On my last test, I had trouble with a couple questions because I wasn’t totally clueful about how angular momentum works in atoms. I should fix this.

Orbital angular momentum should come naturally: Our standard way of understanding electron states in atoms is with *orbitals*, which are chosen to have well-defined angular momentum quantum number *l* and magnetic quantum number *m*. (Essentially, *l* is the magnitude of the angular momentum vector and *m* is the *z*-component of the angular momentum. Of course this is kind of a lie but whatever.)

But every electron has not only orbital angular momentum from its motion around the nucleus, but also spin angular momentum, which is intrinsic. This spin always has *l _{s}*=1/2, so along any given axis

*m*=±1/2.

_{s}The question is: what happens when we add **l**+**s**? That is, what is the *total* angular momentum of the electron? We’ll call this **j**. By looking at vector magnitudes, we see that the angular momentum quantum number *j* must lie between \(l+\frac{1}{2}\) and \(|l-\frac{1}{2}|\) (note the careful absolute value…). Since *j* has to be an integer or half-integer, we see that we must have *j*=*l*±1/2, unless *l*=0 in which case *j*=1/2. For each of these *j* values, we have the standard set of *m _{j}* values from +

*j*to -

*j*.

What if we have an atom with a bunch of electrons in it? What does its angular momentum look like? Well, if a subshell (fixed *n* & *l*) is full, we know that it contributes neither orbital nor spin angular momentum — everything cancels out. So we need only look at partially-filled subshells. We fill such subshells according to Hund’s rules. Rules 1 & 2 tell you how to place your spins, rule 3 tells you how to get a total angular momentum *J* from this.

__Ex: a P subshell__

m_{l} |
L |
S |
J |
||||
---|---|---|---|---|---|---|---|

+1 | 0 | -1 | |||||

^{2}P_{1/2} |
↑ | 1 | 1/2 | 1/2 | (less than half filled; lowest J value is lower-energy) | ||

^{3}P_{0} |
↑ | ↑ | 1 | 1 | 0 | (“) | |

^{4}S_{3/2} |
↑ | ↑ | ↑ | 0 | 3/2 | 3/2 | (half filled; L=0, only one possibility) |

^{3}P_{2} |
↑↓ | ↑ | ↑ | 1 | 1 | 2 | (more than half filled; highest J value is lower-energy) |

^{2}P_{3/2} |
↑↓ | ↑↓ | ↑ | 1 | 1/2 | 3/2 | (“) |

^{1}S_{0} |
↑↓ | ↑↓ | ↑↓ | 0 | 0 | 0 | (“) |

We denote electron configurations with term symbols. They look like ^{2S+1}*L _{J}*.

*L*is given in “spectroscopic notation” (S/P/D/F/etc.). The “2

*S*+1″ part is called

*spin multiplicity*. A

**singlet**state is one with total spin 0, a

**doublet**state is one with total spin 1/2, a

**triplet**state is one with total spin 1, etc. We’re counting the number of possible

*m*values here, which if

_{S}*l*=0 is the number of split energy levels in the presence of a magnetic field.

Example: The base state of neutral helium is 1s^{2}, with spins like ↑↓ in the 1s subshell. This has zero total spin and zero total orbital spin, so it’s ^{1}S_{0}, a singlet. The first excited state of helium is 1s^{1}2s^{1}, with two up spins. This has total spin 1 and zero total orbital spin, so it’s ^{3}S_{1}, a triplet. The next excited state is 1s^{↑}2s^{↓} with term symbol ^{1}S_{0}, a singlet. Hund’s rules are what tell us this is a higher-energy excited state.