What Is the Frauchiger-Renner Theorem?

[atyy]

The rest of your post requires a longer answer which I'll only have time for over the weekend. To help though what do you mean by "formally complete". That it isn't self-contradictory, even though it leaves somethings unexplained like measurement?

Formally complete, in the sense that it is (or can be) consistent after one has defined one classical/quantum cut. However it is not complete from the larger point of that the observer has a special status in quantum mechanics. Here one views QM as a complete theory that is emergent from a more complete theory, just as Newtonian mechanics is a complete theory emergent from more complete theories like special relativity and quantum mechanics.

Also do you take a perspectival view of Copenhagen, i.e. the measurement outcomes aren't fully objective but associated with the agent observing them, relational to some degree.

Within QM, outcomes are fully objective to all agents on the classical side of the classical/quantum cut. I'm not sure what you mean by perspectival, but we have to make only one classical/quantum cut - and things on the other side of the cut cannot be granted the status of agent or observation outcome.

 

[DarMM]

To be clearer then, in a Wigner's friend type set up, do Wigner and the friend have the same cut or different ones?

 

[atyy]

To be clearer then, in a Wigner's friend type set up, do Wigner and the friend have the same cut or different ones?

Wigner is classical, and the friend is quantum.

 

[DarMM]

From Wigner's perspective I get that, but do you think the friend should model themselves as quantum as well?

 

[atyy]

From Wigner's perspective I get that, but do you think the friend should model themselves as quantum as well?

The friend is not an observer and cannot model himself.

Edit: The question of whether the cut can be consistently shifted is interesting. However, I don't think it is in formal QM, since making the cut itself already requires subjectivity. However, I do like Hay and Peres's https://arxiv.org/abs/quant-ph/9712044. I think the question more generally requires a more complete theory like Bohmian Mechanics to be correctly dealt with.

 

[DarMM]

Okay you would be saying the presence of Wigner's measuring device alters the context of the situation and means the friend could not be placed on the classical side of the cut, even by himself.

The highly unusual confined nature of the friend and the contextual effects of Wigner's device means the friend evolves nonclassically?

I think this is legitimate, as an issue I always have with these set ups is what has to actually be done to get a Wigner's friend scenario is very extreme.

I'll need to think a bit.

 

[atyy]

Okay you would be saying the presence of Wigner's measuring device alters the context of the situation and means the friend could not be placed on the classical side of the cut, even by himself.

The highly unusual confined nature of the friend and the contextual effects of Wigner's device means the friend evolves nonclassically?

I think this is legitimate, as an issue I always have with these set ups is what has to actually be done to get a Wigner's friend scenario is very extreme.

I'll need to think a bit.

Yes, the friend is just the same as a Schroedinger's cat or a qubit.

 

[Demystifier]

Before I go digging into Bohr's papers and those of the old Copenhagen group, I would just like to narrow the search for what you want. Which do you think Bohr for example didn't hold:

1. Observations are objective events, not agent experiences
2. There are no hidden variables, i.e. QM is complete
3. QM may be universally applied to any system (note this is not the same as saying there is no Heisenberg cut)
4. The wavefunction is not an ontic object

My problem with this list is that 2 and 4 contradict 1 in a rather trivial way. 2 implies that there is no ontology which is not $\psi$. 4 says that there is no ontology which is $\psi$. Hence 2 and 4 together say that there is no any ontology at all. Yet 1 says that there is some ontology. A contradiction!

 

[DarMM]

My problem with this list is that 2 and 4 contradict 1 in a rather trivial way. 2 implies that there is no ontology which is not $\psi$. 4 says that there is no ontology which is $\psi$. Hence 2 and 4 together say that there is no any ontology at all. Yet 1 says that there is some ontology. A contradiction!

Well I would say (2.) means that your ontology doesn't involve a world that admits an objective description with mathematical terms. It might be non-mathematical, or it might be mathematical but completely reltional. Bohr thought the former, Schrödinger thought the latter.

Regardless it seems to me pretty clear that many believed 1-4. I understand why you think they are daft, I'm not disagreeing with you on this, but they're there in Bohr, Heisenberg and others writings. As silly as one might find them, there was no proof they were self-contradictory until Masanes.

