Undergrad What Is the Frauchiger-Renner Theorem?

[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?

 

[Demystifier]

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

In that case, according to BM, there is never collapse for Alice.

 

[atyy]

In that case, according to BM, there is never collapse for Alice.

OK, but I don't think there is any contradiction to what I wrote in post #103 which was according to Copenhagen.

 

[David Byrden]

1. The epistemological interpretation seems to be contradicted by Bell's proof.
2. The physical interpretation seems to violate causality (no effects can travel faster than light).

1. Yes, it is.
2. No, it doesn't. Not if you interpret QM in a "many worlds" way, remembering that the "worlds" are local.Before her measurement, Alice is in a "world" where both outcomes are in superposition (or, to look at it another way, no information about either outcome has reached her).
The measurement entangles Alice with the particle's wave function, which immediately splits her into "Alice up" and "Alice down" and they promptly decohere and can never again communicate.
"Alice up" is now guaranteed to find "Bob down" when she makes her way over there, or sends a message, or whatever. Her wave function and "Bob down" are lobes of the same wave function. But nothing has moved faster than light. There is also a "Bob up" who is utterly unreachable by "Alice up" but can meet up with "Alice down".
Whether you wish to consider the alternative Alice and Bob as real and made of meat, or as mere potentialities implicit in the wave function and no more solid than the square root of minus one, is up to you.

 

[Demystifier]

OK, but I don't think there is any contradiction to what I wrote in post #103 which was according to Copenhagen.

What you fail to realize is that there is no Copenhagen interpretation. There are several different interpretations that people call "Copenhagen". You are referring to one particular version of Copenhagen (say the Landau-Lifshitz version), but the FR-Masanes theorem rules out another version of Copenhagen, a version that denies the objective collapse but accepts objective measurement outcomes. The theorem is not applicable to "your" LL version of Copenhagen.

 

[atyy]

What you fail to realize is that there is no Copenhagen interpretation. There are several different interpretations that people call "Copenhagen". You are referring to one particular version of Copenhagen (say the Landau-Lifshitz version), but the FR-Matsas theorem rules out another version of Copenhagen, a version that denies the objective collapse but accepts objective measurement outcomes. The theorem is not applicable to "your" LL version of Copenhagen.

That's fine. But did anyone believe the FR-Masanes version of Copenhagen anyway?

 

[Demystifier]

Not if you interpret QM in a "many worlds" way, remembering that the "worlds" are local.

In many-worlds, the worlds are neither non-local nor local. They are alocal: http://de.arxiv.org/abs/1703.08341

 

[Demystifier]

That's fine. But did anyone believe the FR-Masanes version anyway?

Yes, that's what I would like to know too. It seems to me that @vanhees71 believes something like that. If I'm right, it seems that we finally have a theorem against him.

 

[atyy]

Yes, that's what I would like to know too. It seems to me that @vanhees71 believes something like that. If I'm right, it seems that we finally have a theorem against him.

I thought he converted to BM?

 

[Demystifier]

I thought he converted to BM?

I think BM is now his second best interpretation. Maybe when he learns about the FR-Masanes-Leifer theorem it will become his first.

 

[stevendaryl]

1. Yes, it is.
2. No, it doesn't. Not if you interpret QM in a "many worlds" way, remembering that the "worlds" are local.

Yes, I usually make the exception for MW interpretations. Somewhere someone mentioned the assumption that measurements have unique results.

 

[Demystifier]

@DarMM there is one additional question that I would like to discuss with you. Do we really need the undoing of measurement in the FR-Masanes-Leifer theorem? Or can we achieve the same just by preparing another copy of the system?

Let me explain. The basic common scheme in all these thought experiments is the following:
1. First prepare the system in the state $|\Psi\rangle$.
2. Then perform a measurement described by a unitary operation $U|\Psi\rangle$.
3. After that undo the measurement by acting with $V=U^{-1}$, which gives $VU|\Psi\rangle=|\Psi\rangle$.
4. Finally perform a new measurement $U'|\Psi\rangle$.

But for the sake of proving the theorem, it seems to me that we don't really need the step 3. Instead, we can perform:

3'. Prepare a new copy of the state $|\Psi\rangle$.

After that, 4. refers to this new copy. Note that $|\Psi\rangle$ is a known state, so the no-cloning theorem is not an obstacle for preparing a new copy in the same state.

The only problem I see with this is the following. The state $|\Psi\rangle$ is really something of the form
$$|\Psi\rangle=|\psi\rangle |{\rm detector \;\; ready}\rangle$$
which involves not only a simple state $|\psi\rangle$ of the measured system, but also a complex state $|{\rm detector \;\; ready}\rangle$ of the macroscopic detector. In practice it is very very hard to have a control under all microscopic details of the macroscopic detector, meaning that it is very very hard to prepare two identical copies of $|\Psi\rangle$. Nevertheless, it is not harder than performing the operation $V$, which also requires a control under all microscopic details of the macroscopic detector to ensure that $V$ is exactly the inverse of $U$. So for practical purposes, 3.' is as hard as 3. Yet the advantage of 3.' over 3. is that it is more intuitive conceptually.

