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Winter 2013 Chem 356: Introductory Quantum Mechanics Chapter 9 – Many Electron Atoms ........................................................................................................... 121 Many-­‐Electron Atoms .......................................................................................................................... 121 A: Hartree-­‐Fock: Minimize the Energy of a Single Slater Determinant. ............................................... 126 Hartree-­‐Fock Iteration Scheme ........................................................................................................... 127 Chapter 9 – Many Electron Atoms Many-­‐Electron Atoms Hamiltonian for He Z = 2 Ĥ = −
!2 2 !2 2 !2 2
Ze2
Ze2
e2
∇1 −
∇2 −
∇α −
−
+
2me
2me
2mα
4πε 0 r1 4πε 0 r2 4πε 0 r1,2
= Tˆ + Vˆ Ne + Vˆ ee 1) Simplification: neglect kinetic energy of heavy nucleus. Assume nucleus is fixed in origin (exact: separate center of mass motion) Tˆ = Tˆ el + T N ; neglect TˆN , only consider electronic problem 2) Simplify units. We will use atomic units Atomic Units Mass me = 1 a.u. 9.1 ⋅ 10−31 kg Charge e = 1 a.u. 1.6 ⋅ 10−19 C Length a0 = 1a.u. 4πε 0 ! 2
a0 =
= 5.3⋅10−11 m 2
mee
Angular Momentum ! = 1 a.u. ! = 1.05⋅10−36 Js Eh = 1 a.u. e2
Eh =
= 4.36 ⋅10−18 J 4πε 0 a0
Energy (Hartree) In atomic units (a.u.) e = me = ! = a0 = 4πε 0 = 1 a.u. This simplifies equations, as all the constants are gone. Chapter 9 – Many Electron Atoms 121 Winter 2013 Chem 356: Introductory Quantum Mechanics In electronic structure theory, everyone uses a.u. and converts to ‘physical’ units as the need arises Now we wish to solve 1
1
2 2 1
Ĥ = − ∇12 − ∇ 22 − − + 2
2
r1 r2 r12
! !
! !
Ĥψ ( r1 , r2 ) = Eψ ( r1 , r2 ) (For the moment we neglect spin) Let us be more general and write down the Hamiltonian for an N -­‐electron atom N
N
1
Z N
1
! !
Ĥ = ∑ − ∇ 2i − ∑ ! + ∑ ∑ where rij = ri − rj 2
i=1
i=1 ri
i=1 j<i rij
! !
!
The wavefunction ψ ( r1 , r2 ,....rN ) has 3 N coordinates This is a very complicated problem. However, if we could neglect the V el , 1
terms, the problem would rij
be simple. Ĥ = ĥ(1) + h(2) + .....ĥ( N ) 1
Z
ĥ(i) = − ∇ 2i − ! 2
ri
Try product wavefunction !
!
!
ψ = ϕ a ( r1 )ϕ b ( r2 )....ϕ z ( rN ) !
!
Solve h( r1 )ϕ a (r1 ) = ε aϕ a ( r1 ) one-­‐electron orbitals, orbital energies Moreover, each problem for each electron is the same: only the name of the variable is changed. This would simply be Hydrogen like solutions. ! !
!
!
!
ψ ( r1....rM ) = ϕ a ( r1 )ϕ b ( r2 )....ϕ z ( rM ) E = ε a + ε b + ε c + ....ε z Trivial! Question: what would the lowest energy state be? Chapter 9 – Many Electron Atoms 122 Winter 2013 Chem 356: Introductory Quantum Mechanics Answer: It is tempting to fill up these orbital levels with 2 electrons, of either spin. That is not the lowest energy solution to Ĥψ = Eψ ! What is? Put all electrons in the same lowest energy orbital E = ε1 + ε1 + ε1....Nε1 = Nε1 We all know this is physically incorrect, and need to invoke the Pauli principle. Let us try to formulate it, and understand where it originates. Let us go back to basics: If an operator  commutes with the Hamiltonian ⎡ Hˆ , Aˆ ⎤ = 0 and ψ is an ⎣
eigenstate of Ĥ then: ( )
⎦
( )
ˆ ˆ ψ = AE
ˆ ψ = E Aˆψ Hˆ Aˆψ = AH
Hence Âψ is also an eigenstate with the same energy. Special types of operators that commute with the Hamiltonian are the permutations of coordinates. Eg. ⎡⎛
⎤
1⎞
P̂12 ⎢⎜ h(1) + h(2) + ⎟ ψ (1,2) ⎥ r12 ⎠
⎢⎣⎝
⎥⎦
= ⎜ h(2) + h(1) +
ˆ ψ (1, 2) = HP
12
Or ˆ ˆ This is true for any permutation, true for many electron Pˆ12 Hˆ = HP
12
⎛
⎝
1 ⎞
⎟ψ (2,1) r21 ⎠
systems. Special permutations are transpositions Tij , which interchange coordinates of particles i and j. These operators are Hermitian, and TijTij = 1, interchanging twice does nothing. Ĥ and Tij have a set of common eigenfunctions, since they commute Tˆijψ (1,...i, j , M ) = λψ Chapter 9 – Many Electron Atoms 123 Winter 2013 Chem 356: Introductory Quantum Mechanics Tˆij (Tˆijψ ) = Tˆij λψ = λ 2ψ = ψ Hence λ 2 = 1 → λ = 1,
