10
$\begingroup$

When I read descriptions of the many-worlds interpretation of quantum mechanics, they say things like "every possible outcome of every event defines or exists in its own history or world", but is this really accurate? This seems to imply that the universe only split at particular moments when "events" happen. This also seems to imply that the universe only splits into a finite number of "every possible outcome".

I imagine things differently. Rather than splitting into a finite number of universes at discrete times, I imagine that at every moment the universe splits into an uncountably infinite number of universes, perhaps as described by the Schrödinger equation.

Which interpretation is right? (Or otherwise, what is the right interpretation?) If I'm right, how does one describe such a vast space mathematically? Is this a Hilbert space? If so, is it a particular subset of Hilbert space?

Improve this question
$\endgroup$
2
  • 10
    $\begingroup$ No one has any justifiable unique answers to such questions. The many-worlds interpretation isn't an actual theory of physics, an actual set of rules, ideas, or equations. It's just a vague and, when looked with any precision, meaningless and vacuous philosophical paradigm. Obviously, proper quantum mechanics doesn't imply any splitting whatsoever. Any rule when a splitting occurs is bound to be unnatural. The only "splitting" that proper QM allows is an approximate one, given by decoherence: the moment when the chances of parts of $\psi$ to "re-interfere" in the future are negligible. $\endgroup$
    – Luboš Motl
    Jul 21, 2012 at 7:11
  • $\begingroup$ @LubošMotl your statement that "Obviously, proper quantum mechanics doesn't imply any splitting whatsoever." I don't really understand in this contex. They are not explaining splitting, but the state vector reduction/collapse of the wavefunction. I agree that the many-world interpretation is pysically flawed and has no mathematical basis as a theory. However, interpretations like the Many-Minds/multi-consciousness interpretation do. Moreover, this particular theory is complete, well defined and cannot be disprooved from a physical stand-point. Of course, this does not make it correct! $\endgroup$
    – MoonKnight
    Jul 21, 2012 at 10:43

4 Answers 4

Reset to default
2
$\begingroup$

The Many Worlds interpretation is popularly misunderstood. The wave function itself contains a spectrum of universes, one corresponding to each eigenvalue for a given operator. The "splitting" of the "many worlds" is represented by the time evolution of the wave function described by the Schrodinger equation. As Lubos mentions above, these "universes" only become separate through decoherence.

Consider, for example, a wave function in the position-basis given by a delta-function at x=0. This represents one universe. Now time-evolve the wave function using the schrodinger equation. The delta-function has now spread-out a bit. It is peaked at x=0, but has non-zero values at x=+1 and x=-1. This represents the existence of universes in which the position of the particle is at x=0, x=+1, and x=-1. In some sense there are "more" universes at x=0 than at x=+-1, because the wave function is more highly peaked at x=0. This is where some of the difficulty in the Many Worlds interpretation comes in: what ontology to use to describe the "splitting", "how many universes" are at x=0 vs x=+-1, and so on. The main point I want to make is that the "splitting" is just an interpretation of what is happening with the evolution of the wave function according to the schrodinger equation. Nothing "more" is actually happening. You model the "splitting" using the tried-and-true schrodinger evolution of the wave function.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ You imply that there is a spectrum (a countable infinity) of "possible universes". But is it actually a continuum (an uncountable infinity) of "possible universes"? Can the delta-function have non-zero values at locations everywhere between 0 and +/-1? Or maybe a better example (since I don't understand the delta-function), in the double-slit experiment, can't a particular photon hit the detector plane at any point on the plane? (<- thus uncountable infinite possible universes) $\endgroup$
    – John Berryman
    Jul 21, 2012 at 13:04
  • 1
    $\begingroup$ @John Berryman The word 'spectrum' does not imply a countable infinity. It is a continuum representing an uncountably infinite number of universes in the example I gave. You can think of a delta function like a very narrow spike. The schrodinger equation time-evolves a narrow spike into a wider and wider gaussian shape. In the example, in order to keep things simple, I approximated this as {-1,0,1) (a very rough approximation, but serves to illustrate the point). $\endgroup$
    – user1247
    Jul 21, 2012 at 19:05
  • $\begingroup$ When we analyze the electric current in an ordinary circuit using classical physics, we can if we like use a Fourier analysis, and that is a useful thing to do. It does not follow that there is a spectrum of universes, each with a particular value of the frequency. $\endgroup$
    – Andrew Steane
    Jun 28, 2021 at 8:19
  • $\begingroup$ @AndrewSteane, I see an assertion here, not an argument (other, perhaps, an argument from incredulity). From an Everettian perspective, fourier decomposition of a classical wave is certainly suggestive of a multiverse interpretation, however you don't hear about this much because in classical physics there is no entanglement and tensor product space. $\endgroup$
    – user1247
    Jun 28, 2021 at 15:15
1
$\begingroup$

Many worlders won't tell you this dirty little secret but how often splitting happens, and how many worlds there are, depends upon the choice of coarse graining, and the coarse graining resolution. No, it's not possible to ramp up the coarse graining all the way to the finest levels because a decoherence/coherence threshold would be crossed. And no, there is no canonical coarse graining either.

The preferred basis depends upon the environment. Always. What is the preferred basis for a closed self-contained universe?

