he Hartree-Fock Equations Our goal is to construct the best single determinant wave function for a system of electrons. We write our trial function as a determinant of spin orbitals = A ψ,,... ϕ ϕ ϕ, where ϕ ϕ = δ, A is the antisymeterizing operator, i i and we want to determine the orbitals { }, (,, ) ψ (,, ) E = ψ H H = f () i + g(, i ) i i ϕ = so that the energy E is a minimum. Recall that the Hamiltonian is the sum of one and two body operators i< Using the Slater-Condon rules the energy can be written in terms of integrals over the spin orbitals, ( ) E = ψ H ψ = ϕ f ϕ + ϕ () ϕ () g(, ) P ϕ () ϕ () P i i i i where is a transposition operator that interchanges the coordinates and. We want to find those orbitals for which the energy is stationary to first order. his means that if the orbitals we seek are { } { }, ϕi δϕ i i = i i, ϕ = and we incremented each by an infinitesimal amount + and recalculated the energy it should not change. Accordingly if we increment or vary each of the ϕ i independently, we would write (,, ) E ϕ + δϕ ϕ + δϕ ϕ + δϕ = ϕ + δϕ f ϕ + δϕ i i i i ( ϕ ) ( i δϕi ϕ δϕ g P ) ( ϕi δϕi)( χ δϕ ) + + + (, ) + +. he one-electron contribution becomes J. F. Harrison January 005
{ ϕ i f ϕi + δϕi f ϕi + ϕi f δϕi + δϕi f δϕi } or { ϕi f ϕi + δϕi f ϕi + ϕi f δϕi } + ( δ ) 0. In a similar way, we may expand the two-electron terms { ϕ ϕ G ϕ ϕ + ϕ δϕ G ϕ ϕ + δϕ ϕ G ϕ ϕ i i i i i i } + ϕϕ G ϕδϕ + ϕϕ G δϕϕ + 0 δ, i i i i ( ) where G, = g, P. We may write the restricted sum and so as the less restricted sum ϕ δϕ G ϕ ϕ = ' ϕ δϕ G ϕ ϕ, i i i i ' where the prime on the sum means that i. Since i and are dummy indices, this term may be written as and, of course, ' ϕ δϕ G ϕ ϕ ' δϕ δϕ G ϕ ϕ i i i i ' ϕ ϕ G ϕ δϕ = ' ϕ ϕ G δϕ ϕ. i i i i So, if we form the difference (,, ) (,, ) δ E = E ϕ + δϕ ϕ + δϕ ϕ + δϕ E ϕ ϕ ϕ J. F. Harrison January 005
we have the first order change in energy { i i i i } δe = δϕ f ϕ + ϕ f δϕ { δϕ iϕ ϕiϕ ϕiϕ δϕiϕ } + ' G + G δe = δϕi f + ' ϕ G ϕ dτ ϕi = () (,). () + ϕi f + ' ϕ G(,) ϕ dτ δϕi = ote that we may extend the summation ' to because the term vanishes = identically. For compactness, define the operator = () = τ ϕ (,)( ) ϕ V d g P and { i i i i } δe = δϕ f + V ϕ + ϕ f + V δϕ. V is called the Hartree-Fock Potential. We further define the Fock operator as and write δ E as F = f + V { i i i i } δe = δϕ F ϕ + ϕ F δϕ. We must not vary δϕ independently of δϕ, since they are coupled by the constraints, i ϕ ϕ = δ. i i o comply with these constraints, we use the technique of agrangian multipliers. Recall, to vary the functional F( ϕ ( x) ) with respect to ( x) ( ) 0, ϕ subect to G ϕ x = we first form the independent variations of F and G, i. e., δ F and δ G, and then, add a multiple of the variation of the constraint to the variation of the functional to form δ F + λδ G. J. F. Harrison January 005 3
One then chooses ϕ ( x) such that this expression is zero, i. e., solve δ F + λδ G = 0. We then obtain ϕ ( x, λ ) and fix λ by requiring ( ) G ϕ x, λ = 0. λ is called a agrangian multiplier. In this problem we have constraints of the form ϕi ϕ = δiand each has the variation δ ϕ ϕ = ϕ + δϕ ϕ + δϕ ϕ ϕ i i i i = ϕ δϕ + δϕ ϕ + 0 δ. i i Introduce the agrangian multipliers λ i and subtract all the constraint variations to δ E, resulting in i, { i ( i i )} δε λ ϕ δϕ + δϕ ϕ i, ( i i i i ) = δe λ ϕ δϕ + λ δϕ ϕ ow, using the fact the Fock operator is Hermetian we can write δ E as. i { i i i i } δe = δϕ F ϕ + δϕ F ϕ. We may also rearrange the constraint variations as λ δ ϕ ϕ = λ δϕ ϕ + λ δϕ ϕ i i i i i i allowing us to write. J. F. Harrison January 005 4
