QM objects do not have properties until measured?

[atyy]

People have been using the words "simplicial" and "non-simplicial" in this thread without defining them. rubi says that Bell's assumption about the existence of a parameter $\lambda$ is equivalent to the assumption that the underlying theory is simplicial. I take that to mean that for every situation, there is a "best", most-informative description of the situation? Or what does it mean? (I know what a simplex is, but how simplices relate to Bell's argument is unclear).

In both classical and quantum state space, the pure states are those that are not a statistical mixture of anything else (which is why we can consider the quantum state to be real, with the pure state being the state of a single system). In classical physics, each mixture is a unique statistical mixture of pure states (ie. state space is a simplex, where pure states are the pointy things). In quantum physics, a mixed density matrix can arise from more than one statistical mixture of pure states (state space is not a simplex, more like a sphere).

 

[rubi]

I hope I do not misunderstand this, but I regard it as quite intuitive. Momentum is directly proportional to the velocity, which is a measurement of change of position. A particle at any instant in time has a position, but no velocity.

A classical particle with a differentiable trajectory has a both a position and a velocity at each instant of time.

But my difficulty is that I have no idea what it means to reject realism.

Personally, I don't like the word "realism". I think Werner's term "classicality" is much better. It essentially means that it is in principle possible to simultaneously assign real numbers to all observable quantities. It's of course plausible apriori that this should be possible, but QM and experiment teaches us that this must be given up at least in some situations (such as different spin directions of a particle). So for me it's not a big leap to give it up for the remaining situations.

People have been using the words "simplicial" and "non-simplicial" in this thread without defining them. rubi says that Bell's assumption about the existence of a parameter $\lambda$ is equivalent to the assumption that the underlying theory is simplicial. I take that to mean that for every situation, there is a "best", most-informative description of the situation? Or what does it mean? (I know what a simplex is, but how simplices relate to Bell's argument is unclear).

Since we're discussing Werner's reply, I'm using the definition Werner gave in his paper, which I quoted in post #64. However, it is equivalent to the requirement that all observables can be modeled as random variables on one probability space.

Yes, there are notions between (1) and (2) but at least at present, they are not used in the derivation of a Bell inequality. At present there are 2 important routes to a bell inequality, and they use (1) and (2). In using (1) we have to supplement it with another assumption, eg. no randomness, while in using (2) there is no need for any additional assumption. So when people say they are giving up something to preserve locality, at present, they mean locality in the sense of (1). Maudlin doesn't dispute this.

A locality notion that applies to QM shouldn't be able to prove Bell's theorem, since QM violates the inequality, so it's no surprise that none such notion is used in any proof of the theorem. I don't think people preserving locality mean notion (1). They mean that there is no spooky action at a distance, i.e. all causal influences travel at most at the speed of light.

Apparently Werner writes "Naturally, I have taken “physical state" here in the sense of the operational approach, as the quantity which allows us to determine the probabilities for all subsequent operations and measurements (“epistemic" rather than “ontic")."

If the epistemic state is taken to be the physical state, isn't that the same as taking the wave function to be physical?

No, the wave function acts only as a container of information. It's as physical as the probability distribution $p_i=\frac{1}{6}$ for the throw of a die. There is no thing out there, called probability, such that if you measure it, you will get the value $\frac{1}{6}$.

 

[atyy]

A locality notion that applies to QM shouldn't be able to prove Bell's theorem, since QM violates the inequality, so it's no surprise that none such notion is used in any proof of the theorem. I don't think people preserving locality mean notion (1). They mean that there is no spooky action at a distance, i.e. all causal influences travel at most at the speed of light.

But then one needs the notion of a non-real cause.

If one thinks of Maudlin's claim that the world is non-local, it is hard to criticize him for not stating a reality assumption. The world is real in common language, so he is just saying that reality is nonlocal. So the reality assumption is clearly stated.

 

[rubi]

But then one needs the notion of a non-real cause.

One just needs the notion of a cause. It's a perfectly valid concept on all Lorentzian spacetimes, independent of the geometry of the state space.

If one thinks of Maudlin's claim that the world is non-local, it is hard to criticize him for not stating a reality assumption. The world is real in common language, so he is just saying that reality is nonlocal. So the reality assumption is clearly stated.

