D.54 De­riva­tion of the Hartree-Fock equa­tions

This note de­rives the canon­i­cal Hartree-Fock equa­tions. It will use some lin­ear al­ge­bra; see the No­ta­tions sec­tion un­der ma­trix for some ba­sic con­cepts. The de­riva­tion will be per­formed un­der the nor­mally stated rules of en­gage­ment that the or­bitals are of the form $\pe{n}//u//$ or $\pe{n}//d//$. So the spins are cho­sen, and only the spa­tial or­bitals $\pe{n}////$ are to be found.

The de­riva­tions must al­low for the fact that in re­stricted Hartree-Fock, it is re­quired that pairs of spin-up and spin-down or­bitals have the same spa­tial or­bital. So there are three pos­si­ble kinds of spa­tial or­bitals. A spa­tial or­bital may pro­duce a sin­gle un­paired spin or­bital that is spin-up, or a sin­gle un­paired spin or­bital that is spin-down, or a pair of spin-up and spin-down or­bitals with the same spa­tial or­bital. These three types of spa­tial or­bitals will be re­ferred to as un­paired spin-up, un­paired spin-down, and re­stricted. Note that these names do not re­fer to prop­er­ties of the spa­tial or­bits them­selves, of course, but to the prop­er­ties of the spin or­bits that these spa­tial or­bitals pro­duce.

As­sume that there are $N_{\rm {u}}$ spin-up spa­tial or­bitals, $N_{\rm {d}}$ spin-down ones, and $N_{\rm {r}}$ re­stricted ones. The to­tal num­ber of spa­tial or­bitals, call it N, is then

$$N = N_{\rm {u}} + N_{\rm {d}} + N_{\rm {r}}$$

and that is the to­tal num­ber of un­known spa­tial or­bitals to find. A cor­re­spond­ing num­ber of $N$ equa­tions will be needed for them.

How­ever, the to­tal num­ber of spin or­bitals, $I$, is larger than $N$ by an ad­di­tional amount $N_{\rm {r}}$, be­cause the re­stricted spa­tial or­bitals ap­pear in both spin-up and spin-down ver­sions. That makes the math­e­mat­ics messy.

Things be­come a bit eas­ier if the or­der­ing of the or­bitals is spec­i­fied a pri­ori. The or­der­ing makes no dif­fer­ence phys­i­cally. So it will be as­sumed that the spa­tial or­bitals are or­dered with the un­paired spin-up ones first, the un­paired spin-down ones sec­ond, and the re­stricted ones last. The or­der­ing of the spin or­bitals will be the same as that of the spa­tial or­bitals, but with the re­stricted or­bitals at the end ap­pear­ing twice; first in the spin-up ver­sions and then in the spin-down ver­sions.

To find the spa­tial or­bitals, the vari­a­tional method as dis­cussed in chap­ter 9.1.3 says that the ex­pec­ta­tion en­ergy $\left\langle{E}\right\rangle$ must be un­changed un­der small changes in the or­bitals, pro­vided that the or­bitals re­main or­tho­nor­mal. To eas­ily en­force that or­tho­nor­mal­ity con­straint re­quires that terms are added to the change in or­bitals that pe­nal­ize for any go­ing out of bounds.

To do so, first note that $\left\langle{E}\right\rangle$ can be con­sid­ered to be a real func­tion from the real and imag­i­nary parts of the spa­tial or­bitals, and both these parts are real func­tions. The con­di­tion that any spa­tial or­bital $\pe{n}////$ must be nor­mal­ized means that the in­ner prod­uct of the or­bital with it­self must be 1,

$$\langle\pe n////\vert\pe n////\rangle=1$$

This con­di­tion is real too. How­ever, the con­di­tion that any spa­tial mode $\pe{n}////$ must be or­thog­o­nal to any other spa­tial mode $\pe{\underline n}////$ means that the in­ner prod­uct of the two modes must be zero,

$$\langle\pe n////\vert\pe{\underline n}////\rangle=0$$

In gen­eral this con­di­tion has both a real and an imag­i­nary com­po­nent. But it can be writ­ten as two real con­di­tions;

$${\textstyle\frac{1}{2}}\Big(\langle\pe n////\vert\pe{\under... ...e - \langle\pe{\underline n}////\vert\pe n////\rangle\Big)=0.$$

The rea­son is that if you swap the sides in an in­ner prod­uct, you get the com­plex con­ju­gate; there­fore the first equa­tion above is the real part of the in­ner prod­uct and the sec­ond the imag­i­nary part.

