If we solve the time-independent Schrödinger equation for any atom by considering only the electrostatic potential, an electron has the same probability of occupying the $p_{x}$ orbital as it does the $p_{y}$ and $p_{z}$ orbitals, because the electrostatic force is a central / spherically symmetric force which depends only on $r$, and the wavefunctions of the $p_{x}$ , $p_{y}$ , $p_{z}$ orbitals only differ in terms of the angular coordinates $(\theta,\phi)$.

However, a nucleus has a quantum spin which introduces a magnetic field, and the symmetry is broken. If we take into account the magnetic field of the nucleus, then the electromagnetic force which the nucleus exerts on electrons occupying the $p_{x}$, $p_{y}$, and $p_{z}$ orbitals is no longer central. So there should be a difference between the energies of the wavefunctions of the $p_{x}$, $p_{y}$, and $p_{z}$ orbitals.

How significant is this effect? In atoms with a small atomic number or with a small nucleus spin the effect is negligible and can be only observed near absolute zero. However shouldn't it become significant for atoms with bigger atomic number or with a large nuclear spin?

  • $\begingroup$ The degeneracy is only valid for one-electron atoms/ions. $\endgroup$
    – ananta
    May 16, 2023 at 21:00
  • 5
    $\begingroup$ As I understand it, when you start taking interactions like these into account (what you're describing here is hyperfine splitting), it's no longer valid to refer to $p_x$, $p_y$, $p_z$ orbitals anymore (which, as you note, are solutions to the Schrodinger equation when you neglect such factors). But yes, my understanding is that there is a lifting of degeneracy. (See the linked article.) || @ananta That's not true, what you're describing sounds more like degeneracy within a given $n$, and OP is asking about degeneracy within a given $\ell$. $\endgroup$ May 16, 2023 at 22:09
  • $\begingroup$ Should we be asking questions about $p$ orbitals when all we have in one-electron system is an $s$ orbital? $\endgroup$ May 17, 2023 at 9:33
  • 1
    $\begingroup$ You can excite the electron of a one-electron atoms into a $p$ orbital or any orbital at all. We can distinguish all the subshell states in the hydrogen spectrum when fine splitting (relativistic, nuclear spin interactions, etc) is taken into account. $\endgroup$ May 17, 2023 at 17:32


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.