 

[DarMM]

Basically @Demystifier and @atyy , the "problem" is that both of you already hold non-self-contradictory positions* that Masanes's proof doesn't affect and consider the positions it does refute as obviously wrong. However there was no clear proof they were wrong until now and they are the Copenhagen of Bohr and others of early QM.

*e.g. @atyy you have a similar view to Bub, QM must be used from the perspective of a final ultimate classical user and rejects the idea that the Wigner's friend set up has the friend capable of being considered on the classical side, even by himself. I think it's a very clear resolution and you seem to consider it obvious, but note that Bub had to publish a paper about it, it's not a widely known position.

 

[DarMM]

Yes, the friend is just the same as a Schroedinger's cat or a qubit.

Okay sorry for all the questions! I think this is the last one, what about the situation makes this true for you explicitly? Basically do you think that Wigner completely confining his friend to a totally sealed lab and the physical presence of Wigner's equipment causes quantum effects to propagate up to the macroscopic scale for the friend invalidating him considering himself or his equipment as classical?

I can imagine this, Wigner's device would be pretty "extreme" equipment whatever it is.

 

[martinbn]

My problem with this list is that 2 and 4 contradict 1 in a rather trivial way. 2 implies that there is no ontology which is not ψ\psi. 4 says that there is no ontology which is ψ\psi. Hence 2 and 4 together say that there is no any ontology at all. Yet 1 says that there is some ontology. A contradiction!

I don't understand this. How does 2. imply that there is no ontology other than $\psi$? In fact, in my opinion, it is an extreme abuse of language to say that $\psi$ is ontology. What does it even mean?

 

[DarMM]

What does it even mean?

It's true in Many-Worlds and Bohmian mechanics, the state space of ontic objects $\Lambda$ has the form $\Lambda = \mathcal{A} \times \mathcal{H}$ with $\mathcal{H}$ the quantum Hilbert space. So the ontology has $\psi$ as an element.

 

[martinbn]

It's be true in Many-Worlds and Bohmian mechanics, the state space of ontic objects $\Lambda$ has the form $\Lambda = \mathcal{A} \times \mathcal{H}$ with $\mathcal{H}$ the quantum Hilbert space. So the ontology has $\psi$ as an element.

May be I don't know what the word means. But for me "ontology" is related to existence/being. So something like an element in a Hilbert space cannot have ontology. It makes no sense to say that it exists. What exists is the particles/fields, not $\psi$.

 

[Demystifier]

I don't understand this. How does 2. imply that there is no ontology other than $\psi$?

Let me try to answer this step by step. (2) says that "There are no hidden variables, i.e. QM is complete". It really means that in the mathematical description of the world, we don't need anything else except $\psi$. I guess you agree so far.

Now consider ontology. It's a philosophical term, but its meaning is not essential here. All what we explicitly need to prove what I want to prove is the following assumption:
Assumption 1: Ontology can be described mathematically and we need it to describe the world.
Then, from (2) and Assumption 1, it follows that there is no ontology which is not $\psi$. Is it clear now?

So if you want to avoid my conclusion, you must deny Assumption 1, i.e. you must hold that either ontology cannot be described mathematically, or that we don't need ontology to describe the world. The first option looks like mysticism to me, which is logically legitimate but not scientific in spirit. The second option implies that, in order to describe the world, we don't need to believe that the Moon exists when nobody observes it, which is also logically legitimate, but contradicts common sense.

 

[Demystifier]

May be I don't know what the word means. But for me "ontology" is related to existence/being. So something like an element in a Hilbert space cannot have ontology. It makes no sense to say that it exists.

I would basically agree with that.

What exists is the particles/fields, not $\psi$.

A priori, it is not obvious that particles or fields exist. What is obvious is that things such as the Moon exist. But the Moon must be made of something more elementary, which could perhaps be particles, fields, strings, or something else. In my "Bohmian mechanics for instrumentalists" I explain why particles, as objects with well defined positions at all time, are the most natural possibility.

 

[martinbn]

Let me try to answer this step by step. (2) says that "There are no hidden variables, i.e. QM is complete". It really means that in the mathematical description of the world, we don't need anything else except ψ\psi. I guess you agree so far.

