Undergrad Quantum theory - Nature Paper 18 Sept

[stevendaryl]

To what extent does the argument shed light on any of the actual interpretations of QM? It seems to me that it doesn't actually cause problems for a "collapse" interpretation. To quote the basic inferences again:

Let's write down what each observer can reason about the others, based on their observations:

1. If $F$ measures +1/2, then it means that it is impossible that $\overline{F}$ got result $\overline{h}$. That's because there is no overlap between the final state and $|\overline{h}\rangle |\frac{+1}{2}\rangle$. So $F$ concludes that if he got +1/2, $\overline{F}$ must have gotten $\overline{t}$
2. If $\overline{W}$ measures $\overline{ok}$, then it means that is impossible that $F$ got -1/2. That's because there is no overlap between the final state and $|\overline{ok}\rangle |\frac{-1}{2}\rangle$. So $W$ concludes that if he got $\overline{ok}$ then $F$ must have gotten +1/2.
3. If $\overline{F}$ gets $\overline{t}$, then it is impossible that $W$ got $ok$. That's because there is no overlap between the final state and $|\overline{t}\rangle |ok\rangle$. So $\overline{F}$ concludes that if he got $\overline{t}$, then $W$ got $fail$.

These inferences are true of the initial state of the system. If you allow for measurements to affect the thing that is being measured (collapse), then the inferences no longer hold:

If $\overline{W}$ measures $\overline{ok}$, then that places the composite system into the state $\langle \overline{ok}, \frac{+1}{2}\rangle$. $F$ measures +1/2, and the composite system is in the same state. If then $\overline{F}$ measures his system's state, he can get $\overline{h}$ or $\overline{t}$, with equal probability. So that means that 1. above is wrong: $F$ can't conclude that $\overline{F}$ will get $\overline{t}$. He could have concluded that in the initial state, but not after $\overline{W}$ messed with it.

Many-Worlds is a little more complicated. If you try to analyze it using Many-Worlds, what you would find is that different observers may disagree about what's true in "the real world". After $\overline{W}$ measures $\overline{ok}$ and $F$ measures $\frac{+1}{2}$, they disagree about whether $\overline{F}$ is in a definite state or not. $\overline{W}$ thinks he's in a superposition of $|\overline{h}\rangle$ and $|\overline{t}\rangle$, while $F$ thinks he's definitely in state $|\overline{t}\rangle$. So this thought experiment shows that the counting of "possible worlds" is a little problematic, because you can't consider a possible world to be one where everyone agrees on all macroscopic facts.

The only interpretation that to me seems inconsistent with this thought experiment is the combination of:

1. There is only one world.
2. There is no collapse.
3. There is nothing special about observers/measurements.
4. The Born rule gives the probabilities of all measurement outcomes.

 

[atyy]

There is a recent comment by Scott Aaronson on Frauchiger's and Renner's paper: https://www.scottaaronson.com/blog/?p=3975

I haven't read Frauchiger and Renner's work closely enough, but my knee jerk reaction was similar to what Scott Aronson is saying.

 

[DarMM]

My reading:

1. When the observer $\overline{F}$ obtains the $\overline{t}$ result, he can then infer that the $F$ observer's lab is in the $|fail\rangle$ state and thus the $W$ observer will not get the $okay$ result, as the $|fail\rangle$ state is orthogonal to $|okay\rangle$
2. Via application of the Born Rule in the $P(E) = 1$ (i.e. certain) case and the assumption of recursive conclusions based on other agents' reasoning, $\overline{F}$'s conclusion that $W$ must obtain $fail$ can be transferred to $W$ themselves, who will then predict they must measure $fail$.
   To obtain this conclusion they must pass through the reasoning of $\overline{W}$.
   This results in the paradox as in the original state $W$ can get $okay$. So from reasoning from the initial quantum state, $W$ deduces they can obtain $okay$, but from reasoning via $\overline{W}$ (and thereby obtaining $\overline{F}$'s conclusion) they conclude they cannot.
3. The reasoning $W$ and $\overline{W}$ use ultimately involves considering $F$ and $\overline{F}$ to be in the Hardy paradox state:
   $$\frac{1}{\sqrt{3}}\left(|\overline{h}\downarrow\rangle + |\overline{t}\downarrow\rangle + |\overline{t}\uparrow\rangle\right)$$

