The square of the wavefunction gives probability density of finding an electron somewhere in the orbital.

The text I'm referring to says that the value of probability density is always higher than zero at any finite distance from the nucleus.

My question is, provided that we knew position of an electron at a give time, could we now draw a boundary surface diagram that encloses a region in which probability of finding the electron is 100% for some point in time in future?

This would be possible because the electron cannot travel faster than the speed of light, so we could draw a sphere of radius $3*10^8$ meters centered at the current known position and say that probability of finding the electron after one second will not be beyond the region enclosed by the sphere.

Or is this question invalid because we can't know the position of an electron ever, not even within a finite uncertainty that would allow us to draw such a sphere?

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    $\begingroup$ Your own objection to your idea is right: we can never know the position of an electron. $\endgroup$ Mar 29, 2016 at 11:02
  • $\begingroup$ @IvanNeretin is it because of Hiesenberg's uncertainty principle? (shouldn't that still allow for knowledge of position within a finite uncertainty)? $\endgroup$ Mar 29, 2016 at 11:22
  • $\begingroup$ Well, yes, we might put it this way. Finite uncertainty of position does not imply finite distance. Look again at that Gaussian bell curve (or just about any other probability distribution, for that matter): it surely has finite uncertainty, but it continues to infinity and never quite reaches 0. $\endgroup$ Mar 29, 2016 at 11:33
  • $\begingroup$ @IvanNeretin I think I got it now. I always thought that uncertainty principle related $\Delta x$ and $\Delta p$ where these give a range of values that velocity or position may take (i.e. x will lie with 100% probability between $x - \Delta x$ and $x + \Delta x$ ) but actually it relates $\sigma_x$ and $\sigma_p$ where these are the standard deviations in the probability distribution. Am I correct? $\endgroup$ Mar 31, 2016 at 6:53
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    $\begingroup$ That's right, these are the standard deviations. There is no 100% probability interval. $\endgroup$ Mar 31, 2016 at 7:17

3 Answers 3


The (Wave)functional forms for the surface which contains the 100% probability for finding the electron extends to the literal edge of the universe. In the context of 'wave' it is nonsensical to measure an electron's position, and hence its speed may not be determined.

In the context of particle, an electron, having nonzero mass, may not be accelerated to the speed of light.


An electron can be prepared in any (proper wave function) state where its confined to an arbitrarily small volume. In such a state its momentum wave function is in general necessarily diffuse; this diffuseness is regulated via the uncertainty principle. When its energy is then really measured, the originally prepared wave function collapses into a certain eigenfunction of the Hamiltonian of the system with a probability proportional to the square abs value of the complex coefficient of the proportion that eigenfunction contributed to the originally prepared state. Also such eigenfunctions can be very strongly localized (e.g. particle in a box). Moreover in quantum mechanics the electron does not "travel" with some speed between some points in space, it has no trajectory. The wave functions much more show some time evolution which neither in the non-relativistic (Schrödinger eq.) nor the relativistic case (Dirac eq.) would inhibit it to "appear" (get probability amplitude) suddenly at places which it could not reach if it would be a particle obeying classic equations of motions. I think this can be seen in analogy to the famous tunneling where the electron can appear at classical "forbidden" regions.

Comment: I have edited the answer, the position measurement results in a complete localization of the electron; so I have replaced it with energy measurement to get the argument working.


The probability density has some value howsoever small it may be, at any finite distance from the nucleus. It is therefore not possible to draw a boundary surface diagram of rigid size in which the probability of finding the electron is 100 percent.

  • $\begingroup$ As OP points out it's not taking into consideration speed of light - orbital functions don't do this because they're only approximate descriptions. $\endgroup$
    – Mithoron
    Aug 15, 2017 at 23:02

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