Wigner's Friend and Incompatibility

[DarMM]

TL;DR

Wigner's friend seems to lead to certainty in two complimentary contexts

Summary: Wigner's friend seems to lead to certainty in two complimentary contexts

This is probably pretty dumb, but I was just thinking about Wigner's friend and wondering about the two contexts involved.

The basic set up I'm wondering about is as follows:

The friend does a spin measurement in the $\left\{|\uparrow_z\rangle, |\downarrow_z\rangle\right\}$ basis, i.e. of $S_z$ at time $t_1$. And let's say the particle is undisturbed after that.

For experiments outside the lab Wigner considers the lab to be in the basis:
$$\frac{1}{\sqrt{2}}\left(|L_{\uparrow_z}, D_{\uparrow_z}, \uparrow_z \rangle + |L_{\downarrow_z}, D_{\downarrow_z}, \downarrow_z \rangle\right)$$

He then considers a measurement of the observable $\mathcal{X}$ which has eigenvectors:
$$\left\{\frac{1}{\sqrt{2}}\left(|L_{\uparrow_z}, D_{\uparrow_z}, \uparrow_z \rangle + |L_{\downarrow_z}, D_{\downarrow_z}, \downarrow_z \rangle\right), \frac{1}{\sqrt{2}}\left(|L_{\uparrow_z}, D_{\uparrow_z}, \uparrow_z \rangle - |L_{\downarrow_z}, D_{\downarrow_z}, \downarrow_z \rangle\right)\right\}$$
with eigenvalues $\{1,-1\}$ respectively.

At time $t_2$ the friend flips a coin and either he does a measurement of $S_z$ or Wigner does a measurement of $\mathcal{X}$

However if the friend does a measurement of $S_z$ he knows for a fact he will get whatever result he originally got. However he also knows Wigner will obtain the $1$ outcome with certainty.

However $\left[S_{z},\mathcal{X}\right] \neq 0$. Thus the friend seems to be predicting with certainty observables belonging to two separate contexts. Which is not supposed to be possible in the quantum formalism.

What am I missing?

 

[akvadrako]

Maybe it's a minor thing, but why does the coin flip matter? Seems like it creates two contexts and Wigner needs to first measure heads/tails, so the other version of the friend is a counterfactual.

 

[DarMM]

Maybe it's a minor thing, but why does the coin flip matter? Seems like it creates two contexts and Wigner needs to first measure heads/tails, so the other version of the friend is a counterfactual.

It doesn't matter, it's just a shorthand for the Friend can consider one or the other.

 

[Morbert]

It should be ok for Wigner's friend to use one context to reason that Wigner's measurement will yield $1$, and another context to reason that his next measurement will yield (say) $S_z = \uparrow$. Both of these properties have a subspace associated with them. I think he only runs into trouble if he tries to compute probabilities/reason about the property $(S_z=\uparrow) \land 1$ which has no subspace associated with it.

 

[akvadrako]

It doesn't matter, it's just a shorthand for the Friend can consider one or the other.

But those are two orthogonal futures - one where Wigner measures first and one where the friend measures first - so they don't need to be compatible. The more interesting case to consider is Wigner measures first. Then what happens from the friend's perspective? He'll now be uncertain about the result of his remeasurement, because Wigner has changed the state of his qubit.

 

[DarMM]

But those are two orthogonal futures - one where Wigner measures first and one where the friend measures first - so they don't need to be compatible

They're certainly not compatible. The problem is that they aren't compatible and yet $P(E) = 1$ is assigned to both prior to their occurrence.

 

[DarMM]

It should be ok for Wigner's friend to use one context to reason that Wigner's measurement will yield $1$, and another context to reason that his next measurement will yield (say) $S_z = \uparrow$. Both of these properties have a subspace associated with them. I think he only runs into trouble if he tries to compute probabilities/reason about the property $(S_z=\uparrow) \land 1$ which has no subspace associated with it.

In non-encapsulated QM two events $E,F$ that don't commute can't usually have $P(E) = P(F) = 1$. Is this then a special feature of histories with encapsulation?

 

[akvadrako]

They're certainly not compatible. The problem is that they aren't compatible and yet $P(E) = 1$ is assigned to both prior to their occurrence.

