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I have some doubts regarding my personal interpretation that i was contemplating about in the context of Wigner's friend experiment (also tested in the laboratory).Could it be that a system is always in a superposition, and when we perform a measurement, we obtain a definite value due to the interaction, but after it, the system returns to a superposition? For Wigner, who will check if superposition exists for him after his friend's measurement, he will find the system again in a superposition. If they were to measure at the same time, they would see the same definite result instead. Wigner and his friend might have different measurements, but this wouldn't imply a different reality, only their knowledge of it. Please note that i am not a physicist and i do not intend to propose any theory or interpretation because i certantly don’t have the knowledge required. I am just expressing my curiosity.

Marco Fabbri
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It is unclear to me what you think the Wigner's Friend is about so I'm going to describe it and consider the implications of what you have suggested.

Suppose you have a sealed box with an atom, a laboratory and a person who happens to be a friend of Wigner and Wigner is outside the box. By sealed I mean it is completely impossible for any information, fields or matter to get out of the box: no light, no sound, no anything else.

The atom $A$ has two possible states: spin up or spin down each with probability of 1/2, which we write as $\tfrac{1}{\sqrt{2}}(|\uparrow\rangle_A+|\downarrow\rangle_A)$. Wigner's friend $F$ measures the atom and remembers the result and if we assume the joint system of the atom and the friend obeys quantum mechanics the state is then: $$\tfrac{1}{\sqrt{2}}(|\uparrow\rangle_A|\uparrow\rangle_F+|\downarrow\rangle_A|\downarrow\rangle_F).$$

It is common for physicists to say that when a measurement takes place the laws of QM somehow stop operating and only one outcome happens. The operative word here is somehow, as in most physicists can't be bothered to work out the details and assume it doesn't matter. So then Wigner measures the state of the atom and the friend and somehow he sees one of the two possible options and that is the only state. But the question is why we can't treat Wigner $W$ according to the laws of quantum mechanics and just say the state after his measurement is $$\tfrac{1}{\sqrt{2}}(|\uparrow\rangle_A|\uparrow\rangle_F|\uparrow\rangle_W+|\downarrow\rangle_A|\downarrow\rangle_F|\downarrow\rangle_W).$$

You ask

Could it be that a system is always in a superposition, and when we perform a measurement, we obtain a definite value due to the interaction, but after it, the system returns to a superposition?

How would the system you're observing know whether you're looking at it to flip back and forth between those two options?

You further ask:

For Wigner, who will check if superposition exists for him after his friend's measurement, he will find the system again in a superposition.

In the real world, if Wigner observes the friend and the atom, he will find the same result as he found the first time if he repeats the measurement. If the state flipped back to being a superposition when he stopped looking at it what would stop him from seeing the outcome from changing?

It is difficult to come up with a consistent theory that is also testable and correct. It is okay not to have answers to questions like the one I asked as long as you don't pretend to know more than you really do. I have another answer that links to papers about some of the options and how to test them:

Is many-worlds interpretation only a philosophical matter?

alanf
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    I am reminded of Henri Poincare's ideas on the philosophy of science; what good is it to have a theory that is likely not testable and adds complication while not really being useful? That doesn't sound like good philosophy. Everything is not entangled, something must cause decoherence or there would be no reality of any kind. Many worlds, what a crock! – Albertus Magnus Jan 03 '24 at 19:06
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    @AlbertusMagnus Your interpretation is less satisfactory than many words to me, as the observer is something outside the theory. In your interpretation QM is definitely incomplete. – Pato Galmarini Jan 03 '24 at 19:25
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    @Pato Galmarini Yes. QM is incomplete, and even though it would be nice to have a complete theory, there is no reason for reality to reflect human hopes. Incompleteness is the human condition, epistemically speaking, and QM is a poignant reminder of our inability to have such comprehensive knowledge. "We must know, we shall know"; I have always liked such optimism, however, it just isn't in humanity to know more than traditional QM tells us. I do hope I am proven wrong, but I doubt it! – Albertus Magnus Jan 03 '24 at 21:13
  • @AlbertusMagnus What you say might or might not be true. May be there is nothing deeper than QM in the laws of physics. If so then I think many world is the only self consistent and complete interpretation that we know of. I hope it is not the case and we can find something more satisfactory. I also dislike many worlds, for reasons that I cannot fully articulate. – Pato Galmarini Jan 03 '24 at 21:26
  • @AlbertusMagnus Decoherence is a process that happens when one quantum system interacts with other quantum systems and transfers information to them

