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Why do physicists assume this? Is it a proven fact that wave function extends to infinity or just a theory? Would it make sense if they didn't extend to infinity?

Nick V
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    Think about this: what assumption would you suggest as an alternative? – David Z Dec 27 '14 at 00:00
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    That they don't extend to infinity, that at eventually some point it hits 0. – Nick V Dec 27 '14 at 00:03
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    That's just to simplify the math. You can put everything into a perfect box every time you do a calculation and then end up with a very complicated expression for the effects on the boundaries and then run a limit for the boundaries extending towards infinity... and you would get the same result for a lot more work. It's not worth it. In practice all those terms on the boundary are vanishing very quickly, so why bother? – CuriousOne Dec 27 '14 at 00:06
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    @DavidZ Actually, I think that that is a pretty interesting question. The most obvious change to make would be to quantize probability space so that the value of the wavefunction would would always be an integer multiple of some minimum value. We would still want to require normalizability, but now the wave function would not extend to infinity in any nomalizable state. Instead, it would be exactly zero after a certain point which would imply that the rest of the wavefunction would have a slightly larger value than in the continuous case. – Geoffrey Dec 27 '14 at 07:05
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    Physical facts are never proven. Also, note that the wavefunction is not an observable, so "facts" about it are of dubious physical content. – ACuriousMind Dec 27 '14 at 17:47

5 Answers5

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Is it a proven fact that wave function extends to infinity or just a theory?

The mathematical representations of the wavefunctions extends to infinity since there are no boundary conditions to limit the distance. It is just a theory since we cannot go to infinity to test the wavefunctions.

Would it make sense if they didn't extend to infinity?

According to the mathematical models, no.

LDC3
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First of all, remember we are talking physics here, and in physics zero and so small we cannot measure it are just the same thing.

Having said that, it is not true that one always assume that the "electron wave function extends to infinity". One of the most common models is the so-called particle in a box, in which we impose the particle (electron or whatever you like) to be confined in a given volume $V$. This means that its wave function is zero outside that volume $V$.

It turns out that if you do the math for a particle confined in a box of volume $V$, with $V$ very large, all the predictions are identical to those coming from the model in which we assume the electron wave functions to be infinitely extended.

This may sound strange, but it is due to the fact that even those "infinitely extended wave functions" are not really describing the objects you would measure in the real world. They are just used as a basis (just like a basis of a vector space), describing through superposition the "real" wave functions of the objects of interest.

If you want to describe an electron travelling through space in a way that resembles what you would expect from a particle in the classical sense, you have to use a wave packet, which through an appropriate superposition of (for example) infinitely extended wave functions realizes a finitely extended wave function travelling at a given velocity (as you can see for example here)

Answer to the comments:

It's true that the particle in a box has a finite wave function, but it's also not a real potential - it requires that there is an infinite force at two locations, which can't happen.

This is true, of course. That of an infinitely high potential barrier is an approximation of how things really are. However, there are two points to notice here:

  1. It can be a very good approximation, or just a first approximation used to get a qualitative feeling of what's going on in a more complex system. Either way the formalism itself has no problem dealing with it, and just to repeat this point, physically infinitely high and very very high are the same thing.
  2. Even if there are no "walls" or things actually confining the particles, we can always assume there is some very large volume $V$ from which the particles will never escape (we just take $V$ to be larger than any characteristic property of the system), thus using a particle in a box model. This is done mainly for formal reasons: for one, it allows to count the number of states and talk of density of states.

why is it that particles must be described by finitely extended functions?

It really depends of what you mean with "particle". In the classical sense, a particle is an object moving in space with a more or less defined position. To describe something like this in quantum mechanics you need a wave packet with a wave function non vanishing only in some finite region of space. Again, that wave function can be considered to abruptly vanish at some point making the particle really confined in some region, but this is mathematically very "unnatural" to describe. It is way easier to just take the wave function to be vanishingly small outside the given region because this is, as said above, physically indistinguishable from the former case.

glS
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  • Amen to that..finely... – Žarko Tomičić Dec 27 '14 at 11:56
  • I have a little problem with this. It's true that the particle in a box has a finite wave function, but it's also not a real potential - it requires that there is an infinite force at two locations, which can't happen. I find it interesting what you say (I had the same question as OP) but I would be interested in evidence... why is it that particles must be described by finitely extended functions? – doublefelix Dec 27 '14 at 15:22
  • @user31415926535897932384626433 edited to address your points – glS Dec 27 '14 at 17:32
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It does not extend to infinity, that does not make sense. It is possible to have such a solution for a Schrödinger equation but it is not physical. You have to have a solution that is in the form of finite wave packets instead. And you can achieve this by linear superposition of many wave functions that have suitable form. If the wave packet was infinite then the probability density would be infinite too and this is unacceptable. That is why we use only square integrable functions when describing physical systems.

