Fascinating - a well-posed question with non-trivial implications. Thank you!

(my answer is couched in terms of my QM knowledge that goes no further than a stock BSc in Physics from 20 years ago)

Half of me favours 1.

Namely that the wave function is in some manner an incompete representation of the system. The integral in an unbounded and ever-lasting universe is a fiction - there must be something missing from the model of a quantum system that is not yet understood well.

Half of me favours 3 / 4. They are conceptually similar - a universe that is bounded in time has a finite extent. The idea that the boundaries of the universe inevitably create its contents is satisfying in more than a naïve big-bang kind of way.

Depending on how I observe myself I can be found in one of the two above states.

The other half of me :) also favours 5. As it removes all responsibility of having to argue what it is.

James,

Thanks for a very interesting answer

(This was a bit of a counterpart and kindred question to my other question about the wave function of the universe, by the way.)

I believe that we can only truly ever answer by analysing the wider context first - and the question is likely unanswerable if its wider context is not elucidated first.

The wider context is the relationship of math to reality. i.e., we need to understand whether math is only a tool - a human-made descriptive language depicting what we perceive, or alternatively whether it's a reflection (in imperfect human terms) of something far deeper - an ultimate reality which manifests itself in our perceived apparent reality, and which at bottom is purely mathematical and human independent.

Some theoretical physicists espouse the former view, and others the latter.

1- Let's take the latter view first - math is (a human representation of) ultimate reality.

This would explain in a fell swoop, without further ado, a number of puzzles - such as why the very baffling Bell's inequality has been demonstrated in the lab, why e=mc² actually works (the conformity of nature to math used to baffle Einstein), etc. etc.

This would also immediately invalidate answer Nb. 1.

Now in that case it would mean that the only reality of things lies in fact ... in their mathematical wave functions (and physical manifestations of reality is just the way wave functions happen to precipitate and build the material world.)

This leads to a host of consequences - such as, reality demonstrably cannot manifest itself within an infinite universe (It takes a bit of calculations, which I explored in my thesis , but also in layman's terms in its WIP popular science rendition (the treatment of which is a bit longish, but compelling - even experimentally so, by testing some of the downstream physical consequences.)

Combining this with considerations of the wave function of the Universe, which appears to be 'open' , this would seem to demonstrate that the universe is actually made up of an infinite stack of finite universes (and there are many quite distinct ways to accommodate the whole stack in a quantized multiverse - none of which, I hasten to add, being "turtles all the way down" :-)

Exploring where that leads is rather interesting.

2- Now let's take the other view - math is just a language, nothing more profound.

This leads - to begin with - to many questions on the nature of reality still being unanswered (such as, again, Bell's theorem)

It also leads to a situation where all the reliable pegs in the ground vanish - in math formulas and equations, we must now arbitrarily pick and choose what applies and what does not - what is just an artifact, and what we should deem valid

(Incidentally, this approach has historically proved sterile. Two examples: 1- Einstein was not the first who wrote out the equation leading to e=mc² - but he was the first to assign a physical reality to the leftover terms - which previous physicists had just dismissed as mere math artifacts without a further thought. 2- The same is true of so-called imaginary numbers (square root of -1) which have been proved since, in many very real world engineering applications, to be extremely rooted in reality .... and so on.)

So my answer would end up being 2, 3, 4 and 5 together.

In that context 5 means that the consequence of accepting any explanation other than 1 unavoidably leads to further, very interesting features of reality (which I also explore in the above-cited text): 'real' i.e. finite non-empty universes may exist, for instance, but seem to be open 'ad infinitum' - nature's way, mayhap, of preserving overall spacetime infinity?

Your thoughts? ....

Hi Chris,

I would choose a completely different answer: a plane wave is a basis function that is used to build up physically meaningful wave functions, i.e., wave packets: Even in infinite space you can build square integrable functions by the superposition of non-normalizable plane waves, that is the essence of Fourier theory.

Best regards

Herbert

Hi Herbert,

Not sure I follow you here -

1- If you're talking about expanding into a Fourier series, the conditions for doing so are not met:

1a- One of the conditions for legally expanding into a Fourier series is precisely that the function be integrable. By expanding a priori that way you are in fact positing as a presumptive hypothesis precisely the result you seek .

