I What Is the Frauchiger-Renner Theorem?

[DarMM]

I find Murray's take on it much more lucid than the usual, IMHO overly sensationalist, takes on it like Henry Stapp:

This has nothing to do with physics as such, but I love in those videos how Gell-Mann attempts a native like pronunciation at everybody's names, rather than just saying them in English phonetics. You can see he really loves languages.

 

[atyy]

Really as @atyy and Bub say, there is a quantum-classical cut. From Asher and Peres, this can be shifted a bit but not indefinitely.

The theory of indirect measurement, measurement ancilla etc is also about being able to shift the cut. But when the cut is shifted, what is real is also shifted.

One can see the different ideas of reality on each side of the cut in figure 1 of https://arxiv.org/abs/0706.1232 where the measurement outcomes are treated as real invariant events in the sense of classical special relativity, but the quantum state has no reality in that sense, eg. collapse is in a different plane of simultaneity in every Lorentz frame.

There is also a little bit of implicit (not explicit) discussion of the issue in https://arxiv.org/abs/quant-ph/0509061, where Einstein mentions the option to make Alice not real in an EPR experiment, to deny the nonlocality of quantum mechanics.

Another point might be the general unreasonableness of the concept of a superobserver, it might be like taking an arbitrarily large observer in General Relativity and ignoring the fact that as they get larger they'd distort the spacetime. However I haven't thought enough about that.

Interesting thought. So the relationship between BM (a full description of reality, where superobservers that can unitarily reverse measurements are allowed) and Copenhagen (measurements must be irreversible) is analogous to the relationship between full GR and the approximation with geodesics? There has been some speculation that the measurement problem has to be solved for quantum gravity to be solved (my very, very free reading of standard comments in string theory). Because Copenhagen QM needs an observer, and the observer needs a stable place to stand, that explains why quantum gravity has so far only been defined in AdS space, which provides a boundary for the observer to stand.

 

[DarMM]

@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.?

Sorry for the late reply only got around to this now.

I would view that set up as a set of four measurements on the four particle state $|\Psi,\Psi\rangle$, a product state of two Bell pairs.

In that situation $E(a,d)$ and $E(b,c)$ would vanish and there is no contradiction with Fine's theorem.

Or so it seems to me.

 

[Demystifier]

Sorry for the late reply only got around to this now.

I would view that set up as a set of four measurements on the four particle state $|\Psi,\Psi\rangle$, a product state of two Bell pairs.

In that situation $E(a,d)$ and $E(b,c)$ would vanish and there is no contradiction with Fine's theorem.

Or so it seems to me.

Interesting argument, but I disagree because I wouldn't say that $E(a,d)$ and $E(b,c)$ would vanish. For definiteness, let me concentrate on $E(a,d)$. Clearly, it does not vanish if $a$ and $d$ are measured on the same copy of $|\Psi\rangle$. In the original version of the thought experiment, the non-vanishing $E(a,d)$ corresponds to an experimental procedure in which $a$ and $d$ are measured without undoing any of the measurements. It is only $corr(a,d)$ (which is different from $E(a,d)$) that involves undoing of measurements.

 

[DarMM]

Interesting argument, but I disagree because I wouldn't say that $E(a,d)$ and $E(b,c)$ would vanish. For definiteness, let me concentrate on $E(a,d)$. Clearly, it does not vanish if $a$ and $d$ are measured on the same copy of $|\Psi\rangle$. In the original version of the thought experiment, the non-vanishing $E(a,d)$ corresponds to an experimental procedure in which $a$ and $d$ are measured without undoing any of the measurements. It is only $corr(a,d)$ (which is different from $E(a,d)$) that involves undoing of measurements.

Sorry yes, I should say $corr(a,d)$.
$E(a,d)$ being the Bell correlations with those same angles $a$ and $d$, which would be $corr(a,d)$'s numerical value if you believe in a interpretation satisfying Masanes's conditions.

So let me be more clear.

