Sub­sec­tions


9.3 The Hartree-Fock Ap­prox­i­ma­tion

Many of the most im­por­tant prob­lems that you want to solve in quan­tum me­chan­ics are all about atoms and/or mol­e­cules. These prob­lems in­volve a num­ber of elec­trons around a num­ber of atomic nu­clei. Un­for­tu­nately, a full quan­tum so­lu­tion of such a sys­tem of any non­triv­ial size is very dif­fi­cult. How­ever, ap­prox­i­ma­tions can be made, and as sec­tion 9.2 ex­plained, the real skill you need to mas­ter is solv­ing the wave func­tion for the elec­trons given the po­si­tions of the nu­clei.

But even given the po­si­tions of the nu­clei, a brute-force so­lu­tion for any non­triv­ial num­ber of elec­trons turns out to be pro­hib­i­tively la­bo­ri­ous. The Hartree-Fock ap­prox­i­ma­tion is one of the most im­por­tant ways to tackle that prob­lem, and has been so since the early days of quan­tum me­chan­ics. This sec­tion ex­plains some of the ideas.


9.3.1 Wave func­tion ap­prox­i­ma­tion

The key to the ba­sic Hartree-Fock method is the as­sump­tions it makes about the form of the elec­tron wave func­tion. It will be as­sumed that there are a to­tal of $I$ elec­trons in or­bit around a num­ber of nu­clei. The wave func­tion de­scrib­ing the set of elec­trons then has the gen­eral form:

\begin{displaymath}
\Psi({\skew0\vec r}_1,S_{z1},{\skew0\vec r}_2,S_{z2},\ldots,{\skew0\vec r}_i,S_{zi},\ldots
{\skew0\vec r}_I,S_{zI})
\end{displaymath}

where ${\skew0\vec r}_i$ is the po­si­tion of the elec­tron num­bered $i$, and $S_{zi}$ its spin (i.e. in­ter­nal an­gu­lar mo­men­tum) in a cho­sen $z$-​di­rec­tion. Re­call that while the po­si­tion of an elec­tron can be any­where in three-di­men­sion­al space, its spin com­po­nent $S_z$ can have only two mea­sur­able val­ues: $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\hbar$ or $-\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\hbar$. Be­cause of the fac­tor $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, an elec­tron is a “par­ti­cle of spin one-half”. Such a par­ti­cle is also called a “spin dou­blet” be­cause of the two pos­si­ble spin val­ues.

The square mag­ni­tude of the wave func­tion above gives the prob­a­bil­ity for the elec­trons $i=1,2,\ldots,I$ to be near the po­si­tion ${\skew0\vec r}_i$, per unit vol­ume, with spin com­po­nent $S_{zi}$.

Of course, what the wave func­tion is will also de­pend on where the nu­clei are. How­ever, in this sec­tion, the nu­clei are sup­posed to be at given po­si­tions. There­fore to re­duce the clut­ter, the de­pen­dence of the elec­tron wave func­tion on the nu­clear po­si­tions will not be shown ex­plic­itly.

Hartree-Fock ap­prox­i­mates the wave func­tion above in terms of sin­gle-elec­tron wave func­tions. Each sin­gle-elec­tron wave func­tion takes the form of a prod­uct of a spa­tial func­tion $\pe////$ of the elec­tron po­si­tion ${\skew0\vec r}$, times a func­tion of the elec­tron spin com­po­nent $S_z$. The spin func­tion is ei­ther taken to be ${\uparrow}$ or ${\downarrow}$; by de­f­i­n­i­tion, func­tion ${\uparrow}(S_z)$ equals 1 if the spin $S_z$ is $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\hbar$, and 0 if it is $-\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\hbar$. Con­versely, func­tion ${\downarrow}(S_z)$ equals 0 if $S_z$ is $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\hbar$ and 1 if it is $-\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\hbar$. Func­tion ${\uparrow}$ is called spin-up and ${\downarrow}$ spin-down.

A com­plete sin­gle-elec­tron wave func­tion is then of the form

\begin{displaymath}
\pe/{\skew0\vec r}/b/z/
\end{displaymath}

where ${\updownarrow}$ is ei­ther ${\uparrow}$ or ${\downarrow}$. Such a sin­gle-elec­tron wave func­tion is called an “or­bital” or more ac­cu­rately a “spin or­bital.” The rea­son is that peo­ple tend to think of the sin­gle-elec­tron wave func­tion as de­scrib­ing a sin­gle elec­tron be­ing in a par­tic­u­lar or­bit around the nu­clei with a par­tic­u­lar spin. Wrong, of course: the elec­trons do not have well-de­fined po­si­tions on these scales, so you can­not talk about or­bits But peo­ple do tend to think of the spa­tial or­bitals $\pe/{\skew0\vec r}///$ that way any­way.

For sim­plic­ity, it will be as­sumed that the spin or­bitals are taken to be nor­mal­ized; if you in­te­grate the square mag­ni­tude of $\pe/{\skew0\vec r}/b/z/$ over all pos­si­ble po­si­tions ${\skew0\vec r}$ of the elec­tron and sum over the two pos­si­ble val­ues of its spin $S_z$, you get 1. Phys­i­cally that merely ex­presses that the elec­tron must be at some po­si­tion and have some spin for cer­tain (prob­a­bil­ity 1). The in­te­gral plus sum com­bi­na­tion can be ex­pressed us­ing the con­cise bra[c]ket no­ta­tion from chap­ter 2;

\begin{displaymath}
\left\langle\vphantom{\pe//b//}\pe//b//\hspace{-\nulldelimi...
...right.\!\left\vert\vphantom{\pe//b//}\pe//b//\right\rangle = 1
\end{displaymath}

Such a bracket, or in­ner prod­uct, is equiv­a­lent to a dot prod­uct for func­tions.

If there is more than one elec­tron, as will be as­sumed in this sec­tion, a sin­gle spin or­bital $\pe//b//$ is not enough to cre­ate a valid wave func­tion for the com­plete sys­tem. In fact, the “Pauli ex­clu­sion prin­ci­ple” says that each of the $I$ elec­trons must go into a dif­fer­ent spin or­bital, chap­ter 5.7. So a se­ries of or­bitals is needed,

\begin{displaymath}
\pe 1//b//, \pe 2//b//, \ldots, \pe n//b//, \ldots, \pe N//b//
\end{displaymath}

where the num­ber of or­bitals $N$ must be at least as big as the num­ber of elec­trons $I$.

It will be as­sumed that any two dif­fer­ent spin or­bitals $\pe{n}//b//$ and $\pe{{\underline n}}//b//$ are taken to be or­thog­o­nal; by de­f­i­n­i­tion this means that their bracket is zero:

\begin{displaymath}
\left\langle\vphantom{\pe {\underline n}//b//}\pe n//b//\hs...
...}//b//\right\rangle = 0 \quad\mbox{if}\quad n\ne{\underline n}
\end{displaymath}

In short, it is as­sumed that the set of spin or­bitals is or­tho­nor­mal; mu­tu­ally or­thog­o­nal and nor­mal­ized.

Note that the bracket above can be writ­ten as a prod­uct of a spa­tial bracket and a spin one:

\begin{displaymath}
\left\langle\vphantom{\pe {\underline n}//b//}\pe n//b//\hs...
...m{{\updownarrow}_n}{\updownarrow}_{\underline n}\right\rangle
\end{displaymath}

So for dif­fer­ent spin or­bitals to be or­thog­o­nal, ei­ther the spa­tial states or the spin states must or­thog­o­nal; they do not both need to be or­thog­o­nal. (To ver­ify the ex­pres­sion above, just write the first bracket out in terms of a spa­tial in­te­gral over ${\skew0\vec r}$ and a sum over the two val­ues of $S_z$ and re­order terms.)

Note also that the spin states ${\uparrow}$ and ${\downarrow}$ are an or­tho­nor­mal set:

\begin{displaymath}
\left\langle\vphantom{{\uparrow}}{\uparrow}\hspace{-\nullde...
...\!\left\vert\vphantom{{\downarrow}}{\uparrow}\right\rangle = 0
\end{displaymath}

So if the spin states are op­po­site, the spa­tial states do not need to be or­thog­o­nal. In fact, the spa­tial states can then be the same.

The base Hartree-Fock method uses the ab­solute min­i­mum num­ber of or­bitals $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ $I$. In that case, the sim­plest you could do to cre­ate a sys­tem wave func­tion is to put elec­tron 1 in or­bital 1, elec­tron 2 in or­bital 2, etcetera. That would give the sys­tem wave func­tion

\begin{displaymath}
\pe1/{\skew0\vec r}_1/b/z1/ \pe2/{\skew0\vec r}_2/b/z2/ \pe3/{\skew0\vec r}_3/b/z3/ \ldots \pe I/{\skew0\vec r}_I/b/zI/
\end{displaymath}

A prod­uct of sin­gle-elec­tron wave func­tions like this is called a “Hartree prod­uct.”

But a sin­gle Hartree prod­uct like the one above is phys­i­cally not ac­cept­able as a wave func­tion. The Pauli ex­clu­sion prin­ci­ple is only part of what is needed, chap­ter 5.7. The full re­quire­ment is that a sys­tem wave func­tion must be an­ti­sym­met­ric un­der elec­tron ex­change: the wave func­tion must sim­ply change sign when any two elec­trons are swapped. But if, say, elec­trons 1 and 2 are swapped in the Hartree prod­uct above, it pro­duces the new Hartree prod­uct

\begin{displaymath}
\pe1/{\skew0\vec r}_2/b/z2/ \pe2/{\skew0\vec r}_1/b/z2/ \pe3/{\skew0\vec r}_3/b/z3/ \ldots \pe I/{\skew0\vec r}_I/b/zI/
\end{displaymath}

That is a fun­da­men­tally dif­fer­ent wave func­tion, not just mi­nus the first Hartree prod­uct; or­bitals $\pe1//b//$ and $\pe2//b//$ are not al­lowed to be equiv­a­lent.

To get a wave func­tion that does sim­ply change sign when elec­trons 1 and 2 are swapped, you can take the first Hartree prod­uct mi­nus the sec­ond one. That solves that prob­lem. But it is not enough: the wave func­tion must also sim­ply change sign if elec­trons 1 and 3 are swapped. Or if 2 and 3 are swapped, etcetera.

