Quantum theory - Nature Paper 18 Sept

[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.

 

[Demystifier]

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?

Yes, which helps me to give a better answer to the question by @stevendaryl (see my next post).

 

[Demystifier]

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?

This common knowledge is a knowledge of the state $|\psi(t_0)\rangle$ at some time $t_0$. But observation is performed at some later time $t_{\rm obs}>t_0$, so observation determines $|\psi(t_{\rm obs})\rangle$. The act of observation changes the state (that's called quantum contextuality), so $|\psi(t_{\rm obs})\rangle \neq |\psi(t_0)\rangle$. That's how observations acquire new information.

 

[eloheim]

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?

I wish I could be more helpful than this, but did the specific experimental procedure in the Nature Communications paper, with its step-by-step (timing) recipe, do anything to address the problems you see with being able to actually perform an experiment like this in the real world? Here are a couple quotes that might be trying to answer such objections but I'm not really sure:

In the Gedankenexperiment proposed in this article, multiple
agents have access to different pieces of information, and draw
conclusions by reasoning about the information held by others. In
the general context of quantum theory, the rules for such nested
reasoning may be ambiguous, for the information held by one
agent can, from the viewpoint of another agent, be in a superposition
of different “classical” states. Crucially, however, in the
argument presented here, the agents’ conclusions are all restricted
to supposedly unproblematic “classical” cases. For example, agent
W only needs to derive a statement about agent F in the case
where, conditioned on his own information $\overline{w}$, the information z
held by F has a well-defined value (Table 3). Nevertheless, as we
have shown, the agents arrive at contradictory statements.

And:

Another noticeable difference to earlier no-go results is that the
argument presented here does not employ counterfactual reasoning.
That is, it does not refer to choices that could have been
made but have not actually been made. In fact, in the proposed
experiment, the agents never make any choices (also no delayed
ones, as e.g., in Wheeler’s “delayed choice” experiment[63]). Also,
none of the agents’ statements refers to values that are no longer
available at the time when the statement is made (cf. Table 3).

Anyone find this applicable to the objections at hand? Sorry, I can't decide. lol

 

[stevendaryl]

This common knowledge is a knowledge of the state $|\psi(t_0)\rangle$ at some time $t_0$. But observation is performed at some later time $t_{\rm obs}>t_0$, so observation determines $|\psi(t_{\rm obs})\rangle$. The act of observation changes the state (that's called quantum contextuality), so $|\psi(t_{\rm obs})\rangle \neq |\psi(t_0)\rangle$. That's how observations acquire new information.

Well, that's why I said that it's not a paradox for collapse models.

I do not exactly agree with the example of polarizing filters. I would say that a polarizing filter is not an observation. Seeing that a photon passed through a filter is an observation.

 

[stevendaryl]

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.

In the paper under discussion, there are the following observations:

1. $\overline{W}$ observes $\overline{ok}$
2. $F$ observes $+1/2$
3. $\overline{F}$ observes $\overline{t}$
4. $W$ observes $fail$

1&2 correspond to commuting projection operators, as do 2&3, as do 3&4. But 1&3 don't commute, and neither do 2&4.

 

[Demystifier]

Well, that's why I said that it's not a paradox for collapse models.

But even in the minimal interpretation without explicit collapse, there is some kind of "information update" postulate that effectively does the same thing. Something like that exists in all interpretations (in MWI it is split, in BM it is conditional wave function, in CH it is non-classical logic, ...), which is why all interpretations are consistent, which indeed is what the paper claims.

 

[stevendaryl]

In the paper under discussion, there are the following observations:

1. $\overline{W}$ observes $\overline{ok}$
2. $F$ observes $+1/2$
3. $\overline{F}$ observes $\overline{t}$
4. $W$ observes $fail$

1&2 correspond to commuting projection operators, as do 2&3, as do 3&4. But 1&3 don't commute, and neither do 2&4.

In a many-worlds ontology, you can think of an observer's "world" to be the projection of the universal wavefunction using the projection operator corresponding to the observer's current state of knowledge. So there is the world in which $\overline{W}$ observes $\overline{ok}$. In that world, $F$ observes $+1/2$. So there are the following "worlds":

1. $W_1$: A world in which $\overline{W}$ observes $\overline{ok}$, and $F$ observes $+1/2$ and $\overline{F}$ and $W$ are in indefinite states (like Schrodinger's cat)
2. $W_2$: A world in which $F$ observes $+1/2$ and $\overline{F}$ observes $\overline{t}$ and $\overline{W}$ and $W$ are in indefinite states.
3. $W_3$: A world in which $\overline{F}$ observes $\overline{t}$ and $W$ observes $fail$ and $\overline{W}$ and $F$ are in indefinite states.

Even within one world, different observers thus disagree about which world they are. In world $W_1$, $F$ believes that he is in world $W_2$.

Quantum mechanics is most coherent when all observers agree about all macroscopic facts, but this example is specifically designed to violate that.

 

[stevendaryl]

But even in the minimal interpretation without explicit collapse, there is some kind of "information update" postulate that effectively does the same thing.

I don't see how updating can make any sense unless you believe either that (1) observations change the things that are being observed, or (2) there is extra information above and beyond the wave function. So collapse or hidden variables. What does updating mean in the minimal interpretation?

 

[Demystifier]

I don't see how updating can make any sense unless you believe either that (1) observations change the things that are being observed, or (2) there is extra information above and beyond the wave function. So collapse or hidden variables.

I agree.

What does updating mean in the minimal interpretation?

Minimal interpretation is agnostic on that. That's why it is called minimal. In fact, minimal interpretation is the same as shut-up-and-calculate. ... The only problem with minimal interpretation is that some of its adherents (we have one on this forum) do not just shut up but try to explain what it "really" means, and in such attempts easily fall into a contradiction. When you tell them that it's a contradiction, then they reply that it's philosophy without empirical consequences. And few hours later they forget all that and repeat the same contradiction again. But otherwise, when such guys do the "shut-up-and-calculate" stuff, they are great. (I think you know who I am talking about.)

 

[stevendaryl]

Minimal interpretation is agnostic on that. That's why it is called minimal. In fact, minimal interpretation is the same as shut-up-and-calculate. ...

I think that there is some slight disagreement about what the minimal interpretation means. There is a distinction between having a minimum ontology and having minimal assumptions. Minimal assumptions are in some sense maximalist about ontologies: many ontologies are possible, nothing is ruled out.

The minimal ontology might be something like this:

1. There is only one world.
2. Quantum mechanics works the same on any system (microscopic or macroscopic, measurement devices or not)
3. The probabilities of measurement results are given by the Born rule.

I actually don't think the minimal ontology is self-consistent. You can't have the Born rule being true in general if there is nothing special about measurements.