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So here is an image of the third lowest energy eigenfunction of an electron in a hydrogen atom:

Hydrogen atomic orbital

Image from https://i.sstatic.net/oTZEg.jpg

I understand well the eigenfunctions given by Schrodinger's equation for other types of potential energies, but the application to the atom is throwing me off, because of prior knowledge.

For an oscillating potential, or a constant potential, or some other variation, the qualities of the wavefunction make sense. The particle can be found anywhere inside the "box", and it is less likely to be found in areas of greater kinetic energy.

But isn't the pictured energy eigenfunction implying that there is a small chance for an electron with E_2 energy to be found in the first energy shell? I'm not sure how to interpret that. Isn't it also implying that the electron could be found anywhere between the wave functions maxima except for one discrete location where it equals zero?

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  • $\begingroup$ I'm confused by your question. I'm pretty sure "first energy shell" means the ground state of the hydrogen atom, which is orthogonal to the state in the linked picture, so no: if the electron happens to be in the state you are showing, then there is no probability of being in the ground state. As for your last statement: yes, the electron can be found anywhere except for those discrete locations (of course, you can't measure a position precisely, so really there is some finite probability of finding the electron near those points). $\endgroup$
    – march
    Commented Nov 11, 2015 at 6:15
  • $\begingroup$ What do you mean orthagonal? The eigenvectors are orthagonal? $\endgroup$
    – user97626
    Commented Nov 11, 2015 at 6:33
  • $\begingroup$ What I'm saying is that the linked graph is a function of radius, no? Also, no I'm saying the opposite. I understand that the possible energies are only discrete locations, but the energy eigenfunction above seems to suggest otherwise, that there are discrete locations where it cannot be, but a small probability everywhere else. $\endgroup$
    – user97626
    Commented Nov 11, 2015 at 6:35
  • $\begingroup$ Right, the eigenfunctions are orthogonal. Energy shells correspond to eigenfunctions of the Hamiltonian for the hydrogen atom. The eigenfunction that you've shown is not the ground state, and so it must be orthogonal to the ground state. Hence, if the electron is in this state, there is no probability of the electron being in the ground state. $\endgroup$
    – march
    Commented Nov 11, 2015 at 6:35
  • $\begingroup$ What could you possibly mean by "the possible energies are only discrete locations"? Do you mean that if the electron is in an energy shell, then it is at a discrete location? This is not true! Being in an energy shell means the electron is in one of the eigenstates of the Hamiltonian, and yes: these functions are smeared out in space, so that if the electron has a well-defined energy, there is a probability of finding it at almost-all different radii. $\endgroup$
    – march
    Commented Nov 11, 2015 at 6:38

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One has to keep in mind

1) that it is the complex conjugate square of the eigenfunction that gives the probability of finding the electron with energy E at a specific radius.

2) There are no fixed orbits in the quantum mechanical solution, only a locus of probability called orbital

3)orbitals overlap in space, it is the energy that is keeping the electron on a specific wavefunction.

What happens if you have an electron in a higher energy level ? (please note that orbits are drawn for convenience, orbitals would make a messy graph) It has a probability of decaying to a lower one. It is only the electron in the ground state that stays grounded.

But isn't the pictured energy eigenfunction implying that there is a small chance for an electron with E_2 energy to be found in the first energy shell?

Yes, it can overlap with the orbital of a lower energy state.

I'm not sure how to interpret that.

it is called probability of deexcitation, by emitting a photon.

Isn't it also implying that the electron could be found anywhere between the wave functions maxima except for one discrete location where it equals zero?

Yes, it is how probabilities work. Have a look at atomic orbitals.

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  • $\begingroup$ Thank you very much for your reply. In my physics class, it was never really stressed that discrete energy did not mean discrete radial distance from the nucleus. In fact, with the lack of clarification plus the way energy diagrams look, it seems purposefully implied that the radial distances were discrete as with energy. Thank you for the answer. So an electron of energy $E_2$ can be found at the same radial location as an electron with energy $E_1$, at least monetarily, as this does not disagree with the Pauli Exclusion Principle? That only requires that their spin and energy differ, yea? $\endgroup$
    – user97626
    Commented Nov 11, 2015 at 22:33
  • $\begingroup$ Though, you say that the orbital radii can overlap. Does this mean that the probability functions can overlap, but if the electron with energy $E_2$ actually physically ends up near the Bohr radius for $E_1$, it will transition to the ground state? Or can an electron with energy $E_2$ end up near the Bohr radius for $E_1$ without transitioning? Basically I'm asking about the correlation between $E$ and $r$. Bohr makes it seem like they are basically synonymous. Schrodinger's equation solution for hydrogen seem to imply otherwise. $\endgroup$
    – user97626
    Commented Nov 11, 2015 at 23:00
  • $\begingroup$ My essential question is this: Is the probability of an electron of energy $E_2$ being located near the Bohr radius $a_0$ in direct correlation with the probability of an electron of energy $E_2$ transition to energy state $E_1$? Seems that it is. Sorry for so many comments $\endgroup$
    – user97626
    Commented Nov 12, 2015 at 0:24
  • $\begingroup$ The transition probability will have to be calculated using the wavefunction solutions. These depend both on delta(E) between the states and on the probability of overlap in space. The correlation comes through the wave functions and it is probabilistic. In order to transition there should be overlap in space of the two wave functions and also there should be the delta(E) that is required. A lower energy state does not have the energy to climb to a higher one even though the orbitals overlap. $\endgroup$
    – anna v
    Commented Nov 12, 2015 at 5:35
  • $\begingroup$ Probabilistic means for example :out of a million overlaps in space 1 energy transition happens. This probability can be calculated using a quantum mechanical model with the wavefunctions of the states in transition. $\endgroup$
    – anna v
    Commented Nov 12, 2015 at 5:41

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