Since we now have a com­pletely real prob­lem in real in­de­pen­dent vari­ables, the penalty fac­tors (the La­grangian mul­ti­pli­ers) in the prob­lem will be real too. For rea­sons ev­i­dent in a sec­ond, the penalty fac­tor for the nor­mal­iza­tion con­di­tion above will be called $\epsilon_{nn}$, while the ones for the two real or­thog­o­nal­ity con­di­tions will be called $2\epsilon_{n{\underline n},r}$ and $2\epsilon_{n{\underline n},i}$, re­spec­tively. To avoid en­forc­ing the same or­thog­o­nal­ity con­di­tion twice, it is here as­sumed that ${\underline n}>n$.

The rea­son for these no­ta­tions is that in terms of them, the pe­nal­ized vari­a­tional con­di­tion that the spa­tial or­bitals must sat­isfy, chap­ter 9.1.3, takes the sim­ple form

$$\delta \left\langle{E}\right\rangle - \sum_{n=1}^N \sum_{{... ...} \delta \langle\pe n////\vert\pe{\underline n}////\rangle = 0$$

where $\delta$ de­notes a small change in the fol­low­ing quan­tity, ${\underline n}$ is now al­lowed to be both smaller or larger than $n$, and $\epsilon_{n{\underline n}}$ is a Her­mit­ian ma­trix, mean­ing that $\epsilon_{{\underline n}{}n}=\epsilon_{n{\underline n}}^*$

Note how­ever that two spa­tial or­bitals do not have to be or­thog­o­nal if one is a un­paired spin-up one and the other an un­paired spin-down one. In that case the spins take care of or­thog­o­nal­ity. This can be ac­co­mo­dated by stip­u­lat­ing that the penalty fac­tors of the cor­re­spond­ing con­straints are zero,

$$\epsilon_{n{\underline n}}=0 \quad\mbox{if $\pe{n}////$ is spin-up and $\pe{n}////$ is spin-down, or vice versa}$$

Next the vari­a­tional con­di­tion is to be eval­u­ated for a small change $\delta\pe{m}////$ in a sam­ple spa­tial wave func­tion $\pe{m}////$ where $m$ is no larger than $N$. This is straight­for­ward for the in­ner prod­ucts in the penalty terms. How­ever, the ex­pec­ta­tion value of en­ergy $\left\langle{E}\right\rangle$ was ob­tained in chap­ter 9.3.3 in terms of the spin, rather than spa­tial or­bitals:

$$\begin{eqnarray*} \left\langle{E}\right\rangle & = &\sum_{n=1}^I \langle\pe n//... ...angle{\updownarrow}_n\vert{\updownarrow}_{\underline n}\rangle^2 \end{eqnarray*}$$

(From here on, the ar­gu­ment of the first or­bital of a pair in ei­ther side of an in­ner prod­uct is taken to be the first in­ner prod­uct in­te­gra­tion vari­able ${\skew0\vec r}$ and the ar­gu­ment of the sec­ond or­bital is the sec­ond in­te­gra­tion vari­able ${\underline{\skew0\vec r}}$)

Tak­ing that into ac­count, the vari­a­tional con­di­tion for the $\delta\pe{m}////$ takes the messy form

$$\begin{eqnarray*} && [2?] \Big( \langle\delta\pe m////\vert h^{\rm e}\vert\p... ...N \epsilon_{nm} \langle\pe n////\vert\delta\pe m////\rangle = 0 \end{eqnarray*}$$

Here $[2?]$ means to in­sert a fac­tor 2 there if $m$ is one of the re­stricted spa­tial or­bitals, be­cause each of the two cor­re­spond­ing spin or­bitals pro­duces a term like that. And $[\langle{\updownarrow}_.\vert{\updownarrow}_.\rangle^2?]$ means leave away this in­ner prod­uct if $m$ is one of the re­stricted spa­tial or­bitals, be­cause ex­actly one of the two cor­re­spond­ing spin or­bitals has that in­ner prod­uct equal to one, and the other has it zero.