Conditionally, dependent on the result of the coin flip. There is no history that contains both those results.

EDIT: I mean there is no future that contains both those probabilities.

 

[DarMM]

Conditionally, dependent on the result of the coin flip. There is no history that contains both those results.

Certainly there isn't since they don't commute/aren't compatible. However in "regular" QM we similarly can't have $A$ and $B$ occur in one history if $[A,B] \neq 0$. However you still can't assign $P(A) = P(B) = 1$ to them. Incompatible events don't typically have simultaneous certainty assignments.

 

[akvadrako]

Certainly there isn't since they don't commute/aren't compatible. However in "regular" QM we similarly can't have $A$ and $B$ occur in one history if $[A,B] \neq 0$. However you still can't assign $P(A) = P(B) = 1$ to them. Incompatible events don't typically have simultaneous certainty assignments.

Okay, sorry for interceding - I don't see the issue - I would expect conditional probabilities to both be able to be 1. Like if you measure up, then $P(\uparrow) = 1$, measure down, $P(\downarrow) = 1$.

 

[Morbert]

In non-encapsulated QM two events $E,F$ that don't commute can't usually have $P(E) = P(F) = 1$. Is this then a special feature of histories with encapsulation?

I think it is, insofar as not everyone in the foundations community would be cool with using a multiplicity of sample spaces to make "realistic" claims. I.e. It's the case that there is a probabilistic sample space such that $P(A) = 1$, and another sample space such that $P(B) = 1$, but no single sample space that gives us probabilities for both, and this is a sticking point for criticisms of CH.

 

[charters]

You are letting Friend use different states/initial conditions for the two predictions, so maybe its not that surprising you can circumvent the standard rules. I think this is just the self-consistency question, restated. Really, if F is going to assign probability 1 to the |+> outcome for W, then F isn't actually claiming there was ever a singular outcome to the Sz measurement.

 

[DarMM]

Okay, sorry for interceding - I don't see the issue - I would expect conditional probabilities to both be able to be 1. Like if you measure up, then $P(\uparrow) = 1$, measure down, $P(\downarrow) = 1$.

Of course this is true, but these relate to outcomes for $S_z$ conditioned on a previous $S_z$. The problem is two projectors for a $\mathcal{X}$ outcomes and a $S_z$ outcome having certainty.

 

[Morbert]

You are letting Friend use different states/initial conditions for the two predictions, so maybe its not that surprising you can circumvent the standard rules. I think this is just the self-consistency question, restated. Really, if F is going to assign probability 1 to the |+> outcome for W, then F isn't actually claiming there was ever a singular outcome to the Sz measurement.

Hmm, In the CH formalism, the friend should be free to employ whichever boolean event algebra/framework is fit for purpose, without having to commit to one as correct. This might be a departure from conventional QM where a measurement context selects the right framework.

 

[DarMM]

You are letting Friend use different states/initial conditions for the two predictions, so maybe its not that surprising you can circumvent the standard rules. I think this is just the self-consistency question, restated. Really, if F is going to assign probability 1 to the |+> outcome for W, then F isn't actually claiming there was ever a singular outcome to the Sz measurement.

I think it is, insofar as not everyone in the foundations community would be cool with using a multiplicity of sample spaces to make "realistic" claims. I.e. It's the case that there is a probabilistic sample space such that $P(A) = 1$, and another sample space such that $P(B) = 1$, but no single sample space that gives us probabilities for both, and this is a sticking point for criticisms of CH.

Thanks to you both.

I guess this is just the fact that (in a Copenhagen style view) massive observers can witness facts incompatible with my (as a "lower" observer) entire macroscopic environment. Using an example from Omnes where he has such a Wigner be a superpowerful alien arrive from somewhere previously sealed from us, the alien would be led to contradictions if he assumed things like the Earth or Sun factually existed since the alien is using a basis complimentary to our entire macroreality!

 

[DarMM]

then F isn't actually claiming there was ever a singular outcome to the Sz measurement

Is it true that he is denying there was a singular $S_z$ outcome? That seems to say Wigner's friend requires Many Worlds. Is it not possible to read this in a Copenhagen manner (no matter how daft one thinks this is) that Wigner and the Friend have incompatible experiences that cannot be reasoned about together.

