0
$\begingroup$

Since I don't know an associated operator to the principal quantum number $n$, I don't know when it is a good quantum number. By 'good quantum number' I mean a quantum number that is conserved over time.

For instance in which of the following Hamiltonians is $n$ a good quantum number?

  1. H = KE + Radial PE (Central Field)
  2. H = KE + Radial PE + Residual Electrostatic PE (Electron Repulsions included)
  3. H = KE + Radial PE + Residual Electrostatic PE + Spin-Orbit PE (Spin Orbit effects included)
  4. H = KE + Radial PE + Residual Electrostatic PE + Spin-Orbit PE + Magnetic Field (Magnetic Field included)
Improve this question
$\endgroup$
8
  • $\begingroup$ I'm not sure we can associate $n$ with the entire atom, only with particular electrons. (I'm speaking in a very rough sense.) You then need to look at whether a given Hamiltonian can impart energy onto those electrons. Ultimately this becomes very complicated to explicitly write out for more than one electron around the nucleus. The whole scheme of just caring about L and S is particular cases is a valence-like approximation, at which point you just examine the angular momenta of the electrons you care about. $\endgroup$ – Jonathan Jeffrey Feb 7 at 15:03
  • $\begingroup$ Because those electrons are typically restricted to the valence electrons, they have a defined $n$. As soon as you start coupling states (e.g. in coherent Rabi flopping, by e.g. shining resonant light), $n$ might no longer be a good quantum number of the system, if you couple states with different $n$. $\endgroup$ – Jonathan Jeffrey Feb 7 at 15:07
  • $\begingroup$ Thanks, I guess I meant when is the set of principal quantum numbers conserved as a set maybe? $\endgroup$ – Alex Gower Feb 7 at 15:09
  • $\begingroup$ In none of the cases you described are you adding or removing energy to the system, so the distribution of electrons among $n$ stays the same. Interchange of electrons among these occupied $n$ only causes a global phase change of $-1$, and from a second quantized view doesn't do anything at all, so I see $n$ being good quantum numbers here. (Again, very, very, rough explanation.) $\endgroup$ – Jonathan Jeffrey Feb 7 at 15:14
  • 1
    $\begingroup$ Yes but I guess in the framework of my question we are just assuming that these are the only Hamiltonians being considered, I think it works. Thaks! $\endgroup$ – Alex Gower Feb 7 at 15:21
0
$\begingroup$

So if we just take the main Hamiltonian H=K.E+radial then this has energy levels which are labelled by the principal quantum number.

In other words, n is the quantum number for the original Hamiltonian

Improve this answer
$\endgroup$
1
  • $\begingroup$ By 'good quantum number' I mean a quantum number that is conserved over time. $\endgroup$ – Alex Gower Feb 7 at 14:49

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.