The flaw in Frauchiger and Renner’s argument

When the Frauchiger-Renner argument first came out I posted a favourable review, where I corrected the mistake in the presentation without even remarking on it. But since the authors decided to insist on the mistake, I feel the need to point it out.

The well-known argument by Frauchiger and Renner about the consistency of quantum mechanics has finally been published (in Nature Communications). With publication came a substantial change to the conclusion of the paper: while the old version claimed that “no single-world interpretation can be logically consistent”, the new version claims that “quantum theory cannot be extrapolated to complex systems” or, to use the title, that “quantum theory cannot consistently describe the use of itself”.

This is clearly bollocks. We need to find out, though, where exactly has the argument gone wrong. Several discussions popped up on the internet to do so, for example in Scott Aaronson’s blog, but to my surprise nobody pointed out the obvious mistake: the predictions that Frauchiger and Renner claim to follow from quantum mechanics do not actually follow from quantum mechanics. In fact, they are outright wrong.

For example, take the first of the predictions that appear on Table 3 of the paper. $\bar{\text{F}}$ measures $r=\text{tails}$ and claims: “I am certain that W will observe $w = \text{fail}$ at time $n$:$31$”. By assumption, though, $\bar{\text{F}}$ is in an isolated laboratory and their measurement is described by a unitary transformation. This implies that the state of lab L at time $n$:$30$ will be given either by
\[ \frac{3}{\sqrt{10}}\ket{\text{fail}}_\text{L} + \frac{1}{\sqrt{10}}\ket{\text{ok}}_\text{L}\quad\text{or}\quad\frac{1}{\sqrt{2}}\ket{\text{fail}}_\text{L} – \frac{1}{\sqrt{2}}\ket{\text{ok}}_\text{L},\]depending on the result of $\bar{\text{W}}$’s measurement. Therefore, it is not certain that W will observe $w = \text{fail}$; this will happen with probability $9/10$ or $1/2$, respectively.

To obtain the prediction the authors write in Table 3, one would need to assume that $\bar{\text{F}}$’s measurement caused a collapse of the state of their laboratory – contrary to the assumption of unitarity. In this case, the state at time $n$:$30$ would in fact be given by
\[ \ket{\text{fail}}_\text{L},\]independently of the result of $\bar{\text{W}}$’s measurement, and W would indeed observe $w = \text{fail}$ with certainty. But then W would never observe $w = \text{ok}$, and the paradox desired by the authors would never emerge.

To make this point more clear, I will describe how precisely the same problem arises in the original Wigner’s friend gedankenexperiment, so that people who are not familiar with Frauchiger and Renner’s argument can follow it. It goes like this:

Wigner is outside a perfectly isolated laboratory, and inside it there is a friend who is going to make a measurement on a qubit. Their initial state is
\[\ket{\text{Wigner}}\ket{\text{friend}}\frac{\ket{0}+\ket{1}}{\sqrt2}.\]If we assume that the measurement of the friend is a unitary transformation, after the measurement their state becomes
\[\ket{\text{Wigner}}\frac{\ket{\text{friend}_0}\ket{0} + \ket{\text{friend}_1}\ket{1}}{\sqrt2}.\]Now the friend is asked to predict what Wigner will observe if he makes a measurement on the qubit. Frauchiger and Renner claim that, using quantum mechanics, the friend can predict that “If I observed 0, then Wigner will observe 0 will certainty”1.

Wait, what? The quantum prediction is clearly that Wigner will observe 0 with probability 1/2. The claimed prediction only follows if we assume that the friend’s measurement caused a collapse.

And both assumptions are fine, actually. If there is no collapse, the prediction of 0 with probability 1/2 is correct and leads to no inconsistency, and if there is a collapse the prediction of 0 with probability 1 is correct and leads to no inconsistency. We only get an inconsistency if we insist that from the point of view of the friend there is a collapse, from the point of view of Wigner there is no collapse, and somehow both points of view are correct.