 

[charters]

Hmm, In the CH formalism, the friend should be free to employ whichever boolean event algebra/framework is fit for purpose, without having to commit to one as correct. This might be a departure from conventional QM where a measurement context selects the right framework.

But the boolean algebra of a un-made/unitarily reversed event can't be appropriate, right?

 

[charters]

Is it not possible to read this in a Copenhagen manner (no matter how daft one thinks this is) that Wigner and the Friend have incompatible experiences that cannot be reasoned about together.

Yes it is possible but since W and F both can still have the experience of talking to each other after the experiment, this seems to imply a weird solipsism where there is a Friend-reality in which Wigner is a p-zombie (who reports agreeing with Friend) and then vice versa in Wigner-reality. This is basically Matt Liefer's point that Copenhagen is actually so much weirder than all the other interpretations, even though it is marketed as the "conservative" or philosophically "down to Earth" position in the common curriculum.

 

[DarMM]

Yes it is possible but since W and F both can still have the experience of talking to each other after the experiment, this seems to imply a weird solipsism where there is a Friend-reality in which Wigner is a p-zombie (who reports agreeing with Friend) and then vice versa in Wigner-reality.

Can $F$ really talk to Wigner after the experiment though? Omnès has a calculation of a realistic superobserver and it would seem to require $10^{10^{18}}$ atoms. Now some take this to mean superobservers simply don't exist, but ignoring that isn't it likely that any actual $\mathcal{X}$ would be so invasive of the friend's atomic structure as to annihilate them.

 

[akvadrako]

Yes it is possible but since W and F both can still have the experience of talking to each other after the experiment, this seems to imply a weird solipsism where there is a Friend-reality in which Wigner is a p-zombie (who reports agreeing with Friend) and then vice versa in Wigner-reality. This is basically Matt Liefer's point that Copenhagen is actually so much weirder than all the other interpretations, even though it is marketed as the "conservative" or philosophically "down to Earth" position in the common curriculum.

That sounds a lot like the world described by Markus Muller, where everyone else is necessarily a p-zombie because of computational complexity.

 

[Morbert]

Yes it is possible but since W and F both can still have the experience of talking to each other after the experiment, this seems to imply a weird solipsism where there is a Friend-reality in which Wigner is a p-zombie (who reports agreeing with Friend) and then vice versa in Wigner-reality. This is basically Matt Liefer's point that Copenhagen is actually so much weirder than all the other interpretations, even though it is marketed as the "conservative" or philosophically "down to Earth" position in the common curriculum.

One problem might be that Wigner talking to his friend, while simultaneously being aware of his measurement result, would constitute a record of both measurements, which CH would forbid. It's likely that Wigner's friend would be rearranged in a very fatal way.

But the boolean algebra of a un-made/unitarily reversed event can't be appropriate, right?

Going to do a few quick calculations re/ macroscopic quantum states and get back to you asap

 

[charters]

Can $F$ really talk to Wigner after the experiment though? Omnès has a calculation of a realistic superobserver and it would seem to require $10^{10^{18}}$ atoms. Now some take this to mean superobservers simply don't exist, but ignoring that isn't it likely that any actual $\mathcal{X}$ would be so invasive of the friend's atomic structure as to annihilate them.

$\mathcal{X}$, if we suspend disbelief regarding the technology, as is necessary to have this discussion at all, yields an in tact version of F's body. We're basically assuming something in the vein of manipulating a living body at the atomic level, with simultaneous perfect inverse bremmstrahlung, etc. So it is all completely ridiculous, but is it so much more ridiculous and removed from daily life than worrying about the unitarity of black hole evaporation? In my opinion, this has to seen as a question of logical consistency in the extreme edge cases, which are important to understand.

 

[charters]

It's likely that Wigner's friend would be rearranged in a very fatal way.

He is severely rearranged, but its not necessarily fatal with the idealized technology. Friend is alive and walking and talking after the fact, though he may be clinically dead during the process, and at some stage reanimated.

