# What are coulombic and exchange energy and how are they determined?

May 19, 2018

A simplified way is to call them ${\Pi}_{c}$ and ${\Pi}_{e}$, the coulombic and exchange components of the pairing energy $\Pi$:

$\Pi = {\Pi}_{c} + {\Pi}_{e}$

[This is gone into more detail here.]

For example, consider the ground state valence configuration of chromium:

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$
$4 s$

$\underbrace{\underline{\uparrow \textcolor{w h i t e}{\downarrow}} \text{ "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" } \underline{\uparrow \textcolor{w h i t e}{\downarrow}}}$
$\text{ "" "" "" "" "" } \textcolor{w h i t e}{/} 3 d$

This has zero coulombic repulsion energy ${\Pi}_{c}$, because no electrons are paired. However, if we focus on the $3 d$ electrons, there are $10$ possible nonredundant exchanges.

Label each electron to get:

${\uparrow}_{1} {\uparrow}_{2} {\uparrow}_{3} {\uparrow}_{4} {\uparrow}_{5}$

Now, the possible exchanges (that result in the same energy) involve all of the $3 d$ electrons:

$1 \leftrightarrow \left\{2 , 3 , 4 , 5\right\}$ $\to$ $4$ exchanges
$2 \leftrightarrow \left\{3 , 4 , 5\right\}$ $\to$ $3$ exchanges
$3 \leftrightarrow \left\{4 , 5\right\}$ $\to$ $2$ exchanges
$4 \leftrightarrow \left\{5\right\}$ $\to$ $1$ exchange

(Try not to double-count; so don't count $1 \leftrightarrow 2$ and then $2 \leftrightarrow 1$; those are the same thing.)

This is true because all of the electrons are all indistinguishable. This results in an exchange energy stabilization of $10 \times {\Pi}_{e}$ (meaning ${\Pi}_{e} < 0$), one for each exchange.

So the pairing energy in total is:

$\Pi = 10 {\Pi}_{e}$

Now, suppose instead, the chromium atom was in this state:

$\underline{\uparrow \downarrow}$
$4 s$

$\underbrace{\underline{\uparrow \textcolor{w h i t e}{\downarrow}} \text{ "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}}$
$\text{ "" "" "" "" "" } \textcolor{w h i t e}{/} 3 d$

Now there is one pair of electrons that ARE repelling each other, so the coulombic repulsion energy is $1 \times {\Pi}_{c}$. However, the exchange energy stabilization now is only $6 \times {\Pi}_{e}$...

So the pairing energy in total is higher than before:

$\Pi = {\Pi}_{c} + 6 {\Pi}_{e}$

This is less stable by ${\Pi}_{c} - 4 {\Pi}_{e}$, and qualitatively shows that chromium should "prefer" the $3 {d}^{5} 4 {s}^{1}$ configuration over the $3 {d}^{4} 4 {s}^{2}$ configuration.