The word "real" is just a placeholder for a technical condition. Rejecting it doesn't mean that the world is not real (whatever that means), but rather that the technical condition is not satisfied. Moreover, this technical condition is rejected by most physicists, so accepting it is non-standard and must be pointed out clearly. Here is a quote from Maudlin's paper: "Unfortunately, many physicists have not properly appreciated what Bell proved: they take the target of his theorem—what the theorem rules out as impossible—to be much narrower and more parochial than it is. Early on, Bell’s result was often reported as ruling out determinism, or hidden variables. Nowadays, it is sometimes reported as ruling out, or at least calling in question, realism. But these are all mistakes. What Bell’s theorem, together with the experimental results, proves to be impossible (subject to a few caveats we will attend to) is not determinism or hidden variables or realism but locality, in a perfectly clear sense. What Bell proved, and what theoretical physics has not yet properly absorbed, is that the physical world itself is non-local."
Maudlin is provably wrong. It has been pointed out to him by an expert. But not even in his reply, he acknowledges that he must assume classicality. It is completely clear that he thinks that this assumption is not needed. Why else wouldn't he just admit it instead of responding polemically to criticism?

 

[atyy]

One just needs the notion of a cause. It's a perfectly valid concept on all Lorentzian spacetimes, independent of the geometry of the state space.

OK, I'll stop discussing Maudlin, since we aren't getting anywhere.

But let's discuss this point - how do you get the notion of a local common cause when the Bell inequalities are violated?

 

[N88]

But my difficulty is that I have no idea what it means to reject realism. … … Sort of by definition, "realistic" means to me "having to do with reality--that is, having to do with the real world". ...

Like you, I have no idea what it means to reject realism! And again, like you: "realistic" means to me "having to do with reality - that is, having to do with the real world". So I trust you are not reading "reject realism" in anything that I have written?

In my experience, most suggestions linking "reject" and "realism" refer to a qualified realism, and sometimes the qualifier is incomplete!

Thus, believing that we live in a quantum world, I reject classical realism. I reject the partially naive realism in EPR. I reject the naive realism on which Bell's theorem is based (see Bell's endorsement of d'Espagnat's naive realism in post #44 above). I therefore reject the local realism [SIC] associated with Bell's theorem: BUT, in this case, the qualifier is incomplete. I am in fact maintaining locality and rejecting the "naive realism" in the (properly qualified) local naive realism of Bell and d'Espagnat.

In post #67 above, Zeilinger rejects a "rather realistic interpretation of information" - by which I guess he agrees with me that Maudlin's view is "rather naively realistic"; like naively accepting that a mirage in the desert is a real lake; like accepting that the local naive realism of EPR, Bell and d'Espagnat is fully realistic.

 

[zonde]

I don't know how you come to this conclusion?

In post #59 I gave two quotes of Werner (http://arxiv.org/abs/1411.2120):

"According to Maudlin, Bell makes no assumption of “realism” or (as I called it in my reply) of “classicality” (in short “C”), or a hidden-variable description."
And this:
"The first issue is the explanation of classicality “C”. I gave a technical definition, the simplex property, ... "

Clearly Werner in the text after first quote is using words "classicality", "realism", "hidden variables" interchangeably.
And then he defines "C" as a simplex property of a state space (technical definition).
So for him it's the same. But not for you.

EDIT: And another quote of Werner:
"The point he thus missed in my explanation is that any description in terms of properties, thought to pertain to the system itself, and independent of the experimental arrangement and the choice of subsequent measurement, presupposes C. Using classical random variables, ontic states, hidden variables, and especially conditional probabilities based on those, presupposes C. And all these things are quite easy to find in the EPR and Bell arguments."

 

[zonde]

As I see things developing here, it seems to me that EPR went for partial naive realism ["if we can predict with certainty"] and Bell (relatedly) worked on full naive realism (see post #44 above): and both variants of naive realism are rendered inapplicable by QM and Bell-tests.

You are missing some details in EPR argument. Phrase: "if we can predict with certainty" does not refer to some form of realism but to prediction of QM. It's QM that says you can predict with certainty outcome of measurement of other entangled particle given measurement result of first entangled particle under the same measurement settings.

 

[N88]

You are missing some details in EPR argument. Phrase: "if we can predict with certainty" does not refer to some form of realism but to prediction of QM. It's QM that says you can predict with certainty outcome of measurement of other entangled particle given measurement result of first entangled particle under the same measurement settings.

