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Is many-worlds interpretation only a philosophical matter? It seems to me that we can't exclude a possible test for this hypothesis. I explain.

For superposition principle each world would follow the Schrodinger equation and then it seems impossible to distinguish if we collapsed in $\psi_i$ a wave function $\psi=\sum_i\psi_i$ or we ended up in a world where the state of the system is $\Psi_i=\psi_{me}(i)\otimes\psi_i$. This would be, at the multiverse level, just one state of the superposition $\Psi=\sum_i\Psi_i$ with all the other possible outcomes and their worlds. Also, if there are interferences enhancing some worlds and suppressing others, this would be testable only by someone experimenting on $\Psi$, at the multiverse level, then again invisible inside any world.

But if the interference cancels out completely some possible world, then we could be able to recognize that by statistical means. Suppose that I am doing a measure that can collapse $\psi$ in an eigenstate $\psi_i$ such that the world $\psi_{me}(i)\otimes\psi_i$ would be canceled out by interference with other worlds. Then, getting the eigenvalue of $\psi_i$ in the measure wouldn't be an evidence contrary to the many world hypothesis?

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  • $\begingroup$ What's the probability of measuring the eigenvalue of $\psi_i$ in the other interpretations? If it's the same, then this doesn't distinguish anything. $\endgroup$ – probably_someone Feb 12 at 17:38
  • $\begingroup$ MWI sets a completely different framework, it says that $\psi$ doesn't collapse at all. I have no idea why it calls itself an interpretation $\endgroup$ – J.Ask Feb 12 at 18:19
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    $\begingroup$ Do you know how to reverse the passage of time? I certainly don't. As such, "un-measuring" something is not at the moment possible to test, and it may never be. What decoherence and the many-worlds interpretation both do is explain why we observe wavefunction collapse while still being time-reversible. $\endgroup$ – probably_someone Feb 12 at 18:43
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    $\begingroup$ The Frauchiger-Renner thought experiment may be relevant. It is claimed that it shows that you cannot hold various interpretations as being equivalent: "'My take is likely to be that it kills wave-function-collapse or single-universe versions of quantum theory, but they were already stone dead'" $\endgroup$ – JimmyJames Feb 12 at 21:47
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    $\begingroup$ "But if the interference cancels out completely some possible world, then we could be able to recognize that by statistical means." So you claim we can use statistics to demonstrate that something unobservable might have existed, except that it doesn't? Well, that still leaves five out of six impossible things to believe before breakfast ;) $\endgroup$ – alephzero Feb 12 at 23:49
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My understanding of this kind of thing has evolved over time. I used to be convinced that interpretations were inherently untestable, but now I think that was an oversimplification.

To make the discussion more concrete, let's consider a set of axioms for quantum mechanics:

  • (1) States are rays in a vector space over the complex numbers.

  • (2) unitary evolution

You could add more to this list (observables are self-adjoint operators, completeness), but these are the main things that are important, and they are also things that everyone agrees on. This is all you need for the most austere versions of MWI, which I'll call MWI-basic.

If you want the Copenhagen interpretation, you need some more axioms:

  • (3) Born rule

  • (4) Measurement collapses the wavefunction.

So from the point of view of this kind of axiomatic development, CI is the same as MWI-basic plus additional axioms. One thing this tells us is that any experiment that disproves MWI-basic must also disprove CI.

It is certainly true that MWI-basic and CI are falsifiable. Any observation that falsifies 1 or 2 falsifies all of QM, and therefore falsifies both MWI-basic and CI.

I think the right way to look at this kind of thing is that CI is an approximation, and the approximation is good when the measuring instrument is macroscopic. When the measuring instrument is mesoscopic, the approximation is not perfect, and this is something that we can see. A nice example is Allahverdyan 2017. They simulate a measurement by a mesoscopic system, and they come up with all kinds of phenomena that actually do happen according to quantum mechanics, but that are not correctly described by CI. For example, there are time scales that emerge from the simulation, whereas (4) says collapse is instantaneous.

There are also more baroque versions of MWI, which we can refer to collectively as MWI-baroque. DeWitt gives a description of what I would call a baroque version of MWI:

This universe is constantly splitting into a stupendous number of branches, all resulting from the measurement-like interactions between its myriads of components. Moreover, every quantum transition taking place on every star, in every galaxy, in every remote corner of the universe is splitting our local world on earth into myriads of copies of itself.

This is also an approximation, and the approximation is not perfect. The approximation is valid if decoherence leads to a set of preferred states that are not "cat states," i.e., not coherent superpositions of different pointer states (like Schrodinger's cat). This approximation is good in the limit of large systems, for which the time scale for decoherence is very short. So MWI-baroque, like CI, is falsifiable, and is in fact false. Like CI, it's false for a mesoscopic measuring device.

