I was introduced to spectroscopic notation as $n^{2S+1} L_J$ (with L = "S, P, D.."). And the meaning that n stands for the principal quantum number (energy level, as known from the hydrogen atom), L being the orbital angular momentum (as known from the hydrogen atom) and J being the total angular Momentum ($\hat{\vec{J}} = \hat{\vec{S}} + \hat{\vec{L}}$), as such also known from the fine-structure of the hydrogen). What gives me riddles is the "multiplicity" ($2S+1$): A single electrons total spin is always $1/2$, but that would mean that we always write "2" for the multiplicity, and that is pretty useless. On top of that, "S" has been called "total spin", which means that there obviously is more than one state involved?

Which brings me to my question: When this notation "$n^{2S+1} L_J$" is not talking about one specific electron state, what is it talking about instead?

  • Does multiplicity mean "all possible states that the electron can be in, that can be labled by the respective n, L, and J?

  • Or is it even so that the notation doesn't describe one single electron / a single electron state, but instead

  • The the possible multi-electron state in the atom, consisting of all single-electron-states with n, L and J?, independent from how many electrons are actually in that state?

  • The the possible multi-electron state in the atom, consisting of all single-electron-states with n, L and J?, taking into account that e.g. the outer shell of the atom has only X electrons present...

I hope somebody can explain those symbols in a way that really is unambigously - sadly most internet resources begin explaining the meaning in terms of the single-electron hydrogen atom, and then transition to atoms without further explaining the multi-electron shenanigans.

  • $\begingroup$ The labeling involves all relevant angular momentum, in case of electron spin this means adding the unpaired electrons. The reason a single electron spin is called a doublet is because it results in two levels (and two spectral lines). It's quite an old source, but Herzberg's Atomic Spectra and Atomic Structure gives a nice introduction. $\endgroup$
    – Paul
    Commented Jan 8, 2023 at 15:53

1 Answer 1


For an atom with more than one electron, this notation refers to angular momentum properties of the complete set of $N$ electrons.

The total spin of the $N$ electrons is represented by the operator $$ \hat{\bf S} \equiv \sum_{i=1}^N \hat{\bf s}_i $$ where $\hat{\bf s}_i$ is the operator representing the spin of the $i$'th electron. Similarly, the total orbital angular momentum is represented by the operator $$ \hat{\bf L} \equiv \sum_{i=1}^N \hat{\bf l}_i. $$ For the total angular momentum of the system we use the operator $\hat{\bf J}$ which can be expressed in terms of the other operators in either of two ways: $$ \hat{\bf J} \equiv \sum_{i=1}^N \hat{\bf j}_i \; = \; \hat{\bf L} + \hat{\bf S} $$ The second way is more useful when considering 'LS coupled' atoms, and these are the ones for which the notation $^{2S+1}L_J$ is appropriate.

The notation $^{2S+1}L_J$ asserts that the set of $N$ electrons is in an eigenstate of the operators $\hat{S}^2,\;\hat{L}^2,\;\hat{J}^2$ with quantum numbers $S,L,J$. The reason to put $(2S+1)$ rather than $S$ in the superscript is largely historical. It comes from counting how many values of $J$ you can have for given values of $L$ and $S$. Whenever $L \ge S$ there are $2S+1$ values of $J$ for the given $L$ and $S$. So the word 'multiplicity' comes from counting these $J$ values (which amounts to counting orthogonal quantum states). Historically it came from counting the number of energy levels in a close group (one where the energy gaps come from 'fine structure' (mostly spin-orbit interaction)) as deduced from spectroscopy.

There is also another thing you can count: the number of $M_S$ values available for a given value of $S$. This is ALWAYS equal to $(2S+1)$ and this is one reason why the notation has survived. So for example we say $^1S$ is a 'singlet' and $^3S$ is a 'triplet'. For $L=0,\;S=1$ there is only one value of $J=1$ and therefore only one energy level in an atom. But in nuclear physics (or in atomic physics with the Zeeman effect) we are interested in how many quantum states can have $J=1$ and there are $3$ (they have $M_J = -1,0,+1$).

In atoms there are a set of further quantum numbers needed to specify the state of the electrons. These further numbers are a set of $n$ and $l$ values which together is called the 'configuration'. The numbers $n$ and $l$ are the principal quantum number and angular momentum quantum number of each of the parts of the total $N$-electron wavefunction. It is often convenient to omit the part of the configuration corresponding to full shells and subshells, and just write down the part corresponding to valence electrons. That is the meaning of the $n$ part of the notation $n ^{2S+1} L_J$. In this example there is just one valence electron and then $L = l$ so you don't have to explicitly mention $l$ (but I think it is clearer to include it; for example for $l=2$ I would write $4d\,^3\!D_1$ not $4\,^3\!D_1$). For an atom in group 2 you would need to include the $l$'s. For example, the notation might then read $2s3p \,^3\!P_0$ for example, or $2p3p \,^3\!P_0$. But even in group 2 for the lower-lying states the first valence electron is usually in an $s$ state and then you get $L = l_2$ so the $l$ values don't have to be explicitly included if $l_1 = 0$ is being taken as understood.

  • $\begingroup$ What about the n-number? Obviously, it's not the energy quantum number of all the electrons in the atom? Or does this notation treat all electrons with n = 1, .... seperately? $\endgroup$ Commented Jan 8, 2023 at 16:15
  • 1
    $\begingroup$ This is the answer I came in here to write, very nice. @Quantumwhisp, I’m actually not familiar with the leading $n$. In the NIST level database, it appears (as a lowercase letter) in state labels for some heavy elements like Fe I, but not in light elements like hydrogen or lithium. $\endgroup$
    – rob
    Commented Jan 8, 2023 at 17:07
  • $\begingroup$ @Quantumwhisp I added the reply for n $\endgroup$ Commented Jan 9, 2023 at 10:34
  • $\begingroup$ "The numbers n and l are the principal quantum number and angular momentum quantum number of each of the parts " - How does this division into "parts" take place? $\endgroup$ Commented Jan 11, 2023 at 1:15
  • $\begingroup$ @Quantumwhisp That's a moderately long story which you can get from an atomic physics textbook. Briefly, one begins by approximating the entire potential (owing to nucleus AND electrons) as a central (spherically symmetric) part plus the rest (called "residual"). Then the $N$-electron Schrodinger equation separates into $N$ separate equations, and the $N$-electron wavefunction can be written as a product of $N$ single-electron functions. $\endgroup$ Commented Jan 11, 2023 at 10:17

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