So you must add more Hartree prod­ucts with swapped elec­trons to the mix. A lot more in fact. There are $I!$ ways to or­der $I$ elec­trons, and each or­der­ing adds one Hartree prod­uct to the mix. (The Hartree prod­uct gets a plus sign or a mi­nus sign in the mix de­pend­ing on whether the num­ber of swaps to get there from the first one is even or odd). So for, say, a sin­gle car­bon atom with $I$ $\vphantom0\raisebox{1.5pt}{$=$}$ 6 elec­trons, writ­ing down the full Hartree-Fock wave func­tion would mean writ­ing down $6!$ $\vphantom0\raisebox{1.5pt}{$=$}$ 720 Hartree prod­ucts. Roughly a thou­sand of them, in short. Of course, writ­ing all that out would be in­sane. For­tu­nately, there is a more con­cise way to write the com­plete wave func­tion; it uses a so-called Slater de­ter­mi­nant,

\begin{displaymath}
\frac{1}{\sqrt{I!}}
\left\vert
\begin{array}{cccccc}
\pe...
...\ldots & \pe I/{\skew0\vec r}_I/b/zI/
\end{array} \right\vert
\end{displaymath} (9.15)

The de­ter­mi­nant mul­ti­plies out to the $I!$ in­di­vid­ual Hartree prod­ucts. (See chap­ter 5.7 and the no­ta­tions sec­tion for more on de­ter­mi­nants.) The fac­tor $1$$\raisebox{.5pt}{$/$}$$/\sqrt{I!}$ is there to en­sure that the wave func­tion re­mains nor­mal­ized af­ter sum­ming the $I!$ Hartree prod­ucts to­gether.

The most gen­eral sys­tem wave func­tion $\Psi$ us­ing only $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ $I$ or­bitals is any co­ef­fi­cient $a$ of mag­ni­tude 1 times the above Slater de­ter­mi­nant. How­ever, dis­play­ing the Slater de­ter­mi­nant fully as above is still a lot to write and read. There­fore, from now on the Slater de­ter­mi­nant will be ab­bre­vi­ated as in

\begin{displaymath}
\Psi = a {\left\vert{\rm det}\;\pe 1//b//,\pe 2//b//,\ldots,\pe n//b//,\ldots,\pe I//b//\right\rangle} %
\end{displaymath} (9.16)

where ${\left\vert{\rm det}\;\ldots\right\rangle}$ is the Slater de­ter­mi­nant.

It is im­por­tant to re­al­ize that us­ing the min­i­mum num­ber of sin­gle-elec­tron func­tions will un­avoid­ably pro­duce an er­ror that is math­e­mat­i­cally speak­ing not small {N.16}. To get a van­ish­ingly small er­ror, you would need a large num­ber of dif­fer­ent Slater de­ter­mi­nants, not just one. Still, the re­sults you get with the ba­sic Hartree-Fock ap­proach may be good enough to sat­isfy your needs. Or you may be able to im­prove upon them enough with post-Hartree-Fock meth­ods.

But none of that would be likely if you just se­lected the sin­gle-elec­tron func­tions $\pe1//b//$, $\pe2//b//$, ...at ran­dom. The clev­er­ness in the Hartree-Fock ap­proach will be in writ­ing down equa­tions for these sin­gle-elec­tron wave func­tions that pro­duce the best ap­prox­i­ma­tion pos­si­ble with a sin­gle Slater de­ter­mi­nant.

Re­call the ap­prox­i­mate so­lu­tions that were writ­ten down for the elec­trons in atoms in chap­ter 5.9. These so­lu­tions were re­ally sin­gle Slater de­ter­mi­nants. To im­prove on these re­sults, you might think of try­ing to find more ac­cu­rate ways to av­er­age out the ef­fects of the neigh­bor­ing elec­trons than just putting them in the nu­cleus as that chap­ter es­sen­tially did. You could smear them out over some op­ti­mal area, say. But even if you did that, the Hartree-Fock so­lu­tion will still be bet­ter, be­cause it gives the best pos­si­ble ap­prox­i­ma­tion ob­tain­able with any sin­gle de­ter­mi­nant.

That as­sumes of course that the spins are taken the same way. Con­sider that prob­lem for a sec­ond. Typ­i­cally, a non­rel­a­tivis­tic ap­proach is used, in which spin ef­fects on the en­ergy are ig­nored. Then spin only af­fects the an­ti­sym­metriza­tion re­quire­ments.

Things are straight­for­ward if you try to solve, say, a he­lium atom. The cor­rect ground state takes the form

\begin{displaymath}
\Psi_{\rm {He}}({\skew0\vec r}_1,{\skew0\vec r}_2) \times
...
...row}(S_{z2}) -{\downarrow}(S_{z1}){\uparrow}(S_{z2})}{\sqrt2},
\end{displaymath}

The fac­tor $\Psi_{\rm {He}}({\skew0\vec r}_1,{\skew0\vec r}_2)$ is the spa­tial wave func­tion that has the ab­solutely low­est en­ergy, re­gard­less of any an­ti­sym­metriza­tion con­cerns. This wave func­tion must be sym­met­ric (un­changed) un­der elec­tron ex­change since the two elec­trons are iden­ti­cal and the ground state is unique. The an­ti­sym­metriza­tion re­quire­ment is met be­cause the spins com­bine to­gether as shown in the sec­ond fac­tor above; this fac­tor changes sign when the elec­trons are ex­changed. So the spa­tial state does not have to change sign.

The com­bined spin state shown in the sec­ond fac­tor above is called the sin­glet state, chap­ter 5.5.6. In the sin­glet state the two spins can­cel each other com­pletely: the net elec­tron spin is zero. If you mea­sure the net spin com­po­nent in any di­rec­tion, not just the cho­sen $z$-di­rec­tion, you get zero.

Based on the ex­act he­lium ground state wave func­tion above, you would take the Hartree-Fock ap­prox­i­ma­tion to be of the form

\begin{displaymath}
{\left\vert{\rm det}\;\pe1//u//,\pe2//d//\right\rangle}
\end{displaymath}

and then you would make things eas­ier for your­self by pos­tu­lat­ing a pri­ori that the spa­tial or­bitals are the same, $\pe1////$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pe2////$. Lo and be­hold, when you mul­ti­ply out the Slater de­ter­mi­nant,

\begin{displaymath}
\frac{1}{\sqrt{2}}
\left\vert
\begin{array}{cc}
\pe1/{\s...
...2/u/z2/ & \pe1/{\skew0\vec r}_2/d/z2/
\end{array} \right\vert
\end{displaymath}

you get

\begin{displaymath}
\pe1/{\skew0\vec r}_1///\pe1/{\skew0\vec r}_2///
\times \f...
...arrow}(S_{z2})-{\downarrow}(S_{z1}){\uparrow}(S_{z2})}{\sqrt2}
\end{displaymath}

This au­tomag­i­cally re­pro­duces the cor­rect sin­glet spin state! (The ap­prox­i­ma­tion comes in be­cause the ex­act spa­tial ground state, $\Psi_{\rm {He}}({\skew0\vec r}_1,{\skew0\vec r}_2)$ is not just the prod­uct of two sin­gle-elec­tron func­tions as in Hartree-Fock.) And you only need to find one spa­tial or­bital in­stead of two.

As dis­cussed in chap­ter 5.9, a beryl­lium atom has two elec­trons with op­po­site spins in the 1s shell like he­lium, and two more in the 2s shell. An ap­pro­pri­ate Hartree-Fock wave func­tion would be

\begin{displaymath}
{\left\vert{\rm det}\;\pe1//u//,\pe1//d//,\pe3//u//,\pe3//d//\right\rangle}
\end{displaymath}

in other words, two pairs of or­bitals with the same spa­tial states and op­po­site spins. Sim­i­larly, Neon has an ad­di­tional 6 paired elec­trons in a closed 2p shell, and you could use 3 more pairs of or­bitals with the same spa­tial states and op­po­site spins. The num­ber of spa­tial or­bitals that must be found in such so­lu­tions is only half the num­ber of elec­trons. This pro­ce­dure is called the “closed shell Re­stricted Hartree-Fock (RHF)” method. It re­stricts the form of the spa­tial states to be pair-wise equal.

But now look at lithium. Lithium has two paired 1s elec­trons like he­lium, and an un­paired 2s elec­tron. For the third or­bital in the Hartree-Fock de­ter­mi­nant, you will now have to make a choice: whether to take it of the form $\pe3//u//$ or $\pe3//d//$. Lets as­sume you take $\pe3//u//$, so the wave func­tion is

\begin{displaymath}
{\left\vert{\rm det}\;\pe1//u//,\pe2//d//,\pe3//u//\right\rangle}
\end{displaymath}

You have in­tro­duced a bias in the de­ter­mi­nant: there is now a real dif­fer­ence be­tween the spa­tial or­bitals $\pe1////$ and $\pe2////$: $\pe1//u//$ has the same spin as the third spin or­bital, but $\pe2//d//$ the op­po­site.

If you find the best ap­prox­i­ma­tion to the en­ergy among all pos­si­ble spa­tial or­bitals $\pe1////$, $\pe2////$, and $\pe3////$, you will end up with or­bitals $\pe1////$ and $\pe2////$ that are not the same. Al­low­ing for them to be dif­fer­ent is called the “Un­re­stricted Hartree-Fock (UHF)” method. In gen­eral, you no longer re­quire that equiv­a­lent spa­tial or­bitals are the same in their spin-up and spin down ver­sions. For a big­ger sys­tem, you will end up with one set of or­tho­nor­mal spa­tial or­bitals for the spin-up or­bitals and a dif­fer­ent set of or­tho­nor­mal spa­tial or­bitals for the spin-down ones. These two sets of or­tho­nor­mal spa­tial or­bitals are not mu­tu­ally or­thog­o­nal; the only rea­son the com­plete spin or­bitals are still or­tho­nor­mal is be­cause the two spins are or­thog­o­nal, $\left\langle\vphantom{{\downarrow}}{\uparrow}\hspace{-\nulldelimiterspace}\hspace{.03em}\right.\!\left\vert\vphantom{{\uparrow}}{\downarrow}\right\rangle $ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.

If in­stead of us­ing un­re­stricted Hartree-Fock, you in­sist on de­mand­ing that the spa­tial or­bitals for spin up and down do form a sin­gle set of or­tho­nor­mal func­tions, it is called “open shell Re­stricted Hartree-Fock (RHF).” In the case of lithium, you would then de­mand that $\pe2////$ equals $\pe1////$. Since the best (in terms of en­ergy) so­lu­tion has them dif­fer­ent, your so­lu­tion is then no longer the best pos­si­ble. You pay a price, but you now only need to find two spa­tial or­bitals rather than three. The spin or­bital $\pe3//u//$ with­out a match­ing op­po­site-spin or­bital counts as an open shell. For ni­tro­gen, you might want to use three open shells to rep­re­sent the three dif­fer­ent spa­tial states 2p$_x$, 2p$_y$, and 2p$_z$ with an un­paired elec­tron in it.