1b -Furthermore - the separate Dirichlet criteria are not necessarily satisfied everywhere within the spacetime considered, and you'd be a priori able to expand the function in spacetime subsets only, which kind of defeats the purpose.....

2- If you're talking about doing a Fourier transform, the non integrable parts do not go away under the Fourier transform and the problem of non integrability remains

Or did I miss something?

Yes. you miss something.

I was talking about Fourier transformation, _not_ about Fourier series. Furthermore, I did _not_ fall into the trap described in 1a. I was not claiming that the plane wave has a Fourier transformation (it has in terms of a delta distribution, but that is a different story :)) but that square integrable functions, i.e. physically meaningful wave functions have a Fourier integral representation .... which is perfectly correct. Point 2 is simply wrong. Please have a look at http://en.wikipedia.org/wiki/Parseval%27s_theorem. I suggest that you look at any functional analysis textbook that treat the Fourier transformation. Sorry for sounding a bit harsh, but the math is not to be discussed, it is proven stuff for decades.

Best regards

Herbert

Herbert,

I've studied math extensively as my profile probably shows, and I do not necessarily rely on Wikipedia for scientific information. i still have all my textbooks on my shelves, by the way. Being extremely good at maths is one of the conditions to win admission to a so-called 'Grande Ecole', for instance.

I do not understand quite your second response, so bear with me:

1- My point 1a has nothing to do with Fourier transforms, it has to do with expansion into a Fourier series, which is not the same thing at all. So I'm not sure why your reply to point 1a draws in Fourier transforms?

My point was that to expand a function f(x) into a Fourier series, f(x) needs to be integrable in the first place: you cannot implicitly assume a property to demonstrate that property.

2- Or you are talking of other wave functions ?

Wave functions, even complex ones, are usually all perfectly integrable EXCEPT with infinite integration boundaries, which is the point I make, and which you do not address at all.

The non integrability over infinite integration boundaries arises because of the existence of cos(ө)-like terms which takes on determinate, finite values for any finite ө but are indeterminate with ө infinite (to put it in simpler terms, cos(∞) is meaningless: my point EXACTLY.

Did you understand that that was my point ?

3- Now point 2 above - transformation of a non integrable function by a Fourier transform - just displaces the problem without solving it.

.

When you Fourier transform , the infinities and the trigonometric terms remain although they appear differently. The cos(∞) like terms remain, and remain meaningless.

So i do not understand what you mean, and nothing in the wikipedia link you indicate gainsays my points

As for your slightly ungrammatical comment "the math is not to be discussed, it is proven stuff for decades" it would be clearer if your exposition of exactly what you meant were precise rather than couched in generalities - if you explained exactly how you apply the Fourier transform and how it helps.

I do not believe you understood my question at all, perhaps you should have read the question. The integrability of wavefunctions is not at issue (it is obvious that they are integrable, otherwise the associated particles or entities would not have a probability of existing, as they do, in the material world), but the whole point is the consequences of the cos(∞) terms and what they mean for interpreting the reality, or otherwise, of an infinite universe.

Rather than speaking in a vacuum, Herbert, here's a simple exercise for you:

Take the very simple wave function ψ of a free electron

Calculate the numerical value of ∫ψψ* in infinite space

Thanks

PS - Simplest reduced ultimately equivalent problem :

Calculate ∫sin(x) dx

with one or two infinite integration boundaries set at ∞.

Result is : indeterminate

QED

When you say that the WF of a free electron isn't integrable in an infinite space, I guess you are referring to a single sinus (or imaginary exponent, let's keep with real part for simplicity) wave out to infinity? That wavefunction is not only a free electron, it is a free electron with an exactly defined momentum (the frequency of the sinus). When people talk about "free electron" it is normally not required that it is in an eigenstate of momentum, which may be the cause of confusion above. Not sure.

A real physical electron will never have perfectly defined momentum though but will be a superposition of sinuses that form a wave packet (localised in space), which is perfectly integrable as it approaches zero for long distance. You will have an equally integrable distribution in momentum from the fourier transform. As you continue to measure the momentum more and more exactly, you will spread out the wave packet (according to uncertainty principle), and the amplitude of the wave will decrease as the normalisation (probability) is conserved.

