Would studying MWI be a waste of time?

[zonde]

That's only the case if you make the assumption of collapse, and that's contradicting the fundamental assumptions of locality and microcausality built into QED (and all the Standard Model of HEP physics).

This is very interesting sentence.
Lets make two definitions of "locality":
1) measurement result at one place is unaffected by things happening at spacelike separated place and vice versa
2) statistical properties of measurements commute for measurements made at spacelike separated places
The first part of statement (collapse contradicts locality) is true when first meaning of "locality" is used, but false with second meaning.
But second part of the statement (locality is fundamental assumption built into QED) is true when second meaning of "locality" is used, but false with first meaning.
I guess I need laws of QT to find out if this statement is wrong or not.

 

[A. Neumaier]

Then the concept of ontology doesn't make any sense

Only the concept of ''ontology of exact values''. There are more things that can exist - not only values!

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.

 

[stevendaryl]

That's true, but this "flip" is not due to "spooky actions at a distance". That's only the case if you make the assumption of collapse, and that's contradicting the fundamental assumptions of locality and microcausality built into QED (and all the Standard Model of HEP physics). The only conclusion can be to give up the collapse assumption and live with the minimal interpretation.

I'm just saying that in my opinion, that position is incoherent. You're taking parts of the collapse interpretation and rejecting any implications from it.

 

[vanhees71]

I don't dismiss the concept of ontology as meaningless, but perhaps I have another understanding about ontology than you. For me ontic are the real things in nature, and thus observables are defined by measurement procedures with real-world apparati. I also don't think that I'm too confused about interpretation of QT, also I consider many interpretations to belong rather to esoterics, entertainment for Sci Fi fans, etc.

 

[vanhees71]

I'm just saying that in my opinion, that position is incoherent. You're taking parts of the collapse interpretation and rejecting any implications from it.

Where are there collapse assumptions? I'm just agnostic about what the state of the quantum system might be after measurement. This you can't say in generality, because it depends on how you measure some observable on it.

 

[vanhees71]

This is very interesting sentence.
Lets make two definitions of "locality":
1) measurement result at one place is unaffected by things happening at spacelike separated place and vice versa
2) statistical properties of measurements commute for measurements made at spacelike separated places
The first part of statement (collapse contradicts locality) is true when first meaning of "locality" is used, but false with second meaning.
But second part of the statement (locality is fundamental assumption built into QED) is true when second meaning of "locality" is used, but false with first meaning.
I guess I need laws of QT to find out if this statement is wrong or not.

But 1) follows from 2) (linked-cluster principle; see Weinberg, QT of Fields, vol. I)

 

[zonde]

But 1) follows from 2)

Certainly not. We can consider a toy model. We have a series of boxes where each box contains two identical balls except that one ball is red but the other one is blue.
Alice without looking takes one ball from each box. Bob gets the other one. Clearly which ball Alice takes from the box affects which ball will be left for the Bob. So Alice's action affects Bob's result. But statistical properties for Alice's and Bob's "measurements" do not depend in what order they draw the ball. So this example would satisfy 2) but won't satisfy 1) (if they somehow draw the balls non-localy).

 

[Demystifier]

I don't dismiss the concept of ontology as meaningless, but perhaps I have another understanding about ontology than you. For me ontic are the real things in nature, and thus observables are defined by measurement procedures with real-world apparati.

If I understood you correctly, you say that anything which can in principle be observed by measurement procedures with real-world apparati - is ontology. Even if at the moment we don't have such an apparatus, but will have in the future, it is already ontology - now. And if such an apparatus is impossible even in principle, it is not ontology. Is that correct?

If so, is there any difference, in your understanding, between ontology and observables?

 

[vanhees71]

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

 

[vanhees71]

If I understood you correctly, you say that anything which can in principle be observed by measurement procedures with real-world apparati - is ontology. Even if at the moment we don't have such an apparatus, but will have in the future, it is already ontology - now. And if such an apparatus is impossible even in principle, it is not ontology. Is that correct?

If so, is there any difference, in your understanding, between ontology and observables?

Ok, I'm quite sure, I don't understand what you mean by ontology here. To my (admittedly limited) knowledge and understanding of philosophy it's studying the question of existence. According to the natural sciences existing is something that's objectively observable. So for something to exist it must be observable. In physics many things are even quantifiable and then you assign values to observables. According to minimally interpreted QT, not all observables can simultaneously have determined values. Where is a problem with "ontology" here? Only, because the value of an observable is indetermined it doesn't mean that it doesn't exist. That's all I'm saying.

 

[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.