    https://arxiv.org/abs/0707.2832

    https://arxiv.org/abs/1111.2189

    It doesn't require QM to be incomplete.

     – alanf Jan 03 '24 at 23:27
  • @alanf You are right, I used the wrong word. I intended to write disentanglement. sorry about that. – Albertus Magnus Jan 04 '24 at 01:44
  • @AlbertusMagnus Decoherence changes states of open entangled systems to a disentangled state https://arxiv.org/abs/quant-ph/0312068 so how can disentanglement be a reason to say QM is incomplete? – alanf Jan 04 '24 at 09:08
  • @alanf I do not maintain that disentanglement is the reason that QM is incomplete. My statement about disentanglement is in reference to your claim that Wigner and friend are necessarily entangled with the system under observation. And even if this were so, it seems odd to me that Wigner would become entangled unless there were yet a third observer who had yet to make measurements. I am interested in why you made that claim. – Albertus Magnus Jan 04 '24 at 11:53
  • @AlbertusMagnus I didn't say that Wigner is necessarily entangled. He might become disentangled as a result of decoherence. However, even when a system decoheres a proper explanation of its behaviour may require a quantum mechanical treatment, see https://physics.stackexchange.com/a/795327/28512 – alanf Jan 04 '24 at 14:22
  • @alanf I see. Of course, at the most fundamental level, a quantum mechanical treatment will always be needed. Just for the record, I do feel that QM is incomplete, mere opinion\conjecture that my sentiment is; however, thank you for your stimulating and intriguing engagement on this topic. If it is not too much, can you suggest a good intro to the Wigner's friend scenario, as this interpretive element of QM happens to be one with which I am less familiar. – Albertus Magnus Jan 04 '24 at 15:03
  • This paper is okay as an introduction to Wigner's Friend https://arxiv.org/abs/1804.00749 – alanf Jan 04 '24 at 15:36
  • A sealed box as required by this experiment is physically impossible. You can't shield gravity for example. So it does not make much sense to elaborate about the implications of this experiment. It's like discussing levitating monks or flying carpets. – Andrei Jan 05 '24 at 10:29
  • @Andrei The laws of physics don't rule out getting arbitrarily close to isolating a system, e.g. - by moving it away from other masses. This will reduce the probability of producing any detectable effect by gravitational interactions to be as small as you like. So the box will evolve unitarily to high accuracy just as a photon does in an interferometer https://physics.stackexchange.com/questions/119743/qm-why-is-reflection-of-a-photon-not-a-measurement/119749#119749 – alanf Jan 05 '24 at 23:15
  • @alanf, The difference between a photon in an interferometer and your isolated lab is that the state of that photon is not a macroscopic state. There is no way you can extract information about that photon from a distance without measuring it directly. So, as long as you don't let other particles pass through photon's path it remains in superposition. Once the measurement inside your isolated lab takes place, the result can be inferred from virtually any distance by, say, a measurement of the gravitational field. To cont. – Andrei Jan 08 '24 at 06:55
  • @alanf, Cont: You can put a small rock in orbit around the lab at any distance you want and, by observing its trajectory, you can determine the mass distribution inside the lab and see if the pointer went to the right or left. The only way to avoid this is to send the rock to infinity, but in this case you do not fulfill the requirements of the experiment. There should be a Wigner at some finite distance from the lab. – Andrei Jan 08 '24 at 06:58
  • @Andrei The further away the rock the weaker the interaction and the smaller the probability that the state of the rock will change measurably. So you can reduce the probability of detection to any desired level and so reduce decoherence. Do you have an actual explanation, preferably in a paper, as to why this is wrong? – alanf Jan 08 '24 at 09:21
  • @alanf, we are not speaking about probabilities here, the rock (and the pointer) are large enough for the uncertainty to be irrelevant, we are in a classical regime. Once you measure the trajectory for some time you can infer the result inside. I don't know a paper discussing this but I guess the burden is on you to explain why a classical measurement of mass distribution cannot be done accurately in this case. – Andrei Jan 08 '24 at 10:59
  • @Andrei When motion should be considered classical is determined by decoherence even for orbital mechanics https://arxiv.org/abs/quant-ph/0605249 and that "classicality" is only an approximation. If two masses of 1kg $10^7$m apart the gravitational potential energy will be approximately $10^{-17}$J which is of a similar order of magnitude to the energy of a single X-ray photon. So that situation is similar to that of the mirror. – alanf Jan 08 '24 at 13:31
  • I am not sure why this energy is relevant. Let's say that if the pointer goes left the rock's orbit would shift towards region A, otherwise towards some region B. Wigner will then look at those two regions and see where the rock arrives. He does not need to continuously observe the rock so that it's orbit would be perturbed as a result of photons colliding with it. He just waits for the rock to enter one region and then it does not matter if it is perturbed. – Andrei Jan 08 '24 at 14:12
  • The energy is relevant because the rock is in a mixed state that has a spread of different energies (and trajectories) just as with the mirror. The gravitational interaction changes that spread. The size of that change is very small and so the probability of detection is negligible. The idea that the rock is on a single trajectory is always only an approximation and in this case because the amount of change involved is so small it's a bad approximation. – alanf Jan 08 '24 at 14:25
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This experiment cannot be done, even in principle, because it is impossible to isolate the lab with the friend. It is always possible to get the relevant information regarding the outcome of the measurement from the outside. Any macroscopic change inside the lab (required for a successful measurement) will be accompanied by a change in the gravitational, electric and magnetic fields associated with that change, fields which could be measured from outside.