Ellie
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Žarko Tomičić
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  • Just fixed some typos, hope you don't mind! As for the content, it is probably rather terse for the OP I imagine, so just as a suggestion, feel free to expand a little bit more on your main points. – Ellie Dec 27 '14 at 00:17
  • But then how come physicists say that it extends to infinity? Are you saying that they are wrong or that I am misunderstanding it? It clearly says everywhere that wave functions extend to infinity. – Nick V Dec 27 '14 at 00:17
  • You are misunderstanding....same as I did. Textbooks often do not explain this very well, and neither did I obviously :-) – Žarko Tomičić Dec 27 '14 at 00:35
  • So, you can not have function that is infinite, when you integrate it over infinity it is infinity so electron would not even move but it would be everywhere. – Žarko Tomičić Dec 27 '14 at 00:37
  • @ŽarkoTomičić can you provide evidence or a source? I learned from my physics professor in Quantum that all real wave functions extend to infinity. Where is the contradiction? A function can extend to inf and still be square integrable. It is not necessarily infinite when integrated from -inf to +inf. – doublefelix Dec 27 '14 at 00:37
  • It can be in a definite space but wave functions have to be integrable over infinity...as I said before, solutions of the schro equation can be of such form but these solutions are not physical. So I see common sense is not a proof? Authority works better..Walter Greiner Quantum Mechanics An introduction, go through the first 100 pages or so.. – Žarko Tomičić Dec 27 '14 at 00:55
  • http://en.wikipedia.org/wiki/Square-integrable_function – Žarko Tomičić Dec 27 '14 at 00:56
  • http://physics.stackexchange.com/questions/149001/how-to-know-if-a-wave-function-is-physically-acceptable-solution-of-a-schr%C3%B6dinge – Žarko Tomičić Dec 27 '14 at 00:59
  • And -1 is what I get... :-) – Žarko Tomičić Dec 27 '14 at 01:05
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    Well, you do have an objectively wrong statement in your answer: that if a wavepacket is infinite (which I take to mean $\psi(x) \neq 0$ for arbitrarily large $x$) the probability is infinite. I couldn't say whether that's the reason you were downvoted, but it seems plausible. – David Z Dec 27 '14 at 01:17
  • I could hardly imagine that a wave packet could be considered as infinite, and to be a source of confusion for the man asking this question. I supposed he meant plane wave...something like sin(x)...and just tried to say that the function needs to be square integrable.. – Žarko Tomičić Dec 27 '14 at 01:45
  • So sorry about that, I just meant that it has to converge to zero very quickly, the wave packet. – Žarko Tomičić Dec 27 '14 at 02:12
  • especially in QM, "common sense" is absolutely not a proof ;) – BjornW Dec 27 '14 at 11:35
  • common sense in the sense of applying logic..not in a sense of classical way of thinking... – Žarko Tomičić Dec 27 '14 at 11:57
  • I'm the one who downvoted, and it's just because it's not true that wave functions that extend to $\infty$ aren't integrable. Take a gaussian for example. I wasn't trying to be rude, it was just because I think it was causing confusion and I didn't want the asker to misunderstand. – doublefelix Dec 27 '14 at 15:16
  • It is not a problem, man...but I believe that gaussian, although it is not zero, it converges very quickly. But ok, it could be a source of confusion. So what would you say about real electrons in space? Are they described by a simple plane wave? I was under the impression that this was a point. – Žarko Tomičić Dec 27 '14 at 15:19
  • As far as I know, our understanding is that real electrons DO extend to $\infty$, first because I have never seen any finitely-extending wave function (other than ones whose potential has infinite force at some point, which isn't physical), and second because I asked this question to my Quantum professor at NYU and he said that all wave fns extend infinitely. I don't like using authority as evidence, I'm only using it because it's all I have. I'd be interested in some solid evidence. – doublefelix Dec 27 '14 at 15:26
  • Ok Pie user :) as I commented, I agree, but must say, in infinity, value of a wave packet is very, very small. AND plane waves are not sqr intgrble...simple plane wave of the form sin(kx-wt). Nevermind. Sorry to nag this much, lets just close this topic for now. – Žarko Tomičić Dec 27 '14 at 15:31
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Take a Gaussian function as an example of a wave function. It extends to infinity, but the probability associated to finding the particle away from its mean position approaches zero rapidly, even though it is never strictly zero.

If all the universe was a stationary single electron, would you make any sense to impose any boundary to its wave function? If there was a limited-energy barrier, would it impose a strict boundary? The only case the wave function does not extend to infinity is when it is in a infinite potential well.

Ruslan
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toliveira
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If you present a model of a finite electric field of the electron, one has to determine its effective diameter. Such a model could include charged particles (electron, proton, positron, ...) and photons that have the same elementary quanta. There has to be a negative and a postives quanta. Each electron at equal potential points will contain then the same number of quanta, and thus have the identical effective diameter.

There have to be two types of quanta in the model, they could built strings respectively field lines. For electrons or antiprotons at the ends of such field lines sit negative quantum and for protons or positrons at the ends sit positive quantum.

For the emission of photons an equal number of positive and negative quantum leaves the particle which is responsible for the emission. Photons are then composite particles and there is a finite number of energy levels within an given energy interval.

It's a model only.

HolgerFiedler
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