$corr(a,d)$ in Masanes's set up would have the value $E(a,d)$ if you believe in an interpretation satisfying his conditions. It doesn't have to in all interpretations as we have already discussed, Bohmian Mechanics and Retrocausal theories could have $corr(a,d) \neq E(a,d)$. It is the fact that $corr(a,d) = E(a,d)$ that ultimately leads to the contradiction with Fine's theorem.

In the version you propose $a,d$ are measurements on separate Bell pairs, not a measurement on the same Bell pair after one has been measurement reversed. Thus $corr(a,d) = 0$ due to it being product state and hence there is no contradiction with Fine's theorem.

 

[Demystifier]

In the version you propose $a,d$ are measurements on separate Bell pairs, not a measurement on the same Bell pair after one has been measurement reversed. Thus $corr(a,d) = 0$ due to it being product state and hence there is no contradiction with Fine's theorem.

Fine, but then I think that $corr(a,d) = 0$ also in the Masanes's version. In particular, $corr(a,d) = 0$ is compatible with Eq. (31). Is there anything in the paper that contradicts my claim that $corr(a,d) = 0$ in the Masanes's setup?

 

[DarMM]

Fine, but then I think that $corr(a,d) = 0$ also in the Masanes's version. In particular, $corr(a,d) = 0$ is compatible with Eq. (31). Is there anything in the paper that contradicts my claim that $corr(a,d) = 0$ in the Masanes's setup?

Once Alice has reversed Carol's measurement, then the entanglement between the two photons is restored and thus Alice's measurement on the first photon and Dan's measurement on the second photon proceed as if Carol hadn't done anything, thus it is a typical Bell measurement and $corr(a,d) = E(a,d) = -cos(a - d)$.

This is unlike your case where Alice is performing a measurement on the first photon of a separate pair to Dan rather than the first photon of the same pair and thus you'd expect no correlation.

 

[Demystifier]

Once Alice has reversed Carol's measurement, then the entanglement between the two photons is restored and thus Alice's measurement on the first photon and Dan's measurement on the second photon proceed as if Carol hadn't done anything, thus it is a typical Bell measurement and $corr(a,d) = E(a,d) = -cos(a - d)$.

But we agreed that $corr(i,j) \neq E(i,j)$ for some $i,j$. So if $corr(a,d) = E(a,d)$, then for which $i,j$ do we have $corr(i,j) \neq E(i,j)$?

 

[DarMM]

But we agreed that $corr(i,j) \neq E(i,j)$ for some $i,j$. So if $corr(a,d) = E(a,d)$, then for which $i,j$ do we have $corr(i,j) \neq E(i,j)$?

$corr(i,j) \neq E(i,j)$ for some $i,j$ only for interpretations breaking one of Masanes's assumptions.

 

[Demystifier]

$corr(i,j) \neq E(i,j)$ for some $i,j$ only for interpretations breaking one of Masanes's assumptions.

Fine, so let us consider one such interpretation. For definiteness, let it be Bohmian mechanics. Then for which $i,j$ do we have $corr(i,j) \neq E(i,j)$?

 

[DarMM]

Fine, so let us consider one such interpretation. For definiteness, let it be Bohmian mechanics. Then for which $i,j$ do we have $corr(i,j) \neq E(i,j)$?

I don't know, I don't know Bohmian Mechanics well enough to carry out the computation. This paper here: https://arxiv.org/abs/1809.08070, shows how it avoids the original Frauchiger-Renner argument. Probably a similar calculation will show what occurs in Masanes's version, an empty wave comes along and causes $corr(i,j) \neq E(i,j)$ for some $i,j$.

However I don't think it matters because Bohmian Mechanics doesn't obey the assumptions of the theorem and thus isn't susceptible to it. It's not a theorem that shows anything about Bohmian Mechanics.

What it does counter are views like those of Healey himself (which is why Healey is concerned with it), Cutless Objective Copenhagen views, i.e. views that want QM to be an objective generalized probability calculus and no more.