Given the above, the superposition ends when the friend performs its measurement for any observer, including Wigner.

Andrei
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  • Yes, I am of this opinion also, collapse of the wave function dissolves the superposition, from this point unitary time development is the only relevant factor. – Albertus Magnus Jan 05 '24 at 15:20
  • Just wondering. Short of doing the experiment near the center of a black hole, how important would gravity be to QM? Low gravity introduces phase factors doesn't it? This should destroy the superposition? – Albertus Magnus Jan 05 '24 at 15:35
  • @AlbertusMagnus, I don't understand the question. As far as I know there is no theory describing what happens at the center of a black hole. In low gravity regime you can treat the gravitational force in the same way as Coulomb's. Why would it destroy the superposition? – Andrei Jan 05 '24 at 22:41
  • The reason I ruled out black holes and inquired about low gravity is because I know that we have no quantum theory of gravity. The reason for the question, in the first place, was that I mistakenly read into your post that external fields would interact with the superposition and destroy it. I see now that you were only trying to say that there would be no way to hide any observations from Wigner. I will try to exercise more care in the future, apologies. – Albertus Magnus Jan 06 '24 at 02:18
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An undisturbed quantum system remains in superposition. When it is observed it collapses to a specific state, whereafter the system evolves according to the time development of the Schrodinger equation, unless of course that it is continually observed.

If Wigner's friend makes an observation at some time $t$ he will find the system in an eigenstate $A$. If Wigner has not yet made an observation then the system is for him, in a state dictated by unitary time development from the state $A$. If at some time $t_1$, after $t$, Wigner decides to make a measurement; Wigner will find the system to be in an eigenstate $B$. If both Wigner and his friend decide to measure together at time $t_1$, then they will both observe the system to be in state $B$.

Who ever disturbs the system destroys the superposition.

Albertus Magnus
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