At this point there are a few possible objections, two of which:

1. What Scott Aaronson said. $\overline{F}$'s conclusion in typical QM can be considered to have the appended proviso:
   "provided $\overline{W}$ doesn't measure me".
   Since $\overline{F}$ is entangled with $F$, any observations by $\overline{W}$ should update conclusions about $F$.
   $W$ can only obtain $\overline{F}$'s conclusions after hearing the result of a $\overline{W}$ measurement, the exact scenario that would invalidate $\overline{F}$'s reasoning
2. Can you actually consider $F$ and $\overline{F}$ to be in the Hardy paradox state? This is Bub's point I believe.
   If they actually cause measurement/macro outcomes they are not in such an entangled state, thus you cannot use:
   $$\frac{1}{\sqrt{3}}\left(|\overline{h}\downarrow\rangle + |\overline{t}\downarrow\rangle + |\overline{t}\uparrow\rangle\right)$$
   but only something like:
   $$|\overline{h}\downarrow\rangle$$
   And then there is no inconsistent conclusions.
   If the Hardy state is valid to use on them, then there isn't macroscopic events associated with them, thus no "if $\overline{F}$ got $\overline{t}$".

 

[Michael Price]

Where do you see that? In the Nature article the main result seems to be "Theorem 1. Any theory that satisfies assumptions (Q), (C), and (S) yields contradictory statements when applied to the Gedankenexperiment of Box 1."

(S) is the single-world assumption

You've hit the nail on the head, and there is little more that need be said. Only many worlds survives the triple-condition-test, because MW does not assume (S) and hence can retain (Q) and (C), both of which seem essential. Why the paper never mentions many worlds is strange and clearly intentional. Weird.

 

[stevendaryl]

You've hit the nail on the head, and there is little more that need be said. Only many worlds survives the triple-condition-test, because MW does not assume (S) and hence can retain (Q) and (C), both of which seem essential. Why the paper never mentions many worlds is strange and clearly intentional. Weird.

It does:

Theories that violate Assumption (S)
Although intuitive, (S) is not implied by the bare mathematical formalism of quantum mechanics. Among the theories that abandon the assumption are the “relative state formulation” and “many-worlds interpretations”6,45,46,47,48. According to the latter, any quantum measurement results in a branching into different “worlds”, in each of which one of the possible measurement outcomes occurs. Further developments and variations include the “many-minds interpretation”49,50 and the “parallel lives theory”51.

 

[Michael Price]

It does:

Theories that violate Assumption (S)
Although intuitive, (S) is not implied by the bare mathematical formalism of quantum mechanics. Among the theories that abandon the assumption are the “relative state formulation” and “many-worlds interpretations”6,45,46,47,48. According to the latter, any quantum measurement results in a branching into different “worlds”, in each of which one of the possible measurement outcomes occurs. Further developments and variations include the “many-minds interpretation”49,50 and the “parallel lives theory”51.

.
Thanks for the correction. Obviously I didn't pay as close attention as I thought.

 

[DarMM]

Only many worlds survives the triple-condition-test,

Not so, several other interpretations survive it, as the section on interpretations states. Also dropping inter-agent consistency or absolute holding to the Born rule doesn't seem any more essential to me than a single world.

For example Bohmian Mechanics, according to the paper, violates the Born rule in this situation. However it's a very extreme situation.

 

[Michael Price]

Not so, several other interpretations survive it, as the section on interpretations states. Also dropping inter-agent consistency or absolute holding to the Born rule doesn't seem any more essential to me than a single world.

For example Bohmian Mechanics, according to the paper, violates the Born rule in this situation. However it's a very extreme situation.

I don't quite get your example. I thought Louville's theorem holds in a classical phase space for Bohmian Mechanics, which implies standard probability theory? But then I have always have problems with BM as a 'proper' interpretation since it contains the Universal Wavefunction and hence MW.