Update: After a long discussion with Renato, I think I understand his point of view. He thinks that this assumption of “collapse and no collapse” is just part of quantum mechanics, so it doesn’t need to be stated separately. Well, I think this is one hell of an unstated assumption, and in any case hardly part of the consensus about quantum mechanics. More technically, I think Frauchiger and Renner’s formalization of quantum mechanics — called [Q] — does not imply “collapse and no collapse”, it is too vague for that, so there is really a missing assumption in the argument.

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41 Responses to The flaw in Frauchiger and Renner’s argument

  1. David Byrden says:

    Please post your calculations and explain how you arrived at your result.

  2. Mateus Araújo says:

    Dear David,

    I didn’t want to do this because I already do the calculations in detail in my previous post about the Frauchiger-Renner argument, and the current conclusions can be easily extracted from it. But since you asked nicely, here it is:

    The coin starts in the state $\frac1{\sqrt3}\ket{\text{heads}}_R + \sqrt{\frac23}\ket{\text{tails}}_R$. Assuming that $\bar{\text{F}}$’s measurement doesn’t cause a collapse, the state of their lab $\bar{\text{L}}$ becomes $\frac1{\sqrt3}\ket{\bar{\text{h}}}_{\bar{\text{L}}} + \sqrt{\frac23}\ket{\bar{\text{t}}}_{\bar{\text{L}}}$. Then $\bar{\text{F}}$ sends the spin system to F, who measures it in the computational basis. This second measurement is also assumed to be unitary, mapping the joint state of both labs to the Hardy state $\frac1{\sqrt3}\ket{\bar{\text{h}}}_{\bar{\text{L}}}\ket{0}_\text{L} + \frac1{\sqrt3}\ket{\bar{\text{t}}}_{\bar{\text{L}}}\ket{0}_\text{L} + \frac1{\sqrt3}\ket{\bar{\text{t}}}_{\bar{\text{L}}}\ket{1}_\text{L}$. Now $\bar{\text{W}}$ makes a measurement of the lab $\bar{\text{L}}$ in the basis $\ket{\bar{\text{ok}}}_{\bar{\text{L}}} = \frac1{\sqrt2}(\ket{\bar{\text{h}}}_{\bar{\text{L}}} – \ket{\bar{\text{t}}}_{\bar{\text{L}}})$, $\ket{\bar{\text{fail}}}_{\bar{\text{L}}} = \frac1{\sqrt2}(\ket{\bar{\text{h}}}_{\bar{\text{L}}} + \ket{\bar{\text{t}}}_{\bar{\text{L}}})$. This measurement is not assumed to be unitary. If $\bar{\text{W}}$ obtains result $\bar{\text{ok}}$, the state collapses to $\ket{\bar{\text{ok}}}_{\bar{\text{L}}}\ket{1}_\text{L}$, and if $\bar{\text{W}}$ obtains result $\bar{\text{fail}}$, the state collapses to $\ket{\bar{\text{fail}}}_{\bar{\text{L}}}(\frac2{\sqrt5}\ket{0}_\text{L} + \frac1{\sqrt5}\ket{1}_\text{L})$. Now you just need to rewrite the state of the lab L in the basis $\ket{\text{ok}}_{\text{L}} = \frac1{\sqrt2}(\ket{0}_{\text{L}} – \ket{1}_{\text{L}})$, $\ket{\text{fail}}_{\text{L}} = \frac1{\sqrt2}(\ket{0}_{\text{L}} + \ket{1}_{\text{L}})$, and you obtain the result I claimed.

  3. David Byrden says:

    Thank you. Now I see what you did there. You renormalised the state equations after the /W measurement.

    It’s worth mentioning that if you rewrite the system state just before /W’s measurement, putting lab /L into the measurement basis, you will immediately see that /W has a one-in-six chance of measuring “/ok”.

    Combine that with the subsequent one-in-two chance of W measuring “ok” and you get the one-in-twelve chance for “ok, /ok” that is mentioned in Renner and Frauchiger.

  4. David Byrden says:

    I’m rather disappointed that you say “nobody pointed out the obvious mistake”. I did point it out in a few of the online discussions. Agent /F, I said, ignores the fact that she is in a superposition.

    Now, let’s think about that mistaken assumption of hers. It’s interesting.

