# I "Single-world interpretations... cannot be self-consistent"

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1. May 18, 2016

### Truecrimson

A new preprint by Daniela Frauchiger and Renato Renner argues that any interpretation of quantum theory that posits only a single world gives contradictory predictions provided that an observer (which makes predictions) can be observed (Wigner's friend scenarios). This might be of interest to participants in this subforum. Have anyone read this? What do you think?

This certainly sounds like a very big claim so I've been trying to read the preprint to find out about all the fine prints. Below is my attempt to summarize their basic argument based on my first pass at the preprint. I simplify it a bit which may leave some room for ambiguity.

There are 4 players. Wigner (W), his assistant (A), his friend 1 (F1) and friend 2 (F2). I can think of them all as robots that do quantum experiments, record the outcomes in some physical states, and process that information to make predictions. No consciousness is required, and in particular, I don't have to assume that macroscopic conscious human beings can be put into a superposition.

Step 1 F1 observes one of two possible outcomes from measuring the state $$\sqrt{\frac{1}{3}} |H\rangle + \sqrt{\frac{2}{3}} |T\rangle$$ in the "Head" $|H\rangle$ or "Tail" $|T\rangle$ basis. F1 then coherently prepares a spin down state $|\downarrow \rangle$ if she observes Head or $|\rightarrow \rangle = \sqrt{\frac{1}{2}} = |\uparrow \rangle + |\downarrow \rangle$ if she observes Tail, resulting in the entangled state $$| \psi \rangle = \sqrt{\frac{1}{3}} ( | H \rangle | \downarrow \rangle + | T \rangle | \downarrow \rangle + | T \rangle | \uparrow \rangle ).$$ She sends the spin state to F2.

Step 2 F2 measures the spin in the Z basis (up or down).

Step 3 A projectively measures F1 (the whole laboratory) and declares success if he gets the outcome $$\sqrt{\frac{1}{2}} (| H \rangle - | T \rangle )$$ or failure if he gets the orthogonal outcome.

Step 4 W projectively measures F2 and declares success if he gets the outcome $$\sqrt{\frac{1}{2}} (| \downarrow \rangle - | \uparrow \rangle )$$ or failure if he gets the orthogonal outcome.

The preprint analyzes the situation where both A and W succeed. This is possible because of the nonzero overlap of the projection operator $$\frac{1}{2} (| H \rangle - | T \rangle ) \otimes (| \downarrow \rangle - | \uparrow \rangle )$$ and the state $$|\psi \rangle = \sqrt{\frac{1}{3}} [ ( | H \rangle + | T \rangle ) | \downarrow \rangle + | T \rangle | \uparrow \rangle ] .$$ It is the rightmost term that contributes to the nonzero overlap. Given that both A and W succeed, A infers that the spin must be in the up state, which means that F1 observed the outcome Tail. But if the result comes out tail then Wigner must fail! because projecting the state $|\psi \rangle$ onto $$\sqrt{\frac{1}{2}} |T \rangle ( | \uparrow \rangle + | \downarrow \rangle )$$ means that Wigner will never succeed. Q.E.D.

The step that seems problematic to me is the retrodiction that the spin has a definite (up) value. This seems to be reminiscent of Aharonov and Vaidman's Three-Box Paradox.

2. May 18, 2016

### StevieTNZ

Thanks for sharing! I've had correspondence in the past with Renato Renner, regarding other papers he has published.

3. May 18, 2016

### atyy

Another interesting paper might be

http://arxiv.org/abs/quant-ph/9712044
Quantum and classical descriptions of a measuring apparatus
Ori Hay, Asher Peres
"This article examines whether these two different descriptions are mutually consistent. It is shown that if the dynamical variable used in the first apparatus is represented by an operator of the Weyl-Wigner type (for example, if it is a linear coordinate), then the conversion from quantum to classical terminology does not affect the final result. However, if the first apparatus encodes the measurement in a different type of operator (e.g., the phase operator), the two methods of calculation may give different results."