 

[PeterDonis]

I have always have problems with BM as a 'proper' interpretation since it contains the Universal Wavefunction and hence MW.

I don't see the implication here. BM has the wave function, but it does not claim that the wave function is the complete state of the system. MW requires the latter.

 

[DarMM]

I don't quite get your example. I thought Louville's theorem holds in a classical phase space for Bohmian Mechanics, which implies standard probability theory?

Bohmian mechanics reproduces the Born rule via a Louville type theorem called quantum equilibrium. Others here are more knowledgeable on Bohmian Mechanics and can give more details, but it does reproduce the Born rule.

 

[eloheim]

Let's write down what each observer can reason about the others, based on their observations:

1. If $F$ measures +1/2, then it means that it is impossible that $\overline{F}$ got result $\overline{h}$. That's because there is no overlap between the final state and $|\overline{h}\rangle |\frac{+1}{2}\rangle$. So $F$ concludes that if he got +1/2, $\overline{F}$ must have gotten $\overline{t}$
2. If $\overline{W}$ measures $\overline{ok}$, then it means that is impossible that $F$ got -1/2. That's because there is no overlap between the final state and $|\overline{ok}\rangle |\frac{-1}{2}\rangle$. So $W$ concludes that if he got $\overline{ok}$ then $F$ must have gotten +1/2.

Hey on post #14 in this thread, I'm assuming that the $W$, in the last sentence of bullet point number 2, should be $\overline{W}$, right? I only point it out because the post is detailed and amazing, and I'm too dull at this stuff to know immediately that that's the case. Anyway great job!

 

[stevendaryl]

Hey on post #14 in this thread, I'm assuming that the $W$, in the last sentence of bullet point number 2, should be $\overline{W}$, right? I only point it out because the post is detailed and amazing, and I'm too dull at this stuff to know immediately that that's the case. Anyway great job!

Yes, you're right.

 

[Demystifier]

So it's not really Hermitian operators that are measurable. It's macroscopic facts about the world. You can try to be clever and set up an interaction between systems so that this or that value of the operator leads to this or that macroscopic fact about your measurement device. But there is no guarantee that you'll be able to do that for all possible operators.

I agree with this. But there is no any general principle that forbids to do that for all possible operators (except the superselection rules, which would require a separate discussion). In the absence of such a general principle, we say that any hermitian operator is measurable in principle.

 

[Strilanc]

Scott Aaronson did a good blog post explaining what he thinks (and what I think) is the hole in the argument: https://www.scottaaronson.com/blog/?p=3975

In order to perform the compound measurement described in the experiment, you have to uncompute Alice and Bob's thoughts, do the measurement on a single qubit, then recompute the thoughts. During the recomputation, when Alice or Bob is thinking "My state is X therefore the other state is Y", the reasoning is not valid because they no longer started in the 00+01+10 state. This breaks the "if I'm correctly certain they're correctly certain of X, then I'm certain of X" chain, because "they" are no longer correctly certain of X.

 

[DarMM]

In order to perform the compound measurement described in the experiment, you have to uncompute Alice and Bob's thoughts, do the measurement on a single qubit, then recompute the thoughts.

I understand Aaronson's reasoning, but I don't quite grasp this, probably just missing something.

Alice and Bob are the superobservers corresponding to $\overline{W}$ and $W$ in Frauchiger-Renner's Nature paper. Where does performing their two measurements require uncomputing their thoughts and measuring a single qubit. I would have thought for each of them independently they are simply measuring an alternate outcome basis for the respective labs under their control. However they themselves are left alone.

If you are referring to $\overline{F}$ and $F$ instead, even there is one uncomputing their thoughts? I would have thought for example that $W$ measures some strange observable $Q$ that places the $L$ lab containing $F$ into the $\{|okay\rangle,|fail\rangle\}$ basis. However I'm not sure how that corresponds to uncomputing $F$'s thoughts measuring a qubit and recomputing them. Again most likely I am missing something.