Note that the dif­fer­ence be­tween ${\underline n}$ and $n$ can from now on be ig­nored; the name of a sum­ma­tion vari­able makes no dif­fer­ence for the re­sult, and there are no longer name con­flicts in the in­di­vid­ual terms. Note also that the sums over $n$ (or ${\underline n}$) with up­per limit $I$ in­clude the re­stricted spa­tial or­bitals twice, once for each spin di­rec­tion.

The sec­ond term in each row in the ex­pres­sion above is just the com­plex con­ju­gate of the first. These sec­ond terms can be thrown out us­ing the same trick as in chap­ter 9.1.3. (In other words, av­er­age with the same equa­tion with $\delta\pe{m}////$ re­placed by $-{\rm i}\delta\pe{m}////$ and di­vided by ${\rm i}$.) And the in­te­grals with the fac­tors ${\textstyle\frac{1}{2}}$ are pair­wise the same; the dif­fer­ence is just a name swap of the in­ner prod­uct in­te­gra­tion vari­ables. So all there is re­ally left is

$$\begin{eqnarray*} && [2?] \langle\delta\pe m////\vert h^{\rm e}\vert\pe m////\r... ...{n=1}^N \epsilon_{mn}\langle\delta\pe m////\vert\pe n////\rangle \end{eqnarray*}$$

Now write out the in­ner prod­uct over the first po­si­tion co­or­di­nate ${\skew0\vec r}$, be­ing the ar­gu­ment of $\delta\pe{m}////$, for all terms:

$$\begin{eqnarray*} \lefteqn{\int_{{\rm all}\;{\skew0\vec r}}\delta\pe m////\stru... ...psilon_{mn}\pe n//// \\ && \bigg) { \rm d}^3{\skew0\vec r}= 0 \end{eqnarray*}$$

If this in­te­gral is to be zero for what­ever is $\delta\pe{m}////$, then the terms within the paren­the­ses must be zero. (Oth­er­wise just take $\delta\pe{m}////$ pro­por­tional to the par­en­thet­i­cal ex­pres­sion; you would get the norm of the ex­pres­sion, and that is only zero if the ex­pres­sion is.)

Un­avoid­ably then, the fol­low­ing equa­tions, one for each value of $m$, must be sat­is­fied:

$$[2?]h^{\rm e}\pe m//// + [2?] \sum_{n=1}^I \langle\pe n////\... ... m////\rangle\pe n//// = \sum_{n=1}^N \epsilon_{mn}\pe n////$$

This can be cleaned up a bit by di­vid­ing by [2?]:

$$\fbox{$\displaystyle h^{\rm e}\pe m//// + \sum_{n=1}^I \l... ...space{-4pt} \right\} \sum_{n=1}^N \epsilon_{mn}\pe n//// $}$$ (D.35)

These are the gen­eral Hartree-Fock equa­tions, one for each $m$ $\raisebox{-.3pt}{$\leqslant$}$ $N$. The up­per value be­tween braces ap­plies if the spa­tial or­bital $\pe{m}////$ is not a re­stricted one; oth­er­wise the lower value ap­plies. Re­call that the sums with up­per limit $I$ in­clude the re­stricted spa­tial or­bitals twice. And that $\epsilon_{mn}$ is zero if spa­tial or­bital $\pe{m}////$ is un­paired spin-up and $\pe{n}////$ un­paired spin-down or vice-versa. For such in­dex val­ues, $\langle{\updownarrow}_m\vert{\updownarrow}_n\rangle$ is zero too.