So is there any reason why would theorem lose its power if we used 3'. instead of 3.?

 

[atyy]

Some thoughts and questions.

1. Is the claim reasonable that Bohr's version of Copenhagen is refuted, in the light of Haag's statement in his textbook: In Bohr's discussion the time asymmetry appears as obvious. For instance: "The irreversible amplification effects on which the registration of the existence of atomic objects depends reminds us of the essential irreversibility inherent in the very concept of observation" [Bohr 58].

2. In the Healy third argument version of Copenhagen, if we undo the measurement, including the state of the measuring apparatus, then is the record of the measurement outcome lost? In other words, if Alice undoes Carol's "unitary measurement", then is Carol's measurement outcome available to Alice? If it isn't, then this would be another way of saying there is no P(a,b,c,d) for Alice.

3. Does the Healy third argument scenario only make sense in BM? In Copenhagen QM, a measurement must FAPP be irreversible to the observer, and there is no level beyond FAPP. In BM, a measurement is also usually FAPP irreversible. However, since BM has an additional layer, a measurement is reversible in principle. So this is where BM can make predictions where QM is silent, and is a way in which BM differs from QM.

 

[DarMM]

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.

$P(a,b,c,d)$ is the "objective" probability of occurrences of outcomes of the experiment, i.e. simply the frequencies that actually occur. It's not the predicted frequencies of outcomes for any particular observer since nobody in the set up can observe all four of $a,b,c,d$.

Plus this is the sticking point for me, how does Alice know when they've a definite outcome? What condition do you use to switch from a pure state to a mixed state if you're not looking at their laboratory? She models Carol's lab under unitary evolution up to what point? Decoherence?

That's fine. But did anyone believe the FR-Masanes version of Copenhagen anyway?

Brukner, Zeilinger, Healey himself, the older Quantum Bayesians (i.e. not QBism). Anybody who thought QM was purely a probability calculus for objective outcomes.

In the Healy third argument version of Copenhagen, if we undo the measurement, including the state of the measuring apparatus, then is the record of the measurement outcome lost? In other words, if Alice undoes Carol's "unitary measurement", then is Carol's measurement outcome available to Alice? If it isn't, then this would be another way of saying there is no P(a,b,c,d) for Alice.

Alice has no access to $c$ for this reason, but it's not a reason $P(a,b,c,d)$ doesn't exist as it is simply the probability of the outcomes to occur in one run of the experiment, even if nobody has access to all four outcomes.

In Copenhagen QM, a measurement must FAPP be irreversible to the observer, and there is no level beyond FAPP

However is it irreversible to a superobserver with full control over the observer's environment? That's who performs the reversing, not the observer themselves.

 

[DarMM]

Yes, it is.

I genuinely don't understand how the epistemological view of $\psi$ is ruled out by Bell's theorem. I've never seen this expressed in Quantum Foundations papers. The whole motivation of the PBR theorem is to provide constraints on epistemological views when previous theorems didn't, that's what's important about it.

 

[atyy]

$P(a,b,c,d)$ is the "objective" probability of occurrences of outcomes of the experiment, i.e. simply the frequencies that actually occur. It's not the predicted frequencies of outcomes for any particular observer since nobody in the set up can observe all four of $a,b,c,d$.

Plus this is the sticking point for me, how does Alice know when they've a definite outcome? What condition do you use to switch from a pure state to a mixed state if you're not looking at their laboratory? She models Carol's lab under unitary evolution up to what point? Decoherence?

What an observer designates as an objective outcome is subjective and up to the good taste of the observer. This is due to the subjective drawing of the classical/quantum cut. Things on the classical side are objective, and things on the quantum side are not.

If Alice believes Carol has made a measurement, then that means that Alice has granted Carol the same observer status as herself (Alice).

Brukner, Zeilinger, Healey himself, the older Quantum Bayesians (i.e. not QBism). Anybody who thought QM was purely a probability calculus for objective outcomes.

But one must also add that objective outcomes are subjective.

Alice has no access to $c$ for this reason, but it's not a reason $P(a,b,c,d)$ doesn't exist as it is simply the probability of the outcomes to occur in one run of the experiment, even if nobody has access to all four outcomes.

If Alice believes an outcome has occurred, but she has no access to it, then she must collapse her wave function.

However is it irreversible to a superobserver with full control over the observer's environment? That's who performs the reversing, not the observer themselves.

It is traditionally believed that measurements must be irreversible, otherwise there will be a contradiction. The objectivity of an outcome is stored in the mind of the observer. If the observer's mind is reversed, the subjectively designated objective outcome is lost.

A similar view is stated by Peres in his textbook (p376): "Consistency thus requires the measuring process to be irreversible. There are no superobservers in our physical world."