 

[Demystifier]

Scott Aaronson did a good blog post explaining what he thinks (and what I think) is the hole in the argument: https://www.scottaaronson.com/blog/?p=3975

In order to perform the compound measurement described in the experiment, you have to uncompute Alice and Bob's thoughts, do the measurement on a single qubit, then recompute the thoughts. During the recomputation, when Alice or Bob is thinking "My state is X therefore the other state is Y", the reasoning is not valid because they no longer started in the 00+01+10 state. This breaks the "if I'm correctly certain they're correctly certain of X, then I'm certain of X" chain, because "they" are no longer correctly certain of X.

Yes, I had similar objections on the earlier paper by the same authors. See e.g.
"Single-world interpretations... cannot be self-consistent"
"Single-world interpretations... cannot be self-consistent"
"Single-world interpretations... cannot be self-consistent"
"Single-world interpretations... cannot be self-consistent"
"Single-world interpretations... cannot be self-consistent"

I've also noted the relation with the Hardy state here
"Single-world interpretations... cannot be self-consistent"

 

[Demystifier]

For example Bohmian Mechanics, according to the paper, violates the Born rule in this situation.

What is at stake here is the Born rule in the subsystem. Is probability of a subsystem given by $|\psi|^2$?

In standard QM the answer is no, simply because there is no such thing as $\psi$ for a subsystem. A subsystem is not in a pure state $|\psi\rangle$, but in a mixed state $\rho\neq|\psi\rangle\langle\psi|$.

Surprisingly, the paper talks a lot about subsystems, but, as far as I can see, it never mentions mixed states. It only talks about pure states in coherent superpositions.

What can Bohmian mechanics say about subsystems? For simplicity, suppose that the full system consists of two particles with positions $x_1$ and $x_2$. The full wave function is $\Psi(x_1,x_2,t)$. But in Bohmian mechanics there are also trajectories $X_1(t)$, $X_2(t)$, so one can define the wave function of the first subsystem as
$$\psi_1(x_1,t)=\Psi(x_1,X_2(t),t)$$
This is called conditional wave function and it does not have an analog in standard QM. In general, this conditional wave function does not satisfy Schrodinger equation, despite the fact that $\Psi(x_1,x_2,t)$ does. Indeed, according to Bohmian mechanics, it is this violation of Schrodinger equation that creates the illusion of wave function collapse in a subsystem.

Finally, what about the Born rule for conditional wave function? Well the probability is (up to a normalization factor) equal to $|\psi_1(x_1,t)|^2$. This looks like a Born rule, but this is not really the standard Born rule given by $|\Psi(x_1,x_2,t)|^2$. Particularly interesting case is when $\Psi$ is an energy-eigenstate, in which case $|\Psi(x_1,x_2,t)|^2=|\Psi(x_1,x_2)|^2$ is time-independent. Despite this time-independence in standard QM, the Bohmian $|\psi_1(x_1,t)|^2$ depends on time.

 

[stevendaryl]

I don't understand

Yes, I had similar objections on the earlier paper by the same authors. See e.g.
"Single-world interpretations... cannot be self-consistent"
"Single-world interpretations... cannot be self-consistent"
"Single-world interpretations... cannot be self-consistent"
"Single-world interpretations... cannot be self-consistent"
"Single-world interpretations... cannot be self-consistent"

I've also noted the relation with the Hardy state here
"Single-world interpretations... cannot be self-consistent"

I'm not sure I understand the objections. The states $|\overline{h}\rangle, |\overline{t}\rangle, |\overline{ok}\rangle, |\overline{fail}, |\frac{+1}{2}\rangle, |\frac{-1}{2}\rangle, |ok\rangle, |fail\rangle$ are states of the entire labs, not just the particle and/or coin that determined them. What's being measured by $\overline{W}$ and $W$ are the states of the whole labs.

Realistically, there shouldn't just be a single state corresponding to $|\overline{h}\rangle$, etc. By definition of a macroscopic object, there are many, many states consistent with a macroscopic description such as "Observer $\overline{F}$ sees a heads". I'm not sure whether this complication affects the conclusions, or not.

 

[stevendaryl]

What is at stake hear is the Born rule in the subsystem. Is probability of a subsystem given by $|\psi|^2$?

In standard QM the answer is no, simply because there is no such thing as $\psi$ for a subsystem. A subsystem is not in a pure state $|\psi\rangle$, but in a mixed state $\rho\neq|\psi\rangle\langle\psi|$.