So my current view on this is that we should stop talking about the Copenhagen and many-worlds "interpretations" and start talking about the "Copenhagen approximation" and the "splitting approximation" (the latter meaning MWI-baroque approximation).

Allahverdyan, Balian, and Nieuwenhuizen, "A sub-ensemble theory of ideal quantum measurement processes," 2017, https://arxiv.org/abs/1303.7257

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  • $\begingroup$ Thanks for the answer and the link. I'll meditate one ant try to understand the other. But I've to say I don't completely agree with your claim that any experiment that disproves MWI-basic must also disprove CI. The fact is that a theory as much austere as the MWI-basic would never be testable, because it doesn’t say anything about what a measure is. A slightly less austere version of it, that assumed that the measures we do happen (in some way) at the single world level, is already, in my opinion, distinguishable from CI by an ideal experiment of the kind I wrote above $\endgroup$ – J.Ask Feb 12 at 21:29
  • $\begingroup$ @J.Ask: The fact is that a theory as much austere as the MWI-basic would never be testable, because it doesn’t say anything about what a measure is. I observe the energy levels of the hydrogen atom. This is a test of MWI-basic and doesn't depend in any way on the assumption that we have an inner product or a probability measure. $\endgroup$ – Ben Crowell Feb 12 at 22:15
  • $\begingroup$ But you said any experiment. A fail of this test I agree that it would dispel both MWI-basic and CI, but I wonder if there are tests capable of dispel just MWI (in its slightly less basic versions) $\endgroup$ – J.Ask Feb 12 at 22:21
  • $\begingroup$ @J.Ask: I'm claiming that CI is false. Therefore it's of no interest to look for some hypothetical test that disproves some version of MWI without disproving CI. $\endgroup$ – Ben Crowell Feb 13 at 6:10
  • $\begingroup$ It's important to make it clear, as you did in the last comment, that you're "claiming that CI is false"; and it's equally important to make it clear that your earlier statement, "So MWI-baroque... is in fact false" is also your claim, asserted without proof. $\endgroup$ – S. McGrew Feb 13 at 23:16
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This question takes for granted an idea that is common among physicists, but is also false. The idea is that there are multiple interpretations of quantum mechanics that all make the same predictions. In reality, the so-called interpretations fall into three categories.

(1) Alternatives to quantum mechanics that make different predictions, such as the pilot wave theory and spontaneous collapse theories like GRW:

https://arxiv.org/abs/1204.4325

https://arxiv.org/abs/1510.03508

(2) Quantum mechanics without any modifications, which implies the existence of a structure that can sometimes be approximated as a collection of parallel universes (the Everett interpretation). Experimental tests can distinguish between the Everett interpretation and alternatives like pilot wave and GRW:

https://arxiv.org/abs/1508.02048

(3) Theories that are too vague to work out their implications, with the result that they are not testable, such as the Copenhagen and statistical interpretations of quantum mechanics. Such theories fudge the issue of what exists in reality and so can't be used to make testable predictions since they are basically the same as saying "quantum mechanics applies, except when it doesn't".

The interpretations that are philosophical in a bad sense, i.e. - in the sense of being useless talk that obfuscates real problems, are in category (3) not in category (2).

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  • $\begingroup$ I've always considered the term "waveform collapse" to mean "and then something happens" but if pilot wave and GRW make different predictions, why wouldn't that be testable? $\endgroup$ – JimmyJames Feb 13 at 16:50
  • $\begingroup$ @JimmyJames They are testable. That was bad phrasing on my part. I've made an edit. $\endgroup$ – alanf Feb 14 at 19:16
  • $\begingroup$ Excellent summary of how to tackle the quantum foundations question - thanks. $\endgroup$ – Bruce Greetham Feb 16 at 16:15
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The very word interpretation means that it uses the same mathematics and interprets it in words differently. This means that there cannot be a difference in the calculated values in any experiment carried out in our labs, or observations fitted with the same mathematics.

It is futile to try and find either a validation or a falsification, as the mathematical structure is the same .

