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.

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17 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.

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