For the purpose of this conversation, yes, I agree. Strictly speaking there may be more things needed, say equations, a choice of Hilbert space representation, specific operators, boundary conditions etc.

Now consider ontology. It's a philosophical term, but its meaning is not essential here. All what we explicitly need to prove what I want to prove is the following assumption:
Assumption 1: Ontology can be described mathematically and we need it to describe the world.
Then, from (2) and Assumption 1, it follows that there is no ontology which is not ψ\psi. Is it clear now?

No, not at all clear. As I said we might be using the words differently. Let me give you an example. Classical mechanics of several particles. The ontology of the theory is that there are several particles. That's what exists in the physical world. Their behavior may be described by a function (plus possibly other things as above), but it is meaningless and abuse of language to say that the ontology is the function. If your argument was correct, it would imply that in this example the only ontology is the function. Which is absurd.

It seems to me that by ontology you mean the minimum of mathematical apparatus that is needed to describe the world within a given theory.

 

[Demystifier]

Their behavior may be described by a function (plus possible other things as above), but it is meaningless and abuse of language to say that the ontology is the function. If your argument was correct, it would imply that in this example the only ontology is the function. Which is absurd.

Ah, I see what is your problem. Suppose that the ontology is the particle with a well defined position. I think you are fine with that. That position can be described by 3 numbers (x,y,z), but it doesn't mean that those 3 numbers are ontology. The ontology is the position itself, not our mathematical coordinatization of that position. Is it what you are saying?

Well, strictly speaking I think that you right, but that problem can be cured relatively easy. Consider a physical ontological object $\tilde{O}$ (in the case above it is a particle with a well defined position in physical space). Let its all mathematically describable properties be described by some mathematical object $O$ (in the case above it is the numbers (x,y,z)). When we say that $O$ is ontology, it is just an imprecise manner of speak, which really means that $\tilde{O}$ is ontology.

So when someone says that the wave function $\psi$ is ontology, it really means that there is an ontological object $\tilde{O}$ such that its all mathematically describable properties can be described by $O=\psi$.

Does it make more sense now?

EDIT: This abuse of language is similar to calling the numbers $(x,y,z)$ a vector, which really means that the object $x{\bf e}_x+y{\bf e}_y+z{\bf e}_z$ is a vector. Physicists usually do not have problems with calling $(x,y,z)$ a vector, which often annoys mathematicians.

 

[DarMM]

May be I don't know what the word means. But for me "ontology" is related to existence/being. So something like an element in a Hilbert space cannot have ontology. It makes no sense to say that it exists. What exists is the particles/fields, not $\psi$.

@Demystifier has already stated this, but basically in a $\psi$-ontic view $\psi$ directly describes the physical properties of the objects of the theory. In a $\psi$-epistemic view, $\psi$ instead describes an agent's knowledge of those properties.

 

[atyy]

Okay sorry for all the questions! I think this is the last one, what about the situation makes this true for you explicitly? Basically do you think that Wigner completely confining his friend to a totally sealed lab and the physical presence of Wigner's equipment causes quantum effects to propagate up to the macroscopic scale for the friend invalidating him considering himself or his equipment as classical?

I can imagine this, Wigner's device would be pretty "extreme" equipment whatever it is.

Well, the friend can model himself as classical, but then Wigner will also model the friend as classical. Basically, we only allow one observer - Wigner. Wigner can subjectively choose where to put the classical/quantum boundary (ie. whether it includes the friend or not). However, what Wigner regards as an objective measurement outcome differs depending on whether the friend is classical or quantum. If Wigner regards the friend as classical, then the quantum state used by Wigner will collapse when the friend makes the measurement (ie. the friend's measuring apparatus is also Wigner's measuring apparatus).

 

[atyy]

I don't think Healey's third argument is correct. If Alice knows that Dan has recorded a definite outcome, then Dan has collapsed the state. So Eq 23 is not correct. It should be a proper mixture, not a pure state.

 

[stevendaryl]

I don't understand this. How does 2. imply that there is no ontology other than $\psi$? In fact, in my opinion, it is an extreme abuse of language to say that $\psi$ is ontology. What does it even mean?

I think that the issue is whether the wave function is subjective--that is, its value reflects the knowledge of the observer--or objective--it has an actual value, even if the observer may not know what that value is.

