A.34 Ex­pla­na­tion of Hund’s first rule

Hund’s first rule of spin-align­ment ap­plies be­cause elec­trons in atoms pre­fer to go into spa­tial states that are an­ti­sym­met­ric with re­spect to elec­tron ex­change. Spin align­ment is then an un­avoid­able con­se­quence of the weird an­ti­sym­metriza­tion re­quire­ment.

To un­der­stand why elec­trons want to go into an­ti­sym­met­ric spa­tial states, the in­ter­ac­tions be­tween the elec­trons need to be con­sid­ered. Sweep­ing them be­low the car­pet as the dis­cus­sion of atoms in chap­ter 5.9 did is not go­ing to cut it.

To keep it as sim­ple as pos­si­ble, the case of the car­bon atom will be con­sid­ered. As the crude model of chap­ter 5.9 did cor­rectly de­duce, the car­bon atom has two 1s elec­trons locked into a zero-spin sin­glet state, and sim­i­larly two 2s elec­trons also in a sin­glet state. Hund’s rule is about the fi­nal two elec­trons that are in 2p states. As far as the sim­ple model of chap­ter 5.9 was con­cerned, these elec­trons can do what­ever they want within the 2p sub­shell.

To go one bet­ter than that, the cor­rect in­ter­ac­tions be­tween the two 2p elec­trons will need to be con­sid­ered. To keep the ar­gu­ments man­age­able, it will still be as­sumed that the ef­fects of the 1s and 2s elec­trons are in­de­pen­dent of where the 2p elec­trons are.

Call the 2p elec­trons $\alpha$ and $\beta$. Un­der the stated con­di­tions, their Hamil­ton­ian takes the form

$$H_\alpha + H_\beta + V_{\alpha \beta}$$

where $H_\alpha$ and $H_\beta$ are the sin­gle-elec­tron Hamil­to­ni­ans for the elec­trons $\alpha$ and $\beta$, con­sist­ing of their ki­netic en­ergy, their at­trac­tion to the nu­cleus, and the re­pul­sion by the 1s and 2s elec­trons. Note that in the cur­rent analy­sis, it is not re­quired that the 1s and 2s elec­trons are treated as lo­cated in the nu­cleus. Lack of shield­ing can be al­lowed now, but it must still be as­sumed that the 1s and 2s elec­trons are un­af­fected by where the 2p elec­trons are. In par­tic­u­lar, $H_\alpha$ is as­sumed to be be in­de­pen­dent of the po­si­tion of elec­tron $\beta$, and $H_\beta$ in­de­pen­dent of the po­si­tion of elec­tron $\alpha$. The mu­tual re­pul­sion of the two 2p elec­trons is given by $V_{\alpha\beta}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^2$$\raisebox{.5pt}{$/$}$​$4\pi\epsilon_0\vert{\skew0\vec r}_\alpha-{\skew0\vec r}_\beta\vert$.

Now as­sume that elec­trons $\alpha$ and $\beta$ ap­pro­pri­ate two sin­gle-elec­tron spa­tial 2p states for them­selves, call them $\psi_1$ and $\psi_2$. For car­bon, $\psi_1$ can be thought of as the 2p$_z$ state and $\psi_2$ as the 2p$_x$ state, The gen­eral spa­tial wave func­tion de­scrib­ing the two elec­trons takes the generic form

$$a\psi_1({\skew0\vec r}_1)\psi_2({\skew0\vec r}_2) + b\psi_2({\skew0\vec r}_1)\psi_1({\skew0\vec r}_2).$$

The two states $\psi_1$ and $\psi_2$ will be taken to be or­tho­nor­mal, like p$_z$ and p$_x$ are, and then the nor­mal­iza­tion re­quire­ment is that $\vert a\vert^2+\vert b\vert^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.

The ex­pec­ta­tion value of en­ergy is

$$\langle a\psi_1\psi_2 + b\psi_2\psi_1 \vert H_\alpha + H_\... ...+ V_{\alpha\beta} \vert a\psi_1\psi_2 + b\psi_2\psi_1\rangle.$$

That can be mul­ti­plied out and then sim­pli­fied by not­ing that in the var­i­ous in­ner prod­uct in­te­grals in­volv­ing the sin­gle-elec­tron Hamil­to­ni­ans, the in­te­gral over the co­or­di­nate un­af­fected by the Hamil­ton­ian is ei­ther zero or one be­cause of or­tho­nor­mal­ity. Also, the in­ner prod­uct in­te­grals in­volv­ing $V_{\alpha\beta}$ are pair­wise the same, the dif­fer­ence be­ing just a change of names of in­te­gra­tion vari­ables.