When people talk about a mixed state, they mean two different things: (1) The mixed state resulting from ignorance about what the actual pure state is, (2) the mixed state that results from starting with a pure state for a composite system and tracing over all subsystems except the one of interest.

I don't see why it's necessary to consider mixed states (as opposed to superpositions) in this thought experiment.

 

[Demystifier]

What's being measured by $\overline{W}$ and $W$ are the states of the whole labs.

This measurement necessarily changes the state (of the lab). The issue is the following. Once you measure it, can you later unmeasure it and turn back into the initial state before measurement? I would say no, while the authors seem to saying yes.

 

[stevendaryl]

This measurement necessarily changes the state (of the lab). The issue is the following. Once you measure it, can you later unmeasure it and turn back into the initial state before measurement? I would say no, while the authors seem to saying yes.

I don't understand why it's necessary to "unmeasure". As I said in my summary, the point seems to be that the composite system of the two labs evolve into a state, where the following inferences seem to hold:

1. If $\overline{W}$ measures $\overline{ok}$, then he concludes that $F$ measures $+1/2$
2. If $F$ measures $+1/2$, then he concludes that $\overline{F}$ measures $\overline{t}$
3. If $\overline{F}$ measures $\overline{t}$, then he concludes that $W$ measures $fail$

Each of these separately is justified by a "collapse" model, where a measurement forces a system into an eigenstate of the operator being measured. The issues it seems to me is how to understand them without assuming collapse, and how to combine them. Combining them in the most straight-forward way leads to a false conclusion.

I don't see how "unmeasuring" is relevant.

 

[Demystifier]

I don't see why it's necessary to consider mixed states (as opposed to superpositions) in this thought experiment.

You don't need to talk about mixed states as long as you talk only about the full system. But the authors ask whether it is consistent to apply QM to a subsystem. If you want to apply standard QM to a subsystem, then you must use mixed states.

 

[stevendaryl]

You don't need to talk about mixed states as long as you talk only about the full system. But the authors ask whether it is consistent to apply QM to a subsystem. If you want to apply standard QM to a subsystem, then you must use mixed states.

I guess I don't see that it's necessary to apply QM to a subsystem in this thought experiment. Instead, it's a matter of one person deducing that another person is in a state in which he deduces a further fact. $\overline{W}$ deduces that $F$ is in a state in which he deduces that $\overline{F}$ is in a state in which he deduces that $W$ gets some particular result.

 

[stevendaryl]

I guess I don't see that it's necessary to apply QM to a subsystem in this thought experiment. Instead, it's a matter of one person deducing that another person is in a state in which he deduces a further fact. $\overline{W}$ deduces that $F$ is in a state in which he deduces that $\overline{F}$ is in a state in which he deduces that $W$ gets some particular result.

There's definitely weird about the whole business. We start off saying that the whole composite system is in the state $|final\rangle$. Then we say "If $\overline{W}$ observes $\overline{ok}$, then he deduces something or other about the state. But we've already said what the state is, and all the participants know it. So how is it possible to acquire more information through observations?

To make sense of somebody deducing something from an observation, I think we have to assume some kind of ontology. In a collapse model, an observation actually changes the state, so you're not actually learning about the state of the system before the observation, you're learning about the state immediately after the observation. In a many-worlds type ontology, an observation locates you on one (or maybe a set) of possible worlds. In a hidden-variable model, the observation gives information above and beyond what was specified by the wave function.

It seems to me that the lesson is that you need some kind of ontology in order to make sense of deductions from observation.

 

[Demystifier]

There's definitely weird about the whole business. We start off saying that the whole composite system is in the state $|final\rangle$. Then we say "If $\overline{W}$ observes $\overline{ok}$, then he deduces something or other about the state. But we've already said what the state is, and all the participants know it. So how is it possible to acquire more information through observations?

How can someone know that the system is in the state $|final\rangle$ if one didn't observe it?

Who is "we" in your sentence "we've already said what the state is"? Is it the third observer? Or a pure theoretician? How can a theoretician know what the state in the actual laboratory is?