I miss no details in EPR's argument. EPR defined their "elements of physical reality" with a qualifying IF: If, under QM's "prediction with certainty" (which I fully accept). I took their definition to be "partial naive realism" because they made no mention of "elements of physical reality" in the absence of that certainty.

Thus EPR have "elements of physical reality" corresponding to an outcome in the case of certainty (partial naive classicality) whereas (by way of comparison), Bell-d"Espagnat have "naively realistic" elements of physical reality under all conditions (total naive classicality), QM certain or QM uncertain.

 

[Quandry]

A classical particle with a differentiable trajectory has a both a position and a velocity at each instant of time.
##.

Probably not on topic here - but you have defined the particle as having differentiable trajectory and therefore you have defined it as having velocity. Take a particle about which you know nothing (therefore cannot presume velocity). Velocity is defined as the rate of change of position over a period time. If you determine the exact position of a particle at an instant of time there is no period of time and no change of position therefore no velocity.
Classically you can determine an average velocity between two positions over a defined period of time, but you can never know the velocity at any specific point in time

 

[atyy]

Probably not on topic here - but you have defined the particle as having differentiable trajectory and therefore you have defined it as having velocity. Take a particle about which you know nothing (therefore cannot presume velocity). Velocity is defined as the rate of change of position over a period time. If you determine the exact position of a particle at an instant of time there is no period of time and no change of position therefore no velocity.
Classically you can determine an average velocity between two positions over a defined period of time, but you can never know the velocity at any specific point in time

A non-relativistic quantum particle can have a trajectory with position and velocity - what it cannot have is position and canonically conjugate momentum. This just means the equation of motion is different from that in Newtonian physics.

 

[rubi]

how do you get the notion of a local common cause when the Bell inequalities are violated?

Let me try:
Let $\Psi$ be the quantum state. Let $X_A$ be an observable localized in the region $A$ and $P$ be a projector of it. Let $x\in A$. We say that $y\sim_P x$ ("$y$ causes $x$ to have the property $P$"), if there is a Lorentz transform $\Lambda$ such that $\Lambda A\subseteq I^-(A)$ and $y\in\Lambda A$ and for $P_\Lambda := U(\Lambda) P U(\Lambda)^\dagger$, we have that $\neg (P_\Lambda \Psi = \Psi) \Rightarrow \neg (P \Psi = \Psi)$.

Now you can say that there is a common cause for $x$ having the property $P_x$ and $y$ having the property $P_y$, if there exists $z$ such that $z\sim_{P_x} x$ and $z\sim_{P_y} y$.

In post #59 I gave two quotes of Werner (http://arxiv.org/abs/1411.2120):Clearly Werner in the text after first quote is using words "classicality", "realism", "hidden variables" interchangeably.
And then he defines "C" as a simplex property of a state space (technical definition).
So for him it's the same. But not for you.

EDIT: And another quote of Werner:
"The point he thus missed in my explanation is that any description in terms of properties, thought to pertain to the system itself, and independent of the experimental arrangement and the choice of subsequent measurement, presupposes C. Using classical random variables, ontic states, hidden variables, and especially conditional probabilities based on those, presupposes C. And all these things are quite easy to find in the EPR and Bell arguments."

Werner gave a perfectly fine technical condition for $C$ that I quoted earlier. This is the condition one needs. It also includes the way Bell defined his hidden variables. However, one can of course add hidden variables to a non-simplicial theory without making the state space a simplex. One just can't represent them on a single probability space.

 

[zonde]

I miss no details in EPR's argument. EPR defined their "elements of physical reality" with a qualifying IF: If, under QM's "prediction with certainty" (which I fully accept). I took their definition to be "partial naive realism" because they made no mention of "elements of physical reality" in the absence of that certainty.

Thus EPR have "elements of physical reality" corresponding to an outcome in the case of certainty (partial naive classicality) whereas (by way of comparison), Bell-d"Espagnat have "naively realistic" elements of physical reality under all conditions (total naive classicality), QM certain or QM uncertain.

Ok, I now understand that emphasis was on word "partial". But then the rest does not make sense. You said:

As I see things developing here, it seems to me that EPR went for partial naive realism ["if we can predict with certainty"] and Bell (relatedly) worked on full naive realism (see post #44 above): and both variants of naive realism are rendered inapplicable by QM and Bell-tests.

That seems to put me firmly in the camp of those who reject the classicality in EPR-Bell in favour of locality.