The sim­pli­fied ex­pec­ta­tion en­ergy is then:

$$E_{\psi_1} + E_{\psi_2} + \langle\psi_1\psi_2\vert V_{\alp... ...ngle\psi_1\psi_2\vert V_{\alpha\beta}\vert\psi_2\psi_1\rangle.$$

The first two terms are the sin­gle-elec­tron en­er­gies of states $\psi_1$ and $\psi_2$. The third term is the clas­si­cal re­pul­sion be­tween be­tween two elec­tron charge dis­tri­b­u­tions of strengths $\vert\psi_1\vert^2$ and $\vert\psi_2\vert^2$. The elec­trons min­i­mize this third term by go­ing into spa­tially sep­a­rated states like the 2p$_x$ and 2p$_z$ ones, rather than into the same spa­tial state or into greatly over­lap­ping ones.

The fi­nal one of the four terms is the in­ter­est­ing one for Hund’s rule; it de­ter­mines how the two elec­trons oc­cupy the two states $\psi_1$ and $\psi_2$, sym­met­ri­cally or an­ti­sym­met­ri­cally. Con­sider the de­tailed ex­pres­sion for the in­ner prod­uct in­te­gral ap­pear­ing in the term:

$$\langle\psi_1\psi_2\vert V_{\alpha\beta}\vert\psi_2\psi_1\r... ...ec r}_1) {\,\rm d}^3{\skew0\vec r}_1{\rm d}^3{\skew0\vec r}_2$$

where $f({\skew0\vec r}_1,{\skew0\vec r}_2)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_2({\skew0\vec r}_1)\psi_1({\skew0\vec r}_2)$.

The sign of this in­ner prod­uct can be guessti­mated. If $V_{\alpha\beta}$ would be the same for all elec­tron sep­a­ra­tion dis­tances, the in­te­gral would be zero be­cause of or­tho­nor­mal­ity of $\psi_1$ and $\psi_2$. How­ever, $V_{\alpha\beta}$ fa­vors po­si­tions where ${\skew0\vec r}_1$ and ${\skew0\vec r}_2$ are close to each other; in fact $V_{\alpha\beta}$ is in­fi­nitely large if ${\skew0\vec r}_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew0\vec r}_2$. At such a lo­ca­tion $f({\skew0\vec r}_1,{\skew0\vec r}_2)f^*({\skew0\vec r}_2,{\skew0\vec r}_1)$ is a pos­i­tive real num­ber, so it tends to have a pos­i­tive real part in re­gions it re­ally counts. That means the in­ner prod­uct in­te­gral should have the same sign as $V_{\alpha\beta}$; it should be re­pul­sive.

And since this in­te­gral is mul­ti­plied by $a^*b+b^*a$, the en­ergy is small­est when that is most neg­a­tive, which is for the an­ti­sym­met­ric spa­tial state $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$$b$. Since this state takes care of the sign change in the an­ti­sym­metriza­tion re­quire­ment, the spin state must be un­changed un­der par­ti­cle ex­change; the spins must be aligned. More pre­cisely, the spin state must be some lin­ear com­bi­na­tion of the three triplet states with net spin one. There you have Hund’s rule, as an ac­ci­den­tal byprod­uct of the Coulomb re­pul­sion.

This leaves the philo­soph­i­cal ques­tion why for the two elec­trons of the hy­dro­gen mol­e­cule in chap­ter 5.2 the sym­met­ric state is en­er­get­i­cally most fa­vor­able, while the an­ti­sym­met­ric state is the one for the 2p elec­trons. The real dif­fer­ence is in the ki­netic en­ergy. In both cases, the an­ti­sym­met­ric com­bi­na­tion re­duces the Coulomb re­pul­sion en­ergy be­tween the elec­trons, and in the hy­dro­gen mol­e­cule model, it also in­creases the nu­clear at­trac­tion en­ergy. But in the hy­dro­gen mol­e­cule model, the sym­met­ric state achieves a re­duc­tion in ki­netic en­ergy that is more than enough to make up for it all. For the 2p elec­trons, the re­duc­tion in ki­netic en­ergy is nil. When the pos­i­tive com­po­nent wave func­tions of the hy­dro­gen mol­e­cule model are com­bined into the sym­met­ric state, they al­low greater ac­cess to fringe ar­eas far­ther away from the nu­clei. Be­cause of the un­cer­tainty prin­ci­ple, less con­fined elec­trons tend to have less in­de­ter­mi­nacy in mo­men­tum, hence less ki­netic en­ergy. On the other hand, the 2p states are half pos­i­tive and half neg­a­tive, and even their sym­met­ric com­bi­na­tion re­duces spa­tial ac­cess for the elec­trons in half the lo­ca­tions.