If you use un­re­stricted Hartree-Fock in­stead, you will need to com­pute more spa­tial func­tions, and you pay an­other price, spin. Since all spin ef­fects in the Hamil­ton­ian are ig­nored, it com­mutes with the spin op­er­a­tors. So, the ex­act en­ergy eigen­func­tions are also, or can be taken to be also, spin eigen­func­tions. Re­stricted Hartree-Fock has the ca­pa­bil­ity of pro­duc­ing ap­prox­i­mate en­ergy eigen­states with well de­fined spin. In­deed, as you saw for he­lium, in re­stricted Hartree-Fock all the paired spin-up and spin-down states com­bine into zero-spin sin­glet states. If any ad­di­tional un­paired states are all spin up, say, you get an en­ergy eigen­state with a net spin equal to the sum of the spins of the un­paired states. This al­lows you to deal with typ­i­cal atoms, in­clud­ing lithium and ni­tro­gen, very nicely.

But a true un­re­stricted Hartree-Fock so­lu­tion does not have cor­rect, def­i­nite, spin. For two elec­trons to pro­duce states of def­i­nite com­bined spin, the co­ef­fi­cients of spin-up and spin-down must come in spe­cific ra­tios. As a sim­ple ex­am­ple, an un­re­stricted Slater de­ter­mi­nant of $\pe1//u//$ and $\pe2//d//$ with un­equal spa­tial or­bitals mul­ti­plies out to

\begin{displaymath}
{\left\vert{\rm det}\;\pe1//u//,\pe2//d//\right\rangle} =
...
...ew0\vec r}_2///{\downarrow}(S_{z1}){\uparrow}(S_{z2})}{\sqrt2}
\end{displaymath}

or, writ­ing the spin com­bi­na­tions in terms of sin­glets (which change sign un­der elec­tron ex­change) and triplets (which do not),

\begin{eqnarray*}
& & \frac{\pe1/{\skew0\vec r}_1///\pe2/{\skew0\vec r}_2///+\p...
...wnarrow}(S_{z2})+{\downarrow}(S_{z1}){\uparrow}(S_{z2})}{\sqrt2}
\end{eqnarray*}

So the spin will be some com­bi­na­tion of the sin­glet state, the first term, and a triplet state, the sec­ond. And the pre­cise com­bi­na­tion will de­pend on the spa­tial lo­ca­tions of the elec­trons to boot. Now while the sin­glet state has net spin 0, triplet states have net spin 1. So the net spin is un­cer­tain, ei­ther 0 or 1, even though it should not be. (Spin 1 im­plies that the mea­sured com­po­nent of the spin in any di­rec­tion must be one of the triplet of val­ues $\hbar$, 0, or $-\hbar$. For the par­tic­u­lar triplet state shown above, the com­po­nent of spin in the $z$-di­rec­tion hap­pens to be zero. But the net spin is not; in a di­rec­tion nor­mal to the $z$-di­rec­tion, the spin will be mea­sured to be ei­ther $\hbar$ or $-\hbar$.) How­ever, de­spite the spin prob­lem, it may be noted that un­re­stricted wave func­tions are com­monly used as first ap­prox­i­ma­tions of dou­blet (spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$) and triplet (spin 1) states any­way [45, p. 105].

To show that all this can make a real dif­fer­ence, take the ex­am­ple of the hy­dro­gen mol­e­cule, chap­ter 5.2, when the two nu­clei are far apart. The cor­rect elec­tronic ground state is

\begin{displaymath}
\frac{\psi_{\rm {L}}({\skew0\vec r}_1)\psi_{\rm {R}}({\skew...
...arrow}(S_{z2})-{\downarrow}(S_{z1}){\uparrow}(S_{z2})}{\sqrt2}
\end{displaymath}

where $\psi_{\rm {L}}({\skew0\vec r}_1)\psi_{\rm {R}}({\skew0\vec r}_2)$ is the state in which elec­tron 1 is around the left pro­ton and elec­tron 2 around the right one, and $\psi_{\rm {R}}({\skew0\vec r}_1)\psi_{\rm {L}}({\skew0\vec r}_2)$ is the same state but with the elec­trons re­versed. Note that, like for the he­lium atom, the spa­tial state is sym­met­ric un­der elec­tron ex­change. How­ever, it is not just a prod­uct of two sin­gle-elec­tron func­tions but a sum of two of such prod­ucts. Note also that the cor­rect spin state is the sin­glet one with zero net spin, just like for the he­lium atom. It takes care of the an­ti­sym­metriza­tion re­quire­ment.

Now try to ap­prox­i­mate this so­lu­tion with a re­stricted closed shell Hartree-Fock wave func­tion of the form

\begin{displaymath}
{\left\vert{\rm det}\;\pe1//u//,\pe1//d//\right\rangle}
\end{displaymath}

Mul­ti­ply­ing out the de­ter­mi­nant gives

\begin{displaymath}
\pe1/{\skew0\vec r}_1///\pe1/{\skew0\vec r}_2///
\times \f...
...arrow}(S_{z2})-{\downarrow}(S_{z1}){\uparrow}(S_{z2})}{\sqrt2}
\end{displaymath}

Note that you do get the cor­rect sin­glet spin state. But $\pe1////$ will be some­thing like $(\psi_{\rm {L}}+\psi_{\rm {R}})$$\raisebox{.5pt}{$/$}$$\sqrt2$; the en­ergy of ei­ther elec­tron is low­est when it is near one of the nu­clei. If you mul­ti­ply out the re­sult­ing spa­tial wave func­tion, the terms in­clude $\psi_{\rm {L}}\psi_{\rm {L}}$ and $\psi_{\rm {R}}\psi_{\rm {R}}$, in ad­di­tion to the cor­rect $\psi_{\rm {L}}\psi_{\rm {R}}$ and $\psi_{\rm {R}}\psi_{\rm {L}}$. That pro­duces a 50/50 chance that the two elec­trons are found around the same nu­cleus. That is all wrong, since the elec­trons re­pel each other: if one elec­tron is around the left nu­cleus, the other elec­tron should be around the right nu­cleus. The com­puted en­ergy, which should be that of two neu­tral hy­dro­gen atoms far apart, will be much too high due to elec­tron-elec­tron re­pul­sion.

(For­tu­nately, at the nu­clear sep­a­ra­tion dis­tance cor­re­spond­ing to the ground state of the com­plete mol­e­cule, the er­rors are much less, [45, p. 166]. Note that if you put the two nu­clei com­pletely on top of each other, you get a he­lium atom, for which Hartree-Fock gives a much more rea­son­able elec­tron en­ergy. Only when you are break­ing the bond, dis­so­ci­at­ing the mol­e­cule, i.e. tak­ing the nu­clei far apart, do you get into ma­jor trou­ble.)

If in­stead you would use un­re­stricted Hartree-Fock, say

\begin{displaymath}
{\left\vert{\rm det}\;\pe1//u//,\pe2//d//\right\rangle}
\end{displaymath}

you should find $\pe1////$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{\rm {L}}$ and $\pe2////$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{\rm {R}}$ (or vice versa), which would pro­duce a wave func­tion

\begin{displaymath}
\frac{
\psi_{\rm {L}}({\skew0\vec r}_1)\psi_{\rm {R}}({\sk...
...w0\vec r}_2){\downarrow}(S_{z1}){\uparrow}(S_{z2})}{\sqrt{2}}.
\end{displaymath}

In both terms, if the first elec­tron is around the one nu­cleus, the sec­ond elec­tron is around the other. So this pro­duces the cor­rect en­ergy, that of two neu­tral hy­dro­gen atoms. But the spin is now all wrong. It is not a sin­glet state, but the com­bi­na­tion of a sin­glet and a triplet state al­ready writ­ten down ear­lier. Lit­tle in life is ideal, is it?

(Ac­tu­ally there is a dirty trick to fix this. Note that which of the two or­bitals you give spin-up and which spin-down is phys­i­cally im­ma­te­r­ial. So there is a triv­ially dif­fer­ent so­lu­tion

\begin{displaymath}
{\left\vert{\rm det}\;\pe1//d//,\pe2//u//\right\rangle}
\end{displaymath}

If you take a 50/50 com­bi­na­tion of the orig­i­nal Slater de­ter­mi­nant and mi­nus the one above, you get the cor­rect sin­glet spin state. And the spa­tial state will now be the cor­rect av­er­age of $\psi_{\rm {L}}\psi_{\rm {R}}$ and $\psi_{\rm {R}}\psi_{\rm {L}}$ to boot. This spa­tial state is more ac­cu­rate than just two neu­tral atoms if the dis­tance be­tween the nu­clei de­creases, chap­ter 5.2. All this for free! This sort of dirty trick in Hartree-Fock is called a “spin adapted con­fig­u­ra­tion.” It is usu­ally used to deal with a few open shells in an oth­er­wise closed-shell re­stricted Hartree-Fock con­fig­u­ra­tion.)

All of the above may be much more than you ever wanted to hear about the wave func­tion. The pur­pose was mainly to in­di­cate that things are not as sim­ple as you might ini­tially sup­pose. As the ex­am­ples showed, some un­der­stand­ing of the sys­tem that you are try­ing to model def­i­nitely helps. Or ex­per­i­ment with dif­fer­ent ap­proaches.

Let’s go on to the next step: how to get the equa­tions for the spa­tial or­bitals $\pe1////,\pe2////,\ldots$ that give the most ac­cu­rate ap­prox­i­ma­tion of a multi-elec­tron prob­lem. The ex­pec­ta­tion value of en­ergy will be needed for that, and to get that, first the Hamil­ton­ian is needed. That will be the sub­ject of the next sub­sec­tion.


9.3.2 The Hamil­ton­ian

The non­rel­a­tivis­tic Hamil­ton­ian of the sys­tem of $I$ elec­trons con­sists of a num­ber of con­tri­bu­tions. First there is the ki­netic en­ergy of the elec­trons; the sum of the ki­netic en­ergy op­er­a­tors of the in­di­vid­ual elec­trons:

\begin{displaymath}
{\widehat T}^{\rm E}
= - \sum_{i=1}^I \frac{\hbar^2}{2m_{\...
...\partial y_i^2} +
\frac{\partial^2}{\partial z_i^2}
\right).
\end{displaymath} (9.17)

Next there is the po­ten­tial en­ergy due to the am­bi­ent elec­tric field that the elec­trons move in. It will be as­sumed that this field is caused by $J$ nu­clei, num­bered us­ing an in­dex $j$, and hav­ing charge $Z_je$ (i.e. there are $Z_j$ pro­tons in nu­cleus num­ber $j$). In that case, the to­tal po­ten­tial en­ergy due to nu­cleus-elec­tron at­trac­tions is, sum­ming over all elec­trons and over all nu­clei:

\begin{displaymath}
V^{\rm NE}
= - \sum_{i=1}^I \left(\sum_{j=1}^J
\frac{Z_j e^2}{4\pi\epsilon_0} \frac{1}{r^{\rm n}_{ij}}\right)
\end{displaymath} (9.18)

where $r^{\rm n}_{ij}\equiv\vert{\skew0\vec r}_i-{\skew0\vec r}^{\,\rm {n}}_j\vert$ is the dis­tance be­tween elec­tron num­ber $i$ and nu­cleus num­ber $j$, and $\epsilon_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 8.85 10$\POW9,{-12}$ C$\POW9,{2}$/J m is the per­mit­tiv­ity of space.