EPR arrives at this partial naive realism based explanation in EPR under condition of locality ["without in any way disturbing a system"]. Obviously rejecting "locality" renders EPR reasoning inapplicable (without rejecting realism).

On the other hand realism (in it's proper philosophical sense) can't be rejected if we hold on to scientific approach, as realism (in it's proper philosophical sense) is fundamental to science. Or more specifically science aims to explain reproducible certainty. And we favor such explanations over other types of explanations.
So we (should) favor non-local explanation of reproducible certainty over local non-explanation of reproducible certainty.

 

[rubi]

On the other hand realism (in it's proper philosophical sense) can't be rejected if we hold on to scientific approach, as realism (in it's proper philosophical sense) is fundamental to science. Or more specifically science aims to explain reproducible certainty. And we favor such explanations over other types of explanations.
So we (should) favor non-local explanation of reproducible certainty over local non-explanation of reproducible certainty.

You are confusing a loose philosophical concept ("realism") with a sharp technical condition ("classicality") on the mathematical description of reality. Just because we reject a certain mathematical way to describe reality, it doesn't mean that we reject reality in the philosophical sense (whatever that means).

 

[atyy]

Let me try:
Let $\Psi$ be the quantum state. Let $X_A$ be an observable localized in the region $A$ and $P$ be a projector of it. Let $x\in A$. We say that $y\sim_P x$ ("$y$ causes $x$ to have the property $P$"), if there is a Lorentz transform $\Lambda$ such that $\Lambda A\subseteq I^-(A)$ and $y\in\Lambda A$ and for $P_\Lambda := U(\Lambda) P U(\Lambda)^\dagger$, we have that $\neg (P_\Lambda \Psi = \Psi) \Rightarrow \neg (P \Psi = \Psi)$.

Now you can say that there is a common cause for $x$ having the property $P_x$ and $y$ having the property $P_y$, if there exists $z$ such that $z\sim_{P_x} x$ and $z\sim_{P_y} y$.

But wouldn't this work even if we take the wave function to be real?

 

[rubi]

But wouldn't this work even if we take the wave function to be real?

This definition of course only applies to quantum objects. If the wave function itself is a real classical object, then the classical definition applies to it. (By the way, I don't really know what the idea of a real wave function is supposed to mean if the you don't have a quantum system consisting of one particle. Already in the case of 2 particles, the wave function depends on 7 coordinates ($t, x_1, y_1, z_1, x_2, y_2, z_2$) and not on 4 as it would have to if it were supposed to be a field on spacetime.)

 

[atyy]

This definition of course only applies to quantum objects. If the wave function itself is a real classical object, then the classical definition applies to it. (By the way, I don't really know what the idea of a real wave function is supposed to mean if the you don't have a quantum system consisting of one particle. Already in the case of 2 particles, the wave function depends on 7 coordinates ($t, x_1, y_1, z_1, x_2, y_2, z_2$) and not on 4 as it would have to if it were supposed to be a field on spacetime.)

In what you wrote, is $\Psi$ a pure quantum state (ie. a ray in Hilbert space)?

The wave function is always in Hilbert space. If one wants to, one can attach a copy of Hilbert space to every point on a spatial slice of spacetime (won't make any change to the predictions, but will make collapse manifestly nonlocal). So when one is saying that the wave function is real, one regards Hilbert space as real.

Edit: in fact, since the wave function is always in Hilbert space, it is always manifestly nonlocal.

 

[Quandry]

A non-relativistic quantum particle can have a trajectory with position and velocity - what it cannot have is position and canonically conjugate momentum. This just means the equation of motion is different from that in Newtonian physics.

True, but it does not change the fact that at any point in time ΔT = 0

 

[rubi]

In what you wrote, is $\Psi$ a pure quantum state (ie. a ray in Hilbert space)?

Yes.

The wave function is always in Hilbert space. If one wants to, one can attach a copy of Hilbert space to every point on a spatial slice of spacetime (won't make any change to the predictions, but will make collapse manifestly nonlocal). So when one is saying that the wave function is real, one regards Hilbert space as real.

If a copy of wave function at some time $t$ is a physical object attached at each point of space, then I should be able to access it completely from every point of the universe, so by performing an experiment here in front of my computer, I can get all information I want about the state of the Andromeda galaxy (since a copy of all that information is supposed to be available here). That sounds very strange to me.

