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The insight to understanding why electrons do not spiral into the nucleus was provided by the uncertainty relation between position and momentum - at university, this was described to me as follows:

"If we measure the location of the electron in a Hydrogen atom, we will find it somewhere within this vicinity represented by the square of the wave function, which should be interpreted as a probability density."

However, there seems to be a contradiction here: If the probability distribution applies only to measurements, then it seems when I'm not performing a measurement on the electron, it's unclear why it's in a stable configuration remaining uncaptured by the positive nucleus. Or put another way, is the electron's interaction with the nucleus in some manner performing a continuous measurement on the electron (and vice versa)?

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  • $\begingroup$ That’s Bohr’s hypothesis. This has nothing to do, with measurement or uncertainty. $\endgroup$
    – ZeroTheHero
    Commented Jul 23, 2022 at 1:31
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    $\begingroup$ Related: Why electrons in an atom don't radiate photons $\endgroup$
    – Sandejo
    Commented Jul 23, 2022 at 3:30
  • $\begingroup$ @ZeroTheHero this question and the answers to it seem to contradict your statement: physics.stackexchange.com/questions/206382/… $\endgroup$
    – JPattarini
    Commented Jul 30, 2022 at 17:14
  • $\begingroup$ @JPattarini I don’t follow. The stability of the atom is not related to measurements. If it were we could make the atom unstable by measuring (some of) its properties. $\endgroup$
    – ZeroTheHero
    Commented Jul 30, 2022 at 19:17

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The electron's interaction with the nucleus is not performing a continuous measurement on the electron (and vice versa).

A measurement does correspond to an interaction between the measured system and something else. However, in the case of electron and nucleus, the system is made by both electron and nucleus. The modulus squared of the wavefunction of the electron interacting with the nucleus in an energy eigenstate provides the density probability of position measurement of the electron in that state. However, such position measurement does not coincide with the interaction with the nucleus. In an ideal experiment, one has to send a third particle to interact with the electron to determine its position.

The idea that the electron is not spiraling into the nucleus due to the Uncertainty Principle is wrong. The simplest way to see it is by considering that the classical spiraling is due to the radiation of electromagnetic waves by an accelerated electron. The derivation of uncertainty relations (UR) only uses the quantum description of the electron, without any relation with electrodynamic fields or photons.

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  • $\begingroup$ I don’t understand how the uncertainty relation can be said to not be in play here, since Feynman derives the atomic radius in part by using this relationship, discussed here: physics.stackexchange.com/questions/206382/… $\endgroup$
    – JPattarini
    Commented Jul 30, 2022 at 17:16
  • $\begingroup$ @JPattarini If principle A implies B and C, it is wrong to say that C is a consequence of B, unless A and B are equivalent. The uncertainty principle (UP) is a consequence of the principles of QM, but it is not possible to derive the principle from the UP. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Jul 30, 2022 at 20:50

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