myntfalskj4

2022-07-08

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.

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.

Tatiana Gentry

Beginner2022-07-09Added 10 answers

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,

$u(x,0)=\int \frac{dk}{2\sqrt{\pi}}{e}^{ikx}{e}^{-(k-\overline{k}{)}^{2}/4}={e}^{-{x}^{2}+i\overline{k}x},$

is seen to be

$\begin{array}{rl}u(x,t)& =\frac{1}{\sqrt{1+2it}}{e}^{-\frac{1}{4}{\overline{k}}^{2}}\text{}{e}^{-\frac{1}{1+2it}{(x-\frac{i\overline{k}}{2})}^{2}}\\ & =\frac{1}{\sqrt{1+2it}}{e}^{-\frac{1}{1+4{t}^{2}}(x-\overline{k}t{)}^{2}}\text{}{e}^{i\frac{1}{1+4{t}^{2}}((\overline{k}+2tx)x-\frac{1}{2}t{\overline{k}}^{2})}\\ & ={e}^{i\overline{k}x-it{\overline{k}}^{2}/2}\text{}\text{}\frac{{e}^{-\frac{1-2it}{1+4{t}^{2}}(x-\overline{k}t{)}^{2}}\text{}}{\sqrt{1+2it}}\text{}.\end{array}$

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 $\overline{k}$, as it rapidly spreads,

$|u(x,t){|}^{2}=\frac{1}{\sqrt{1+4{t}^{2}}}\text{}{e}^{-\frac{2(x-\overline{k}t{)}^{2}}{1+4{t}^{2}}}\text{}.$

A bit too rapidly...

The width here, $\sqrt{1+4{t}^{2}}\to 2t$, after reinstating the naturalized constants, amounts to

$\mathrm{\Delta}x=A\sqrt{1+(\hslash t/m{A}^{2}{)}^{2}},$

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 $\overline{k}$, as posited; this distribution has not changed. The consequent spread in detection times, non-relativistically, will be

$\mathrm{\Delta}t=\frac{\mathrm{\Delta}x}{\overline{v}}=\frac{t}{A\overline{k}}\text{}.$

Where did they come from (in x; the momenta in 3D specify the direction)? Who knows.

Numbers : Prompted by the question, let's summarize the basic numerics of the spread. Define a characteristic time

$\tau ={A}^{2}\frac{m}{\hslash},\phantom{\rule{2em}{0ex}}\u27f9\phantom{\rule{2em}{0ex}}\mathrm{\Delta}x=A\frac{t}{\tau}.$

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?

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,

$u(x,0)=\int \frac{dk}{2\sqrt{\pi}}{e}^{ikx}{e}^{-(k-\overline{k}{)}^{2}/4}={e}^{-{x}^{2}+i\overline{k}x},$

is seen to be

$\begin{array}{rl}u(x,t)& =\frac{1}{\sqrt{1+2it}}{e}^{-\frac{1}{4}{\overline{k}}^{2}}\text{}{e}^{-\frac{1}{1+2it}{(x-\frac{i\overline{k}}{2})}^{2}}\\ & =\frac{1}{\sqrt{1+2it}}{e}^{-\frac{1}{1+4{t}^{2}}(x-\overline{k}t{)}^{2}}\text{}{e}^{i\frac{1}{1+4{t}^{2}}((\overline{k}+2tx)x-\frac{1}{2}t{\overline{k}}^{2})}\\ & ={e}^{i\overline{k}x-it{\overline{k}}^{2}/2}\text{}\text{}\frac{{e}^{-\frac{1-2it}{1+4{t}^{2}}(x-\overline{k}t{)}^{2}}\text{}}{\sqrt{1+2it}}\text{}.\end{array}$

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 $\overline{k}$, as it rapidly spreads,

$|u(x,t){|}^{2}=\frac{1}{\sqrt{1+4{t}^{2}}}\text{}{e}^{-\frac{2(x-\overline{k}t{)}^{2}}{1+4{t}^{2}}}\text{}.$

A bit too rapidly...

The width here, $\sqrt{1+4{t}^{2}}\to 2t$, after reinstating the naturalized constants, amounts to

$\mathrm{\Delta}x=A\sqrt{1+(\hslash t/m{A}^{2}{)}^{2}},$

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 $\overline{k}$, as posited; this distribution has not changed. The consequent spread in detection times, non-relativistically, will be

$\mathrm{\Delta}t=\frac{\mathrm{\Delta}x}{\overline{v}}=\frac{t}{A\overline{k}}\text{}.$

Where did they come from (in x; the momenta in 3D specify the direction)? Who knows.

Numbers : Prompted by the question, let's summarize the basic numerics of the spread. Define a characteristic time

$\tau ={A}^{2}\frac{m}{\hslash},\phantom{\rule{2em}{0ex}}\u27f9\phantom{\rule{2em}{0ex}}\mathrm{\Delta}x=A\frac{t}{\tau}.$

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?

Holetaug

Beginner2022-07-10Added 8 answers

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 $c{m}^{3}$ (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.

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 $c{m}^{3}$ (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.

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