 

[atyy]

If a copy of wave function at some time $t$ is a physical object attached at each point of space, then I should be able to access it completely from every point of the universe, so by performing an experiment here in front of my computer, I can get all information I want about the state of the Andromeda galaxy (since a copy of all that information is supposed to be available here). That sounds very strange to me.

And by doing a measurement, Alice can instantly change the wave function at Bob's location, even though they are spacelike-separated. So if a pure quantum state is the complete information about the state of a system (eg. the entangled particles of Alice and Bob), then the quantum formalism is manifestly nonlocal.

 

[rubi]

And by doing a measurement, Alice can instantly change the wave function at Bob's location, even though they are spacelike-separated. So if a pure quantum state is the complete information about the state of a system (eg. the entangled particles of Alice and Bob), then the quantum formalism is manifestly nonlocal.

Yes, but only if you make this strange attachment of quantum states to points in spacetime. It's like attaching the probability distribution $p_i=\frac{1}{6}$ to every point of space and then after finding that the die shows the number 5, changing it to $p_i = \delta_{i5}$ everywhere. Of course, there is not really a physical object called probability distributions that changes everywhere in the universe as soon as I look at the die. If I don't consider the probability distribution to be a physical object, nothing non-local happens. Of course, if you claim that there is actually a physical object with that property, then it changes non-locally, but why would you do that? Isn't it absurd?

 

[atyy]

Yes, but only if you make this strange attachment of quantum states to points in spacetime. It's like attaching the probability distribution $p_i=\frac{1}{6}$ to every point of space and then after finding that the die shows the number 5, changing it to $p_i = \delta_{i5}$ everywhere. Of course, there is not really a physical object called probability distributions that changes everywhere in the universe as soon as I look at the die. If I don't consider the probability distribution to be a physical object, nothing non-local happens. Of course, if you claim that there is actually a physical object with that property, then it changes non-locally, but why would you do that? Isn't it absurd?

But no prediction of the theory actually changes (no matter how absurd it is). So why would there be any problem?

And it seems that what you proposed for defining a local common cause would work here too. So even though the wave function is real, and eveything is manifestly nonlocal, we have no problem defining a common cause and keeping locality. So this is not an example of giving up realism to preserve locality - we can have locality in your definition, regardless of whether the wave function is real or not.

 

[rubi]

But no prediction of the theory actually changes (no matter how absurd it is). So why would there be any problem?

There is no problem. It's just not very reasonable.

And it seems that what you proposed for defining a local common cause would work here too. So even though the wave function is real, and eveything is manifestly nonlocal, we have no problem defining a common cause. So this is not an example of giving up realism to preserve locality - we can have locality in your definition, regardless of whether the wave function is real or not.

No, you have added an additional physical object to the formalism. The formalism without a real existing object called wave function is local, but since your new formalism contains an additional physical object that evolves non-locally, the new formalism becomes non-local. (Of course, if you add something non-local to a local theory, the new theory will contain non-local elements.)

 

[atyy]

No, you have added an additional physical object to the formalism. The formalism without a real existing object called wave function is local, but since your new formalism contains an additional physical object that evolves non-locally, the new formalism becomes non-local. (Of course, if you add something non-local to a local theory, the new theory will contain non-local elements.)

OK, but then how can the wave function be a "cause" if it is not physical?

 

[rubi]

OK, but then how can the wave function be a "cause" if it is not physical?

The wave function isn't a cause. The events $x$, $y$, $z$ in spacetime are causes (or effects). If you roll a die, then the event in spacetime where you rolled the die is the cause and an effect is an event in spacetime, where the die shows the number 5. The probability distribution $p_i=\frac{1}{6}$ didn't cause anything. It's just a container of information about what could happen.

 

[zonde]

However, one can of course add hidden variables to a non-simplicial theory without making the state space a simplex. One just can't represent them on a single probability space.

Why do you think that we can't represent them on a single probability space given locality?
Outcomes (sample spaces) are the same. So it leaves measurement settings. But if we enforce locality then measurement settings at one end can't have any effect at the other end. So different measurement settings would have to be modeled in the same probability space. What other possible reason do you see why we can't represent them on a single probability space?

 

[atyy]

The wave function isn't a cause. The events $x$, $y$, $z$ in spacetime are causes (or effects). If you roll a die, then the event in spacetime where you rolled the die is the cause and an effect is an event in spacetime, where the die shows the number 5. The probability distribution $p_i=\frac{1}{6}$ didn't cause anything. It's just a container of information about what could happen.