Note that the gen­eral Hartree-Fock equa­tion above in­cludes $N$ eigen­val­ues $\epsilon_{mn}$. The canon­i­cal equa­tions in­clude just a sin­gle eigen­value $\epsilon_m$. So to get the canon­i­cal Hartree-Fock equa­tions, the sum in the right hand side must be fur­ther sim­pli­fied to the form $\epsilon_m\pe{m}////$.

The re­stricted closed-shell Hartree-Fock case will be done first, since it is the eas­i­est one. Every spa­tial or­bital is re­stricted, so the lower choice in the curly brack­ets al­ways ap­plies. The sum­ma­tion up­per lim­its $I$, be­ing the num­ber of spin or­bitals, can be re­duced to the num­ber of spa­tial or­bitals $N$ by adding a fac­tor 2. We can also get rid of the fac­tor ${\textstyle\frac{1}{2}}$ in front of the $\epsilon_{mn}$ by sim­ply re­defin­ing them by that fac­tor. So for re­stricted closed-shell Hartree-Fock

$$h^{\rm e}\pe m//// + 2 \sum_{n=1}^N \langle\pe n////\vert ... ... m////\rangle\pe n//// = \sum_{n=1}^N \epsilon_{mn}\pe n////$$

Now the rea­son why all these $\epsilon_{mn}$ are there is be­cause the set of $N$ spa­tial or­bitals that gives the low­est en­ergy state are not unique. The equa­tion above ap­plies to a typ­i­cal set. Only a spe­cial set will get rid of the $\epsilon_{mn}$ for $n$ not equal to $m$, leav­ing only $\epsilon_{mm}$, which can then be de­fined to be $\epsilon_m$.

Each or­bital in the spe­cial set will be some com­bi­na­tion of the or­bitals in the typ­i­cal set above. In par­tic­u­lar, any or­bital in the spe­cial set, call it $\overline\pe\nu////$, will be a lin­ear com­bi­na­tion of the or­bitals $\pe{n}////$ in the typ­i­cal set as fol­lows:

$$\overline\pe\nu//// \equiv \sum_{n=1}^N c_{n,\nu} \pe n//// \qquad\mbox{for any $\nu=1,2,\ldots,N$}$$

where the num­bers $c_{1,\nu},c_{2,\nu},\ldots$ are the mul­ti­ples of the typ­i­cal or­bitals $\pe1////$, $\pe2////$, .... The com­plete set of num­bers $c_{n.\nu}$ for all pos­si­ble val­ues of both $n$ and $\nu$ can be writ­ten as a ma­trix, a ta­ble of num­bers. This ma­trix will be in­di­cated by $C$. The first in­dex in $c_{n,\nu}$, $n$, says what row in $C$ that co­ef­fi­cient is in, and the sec­ond in­dex, $\nu$, what col­umn.

The mul­ti­ples $c_{n,\nu}$ can­not be ar­bi­trary, be­cause the spe­cial or­bitals must still be or­tho­nor­mal. As noted ear­lier, they will be if they are nor­mal­ized (so the in­ner prod­uct of any or­bital with it­self is 1), and mu­tu­ally or­thog­o­nal (so the in­ner prod­uct of any or­bital with any other one is zero). In short, the re­quire­ment is that

$$\langle\overline\pe\mu////\vert\overline\pe\nu////\rangle=\delta_{\mu\nu}$$

where $\delta_{\mu\nu}$ is one if $\mu$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\nu$, and zero oth­er­wise. The set of num­bers $\delta_{\mu\nu}$ is called the “Kro­necker delta” or “unit ma­trix” or “iden­tity ma­trix.” (The iden­tity ma­trix is for ma­tri­ces what the num­ber 1 is for nor­mal num­bers; mul­ti­ply­ing an ar­bi­trary ma­trix or vec­tor by the iden­tity ma­trix does not change that ma­trix or vec­tor.)

