- 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

where

The square magnitude of the wave function above gives the probability
for the electrons

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 spin-up

and spin-down.

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

where

orbitsBut people do tend to think of the

spatial orbitals

For simplicity, it will be assumed that the spin orbitals are taken to
be normalized;

if you integrate the square magnitude
of

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

where the number of orbitals

It will be assumed that any two different spin orbitals orthogonal;

by
definition this means that their bracket is zero:

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:

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

Note also that the spin states

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

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

That is a fundamentally different wave function, not just minus the first Hartree product; orbitals

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 Slater determinant,

(9.15) |

The most general system wave function

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

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

The factor

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

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

and then you would make things easier for yourself by postulating a priori that the spatial orbitals are the same,

you get

This automagically reproduces the correct singlet spin state! (The approximation comes in because the exact spatial ground state,

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

2pshell, 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

You have introduced a bias in the determinant: there is now a real difference between the spatial orbitals

If you find the best approximation to the energy among all
possible spatial orbitals

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

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

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

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

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

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

Multiplying out the determinant gives

Note that you do get the correct singlet spin state. But

(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

you should find

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

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

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

9.3.2 The Hamiltonian

The nonrelativistic Hamiltonian of the system of

(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

(9.18) |

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

(9.19) |

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,

and

Note that

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

where

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

The combined single-electron energy for all

It is just as if you had electron 1 in state

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:”

To understand the Coulomb integrals better, the inner product above
can be written out explicitly as an integral, while also expanding

The integrand equals the probability of an electron in state

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

It is an interaction of the possibility that the first electron is in state

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

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

It is actually somewhat tricky to prove that the

(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

However, the normal unrestricted spin-up or spin-down orbitals, in which either

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

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

So, if there were no electron-electron repulsions, i.e.

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

The first term in the Fock operator is the single-electron
Hamiltonian. The mischief is in the innocuous-looking second term

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
potential

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

is only an operator after you have
found the orbitals 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

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

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”

To verify the theorem, a suitable equation for

However,

So by the removal of the electron from

(read: and)
orbital

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

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

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

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

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

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

(9.30) |

However, in the case that

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

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

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

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

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

After the

You might want to try to start small. If you include just one more
orbital

where the coefficients

The additional

is like a state where you excited an electron out of the lowest state

(However, note that if you really wanted to satisfy the variational
requirement

It may seem that this must be a winner: as much as

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 doubly

excited
determinants, in which two different original states are replaced by
excited ones like

Note that you can form

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.)