    You describe it as the lab being collapsed into the “tails” state. I don’t think that an isolated lab can unilaterally “collapse”. What would make it collapse? A measurement? That concept is part of the Copenhagen interpretation and we have no reason to think it real.

    The superposition inside lab /L exists relative to the surrounding world. Agent /F cannot detect her own superposition unilaterally, and (as Renner has pointed out in another forum) she cannot measure the outside world in any way to verify that she is in a superposition.

    Her own lab /L will “collapse” as she assumed only if it’s imperfectly sealed, which makes it entangled with the outside world. And then, the outside world will also go into the superposition. Rather than saying the lab “collapses”, we could say that the world “explodes”……

    Sorry. I was just musing there. But I do have a question about the setup of this experiment:

    How can agent /F send her qubit to agent F ?

    Let’s assume that the perfectly isolating lab is a practical possibility. Inside the lab, agent /F really is in a superposition of two states. And since she’s a macroscopic object, those states have decohered.

    Now, both of her must send a qubit. As far as I understand it, those two superpositions of the qubit must be in principle indistinguishable (e.g. if agent F reads both of them as “down”, she must not be able to deduce the state of agent /F).

    Any difference between them, e.g. a minute difference in their timing, would allow F to deduce the state of /F. That’s entanglement but it’s not the same entanglement we’re trying to achieve.

    Can this be achieved in principle? In practice?

    Assume that she manually configures and sends the qubit. If she uses an atomic clock to time the sending of the qubit (which would be e.g. a photon rather than a macroscopic object) then she would have achieved “recoherence” after her decoherence. Is that possible?

    Or, alternatively, could she plug the “coin” into quantum equipment that directly translates it into an outgoing photon? That pathway could remain coherent, even if the rest of her lab didn’t. But then, could she read the “coin” at all? Wouldn’t she be copying a quantum state and violating the No Cloning Theorem?

  5. Mateus Araújo says:

    Dear David,

    At the time I wrote that your comments in my blog and Scott Aaronson’s blog were not up yet. If you did post it somewhere else before I posted my comment on Scott Aaronson’s blog I’m sorry, I haven’t seen it.

    About the possibility of $\bar{\text{F}}$ sending the qubit to F: In principle there is no problem with that. You just need to start with the state $(\frac1{\sqrt3}\ket{\bar{\text{h}}}_{\bar{\text{L}}} + \sqrt{\frac23}\ket{\bar{\text{t}}}_{\bar{\text{L}}})\ket{0}$ and apply a controlled-Hadamard to it. It is just a unitary transformation, so perfectly kosher. In practice, it would be rather difficult, but the whole experiment is extremely difficult to realize anyway.

  6. David Byrden says:

    Sorry if I didn’t make myself clear. The transformation isn’t the problem. Let’s implement the transformation as a machine on the pathway between the labs. So, agent /F will send something (perhaps a polarised photon) in her OWN quantum state, and it will get Hadamarded en route.

    My question is about what just happened? Agent /F sent a qubit in her own state. And she’s still in her own state. There are now two copies of her state. Does that violate the No Cloning Theorem?

    That question would arise if and only if the coin directly and physically creates the outgoing qubit. There’s an alternative. Agent /F could read the coin and manually set up the outgoing qubit. Even if she does a perfect job of it, that’s not “cloning”.

    But then the two superposed, decohered copies of her would have to send indistinguishable qubits. Is THAT possible in principle?

    I don’t see a third way to set this up so long as the labs are macroscopic.

  7. Mateus Araújo says:

    There is no cloning going on in any scenario. Cloning is a transformation that takes $\ket{\psi}\ket{0}$ to $\ket{\psi}\ket{\psi}$. What the argument asks for is to transform $(\frac1{\sqrt3}\ket{\bar{\text{h}}}_{\bar{\text{L}}} + \sqrt{\frac23}\ket{\bar{\text{t}}}_{\bar{\text{L}}})\ket{0}$ into $\frac1{\sqrt3}\ket{\bar{\text{h}}}_{\bar{\text{L}}}\ket{0} + \sqrt{\frac23}\ket{\bar{\text{t}}}_{\bar{\text{L}}}\ket{+}$, which is just a controlled Hadamard, as I said before. Since it is a unitary transform, in particular it does not do any cloning.