Improve this answer
$\endgroup$
4
  • 2
    $\begingroup$ A more accurate answer than "The preferred basis depends upon the environment. Always." would be that the supporters of the MWI haven't yet described any other mechanism by which it could arise, just as they haven't yet shown how the Born rule would emerge even for a finite system. $\endgroup$
    – Niel de Beaudrap
    Jul 21, 2012 at 11:52
  • $\begingroup$ Nice answer. I find that many worlders do actively sweep this under the rug as often as possible. You could improve it by attempting some calculation at some level - for instance merely assuming that there is only one observer - you - in the Universe, how many universes split per second? I can see the coarse graining being undefined even for that! $\endgroup$
    – Tom Andersen
    Jan 2, 2015 at 16:02
  • $\begingroup$ Many worlders won't tell you this dirty little secret but how often splitting happens, and how many worlds there are, depends upon the choice of coarse graining, and the coarse graining resolution. This shows a lack of understanding of MWI. There is no discrete branching in MWI. MWI simply says there's unitary evolution, and that's all, folks. The image of discrete branching comes from popularizations. $\endgroup$
    – user4552
    Dec 28, 2016 at 19:32
  • $\begingroup$ @user4552 "unitary evolution, and that's all, folks" is what is claimed concerning the evolution, but a theory must connect that evolution to what is experienced by the human community, namely that the cat either lived or died, the Geiger counter clicked, or did not, etc. Also that there is such a thing as a "cat" and a "counter". All this requires lengthy analysis of what the state vector means, and when can a quantum amplitude be squared, and what is a probability, etc. It is here that the whole thing becomes unwieldy, not simple, and much like unobservable multiple universes. $\endgroup$
    – Andrew Steane
    Jun 28, 2021 at 8:29
1
$\begingroup$

An interesting many-worlds model is the "many interacting worlds" (MIW) in which there is no "splitting" (creation or destruction of worlds) at all. Rather, worlds influence each other when they are near each other in configuration space (their constituent particles are all nearly at the same position and orientation). Each world is strictly Newtonian, except for this interaction which can (amazingly to me) reconstruct the well-known Schordinger quantum phenomenon, including the uncertainty principle and interference patterns, etc. See http://dx.doi.org/10.1103/PhysRevX.4.041013

A conceptual burden one must bear in holding onto this MIW model is the unknown inter-world interaction force, which must be repulsive and satisfy certain constraints but is otherwise open to be freely chosen.

On the other hand, there is no need to deal with "collapse," so a matter of taste if this burden is a net win.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Welcome on Physics SE :) Please include the full citation of the article you have provided a link for - journals do sometimes change their URLs and then this will be very helpful. Also, it helps if you explain abreviations you are using :) $\endgroup$
    – Sanya
    Dec 15, 2016 at 16:48
  • 1
    $\begingroup$ Sanya, I've replaced the URL with the (theoretically "permanent") DOI link and defined the meaning of MIW at first use. $\endgroup$
    – RBSmith
    Dec 27, 2016 at 15:13
1
$\begingroup$

The answer is 'it depends'.😊 It depends on what you mean by "world". Personally, I prefer the word "timeline" , but it makes no difference. And before I give my answer I should state that Hugh Everett, the formulator of MWI, reckoned there were an uncountable infinity of worlds (or timelines). Unfortunately he did not state his reasoning for this, since it was just a remark given in passing, in an answer to a conference question, c.1962.

Everett did state in his thesis that the branching of an observer was limited by the observer's entropy or memory capacity. Since these are finite quantities we would expect the branching to be finite, although usually very large. (Unless of course the universe is infinitely large, in which case its entropy is as well, but I don't think Everett was referring to this.)

There is also Boltzmann's S=k log W to consider, which again suggest the number of worlds (now each considered as a microstate) is finite. Although the infra red divergences suggest it may be infinite, since the number of real photons emitted during Bremsstrahlung is infinite - although only a finite number can be measured..

So, on balance, I would say the branching is finite, if we think in terms of the number of possible distinct configurations of any system, including any observer such as you and me. But the number of branches of the totality of existence could well be infinite.

How do we model splitting mathematically? After a measurement the observer can be deemed to be split if the overlap (which has a precise technical definition) between the different observer states is negligible.

Improve this answer
$\endgroup$
3
  • $\begingroup$ The upvote's for the 1st paragraph: Although, by itself, it's pretty shaky as a reference, it gives a timespan around which I can dig further. $\endgroup$
    – Edouard
    Jun 26, 2021 at 17:14
  • $\begingroup$ Re my previous comment: "Everett" and "'1961" OR "1962" OR "1963" brings up over 82,000 references, although one of them is to a recording (unaccessible to me) of a conference. $\endgroup$
    – Edouard
    Jun 26, 2021 at 17:20
  • 1
    $\begingroup$ It was at a private conference of leading physicists, on the topic of the measurement problem, October 1962, at Xavier University, Cincinnati, Ohio, that Everett agreed with a comment by Podolsky, that MWI implied a "non-denumerable infinity of worlds". Peter Byrne details it in his Everett biography, The Many Worlds of Hugh Everett III. $\endgroup$
    – Michael C Price
    Jun 27, 2021 at 20:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.