in the vague informal sense.

They all use something precisely defined. And, given the theorems, this gives some sufficiently clear restrictions about what makes sense and what does not make sense.

In particular, if one adopts a many-worlds ontology it becomes quite fair to say that all possible results exist prior to measurement, just like a classical wave packet "contains" all frequencies in the support of its momentum space representation.

My problem with many worlds is that it makes no sense at all. Neither to talk about existence, nor about causality or probability makes any sense in MWI.

I wouldn't call pilot wave interpretations either "fine" or "unproblematic", but that is neither here nor there.

In comparison with the complete ... of MWI and the many serious problems of other interpretations with measurement and collapse and so on dBB is completely unproblematic. One may not like it for various reasons. I have also some things which I don't like in dBB theory. But problems? Sorry, not seen.

 

[LeandroMdO]

They all use something precisely defined.

That's what I said.

My problem with many worlds is that it makes no sense at all. Neither to talk about existence, nor about causality or probability makes any sense in MWI.

I don't think any of the existing attempts at a formulation of Everett's ideas make much sense either, but that doesn't mean that a sensible, precise formulation couldn't be found in the future. There's certainly nothing logically incoherent about replacing a collapse postulate by a split postulate. At the very least, entertaining the possibility has improved our understanding of the foundations of quantum mechanics. That makes it useful to think about.

In comparison with the complete ... of MWI and the many serious problems of other interpretations with measurement and collapse and so on dBB is completely unproblematic. One may not like it for various reasons. I have also some things which I don't like in dBB theory. But problems? Sorry, not seen.

There are many problems, such as its incompatibility with relativity and the unwarranted assumption of quantum equilibrium. It's not even clear that it correctly reproduces quantum mechanical results in every experiment. It's not clear what happens to the wavefunction once particles are emitted/absorbed, etc. It's not clear how to implement massless vector particles. You can say you like the idea, but to say that it's unproblematic is definitely premature. But as I said, that's neither here nor there. This thread is not about pilot wave theories.

 

[Denis]

I don't think any of the existing attempts at a formulation of Everett's ideas make much sense either, but that doesn't mean that a sensible, precise formulation couldn't be found in the future.

Wishful thinking is nothing scientists have to care about. Once it does not exist today, it is irrelevant.

There's certainly nothing logically incoherent about replacing a collapse postulate by a split postulate.

Fine, but given that a collapse postulate has a lot of problems itself, this is not really an argument in favor of splits. I think, MWI has only heavily increased quantum confusion.

There are many problems, such as its incompatibility with relativity and the unwarranted assumption of quantum equilibrium.

The conflict with a fundamental interpretation of relativity is unavoidable for every realistic interpretation (given Bell's theorem), so this is not a serious problem. In a weak interpretation of relativity, which allows for a hidden preferred frame, there is no problem with this.
Quantum equilibrium is not a problem given Valentini's subquantum H theorem.

It's not clear what happens to the wavefunction once particles are emitted/absorbed, etc. It's not clear how to implement massless vector particles.

Of course, one has to use dBB field theory. In this case, particles are nothing but phonons, with no fundamental importance. Given that for dBB field theory a lattice regularization is fine even as a fundamental theory, because relativistic symmetry is not a fundamental value, it has even less problems than usual RQFT, which has to care about the limit. Models for fermionic fields exist too.

Ok, this is off-topic here, but once you mentioned these as problems, I think I have to explain why I do not consider them as problematic. Anyway, these are problems of application to particular quantum theories. Which is, of course, a point - the straightforward construction works for canonical Hamiltonians $H= \frac12 p^2 + V(q)$ and some generalizations, but not for a completely general Hamiltonian. But relativistic field theory fits nicely into this scheme, so that this is much less problematic than it seems.

And the value of MWI depends, of course, on the available alternatives. So, it is not completely off-topic here too.

 

[LeandroMdO]

Wishful thinking is nothing scientists have to care about. Once it does not exist today, it is irrelevant.

If science actually proceeded that way, it would stop.

Fine, but given that a collapse postulate has a lot of problems itself, this is not really an argument in favor of splits. I think, MWI has only heavily increased quantum confusion.

Not true at all---attempting to understand quantum mechanics from a purely unitary time evolution perspective has clarified what is required of an axiomatic formulation and what isn't. It is now clear that Born's rule is the unique probability measure in Hilbert space given fairly weak assumptions. Zurek's work on einvariance is interesting regardless of whether one buys into MWI, etc. Regardless of MWI's status as a fundamental theory of reality, it is clear that investigating it has been useful and illuminating. The same could be said of string theory, for example.

The conflict with a fundamental interpretation of relativity is unavoidable for every realistic interpretation (given Bell's theorem), so this is not a serious problem.

Not really, but like I said, this thread is not about Bohmian mechanics. If you want to discuss that please start a new thread.

 

[zonde]

You should feel uneasy, because it's not actually true. What's really going on is a lot more complicated. Basically, because in order to be metastable, the measuring device has to make irreversible changes. You can't represent irreversible changes using simple state vectors.

I would say that (FAPP-)irreversibility of amplification comes from non-linearity of the process. But can QM model non-linearity using only unitary evolution?

Well, photons can't really be treated using QM without going to QFT, so let's use an electron instead. So in that case, you send an electron through a Stern-Gerlach device, and a spin-up electrons is deflected to the left, where it hits one detector and a spin-down electrons is deflected right, where it hits a different detector.

