0
$\begingroup$

I was studying the applications of the Schröedinger wave function in the Bohr's atom.

For what I understood, the $\psi$ should only depend on r and not on $\theta$ and $\phi$. Does that mean that $\psi(r, \theta, \phi)=R(r)$?

$\endgroup$
2
$\begingroup$

In general the wavefunction depends on all the coordinates. However, there are special states called $s$ states, without any angular momentum, where it depends only on $r$.

$\endgroup$
  • $\begingroup$ And there is a symmetry in the $1/r^2$ potential that makes that other values of the angular momentum quantum number have the same energy. $\endgroup$ – Pieter Nov 29 '19 at 0:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.