So in the limit of deciding momentum exactly, you will get a pure sin wave with amplitude approaching zero (with non-zero amplitude, it would correspond to infinitely many electrons spread out in the infinite space), but as it is not possible to measure momentum exactly, this limit will never be reaching in reality. So in that way, it can be classified as a mathematical artifact, as you suggested. In other words (as you were talking about the "wave function of the universe") you will never see such an electron in any (reasonable) universe, so we don't need to bother. :)

They are still very useful as mathematical tools (and in teaching QM!) as eigenfunctions of momentum in both free and bound particles, but you will never see a physical particle in that exact state. If anything, the fact that you cannot normalise a free electron with exact momentum in an infinite space is a sign that it is impossible to measure momentum exactly.

Not sure if I addressed your problem. Hope it helped.

Christoffer,

Yes, it helps in part, thanks, because you understand both the problem and the fact that the answer is an issue of interpretation - the original point exactly.

To reword in simple terms, the issue can be reworded in a simple scenario: if you created a finite empty Universe of any size, within that Universe the integration would be determinate and the numercial result would be 1, meaning that the likelihood of the electron being somewhere in the Universe is 1 (of course however, in a real Universe however the electron can never have undergone full decoherence and can never be wholly 'free', but that's another tale.)

Now if you created instead an infinite universe, then the numerical value of the integration within those boundaries is *unascertainable* - regardless of momentum by the way.

Now as you correctly hint, the issue - as so often is the case in quantum physics - is one of interpretation: how do you interpret this unascertainability ?

You may interpret as a mathematical artifact, which is legitimate, and many will choose that option.

However I'm more in keeping with Tegmark et al., who believe that math represent a far deeper reality. That was my question exactly: how do you interpret.

As for the comments by HH, upon reflection I believe they arose because of a vocabulary misunderstanding. In my language, a a)-perfectly integrable function may be called b)-not integrable within boundaries if a numerical result within these specific boundaries is inherently not ascertainable.

I believe that HH may have confused or coalesced a) and b). I am changing the vocabulary in my text from 'non integrable' to ' unascertainable' to forestall the possibility of confusion.

Dear Chris,

The non-integrability of a sinus function over infinity does not rule out infinite universes, in space or time. That is not a matter of interpretation. No physical particle will have that wavefunction, so it is not a problem for an infinite universe.

It is a very useful mathematical tool in calculations, and there are many (equivalent) ways of interpreting the role of this mathematical tool, but no interpretation of this mathematical tool changes the physics of the calculation. No matter how you interpret the mathematics, they will always give you the same numbers out for your observables. In principle, you can do any calculation on any physical state without ever using that function.

Maybe the part you missed when you posed the original question was that no "real life" particle will have a sinus over the entire real axis as wavefunction, infinite universe or not. A real life free electron is always described by a localised (and thus normalisable) wavefunction.

Hope I hit a bit closer to the heart of your question this time. :)

Christoffer,

Not sure I agree - if you actually do have an infinite universe, and if you have any object in it, with associated wave function ψ , then ∫ψψ*dσ should be equal to 1 , because the probability of this object being somewhere, anywhere, in that Universe is 1.

If the math tells you however that ∫ψψ*dσ is meaningless - as it does with integration boundaries set at ∞ , then you have a non-trivial interpretation problem.

Besides - nothing prevents a wavefunction from spreading out - as it actually does, because in an isotropic universe the free object may theoretically be found anywhere - there are no forbidden zones, and every bit of space is equivalent in terms of the likelihood of finding your object - equally "likely".

You may more or less arbitrarily decree that it means this or that, at least the way I see it, but that's not very satisfactory. There are quite a number of distinct possible interpretations that are compatible with the math, as they must, but they remain interpretations.

The math tells you, in effect, that the probability of finding the object anywhere in the universe is mathematically meaningless. You could interpret that by saying that an infinite Universe is impossible, or that an infinite Universe is necessarily anisotropic, or that matter 'dissipates' in an infinite Universe, or any number of other more or less exotic or even fanciful interpretations: my question exactly.