    But then the two superposed, decohered copies of her would have to send indistinguishable qubits. Is THAT possible in principle?

    Yes, in principle there is no problem with that. You just need both copies to implement precisely the same operation. Even if they are slightly different, this just results in a slight perturbation of the final state, nothing catastrophic for the argument.

  8. David Byrden says:

    I still don’t understand. You say

    >> “the argument asks … to transform… (some stuff) into (other stuff) which is just a controlled Hadamard”

    All right, so let’s build a quantum pathway inside lab /L, as follows :

    “coin” -> Hadamard circuit -> output

    Now, correct me if I’m wrong. I think there should be no “measurement” anywhere along this quantum path. Otherwise, the state would “collapse” and the output would be altered.

    Therefore, agent /F, her table, her chair and her coffee…. will not enter the superposition state.

    Am I wrong there?

  9. Mateus Araújo says:

    I really don’t see what you mean. Can you state your point with equations? You can use LaTeX normally on the comments.

  10. David Byrden says:

    Equations won’t make my point. I am not disputing that a unitary can convert the “coin” state into the slightly different state of the qubit “S”.

    My point is that “S” physically leaves laboratory /L and travels to laboratory L, in order to entangle L with the quantum state of the “coin”.

    I’m not assuming that “S” is an electronic device containing a qubit. It could be something as simple as a photon. But it must physically exit from /L and travel to L.

    So, there will exist something (probably a photon) outside of lab /L, bearing a quantum state that originated in the “coin”.

    And we ALSO need agent /F to read the “coin” so that she, and her lab, will go into a superposition.

    How would you implement this?

  11. Mateus Araújo says:

    But that’s the whole assumption behind the gedankenexperiment! That agent $\bar{\text{F}}$ does make a measurement, and that it is a unitary transformation, it doesn’t cause any collapse. Now, both of the superposed copies of $\bar{\text{F}}$ must prepare S with the state depending on the result of their measurement. If $\bar{\text{F}}$ got heads, they prepare S in the state $\ket{0}$, and in case of tails, $\ket{+}$. This can be done by producing a single photon (before the measurement), and putting a wave plate in its path with the orientation depending on the result of the measurement.

    Now, the delicate thing is, that in general one would entangle more degrees of freedom with the result of the measurement than just the polarisation (for example, the time of emission, as you mentioned). If this happened the state of the photon would become maximally entangled with the state of $\bar{\text{F}}$, and this would ruin the experiment. But there is nothing fundamental about this. One can synchronize very well the time of emission (for example by doing it before the measurement), and take care that nothing changes in the photon except the polarisation. Yes, it is difficult, but this gedankenexperiment is not about what is easy, but what is in principle possible.

  12. David Byrden says:

    Thank you. Your implementation seems very plausible.
    So, my concerns about the quantum state being measured twice are not relevant.

  13. David Byrden says:

    I’m returning to this after some more study, because I have new thoughts about it.

    When last I wrote, I was worried about the qubit that goes from /L to L.
    I thought that both copies of it (from the two superposed copies of the lab) would need to “decohere” into a single copy, such that if you got a “down” reading in lab L, you would be unable (in principle) to know the state of lab /L.

    I can see now that’s not necessary.

    It’s not necessary to “recohere” the qubits from the two superposed copies of /L. It doesn’t matter if they are incoherent. It doesn’t matter if lab L is fully aware of the state of lab /L. The overall system will still end up in the desired configuration (three states with 1/3 probability each).

    So, your proposed mechanism of controlled wave plates, is not necessary for the qubit S.

    But I’m also worried about the final measurements made by W and /W.

    You see, the experiment requires these external agents to measure the labs in a basis called “ok / fail “.
    This is not the natural basis of the lab contents. With this basis, the agents are in superposition after measurement! This basis cannot possibly be used by somebody opening the lab and examining its contents; instead, the lab must somehow send its state out, encoded into something, for measurement.

    Now, I’m looking at your proposed mechanism of controlled wave plates. Can it be used in this situation? Can an agent, within a lab, encode her state into a photon by polarising it, such that the two superposed copies of the photon (from two superposed copies of the agent) will combine into a single photon?

