When we discuss about configurations we specify n, l, m quantum numbers for the individual electrons.

My question is: why when we pass from configurations to atomic terms in order to use the total angular momentum we don't use anymore the n quantum number? Does exist an 'equivalent' number for the terms or is conceptual wrong to think about it?

I know the formal procedure for the change of basis set from uncoupled to coupled basis and i understand why this is important in terms of commutation properties of hamiltonian, but i simply don't see anymore this n quantum number, so i'm a bit confused about it.

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    $\begingroup$ The quantum number n does not contribute to angular momentum, in contrary to the others. $\endgroup$
    – Poutnik
    Commented Aug 24, 2023 at 17:20

1 Answer 1


As Poutnik points out in the comments, the angular momentum of an electron does not depend on its quantum number $n$; it only depends on $l$ and $m_l$.

Thus, for example, $\mathrm{1s}^1$ and $\mathrm{2s}^1$ configurations (which might be, for example, the ground and first excited state of hydrogen) both correspond to the same term symbol $^2\!S_{1/2}$.

Clearly, this is fine if all the states you care about have different angular momentum properties. But there are cases where this isn't fine, and you need to distinguish. I don't know if there is 'official' notation specified out there, but what people usually do is to specify the orbital configuration along with the term symbol, or $n$ along with the term symbol (if that provides enough information).

Both of these possibilities are mentioned in the following extract from Hollas' Modern Spectroscopy, 4th ed. (p 246):

In the sodium atom pairs of $^2\!P_{1/2}$, $^2\!P_{3/2}$ states result from the promotion of the 3s valence electron to any $np$ orbital with $n > 2$. It is convenient to label the states with this value of $n$, as $n{}^2\!P_{1/2}$ and $n{}^2\!P_{3/2}$, the n label being helpful for states that arise when only one electron is promoted and the unpromoted electrons are either in filled orbitals or in an $s$ orbital. The $n$ label can be used, therefore, for hydrogen, the alkali metals, helium and the alkaline earths. In other atoms it is usual to precede the state symbols by the configuration of the electrons in unfilled orbitals, as in the $2p3p~{}^1\!S_0$ state of carbon.

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    $\begingroup$ this is what i was searching for. Let me be sure to have well understand: so if i wanna put the energies in increasing order i write down the configurations in increasing order and for each configuration i write its terms in increasing order? $\endgroup$
    – Chemistry.
    Commented Aug 25, 2023 at 11:33
  • 1
    $\begingroup$ @Chemistry. Yup, pretty much. $\endgroup$ Commented Aug 25, 2023 at 13:05

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