But nothing in your definition would fail if I made the wave function a real physical object.

 

[rubi]

Why do you think that we can't represent them on a single probability space given locality?
Outcomes (sample spaces) are the same. So it leaves measurement settings. But if we enforce locality then measurement settings at one end can't have any effect at the other end. So different measurement settings would have to be modeled in the same probability space. What other possible reason do you see why we can't represent them on a single probability space?

No, this doesn't follow. It's just a hard mathematical fact that non-commuting observables can't be modeled as random variables on one probability space and quantum theory just happens to be a theory with non-commuting observables. It's not the outcomes that need to represented on one probability space. Also the hidden variables need to represented on that space. (Moreover, I don't have the burden of proof. You are the one who claims that locality implies a simplicial state space, so you are the one who has the obligation to prove it.)

But nothing in your definition would fail if I made the wave function a real physical object.

Of course the definition fails, since it can be applied only to objects that are described by quantum theory. Your real wave function is not such an object (it is not represented by an observable on a Hilbert space). It's an additional object, external to the Hilbert space description, so the definition can't be applied to it.

 

[martinbn]

The wave function is always in Hilbert space. If one wants to, one can attach a copy of Hilbert space to every point on a spatial slice of spacetime (won't make any change to the predictions, but will make collapse manifestly nonlocal). So when one is saying that the wave function is real, one regards Hilbert space as real.

Can you explain more? What does this mean?

 

[atyy]

Of course the definition fails, since it can be applied only to objects that are described by quantum theory. Your real wave function is not such an object (it is not represented by an observable on a Hilbert space). It's an additional object, external to the Hilbert space description, so the definition can't be applied to it.

The wave function is still an object in Hilbert space. It's just that there is a copy of Hilbert space at every point in space.

 

[rubi]

The wave function is still an object in Hilbert space. It's just that there is a copy of Hilbert space at every point in space.

But it's not modeled as an observable on Hilbert space. That's what you need in order to get the projections I used in my definition. In a quantum theory, every physical object has a corresponding self-adjoint operator that models it. This is not the case for the "physical object" called wave function. The wave function in your theory of real wave functions is not a quantum object itself.

 

[zonde]

No, this doesn't follow. It's just a hard mathematical fact that non-commuting observables can't be modeled as random variables on one probability space and quantum theory just happens to be a theory with non-commuting observables.

Do I have to understand this as counterexample to my considerations? But whether or not QM is local is the topic of current discussion. So this can't be viewed as counterexample.

(Moreover, I don't have the burden of proof. You are the one who claims that locality implies a simplicial state space, so you are the one who has the obligation to prove it.)

I am not trying to prove anything. I am trying to understand your position.
If you want proof that QM is non-local then tell me and I will continue discussion from your response in post #20.

 

[martinbn]

The wave function is still an object in Hilbert space. It's just that there is a copy of Hilbert space at every point in space.

And? If the spatial slices are $\mathbb R^3$, then you consider $\mathbb R^3\times\mathcal H$. Fine, but what do you do with it? By the way is there a particular reason why you write "Hilbert space" without any article.

 

[rubi]

Do I have to understand this as counterexample to my considerations? But whether or not QM is local is the topic of current discussion. So this can't be viewed as counterexample.

Bell proved that classical local theories obey an inequality that is violated. He proved nothing about non-classical theories. We don't have any tool to decide upon the locality of quantum mechanics. A counterexample to your claim that non-simplicial states can't have hidden-variable descriptions is contained in this paper (section 4).

If you want proof that QM is non-local then tell me and I will continue discussion from your response in post #20.

None such proof exists. The book I mentioned in post #20 debunks all of them, be it a probabilistic proof or a proof using relative frequencies. I don't have enough time to discuss why proven mathematical theorems are not false. If you have another opinion, then please publish a paper, then we can discuss it.

 

[atyy]

But it's not modeled as an observable on Hilbert space. That's what you need in order to get the projections I used in my definition. In a quantum theory, every physical object has a corresponding self-adjoint operator that models it. This is not the case for the "physical object" called wave function. The wave function in your theory of real wave functions is not a quantum object itself.

OK. So in your proposal, in the Bell test, you would like to say that the preparation causes the correlations. But that means that the measurement choice of either Alice ore Bob is not a cause of the result?

(Typically, we say Alice's result is caused by the preparation as well as her measurement choice.)