Sub­sti­tut­ing in the ex­pres­sion for the spe­cial or­bitals above, mak­ing sure not to use the same name $\nu$ for two dif­fer­ent in­dices, the re­quire­ment be­comes

$$\sum_{m=1}^N \sum_{n=1}^N \langle c_{m,\mu}\pe m////\vert c_{n,\nu}\pe n//// \rangle=\delta_{\mu\nu}$$

or not­ing that num­bers come out of the left side of an in­ner prod­uct as com­plex con­ju­gates,

$$\sum_{m=1}^N \sum_{n=1}^N c^*_{m,\mu} c_{n,\nu} \langle\pe m////\vert\pe n////\rangle=\delta_{\mu\nu}$$

Now since the set of typ­i­cal or­bitals $\pe{n}////$ are al­ready or­tho­nor­mal, the in­ner prod­uct in the re­quire­ment above is only nonzero when $m$ is $n$, and then it is one. So drop­ping the zero terms that have $m\ne{}n$, the re­quire­ment on the co­ef­fi­cients sim­pli­fies to

$$\sum_{n=1}^N c^*_{n,\mu} c_{n,\nu} = \delta_{\mu\nu}$$

What does that mean? Well, for given val­ues of $\mu$ and $\nu$, con­sider the co­ef­fi­cients $c_{n,\mu}$ to form a vec­tor $\vec{u}_\mu$, where $n$ in­di­cates the com­po­nent num­ber of that vec­tor. Sim­i­larly, con­sider the co­ef­fi­cients $c_{n,\nu}$ to form a vec­tor $\vec{u}_\nu$. Then the left hand side in the re­quire­ment above is the in­ner (or dot, if real) prod­uct of these two vec­tors. So the set of vec­tors must be or­tho­nor­mal, just like the spe­cial or­bitals must be or­tho­nor­mal. So the ma­trix of co­ef­fi­cients $C$ must con­sist of or­tho­nor­mal vec­tors. Math­e­mati­cians call such ma­tri­ces “uni­tary,” rather than or­tho­nor­mal, since it is eas­ily con­fused with unit, and that keeps math­e­mati­cians in busi­ness ex­plain­ing all the con­fu­sion.

The Her­mit­ian ad­joint ma­trix $C^\dagger$ of $C$ is de­fined as the ma­trix you get by swap­ping the or­der of the in­dices of the el­e­ments of $C$ and adding a com­plex con­ju­gate. So by de­f­i­n­i­tion the fac­tor $c^*_{n,\mu}$ in the re­quire­ment above equals the co­ef­fi­cient $c^\dagger_{\mu,n}$ of $C^\dagger$. And ma­trix mul­ti­pli­ca­tion is de­fined such that then the sum over $n$ in the re­quire­ment gives ex­actly the co­ef­fi­cients of the prod­uct $C^\dagger{}C$. So the re­quire­ment above can be writ­ten as

$$C^\dagger C = I$$

where $I$ is the unit ma­trix. That means $C^\dagger$ is the in­verse ma­trix to $C$, $C^\dagger$ $\vphantom0\raisebox{1.5pt}{$=$}$ $C^{-1}$. Then you also have that $C$ is the in­verse of $C^\dagger$, $CC^\dagger$ $\vphantom0\raisebox{1.5pt}{$=$}$ $I$, which writes out to

$$\sum_{\nu=1}^N c_{n,\nu} c^*_{m,\nu} = \delta_{mn}.$$

This can be used to find the typ­i­cal or­bitals in terms of the spe­cial ones. To do so, pre­mul­ti­ply the ex­pres­sion for the spe­cial or­bitals as given ear­lier by $c^*_{m,\nu}$ and sum over $\nu$:

$$\sum_{\nu=1}^N c^*_{m,\nu} \overline\pe\nu//// = \sum_{\nu=1}^N c^*_{m,\nu} \sum_{n=1}^N c_{n\nu} \pe n////$$

As seen above, the sum over $\nu$ in the right hand side is just $\delta_{mn}$, so in the sum over $n$, only the term with $n$ equal to $m$ is nonzero:

$$\sum_{\nu=1}^N c^*_{m,\nu} \overline\pe\nu//// = \pe m////$$

That then gives any typ­i­cal or­bital $\pe{m}////$ in terms of a sum of the spe­cial or­bitals $\overline\pe\nu////$.