    Once again, I see a possible violation of the No Cloning Theorem, but I don’t understand it well enough to wield it here, so let’s ignore that.

    Because I see another problem.

    For the two copies of the photon to recombine, is interference. Their wave functions would add.

    But we’ve just given them orthogonal polarities. Doesn’t that preclude them interfering as we want?

  14. Mateus Araújo says:

    Actually, you do need both “copies” of the qubit sent from $\bar{\text{L}}$ to L to be non-orthogonal or, to use your words, for them to recohere. Otherwise the state between $\bar{\text{L}}$ and the qubit, and consequently between $\bar{\text{L}}$ and L, will be maximally entangled, not the Hardy state, and this would ruin the argument.

  15. Mateus Araújo says:

    I think you are confused about how the photons “recombine”. It is not interference, it is just that the state of the photon factors out of the entangled state. Let me explain. You have two copies of some agent A, call them $\ket{A_0}$ and $\ket{A_1}$, and they are superposed in the state $(\ket{A_0}+\ket{A_1})/\sqrt{2}$. Now each copy prepares a photon, both with the same polarisation $\ket{H}$. The state then becomes $\frac{1}{\sqrt{2}}(\ket{A_0}\ket{H}+\ket{A_1}\ket{H})$, which is mathematically identical to $\frac{1}{\sqrt{2}}(\ket{A_0}+\ket{A_1})\ket{H}$. This is what you called recombination, which is completely different to interference. Now, if the copies of the agent would prepare photons with orthogonal polarisations, you would end up with the state $\frac{1}{\sqrt{2}}(\ket{A_0}\ket{H}+\ket{A_1}\ket{V})$, which is maximally entangled, and indeed does not recombine in the above way.

  16. David Byrden says:

    In order for this experiment to work, agent Fbar must send a qubit from her lab to the second lab.
    She must do this after making her quantum measurement, therefore she will be in a superposition, and two copies of her will be sending the qubit which must be received as a single qubit.
    That qubit, according to the procedure laid out by Renner and Frauchinger, must have the state “down” for one of the copies of Fbar, and the mixed state “up + down” for the other Fbar.

    Is it impossible, as you just said, for these two copies of the qubit to “recombine” ?

    The experiment also requires that both labs be measured in the basis ” ok – fail ” when the setup is complete. This is not their natural basis.

    I therefore suggest that they would need to emit a qubit, in much the same way as above, so that their state could be measured in some other basis than the natural one.

    But, encoding their state would once again require that each copy of the superposed agent put the qubit into a different state.

    Is “recombination” impossible here too? Can this whole experiment exist, in principle?

  17. Mateus Araújo says:

    I prefer to simply use the equations rather than reason in terms of informal concepts like “recombination”. If you insist, though, one can say that if the states are the same one achieves full recombination, if they are different but not orthogonal (like “down” and “up + down”) they recombine only partially, and if they are different and orthogonal (like “down” and “up”) they don’t recombine at all.

    What matters is that all these transformations are unitary, so they are in principle allowed in quantum mechanics.

  18. David Byrden says:


    You said:
    “…recombination, which is completely different to interference”

    Could you explain the difference please?

  19. Mateus Araújo says:

    Sure. Recombination is what I explained in the comment above (keep in mind that this is not a standard word, so if you search you will not find anything about it). Interference is when some components of a state in a superposition cancel out. For example, a photon can be in a superposition of two different paths $\ket{1}$ and $\ket{2}$. If you send them through a beam splitter they will map to $\frac{1}{\sqrt2}(\ket{1}+\ket{2})$ and $\frac{1}{\sqrt2}(\ket{1}-\ket{2})$, respectively, and the $\ket{2}$ component of the first one will cancel out with the $-\ket{2}$ component of the second one. This is the prototypical quantum effect, you’ll find stuff about it everywhere. The standard experiments are the double-slit experiment and the Mach-Zender interferometer.

  20. Daniel Mahler says:

    Hi Mateus,
    Is the probability of (ok, ok) still 1/12?

