### D.54 De­riva­tion of the Hartree-Fock equa­tions

This note de­rives the canon­i­cal Hartree-Fock equa­tions. The de­riva­tion be­low will be per­formed un­der the nor­mally stated rules of en­gage­ment that the or­bitals are of the form or , so that only the spa­tial or­bitals are con­tin­u­ously vari­able. The de­riva­tions al­low for the fact that some spa­tial spin states may be con­strained to be equal.

First, you can make things a lot less messy by a pri­ori spec­i­fy­ing the or­der­ing of the or­bitals. The or­der­ing makes no dif­fer­ence phys­i­cally, and it sim­pli­fies the math­e­mat­ics. In par­tic­u­lar, in re­stricted Hartree-Fock some spa­tial or­bitals ap­pear in pairs, but you can only count each spa­tial or­bital as one un­known func­tion. The eas­i­est way to han­dle that is to push the spin-down ver­sions of the du­pli­cated spa­tial or­bitals to the end of the Slater de­ter­mi­nant. Then the start of the de­ter­mi­nant com­prises a list of unique spa­tial or­bitals.

So, it will be as­sumed that the or­bitals are or­dered as fol­lows:

1.
the paired spa­tial states in their spin-up ver­sion; as­sume there are 0 of them;
2.
un­paired spin-up states; as­sume there are of them;
3.
un­paired spin-down states; as­sume there are of them;
4.
and fi­nally, the paired spa­tial states in their spin-down ver­sion.
That means that the Slater-de­ter­mi­nant wave func­tion looks like:

The to­tal num­ber of un­known spa­tial or­bitals is , and you need a cor­re­spond­ing equa­tions for them.

The vari­a­tional method dis­cussed in chap­ter 9.1 says that the ex­pec­ta­tion en­ergy must be un­changed un­der small changes in the or­bitals, pro­vided that penalty terms are added for changes that vi­o­late the or­tho­nor­mal­ity re­quire­ments on the or­bitals.

The ex­pec­ta­tion value of en­ergy was in chap­ter 9.3.3 found to be:

(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 and the ar­gu­ment of the sec­ond or­bital is the sec­ond in­te­gra­tion vari­able )

The penalty terms re­quire penalty fac­tors called La­grangian vari­ables. The penalty fac­tor for vi­o­la­tions of the nor­mal­iza­tion re­quire­ment

will be called for rea­sons ev­i­dent in a sec­ond. Or­thog­o­nal­ity be­tween any two spa­tial or­bitals and re­quires

where the first con­straint says that the real part of must be zero and the sec­ond that its imag­i­nary part must be zero too. (Re­mem­ber that if you switch sides in an in­ner prod­uct, you turn it into its com­plex con­ju­gate.) To avoid in­clud­ing the same or­thog­o­nal­ity con­di­tion twice, the con­straint will be writ­ten only for . The penalty fac­tor for the first con­straint will be called , and the one for the sec­ond con­straint .

In those terms, the pe­nal­ized change in ex­pec­ta­tion en­ergy be­comes, in the re­stricted case that all unique spa­tial or­bitals are mu­tu­ally or­thog­o­nal,

where is an Her­mit­ian ma­trix, with . The no­ta­tions for the La­grangian vari­ables were cho­sen above to achieve this fi­nal re­sult.

But for un­re­stricted Hartree-Fock, spa­tial or­bitals are not re­quired to be or­thog­o­nal if they have op­po­site spin, be­cause the spins will take care of or­thog­o­nal­ity. You can re­move the er­ro­neously added con­straints by sim­ply spec­i­fy­ing that the cor­re­spond­ing La­grangian vari­ables are zero:

or equiv­a­lently, if , or , .

Now work out the pe­nal­ized change in ex­pec­ta­tion en­ergy due to a change in the val­ues of a se­lected spa­tial or­bital with . It is

OK, OK it is a mess. Sums like are for the re­stricted Hartree-Fock case, in which spa­tial or­bital may ap­pear twice in the Slater de­ter­mi­nant. From now on, just write them as , mean­ing, put in a fac­tor 2 if or­bital ap­pears twice. The ex­cep­tion is for the ex­change in­te­grals which pro­duce ex­actly one nonzero spin prod­uct; write that as , mean­ing take out that prod­uct if the or­bital ap­pears twice.

Next, note that the sec­ond term in each row is just the com­plex con­ju­gate of the first. Con­sid­er­ing as a sec­ond pos­si­ble change in or­bital, as was done in the ex­am­ple in chap­ter 9.1, it is seen that the first terms by them­selves must be zero, so you can just ig­nore the sec­ond term in each row. And the in­te­grals with the fac­tors are pair­wise the same; the dif­fer­ence is just a name swap of the first and sec­ond sum­ma­tion and in­te­gra­tion vari­ables. So all that you re­ally have left is

Now note that if you write out the in­ner prod­uct over the first po­si­tion co­or­di­nate, you will get an in­te­gral of the gen­eral form

If this in­te­gral is to be zero for what­ever you take , then the terms within the paren­the­ses must be zero. (Just take pro­por­tional to the par­en­thet­i­cal ex­pres­sion; you would get the in­te­gral of an ab­solute square, only zero if the square is.) Un­avoid­ably, you must have that

You can di­vide by [2]:

 (D.35)

where you use the lower of the choices be­tween the braces in the case that spa­tial or­bital ap­pears twice in the Slater de­ter­mi­nant, or equiv­a­lently, if . There you have the Hartree-Fock equa­tions, one for each . Re­call that they ap­ply as­sum­ing the or­der­ing of the Slater de­ter­mi­nant given in the be­gin­ning, and that for un­re­stricted Hartree-Fock, is zero if 0 is.