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    $\begingroup$ I know it calls itself an "interpretation", but it seems to me it pushes itself blatantly to the point of changing the math. If there isn't a collapse, then what prevents the worlds to interfere? Maybe I misunderstand what the interpretation is actually saying, but I read on wikipedia that even David Deutsch has proposed a test. I'm going to search the article $\endgroup$ – J.Ask Feb 12 at 17:55
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    $\begingroup$ collapse is another interpretation, although I do not like the term, as the wavefunctionis not a balloon. In the end there are measurable quantities that have to be predicted , and those are the same, because the math is the same. $\endgroup$ – anna v Feb 12 at 18:11
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    $\begingroup$ And how can we see that the math is the same? Is it so trivial that a unitary, continue and deterministic evolution of the supersystem can, somehow, be locally equivalent to the stochastic, discontinue, irreversible evolution predicted by the quantum theory? $\endgroup$ – J.Ask Feb 12 at 18:32
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    $\begingroup$ It is futile to try and find either a validation or a falsification, as the mathematical structure is the same. This doesn't seem quite right to me. The Copenhagen interpretation involves the Born rule and wavefunction collapse. These are mathematical structures, and they are not present in MWI. $\endgroup$ – Ben Crowell Feb 12 at 20:41
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    $\begingroup$ @BenCrowell If it gives different measurement predictions, it is not an interpretation but a different theory. It is the numbers predicted that finally decide.It should not be called an interpretation. $\endgroup$ – anna v Feb 13 at 5:28
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This is a philosophical question instead of a physics question, unless and until someone figures out, and then executes, a way to answer the question by experiment. So far, it seems that nobody has figured out the right experiment -- yet.

IMHO, the essence of the MWI is that really macroscopic objects can be put in a superposition of states. A convincing proof of the MWH would be to set up an interferometer that includes a truly macroscopic apparatus in one path which can only have been in a superposition of states in order to allow the interferometer to produce an interference pattern. It certainly wouldn't be easy, but might be possible.

For example, 1. Set up a Schroedinger's Cat experiment in which the cat's fate is determined by the state of photon A which enters the box from an outside source through a one-way window. The cat and the apparatus inside the box amount to a photon state detector. 2. Let photon A be one of an entangled pair; photon B be another of the same pair. By measuring the state of B, we can predict the fate of the cat without opening the box and without directly measuring the state of A. 3. Set up the source of A and B so that it emits one entangled pair every nanosecond. Put a thick etalon (~ 15 cm) in the path of B, so that when B passes through it, B's wavefunction is split into a string of pulses separated by one nanosecond between pulses.
4. Downstream from the etalon, place a Mach-Zehnder interferometer with a polarizing beamsplitter, a 90 degree polarization rotator in one arm and a non-polarizing beamsplitter; then downstream from the Mach-Zehnder interferometer, place two fast photodetectors: one for each output of the interferometer.
5. Now we need a bit of magic: we need to ensure that it's impossible to look inside the box and determine if the cat is alive or dead. Until a more practical method becomes available, let's just say that the box, along with the cat, is put into stasis and then dropped into a black hole. That way, we can't know the state of any photon B by looking inside the box. 6. Now a bit of difficult engineering: we need a way to load and remove boxes, complete with cats, once every nanosecond. Assume that can be done. 7. So now we can do an experiment in which cats' fates are entangled, one at a time, with photons B. All direct information about the cats' fates is forever unavailable.(See Black Hole Information Paradox for a possible counter-argument) At the “B” end of the experiment, we can gather information, but it's messy. Photons arrive at predictable times and we can measure their states. Every time we measure a photon, it corresponds to a cat, but we don't know if the cat is alive or dead. We might receive multiple photons at the same time, because the photon wavefunctions have been "echoed" multiple times by the etalon. It is impossible to know which photon corresponds to which cat. 8. Here is the kicker: IF each cat is actually in a mixed state, then each photon B will also be in a mixed state. The photons arriving at the photodetectors will interfere differently if each photon is in a definite state, than if each photon is in a mixed state. So analysis of the photon counts at the two detectors should tell us if the cats are in mixed states.

What I don't like about this experiment is that we lose billions of cats and boxes. Maybe a clever person will think of a way to make an electronic stand-in for the cat which can be blindly reset a billion times a second so we don't need to throw anything into a black hole. For example, maybe an apparatus could be made that detects the state of A (destroying A in the process) and then emits a photon C of the same state; and the apparatus returns to its original state. (This is necessary because of the no-hiding theorem.) We can drop photons C into a black hole, or scramble the states and identities of C using the same etalon and interferometer as was used for photons B. But if the right kind of interference is observed at the photodetectors watching the B photons, we can be sure that the apparatus standing in for the cats is indeed in a mixed state -- and pretty sure that the MWI is correct.

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    $\begingroup$ I'm having trouble understanding your example. I am interested in interference of schodinger cat states too, but it's not clear to me exactly how to see their interference fringes. But I'm not exactly sure how the intereference happens here. $\endgroup$ – Steven Sagona Feb 14 at 1:18
  • $\begingroup$ In the example I gave, the interference is indirect: it is at the "A" end, among the particles that are entangled with the cats. If the "A" particles can interfere, they must be in mixed states. Because they are entangled with the cats, the cats, too, must be in mixed states. $\endgroup$ – S. McGrew Feb 14 at 3:38

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