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As far as I understand, the primary reason why an electron doesn't fall into a nucleus is that, when it gets close, it has a sufficiently high probability of having a lot of momentum to get back to its original position. This high probability, in my understanding, arises because the electron is confined to only a small amount of space around the nucleus, and hence, by Heisenberg's Uncertainty Principle,the uncertainty in its momentum is high, which in turn increases the probability of the momentum being large.

In such cases, where the electron "bounces off" or "phases through" the nucleus, the momentum would always have a non-zero magnitude, and hence the kinetic energy would always be non-zero. This way, energy would be gained with each new "bounce". Additionally, the electrons would be constantly accelerating, radiating away light of all frequencies.

Is my premise incorrect, or is it the implication? And in what way is it incorrect?

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  • $\begingroup$ There is another form of the Uncertainty Principle, namely Time-Energy: $\Delta t \Delta E \ge \frac{\hbar}{2}$. Conservation of energy is only true within a timescale $t > \Delta t $; else virtual particles would also violate energy conservation. $\endgroup$ – Sparrow Jun 22 '19 at 1:20
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    $\begingroup$ @Sparrow please no (wrong) answers in comments. Thanks. $\endgroup$ – AccidentalFourierTransform Jun 22 '19 at 1:22
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    $\begingroup$ Possible duplicate of Energy conservation limited by uncertainty principle $\endgroup$ – Aaron Stevens Jun 22 '19 at 2:40
  • $\begingroup$ @Aaron Stevens Totally different questions. I have looked at this question before posting this one. My question is about the momentum-position uncertainty; the question you linked is about the energy-time uncertainty. Even if these are identical, I do not understand how, which is the point of this question. $\endgroup$ – Max Jun 22 '19 at 12:25
  • $\begingroup$ @Sparrow what about the expectation of the electron's energy over a long period of time? Clearly, if the momentum keeps bumping up when it gets close to the nucleus, the expectation value of energy at a given time will grow? $\endgroup$ – Max Jun 22 '19 at 12:32
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Don't take this the wrong way, but it's pretty much incorrect start to finish because it applies classical reasoning to quantum mechanical systems. Sometimes semi-classical reasoning can help give us a bit of intuition, but this is not one of those times.

The real answer to this question is simply that electrons are not little balls orbiting a nucleus like planets around the sun. Electrons bound to an atom do not have well-defined momenta, nor do they have well-defined positions, and the spreads in the possible results of position and momentum measurements obey the uncertainty principle.

Additionally, the conservation of energy takes on a somewhat different character in quantum mechanics because quantum mechanical particles do not generically have well-defined energies either. It's true that for certain systems (i.e. those which can be described by Hamiltonian operators with no explicit time-dependence) the expectation value of the energy is conserved as the state evolves, but this doesn't correspond to the actual energy of the system at any point in time.

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  • $\begingroup$ I understand that both position and momentum are not well-defined. However, one can still measure each of these quantities with some degree of precision given the precision of the other quantity is sacrifice, right? For example, one could measure the effect a particular electron has on the nucleus by measuring the change in potential energy of the nucleus (by introducing an EM field of fixed strength, for example), which would give away its position. I'm sure a similar experiment could be done to determine the momentum. And if we keep adding the KE of the electron based on measured momentum, $\endgroup$ – Max Jun 22 '19 at 12:14
  • $\begingroup$ we'll see that it just keeps growing. Aldo, I do understand that the energy of a system need not always be precisely conserved, but it might fluctuate around a fixed value. However, here, that value (the expectation value) obviously keeps growing (or not?). How is that possible? Finally, since you say that my entire understanding is wrong, can you please explain why the probability distribution of the electron's position isn't centred around, and doesn't have its peak at, the nucleus? $\endgroup$ – Max Jun 22 '19 at 12:18
  • $\begingroup$ @Max "However, one can still measure each of these quantities with some degree of precision given the precision of the other quantity is sacrifice, right?" The HUP, contrary to popular belief, is not a statement of how precisely we can measure something. You seem to still have some classical reasoning in your Quantum mechanics. $\endgroup$ – Aaron Stevens Jun 22 '19 at 12:37
  • $\begingroup$ @Aaron Stevens Yep, that was bad vocabulary from me. But you know what I meant, right? Precision, in this context, = "standard deviation of the quantity". My vocabulary may not be precise (pun intended), but my reasoning is for the most part quantum (if not, where is it classical?) $\endgroup$ – Max Jun 22 '19 at 13:18

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