It seems to me that the lesson is that you need some kind of ontology in order to make sense of deductions from observation.

Yes, that's one class of quantum interpretations, but the authors want to be agnostic on interpretations and prove a theorem which does not rest on the assumption of ontology.

 

[stevendaryl]

How can someone know that the system is in the state $|final\rangle$ if one didn't observe it?

That's a presupposition of the thought experiment.

Yes, that's one class of quantum interpretations, but the authors want to be agnostic on interpretations and prove a theorem which does not rest on the assumption of ontology.

But I don't think the conclusion is independent of ontology.

Mathematically, what they seem to be assuming is something like this:

If the state is $|\psi \rangle$ and $A$ observes $\alpha$, then he can conclude that $B$ will observe $\beta$ provided that:

$\Pi_{B,\beta} \Pi_{A,\alpha} |\psi\rangle = \Pi_{A,\alpha} |\psi\rangle$

where $\Pi_{X,x}$ is the projection operator onto the state in which $X$ has definite observation $x$. (projection operators for a claim have eigenvalue +1 to mean the claim is true, and 0 to mean the claim is false). Stated mathematically, the chaining is obviously false:

1. If $A$ observes $\alpha$, he concludes $B$ will observe $\beta$: $\Pi_{B,\beta} \Pi_{A,\alpha} |\psi\rangle = \Pi_{A,\alpha} |\psi\rangle$
2. If $B$ observes $\beta$, he concludes that $C$ will observe $\gamma$: $\Pi_{C,\gamma} \Pi_{B,\beta} |\psi\rangle = \Pi_{B,\beta} |\psi\rangle$
3. Therefore, if $A$ observes $\alpha$, he concludes that $C$ will observe $\gamma$: $\Pi_{C,\gamma} \Pi_{A,\alpha} |\psi\rangle = \Pi_{A, \alpha} |\psi\rangle$

Mathematically, statement 3 doesn't follow from 1&2. It would follow if 1&2 were true for all $\psi$, but not just from them being true for a particular $\psi$

 

[zonde]

I think the paradox provides argument against no collapse interpretations.
In the final state there are three terms:
$|final\rangle = \frac{1}{\sqrt{3}} |\overline{h}\rangle |\frac{-1}{2}\rangle + \frac{1}{\sqrt{3}} |\overline{t}\rangle |\frac{+1}{2}\rangle + \frac{1}{\sqrt{3}} |\overline{t}\rangle |\frac{-1}{2}\rangle$
Measurement determines between which two terms there will be interference. And interference determines certainty for some outcomes. But there can't be interference between all three terms at the same time. So you can't get certain outcome just by change of perspective.

 

[Demystifier]

That's a presupposition of the thought experiment.

A thought experiment only makes sense if a similar actual experiment is possible, at least in principle. So in an actual experiment, how would that presupposition be acquired?

 

[zonde]

A thought experiment only makes sense if a similar actual experiment is possible, at least in principle. So in an actual experiment, how would that presupposition be acquired?

Take polarized light, split it with PBS so that intensities in output beams are 1/3 and 2/3. In the beam with 2/3 intensity place a wave plate so that polarization is rotated by 45deg.
Will it do?

 

[Demystifier]

1. If $A$ observes $\alpha$, he concludes $B$ will observe $\beta$: $\Pi_{B,\beta} \Pi_{A,\alpha} |\psi\rangle = \Pi_{A,\alpha} |\psi\rangle$
2. If $B$ observes $\beta$, he concludes that $C$ will observe $\gamma$: $\Pi_{C,\gamma} \Pi_{B,\beta} |\psi\rangle = \Pi_{B,\beta} |\psi\rangle$
3. Therefore, if $A$ observes $\alpha$, he concludes that $C$ will observe $\gamma$: $\Pi_{C,\gamma} \Pi_{A,\alpha} |\psi\rangle = \Pi_{A, \alpha} |\psi\rangle$

Mathematically, statement 3 doesn't follow from 1&2.

Mathematical statement 3 does follow from 1 and 2 if different $\Pi$'s commute. And I think that they do commute if they represent different observers.