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The atrx Hartree-Fock equatons The most common method of solvng the Hartree-Fock equatons f the spatal btals s to expand them n terms of known functons, { χ µ } µ= consder the spn-unrestrcted case. We wll expand both the terms of the same bass functons φ = χ µ C µ & γ = χ µ C µ µ= µ= called the bass set. Let s frst & spatal btals n and then determne the expanson coeffcents Cµ & Cµ. ote that these equaltes hold only when the set { χ µ } µ= and { χ µ } µ= s complete and ths usually happens, f ever, when s large are carefully chosen. The qualty of the soluton depends crtcally on the extent to whch the expanson bass s complete. We wll return to ths pont subsequently. Substtutng the expanson of φ & γ nto the unrestrcted Hartree-Fock equatons results n & ˆ χµ µ = ε χµ µ χµ µ = ε χµ µ µ = µ = µ = µ = Fˆ C C F C C * and f one multples by χ ν and ntegrates one obtans F Cµ = ε Δ µν Cµ & F Cµ = ε Δµν Cµ µ = µ = µ = µ = F C = ε Δ C & F C = ε Δ C where we defne the Fock matrces f & spns as ˆ F F & F Fˆ F = = χ χ F = = χ χ ν µ ν µ

Δ s the overlap matrx n the expanson bass Δ = χ χ = ( Δ ) µν µ ν µν and the column vects contanng the expanson coeffcents C = C C C & C = C C C Wrtng the Fock operats n terms of the expanson bass results n F fˆ V V Kˆ fˆ = χ + + χ = χ χ + P νλ µρ P νλ ρµ ν µ ν µ where νλ µρ = χ () χ () χ () χ () dv () dv () and P = P + P wth * * ν λ µ ρ r = λ ρ & = λ ρ = = P C C P C C & P P are elements of the Coulson densty matrx f & spns. Whle the two electron ntegrals n the expanson bass are often wrtten as νλ µρ wth the assumed electron der one also sees the notaton ( λπ ) wth the electron der,.e., dv * * (, ) χ χ χ χ = νλ µρ ( ) ν λ µ ρ r

We wll use both notatons, dstngushng them by ether the bra-ket notaton (der ) the parenthess notaton (der ). By symmetry the Fock matrx f spns has the fm F fˆ V V Kˆ fˆ = χ + + χ = χ χ + P νλ µρ P νλ ρµ ν µ ν µ If we defne the matrces G & G ( νλ µρ νλ ρµ ) & ( νλ µρ νλ ρµ ) = = G P P G P P We may wrte the Fock matrces as F = f + G & F = f + G Knowng the bass functons the one electron term, f µν, s easly calculated whereas the two electron terms, G & G are me complcated as they depend on the expanson coeffcents! C &! C (through the Coulson densty matrces) whch are to be determned. To determne the expanson coeffcents we frst calculate all of the requred one and two electron ntegrals n the expanson bass. Then. Estmate the densty matrces f each spn. Ths s not as dffcult as t mght seem. One could use Huckel btals to fm an approxmate P even set the ntal estmate to zero.. Fm the matrces ( νλ µρ νλ ρµ ) & ( νλ µρ νλ ρµ ) = = G P P G P P 3. Assemble the Fock and overlap matrces 4. Solve the generalzed egenvalue problem F C = ε ΔC & F C = ε ΔC f C & C and ε & ε

5. Construct the Coulson densty matrces, P & P and compare wth the prevous estmate. If they agree wthn a gven tolerance we are fnshed. If they do not, return to step and fm the matrces wth the current vects. Repeat the process untl the nput and output densty matrces are equal wthn the set tolerance. Ths tolerance often depends on the subsequent use of the btals. Once we have the converged Hartree-Fock btals { φ } & { γ } { ε } & { ε } = = the electronc energy may be wrtten as = = and egenvalues ˆ ˆ ˆ ˆ ˆ ˆ E = ψ H ψ = φ f + J + J K φ + γ f + Jˆ + Jˆ Kˆ γ = = And recognzng as above that ˆ ˆ ˆ F f J Jˆ Kˆ = + + and Fˆ = fˆ + Jˆ + Jˆ Kˆ we have ˆ ˆ ˆ ˆ φ φ γ γ = = E = f + F + f + F ( ˆ ) ( ˆ φ φ ε γ γ ε ) E = f + + f + = = Whch n terms of the bass functons and the one electron ntegrals becomes E = P f + ε + ε = = Another convenent fm nvolvng the two electron ntegrals s easly derved E = tracep Γ + tracep ( Γ ) Γ ( Γ ) Γ = = f + G & = Γ = f + G

Closed Shell Systems The matrx Hartree-Fock equatons f a closed shell system are a specal case of the unrestrcted equatons that obtan when = = / and the spatal part of the and spn btals are dentcal, e, φ = γ. One solves FC = ε ΔC where F f G = + and G = P ( νλ µρ νλ ρµ ) wth P / = C C. λ ρ = Once one has the btals and egenvalues the electronc energy may be obtaned from E = P f + E / = tracep Γ = ε ( Γ ) = Γ = f + G