# Can quantum particles spread out over large distances?

While trying to understand quantum mechanics I was wondering about this: since free quantum particles naturally spread out until the wave function collapses (if I understand correctly); does there exist an abundance of extremely spread out particles in outer space where interaction with other particles is rare or do the particles collapse before this happens?

To be more specific:

1. Does it occur often that particles in outer space reach macroscopic spreads of let's say multiple kilometers? Or does quantum decoherence occur before this happens? With spreading I mean $\sigma_x$ or the uncertainty in position.
2. If a particle reaches such a big spread, does this accelerate or inhibit wave function collapse?

A spread out particle covers more area making it interact with more matter but at the same the probability amplitude per area decreases making the chance of interaction smaller.

• I'm sorry if my wording is confusing. I mean the spread in the wave function itself. I used 'wave function' and 'particle' interchangeably here. – user3502079 Sep 6 '17 at 21:13
• I did what I normally do, I rushed reading it. Apologies. I think you are talking about something that is better deal with through field theory But that's a pure naive guess on my part. Anyway, best of luck with your question. – user167453 Sep 6 '17 at 23:20
• I would strongly recommend you read this question and answer about wave function collapse. It should help clear up some misconceptions. – StephenG Sep 13 '17 at 3:05

The quantum wavepacket spreading is underlain by the dispersiveness of the free
Schrödinger equation, which is basically a diffusion equation of sorts. For simplicity, let's consider an electron coming from outer space, and stick to one dimension, with m and ħ set equal to one—we'll reinstate them later. Also, let's not use "collapse" (reserved for observation) to denote wavepacket spreading.

In momentum space, there is no counterintuitive stuff: momenta are conserved and the wave packet profile in momentum space preserves its shape upon propagation: no forces.

Restricting attention to one dimension, the solution to the Schrödinger equation satisfying the Gaussian initial condition starting at the origin with minimal space uncertainty, $$u(x,0) =\int \frac{dk}{2\sqrt{\pi}}e^{ikx}e^{-(k-\bar{k})^2/4}= e^{−x^2+i\bar{k} x},$$
is seen to be \begin{align} u(x,t) &= \frac{1}{\sqrt{1 + 2it}} e^{-\frac{1}{4}\bar{k}^2} ~ e^{-\frac{1}{1 + 2it}\left(x - \frac{i\bar{k}}{2}\right)^2}\\ &= \frac{1}{\sqrt{1 + 2it}} e^{-\frac{1}{1 + 4t^2}(x - \bar{k}t)^2}~ e^{i \frac{1}{1 + 4t^2}\left((\bar{k} + 2tx)x - \frac{1}{2}t\bar{k}^2\right)} \\ &= e^{i\bar{k}x-it\bar{k}^2/2}~~\frac{e^{-\frac{1-2it}{1 + 4t^2}(x - \bar{k}t)^2}~ }{\sqrt{1 + 2it}} ~. \end{align} The leading part is the plane wave corresponding to the "center" of the wave packet, and the trailing part contains the real exponent which demarcates the envelope, so the probability density $|u|^2\sqrt{2/\pi}$ propagates "classically" with group velocity $\bar k$, as it rapidly spreads, $$|u(x,t)|^2 = \frac{1}{\sqrt{1+4t^2}}~e^{-\frac{2(x-\bar{k}t)^2}{1+4t^2}}~.$$ A bit too rapidly...

The width here, $\sqrt{1+4t^2} \to 2t$, after reinstating the naturalized constants, amounts to $$\Delta x=A\sqrt{1+(\hbar t/mA^2)^2},$$ for A the initial width. If we took it to be ångströms, plugging in values for the electron mass, we see a spread to kilometers in a milisec. That is, the normalized probability envelope above has all-but dissolved to a delocalized flapjack, and the electron consists of plane wave components, which is why distant cosmic rays are modeled by plane waves. (1. Yes, multi-multi-multiple kilometers, before detection.)

The electron has not disappeared to nothingness (2. It's all there, so it will keep coming and be detected, ultimately, with the same probability), it is that its precise location has quantum-diffused to all over the place. With some probability, these electrons will come to your detector, normally modeled by plane waves, and will have mostly momenta/velocities close to $\bar k$, as posited; this distribution has not changed. The consequent spread in detection times, non-relativistically, will be $$\Delta t = \frac{\Delta x} {\bar{v}}= \frac{t}{A\bar{k}}~.$$ Where did they come from (in x; the momenta in 3D specify the direction)? Who knows.

