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The probability of an electron found outside the atom is never zero. Consider building an electron detector, it must receive permanent signals from all electrons in the universe, as they can exist everywhere. Of course the probability decreases with increasing distance from the atom, but for the huge amount of atoms in the universe the probability of finding an electron at every point in space cannot be negligible.

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    $\begingroup$ If the probability of winning the lottery is non-zero, how come you don't win the lottery all the time? Because "non-zero" can still be very small. $\endgroup$
    – Jessica
    Jun 11 at 18:10
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    $\begingroup$ Please clarify your definition of "being full of something" I feel like when you do that you may answer your own question. $\endgroup$
    – klm123
    Jun 12 at 14:07
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    $\begingroup$ I think this question is analogous to Olber's paradox except that instead of the light from stars we are talking about the wave function of electrons. $\endgroup$
    – quant
    Jun 12 at 14:16
  • $\begingroup$ The amplitudes of several electrons can add up to zero because they are complex-valued, so non-zero probability everywhere is not actually a given. Of course, in any finite volume the probability will be non-zero but small for the reasons the other answers pointed out. $\endgroup$
    – tobi_s
    Jun 13 at 4:39
  • $\begingroup$ @klm123, nah, the question is fine IMO. From the POV of a lay person I find it very easy to understand how it was meant, and how to give a helpful answer (like the top-voted one right now, or really just the top-voted comment here). $\endgroup$
    – AnoE
    Jun 13 at 8:47

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I think this is a very creative question. You are thinking: There are so many atoms in the universe that all of their electron wavefunctions should overlap and lead to a detection everywhere.

But remember that the radial wavefunction of the electron very roughly goes as $e^{-r/a_0}$, where $a_0$ is the Bohr radius, which is (also very roughly) $10^{-10}$m, and $r$ is the distance. The estimated amount of atoms in the observable universe is something of the order of $10^{80}$, but it does not matter if its some other order close to that. Let's now assume they are distributed not as in the observable universe, but ALL of them are just ONE meter away (which would of course lead to other complications). But still, this means the calculation of detection probability will contain a factor (or even the square of) $10^{80}e^{-1\mathrm{m}/a_0}=10^{80}e^{-10^{10}}(\approx0)$, which for all intents and purposes is zero. You cannot beat the exponential suppression at larger distances even with very large numbers of electrons. Since you seem to be an outside of the box thinker: Keep in mind that even more atoms in the maybe un-observable universe do not matter, since they are so far away that their exponential suppression is even stronger. Does not matter how much you add.

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  • $\begingroup$ Does the Dirac equation make the probability of electrons from outside the observable universe exactly (rather than "effectively") zero ? $\endgroup$ Jun 12 at 7:04
  • $\begingroup$ @BlueRaja-DannyPflughoeft You are most probably right. I did not go into any details about that in my quite heuristical answer. Since the solution for the hydrogen atom is found in non-relativistic quantum mechanics, you do not see this, but of course relativistic QM cares about if events are causally disconnected, i.e. have spacelike separation. $\endgroup$
    – Koschi
    Jun 12 at 8:28
  • $\begingroup$ The answer seems to imply that the probability of detecting a single electron at distance exactly $d$ equals the probability of detecting $10^{80}$ electrons all sitting at distance exactly $80\cdot \ln 10\cdot d = 184\cdot d$. It is impressive how little effect the number of the electrons has. However, we believers in the Dirac sea have way more electrons than atoms in the observable universe, and we have also strategically placed them nearly everywhere, hoping that the detectors will go mad anyway. $\endgroup$ Jun 12 at 17:23
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    $\begingroup$ @BЈовић Very very roughly take the mass of an average star to be the mass of the Sun $\approx10^{33}$ g/star $\times10^{23}$ roughly the number hydrogen atoms/g $\times 10^{12}$ (Sun-like-)stars/galaxy $\times10^{12}$ galaxies/observable universe $\approx10^{80}$. $\endgroup$
    – Koschi
    Jun 13 at 8:04
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    $\begingroup$ @Koschi Ok, I see your logic. Considering how much we really don't know about our universe, that is a good guess :) $\endgroup$ Jun 13 at 8:27
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Let's suppose we are surrounded by an approximately uniform density of hydrogen atoms $\rho_h$, then the number of atoms in a shell around us of radius $r$ and thickness $dr$ is:

$$ dN = \rho_h 4 \pi r^2 dr $$

where if we average over the whole universe $\rho_h$ is about $0.2$ hydrogen atoms per cubic metre.

The electron density of a hydrogen $1s$ orbital at a distance $r$ is:

$$ \rho_e = \frac{1}{\pi a_0^3} e^{-2r/a_0} $$

Then the electron density due to the atoms in our shell at distance $r$ is:

$$ d\rho_t = \frac{\rho_h 4 \pi}{\pi a_0^3} e^{-2r/a_0} $$

Then the total electron density is just the integral of this from $r = 0$ to $\infty$:

$$ \rho_t = \frac{4\rho_h}{a_0^3} \int_0^\infty e^{-2r/a_0} dr $$

And $\int_0^\infty e^{-2r/a_0} = a_0^3/4$ so the whole expression simplifies to:

$$ \rho_t = \rho_h $$

Which is of course exactly what we expect since each hydrogen atom has one electron so the density of electrons has to be the same as the density of the hydrogen atoms i.e. about $0.2$ electrons per cubic metre.

As Koschi says in their answer the electron density falls exponentially with distance and this totally swamps the $r^2$ increase in the number of atoms with distance. The end result is that the electron density is only affected by the atoms very near to our position.

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I think, you are underestimating the effect of the electrons' amplitudes exponential decay.

While the wave function of a bound electron is indeed non-zero everywhere, the amplitude of the wave function falls off exponentially with distance. When you square the amplitude, you get the probability density function of finding the electron somewhere it shouldn't be classically. I.e. the probability also falls off exponentially.

Now, the number of electrons that may tunnel to some place grows with the cube of the distance. This growth is fast, but it's still just a cubic function. The exponential probability fall-off completely dwarfs the cubic growth of available electrons.

As such, the contribution by far-away electrons to finding an electron at some specific place is practically zero.

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The Universe isn't even full of matter, it's more space than matter. From micro to macro this is true. The empty space is far more than what matter actually occupies. The probability of electrons outside an atom is low and being in a specific space is even lower, since probability of finding matter in space is also low and especially since the size and mass of electron is considered negligiable already when compared to protons and neutrons.

Being negligiable doesn't mean it's zero but it does mean that it's impractical, we might not have suitables tools for even measuring such small values or getting anything of practical value from such.

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    $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Jun 12 at 19:37

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