As an example :your statement (quite correct in our Universe) " A real life free electron is always described by a localised (and thus normalisable) wavefunction " seems to be saying that either an infinite Universe is anisotropic, or to posit a priori that an infinite Universe is 'real life' and that hence there are 'real life' objects within it - which is not obvious, and, again, my question exactly.

Can a "real life" electron exist at all in an infinite Universe ? If yes - why ? If no, why? MQE ...

Hi Chris,

it seems that you argue about points that I did not discuss. I do not say that a plane wave is integrable or even square integrable. To avoid such misunderstandings:

1) There is nothing like a free electron: that is only an idealization in a model. In real world there are always interactions.

2) In a real world situation you have to "prepare your wave function" which due to point 1) is always a wave packet

\psi(x) = \int \phi(k) \exp(i k x) d^3 k

and \phi is square integrable. This entails that the wave packet is square integrable. You may think of that as a superposition of plane waves. Thus the wave packet is usually confined to a certain region in k-space or in x-space.

This, by the way, is the answer to your question "Can a "real life" electron exist at all in an infinite Universe ?", since for a real electron wave packet, there is only a limited region that it can interact with due to relativity and causality reasons. Thus, for a "real life" electron, the question of an infinite universe is meaningless.

A lot of the confusion in the present discussion stems from taking into account only idealized models and interpreting them in a sense that neglects the model character.

If you know what you are doing mathematically, such idealizations can help, and Dirac delta distributions and similar "strange" (from a classical function point of view) things can be given a well-defined mathematical meaning.

Be certain that there are many people who have the textbooks not only on the shelve but even read them, including me.

This is the simple story. And also Christoffer Flensburgs answers point into the same direction, it seems to me....

Best regards

Herbert

Chris,

Hmm, I am now starting to think that you do not use the term "free particle" in the same way that I do. In my education and research, "free particle" has always meant a particle that is not feeling any forces, ie moving in a flat potential. So if I go out to open space and fire an electron out into nowhere, that is a (almost, it will still feel some forces from distant stars, radiation etc) free electron. This free electron will still be described by a localised wavefunction. I'm getting the impression that you are talking about something else?

So I will just ignore the word "free particle" and try to address what you are saying. :)

You start from a sinus wavefunction spread out over a infinite universe, and correctly point out that this wavefunction cannot be normalised to any finite value. What I said in my previous reply, is that this mathematical exercise is not relevant, as no single physical particle will have that wavefunction. There is no particle that is spread out evenly over an infinite universe. You could imagine an infinite number of particles spread out in this way though, and they could have such a wavefunction at that point.

If you take an infinite universe, and place a particle in it and let the wavefunction diffuse, it will spread out as you wait (to arbitrarily large width if you wait long enough), but at any point in time it will still be localised. It will never diffuse to infinite width, no matter how long you wait.

For comparison, I can take a finite universe and make up a non-normalisable wavefunction that no physical wavefunction would ever have (say for example 1/r^3 around a point). Would that be a proof that no finite universe can exist?

Bottom line is that there is no requirement for a universe to be able to hold a single particle spread out evenly in space, so such a particle cannot be used to disprove a universes existence.

Dear Kristoffer,

We're at last getting somewhere !

Quote

but at any point in time it will still be localised. It will never diffuse to infinite width, no matter how long you wait.

Unquote

true given speed of light constraints. (You're right that it is unavoidable that we use idealized representations - in the real world you never can have full decoherence. )

My question in essence is: can an infinite Universe exist in reality ? What you just proved is that you can't answer this through analysing the probability of presence of particles introduced somewhere within such a universe. I'm not entirely sure it's the whole answer though.

Herbert,

Fair enough - Quote for a "real life" electron, the question of an infinite universe is meaningless Unquote - this is correct.

It means that matter does not, in finite time, 'see' or 'recognize' (for lack of better words) an infinite universe.

By the sheer fact that an object is located somewhere at some point in time, the Universe becomes anisotropic for that object in finite time . This is perfectly correct (and is also a consequence of speed of light limitations.)

