# Is the quantum number $L$ a combination of the azimuthal and magnetic quantum numbers $(\ell, m_\ell)$?

In the $$J=L+S$$ equation about total angular momentum, which of the four quantum numbers used to describe the electrons and their states is included in the $$L$$?

And does the $$J$$ include the first, principal quantum number?

The values of $$L$$ come from the individual $$\ell$$ values, according to the Clebsch-Gordan series:

$$L = |\ell_1 - \ell_2|, \ldots , \ell_1 + \ell_2.$$

where the dots indicate all "in-between" values with a step size of 1. The values of $$S$$ come from the individual $$s$$ values, in a similar fashion. Note that it is $$s$$, not $$m_s$$.

Finally, the values of $$J$$ come from $$L$$ and $$S$$:

$$J = |L - S|, \ldots, L + S.$$

The principal quantum number $$n$$ does not play any role in this, since it has no direct relation with angular momentum.

Usually it helps to see an example. Consider the case of the ground-state carbon atom; the $$\mathrm{1s}$$ and $$\mathrm{2s}$$ orbitals can be neglected since those are closed-shell, so we only need the two highest-energy electrons, i.e. a $$\mathrm{p^2}$$ configuration.

We therefore have two electrons, both having $$\ell = 1$$ (p-orbital) and $$s = 1/2$$ (because they are electrons, and all electrons are spin-$$1/2$$). The permissible values of $$L$$ are

\begin{align} L &= |\ell_1 - \ell_2|, \ldots , \ell_1 + \ell_2 \\ &= | 1 - 1 |, \ldots, 1 + 1 \\ &= 0 , \ldots, 2 \\ &= 0, 1, 2 \end{align}

The permissible values of $$S$$ are

\begin{align} S &= |s_1 - s_2|, \ldots , s_1 + s_2 \\ &= | 1/2 - 1/2 |, \ldots, 1/2 + 1/2 \\ &= 0 , \ldots, 1 \\ &= 0, 1 \end{align}

The permissible values of $$J$$ depend on what $$L$$ and $$S$$ are, and must be evaluated on a case-by-case basis. So far, we have identified three possible values of $$L$$ and two possible values of $$S$$. Consider the case where $$L = S = 0$$: here,

\begin{align} J &= |L - S|, \ldots , L + S \\ &= |0 - 0|, \ldots, 0 + 0 \\ &= 0 , \ldots, 0 \\ &= 0 \end{align}

So when $$L = S = 0$$, the only permissible value of $$J$$ is also $$0$$. If you've seen term symbols before, the corresponding term symbol is denoted as $$^1\!S_0$$; if you haven't, then all you need to know is that this is a chemist's "shorthand" way of writing $$(J, L, S) = (0, 0, 0)$$, which is the combination we just found.

You will need to work through this for the other five combinations of $$L$$ and $$S$$. This will give you a full series of term symbols, except that there's a catch. The Pauli exclusion principle forbids some of those six combinations: to be precise, the combinations $$(L, S) = (2, 1)$$, $$(1, 0)$$, and $$(0, 1)$$ are not permissible, which leaves only three valid combinations, one of which I've already demonstrated. See Pauli-forbidden term symbols for atomic carbon for more details.

Addendum. As for the $$m_\ell$$ and $$m_s$$ quantum numbers: they do have a use, but in a rather more subtle way. I don't expect this very brief summary to be useful, but I'll try; feel free to ignore it at this stage of your learning if it doesn't make much sense.

Much like how (for one electron) $$\ell = 1$$ implies $$m_\ell = -1, 0, 1$$, for the entire atom each value of $$L$$ is associated with a range of $$M_L$$ values: for example, $$L = 1$$ implies $$M_L = -1, 0, 1$$. Therefore, the $$L = 1$$ term is actually not one single electronic state, but rather a collection of three different values of $$M_L$$, which is why we call it a "term" rather than a state. Likewise, the $$L = 2$$ term actually corresponds to five different values of $$M_L$$. How does this relate to $$m_\ell$$? Well, it turns out that $$M_L$$ is the sum of the individual electron $$m_\ell$$ values. A similar situation exists with $$S$$, $$M_S$$, and $$m_s$$. However, I think that's a topic for another question, if necessary.