Now plug that into the non canon­i­cal re­stricted closed-shell Hartree-Fock equa­tions given ear­lier. Be care­ful not to use the same sum­ma­tion in­dex name twice in the same term; this de­riva­tion will use

$$\pe m//// = \sum_{\nu=1}^N c^*_{m,\nu} \overline\pe\nu//// ... ...//// = \sum_{\kappa=1}^N c^*_{n,\kappa} \overline\pe\kappa////$$

for $\pe{m}////$, the first oc­cur­rence of $\pe{n}////$ in the terms, and the sec­ond oc­cur­rence, re­spec­tively. Pre­mul­ti­ply it all by $C$, i.e. put $\sum_{m=1}^{N}c_{m,\mu}$ in front of each term. That cleans up to

$$h^{\rm e}\overline\pe\mu//// + 2 \sum_{\lambda=1}^N \lang... ..._{m,\mu} \epsilon_{mn} c^*_{n,\lambda} \overline\pe\lambda////$$

Note that the only thing that has changed more than just by sym­bol names is the ma­trix in the right hand side. Now for each sep­a­rate value of $\lambda$, take $c^*_{n\lambda}$ as the $\lambda$-th or­tho­nor­mal eigen­vec­tor of Her­mit­ian ma­trix $\epsilon_{mn}$, call­ing the eigen­value $\epsilon_\lambda$. Then by the de­f­i­n­i­tion of eigen­vec­tor,

$$\sum_{n=1}^N \epsilon_{mn} c^*_{n,\lambda} = \epsilon_\lambda c^*_{m,\lambda}$$

So the right hand side be­comes

$$\sum_{m=1}^{I} \sum_{\lambda=1}^N c_{m,\mu} \epsilon_{\lam... ...\overline\pe\lambda//// = \epsilon_{\mu} \overline\pe\mu////$$

So, in terms of the spe­cial or­bitals de­fined by the re­quire­ment that $c^*_{m,\mu}$ gives the $\mu$-th eigen­vec­tor of $\epsilon_{mn}$, the right hand side sim­pli­fies to the canon­i­cal one.

Since the old typ­i­cal or­bitals are no longer of in­ter­est, the over­lines on the spe­cial or­bitals can be dropped to save typ­ing, and the Greek in­dex names $\mu$ and $\lambda$ can be re­named $n$ and ${\underline n}$. That then fi­nally pro­duces the canon­i­cal closed-shell re­stricted Hartree-Fock equa­tions:

$$\fbox{$\displaystyle h^{\rm e}\pe n//// + 2 \sum_{{\under... ...n////\rangle\pe{\underline n}//// = \epsilon_n \pe n//// $}$$ (D.36)

Note that the left-hand side di­rectly pro­vides a Her­mit­ian Fock op­er­a­tor if you iden­tify it as ${\cal F}\pe{n}////$; there is no need to in­volve spin in the closed-shell re­stricted case. This also pro­vides a much sim­pler ex­pla­na­tion than all the al­ge­bra above why all the ear­lier $\epsilon_{mn}$ with $m\ne{}n$ were not needed; ex­is­tence of a set of or­tho­nor­mal eigen­func­tions of a Her­mit­ian op­er­a­tor is au­to­matic. So there is no fun­da­men­tal need to en­force that sep­a­rately through La­grangian mul­ti­pli­ers.