Unfortunately it does not solve the issue of whether an infinite universe is possible at all. I was trying to explore different modes of object introduction within or into an infinite, empty universe (such as by creation of real particles from quantum foam, for instance) but there seem to be intractable issues there as well .....

I also have this lingering suspicion that we are actually missing something. For starters, the laws of physics (if not of maths) may be quite different in an arbitrarily different universe.

Chris

I want to answer the question "can an infinite Universe exist in reality ?" in relation to our universe using logic rather than any mathematical rigour.

We all know our universe is expanding, it can't be infinite if it is expanding! can infinity expand ? Of course not otherwise it would not be called infinity in the fisrt place! Our universe is only expanding because it is finite. Does this make any sense to you Chris?

Issam, I'm actually talking of another universe, not ours, which is finite.

There are several ways that another, actually infinite universe could exist and still not interfere with ours (which thus would be separate, as it were, from infinity itself ...)

What I'm trying to apprehend is whether an infinite universe - within a larger multiverse or even metaverse - is theoretically possible at all.

Most signs seem to point to a no answer, but not quite conclusively.

Chris, I understand perfectly what you are trying to find out. If my argument is correct then your infinite universe, according to my logic, can't be an expanding universe and therefore must be created by an entirely different process than the Big Bang!

Issam,

you are exactly right.

This could be one of the very things that we all missed in our above argument - that in fact , in an infinite Universe, if it exists or can exist, then infinite time is already here !

Then we'd have to revisit all of our arguments ...... with an integral that has now become:

∫ψψ*dσ

with integration boundaries in time respectively at - ∞ and t (t= say, now), and we're back where we started ....

I am sorry to dissapoint you Chris and I am glad you reached the same conclusion without me having to explain it!!

Ok Chris, good that you feel some conclusion on the evenly distributed single particle. Let me just point out that the speed of light is not necessary, but it is rather a consequence of conservation of probability, which is an essential property already of the Schrodinger equation (so non-relativistic, no speed of light). A physical particle (which is normalised to a probability of 1) that evolve in time according to the schrodinger equation will conserve it's probability, so that it will still be a probability of 1 to find it in the future. As you pointed out, this property rules out that the evolution in time according to the schrodinger equation will take the particle to a non-normalisable wave function. A similar argument can be made in more modern quantum field theory (ie relativistic quantum mechanics), and it would not depend on a limited speed of light.

I think I'll drop out of the discussion at this point, but let me just leave two remarks first:

- The integral of a wavefunction over time does not correspond to any physical observable, and thus do not have to be finite. Compare to the integral of the absolute square of the wavefunction which corresponds to the probability to find the particle anywhere and is conserved in classical quantum mechanics. (more advanced particle physics breaks the conservation of probability as new particles can be emitted and absorbed, but that is a different story.)

- An infinite universe can still expand (in the sense that the distance between any two points increase over time). If this is not clear, read up a bit on your general relativity and cosmology. ;) If you actually look at the spaces and metrics in general relativity, it is very straight forward. Otherwise imagine a big elastic carpet or something with an ant on it. If you stretch it, the ant will see everything moving away from each other, and the ant does not have to see the edge of the carpet to do this observation. Then it is a small step to an infinite carpet that is still being stretched at every point. The "there is no space to expand" argument doesn't make much sense, as a two times as big infinite universe is still an equally infinitely large universe, in the same way that there are just as many integer numbers as there are even numbers. It can seem counterintuitive at first, but I will let someone else explain the details of this.

good luck and have fun with the discussion. :)

Totally agree Christoffer

Chris

When I said an infinity can't expand, it is not because there is no space to expand into, it is because it defy the meaning of infinity!

Infinity isn't something we can see or experience, it baffles us easily. Infinity is a concept not a number, so we can't do arithmetic with it. We use the word or symbol for infinity because we have no other way to describe the concept of infinity, since it is beyond our perceptions.

Consider the set of all positive even numbers, or 2, 4, 6, 8, .... There is an infinite amount of these (call them elements of the set), since we can keep on counting them forever. Now think of the regular counting numbers, or natural numbers: 1, 2, 3, 4, .... It would seem that there are twice as many numbers in this set, so it would be twice as big as infinity. However, we can prove that there are the same number of elements in each set. Take each of the counting numbers and double them. You get the positive even integers. So for every element in the set of natural numbers, double it to get its corresponding element in the set of positive even integers. This is what is called a one-to-one correspondence, and if you can draw a one-to-one correspondence between the elements of two sets, then they have the same size.