 

[martinbn]

I think that the issue is whether the wave function is subjective--that is, its value reflects the knowledge of the observer--or objective--it has an actual value, even if the observer may not know what that value is.

But I still don't understand Demistifier's argument. I don't see how 2. and 4. contradict 1.

 

[stevendaryl]

But I still don't understand Demistifier's argument. I don't see how 2. and 4. contradict 1.

I'm not sure I understand that specific argument. But EPR-type correlations seem to me to make it difficult to understand how the wave function can be epistemological (if that's the antonym of ontological).

I've stated this simple argument several times before, but I don't know of a good answer to it.

Let's take the case where Alice and Bob are measuring spins of anti-correlated spin-1/2 particles. To make it definite, let's assume that they are both planning to measure spin along the z-axis. Let's pick an inertial coordinate system in which Alice and Bob are both at rest, and assume that Bob is farther from the source of twin particles than Alice, so he measures his particle's spin slightly later than Alice does (although the two measurements have a spacelike separation).

1. Immediately before Alice performs her measurement, she would rate the probabilities for Bob's results to be 50/50 spin-up or spin-down.
2. Immediately after Alice performs her measurement and gets the result "spin-up", she knows with 100% certainty that Bob will get the result "spin-down".
3. So the statement "Bob will get spin-down" goes from being uncertain with 50/50 probability to true.
4. So it seems that Bob's situation (as understood by Alice) makes a nearly-instantaneous change.
5. There are two different possible interpretations of this sudden change. They are (1) epistemological, or (2) physical. (Maybe there are more than two possibilities, but I don't know of others.) Under the epistemological interpretation, the fact that Bob will get spin-down was true BEFORE Alice performed her measurement, and Alice's measurement simply allowed her to know this. Under the physical interpretation, the fact that Bob will get spin-down wasn't true before Alice performed her measurement, but became true as a side-effect of her measurement.
6. The epistemological interpretation seems to be contradicted by Bell's proof.
7. The physical interpretation seems to violate causality (no effects can travel faster than light).

I don't see a satisfactory way out if you accept Bell's proof and you also accept FTL limitations for effects. There are more exotic ways out, and they are Many-Worlds (Bob's result doesn't have a definite value) and superdeterminism, but those are unsatisfactory for other reasons.

 

[1977ub]

... The physical interpretation seems to violate causality (no effects can travel faster than light).
I don't see a satisfactory way out if you accept Bell's proof and you also accept FTL limitations for effects. There are more exotic ways out, and they are Many-Worlds (Bob's result doesn't have a definite value) and superdeterminism, but those are unsatisfactory for other reasons.

Can you articulate or link your favorite reasons why SD is unsatisfactory? Thank you.

 

[RUTA]

I'm not sure I understand that specific argument. But EPR-type correlations seem to me to make it difficult to understand how the wave function can be epistemological (if that's the antonym of ontological).

I've stated this simple argument several times before, but I don't know of a good answer to it.

Let's take the case where Alice and Bob are measuring spins of anti-correlated spin-1/2 particles. To make it definite, let's assume that they are both planning to measure spin along the z-axis. Let's pick an inertial coordinate system in which Alice and Bob are both at rest, and assume that Bob is farther from the source of twin particles than Alice, so he measures his particle's spin slightly later than Alice does (although the two measurements have a spacelike separation).

1. Immediately before Alice performs her measurement, she would rate the probabilities for Bob's results to be 50/50 spin-up or spin-down.
2. Immediately after Alice performs her measurement and gets the result "spin-up", she knows with 100% certainty that Bob will get the result "spin-down".
3. So the statement "Bob will get spin-down" goes from being uncertain with 50/50 probability to true.
4. So it seems that Bob's situation (as understood by Alice) makes a nearly-instantaneous change.
5. There are two different possible interpretations of this sudden change. They are (1) epistemological, or (2) physical. (Maybe there are more than two possibilities, but I don't know of others.) Under the epistemological interpretation, the fact that Bob will get spin-down was true BEFORE Alice performed her measurement, and Alice's measurement simply allowed her to know this. Under the physical interpretation, the fact that Bob will get spin-down wasn't true before Alice performed her measurement, but became true as a side-effect of her measurement.
6. The epistemological interpretation seems to be contradicted by Bell's proof.
7. The physical interpretation seems to violate causality (no effects can travel faster than light).

