β(n α +1)φ 2 α(2)!φ 1

Transcription

1 Spin Contamination in Unrestricted Hartree-Fock Wavefunctions An unrestricted Hartree-Fock (UHF) wavefuntion with α α spins and β β spins has the form ψ (1,,!, ) = Aϕ 1 α(1)ϕ α()!ϕ α α( α )φ 1 β( α +1)φ β( α + )!φ β β( α + β ) where the α spin spatial orbitals { ϕ i } α and the β spin spatial orbitals φi { } β are eigenfunctions of different Fock operators. To show that this is an eigenfunction of Ŝ z we form Ŝ z ψ = Ŝ z (i) Aϕ 1 α(1)ϕ α()!φ 1 β( α +1)φ β( α + )! and since Ŝ z is symmetric with respect to electron permutations it commutes with Ŝ z ψ = A Ŝ z (i) ϕ 1 α(1)ϕ α()!φ 1 β( α +1)φ β( α + )! and since α & β are Eigenfunctions of Ŝ z (i) with eigenvalues ± 1 Ŝ z ψ = 1 ( α β )ψ we have To show that ψ is not an eigenfunction of Ŝ we evaluate the expectation value Ŝ = ψ Ŝ ψ = ψ S(i) i S( j) ψ = ψ Ŝ (i) ψ + ψ j=1 i> j A S(i) i Ŝ(! j) ψ Since Ŝ (i) is a one electron operator its expectation value is ψ Ŝ (i) ψ = ϕ i α Ŝ ϕ i α + φ i β Ŝ φ i β = 3 4 ( α + β ) = 3 4 α β To evaluate the two electron terms we will use the particle density matrix discussed earlier. ψ S(i) i Ŝ(! j) ψ = i> j 1 ' ' 1 S(1) i! Ŝ()Γ(1, / 1 ', ' )dτ (1,) where after the spin operators operate on coordinates 1& one changes the primed coordinates to 1&. It s convenient to use the operator equivalent

2 S(1) i Ŝ()! = 1 ( 4 1),first noted by Dirac, where P1 interchanges the SPI variables. So we have S(1) i Ŝ()Γ(1,! / 1 ', ' )dτ (1,) = 1 1 ' ' 1 Γ(1, / 1 ', ' )dτ (1,) ' ' 1 and from the definition of the two-particle density matrix Γ(1, / 1,)dτ (1,) = ( 1) To evaluate the remaining term we recognize that the two-particle density matrix for a single determinant wave function can be written as products of the one particle density matrix Γ( 1, / 1 ', ' ) = 1 ( γ (1/ 1' )γ ( / ' ) γ ( / 1 ' )γ (1/ ' )) where for the UHF wavefunction Γ(1, / 1,)dτ (1,) γ (1/ 1 ' ) = P α (1,1 ' )α(1)α(1 ' ) + P β (1,1 ' )β(1)β(1 ' ) with α P α (1,1 ' ) ϕ i (1)ϕ i (1 ' ) & P β (1,1 ' ) φ i (1)φ i (1 ' ) and with these definitions β γ (1/ 1 ' )γ ( / ' ) = ( P α (1,1 ' )α(1)α(1 ' ) + P β (1,1 ' )β(1)β(1 ' ))( P α (,' )α()α( ' ) + P β (, ' )β()β( ' )) Then operating with on the spin coordinates γ (1/ 1 ' )γ ( / ' ) = ( P α (1,1 ' )α()α(1 ' ) + P β (1,1 ' )β()β(1 ' ))( P α (,' )α(1)α( ' ) + P β (, ' )β(1)β( ' )) Changing the primes γ (1/ 1 ' )γ ( / ' )dτ (1,) = P α (1,1)α()α(1) + P β (1,1)β()β(1) 1 ', ' 1, Doing the spin integrations ( )( P α (,)α(1)α() + P β (,)β(1)

3 γ (1/ 1 ' )γ ( / ' )dτ (1,) = P α (1,1)P α (,) + P β (1,1)P β (,) dv (1,) 1 ', ' 1, and then the spatial integrations γ (1/ 1 ' )γ ( / ' )dτ (1,) = α + β 1 ', ' 1, In a similar way ( ) γ ( / 1 ' )γ (1/ ' ) = ( P α (,1 ' )α()α(1 ' ) + P β (,1 ' )β()β(1 ' ))( P α (1,' )α(1)α( ' ) + P β (1, ' )β(1)β( ' )) operating with on the spin coordinates γ ( / 1 ' )γ (1/ ' ) = ( P α (,1 ' )α(1)α(1 ' ) + P β (,1 ' )β(1)β(1 ' ))( P α (1,' )α()α( ' ) + P β (1, ' )β()β( ' )) Changing the primes ( )( P α (1,)α()α() + P β (1,)β()β( γ (,1')γ (1,')dτ (1,) = P α (,1)α(1)α(1) + P β (,1)β(1)β(1) 1',' 1, doing the spin integration ( ) γ (,1')γ (1,')dτ (1,) = P α (,1)P α (1,) + P α (,1)P β (1,) + P β (,1)P α (1,) + P β (,1)P β (1,) 1',' 1, and the spatial integrations dv and likewise ( P α (,1)P α (1,) ) α α dv (1,) = ϕ i ϕ j = ( P β (,1)P β (1,) ) j=1 β β dv (1,) = φ i φ j = whereas we have ( P α (,1)P β (1,) ) j=1 α β α α δ ij = α j=1 β β δ ij = β j=1 dv (1,) = ϕ i φ j = Δ ij j=1 α β ( ) j=1

