PX262 - J1 - electrons in atoms and molecules

i ℏ \frac{\partial ψ}{\partial t} = (\frac{- ℏ^{2}}{2 m} \nabla^{2} + V (x, y, z)) ψ

the stationary states: $ψ = ϕ_{λ} \exp (i E_{λ} t / ℏ)$
if the potential is spherically symmetric, $V (x, y, z) = V (r)$ , use spherical polar coordinates
eg: for hydrogen:

V (r) = - \frac{e^{2}}{4 π ε_{0} r}

\hat{H} ϕ_{n l m_{l}} (r, θ, ρ) = E_{n} ϕ_{n l m_{l}} (r, θ, φ)

ϕ_{n l m_{l}} (r, θ, ρ) = R_{n l} (r) Y (θ, φ)

$\hat{H}$ , ${\hat{L}}^{2}$ , ${\hat{L}}_{z}$ commute, which gives the conservation of angular momentum
electron in a hydrogen atom:

\begin{matrix} {\hat{L}}^{2} Y_{l m_{l}} (θ, φ) = l (l + 1) ℏ^{2} Y_{l m_{l}} (θ, φ) \\ {\hat{L}}_{z} Y_{l m_{l}} (θ, φ) = m_{l} ℏ Y_{l m_{l}} (θ, φ) \end{matrix}

reminder: $l \in Z$ up to $n - 1$ , $m_{l} \in [- l, l]$
graft on spin (electrons = $1 / 2$ , fermions): $ϕ_{n l m_{l} ↑} (r, θ, φ), ϕ_{n l m_{l} ↓} (r, θ, φ)$
for $n = 2 :$
- $l = 0, m_{l} = 0, ↑↓$
- $l = 1, m_{l} = - 1, 0, 1, ↑↓$
- this is eight fold degenerate
for a many-electron system, electrons cannot occupy the same state in space unless they have different spin states - pauli exclusion principle
using the hydrogen atom to set up a model of other atoms:

\begin{matrix} V (r) = - \frac{e^{2}}{4 π ε_{0} r} \\ E_{n} = - \frac{13.6}{n^{2}} eV \end{matrix}

the model of an atom (Z-atomic number) has each of its $Z$ electrons moving in a potential set up by the positively charged nucleus and a symmetric cloud of negative charges from the other $(Z - 1)$ electrons
the net potential falls off more quickly than straight coulombic potential, ie. $- \frac{Z e^{2}}{4 π ε_{0} r}$ close to the nucleus, and $- \frac{e^{2}}{4 π ε_{0} r}$ far away
so an electron in a state which is distributed, on average, close to the nucleus will 'see' a higher charge on the nucleus than one in a state distributed further away
this removes degeneracy of energy levels with respect to the angular momentum
the order of the energies is such that the s-orbitals $l = 0$ have lower energies than the p-orbitals $(l = 1)$ and so on
example: for sodium $Z = 11$

\begin{matrix} n = 1 & l = 0 & 1 s^{2} & ↑↓ & 2 \\ n = 2 & l = 0 & 2 s^{2} & ↑↓ & 2 \\ l = 1 & 2 p^{6} & ↑↓ & 6 \\ n = 3 & l = 0 & 3 s^{1} & 1 & (valence) \end{matrix}

\begin{matrix} n = 1 & l = 0 & 1 s^{2} & ↑↓ & 2 \\ n = 2 & l = 0 & 2 s^{2} & ↑↓ & 2 \\ l = 1 & 2 p^{6} & ↑↓ & 6 \\ n = 3 & l = 0 & 3 s^{1} & ↑↓ & 1 \\ l = 1 & 3 p^{6} & ↑↓ & 6 \\ l = 2 & 3 d^{6} & ↑↓ & 6 / 7 \\ n = 4 & l = 1 & 4 s^{2} & ↑↓ & 2 / 1 \end{matrix}