  21. Mateus Araújo says:

    Hi Daniel,

    Sure. The probability that $\bar{\text{W}}$ obtains $\bar{\text{ok}}$ is 1/6, and the probability that W obtains ok given that $\bar{\text{W}}$ obtained $\bar{\text{ok}}$ is 1/2, so the probability of $(\bar{\text{ok}}, \text{ok})$ is 1/12.

  22. Daniel Mahler says:

    Ok. I am trying to model that experiment as a quantum circuit.
    The initial part of the circuit behaves as expected, but the final ok-ok probability is 1/4.The circuit is here
    The order of the quibis is
    – coin
    – Alice’s Friend
    – spin
    – Bob’s friend
    – Alice
    – Bob

    Initially I am only modelling the physical setaup and process, not the agents mental processes that make up the bulk of F&R’s argument.
    I am taking the Everett, measurement as entanglement, perspective using controlled quantum gates to represent measurements.
    The coin probability is only approximately 1/3 since it needs to be prepared with standard gates;
    I do not believe that makes a significant difference.

    I would appreciate your feedback

  23. Mateus Araújo says:

    Hi Daniel,

    If the amplitudes of the coin are not exactly $\sqrt{\frac23}$, $\sqrt{\frac13}$ it is not a problem, the paradox still works. It might be a problem if they are complex, as it is important how they interfere.

    But what is wrong with your circuit is that the Wigners are measuring only the friends. For the paradox to work they must measure the whole laboratories, i.e., $\bar{\text{F}}$ together with the coin and F together with the spin.

    Notice that if you just want to get the correct probabilities it doesn’t matter if the friends exist, you can just use the coin and the spin as if they are $\bar{\text{F}}$ and F, and have the Wigners make measurements on them.

  24. Daniel Mahler says:

    Thanks Mateus! That agrees with other feedback I have received.

  25. Daniel Mahler says:

    This does look right Thanks again Mateus!! Now I just need to figure out how to represent simultaneous measurement in quantum circuits in order to model the agents, since I ultimately want to model the whole experiment, including beliefs inferences & communications :)
    I took a shot at it with, but the numbers do not agree. I guess that double conttrol construction is not the right way to represent simultaneous measurement.

  26. Daniel Mahler says:

    I celebrated too soon. Little more instrumentation shows while the ok-ok probability seems right, the individual OK probabilities look nothing like 1/6 & 1/2.
    The second ok probability though does look like 1-1/6 …

  27. Mateus Araújo says:

    Well, these are conditional and marginal probabilities; if you look at the individual outcomes, the probability of $(\bar{\text{ok}}, \text{ok})$ is 1/12, the probability of $(\bar{\text{fail}}, \text{fail})$ is 9/12, and the other two possibilities have probability 1/12 each.

  28. Mateus Araújo says:

    To measure the whole laboratory (of F together with the spin, for example) in the superposition basis, what you need is not the double control, but a Bell state measurement. It is just a CNOT followed by a Hadamard.

  29. Daniel Mahler says:

    Hi Mateus,

    Thanks for your help! I think I have found an interesting way to look at the F-R thought experiment using time reversed quantum circuits:

  30. Max Madera says:

    Boas festas Mateús e continue assim com o blog

  31. Mateus Araújo says:

    Valeu, você também!

  32. Roger Schafir says:

    Can I comment on this, a few months after your post?

    I think that Frauchiger and Renner’s argument is indeed flawed, but I’m not sure it’s exactly the flaw you point out about the state of W. But first, this matter of collapse and no-collapse. As I read Frauchiger and Renner, their basic assumption is that the state of a quantum system is different relative to different observers, because they regard different information possessed by the observers as changing the state.

    So far as I know, no version of quantum mechanics, in any interpretation, allows this. The state is an absolute thing, and if the observer doesn’t know what the state is, this is merely classical-style ignorance – it’s not like the unknowability (or non-existence) of the values of non-commuting observables at the same time. But Frauchiger and Renner suppose that F-bar’s measurement of the coin sends the coin plus particle into one of two possible product states relative to F-bar, but into an entangled state relative to W-bar. However they do seem to regard any measurement by an observer as a collapse relative to that observer, and not a unitary evolution, contrary to some of the speculations on your blog. (But see my Addendum, below.)

