Hartree-Fock and Density Functional Theory

Atomic Units It is advantageous to use simplified units. If we set , then the Schrödinger equation for the hydrogen atom becomes:

To find the ground-state 1s solution, write , then

The solution is and . Thus radius of atom is 1 a.u.

The unit of length is then 1 a.u. = Bohr radius = 0.529 Å and the unit of energy

1 a.u. = 27.214 eV.

The many-body Schrödinger equation for the electrons in the field of fixed nuclei is then:

Here . The full Hamiltonian must include the kinetic energy of the ions.

Bohn-Oppenheimer Approximation

Write the total wave-function as

where is the wave-function of the ions alone and the wave-function of the electrons for infinitely massive ions, ie . Thus for finite nuclear masses, the term modulates the wavefunction. The next step is to substitute into the Schr{ödinger equation, multiply by and integrate over r. This gives:

Here,

The term on the right hand side of the first equation is zero when the state is non-degenerate as then we can always choose to be real. For then, RHS is

and integral is unity from the normalisation condition. This proof fails when state is degenerate or almost degenerate. We are usually interested in the ground state of a system is usually non-degenerate. The W term is also very small (actually zero for infinitely massive nuclei) and usually neglected. We are now left with the equation:

where

Hartree-Fock Theory

The Hartree-Fock assumption is :

where are set of orthonormal one-electron orbitals to be determined. is a spin function taking care of the internal degrees of freedom of the electron. Thus s take two values (up-, down-, or 1, 2) while is a quantum number - again (the z-components of the spin ), or possibly up or down. The spin functions satisfy:

The sum being over 2-values of s.

To derive the Hartree Fock equations for a H molecule, write the wave function as

Then the energy expectation is

This is, using atomic units and remembering the denominator is unity as the orbitals are orthonormal (including spin functions),

The average energy is then

Here,

Thus, summing over ,

The penultimate term is the exchange energy. Now, we use the variational principle and seek to minimise E subject to orthonormal spin-orbitals . Now, we use the variational principle and seek to minimise E subject to orthonormal .

To do this, we introduce a mathematical technique useful to find the maximum or minimum of functions with constraints. Sometimes we want to maximise a function f(x, y) subject to some relation like a x + b y =c where a, b and c are constants. Clearly, we could eliminate y as

and hence we maximise f(x, c/b -ax/b) with respect to x. This is maximum when

We can also do this by a method introduced by Lagrange. Introduce the function

Then maximise F with respect to x, y and z independently. We get

The last equation recovers the constraint and z can be eliminated from the first two giving

which is the same as the direct method. This can be generalised to non-linear constraints like g(x, y) = 0. For an extremal value, must be zero but dx and dy are not arbitrary as

Thus stationary point satisfies,

But Lagrange's method used to maximise f = F - z g gives the same equations:

From the first two we get, as before, . \

We minimise wrt the function:

We then get the Hartree-Fock equations:

are the Hartree and exchange potentials. The last involves a sum over other occupied orbitals whose spin is the same as . The electron density is:

Now can carry out a unitary transformation on the Slater determinant which diagonalises and then right hand side of the differential equation vanishes. Once the orbitals are known, can be evaluated.

The exchange potential can be rewritten as:

Two simple properties of are:

These relations show that the exchange forces the electrons away from resulting in an exchange hole. The total exchange charge density is -1. The total energy E can be found by multiplying the Hartree-Fock equations by and integrating. This gives:

Notice that the interaction terms must be subtracted from the sum of energy eigenvalues.

Problem: prove Koopmann's theorem which states that the difference in energy between two configurations differing by the occupation of an orbital , while all the other orbitals are unchanged, is . This allows us to interpret as the ionisation energy for the electron.

The Homogeneous Electron Gas.

Let us now apply the theory to jellium where the ions form a uniform background. Then a solution of the Hartree-Fock equations is

Here, stands for the quantum numbers . The charge density, n, is then uniform and the Hartree term cancels the electron-ion term leaving only the kinetic energy and exchange terms only. The exchange potential for either spin is independent of and hence the energy levels

This is independent of and depends on the magnitude of only.

Now, define , and note the electron density is . Then, we show below:

Notice that tends to 1 as , and to 0.5 as . tends to as tends to 1. The average of F is , ie

The total energy is

The total energy density is then the sum of the kinetic and exchange energy densities. This is

The last term is written where

is called the exchange energy density.

Evaluation of exchange energy

Here we have used integration by parts:

Hence

Notice that the density of states, per unit energy, for each spin is

Now, show that as . Hence density of states of a metal is zero at the Fermi level. This is incorrect and is due to absence of correlation in Hartree-Fock theory.

Density Functional Theory

Hohenberg and Kohn showed there is a 1-1 correspondence between the ground state (non-degenerate) wave-function and the electron density defined by

Here, we integrate over all the electrons spatial coordinates and sum over their spin coordinates.

The proof rests on the preliminary result that in the Hamiltonian

then the ground state electron density is in 1-1 correspondence with the external potential . Suppose, this is false, ie there exists two external potentials and having the same n. Then from the variational principle , if and are the corresponding normalised wave-functions, and if is the Hamiltonian with potential and energy , then

But in a similar way we can show

Adding these equations gives us

which is absurd unless the states are degenerate.

This shows that is determined uniquely by n and hence the Schrödinger equation for can be solved in terms of n. Thus is a unique functional of n. We use this to write down a variational equation for E in terms of n.

Define orthonormal orbitals so that the electron density is

Here the orbitals will be determined by the requirement that E is minimised wrt to n. To do this we need an expression for E in terms of n. The theorem we have just proved shows that there must be an expression. Kohn and Sham suggested we use

Here is exchange-correlation energy density which in the case of jellium is:

It is possible to carry out more accurate calculations for jellium leading to more precise values of .

The Kohn-Sham Equations are now derived by a variational principle remembering that the orbitals are orthonormal. Thus the quantity

is minimised wrt to This yields:

Here

The main difference with the Hartree-Fock equations is the exchange-correlation term. Show that for jellium, N(E) is no longer zero at .

An alternative formula for E is found by multiplying by and integrating, followed by a sum over the occupied orbitals .

This formula is useful for a discussion of approximate methods.

The terms and tend to cancel: as they must for jellium. For simple systems, the term acts as an attractive potential -- it gets more positive as the separation between atoms increases-- but the other three terms combine to act as a repulsive potential. It is often assumed that the repulsive one is short ranged and falls off quickly to zero. The attractive bonding term is longer ranged.