But Hay's and Peres's result was not considered to overturn Copenhagen or de Broglie Bohm - it just means that the folklore that the classical quantum cut can be placed anywhere is not absolutely true - and makes sense from both Copenhagen and dBB viewpoints where some "common sense" is needed to say that FAPP we do know what a "macroscopic measurement apparatus" is.

I believe that even von Neumann was aware of this, so the cut at an arbitrary place along the von Neumann chain is not always a classical/quantum cut, but in some cases has to be a quantum/quantum cut. Wiseman and Milburn's textbook also mentions that placing the classical/quantum cut in the wrong place gives experimentally incorrect results.

Last edited: May 18, 2016
4. May 18, 2016

### bhobba

Without looking at the detail it looks highly dubious to me.

In QM you need a framework of system being observed and something doing the observing. Its not a contradiction that different frameworks may give different results. Its one of the weird things about QM, although this is the first paper I have seen that describes such a situation. That said I personally doubt such an experiment can actually be performed, but I will leave that up to experimental types to comment. The difficulty in doing it would seem to explain why in the world around us we never notice such a possibility of different frameworks giving different outcomes.

Thanks
Bill

Last edited: May 19, 2016
5. May 18, 2016

### bhobba

Well Von Neumann proved it rigorously so it's not exactly folklore.

But like you I suspect there are some caveats involved to do with differing frameworks of observer and observed. That would certainly be the decoherent histories viewpoint which requires non mutually contradictory frameworks. That is usually enforced by decoherence, but its possible to perhaps come up with a way around it.

Thanks
Bill

6. May 18, 2016

### atyy

But actually even if one thinks of it from say Schlosshauer's discussion of decoherence, his point is that decoherence plus some additional criteria gives us objective ways to say where we can place the cut (usually must be far out enough), then we can see that decoherence itself suggests the cut cannot be too early.

7. May 19, 2016

### Truecrimson

I agree that where the cut is can't be arbitrary. With decoherence, one can't place the cut too early before decoherence has occurred. The Wigner's friends scenario though doesn't explicitly has decoherence. Then one probably has to have a rule for co-existing viewpoints as bhobba pointed out. (I don't know what the axioms of decoherent histories are.) In particular, the conclusion of Frauchiger and Renner probably comes about by combining viewpoints that you shouldn't combine.

I need to read the paper atyy brought up. Thank you.

8. May 19, 2016

### Demystifier

I had an extensive discussion about that paper with one of the authors, so I think I can tell what is the main problem with their idea. One of their assumptions is that new measurements delete information about the outcomes of previous measurements, and in my opinion this assumption is unjustified.

9. May 19, 2016

### Truecrimson

It's 1 am over here so I need to go to bed. I will think about the paper and this statement some more when I have time. Thank you.

10. May 19, 2016

### bhobba

As I said Von Neumann did prove it rigorously and he is no mean mathematician. But of course not infallible as his supposed proof of no hidden variables showed. I only suspect there is an out.

Thanks
Bill

11. May 19, 2016

### Demystifier

Another important thing about decoherence is that it is FAPP irreversible. Therefore information about outcomes of previous measurements is not deleted by new measurements, in contradiction with the crucial assumption used in the paper under discussion.

12. May 19, 2016

### bhobba

Hmmmm. The delayed choice experiment?

Thanks
Bill

13. May 19, 2016

### Demystifier

The delayed choice is reversible precisely because it does not involve decoherence.

14. May 19, 2016

### bhobba

Got it.

It ties in with a discussion about entanglement and decoherence. Its easy to get the two confused.

Thanks
Bill

15. May 19, 2016

### Demystifier

Yes. Decoherence always involves entanglement, but not the vice versa.

16. May 19, 2016

### Demystifier

The problem with the paper is that they do not explain what does it mean to measure the whole laboratory. Certainly the whole laboratory cannot be in the state
$$\sqrt{\frac{1}{2}} (| H \rangle - | T \rangle )$$
Instead, it should involve something like
$$\sqrt{\frac{1}{2}} (| H \rangle - | T \rangle ) \otimes | {\rm detector \;\; shows\;\;} H-T \rangle$$
where the second factor is a macroscopic state that cannot be easily destroyed by a new measurement.