My understanding is that when you have an infinite universe, conceptually you can't have a larger infinite universe, it will be same size. This is what is meant by infinity. Hence you can't have an expanding infinite universe. I am not an expert in the field and so could be wrong and I very much welcome any expert opinion on this very interesting subject.

ok, so despite promising to drop out, and this being quite far from the original question, let me briefly address the expanding infinite universe. :)

I am by no means an expert, but I have had some courses on general relativity and gravity, so I think I can clear this up.

An expanding universe is not a matter of adding new space to the universe, but a matter of increasing the distance between what is already there. Compare to the ant on the elastic carpet that we pull out to cover a larger area rather than adding more carpets at the side. Let's compare how to observe the two:

- To observe an extra carpet being added at the side, you would have to be aware of the border, and see how it got moved away. If you were standing in the middle of the first carpet and couldn't see all the way to the edge, there is no way for you to know that the carpet got extended. It is also hard to imagine how this would work on an infinite carpet without borders.

- To observe the carpet being pulled to a larger area, you don't need to see the edge. You will notice that the 1cm distance to your neighbour after a while is 2cm. This observation does not need the edge to be there, and it is very possible for the ant to say that the carpet is expanding without knowing if there is an edge or not.

So while the points in space are the same, the distance between any two points increase. That is, the metric on the manifold is changing.

In our world, the expanding universe is observed through the redshift of distant objects. See Hubbles law: http://en.wikipedia.org/wiki/Hubble%27s_law That is, distant stellar objects are moving away from us, and the more distant, the faster they move away. Which is a clear sign of an expanding universe. Notice that we do not observe any "edge of the universe" moving away from us, we only observe what is in the universe move away from us (and from each other). There is an edge of the visible universe, due to the limited speed of light, but that does not imply that the universe ends outside our vision.

There are a lot of argument for and against infinite universes, and I do not know what the current consensus among cosmologists is (infinite, finite, don't know), but I know that you need some knowledge about general relativity to understand them. :)

I agree Christoffer,

But we can't reach an infinite Universe starting from a finite one, it just does not work, on all kinds of grounds (one being that you'd need infinite time).

( Generally speaking by the way, we're bad at apprehending infinity. There are ongoing controversies in math as to whether some infinites are bigger than others (I think personally the ad hoc literature proves that this is the case, but this is far from a consensus view)

There are therefore 3 alternatives:

1- There is, somewhere, a pre existing infinite Universe

or

2- We have to revise the concept of infinity, and an infinite Universe is not accommodating, or 'housing', its infinity the way we'd think. My work points squarely to §2

3- or physical infinities do not exist in the Universe/metaverse, and only exist in math

Dear Chritoffer

It is nice of you to drop in! I was expecting it and your comment is very welcome.

I totally agree that the universe is expanding only because its constituents are moving apart. Now if the boundary of the universe is not changing does that not make the universe finite?

According to general relativity, space can expand faster than the speed of light, although we can view only a small portion of the universe due to the limitation imposed by light speed. Since we cannot observe space beyond the limitations of light (or any electromagnetic radiation), it is uncertain whether the size of the universe is finite or infinite. However, I think we all agree that the observable universe is finite.

The expansion we see in the (observable) universe is an expansion in a( finite) universe.

@ Chris, I totally agree with you that you can't reach an infinite universe from a finite one because it requires infinite time and it is difficult to apprehend infinity.

The big question now is how does an infinite universe form? Or if there is a pre-existing infinite universe how did it form? The process must be incredibly fast if we are to avoid infinite time. Again I am not an expert and these questions are not my turf!

Chris

Almost a year ago you posted this question but I got to it googling for an answer in RG on a question on the meaning of the wavefunction.

I went to the basics and it says there in Born's postulates: a wave-function must be finite ( so that its square integratable), meaning that a free electron can only be described by a wave-packet but not a wave with a single wavenumber