Turn­ing now to the case of (fully) un­re­stricted Hartree-Fock (UHF), you might make the same sim­ple ar­gu­ment as above and be done. But it is worth­while to go through the full math­e­mat­ics any­way, to bet­ter un­der­stand open-shell re­stricted Hartree-Fock later. In the un­re­stricted case, the non canon­i­cal equa­tions are

$$h^{\rm e}\pe m//// + \sum_{n=1}^N \langle\pe n////\vert v^... ... m////\rangle\pe n//// = \sum_{n=1}^N \epsilon_{mn}\pe n////$$

In this case, there are two dif­fer­ent types of spa­tial or­bitals; those ap­pear­ing in spin-up spin or­bitals, and those ap­pear­ing in spin-down spin or­bitals. You can­not just make ar­bi­trary com­bi­na­tions of all these or­bitals. If you com­bine spin-up and spin-down or­bitals, they cor­re­spond to spin or­bitals of un­cer­tain spin. That would make the as­sump­tions used to de­rive the Hartree-Fock equa­tions in­valid.

How­ever, com­bi­na­tions of purely spin-up or­bitals can still be made with­out prob­lems, and so can com­bi­na­tions of purely spin down or­bitals. To do the math­e­mat­ics, the spa­tial or­bitals can be sep­a­rated into two sets. The set of or­bital num­bers $n$ cor­re­spond­ing to spin-up spin or­bitals will be in­di­cated by U, and the set of num­bers $n$ cor­re­spond­ing to spin-down spin or­bitals by D. So you can par­ti­tion (sep­a­rate) the non canon­i­cal equa­tions above into equa­tions for $m\in{\rm {U}}$ (mean­ing $m$ is one of the val­ues in set U),

$$h^{\rm e}\pe m//// + \sum_{n\in{\rm U}} \langle\pe n////\v... ...\rangle\pe n//// = \sum_{n\in{\rm U}} \epsilon_{mn}\pe n////$$

and equa­tions for $m\in{\rm {D}}$,

$$h^{\rm e}\pe m//// + \sum_{n\in{\rm U}} \langle\pe n////\v... ...\rangle\pe n//// = \sum_{n\in{\rm D}} \epsilon_{mn}\pe n////$$

In these two types of equa­tions, the fact that the up and down spin states are or­thog­o­nal was used to get rid of one pair of sums, and an­other pair was elim­i­nated by the fact that there are no La­grangian vari­ables $\epsilon_{mn}$ link­ing the sets, since the spa­tial or­bitals in the two sets are al­lowed to be mu­tu­ally non or­thog­o­nal.

Now sep­a­rately re­place the or­bitals of the up and down states by a mod­i­fied set just like for the re­stricted closed-shell case above, for each us­ing the uni­tary ma­trix of eigen­vec­tors of the $\epsilon_{mn}$ co­ef­fi­cients ap­pear­ing in the right hand side of the equa­tions for that set. It leaves the equa­tions in­tact ex­cept for changes in names, but gets rid of the $\epsilon_{mn}$ for $m$ $\raisebox{.2pt}{$\ne$}$ $n$, leav­ing only $\epsilon_{mm}$ val­ues, call them $\epsilon_m$. Then com­bine the spin-up and spin-down equa­tions again into a sin­gle ex­pres­sion. You get, in terms of re­vised sym­bol names,

$$\fbox{$\displaystyle h^{\rm e}\pe n//// + \sum_{{\underli... ...////\rangle\pe {\underline n}//// = \epsilon_n \pe n//// $}$$ (D.37)

That leaves only the re­stricted open-shell Hartree-Fock method. Here, the par­ti­tion­ing also needs to in­clude the set R of of re­stricted or­bitals be­sides U and D. There is now a prob­lem, be­cause you can­not make com­bi­na­tions of re­stricted or­bitals with spin-up or spin-down or­bitals. That means that the $\epsilon_{mn}$ val­ues where ei­ther $m$ or $n$ is re­stricted and the other is not, can­not be elim­i­nated. So­lu­tions range from just ig­nor­ing the whole thing to prop­erly ac­count­ing for these $\epsilon_{mn}$ val­ues by en­forc­ing that re­stricted and non re­stricted or­bitals must stay or­thog­o­nal as ad­di­tional equa­tions. This (even more) elab­o­rate case will be left to the ref­er­ences that you can find in [46], in par­tic­u­lar [28, pp. 242-253].

Woof.