δe + λ δ ϕ ϕ = i, i i δϕi ϕi λiϕ τ δϕi ϕi λiϕ = = () () () () () () F d F() () () d () τ If we require that this be zero, we recover the equations Fϕ λ ϕ = 0andFϕ λ ϕ = 0, i i i i = = ; from which we deduce λi = λ i i. e., λ i is an element of an Hermetian matrix. Since the two sets of equations are equivalent, we consider only one, i. e., F ϕi = λϕ i = ow, define a row vector ϕ = ( ϕ ϕ ϕ ), equations in a very compact matrix notation, and note that we may write the above set of F ϕ = ϕλ, where Fϕ ( F ϕ FϕFϕ ) and λ is the Hermetian matrix with elements λ i. o continue, we make use of the invariance of a Slater determinant to a unitary transformation of the constituent spin orbitals; i. e., since the normalized Slater determinant ( ) A ϕ ϕ ϕ,, Ψ = has associated with it a matrix, ϕ, i. e., J. F. Harrison January 005 5
() ( ) () ( ) ϕ ϕ ϕ ϕ ϕ ϕ ϕ = ϕ () ϕ ϕ ( ) then, (,, ) Ψ =! ϕ where ϕ is the determinant of ϕ. Define a new set of functions ξ i, by the unitary transformation  ξ k = U ϕ k i =, ; k =, i i = or in matrix notation and ξ = Uϕ ϕ = U ξ ote that if we choose not to consider the electron index as a label we may write the transformation as i i i = =   ξ = U ϕ = ϕ U i =, and the row vectors ξ and ϕ are related by ξ = ϕ U or ϕ = ξ U he wavefunction in terms of the transformed orbitals becomes Ψ (,, ) = ϕ = U ξ = U ξ =!!!! ξ J. F. Harrison January 005 6
and so, the determinant formed from a unitary transformation of the { ϕ } functions is equal to the original determinant. Given this invariance, how are the Hartree-Fock Equations in the ϕ basis related to those in the ξ basis? o find out we substitute ϕ = ξ U into the Hartree-Fock equations. Since we have F = ϕ ϕ ϕλ F ξ U ξu = ξu λ and so F ξu ξ = ξuλu ow, U is an arbitrary unitary matrix and λ is Hermetian. Since a Hermetian matrix may always be diagonalized by a unitary matrix we have U λu = ε where ε is a diagonal matrix containing the eigenvalues of λ. hen, F ξ U ξ = ξε o find F ( ξ U ), we note: So, F f G, d ( ϕ) = + ϕ ϕ τ = () trace ϕ G (,) ϕ = f + () G () = f + trace. ϕ J. F. Harrison January 005 7
= + F ξu f trace ξu G ξu () trace { ξ ξ } = f + U G U () trace { G } ξ () trace { Gξ } = f + U U = f + and so = F ξ U F ( ξ ) or F ξ = εξ. i i i hese are the canonical Hartree-Fock Equations. he ξ i are the molecular (atomic) orbitals and the ε i are called the Hartree-Fock eigenvalues or the one-electron eigenvalues. otal Energy Once one has the HF orbitals, one may calculate the total energy, as follows: (,, ) (,, ) E = Ψ H Ψ () H () = Aϕ ϕ ϕ A ϕ ϕ ϕ () G = ϕ f ϕ + ϕ ϕ, ϕ ϕ + nuclear repulsion i i i i Recall the Hartree-Fock potential V ( ) is given by and so, τ ϕ ϕ V d G, =, = nuclei E = ϕ f + V ϕ + Z Z R, K i i K K< since F = f + V at least two choices present themselves: J. F. Harrison January 005 8
i) f + V = f + ( f + V) = ( f + F), he energy may be written in terms of the Hartree-Fock eigenvalues ( the orbital energies) the one electron matrix elements f i, and the nuclear repulsion energy. ( ) uclei E = ϕ f + F ϕ = ε + f + i i i i K< Z Z ii) Alternatively we may write the energy in terms of the Hartee-Fock eigenvalues, the two electron matrix elements (coulomb and exchange integrals) and the nuclear repulsion energy. then f + V = f + V V = F V, K R K uclei E = ϕ F V ϕ = ε ϕ V ϕ + i i i i i K< Z Z K R K uclei = + εi ϕi () ϕ G(,) ϕi() ϕ dτ (,) = K< Z Z K R K J. F. Harrison January 005 9