And now for the black plague of quan­tum me­chan­ics, the elec­tron to elec­tron re­pul­sions. The po­ten­tial en­ergy for those re­pul­sions is

\begin{displaymath}
V^{\rm EE}= \sum_{i=1}^I \sum_{{\underline i}>i}^I
\frac{e^2}{4\pi\epsilon_0} \frac{1}{r_{i{\underline i}}}
\end{displaymath} (9.19)

where $r_{i{\underline i}}\equiv\vert{\skew0\vec r}_i-{\skew0\vec r}_{\underline i}\vert$ is the dis­tance be­tween elec­tron num­ber $i$ and elec­tron num­ber ${\underline i}$. To avoid count­ing each re­pul­sion en­ergy twice, (the sec­ond time with re­versed elec­tron or­der), the sec­ond elec­tron num­ber is re­quired to be larger than the first.

With­out this re­pul­sion be­tween dif­fer­ent elec­trons, you could solve for each elec­tron sep­a­rately, and all would be nice. But you do have it, and so you re­ally need to solve for all elec­trons at once, usu­ally an im­pos­si­ble task. You may re­call that when chap­ter 5.9 ex­am­ined the atoms heav­ier than hy­dro­gen, those with more than one elec­tron, the dis­cus­sion clev­erly threw out the elec­tron to elec­tron re­pul­sion terms, by as­sum­ing that the ef­fect of each neigh­bor­ing elec­tron is ap­prox­i­mately like can­cel­ing out one pro­ton in the nu­cleus. And you may also re­mem­ber how this out­ra­geous as­sump­tion led to all those wrong pre­dic­tions that had to be cor­rected by var­i­ous ex­cuses. The Hartree-Fock ap­prox­i­ma­tion tries to do bet­ter than that.

It is help­ful to split the Hamil­ton­ian into the sin­gle elec­tron terms and the trou­ble­some in­ter­ac­tions, as fol­lows,

\begin{displaymath}
H = \sum_{i=1}^I h^{\rm e}_i
+ \sum_{i=1}^I \sum_{{\underline i}>i}^I v^{\rm ee}_{i{\underline i}} %
\end{displaymath} (9.20)

where $h^{\rm e}_i$ is the sin­gle-elec­tron Hamil­ton­ian of elec­tron $i$,
\begin{displaymath}
h^{\rm e}_i =
- \frac{\hbar^2}{2m_{\rm e}}\nabla_i^2
+ \s...
...}^J \frac{Z_j e^2}{4\pi\epsilon_0}
\frac{1}{r^{\rm n}_{ij}} %
\end{displaymath} (9.21)

and $v^{\rm ee}_{i{\underline i}}$ is the elec­tron $i$ to elec­tron ${\underline i}$ re­pul­sion po­ten­tial en­ergy,
\begin{displaymath}
v^{\rm ee}_{i{\underline i}} = \frac{e^2}{4\pi\epsilon_0} \frac{1}{r_{i{\underline i}}} %
\end{displaymath} (9.22)

Note that $h^{\rm e}_1$, $h^{\rm e}_2$, ..., $h^{\rm e}_I$ all take the same gen­eral form; the dif­fer­ence is just in which elec­tron you are talk­ing about. That is not sur­pris­ing be­cause the elec­trons all have the same prop­er­ties. Sim­i­larly, the dif­fer­ence be­tween $v^{\rm ee}_{12}$, $v^{\rm ee}_{13}$, ..., $v^{\rm ee}_{(I-1)I}$ is just in which pair of elec­trons you talk about.


9.3.3 The ex­pec­ta­tion value of en­ergy

As was dis­cussed in more de­tail in sec­tion 9.1, to find the best pos­si­ble Hartree-Fock ap­prox­i­ma­tion, the ex­pec­ta­tion value of en­ergy will be needed. For ex­am­ple, the best ap­prox­i­ma­tion to the ground state is the one that has the small­est ex­pec­ta­tion value of en­ergy.

The ex­pec­ta­tion value of en­ergy $\left\langle{E}\right\rangle $ is de­fined as the in­ner prod­uct

\begin{displaymath}
{\left\langle\vphantom{H\Psi}\Psi\hspace{0.3pt}\right\vert}H{\left\vert\Psi\vphantom{\Psi H}\right\rangle}
\end{displaymath}

where $H$ is the Hamil­ton­ian as given in the pre­vi­ous sub­sec­tion. There is a prob­lem with us­ing this ex­pres­sion mind­lessly, though. Take once again the ex­am­ple of the ar­senic atom. There are 33 elec­trons in this atom, so you could try to choose 33 promis­ing sin­gle-elec­tron wave func­tions to de­scribe it. You could then try to mul­ti­ply out the Slater de­ter­mi­nant for $\Psi$, but that pro­duces 33!, or about 4 10$\POW9,{36}$, Hartree prod­ucts. If you put these 33! terms in both sides of the in­ner prod­uct, you get (33!)$\POW9,{2}$ or 7.5 10$\POW9,{73}$ pairs of terms, each pro­duc­ing one in­ner prod­uct that must be in­te­grated. Now since there are 3 co­or­di­nates for each of the po­si­tions of the 33 elec­trons, this means that each term re­quires in­te­gra­tion over 99 scalar co­or­di­nates. Even us­ing only 10 points in each di­rec­tion, that would mean eval­u­at­ing 10$\POW9,{99}$ in­te­gra­tion points for each of the 7.5 10$\POW9,{73}$ pairs of terms. A com­puter that could do that is unimag­in­able. As of 2014, the fastest com­puter in the world can do no more than 10$\POW9,{25}$ float­ing point com­pu­ta­tions if it stays at it for 10 years.

For­tu­nately, it turns out, {D.52}, that al­most all of those in­te­gra­tions are triv­ial since the sin­gle-elec­tron func­tions are or­tho­nor­mal. If you sit down and iden­tify what is re­ally left, you find that only a few three-di­men­sion­al and six-di­men­sion­al in­ner prod­ucts sur­vive the weed­ing-out process.

In par­tic­u­lar, the sin­gle-elec­tron Hamil­to­ni­ans from the pre­vi­ous sub­sec­tion pro­duce only sin­gle-elec­tron en­ergy ex­pec­ta­tion val­ues of the gen­eral form

\begin{displaymath}
\fbox{$\displaystyle
E^{\rm e}_n \equiv {\left\langle\vpha...
...vert\pe n////\vphantom{\pe n////h^{\rm e}}\right\rangle}
$} %
\end{displaymath} (9.23)

If you had only one sin­gle elec­tron, and it was in the spa­tial sin­gle-par­ti­cle state $\pe{n}/{\skew0\vec r}///$, the above in­ner prod­uct would be its en­ergy.

The com­bined sin­gle-elec­tron en­ergy for all $I$ elec­trons is then

\begin{displaymath}
\sum_{n=1}^I E^{\rm e}_n
\end{displaymath}

It is just as if you had elec­tron 1 in state $\pe1////$, elec­tron 2 in state $\pe2////$, etcetera. Of course, that is not re­ally true. An­ti­sym­metriza­tion re­quires that all elec­trons are partly in all states. In­deed, if you look a bit closer at the math, you see that each of the $I$ elec­trons con­tributes an equal frac­tion 1$\raisebox{.5pt}{$/$}$$I$ to each of the $I$ terms above. But it does not make a real dif­fer­ence. With­out elec­tron-elec­tron in­ter­ac­tions, quan­tum me­chan­ics would be so much eas­ier!

But the re­pul­sions are there. The Hamil­to­ni­ans of the re­pul­sions turn out to pro­duce six-di­men­sion­al spa­tial in­ner prod­ucts of two types. The in­ner prod­ucts of the first type are called “Coulomb in­te­grals:”

\begin{displaymath}
\fbox{$\displaystyle
J_{n{\underline n}} \equiv {\left\lan...
...{\pe n////\pe{\underline n}////v^{\rm ee}}\right\rangle}
$} %
\end{displaymath} (9.24)

To un­der­stand the Coulomb in­te­grals bet­ter, the in­ner prod­uct above can be writ­ten out ex­plic­itly as an in­te­gral, while also ex­pand­ing $v^{\rm ee}$:

\begin{displaymath}
\int_{{\rm all}\;{\skew0\vec r}} \int_{{\rm all}\;{\underli...
...{\,\rm d}^3{\skew0\vec r}{\,\rm d}^3{\underline{\skew0\vec r}}
\end{displaymath}

The in­te­grand equals the prob­a­bil­ity of an elec­tron in state $\pe{n}////$ to be found near a po­si­tion ${\skew0\vec r}$, times the prob­a­bil­ity of an elec­tron in state $\pe{\underline n}////$ to be found near a po­si­tion ${\underline{\skew0\vec r}}$, times the Coulomb re­pul­sion en­ergy if the two elec­trons are at those po­si­tions. In short, $J_{n{\underline n}}$ is the ex­pec­ta­tion value of the Coulomb re­pul­sion po­ten­tial be­tween an elec­tron in state $\pe{n}////$ and one in state $\pe{\underline n}////$. Think­ing again of elec­tron 1 in state $\pe1////$, elec­tron 2 in state $\pe2////$, etcetera, the to­tal Coulomb re­pul­sion en­ergy would be

\begin{displaymath}
\sum_{n=1}^I \sum_{{\underline n}>n}^I J_{n{\underline n}}
\end{displaymath}

which is in­deed the cor­rect com­bined sum of the Coulomb in­te­grals.