4 and ( P β (,1)P α (1,) ) α β dv (1,) = ϕ i φ j = Δ ij j=1 α β ( ) j=1 and so α β γ (,1')γ (1,')dτ (1,) = α + β + (Δ ij ) 1',' 1, j=1 The two-electron contribution is then ψ S(i) i Ŝ(! j) ψ = 1 β α α + β Δij i> j j=1 ( ) After completing the square and adding the one electron term we have Ŝ = +1 α β α β Δ ( ij ) j=1 Check 1 ( 1) 4 Although we derived this formula assuming an UHF wavefunction it is general and we can test it using more restrictive functions. For example if we have a closed shell wavefunction with α = β = / and { ϕ i } α = φ { i } then Ŝ = +1 4 = 0 α β β then ( Δij ) = α j=1 = / as required. As another example suppose we have linear triplet methylene, CH ( 3 Σ g ) with the electronic configuration 1σ g σ g σ u 1π x 1 1π y 1 we have α = 5& β = 3 and α β ( Δ ij ) = β and j=1 Ŝ = x3 3 =, corresponding to S(S +1) with S = 1.

5 The UHF function for the ground state of Li is ψ (1,,3) = A1sα(1)1s'β()sα(3) UHF for the Li atom If we expand these orbitals in term of spherical gaussians 11 1s = C 1s,i χ i = C 1s,i 11 γ i π 3/4 e γ i r and use the VTZ basis the resulting UHF orbitals are collected in the table below. i γ i C 1s,i C 1s',i C s,i ote that the 1s &1s' are very similar but not identical as they would be in a restricted open shell wavefunction. The energy of this function is xxx and Ŝ = Using the general formula Ŝ = +1 α β With = 3, α = & β = 1 and α β Δ ( ij ) j=1 α β ( Δ ij ) j=1 = 1s 1s' + s 1s' = ( ) + ( ) = And so Ŝ = as required. Eliminate the spin contamination From the branching diagram discussion we know that 3 electrons in different spin orbitals give rise to two doublets and one quartet and in the Li UHF it must be the quartet that s causing the spin contamination. We can write the UHF function as

6 ψ (1,,3) = C D ψ D + C Q ψ Q where ψ D &ψ Q are eigenfunctions of Ŝ Ŝ ψ D = ψ D = 3 4 ψ D & Ŝ ψ Q = ψ Q = 15 4 ψ Q We project out the quartet component ( ) = 3C D ψ D Ŝ 15 4 ψ (1,,3) = Ŝ 15 4 C D ψ D + C Q ψ Q We can ignore 3C D and simply renormalize the result of Ŝ 15 4 ψ (1,,3). We write Ŝ = Ŝ Ŝ+ + Ŝz + Ŝz and note ( Ŝ z + Ŝ z ) ψ = ψ = 3 4 ψ then Ŝ Ŝ + ψ = Ŝ AŜ+ 1sα(1)1s'β()sα(3) = Ŝ A1sα(1)1s'β()sα(3) and Ŝ A1sα(1)1s'α()sα(3) = A1sβ(1)1s'α()sα(3) + A1sα(1)1s'β()sα(3) + A1sα(1)1s'α()sβ(3) and so Ŝ 15 4 ψ = ψ + A1sβ(1)1s'α()sα(3) + A1sα(1)1s'α()sβ(3) Since the 1s &1s' orbitals are virtually identical the last term on the right is ~0 and ψ D = ψ + A1sβ(1)1s'α()sα(3) = aψ + bψ ' where 5 ψ ' = A1sβ(1)1s'α()sα(3) and a = 5 & b = 1 5. The energy of this function is E D = ψ D Ĥ ψ D = aψ + bψ ' Ĥ aψ + bψ ' = E UHF + ab ψ Ĥ ψ ' where we note

7 ψ Ĥ ψ = ψ ' Ĥ ψ ' The off diagonal term is ψ Ĥ ψ ' = A1sα(1)1s'β()sα(3) Ĥ so A1sβ(1)1s'α()sα(3) = 1sα(1)1s'β() 1 r 1 (1 P 1 ) 1sβ(1)1s'α() E D = E UHF J 1s,1s' > E UHF So the uncontaminated function has a higher energy than the UFH as it should since ψ D is restricted relative to ψ UHF