I have a question about the derivation of the possible atomic terms of an atom. Consider an atom with $N$ electrons in the central-field approximation, with hamiltonian

$$ H_{0} = \sum_{i=1}^{N} \left[ - \frac{1}{2} \nabla ^{2}_{i} + V _{c}(ri)\right] $$

We can write its states as the product of the states of one electron, which are characterized by the quantum numbers $n,\ l,\ m_{l}$ and $m_{s}$. We know that the configurations we get like this are degenerate: all configurations with same values of $n$ and $l$ but differents $m_{l}$ and $m_{s}$ have the same energy. These configurations also have well-defined values ​​of total orbital angular momentum and spin $L$ and $S$, and we can classify them in atomic terms $^{2S + 1}L$, that have the same energy. However, if we consider corrections for the central field approximation, the degeneracy of these terms is broken, and we can determine which ones have higher or lower energy by Hund's rules.

My doubt is the following: thinking from the perspective of perturbation theory, I am considering the various states of the same electronic configuration as unperturbed states:

$$ | \psi^{(0)} \rangle = |n_{1},l_{1},m_{l1},m_{s1}, ..., n_{N},l_{N},m_{lN},m_{sN} \rangle $$

and taking as perturbation a potential equal to $V_{ee} - V_{c}$, where $V_{ee}$ is the Coulomb interaction between the electrons. The use of degenerate perturbation theory generates $|\psi \rangle$ eigenstates that are also functions with well-defined $L,\ S,\ M_{L}$ and $M_{S}$, since the perturbed hamiltonian commutes with the spin and orbital angular momentum operators (I'm ignoring spin-orbit interaction for now). But why do the possible $L,\ S,\ M_{L},\ M_{S}$ values ​​of the disturbed states have to be the same as the values ​​of the undisturbed states? Couldn't the perturbation generate states with well-defined angular moments but with different eigenvalues?

  • $\begingroup$ Afraid in general you must include the spin-orbit coupling term in the Hamiltonian for that porpouse. However, you can distinguish between states of different total spin $\mathrm{S}$, eg. triplet and singlet , since the wavefunction are built using a different set of spin-orbitals. $\endgroup$
    – PAEP
    Jun 17, 2023 at 22:57

1 Answer 1


As you know very well, this angular degeneracy is due to the fact that the potential $V_c=V_c(r)$ is purely radial and the angular part do not affect the energy. The angular solution are degenerate if we do not include interactions like spin-orbit interaction.

The interaction $V_{ee} \propto \frac{e^2}{|\vec{r}'-\vec{r}|}=V_{ee}(r,\phi)$ has an angular dependency. If you solve the equation $H|\psi \rangle = E |\psi \rangle $ with $H = H_0 + V_{ee}$ the angular degeneracy is broken.

If you consider $V_{ee}$ as a perturbation, acting on the state $|\psi^0\rangle$, the correction $\langle \psi^0 |V_{ee}|\psi^0\rangle$ will not break the degeneracy, the eigenvalues will be only shifted compared to the previous equation.

  • $\begingroup$ thanks for the answer, but i still have a doubt. For example, consider the carbon atom in the ground state. In this configuration, we know that there are only 3 possible terms, $^{1}S,\ ^{1}D$ and $^{3}P$. To deduce this, I need to use a central field approximation and the Pauli exclusion principle. However, how do I know that if I consider the electron-electron interaction and solve the Schrodinger equation, the only possible terms will be these three as well? Does it make sense to talk about the Pauli exclusion principle when not using an independent particle model? $\endgroup$
    – AlfredV
    Jun 19, 2023 at 0:48
  • $\begingroup$ I understand your concern, your central field is only radial and leads to levels with only a certain correction and the Pauli principle (PP) the electronic distribution. The most important part of $V_{ee}$ remains radial, the angular part of this potential breaks the angular degeneracy but this splitting is often weak that in a mean field approximation a degeneracy can still noticeable. Therefore approximately the configuration remains and the huge part of the work is to correct the energies. Even when not using an independent model, PP gives the e- distribution in the levels. $\endgroup$
    – M06-2x
    Jun 19, 2023 at 12:59

Your Answer

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

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