Then the whole measuring device setup might be described by a state such as |S_L, S_R, ...\rangle, where S_L describes the state of the left detector, either "on" or "off" ("on" means "detected something" and "off" means "detected nothing so far"), and where S_R describes the state of the right detector, and ... represents other irrelevant degrees of freedom of the measuring device.

So you simply take classical macro states as dimensions of Hilbert space and then specify the state using vector. Well, but it does not mean that it is possible to go from one classical macro state to the other one by unitary evolution. And if it is not possible then this approach is simply meaningless.

So in my terms, the state |\Phi_\emptyset\rangle = |off, off, ...\rangle, |\Phi_u\rangle = |on, off, ...\rangle and |\Phi_d\rangle = |off, on, ...\rangle

You lost state |\Phi_x\rangle = |on, on, ...\rangle. What is it? And if it's meaningless then there is something wrong with this vector space.

Then the assumption is that
|u\rangle \otimes |\Phi_\emptyset\rangle \Longrightarrow |\Phi_u\rangle
|d\rangle \otimes |\Phi_\emptyset\rangle \Longrightarrow |\Phi_d\rangle

$|u\rangle \otimes |\Phi_\emptyset\rangle$ and $|\Phi_u\rangle$ have different dimensions. And again it seems that some inconvenient dimensions disappear.

So why pretending that classical macro states are like quantum states should lead to any meaningful insight?

 

[stevendaryl]

I would say that (FAPP-)irreversibility of amplification comes from non-linearity of the process. But can QM model non-linearity using only unitary evolution?

What do you mean by nonlinearity? Nonlinearity of what?

So you simply take classical macro states as dimensions of Hilbert space and then specify the state using vector.

Obviously, it's a lot more complicated than that. The actual quantum mechanics is quantum field theory, and there are no subsystems--everything is just particles and fields. I'm trying to make a model with enough detail to bring up the important points, but still simple enough to discuss.

Well, but it does not mean that it is possible to go from one classical macro state to the other one by unitary evolution.

I'm not saying that it necessarily is possible. The whole point is to understand to what extent a "minimalist" approach makes sense. I don't think it does. Because if everything is described by quantum mechanics, then it doesn't make sense to single out "measurements" or "macroscopic vs microscopic". Those should in principle be derivable from the microscopic rules.

You lost state |\Phi_x\rangle = |on, on, ...\rangle.

What does "lost state" mean?

$|u\rangle \otimes |\Phi_\emptyset\rangle$ and $|\Phi_u\rangle$ have different dimensions

I didn't specify the dimensions of either one. In reality, what we have is a QFT with potentially infinite number of particles. So any time you write a "composite wave function", you're actually making some kind of approximation. In reality, there aren't subsystems, except in modeling.

I intentionally did not include the state for |u\rangle after the detection, because detection events are often destructive. After the detection, the particle is no longer a separate subystem, because it's often absorbed by the detector.

 

[Demystifier]

You misunderstand me. The Kochen-Specker theorem follows from QT, so it's part of QT, and I don't claim that the indeterminate values of observables of a system in a state, where these observables are indeterminate, in fact have a certain value which is just unknown, but I say that they are "really" (sic!) indeterminate.

Then I have no idea what do you mean when you say that "the value exists but is indeterminate". Can you give some example or analogy from classical physics? (Analogy from anything else, e.g. psychology, which is not quantum physics would also do.)

 

[vanhees71]

I never said "the value exists" but "the observable exists". It's important to read everything carefully. An observable within QT does not need to have a defined/determined value. That's the point of this discussion. There is no analogy from classical physics for that; it's the very essence of quantum physics in contradistinction to classical physics, and that makes this point so difficult to express (note that your posting was already #194 in this thread!).

 

[atyy]

I never said "the value exists" but "the observable exists". It's important to read everything carefully. An observable within QT does not need to have a defined/determined value. That's the point of this discussion. There is no analogy from classical physics for that; it's the very essence of quantum physics in contradistinction to classical physics, and that makes this point so difficult to express (note that your posting was already #194 in this thread!).

Position and momentum observables can simultaneously exist.

Just like the universe can simultaneously exist and not exist.

 

[vanhees71]

All observables always exist simultaneously. Not all observables can take determined values simultaneously in any state.

The universe as a whole is not an observable ;-).

 

[Demystifier]

I never said "the value exists" but "the observable exists". It's important to read everything carefully.

Yes, but it's also important to write carefully. In post #150 you wrote:
"Only, because the value of an observable is indetermined it doesn't mean that it doesn't exist."
I assumed that "it" refers to "value of an observable". Obviously, by "it" you only meant "observable".

So, do you say that the value (in a non-eigen-state) doesn't exist before measurement? Or are you just agnostic about that? How about the value in an eigen-state before measurement?

 

[stevendaryl]

So, do you say that the value (in a non-eigen-state) doesn't exist before measurement? Or are you just agnostic about that?

Bell's theorem and the KS theorem both say that it doesn't.

 

[Denis]

There is no analogy from classical physics for that; it's the very essence of quantum physics in contradistinction to classical physics, and that makes this point so difficult to express.

Sorry, I disagree. Does there exist a reaction of your neighbor if your throw a ... (whatever) at him? No, once you don't do such stupid things, the reaction does not exist too. Is the reaction of your throwing if your throw a ... (whatever) at him observable? Of course, all you have to do to observe it is to do it. I hope you will not try to throw (whatever). But this does not make his reaction unobservable. The distinction between existence (now) and being observable (under certain circumstances) is not only common in classical circumstances but even part of our everyday life.

So, that some of the "observables" in quantum theory may not exist until they are observed is in no way strange, "quantum" or so.