 

[DarMM]

$\mathcal{X}$, if we suspend disbelief regarding the technology, as is necessary to have this discussion at all, yields an in tact version of F's body. We're basically assuming something in the vein of manipulating a living body at the atomic level, with simultaneous perfect inverse bremmstrahlung, etc. So it is all completely ridiculous, but is it so much more ridiculous and removed from daily life than worrying about the unitarity of black hole evaporation? In my opinion, this has to seen as a question of logical consistency in the extreme edge cases, which are important to understand.

I think it is a bit more than just "technology" though. Given what Omnès calculates it seems unlikely that $\mathcal{X}$ is an element of any local algebra $\mathcal{A}\left(\mathcal{O}\right)$, thus it's not just a case of it "existing but being unachievable".

Of course in the OP I was ignoring this and just analyzing the scenario itself, but this is just to state people who object to Wigner's friend aren't doing so just on the basis of it being impossible to pull of practically.

 

[charters]

I think it is a bit more than just "technology" though. Given what Omnès calculates it seems unlikely that $\mathcal{X}$ is an element of any local algebra $\mathcal{A}\left(\mathcal{O}\right)$, thus it's not just a case of it "existing but being unachievable".

Of course in the OP I was ignoring this and just analyzing the scenario itself, but this is just to state people who object to Wigner's friend aren't doing so just on the basis of it being impossible to pull of practically.

For clarity we should separate two issues. 1) Is Wigner's operation possible in practice or in principle? 2) If Wigner's operation is possible in principle, and if we at least pretend it is therefore possible in practice with sufficient tech, is it conceivable Friend is alive and apparently sentient afterwards? I was only saying yes to the latter. I think, at a minimum, this is straighforwardly true in the Masanes type "full unitary reversal" versions of the protocol.

 

[DarMM]

For clarity we should separate two issues. 1) Is Wigner's operation possible in practice or in principle? 2) If Wigner's operation is possible in principle, and if we at least pretend it is therefore possible in practice with sufficient tech, is it conceivable Friend is alive and apparently sentient afterwards? I was only saying yes to the latter. I think, at a minimum, this is straighforwardly true in the Masanes type "full unitary reversal" versions of the protocol.

Agreed, if it is possible in principal you have a form of outcome subjectivity. Crudely speaking in a Copenhagenish view QM is a probability calculus for impressions left in a Classical/Boolean background by Quantum/Non-Boolean systems. If we have a Masanes type superobserver in principal then it has to be acknowledged that this classical background is not in principle unique and there can be other classical backgrounds whose events cannot be logically combined with our own. I sketched a PR-box version of Masanes due to Bub here:
https://www.physicsforums.com/threa...d-type-experiment.968181/page-10#post-6223680
In the Deutsch version of Wigner's friend where the backgrounds later combine this leads to observers with incompatible memories.

Regarding (1) I do think there are strong arguments it's not possible in principle. It seems $\mathcal{X} \notin \mathcal{A}\left(\mathcal{O}\right)$ where $\mathcal{O}$ is the observable horizon.

 

[charters]

Regarding (1) I do think there are strong arguments it's not possible in principle. It seems X∉A(O)X∉A(O)\mathcal{X} \notin \mathcal{A}\left(\mathcal{O}\right) where OO\mathcal{O} is the observable horizon.

Is this the de Sitter horizon?

 

[Morbert]

In non-encapsulated QM two events $E,F$ that don't commute can't usually have $P(E) = P(F) = 1$. Is this then a special feature of histories with encapsulation?

Actually my previous answer might be completely wrong and your intuition correct. If Wigner's friend wants to compute a prediction that Wigner definitely records $1$, he needs a record of his entire lab including himself, which should be incompatible with his record of his own measurement. I don't think he can be aware of both the pure state of his lab and his measurement.

 

[DarMM]

Is this the de Sitter horizon?

I'm not fully conversant on all the horizons in modern cosmology, but if you mean the $t \rightarrow \infty$ limit of the cosmological horizon, i.e. roughly 63 billion light years under current models, then yes.

 

[charters]

I'm not fully conversant on all the horizons in modern cosmology, but if you mean the $t \rightarrow \infty$ limit of the cosmological horizon, i.e. roughly 63 billion light years under current models, then yes.

Right, so this is dependent on the existence of dark energy/accelerated expansion right? Because otherwise the horizon does not asymptote to a finite volume