    Regarding their deduction that F-bar getting a tails implies that W, later on, will get fail, although you are right about the state of W immediately before W’s measurement, I don’t think that they are making their deduction from the state of W. Looking at their statement and paragraph just before their equation (4), they are making their deduction from the fact that tails projects the spinning particle into (1/√2)(|up>+|down>) and this is orthogonal to |OK>, i.e. to (1/√2)(|up>−|down>). Since a quantum system cannot be projected from one state into a state orthogonal to it, and since in this case fail counts as anything other than projection into |OK>, so therefore W must get fail.

    But this shows the flaw. F-bar’s measurement of tails sends the particle into (1/√2)(|up>+|down>) relative to F-bar and F, whereas W is measuring the state of the particle relative to W-bar and W, and by Frauchiger and Renner’s own assumptions states relative to F-bar and F are not the same as states relative to W-bar and W. It is contradictory to suppose that the particle is in a different state for observers inside and outside the labs after the initial coin measurement but then assume that the state is the same for all observers at the later measurement. According to their assumptions, the states would only be the same if information had now been exchanged by the observers inside and outside the labs, and I cannot see that that is the case.

    I grant though that it is ambiguous. Really, as regards the whole paper I think it is legitimate to complain that it is written and presented as if determined to make it as hard to read as possible. Everything is in a pointlessly intricate notation, and the basic argument is relegated to tables which refer you to other tables which refer you to still other tables to discover even what the most basic symbols mean.

    But then, if the paper was easy and clear, perhaps it would not have secured attention.

    Addendum I think that they are also mistaken about the states relative to the observers outside the labs, not only W (as you point out) but even for W-bar immediately after the first measurement. They assume that relative to W-bar the coin and particle are in an entangled state, which according to their assumptions would be the case if W-bar didn’t know that a measurement had taken place inside the lab L-bar. But the thought-experiment assumes that W-bar does know that a measurement took place, and in what basis, but doesn’t know what the result was. Therefore, consistently with what they say in their preliminary discussion of the original Wigner thought-experiment, W-bar should assume that the coin and particle are in a mixed state of |heads>|down> and (1/√2)|tails>(|up>+|down>) with weights 1/3 and 2/3 respectively.

  33. Mateus Araújo says:

    Hi Roger,

    No problem with commenting so late in the game, I still enjoy the subject.

    1 – QBism and Relational Quantum Mechanics do allow quantum states to be relative to the observer; the key assumption behind the argument is precisely that the state is not an absolute thing, and that it collapses and does not collapse simultaneously for different observers. The problem is that Frauchiger and Renner do not list this as an assumption, they just regard it as quantum mechanics itself.

    2 – They also describe measurements as unitary transformations, when viewed from the outside. This is not an “speculation” from my blog, it is a fundamental part of their argument.

    3 – If you grant their collapse/no-collapse assumption, though, their argument holds without problem. From the point of view of F the state is |fail>, and W-bar will observe fail. From the point of view of W-bar, the state is such that it has probability 1/12 of giving the result |ok>, so if it happens we have the contradiction.

    3 – In particular, the state from W-bar’s point of view is not the mixed state you claim in the Addendum. That would be the case if Frauchiger and Renner thought that the collapse is absolute, and W-bar merely ignores the result. No, remember, Frauchiger and Renner think that from W-bar’s point of view F’s measurement is a unitary.

  34. Roger Schafir says:

    Many thanks for your reply. As far as your point (1) is concerned, OK I show my age (76) in not being familiar with those two versions of quantum mechanics, but as you say, Frauchiger and Renner do not appear to be using them anyway.

    Your point (2): “when viewed from the outside” covers at least two different cases, (i) when a measurement is known to have taken place, and in what basis, but the result is unknown, or (ii) it is not known even whether the measurement has taken place. Frauchiger and Renner base their thought-experiment on Wigner’s friend idea, but Wigner was unsure where and whether a measurement had taken place, and famously speculated that it was when it reached a conscious being. But F & R’s thought-experiment assumes it known that a measurement has taken place inside the lab.

    I took this to be different. But I grant, again, that it is ambiguous.