Un­for­tu­nately, that is not the com­plete story for the re­pul­sion en­ergy. Re­call that there are $I!$ dif­fer­ent ways in which you can dis­trib­ute the $I$ elec­trons over the $I$ sin­gle par­ti­cle states. And the an­ti­sym­metriza­tion re­quire­ment re­quires that the sys­tem wave func­tion is an equal com­bi­na­tion of all these $I!$ dif­fer­ent pos­si­bil­i­ties. In terms of clas­si­cal physics, it might still seem that this should make no dif­fer­ence: if any one of these $I!$ pos­si­bil­i­ties is true, then the oth­ers must be un­true. But quan­tum me­chan­ics al­lows states in which the elec­trons are dis­trib­uted in one way to in­ter­act with states in which they are dis­trib­uted in an­other way. That pro­duces the so-called “ex­change in­te­grals:”

\begin{displaymath}
\fbox{$\displaystyle
K_{n{\underline n}} \equiv {\left\lan...
...{\pe n////\pe{\underline n}////v^{\rm ee}}\right\rangle}
$} %
\end{displaymath} (9.25)

Writ­ten out ex­plic­itly, that equals

\begin{displaymath}
\int_{{\rm all}\;{\skew0\vec r}} \int_{{\rm all}\;{\underli...
...{\,\rm d}^3{\skew0\vec r}{\,\rm d}^3{\underline{\skew0\vec r}}
\end{displaymath}

It is an in­ter­ac­tion of the pos­si­bil­ity that the first elec­tron is in state $\pe{n}////$ and the sec­ond in state $\pe{\underline n}////$ with the pos­si­bil­ity that the sec­ond elec­tron is in state $\pe{n}////$ and the first in state $\pe{\underline n}////$. This book likes to call terms like this twi­light terms, since in terms of clas­si­cal physics they do not make sense.

It may be noted that a sin­gle Hartree prod­uct sat­is­fy­ing the Pauli ex­clu­sion prin­ci­ple would not pro­duce ex­change in­te­grals; in such a wave func­tion, there is no pos­si­bil­ity for an elec­tron to be in an­other state. But don't start think­ing that the ex­change in­te­grals are there just be­cause the wave func­tion must be an­ti­sym­met­ric un­der elec­tron ex­change. They, and oth­ers, would show up in any rea­son­ably gen­eral wave func­tion. You can think of the ex­change in­te­grals in­stead as Coulomb in­te­grals with the elec­trons in the right hand side of the in­ner prod­uct ex­changed.

Adding it all up, the ex­pec­ta­tion en­ergy of the com­plete sys­tem of $I$ elec­trons can be writ­ten as

\begin{displaymath}
\fbox{$\displaystyle
\left\langle{E}\right\rangle =
\sum_...
...rrow}_{\underline n}\right\rangle ^2 K_{n{\underline n}}
$} %
\end{displaymath} (9.26)

Note that the above ex­pres­sion sums over all val­ues of ${\underline n}$, not just ${\underline n}>n$. That counts each pair of sin­gle-elec­tron wave func­tions twice, so fac­tors one-half have been added to com­pen­sate. It also adds terms in which ${\underline n}=n$, both elec­trons in the same state, which is not al­lowed by the Pauli prin­ci­ple. But since $J_{nn}=K_{nn}$, these ad­di­tional terms can­cel each other.

Note also the spin in­ner prod­ucts mul­ti­ply­ing the ex­change terms. These are zero if the two states have op­po­site spin, so there are no ex­change con­tri­bu­tions be­tween elec­trons in spin or­bitals of op­po­site spins. And if the spin or­bitals have the same spin, the spin in­ner prod­uct is 1, so the square is some­what su­per­flu­ous.

There are also some a pri­ori things you can say about the Coulomb and ex­change in­te­grals, {D.53}; they are real, and ad­di­tion­ally

\begin{displaymath}
J_{nn} = K_{nn}
\qquad
J_{n{\underline n}} = J_{{\underli...
...}K_{n{\underline n}} \mathrel{\raisebox{-1pt}{$\geqslant$}}0 %
\end{displaymath} (9.27)

Note in par­tic­u­lar that since the $K_{n{\underline n}}$ terms are pos­i­tive, they lower the net ex­pec­ta­tion en­ergy of the sys­tem. So a wave func­tion con­sist­ing of a sin­gle Hartree prod­uct, which pro­duces no ex­change terms, can­not be the state of low­est en­ergy. Even with­out the an­ti­sym­metriza­tion re­quire­ment, you would need Hartree prod­ucts with the elec­trons ex­changed, sim­ply to lower the en­ergy.

It is ac­tu­ally some­what tricky to prove that the $K_{n{\underline n}}$ terms are pos­i­tive and so lower the en­ergy, {D.53}. But there is a sim­ple phys­i­cal rea­son why you might guess that an an­ti­sym­met­ric wave func­tion would lower the elec­tron-elec­tron re­pul­sion en­ergy com­pared to the in­di­vid­ual Hartree prod­ucts from which it is made up. In par­tic­u­lar, the Coulomb re­pul­sion be­tween elec­trons be­comes very large when they get close to­gether. But for an anti-sym­met­ric wave func­tion, un­like for a sin­gle Hartree prod­uct, the rel­a­tive prob­a­bil­ity of elec­trons of the same spin get­ting close to­gether is van­ish­ingly small. That pre­vents any strong Coulomb re­pul­sion be­tween elec­trons of the same spin.

(Re­call that the rel­a­tive prob­a­bil­ity for elec­trons to be at given po­si­tions and spins is given by the square mag­ni­tude of the wave func­tion at those po­si­tions and spins. Now an an­ti­sym­met­ric wave func­tion must be zero wher­ever any two elec­trons are at the same po­si­tion with the same spin, mak­ing this im­pos­si­ble. Af­ter all, if you swap the two elec­trons, the an­ti­sym­met­ric wave func­tion must change sign. But since nei­ther elec­tron changes po­si­tion nor spin, the wave func­tion can­not change ei­ther. Some­thing can only change sign and stay the same if it is zero. See also {A.34}.)

The analy­sis given in this sub­sec­tion can eas­ily be ex­tended to gen­er­al­ized or­bitals that take the form

\begin{displaymath}
\pp n/{\skew0\vec r}//z/ = \pe n+/{\skew0\vec r}/u/z/+\pe n-/{\skew0\vec r}/d/z/.
\end{displaymath}

How­ever, the nor­mal un­re­stricted spin-up or spin-down or­bitals, in which ei­ther $\pe{n+}////$ or $\pe{n-}////$ is zero, al­ready sat­isfy the vari­a­tional re­quire­ment $\delta\left\langle{E}\right\rangle $ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 even if gen­er­al­ized vari­a­tions in the or­bitals are al­lowed, {N.17}.

In any case, the ex­pec­ta­tion value of en­ergy has been found.


9.3.4 The canon­i­cal Hartree-Fock equa­tions

The pre­vi­ous sub­sec­tion found the ex­pec­ta­tion value of en­ergy for any elec­tron wave func­tion de­scribed by a sin­gle Slater de­ter­mi­nant. The fi­nal step is to find the or­bitals that pro­duce the best ap­prox­i­ma­tion of the true wave func­tion us­ing such a sin­gle de­ter­mi­nant. For the ground state, the best sin­gle de­ter­mi­nant would be the one with the low­est ex­pec­ta­tion value of en­ergy. But surely you would not want to guess spa­tial or­bitals at ran­dom un­til you find some with re­ally, re­ally, low en­ergy.

What you would like to have is spe­cific equa­tions for the best spa­tial or­bitals that you can then solve in a me­thod­i­cal way. And you can have them us­ing the meth­ods of sec­tion 9.1, {D.54}. In un­re­stricted Hartree-Fock, for every spa­tial or­bital $\pe{n}/{\skew0\vec r}///$ there is an equa­tion of the form:

\begin{displaymath}
\fbox{$\displaystyle
\begin{array}[b]{l}
\displaystyle
h...
... r}///
= \epsilon_n \pe n/{\skew0\vec r}///
\end{array} $} %
\end{displaymath} (9.28)

These are called the canon­i­cal Hartree-Fock equa­tions. For equa­tions valid for the re­stricted closed-shell and sin­gle-de­ter­mi­nant open-shell ap­prox­i­ma­tions, see the de­riva­tion in {D.54}.

Re­call that $h^{\rm e}$ is the sin­gle-elec­tron Hamil­ton­ian con­sist­ing of the elec­tron's ki­netic en­ergy and po­ten­tial en­ergy due to nu­clear at­trac­tions, and that $v^{\rm ee}$ is the po­ten­tial en­ergy of re­pul­sion be­tween the elec­tron and an­other at a po­si­tion ${\underline{\skew0\vec r}}$:

\begin{displaymath}
h^{\rm e}= - \frac{\hbar^2}{2m_{\rm e}}\nabla^2
- \sum_{j=...
...e r}\equiv\vert{\skew0\vec r}- {\underline{\skew0\vec r}}\vert
\end{displaymath}

So, if there were no elec­tron-elec­tron re­pul­sions, i.e. $v^{\rm ee}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, the canon­i­cal equa­tions above would be sin­gle-elec­tron Hamil­ton­ian eigen­value prob­lems of the form $h^{\rm e}\pe{n}////$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\epsilon_n\pe{n}////$ where $\epsilon_n$ would be the en­ergy of the sin­gle-elec­tron or­bital. This is re­ally what hap­pened in the ap­prox­i­mate analy­sis of atoms in chap­ter 5.9: the elec­tron to elec­tron re­pul­sions were ig­nored there in fa­vor of nu­clear strength re­duc­tions, and the re­sult was sin­gle-elec­tron hy­dro­gen-atom or­bitals.

In the pres­ence of elec­tron to elec­tron re­pul­sions, the equa­tions for the or­bitals can still sym­bol­i­cally be writ­ten as if they were sin­gle-elec­tron eigen­value prob­lems,

\begin{displaymath}
{\cal F}\pe n/{\skew0\vec r}/b/z/ = \epsilon_n\pe n/{\skew0\vec r}/b/z/
\end{displaymath}

where ${\cal F}$ is called the “Fock op­er­a­tor,” and is writ­ten out fur­ther as:

\begin{displaymath}
{\cal F}= h^{\rm e}+ v^{\rm HF}.
\end{displaymath}

The first term in the Fock op­er­a­tor is the sin­gle-elec­tron Hamil­ton­ian. The mis­chief is in the in­nocu­ous-look­ing sec­ond term $v^{\rm {HF}}$. Sup­pos­edly, this is the po­ten­tial en­ergy re­lated to the re­pul­sion by the other elec­trons. What is it? Well, it will have to be the terms in the canon­i­cal equa­tions (9.28) not de­scribed by the sin­gle-elec­tron Hamil­ton­ian $h^{\rm e}$:

\begin{eqnarray*}
v^{\rm HF} \pe/{\skew0\vec r}/b/z/ & = & \sum_{{\underline n}...
.../v^{\rm ee}}\right\rangle} \pe{\underline n}/{\skew0\vec r}/b/z/
\end{eqnarray*}

The de­f­i­n­i­tion of the Fock op­er­a­tor is un­avoid­ably in terms of spin rather than just spa­tial or­bitals: the spin of the state on which it op­er­ates must be known to eval­u­ate the fi­nal term.