I don't see a satisfactory way out if you accept Bell's proof and you also accept FTL limitations for effects. There are more exotic ways out, and they are Many-Worlds (Bob's result doesn't have a definite value) and superdeterminism, but those are unsatisfactory for other reasons.

I'll give my response again, for the readers who might not have seen it. QM is in the business of supplying the distributions of quantum exchanges in 4D (block universe). The change in Alice's knowledge is purely epistemic, since the pattern (QM distribution) is "already there" in the block universe. Unitary evolution simply represents our (necessary) ignorance about what lies in the future.

 

[stevendaryl]

Can you articulate or link your favorite reasons why SD is unsatisfactory? Thank you.

I'm sorry, what is SD?

 

[1977ub]

I'm sorry, what is SD?

I meant 'superdeterminism'

 

[DarMM]

The epistemological interpretation seems to be contradicted by Bell's proof.

I wouldn't say Bell's theorem says anything too strong about the epistemological view. In fact I would say it has the same implications for ontic and epistemic views. It tells you the underlying ontology can't be a non-superdeterministic mathematical* local causal** single world, regardless of whether $\psi$ is part of that ontology or not.

It's the PBR theorem that has stronger implication for epistemic theories, meaning they basically have to go the retro/acausal route or the antirealist (non-mathematical) route.

* I prefer mathematical to realist, as what's actual rejected is that the underlying reality has no mathematical description not that it's not "real"

** Causal meaning not retrocausal or acausal

 

[DarMM]

I don't think Healey's third argument is correct. If Alice knows that Dan has recorded a definite outcome, then Dan has collapsed the state. So Eq 23 is not correct. It should be a proper mixture, not a pure state.

You're assuming an interpretation with some sort of reasonably objective collapse, not that that's wrong, but it doesn't affect Masanes proof.

Under the kind of Copenhagen advocated by Bohr and others since QM is just a calculus of expectations (in modern terminology a Bayesian framework) from the perspective of Alice you would have:
$$|\uparrow\rangle|d-ready\rangle \rightarrow |\uparrow\rangle|d-up\rangle$$
and
$$|\downarrow\rangle|d-ready\rangle \rightarrow |\downarrow\rangle|d-down\rangle$$
you'd have to have:
$$\sqrt{\frac{1}{2}}\left(|\downarrow\rangle + |\uparrow\rangle\right) |d-ready\rangle \rightarrow \sqrt{\frac{1}{2}}\left(|\uparrow\rangle|d-up\rangle + |\downarrow\rangle|d-down\rangle\right)$$

You're saying it should be a mixed state instead, but that implies you should know for some systems`you have yet to observe you should decide whether to apply unitary evolution or collapse. Would you measure this via decoherence?

 

[atyy]

You're assuming an interpretation with some sort of reasonably objective collapse, not that that's wrong, but it doesn't affect Masanes proof.

Under the kind of Copenhagen advocated by Bohr and others since QM is just a calculus of expectations (in modern terminology a Bayesian framework) from the perspective of Alice you would have:
$$|\uparrow\rangle|d-ready\rangle \rightarrow |\uparrow\rangle|d-up\rangle$$
and
$$|\downarrow\rangle|d-ready\rangle \rightarrow |\downarrow\rangle|d-down\rangle$$
you'd have to have:
$$\sqrt{\frac{1}{2}}\left(|\downarrow\rangle + |\uparrow\rangle\right) |d-ready\rangle \rightarrow \sqrt{\frac{1}{2}}\left(|\uparrow\rangle|d-up\rangle + |\downarrow\rangle|d-down\rangle\right)$$

You're saying it should be a mixed state instead, but that implies you should know for some systems`you have yet to observe you should decide whether to apply unitary evolution or collapse. Would you measure this via decoherence?

Dan and Carol are simply Alice's proxies - ie. if Alice knows that Dan or Carol has a definite outcome, then that is enough information for the wave function to collapse from Alice's point of view. Basically, if there is a definite outcome, then there is collapse.

