- 9.3.1 Wave function approximation
- 9.3.2 The Hamiltonian
- 9.3.3 The expectation value of energy
- 9.3.4 The canonical Hartree-Fock equations
- 9.3.5 Additional points

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 electrons in orbit around a
number of nuclei. The wave function describing the set of electrons
then has the general form:

where is the position of electron number , and its spin in a chosen -direction, with measurable values and . Of course, what answer you get for the wave function will also depend on where the nuclei are, but in this section, the nuclei are supposed to be at given positions, so to reduce the clutter, the dependence of the electron wave function on the nuclear positions will not be explicitly shown.

Hartree-Fock approximates the wave function in terms of a set of
single-electron functions, each a product of a spatial
function and a spin state:

where stands for either spin-up, , or spin-down, . (By definition, function equals one if the spin is , and zero if it is , while function equals zero if is and one if it is .) These single-electron functions are called “orbitals” or “spin orbitals.” The reason is that you tend to think of them as describing a single electron being in orbit around the nuclei with a particular spin. Wrong, of course: the electrons do not have reasonably defined positions on these scales. But you do tend to think of them that way anyway.

The spin orbitals are taken to be an orthonormal set. Note that any two spin orbitals are automatically orthogonal if they have opposite spins: spin states are orthonormal so 0. If they have the same spin, their spatial orbitals will need to be orthogonal.

Single-electron functions can be combined into multi-electron functions
by forming products of them of the form

where is the number of the single-electron function used for electron 1, the number of the single-electron function used for electron 2, and so on, and is a suitable numerical constant. Such a product wave function is called a “Hartree product.”

Now if you use enough single-electron functions, with all their
Hartree products, you can approximate any multi-electron wave function
to arbitrarily high accuracy. Unfortunately, using many of them
produces a problem much too big to be solved on even the most powerful
computer. So you really want to use as little of them as possible.
But you cannot use too few either; as chapter 5.7
explained, nature imposes an antisymmetrization

requirement: the complete wave function that you write must change
sign whenever any two electrons are exchanged, in other words when you
replace by and vice-versa for any
pair of electrons numbered and . That is only possible
if you use at least different single-electron functions for your
electrons. This is known as the Pauli exclusion principle: any
group of electrons occupying the minimum of
single-electron functions exclude

an additional
-th electron from simply entering the same functions. The
-th electron will have to find its own single-electron
function to add to the mix.

The basic Hartree-Fock approximation uses the absolute minimum that is
possible, just different single-electron functions for the
electrons. In that case, the wave function can be written as a
single Slater determinant:

(9.14) |

Displaying the Slater determinant fully as above may look impressive,
but it is a lot to read. Therefore, from now on it will be abbreviated
as

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 , , ... at random. The cleverness in the Hartree-Fock approach will be in writing down equations for these single-electron functions that produce the best approximation possible with a single Slater determinant.

This section will reserve the term orbitals

specifically for the single-electron functions that provide the
best single-determinant approximation. In those terms, if the
Hartree-Fock orbitals provide the best single-determinant
approximation, their results will certainly be better than the
solutions that were written down for the atoms in chapter
5.9, because those were really single Slater determinants.
In fact, you could find much more accurate ways to average out the
effects of the neighboring electrons than just putting them in the
nucleus like the section on atoms essentially did. You could smear
them out over some optimal area, say. But the solution you will get
doing so will be no better than you could get using Hartree-Fock.

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. Spin then really only directly affects the antisymmetrization requirements.

Things are straightforward if you try to solve, say, a helium atom. In
the exact ground state, the two electrons are in the spatial wave
function that has the absolutely lowest energy, regardless of any
antisymmetrization concerns. This spatial wave function is symmetric
under electron exchange since the two electrons are identical. The
antisymmetrization requirement is met since the electrons assume the
singlet configuration,

for their combined spins.

The approximate Hartree-Fock wave function for helium you would
correspondingly take to be

and then you would make things easier for yourself by postulating a priori that the spatial orbitals are the same, . Lo and behold, when you multiply out the Slater determinant,

it automagically reproduces the correct singlet spin state. 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
,
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 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 or . Lets assume
you take , so the wave function is

You have introduced a bias in the determinant: there is now a real difference between and : has the same spin as the third spin orbital, and opposite.

If you find the best approximation among all possible orbitals , , and , you will end up with spatial orbitals and 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, 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. In the case of lithium, you would then
demand that equals . 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
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,
2p, and 2p 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.