Note that the above ex­pres­sion did not give an ex­pres­sion for $v^{\rm {HF}}$ by it­self, but only for $v^{\rm {HF}}$ ap­plied to an ar­bi­trary sin­gle-elec­tron func­tion $\pe//b//$. The rea­son is that $v^{\rm {HF}}$ is not a nor­mal po­ten­tial at all: the sec­ond term, the one due to the ex­change in­te­grals, does not mul­ti­ply $\pe//b//$ by a po­ten­tial func­tion, it shoves it into an in­ner prod­uct! The Hartree-Fock po­ten­tial $v^{\rm {HF}}$ is an op­er­a­tor, not a nor­mal po­ten­tial en­ergy. Given a sin­gle-elec­tron func­tion in­clud­ing spin, it pro­duces an­other sin­gle-elec­tron func­tion in­clud­ing spin.

Ac­tu­ally, even that is not quite true. The Hartree-Fock po­ten­tial is only an op­er­a­tor af­ter you have found the or­bitals $\pe1//b//$, $\pe2//b//$, ..., $\pe{\underline n}//b//$, ..., $\pe{I}//b//$ ap­pear­ing in it. While you are still try­ing to find them, the Fock op­er­a­tor is not even an op­er­a­tor, it is just a thing. How­ever, given the or­bitals, at least the Fock op­er­a­tor is a Her­mit­ian one, one that can be taken to the other side if it ap­pears in an in­ner prod­uct, and that has real eigen­val­ues and a com­plete set of eigen­func­tions, {D.55}.

So how do you solve the canon­i­cal Hartree-Fock equa­tions for the or­bitals $\pe{n}////$? If the Hartree-Fock po­ten­tial $v^{\rm {HF}}$ was a known op­er­a­tor, you would have only lin­ear, sin­gle-elec­tron eigen­value prob­lems to solve. That would be rel­a­tively easy, as far as those things come. But since the op­er­a­tor $v^{\rm {HF}}$ con­tains the un­known or­bitals, you do not have a lin­ear prob­lem at all; it is a sys­tem of cou­pled cu­bic equa­tions in in­fi­nitely many un­knowns. The usual way to solve it is it­er­a­tively: you guess an ap­prox­i­mate form of the or­bitals and plug it into the Hartree-Fock po­ten­tial. With this guessed po­ten­tial, the or­bitals may then be found from solv­ing lin­ear eigen­value prob­lems. If all goes well, the ob­tained or­bitals, though not per­fect, will at least be bet­ter than the ones that you guessed at ran­dom. So plug those im­proved or­bitals into the Hartree-Fock po­ten­tial and solve the eigen­value prob­lems again. Still bet­ter or­bitals should re­sult. Keep go­ing un­til you get the cor­rect so­lu­tion to within ac­cept­able ac­cu­racy.

You will know when you have got the cor­rect so­lu­tion since the Hartree-Fock po­ten­tial will no longer change; the po­ten­tial that you used to com­pute the fi­nal set of or­bitals is re­ally the po­ten­tial that those fi­nal or­bitals pro­duce. In other words, the fi­nal Hartree-Fock po­ten­tial that you com­pute is con­sis­tent with the fi­nal or­bitals. Since the po­ten­tial would be a field if it was not an op­er­a­tor, that ex­plains why such an it­er­a­tive method to com­pute the Hartree-Fock so­lu­tion is called a “self-con­sis­tent field method.” It is like call­ing an it­er­a­tive scheme for the Laplace equa­tion on a mesh a “self-con­sis­tent neigh­bors method,” in­stead of point re­lax­ation. Surely the equiv­a­lent for Hartree-Fock, like “it­er­ated po­ten­tial” or po­ten­tial re­lax­ation would have been much clearer to a gen­eral au­di­ence?


9.3.5 Ad­di­tional points

This brief sec­tion was not by any means a tu­to­r­ial of the Hartree-Fock method. The pur­pose was only to ex­plain the ba­sic ideas in terms of the no­ta­tions and cov­er­age of this book. If you ac­tu­ally want to ap­ply the method, you will need to take up a book writ­ten by ex­perts who know what they are talk­ing about. The book by Sz­abo and Ostlund [45] was the main ref­er­ence for this sec­tion, and is rec­om­mended as a well writ­ten in­tro­duc­tion. Be­low are some ad­di­tional con­cepts you may want to be aware of.


9.3.5.1 Mean­ing of the or­bital en­er­gies

In the sin­gle elec­tron case, the or­bital en­ergy $\epsilon_n$ in the canon­i­cal Hartree-Fock equa­tion

\begin{displaymath}
\begin{array}[b]{l}
\displaystyle
h^{\rm e}\pe n/{\skew0\...
...w0\vec r}///
= \epsilon_n \pe n/{\skew0\vec r}///
\end{array}\end{displaymath}

rep­re­sents the ac­tual en­ergy of the elec­tron. It also rep­re­sents the ion­iza­tion en­ergy, the en­ergy re­quired to take the elec­tron away from the nu­clei and leave it far away at rest. This sub­sub­sec­tion will show that in the mul­ti­ple elec­tron case, the “or­bital en­er­gies” $\epsilon_n$ are not or­bital en­er­gies in the sense of giv­ing the con­tri­bu­tions of the or­bitals to the to­tal ex­pec­ta­tion en­ergy. How­ever, they can still be taken to be ap­prox­i­mate ion­iza­tion en­er­gies. This re­sult is known as “Koop­man’s the­o­rem.”

To ver­ify the the­o­rem, a suit­able equa­tion for $\epsilon_n$ is needed. It can be found by tak­ing an in­ner prod­uct of the canon­i­cal equa­tion above with $\pe{n}/{\skew0\vec r}///$, i.e. by putting $\pe{n}/{\skew0\vec r}///^*$ to the left of both sides and in­te­grat­ing over ${\skew0\vec r}$. That pro­duces

\begin{displaymath}
\epsilon_n = E^{\rm e}_n
+ \sum_{{\underline n}=1}^I J_{n{...
...ownarrow}_{\underline n}\right\rangle ^2 K_{n{\underline n}} %
\end{displaymath} (9.29)

which con­sists of the sin­gle-elec­tron en­ergy $E^{\rm e}_n$, Coulomb in­te­grals $J_{n{\underline n}}$ and ex­change in­te­grals $K_{n{\underline n}}$ as de­fined in sub­sec­tion 9.3.3. It can al­ready be seen that if all the $\epsilon_n$ are summed to­gether, it does not pro­duce the to­tal ex­pec­ta­tion en­ergy (9.26), be­cause that one in­cludes a fac­tor ${\textstyle\frac{1}{2}}$ in front of the Coulomb and ex­change in­te­grals. So, $\epsilon_n$ can­not be seen as the part of the sys­tem en­ergy as­so­ci­ated with or­bital $\pe{n}//b//$ in any mean­ing­ful sense.

How­ever, $\epsilon_n$ can still be viewed as an ap­prox­i­mate ion­iza­tion en­ergy. As­sume that the elec­tron is re­moved from or­bital $\pe{n}//b//$, leav­ing the elec­tron at in­fi­nite dis­tance at rest. No, scratch that; all elec­trons share or­bital $\pe{n}//b//$, not just one. As­sume that one elec­tron is re­moved from the sys­tem and that the re­main­ing $I-1$ elec­trons stay out of the or­bital $\pe{n}//b//$. Then, if it is as­sumed that the other or­bitals do not change, the new sys­tem’s Slater de­ter­mi­nant is the same as the orig­i­nal sys­tem’s, ex­cept that col­umn $n$ and a row have been re­moved. The ex­pec­ta­tion en­ergy of the new state then equals the orig­i­nal ex­pec­ta­tion en­ergy, ex­cept that $E^{\rm e}_n$ and the $n$-​th col­umn plus the $n$-​th row of the Coulomb and ex­change in­te­gral ma­tri­ces have been re­moved. The en­ergy re­moved is then ex­actly $\epsilon_n$ above. (While $\epsilon_n$ only in­volves the $n$-​th row of the ma­tri­ces, not the $n$-​th col­umn, it does not have the fac­tor $\frac12$ in front of them like the ex­pec­ta­tion en­ergy does. And rows equal columns in the ma­tri­ces, so half the row in $\epsilon_n$ counts as the half col­umn in the ex­pec­ta­tion en­ergy and the other half as the half row. This counts the el­e­ment ${\underline n}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n$ twice, but that is zero any­way since $J_{nn}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $K_{nn}$.)

So by the re­moval of the elec­tron from (read: and) or­bital $\pe{n}//b//$, an amount of en­ergy $\epsilon_n$ has been re­moved from the ex­pec­ta­tion en­ergy. Bet­ter put, a pos­i­tive amount of en­ergy $-\epsilon_n$ has been added to the ex­pec­ta­tion en­ergy. So the ion­iza­tion en­ergy is $-\epsilon_n$ if the elec­tron is re­moved from or­bital $\pe{n}//b//$ ac­cord­ing to this story.

Of course, the as­sump­tion that the other or­bitals do not change af­ter the re­moval of one elec­tron and or­bital is du­bi­ous. If you were a lithium elec­tron in the ex­pan­sive 2s state, and some­one re­moved one of the two in­ner 1s elec­trons, would you not want to snug­gle up a lot more closely to the now much less shielded three-pro­ton nu­cleus? On the other hand, in the more likely case that some­one re­moved the 2s elec­tron, it would prob­a­bly not seem like that much of an event to the re­main­ing two 1s elec­trons near the nu­cleus, and the as­sump­tion that the or­bitals do not change would ap­pear more rea­son­able. And nor­mally, when you say ion­iza­tion en­ergy, you are talk­ing about re­mov­ing the elec­tron from the high­est en­ergy state.

But still, you should re­ally re­com­pute the re­main­ing two or­bitals from the canon­i­cal Hartree-Fock equa­tions for a two-elec­tron sys­tem to get the best, low­est, en­ergy for the new $I-1$ elec­tron ground state. The en­ergy you get by not do­ing so and just stick­ing with the orig­i­nal or­bitals will be too high. Which means that all else be­ing the same, the ion­iza­tion en­ergy will be too high too.