 

[DarMM]

Hold the phone!

I was just reading this paper here:
https://arxiv.org/pdf/1901.10331.pdf

Which agrees with you that $corr(b,c) = 0$ for Alice's frame (and similar at least one vanishes for every frame) and that no real contradiction is reached even for views like Healey's (which would mean Healey is wrong about his own incorrectness!)

I'd be interested to hear your thoughts.

If that paper is valid, then my original post could be modified to
"Summary of Frauchiger-Renner: Don't bother!"

EDIT:
From a proper read of their paper they basically say once Carol obtains a result, the $b$ result will be correlated with it via the normal predictions for a Bell pair $E(c,b) = -cos(c - b)$.

However once you reverse the measurement, you restore the first particle back to its original state and could result in any value upon Alice's $a$ measurement. This reversal then decouples the value of $b$ from the value of $c$ and thus $E(c,b) = 0$, because for predicting $b$ it is as if $c$ never occurred.

Similarly for $E(a,d)$. Thus there is no contradiction with Fine's theorem.

This seems pretty valid to me and would render Masanes's result fairly powerless.

Open to correction though.

 

[Demystifier]

@DarMM It seems interesting, but I need some time to study it in detail.

 

[DarMM]

No worries. In short I think it validates your example. If you totally reverse the measurement it is like preparing a new state because you've completely undone any effects. So the event $c$ becomes completely detached from the event $b$ and similarly $d$ from $a$. Hence $E(b,c) = E(a,d) = 0$

 

[Demystifier]

No worries. In short I think it validates your example. If you totally reverse the measurement it is like preparing a new state because you've completely undone any effects. So the event $c$ becomes completely detached from the event $b$ and similarly $d$ from $a$. Hence $E(b,c) = E(a,d) = 0$

Did you mean $corr(b,c) = corr(a,d) = 0$?

 

[DarMM]

Ha! Sorry, yes indeed $corr(b,c) = corr(a,d) = 0$.

The papers themselves aren't very careful in making that distinction in their notation just to warn you.

 

[Demystifier]

Ha! Sorry, yes indeed $corr(b,c) = corr(a,d) = 0$.

So, whatever the Masanes theorem proves, the same thing can also be proved without undoing measurements, by using two copies of the state. Do we agree now?

 

[DarMM]

Yes and I think that means Masanes proves very little.

 

[Demystifier]

Yes and I think that means Masanes proves very little.

Yes, I have concluded it already in #44, while my later arguments only served to refine that conclusion. Thank you very much for the discussion, without you I would never understood all that.

 

[Demystifier]

The final summary:

What do the Frauchiger-Renner-Masanes-Leifer-Healey (FRMLH) theorems actually prove?

They just prove contextuality, that is the already known fact that the process of measurement can change the properties of the system.

 

[DarMM]

Yes, I have concluded it already in #44, while my later arguments only served to refine that conclusion. Thank you very much for the discussion, without you I would never understood all that.

Thanks to you as well and also @atyy . I've learned a lot from this thread, including having a much better grasp of Old/Classic Copenhagen. I've come to have a better understanding of the need for the Heisenberg Cut and through discussions here and Bub's papers a better understanding of the strange implications of reversibility and how you have to be careful when reasoning about reversed results.

 

[A. Neumaier]

Just a comment that is often overlooked: Until at least the end of 1927 (Solvay conference), the quantum physicists in Göttingen and Copenhagen had a realistic view of quantum mechanics in which particles were always in stationary states (characterized by energy and momentum) and performed quantum jumps guided by the wave function. Thus in their writing, state = stationary state and not = wave function up to a phase! And only the wave function was sort of epistemic...

I haven't seen my claim explicitly researched on from a comparative historical point of view. But I am doing a historical study myself, and have plenty of detailed evidence, that will be the content of a paper to be finished later this year. Once one realizes what I wrote, many otherwise difficult to understand things get a straightforward sense.

Nice, can't wait to read it!

The essence is now here, rather than in a paper.