====== Spectroscopic Notation ======
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The combination of energy and orbital angular momentum together determine the //orbital// or //shell// in which the electron resides. We may write an electron orbital as 1s, 2s, 2p, 3d, etc. where the letter here is representative of the azimuthal quantum number $\ell$. The letters are historic relics, but persist anyway. (See Table 1.)
| $\ell$  | Letter  | Number of Possible Electrons  | Name  | Shape  |
| 0  | s  | 2  | sharp  | sphere  |
| 1  | p  | 6  | principal  | two dumbbells  |
| 2  | d  | 10  | diffuse  | four dumbbells  |
| 3  | f  | 14  | fundamental  | eight dumbbells  |
| 4  | g  | 18  | -  | -  |
| 5  | h  | 22  | -  | -  |
| Table 1: The letter correspondence for orbital angular momentum quantum number $\ell$. |||||

This notation is sometimes used to present the **configuration** of an atom, or the number of electrons in each orbital of an atom. In such a case, the form is $n\ell^x$, where $n$ is the primary (radial) quantum number and $x$ is the number of electrons in that state. For example, boron's ground state configuration is $1s^2 2s^2 2p^1$. There are five electrons with two in the $n=1$, $\ell =0$ orbital, two in the $n=2$, $\ell =0$ orbital, and one in the $n=2$, $\ell =1$ orbital.

== Term Symbol Notation ==
If we want to include more information, we can use another form of **spectroscopic notation** called the **term symbol** which has the form

$\textrm{n}{}^{2\textrm{s}+1}\textrm{L}_\textrm{j}$

where $n$ is again the principal quantum number, $s$ is the spin quantum number, $L$ is the (now capitalized) letter corresponding the orbital quantum number $\ell$, and $j$ is the total angular momentum $\left( \overrightarrow j = \overrightarrow s + \overrightarrow \ell\right)$. The $n$ term is considered optional and is often omitted. 

To represent boron in this notation, we first drop the full orbitals from the $n$ term, leaving us with $2p^1$. The orbital quantum number of $\ell = 2$ gives us $L$ = $1$ here corresponding to the letter $P$.  The single unpaired electron will occupy an orbital where $m_\ell = +1$ and $m_s = +1/2$.  The total spin angular momentum $S = +1/2$ since there is only the one unpaired electron.  The total angular momentum $J = |L - S| = 1/2$ (remember we're adding vectors here.  It's annoying.) Thus, boron's ground state is represented as ${}^{2s+1}P_{1/2} = {}^{2}P_{1/2}$

=== State Transitions ===
When transitioning from one state to another, the following selection rules must be obeyed:

$\Delta s = 0$

$\Delta \ell = \pm 1$ 

$\Delta j = 0^* \pm 1$

$\Delta m_s = 0$

$\Delta m_j = 0\pm 1$

where the transition from $j=0 \rightarrow j=0$ is not allowed, and where $m_s$ and $m_j$ are the z-projections of the spin and total angular momentum, respectively. These rules are a manifestation of the conservation of angular momentum. Consider that when an electron jumps from one energy level to another, it does so by emitting or absorbing a photon. That photon carries with it one unit of angular momentum $\hbar$. If the spin does not change, then that change in angular momentum must be accounted for by an appropriate change in the orbital angular momentum.