How­ever, there is an­other er­ror of im­por­tance here, the er­ror in the Hartree-Fock ap­prox­i­ma­tion it­self. If the orig­i­nal and fi­nal sys­tem would have the same Hartree-Fock er­ror, then it would not make a dif­fer­ence and $\epsilon_n$ would over­es­ti­mate the ion­iza­tion en­ergy as de­scribed above. But Sz­abo and Ostlund [45, p. 128] note that Hartree-Fock tends to over­es­ti­mate the en­ergy for the orig­i­nal larger sys­tem more than for the fi­nal smaller one. The dif­fer­ence in Hartree-Fock er­ror tends to com­pen­sate for the er­ror you make by not re­com­put­ing the fi­nal or­bitals, and in gen­eral the or­bital en­er­gies pro­vide rea­son­able first ap­prox­i­ma­tions to the ex­per­i­men­tal ion­iza­tion en­er­gies.

The op­po­site of ion­iza­tion en­ergy is “elec­tron affin­ity,” the en­ergy with which the atom or mol­e­cule will bind an ad­di­tional free elec­tron [in its va­lence shell], {N.19}. It is not to be con­fused with elec­troneg­a­tiv­ity, which has to do with will­ing­ness to take on elec­trons in chem­i­cal bonds, rather than free elec­trons.

To com­pute the elec­tron affin­ity of an atom or mol­e­cule with $I$ elec­trons us­ing the Hartree-Fock method, you can ei­ther re­com­pute the $I+1$ or­bitals with the ad­di­tional elec­tron from scratch, or much eas­ier, just use the Fock op­er­a­tor of the $I$ elec­trons to com­pute one more or­bital $\pe{I+1}//b//$. In the later case how­ever, the en­ergy of the fi­nal sys­tem will again be higher than Hartree-Fock, and it be­ing the larger sys­tem, the Hartree-Fock en­ergy will be too high com­pared to the $I$-​elec­tron sys­tem al­ready. So now the er­rors add up, in­stead of sub­tract as in the ion­iza­tion case. If the fi­nal en­ergy is too high, then the com­puted bind­ing en­ergy will be too low, so you would ex­pect $\epsilon_{I+1}$ to un­der­es­ti­mate the elec­tron affin­ity rel­a­tively badly. That is es­pe­cially so since affini­ties tend to be rel­a­tively small com­pared to ion­iza­tion en­er­gies. In­deed Sz­abo and Ostlund [45, p. 128] note that while many neu­tral mol­e­cules will take up and bind a free elec­tron, pro­duc­ing a sta­ble neg­a­tive ion, the or­bital en­er­gies al­most al­ways pre­dict neg­a­tive bind­ing en­ergy, hence no sta­ble ion.


9.3.5.2 As­ymp­totic be­hav­ior

The ex­change terms in the Hartree-Fock po­ten­tial are not re­ally a po­ten­tial, but an op­er­a­tor. It turns out that this makes a ma­jor dif­fer­ence in how the prob­a­bil­ity of find­ing an elec­tron de­cays with dis­tance from the sys­tem.

Con­sider again the Fock eigen­value prob­lem, but with the sin­gle-elec­tron Hamil­ton­ian iden­ti­fied in terms of ki­netic en­ergy and nu­clear at­trac­tion,

\begin{displaymath}
\begin{array}[b]{l}
\displaystyle
- \frac{\hbar^2}{2m_{\r...
...w0\vec r}///
= \epsilon_n \pe n/{\skew0\vec r}///
\end{array}\end{displaymath}

Now con­sider the ques­tion which of these terms dom­i­nate at large dis­tance from the sys­tem and there­fore de­ter­mine the large-dis­tance be­hav­ior of the so­lu­tion.

The first term that can be thrown out is $v^{\rm Ne}$, the Coulomb po­ten­tial due to the nu­clei; this po­ten­tial de­cays to zero ap­prox­i­mately in­versely pro­por­tional to the dis­tance from the sys­tem. (At large dis­tance from the sys­tem, the dis­tances be­tween the nu­clei can be ig­nored, and the po­ten­tial is then ap­prox­i­mately the one of a sin­gle point charge with the com­bined nu­clear strengths.) Since $\epsilon_n$ in the right hand side does not de­cay to zero, the nu­clear term can­not sur­vive com­pared to it.

Sim­i­larly the third term, the Coulomb part of the Hartree-Fock po­ten­tial, can­not sur­vive since it too is a Coulomb po­ten­tial, just with a charge dis­tri­b­u­tion given by the or­bitals in the in­ner prod­uct.

How­ever, the fi­nal term in the left hand side, the ex­change part of the Hartree-Fock po­ten­tial, is more tricky, be­cause the var­i­ous parts of this sum have other or­bitals out­side of the in­ner prod­uct. This term can still be ig­nored for the slow­est-de­cay­ing spin-up and spin-down states, be­cause for them none of the other or­bitals is any larger, and the mul­ti­ply­ing in­ner prod­uct still de­cays like a Coulomb po­ten­tial (faster, ac­tu­ally). Un­der these con­di­tions the ki­netic en­ergy will have to match the right hand side, im­ply­ing

\begin{displaymath}
\mbox{slowest decaying orbitals: }
\pe n/{\skew0\vec r}/// \sim \exp(-\sqrt{-2m_{\rm e}\epsilon_n}r/\hbar + \ldots)
\end{displaymath}

From this ex­pres­sion, it can also be seen that the $\epsilon_n$ val­ues must be neg­a­tive, or else the slow­est de­cay­ing or­bitals would not have the ex­po­nen­tial de­cay with dis­tance of a bound state.

The other or­bitals, how­ever, can­not be less than the slow­est de­cay­ing one of the same spin by more than al­ge­braic fac­tors: the slow­est de­cay­ing or­bital with the same spin ap­pears in the ex­change term sum and will have to be matched. So, with the ex­change terms in­cluded, all or­bitals nor­mally de­cay slowly, rais­ing the chances of find­ing elec­trons at sig­nif­i­cant dis­tances. The de­cay can be writ­ten as

\begin{displaymath}
\pe n/{\skew0\vec r}/// \sim
\exp(-\sqrt{2m_{\rm e}\vert\e...
...on_m\vert _{\rm min,\ same\ spin,\ no\ ss}}r
/\hbar + \ldots)
\end{displaymath} (9.30)

where $\epsilon_m$ is the $\epsilon$ value of small­est mag­ni­tude (ab­solute value) among all the or­bitals with the same spin.

How­ever, in the case that $\pe{n}/{\skew0\vec r}///$ is spher­i­cally sym­met­ric, (i.e. an s state), ex­clude other s-states as pos­si­bil­i­ties for $\epsilon_m$. The rea­son is a pe­cu­liar­ity of the Coulomb po­ten­tial that makes the in­ner prod­uct ap­pear­ing in the ex­change term ex­po­nen­tially small at large dis­tance for two or­thog­o­nal, spher­i­cally sym­met­ric states. (For the in­cur­ably cu­ri­ous, it is a re­sult of Maxwell’s first equa­tion ap­plied to a spher­i­cally sym­met­ric con­fig­u­ra­tion like fig­ure 13.1, but with mul­ti­ple spher­i­cally dis­trib­uted charges rather than one, and the net charge be­ing zero.)


9.3.5.3 Hartree-Fock limit

The Hartree-Fock ap­prox­i­ma­tion greatly sim­pli­fies find­ing a many-di­men­sion­al wave func­tion. But re­ally, solv­ing the “eigen­value prob­lems” (9.28) for the or­bitals it­er­a­tively is not that easy ei­ther. Typ­i­cally, what one does is to write the or­bitals $\pe{n}////$ as sums of cho­sen sin­gle-elec­tron func­tions $f_1,f_2,\ldots$. You can then pre­com­pute var­i­ous in­te­grals in terms of those func­tions. Of course, the num­ber of cho­sen sin­gle-elec­tron func­tions will have to be a lot more than the num­ber of or­bitals $I$; if you are only us­ing $I$ cho­sen func­tions, it re­ally means that you are choos­ing the or­bitals $\pe{n}////$ rather than com­put­ing them.

But you do not want to choose too many func­tions ei­ther, be­cause the re­quired nu­mer­i­cal ef­fort will go up. So there will be an er­ror in­volved; you will not get as close to the true best or­bitals as you can. One thing this means is that the ac­tual er­ror in the ground state en­ergy will be even larger than true Hartree-Fock would give. For that rea­son, the Hartree-Fock value of the ground state en­ergy is called the Hartree-Fock limit: it is how close you could come to the cor­rect en­ergy if you were able to solve the Hartree-Fock equa­tions ex­actly.

In short, to com­pute the Hartree-Fock so­lu­tion ac­cu­rately, you want to se­lect a large num­ber of sin­gle-elec­tron func­tions to rep­re­sent the or­bitals. But don't start us­ing zil­lions of them. The prob­lem is that even the ex­act Hartree-Fock so­lu­tion still has a fi­nite er­ror; a wave func­tion can­not in gen­eral be de­scribed ac­cu­rately us­ing only a sin­gle Slater de­ter­mi­nant. So what would the point in com­put­ing the very in­ac­cu­rate num­bers to ten dig­its ac­cu­racy?


9.3.5.4 Cor­re­la­tion en­ergy

As the pre­vi­ous sub­sub­sec­tion noted, the Hartree-Fock so­lu­tion, even if com­puted ex­actly, will still have a fi­nite er­ror. You might think that this er­ror would be called some­thing like “Hartree-Fock er­ror.” Or maybe “rep­re­sen­ta­tion er­ror“ or sin­gle-de­ter­mi­nant er­ror, since it is due to an in­com­plete rep­re­sen­ta­tion of the true wave func­tion us­ing a sin­gle Slater de­ter­mi­nant.

How­ever, the Hartree-Fock er­ror in en­ergy is called “cor­re­la­tion en­ergy.” The rea­son is be­cause there is a en­er­giz­ing cor­re­la­tion be­tween the more im­pen­e­tra­ble and poorly de­fined your jar­gon, and the more re­spect you will get for do­ing all that in­com­pre­hen­si­ble stuff.

And of course the word er­ror should never be used in the first place, God for­bid. Or those hated non-ex­perts might fig­ure out that Hartree-Fock has an er­ror in en­ergy so big that it makes the base ap­prox­i­ma­tion pretty much use­less for chem­istry.

To un­der­stand what physi­cists are re­fer­ring to with cor­re­la­tion, re­con­sider the form of the Hartree-Fock wave func­tion, as de­scribed in sub­sec­tion 9.3.1. It con­sisted of a sin­gle Slater de­ter­mi­nant. How­ever, that Slater de­ter­mi­nant in turn con­sisted of a lot of Hartree prod­ucts, the first of which was

\begin{displaymath}
\pe1/{\skew0\vec r}_1/b/z1/ \pe2/{\skew0\vec r}_2/b/z2/ \pe3/{\skew0\vec r}_3/b/z3/ \ldots \pe I/{\skew0\vec r}_I/b/zI/
\end{displaymath}

The other Hartree prod­ucts were dif­fer­ent only in the or­der in which the elec­trons ap­pear in the prod­uct. And since the elec­trons are all the same, the or­der does not make a dif­fer­ence: each of these Hartree prod­ucts has the same ex­pec­ta­tion en­ergy. Each also sat­is­fies the Pauli ex­clu­sion prin­ci­ple but, by it­self, not the an­ti­sym­metriza­tion re­quire­ment.