    Your point (3): So have you changed your mind, then, from when you said that F & R’s reasoning was wrong because it ignored the fact that W-bar’s measurement had projected the system out of the entangled state into one or other of two other states?

    My reservation about their argument is their statement (just before their equation (4)) that F-bar getting tails implies that W is in the state (1/√2)(|up>+|down>) just before he makes his measurement. That might be true relative to F-bar and F, but why should it be true relative to W-bar and W? Yet that is what is needed to deduce that W gets a fail.

    Point (4) Certainly I was using “mixed state” in the sense in Dirac’s book of an ensemble in which the particles are in different pure states, but one also uses the term for the result of partial tracing of one particle in an entangled state.

    That is a question of terminology. As for the point about unitarity, there I refer back to my reply to your point (2) above.

    Something different: I see that you have also posted a blog in which you express extreme dislike of counterfactual reasoning in Bell’s theorem. I must say that I will need a lot of persuading to agree with that, for I regard counterfactual reasoning as what frees non-locality from a relatively minor result about hidden-variable theories to a major discovery that quantum mechanics, and not only quantum mechanics but any other theory which gives the same observable results as quantum mechanics, must be non-local.

    But let me read what you have to say.

  35. Mateus Araújo says:

    No no, this is the main problem with the paper, that Frauchiger and Renner are using such a version of quantum mechanics, but are not explicit about it. First because of the inconsistent collapse/no-collapse assumption, and second because they describe measurements done by other observers as unitary. Both are central tenets of QBism.

    I haven’t changed my mind. Their argument is still wrong, because they claim it follows from quantum mechanics, and it does not. The problem I pointed out is that if one assumes that F-bar’s measurement is a unitary, as the setup needs, the probability of W-bar observing ok is 1/12, not 0, and no contradiction arises. But with their additional assumption that F-bar’s measurement causes a collapse for F-bar but is a unitary for W-bar then yes, the argument does follows.

    About Bell’s theorem: I probably agree with your point of view. In my blog post I’m not criticizing counterfactual reasoning per se, I’m expressing extreme dislike about the counterfactual version of Bell’s theorem. Precise because it makes it seem like there is a problem with counterfactual reasoning (there isn’t) and that there is no problem with locality (there is).

  36. Roger Schafir says:

    My reason for why their argument seems wrong, even in terms of their own assumptions about states being relative, is: they deduce from F-bar getting tails that therefore the particle which F gets is in (1/√2)(|up>+|down>), and that is orthogonal to W’s |OK>. But (1/√2)(|up>+|down>) is the state relative to F and not necessarily the state relative to W, which they need it to be in order to say that it can’t be projected into (1/√2)(|up>-|down>) relative to W.

    I have sent a response to your blog about Bell’s theorem and counterfactual definiteness.

  37. Mateus Araújo says:

    No, that part of their argument is fine; passing through each observer’s relative state to get a contradiction is precisely what they need the chain of implications for. Maybe it is more clear in my version of their argument.

  38. Ruth Kastner says:

    Sorry I didn’t see this interesting discussion before I posted my paper on FR:: . I agree there is an equivocation in the FR construction, but I argue that the scenario they present points to a problem in our basic understanding and formulation of quantum theory, which I discuss in the paper Comments welcome, as I am currently adding references and revising.

  39. Ruth Kastner says:

    Re my paper on the FR paradox on philsci archive: I’ve made a few minor corrections (typos) and included some additional arguments. The latest version is here:

  40. David Byrden says:

    Sorry to jump in so again, but I’ve written my own explanation of why R-F’s paper is wrong, and who knows, maybe my way of explaining it will be the best way for some readers. It’s here;

  41. Knud says:

    Dear All,
    only yesterday I read 2 nice papers by Ruth and thus found your blog, and thus, I am really late (this is not my core field of activity). Still, I would like to draw your attention to my proposal why/how FR are wrong, accepting real collapse, not needing counterintuitive TI (while still having much in common), but putting a quantitative label on it: Landauers Limit. The only probably hard to digest point then is that “time” is not a classically valid concept inside isolated quantum systems; please, see:
    an update is in the works, comments would be very much appreciated!
    cheers, Knud

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