How about those , you say? Shouldn’t the right hand side just be ? Ah, you want the canon­i­cal Hartree-Fock equa­tions, not just the plain vanilla ver­sion.

OK, let’s do the re­stricted closed-shell Hartree-Fock case first, then, since it is the eas­i­est one. Every state is paired, so the lower choice in the curly brack­ets al­ways ap­plies, and the num­ber of unique un­known spa­tial states is ​2. Also, you can re­duce the sum­ma­tion up­per lim­its to ​2 if you add a fac­tor 2, since the sec­ond half of the spa­tial or­bitals are the same as the first half. So you get

Now, sup­pose that you de­fine a new set of or­bitals, each a lin­ear com­bi­na­tion of the cur­rent ones:

where the are the mul­ti­ples of the orig­i­nal or­bitals. Will the new ones still be an or­tho­nor­mal set? Well, they will be if

where is the Kro­necker delta, one if , zero oth­er­wise. Sub­sti­tut­ing in the de­f­i­n­i­tion of the new or­bitals, mak­ing sure not to use the same name for two dif­fer­ent in­dices,

Now note that the are or­tho­nor­mal, so to get a nonzero value, must be , and you get

Con­sider to be the com­po­nent in­dex of a vec­tor. Then this re­ally says that vec­tors and must be or­tho­nor­mal. So the ma­trix of co­ef­fi­cients 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.

Call the com­plete ma­trix . Then, ac­cord­ing to the rules of ma­trix mul­ti­pli­ca­tion and Her­mit­ian ad­joint, the or­tho­nor­mal­ity con­di­tion above is equiv­a­lent to where is the unit ma­trix. That means is the in­verse ma­trix to , and then you also have :

Now pre­mul­ti­ply the de­f­i­n­i­tion of the new or­bitals above by ; you get

but the sum over in the right hand side is just and you get

That gives you an ex­pres­sion for the orig­i­nal or­bitals in terms of the new ones. For aes­thetic rea­sons, you might just as well reno­tate to , the Greek equiv­a­lent of , to get

Now plug that into the non­canon­i­cal re­stricted closed-shell Hartree-Fock equa­tions, with equiv­a­lent ex­pres­sions for us­ing what­ever sum­ma­tion vari­able is still avail­able,

and use the re­duc­tion for­mula ,

pre­mul­ti­ply­ing all by , i.e. put be­fore each term. You get

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 value of , take as the -​th or­tho­nor­mal eigen­vec­tor of Her­mit­ian ma­trix , call­ing the eigen­value . Then the right hand side be­comes

So, in terms of the new or­bitals de­fined by the re­quire­ment that gives the eigen­vec­tors of , the right hand side sim­pli­fies to the canon­i­cal one.

Since you no longer care about the old or­bitals, you can drop the over­lines on the new ones, and re­vert to sen­si­ble ro­man in­dices and in­stead of the Greek ones and . You then have the canon­i­cal re­stricted closed-shell Hartree-Fock equa­tions

 (D.36)

that, as noted, as­sume that the Slater de­ter­mi­nant is or­dered so that the ​2 spin-up or­bitals are at the start of it. Note that the left hand side di­rectly pro­vides a Her­mit­ian Fock op­er­a­tor if you iden­tify it as ; there is no need to in­volve spin here.

In the un­re­stricted case, the non­canon­i­cal equa­tions are

In this case the spin-up and spin-down spa­tial states are not mu­tu­ally or­tho­nor­mal, and you want to re­de­fine the group of spin up states and the group of spin down states sep­a­rately.

The term in lin­ear al­ge­bra is that you want to par­ti­tion your ma­trix. What that means is sim­ply that you sep­a­rate the or­bital num­bers into two sets. The set of num­bers of spin-up or­bitals will be in­di­cated as U, and the set of val­ues of spin-down ones by D. So you can par­ti­tion (sep­a­rate) the non­canon­i­cal equa­tions above into equa­tions for (mean­ing is one of the val­ues in set U):

and equa­tions for

In these two 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 link­ing the sets, since the spa­tial or­bitals in the sets are al­lowed to be mu­tu­ally nonorthog­o­nal.

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 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 equiv­a­lent of the for , leav­ing only -equiv­a­lent val­ues. Then com­bine the spin-up and spin-down equa­tions again into a sin­gle ex­pres­sion. You get, in terms of re­vised names,

 (D.37)

In the re­stricted open-shell Hartree-Fock method, the par­ti­tion­ing also needs to in­clude the set P of or­bitals whose spa­tial or­bitals ap­pear both with spin-up and spin-down in the Slater de­ter­mi­nant. In that case, the pro­ce­dure above to elim­i­nate the val­ues for no longer works, since there are co­ef­fi­cients re­lat­ing the sets. This (even more) elab­o­rate case will be left to the ref­er­ences that you can find in [45].

Woof.