17. May 19, 2016

### S.Daedalus

My main issue with the paper is that I don't understand how it's justified to have the value of z be an element of the story of A, before A has carried out any measurement. To A, and hence (to my understanding), in any story about A's experiment, before they perform a measurement on the system F2, the total system F1 + F2 should be in a superposition, with components having both $$|z=+\frac{1}{2}\rangle$$ and $$|z=-\frac{1}{2}\rangle$$ elements; that is, a state to which one could not assign any definite value of z, at least if one wants to keep the eigenvalue-eigenstate link.

Yet, they consider $$(n:20,*,z,*)$$ to be a 'plot point' of A's story (in their notation, this roughly means that at time n:20, there exist some values for the asterisks such that the completed element is part of the set of events that characterize A's experiment). I don't see how that is justified; in particular, A could, instead of the measurement they perform in the paper, perform an interference experiment, which would tell them that indeed a superposed state is present (even after the measurement of F2). So it seems to me more natural to add some rule to the effect that 'quantities only have a definite value in X's story if the system is in an eigenstate of the relevant operator as described by X', which, it seems, would rule out the definiteness of $z$, and block the proof in the paper; thus, with such a rule in place, one could indeed find a consistent single-world interpretation. Or am I way off base here?

18. May 19, 2016

### Giulio Prisco

Very interesting!

The authors provide a simple summary:

Main result (informal version)
There cannot exist a physical theory T that has all of the following properties
:
(QT) Compliance with quantum theory: T forbids all measurement results that are forbidden by standard quantum theory (and this condition holds even if the measured system is large enough to contain itself an experimenter).
(SW) Single-world: T rules out the occurrence of more than one single outcome if an experimenter measures a system once.
(SC) Self-consistency: T's statements about measurement outcomes are logically consistent (even if they are obtained by considering the perspectives of different experimenters).

19. May 19, 2016

### Truecrimson

Let's see if I understood this correctly.

Recall the state before any measurement by F2, A, and W: $$|\psi \rangle = \sqrt{\frac{1}{3}} (|H\rangle |\downarrow \rangle + |T\rangle |\downarrow \rangle + |T\rangle |\uparrow \rangle)$$ After the measurements, there are associated memory states which record the value of the spin and whether both A and W succeeded. For example, $|\uparrow ,\checkmark \rangle$ means that the spin was measured to be up and A and W succeeded. Hence the total state will have 3 terms: $$(|H\rangle +|T\rangle ) |\downarrow \rangle |\downarrow , \times \rangle,$$ $$|T\rangle |\uparrow \rangle |\uparrow ,\checkmark \rangle ,$$ $$|T\rangle |\uparrow \rangle |\uparrow ,\times \rangle$$ Given that A and W succeeded, putting the Heisenberg cut after F2's spin measurement or after A and W's measurements makes no difference. (You just collapse to the state to $|T\rangle |\uparrow \rangle |\uparrow ,\checkmark \rangle$.)

However A and F2 can't infer that because the spin is up, the outcome Tail must had happened because this entails a different set of memory states, one of which may look like this: $$|T\rangle |\rightarrow \rangle |\rightarrow ,\times \rangle .$$ (F2 can make a measurement with 3 POVM elements: $|\uparrow \rangle$, $|\leftarrow \rangle$ and another one to complete the resolution of the identity. This way, she will never confuse the states $|\downarrow \rangle$ and $|\rightarrow \rangle$ but the price to pay is that the third POVM element gives an inconclusive result.) The point is that they have to undo the measurement and

Or even if they can undo the measurement, that memory A and W succeeded will no longer be there. So no one will predict that A and W succeed, and the contradiction cannot be reached.

Does this look remotely right?

Last edited: May 19, 2016
20. May 19, 2016

### Truecrimson

That was the point that I was suspicious of too. But now I wonder if it matters in this case? Sure, it is wrong to place the Heisenberg cut (and infer a definite value of an observable) too early, but conditional upon the success of A and W, this doesn't seem to affect the argument (as I mentioned in #19 above).