Quantification may be fiendishly conducive to errors by dozens of orders of magnitude, however. See Tzara 1988 for reconciliation/reassurance that the short ultra relativistic neutrino burst of the SN1987A supernova did not violate the above standard wavepacket spreading notions, as erroneously claimed: one must compute the spreading in the wavepacket's rest frame and then transform to the terrestrial detector frame! Phew...

Numbers : Prompted by the question, let's summarize the basic numerics of the spread. Define a characteristic time $$\tau=A^2 \frac{m }{\hbar} , \qquad \Longrightarrow \qquad \Delta x= A \frac{t}{\tau} .$$ For an electron and A ~ ångström, $\tau= 10^{-16} s$, whence the above spread to a kilometer in miliseconds. But for an iron nucleus and A in microns, we have, instead, $\tau= 10^{-7}s$. This would amount to only an expansion to 10m in a second. Can you estimate the ages required to expand the probabilistic size of a quantum basketball by such a factor of 10?

• Thanks for the detailed answer! If I understand correctly particles with minimal spatial uncertainty will reach an uncertainty of kilometers almost instantaneously . Is this a (somewat) realistic state though? – user3502079 Sep 13 '17 at 20:16
• Well, QM normally describes short lived processes... a millisecond is a looooong time, unlike picoseconds, etc... Electron interference experiments certainly have to take the spreading into consideration. That's the point of considering slow objects out of nowhere as plane waves. The SN1987A neutrino bursts did, as you see, confound careless "experts" at first.... – Cosmas Zachos Sep 13 '17 at 21:24
• Keep in mind we can only talk about these sorts of wave-functions in non-relativistic processes where quantum mechanics is the correct way to describe the process. If the process is relativistic then you should really use quantum field theory. A naive treatment of a massless particle in quantum mechanics would mean that the dispersion relation becomes E(p) ~ O(p) whereas for a massive particle at low momenta the dispersion relation is E(p) ~ O(p^2). I'm fairly sure this results in no actual spreading of the wave-function. – chuckstables Sep 14 '17 at 0:26
• Of course. Massless particles such as photons are not subject to this dispersive wavepacket discussion. Massive Klein-Gordon particles are, by contrast. But this is not a QFT question; neutrino bursts from supernovae do not necessitate QFT complications to describe. – Cosmas Zachos Sep 14 '17 at 14:11

Maybe someone more knowledgeable than me can make a better comment on the metaphysical understanding of the wavefunction, but the way I was taught and understand it is that the particle isn't per se spread out over space. Ie. the particle does not entirely occupy the space its wavefunction covers. Since the wavefunction is nothing more than a probability density function it simply indicates with what probability the particle can be at point $x$ (1D example) should a measurement take place and, as you put it, collapse it.

Now to answer your two questions I would say: A) decoherence means there is a breakdown of the phase relation between states which, in this example, would occur if the particle interacted with its environment. In the cold solitude of space I would say interaction is rare but remember that space isn't entirely a vacuum. Since in space there exists an atom approximately every $cm^3$ (Source: An Introduction in Astronomy, Textbook by Thomas Arny) I would claim that decoherence through interaction with another particle/system is very probable over short distances even in what is considered interstellar vacuum.

B) If we're assuming the distance between wavefunctions is very large then a smaller overlap between adjacent wavefunctions minimizes their interaction, yes. However, if you increase the functions variance (assuming a gaussian distribution of the particle's position) that pushes more probability farther away from its expectation value increasing the overlap integral between adjacent particles.

• Thank you for your answer! It seems the quoted density of outer space would indeed prevent much macroscopic spreading to answer A. Does increasing the overlap integral mean the chance collapse increase? – user3502079 Sep 6 '17 at 22:07
• Technically speaking wave-functions don't 'interact' with eachother. If you have two particles in your quantum mechanical system then the quantum state of those two particles is given by a wave-function of the spatial coordinates of BOTH particles. The WF squared evaluated at point 1 (for particle 1) and point 2 (for particle 2) gives you the probability density to find particle 1 around point 1 AND to find particle 2 around point 2. You can get marginal PDF's by integrating out the position of the other particle. – chuckstables Sep 14 '17 at 0:34