The Requirement of integer orbitals

Phyzwizz

If there is a cloud of electrons around an atom than why can't there be orbitals between 1 and 2 or between 2 and 3. I know the probability of an electron being between certain nodes decreases as they approach them but why as the probabilities go away from the perfect orbital do they not become fractional orbitals? (just starting to learn this stuff)

TriTertButoxy

If you solve the angular part of the Schrödinger equation in the Coulomb potential (or for any spherically symmetric potential), you'll find that in order to satisfy boundary conditions at [itex]\theta=0[/itex] and [itex]\theta=\pi[/itex] and [itex]\phi=0[/itex] and [itex]\phi=2\pi[/itex], you need to have "integer orbitals" (in your language).

Nabeshin

Simply because there are no (fractional) integers! The orbitals are labeled by their radial quantum number, n, which is an integer. So while an electron in the n=1 orbital has a finite probability of being found at the most probable radius for an electron in the n=2 orbital, and vice versa, they are distinct quantum states.

Jim Kata

Anytime you have a bounded system |U|>|E| the eigenvalues will be integral.

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Phyzwizz	The Requirement of integer orbitals If there is a cloud of electrons around an atom than why can't there be orbitals between 1 and 2 or between 2 and 3. I know the probability of an electron being between certain nodes decreases as they approach them but why as the probabilities go away from the perfect orbital do they not become fractional orbitals? (just starting to learn this stuff)

TriTertButoxy	If you solve the angular part of the Schrödinger equation in the Coulomb potential (or for any spherically symmetric potential), you'll find that in order to satisfy boundary conditions at [itex]\theta=0[/itex] and [itex]\theta=\pi[/itex] and [itex]\phi=0[/itex] and [itex]\phi=2\pi[/itex], you need to have "integer orbitals" (in your language).

Nabeshin Recognitions: Science Advisor	Simply because there are no (fractional) integers! The orbitals are labeled by their radial quantum number, n, which is an integer. So while an electron in the n=1 orbital has a finite probability of being found at the most probable radius for an electron in the n=2 orbital, and vice versa, they are distinct quantum states.

Jim Kata Blog Entries: 1	The Requirement of integer orbitals Anytime you have a bounded system \|U\|>\|E\| the eigenvalues will be integral.