 

[Demystifier]

I don't think Healey's third argument is correct. If Alice knows that Dan has recorded a definite outcome, then Dan has collapsed the state. So Eq 23 is not correct. It should be a proper mixture, not a pure state.

It is assumed that there is no objective collapse. It is an update of agent's knowledge. So if Dan subjectively collapsed the state, it doesn't mean that the subjective collapse refers also to Alice. To make an analogy, if Alice knows that Dab thinks that Angelina Jolie is beautiful, it doesn't mean that Angelina Jolie is beautiful from the point of view of Alice.

 

[atyy]

It is assumed that there is no objective collapse. It is an update of agent's knowledge. So if Dan subjectively collapsed the state, it doesn't mean that the subjective collapse refers also to Alice. To make an analogy, if Alice knows that Dab thinks that Angelina Jolie is beautiful, it doesn't mean that Angelina Jolie is beautiful from the point of view of Alice.

But Dan has a definite outcome - to whom? If Alice acknowledges that Dan has a definite outcome, then Alice must collapse the state. If Alice does not acknowledge that Dan has a definite outcome, then there is no definite outcome for Alice, and there is no P(a,b,c,d) for Alice.

 

[Demystifier]

But Dan has a definite outcome - to whom? If Alice acknowledges that Dan has a definite outcome, then Alice must collapse the state. If Alice does not acknowledge that Dan has a definite outcome, then there is no definite outcome for Alice, and there is no P(a,b,c,d) for Alice.

If my Angelina Jolie counterexample has not convinced you, here is another counterexample: Bohmian mechanics. In BM there is no collapse of the full wave function of the Universe. But BM has a conditional wave function, which does not obey the Schrodinger equation and hence collapses when the measurement is performed. For definiteness, let us model the full wave function as
$$\Psi(x,x_A,x_D,t)$$
where $x$ is the position of the measured particle, $x_A$ are positions of particles constituting the Alice's measurement apparatus and $x_D$ are positions of particles constituting the Dan's measurement apparatus. Then the Alice's conditional wave function is
$$\psi_A(x,x_D,t)=\Psi(x,X_A(t),x_D,t)$$
where $X_A(t)$ are the Bohmian trajectories. Similarly, the Dan's conditional wave function is
$$\psi_D(x,x_A,t)=\Psi(x,x_A,X_D(t),t)$$
Clearly, the collapse of $\psi_D$ does not imply the collapse of $\psi_A$. However, Alice knows that Dan's wave function collapses, so Alice may alternatively use the wave function
$$\psi'_A(x,t)=\Psi(x,X_A(t),X_D(t),t)$$
which does collapse when $\psi_D$ collapses. So which wave function should Alice use? It's up to her. But she must be consistent. A logical contradiction may arise if she mixes conclusions obtained from $\psi_A$ with those obtained from $\psi'_A$.

 

[atyy]

If my Angelina Jolie counterexample has not convinced you, here is another counterexample: Bohmian mechanics. In BM there is no collapse of the full wave function of the Universe. But BM has a conditional wave function, which does not obey the Schrodinger equation and hence collapses when the measurement is performed. For definiteness, let us model the full wave function as
$$\Psi(x,x_A,x_D,t)$$
where $x$ is the position of the measured particle, $x_A$ are positions of particles constituting the Alice's measurement apparatus and $x_D$ are positions of particles constituting the Dan's measurement apparatus. Then the Alice's conditional wave function is
$$\psi_A(x,x_D,t)=\Psi(x,X_A(t),x_D,t)$$
where $X_A(t)$ are the Bohmian trajectories. Similarly, the Dan's conditional wave function is
$$\psi_D(x,x_A,t)=\Psi(x,x_A,X_D(t),t)$$
Clearly, the collapse of $\psi_D$ does not imply the collapse of $\psi_A$. However, Alice knows that Dan's wave function collapses, so Alice may alternatively use the wave function
$$\psi'_A(x,t)=\Psi(x,X_A(t),X_D(t),t)$$
which does collapse when $\psi_D$ collapses. So which wave function should Alice use? It's up to her. But she must be consistent. A logical contradiction may arise if she mixes conclusions obtained from $\psi_A$ with those obtained from $\psi'_A$.

Alice can always just use $\Psi(x,x_A,x_D,t)$.

Also, does what you wrote contradict my points in post #103?