Would studying MWI be a waste of time?

[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.

 

[Denis]

Just like the universe can simultaneously exist and not exist.

Sorry, but this would be the ultimate end of science.

 

[vanhees71]

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?

Scripsi scripsi! I say the value of an observable for a system not prepared in a state of the form
$$\hat{\rho}=\sum_{\beta} p_{\beta} |a,\beta \rangle \langle a,\beta|,$$
where $|a,\beta \rangle$ are a set of eigenvectors to the eigenvalue $a$ of the self-adjoint operator $\hat{A}$, that represents the observable $A$, is indetermined. For me the phrase "The value does not exist" doesn't make any sense.

Of course, if the system is prepared in a state of the form given above, then the observable has the determined value $a$ before observation.

I'm agnostic about what in general happens when measuring the observable on a system not prepared in a state as given above. It depends on the construction of the measurement apparatus. Only for "filter measurements" you end up, by definition, in a state of the above form.

 

[Demystifier]

Of course, if the system is prepared in a state of the form given above, then the observable has the determined value a before observation.

I think now it's the right time to ask you how do you interpret the EPR case. Let Alice and Bob be two spatially separated observers. Let Alice has measurement apparatus which projects a state to either $|A1\rangle$ or $|A2\rangle$. Likewise, let Bob has measurement apparatus which projects a state to either $|B1\rangle$ or $|B2\rangle$.

At time $t_0$ let the state before the measurement be prepared in
$$|A1\rangle|B1\rangle + |A2\rangle|B2\rangle$$
At that moment $t_0$, both $A$ and $B$ are in an indetermined state, do you agree?

At the next moment $t_1$, assume that Alice performs her measurement and founds that the system is in the state $|A1\rangle$. But Bob does not yet perform his measurement. At that time $t_1$, after the measurement by Alice but before the measurement by Bob, can we say that the Bob's system is in the determined state $|B1\rangle$?

If your answer is no, then how is it consistent with your statement quoted above?

If your answer is yes, then what has caused this change (from indeterminate at $t_0$ to determinate at $t_1$) of the $B$-state? If it was caused by the measurement by Alice, then how can it be compatible with locality?

 

[vanhees71]

This I've also answered very often. Here you have the usual entanglement issue, where you have a composed system, for which the values of observables concerning the parts are indetermined. The correlations described by the entangled state are, however, there from the very beginning. So when Alice finds the value $A1$ she knows that Bob must find $B1$ and vice versa. For Bob $B1$ is still undetermined, and he finds the values according to the probabilities when measuring an ensemble. There's no action at a distance (at least not within local microcausal relativistic QFT), i.e., A cannot affect instantaneously the values of the observables of B's part of the system. The correlations are revealed only when A and B compare there measurement protocols.

 

[stevendaryl]

This I've also answered very often.

I would say that you've responded many times, rather than answered many times. Your response is an answer to a different question, which is whether Alice can affect Bob's measurement results. But you never answered the original question, which is:

If Alice has measured her particle to have spin-up (along the z-axis, say), does that mean that Bob's particle's state has a definite value for its z-component of spin (even if he hasn't measured it yet)?

 

[PeterDonis]

Thread closed for moderation.