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 and with unequal spatial
orbitals multiplies out to

or, writing the spin combinations in terms of singlets and triplets,

So, the spin will be some combination of zero spin (the singlet) and spin one (the triplet), and the combination will be different at different locations of the electrons to boot. However, it may be noted that unrestricted wave functions are commonly used as first approximations of doublet and triplet 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

where is the state in which electron 1 is around the left proton and electron 2 around the right one, and is the same state but with the electrons reversed. Note that the spin state is a singlet one with zero net spin.

Now try to approximate it with a restricted closed shell Hartree-Fock
wave function of the form
. The
determinant multiplies out to

Now will be something like ; the energy of the electrons is lowest when they are near the nuclei. But if is appreciable when electron 1 is near say the left nucleus, then is also appreciable when electron 2 is near the same nucleus, since it is the exact same function. So there is a big chance of finding both electrons together near the same nucleus. That is all wrong, since the electrons repel each other: if one electron is around the left nucleus, the other should be around the right one. The computed energy, which should be that of two neutral hydrogen atoms far apart, will be much too high. Note however that you do get the correct spin state. Also, at the nuclear separation distance corresponding to the ground state of the complete molecule, the errors are much less, [45, p. 166]. Only when you are “breaking the bond” (dissociating the molecule, i.e. taking the nuclei apart) do you get into major trouble.

If instead you would use unrestricted Hartree-Fock,
, you
should find and
(or vice versa), which would produce a wave function

This would produce the correct energy, though the spin would now be all wrong. Little in life is ideal, is it?

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 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 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:

(9.16) |

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 nuclei, numbered using an index , and having
charge (i.e. there are protons in nucleus number ).
In that case, the total potential energy due to nucleus-electron
attractions is, summing over all electrons and over all nuclei:

(9.17) |

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

(9.18) |

Without this interaction 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,

and is the electron to electron repulsion potential energy

Note that , , ..., 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 , , ..., , 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 is defined as . There is a problem with using this expression as stands, though. Look once again at the arsenic atom example. There are 33 electrons in it, so you could try to choose 33 promising single-electron functions to describe it. You could then try to multiply out the Slater determinant for , integrate the inner products of the individual terms on a computer, and add it all together. However, the inner product of each pair of terms involves an integration over 99 scalar coordinates. Taking 10 locations along each axis to perform the integration, you would need to compute 10 values for each pair of terms. And there are 33! terms in the Slater determinant, or (33!) = 7.5 10 pairs of terms... A computer that could do that is unimaginable.

Fortunately, it turns out 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 produce only
single-electron energy expectation values of the general form

and that the name of the integration variable does not make any difference: you get the exact same value for electron 1 as for electron or any other. So the value of does not make a difference, and it will just be left away.

If there was just one electron and it was in single-electron state , would be its expectation value of energy. Actually, of course, there are electrons, each partly present in state because of the way the Slater determinant writes out, and each electron turns out to contribute an equal share to the total energy associated with single-electron state .

The Hamiltonians of the pair-repulsions produce six-dimensional inner
products that come in two types. The inner products of the first type
will be indicated by , and they are

If all you had was one electron in single-electron state and a second electron in single-electron state , this would be the expectation potential energy of their interaction. It would be the probability of electron being near and electron being near times the Coulomb potential energy at those positions. For that reason these integrals are called “Coulomb integrals.”

The second type of integrals will be indicated by ,
and they are

The exchange integrals are a reflection of nature doing business in
terms of an unobservable wave function, rather than the observable
probabilities that appear in the Coulomb integrals. They are the
equivalent of the twilight terms that have appeared before in two-state systems.
Written out as integrals, you get

Going back to the original question of the expectation energy of the
complete system of electrons, it turns out that it can be written
in terms of the various inner products above as

If you want to see where all this comes from, the derivations are in
{D.52}. There are also some a priori things you can
say about the Coulomb and exchange integrals, {D.53};
they are real, and additionally

The analysis can easily be extended to generalized orbitals that take
the form

However, the normal unrestricted spin-up or spin-down orbitals, in which either or is zero, already satisfy the variational requirement 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 section 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 there is an equation of the form:

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 is the single-electron Hamiltonian consisting of its
kinetic energy and its potential energy due to nuclear attractions,
and that is the potential energy of repulsion between two
electrons at given locations:

So, if there were no electron-electron repulsions, i.e. 0, the canonical equations above would turn into single-electron Hamiltonian eigenvalue problems of the form where 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,

where is called the “Fock operator,” and is written out further as:

The first term in the Fock operator is the single-electron Hamiltonian. The mischief is in the innocuous-looking second term . 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 (9.27) not described by the single-electron Hamiltonian :

To recover the canonical equations (9.27) from the Fock form, take an inner product with the spin . The definition of the Fock operator is unavoidably in terms of spin rather than spatial single-electron functions: 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
by itself, but only for applied to an
arbitrary single-electron function . The reason is
that is not a normal potential at all: the second term,
the one due to the exchange integrals, does not multiply by
a potential function, it shoves it into an inner product! The
Hartree-Fock potential

is an
operator, not a normal potential energy. Given a
single-electron function, it produces another function.