Now, con­sider what the Born sta­tis­ti­cal in­ter­pre­ta­tion says about the sin­gle Hartree prod­uct above. It says that the prob­a­bil­ity of elec­tron 1 to be within a vicin­ity of vol­ume ${\rm d}^3{\skew0\vec r}_1$ around a given po­si­tion ${\skew0\vec r}_1$ with given spin $S_{z1}$, and elec­tron 2 to be within a vicin­ity of vol­ume ${\rm d}^3{\skew0\vec r}_2$ around a given po­si­tion ${\skew0\vec r}_2$ with given spin $S_{z2}$, etcetera, is given by

\begin{eqnarray*}
&& \left\vert\pe1/{\skew0\vec r}_1/b/z1/\right\vert^2 {\rm d}...
...t^2 {\rm d}^3{\skew0\vec r}_3 \\
&& \quad \quad \quad{} \ldots
\end{eqnarray*}

This takes the form of a prob­a­bil­ity for elec­tron 1 to be in the given state that is in­de­pen­dent of where the other elec­trons are, times a prob­a­bil­ity for elec­tron 2 to be in the given state that is in­de­pen­dent of where the other elec­trons are, etcetera. In short, in a sin­gle Hartree prod­uct the elec­trons do not care where the other elec­trons are. Their po­si­tions are un­cor­re­lated.

Un­cor­re­lated po­si­tions would be OK if the elec­trons did not re­pel each other. In that case, each elec­tron would in­deed not care where the other elec­trons are. Then all Hartree prod­ucts would have the same en­ergy, which would also be the en­ergy of the com­plete Slater de­ter­mi­nant.

But elec­trons do re­pel each other. So, if elec­tron 1 is at a given po­si­tion ${\skew0\vec r}_1$, elec­tron 2 can re­duce its po­ten­tial en­ergy by pre­fer­ring po­si­tions far­ther away from that po­si­tion. It can­not overdo it, as that will in­crease its ki­netic en­ergy too much, but there is some room for im­prove­ment. So the ex­act wave func­tion will have cor­re­la­tions be­tween the po­si­tions of dif­fer­ent elec­trons. Based on ar­gu­ments like that, physi­cists then come up with the term cor­re­la­tion en­ergy.

Not so fast, physi­cists! For one, a Slater de­ter­mi­nant is not a sin­gle Hartree prod­uct but al­ready in­cludes some elec­tron cor­re­la­tions. Also, cor­re­la­tion en­ergy is not the same as “er­ror in en­ergy caused by in­cor­rect cor­re­la­tions.” And “er­ror in en­ergy caused by in­cor­rect cor­re­la­tions” is not the same as “er­ror in en­ergy for an in­cor­rect so­lu­tion, in­clud­ing in­cor­rect cor­re­la­tions.” And the last is what the Hartree-Fock er­ror re­ally is. Note that while there is some rough qual­i­ta­tive re­la­tion be­tween po­ten­tial en­ergy and elec­tron po­si­tion cor­re­la­tions, you can­not find the po­ten­tial en­ergy by pon­tif­i­cat­ing about elec­trons try­ing to stay away from each other. And the cor­rect en­ergy state is found by del­i­cately bal­anc­ing sub­tle re­duc­tions in po­ten­tial en­ergy against sub­tle in­creases in ki­netic en­ergy. The ki­netic en­ergy does not even care about elec­tron cor­re­la­tions. How­ever, the ki­netic en­ergy is wrong too when ap­plied on a sin­gle Slater de­ter­mi­nant.

See note {N.18} for more.


9.3.5.5 Con­fig­u­ra­tion in­ter­ac­tion

Since the base Hartree-Fock ap­prox­i­ma­tion has an er­ror that is far too big for typ­i­cal chem­istry ap­pli­ca­tions, the next ques­tion is what can be done about it. The ba­sic an­swer is sim­ple: use more that $I$ or­bitals, i.e. sin­gle-par­ti­cle wave func­tions. As al­ready noted in sec­tion 5.7, if you in­clude enough or­tho­nor­mal ba­sis func­tions, us­ing all their pos­si­ble Slater de­ter­mi­nants, you can ap­prox­i­mate any func­tion to ar­bi­trary ac­cu­racy.

Af­ter the $I$, (or $I$$\raisebox{.5pt}{$/$}$​2 in the re­stricted closed-shell case,) spa­tial or­bitals have been found, the Hartree-Fock op­er­a­tor be­comes just a Her­mit­ian op­er­a­tor, and can be used to com­pute fur­ther or­tho­nor­mal or­bitals $\pe{I+1}//b//,\pe{I+2}//b//,\ldots$. You can add these to the mix, say to get a bet­ter ap­prox­i­ma­tion to the true ground state wave func­tion of the sys­tem.

You might want to try to start small. If you in­clude just one more or­bital $\pe{I+1}//b//$, you can al­ready form $I$ more Slater de­ter­mi­nants: you can re­place any of the $I$ or­bitals in the orig­i­nal de­ter­mi­nant by the new func­tion $\pe{I+1}//b//$. So you can now ap­prox­i­mate the true wave func­tion by the more gen­eral ex­pres­sion

\begin{eqnarray*}
\Psi
&=& a_0 \Big({\left\vert{\rm det}\;\pe1//b//,\pe2//b//,...
...b//,\pe2//b//,\pe3//b//,\ldots,\pe I+1//b//\right\rangle}
\Big)
\end{eqnarray*}

where the co­ef­fi­cients $a_1,a_2,\ldots$ are to be cho­sen to ap­prox­i­mate the ground state en­ergy more closely and $a_0$ is a nor­mal­iza­tion con­stant.

The ad­di­tional $I$ Slater de­ter­mi­nants are called “ex­cited de­ter­mi­nants”. For ex­am­ple, the first ex­cited state

\begin{displaymath}
{\left\vert{\rm det}\;\pe I+1//b//,\pe2//b//,\pe3//b//,\ldots,\pe I//b//\right\rangle}
\end{displaymath}

is like a state where you ex­cited an elec­tron out of the low­est state $\pe1//b//$ into an el­e­vated en­ergy state $\pe{I+1}//b//$.

(How­ever, note that if you re­ally wanted to sat­isfy the vari­a­tional re­quire­ment $\delta\left\langle{E}\right\rangle $ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 for such a state, you would have to re­com­pute the or­bitals from scratch, us­ing $\pe{I+1}//b//$ in the Fock op­er­a­tor in­stead of $\pe1//b//$. That is not what you want to do here; you do not want to cre­ate to­tally new or­bitals, just more of them.)

It may seem that this must be a win­ner: as much as $I$ more de­ter­mi­nants to fur­ther min­i­mize the en­ergy. Un­for­tu­nately, now you pay the price for do­ing such a great job with the sin­gle de­ter­mi­nant. Since, hope­fully, the Slater de­ter­mi­nant is the best sin­gle de­ter­mi­nant that can be formed, any changes that are equiv­a­lent to sim­ply chang­ing the de­ter­mi­nant's or­bitals will do no good. And it turns out that the $I+1$-​de­ter­mi­nant wave func­tion above is equiv­a­lent to the sin­gle-de­ter­mi­nant wave func­tion

\begin{displaymath}
\Psi = a_0 {\left\vert{\rm det}\;\pe1//b//+a_1\pe{I+1}//b//...
...{I+1}//b//,
\ldots,\pe{I}//b//+a_I\pe{I+1}//b//\right\rangle}
\end{displaymath}

as you can check with some knowl­edge of the prop­er­ties of de­ter­mi­nants. Since you al­ready have the best sin­gle de­ter­mi­nant, all your ef­forts are go­ing to be wasted if you try this.

You might try form­ing an­other set of $I$ ex­cited de­ter­mi­nants by re­plac­ing one of the or­bitals in the orig­i­nal Hartree-Fock de­ter­mi­nant by $\pe{I+2}//b//$ in­stead of $\pe{I+1}//b//$, but the fact is that the in­fin­i­tes­i­mal vari­a­tional con­di­tion $\delta\left\langle{E}\right\rangle $ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 is still go­ing to be sat­is­fied when the wave func­tion is the orig­i­nal Hartree-Fock one. For small changes in wave func­tion, the ad­di­tional de­ter­mi­nants can still be pushed in­side the Hartree-Fock one. To en­sure a de­crease in en­ergy, you want to in­clude de­ter­mi­nants that al­low a nonzero de­crease in en­ergy even for small changes from the orig­i­nal de­ter­mi­nant, and that re­quires dou­bly ex­cited de­ter­mi­nants, in which two dif­fer­ent orig­i­nal states are re­placed by ex­cited ones like $\pe{I+1}//b//$ and $\pe{I+2}//b//$.

Note that you can form $I(I-1)$ such de­ter­mi­nants; the num­ber of de­ter­mi­nants rapidly ex­plodes when you in­clude more and more or­bitals. And a math­e­mat­i­cally con­ver­gent process would re­quire an as­ymp­tot­i­cally large set of or­bitals, com­pare chap­ter 5.7. How big is your com­puter?

Most peo­ple would prob­a­bly call im­prov­ing the wave func­tion rep­re­sen­ta­tion us­ing mul­ti­ple Slater de­ter­mi­nants some­thing like mul­ti­ple-de­ter­mi­nant rep­re­sen­ta­tion, or ex­cited-de­ter­mi­nant cor­rec­tion.. How­ever, it is called con­fig­u­ra­tion in­ter­ac­tion. The rea­son is that every hated non-ex­pert will won­der whether the physi­cist is talk­ing about the con­fig­u­ra­tion of the nu­clei or the elec­trons, and what it is in­ter­act­ing with.

(Ac­tu­ally, con­fig­u­ra­tion refers to the prac­ti­tioner con­fig­ur­ing all those de­ter­mi­nants, no kid­ding. The in­ter­ac­tion is with the com­puter used to do so. Sup­pose you were cre­at­ing the nu­mer­i­cal mesh for some fi­nite dif­fer­ence or fi­nite el­e­ment com­pu­ta­tion. If you called that con­fig­u­ra­tion in­ter­ac­tion in­stead of “mesh gen­er­a­tion,” be­cause it re­quired you to con­fig­ure all those mesh points through in­ter­act­ing with your com­puter, some peo­ple might doubt your san­ity. But in physics, the stan­dards are not so high.)