Actually, even that is not quite true. The Hartree-Fock
potential

is only an operator after you have
found the orbitals , , ...,
, ..., 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 ? If the Hartree-Fock potential 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 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

in the canonical Hartree-Fock equation

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” 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 is needed.
It can be found by taking an inner product of the canonical equation
above with , i.e. by putting to
the left of both sides and integrating over . That
produces

However, can still be viewed as an approximate ionization energy. Assume that the electron is removed from orbital , leaving the electron at infinite distance at rest. No, scratch that; all electrons share orbital , not just one. Assume that one electron is removed from the system and that the remaining electrons stay out of the orbital . 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 and a row have been removed. The expectation energy of the new state then equals the original expectation energy, except that and the -th column plus the -th row of the Coulomb and exchange integral matrices have been removed. The energy removed is then exactly above. (While only involves the -th row of the matrices, not the -th column, it does not have the factor in front of them like the expectation energy does. And rows equal columns in the matrices, so half the row in counts as the half column in the expectation energy and the other half as the half row. This counts the element twice, but that is zero anyway since .)

So by the removal of the electron from

(read: and)
orbital , an amount of energy has been
removed from the expectation energy. Better put, a positive amount of
energy has been added to the expectation energy. So the
ionization energy is if the electron is removed from
orbital 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 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 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 electrons using the Hartree-Fock method, you can either recompute the orbitals with the additional electron from scratch, or much easier, just use the Fock operator of the electrons to compute one more orbital . 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 -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 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,

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 , 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 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

From this expression, it can also be seen that the 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

(9.29) |

However, in the case that is spherically symmetric, (i.e. an s state), exclude other s-states as possibilities for . 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” (9.27) for the orbitals iteratively is not that easy either. Typically, what one does is to write the orbitals as sums of chosen single-electron functions . 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 ; if you are only using chosen functions, it really means that you are choosing the orbitals 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.

9.3.5.4 Configuration interaction

According to the previous subsubsection, 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 bottom line is that the Hartree-Fock solution still has a finite error, because a wave function cannot in general be described accurately using only a single Slater determinant. So what is the point in computing the wrong numbers to ten digits accuracy?

You might think that the error in the Hartree-Fock approximation would
be called something like Hartree-Fock error,

single determinant error,

or “representation
error,“ since it is due to an incomplete representation of the
true wave function. However, the error is called “correlation energy” 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,
{N.18}.

Anyway, in view of the fact that even an exact solution to the Hartree-Fock problem has a finite error, trying to get it exactly right is futile. At some stage, you would be much better off spending your efforts trying to reduce the inherent error in the Hartree-Fock approximation itself by including more determinants. As 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 , (or 2 in the restricted closed-shell case,) orbitals have been found, the Hartree-Fock operator becomes just a Hermitian operator, and can be used to compute further orthonormal orbitals . You can add these to the stew, 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 compute just one more orbital , you can already form more Slater determinants: you can replace any of the orbitals in the original determinant by the new function . So you can now approximate the true wave function by the more general expression

where the coefficients are to be chosen to approximate the ground state energy more closely and is a normalization constant.

The additional Slater determinants are called “excited determinants”. For example, the first excited state is like a state where you excited an electron out of the lowest state into an elevated energy state . (However, note that if you really wanted to satisfy the variational requirement 0 for such a state, you would have to recompute the orbitals from scratch, using in the Fock operator instead of . 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 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 -determinant wave function above is
equivalent to the single-determinant wave function

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 excited determinants by
replacing one of the orbitals in the original Hartree-Fock determinant
by instead of , but the fact is
that the variational condition 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 and .

Note that you can form 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 maybe
excited-determinant correction

or so. However, it is
called configuration interaction,

because every
nonexpert will wonder whether the physicist is talking about the
configuration of the nuclei or the electrons, (actually, it refers to
the practitioner configuring

all those
determinants, no kidding,) and what it is interacting with (with the
person bringing in the coffee, of course. OK.) If you said that you
were performing a configuration interaction

while
actually doing, say, some finite difference or finite element
computation, just because it requires you to specify a configuration
of mesh points, some people might doubt your sanity. But in physics,
the standards are not so high.