λ = −1 From these considerations (plus a bit more solid reasoning) we can deduce there are 2 types of particles: 1) Bosons λ = 1 : The wave function is symmetric under interchange of coordinates (any transposition) ψ (1,...i,... j,...N ) = ψ (1,... j,...i,...N ) 2) Fermions λ = −1 : The wavefunction changes sign under a transposition All spin Tijψ = ψ (1,... j,...i,...N ) = −ψ (1,...i,... j,...N ) 1
particles, like protons and electrons are Fermions. I do not have a simple explanation for this 2
(spin-­‐statistics theorem in quantum field theory). All permutations can be written as a product of transpositions p = TijTklTmj .... either odd or even. Hence action of all Tij determines action of P for any permutation. For electrons wavefunction is antisymmetric under interchange of 2 electron coordinates (This includes the spin degree of freedom) If we return to our orbital product wavefunction ψ = ϕa (1)ϕb (2)...ϕ z ( N ) one can design a clever trick to make ‘antisymmetric’ wavefunctions Put the orbitals in a determinant: ϕ a (1) ϕ b (1)
ϕ a (2) ϕ b (2)
= ϕ a (1)ϕ b (2) − ϕ b (1)ϕ a (2) a b c Or 1 ϕa (1) ϕb (1) ϕc (1) 2 ϕ a (2) ϕb (2) ϕc (2) 3 ϕa (3) ϕb (3) ϕc (3) Interchanging the coordinates of 2 electrons = interchange 2 rows in a determinant. → automatic minus sign. Also: interchange orbitals a and b → introduces minus sign We can simplify notation and write ψ = ϕ a (1)ϕ b (2)ϕ c (3) indicating the diagonal terms only, Then ϕ a (2)ϕ b (1)ϕ c (3) = − ϕ a (1)ϕ b (2)ϕ c (3) Chapter 9 – Many Electron Atoms 124 Winter 2013 Chem 356: Introductory Quantum Mechanics = − ϕ b (1)ϕ a (2)ϕ c (3) Slater determinants easily fold in antisymmetry. If we consider ϕa (1) as a spin-­‐orbital, then in the determinant we flip spin-­‐orbitals (i.e. labels a,b, etc.) by permutations, or we should flip space and spin coordinates upon permutations. Finally : ( h(1) + h(2) + ...h(N )) (ϕ (1)ϕ (2)...ϕ (N )) = (ε a + ε b + ...ε z ) ϕa (1)...ϕz ( N ) If h(1)ϕi (1) = ε iϕi (1) a
b
z
Now we have the Pauli principle too: If two orbitals in a determinant are equal then ψ = 0 ϕ a (1)ϕ b (2)...ϕ a (k)... = − φa (1)....φa (k)... = 0 (2 equal columns a) For this reason we put one electron in each spin-­‐orbital A simple Slater determinant is the appropriate antisymmetric generalization of a product of orbitals. These would be eigenfunctions if Hˆ =
∑ hˆ(i) , a sum of one-­‐electron Hamiltonians. It is not, we have to i
1
worry about Vˆ el = ∑ rij . 2 i≠ j
To deal with Vˆ el one applies the variational procedure in 2 steps. a) Minimize the energy of a single determinant wavefunction. This is the Hartree-­‐Fock self-­‐
consistent field procedure. It leads to an optimal set of orbitals (one-­‐electron wavefunctions) and orbital energies. b) Apply linear variational principle and minimize energy of ∑c Φ
λ
λ
λ
in which Φ λ are Slater determinants. This is called Configuration Interaction. It is important to go behind Hartree-­‐
Fock (single determinant) to get accurate results. Moreover for (nearly) degenerate states one has to use CI to get splitting (see atomic term symbols later on.). There are many other approaches to improve results beyond the basic Hartree-­‐Fock result. Chapter 9 – Many Electron Atoms 125 Winter 2013 Chem 356: Introductory Quantum Mechanics A: Hartree-­‐Fock: Minimize the Energy of a Single Slater Determinant. If we label orbitals in the determinant aσ , to indicate spatial index a , and spin index σ = α , β , then the following integrals are relevant: !
! !
! !
haσ = ∫ dr1ϕ aσ *( r1 )h( r1 )ϕ aσ ( r1 )dr1 σ | σ The J ab integral describes the Coulomb interaction between ρ a (r1 ) = ϕ a (r1 ) And ρ b (r2 ) = ϕ b (r2 ) It is independent of spin. σσ '
The kab
integral arises due to the antisymmetry in Φ , This integral is the Coulomb interaction σσ '
J ab = J ab
=∫
σσ '
kab
=
∫
1
! !
!
dr1 dr2ϕ σa *( r1 )ϕ bσ ' (r2 ) * ϕ σa (r1 )ϕ bσ ' (r2 ) σ | σ σ ' | σ ' r12
1
! !
dr1 dr2ϕ σa (r1 ) *ϕ bσ ' (r2 ) * ϕ bσ ' (r1 )ϕ σa (r2 ) σ | σ ' σ ' | σ r12
∫∫
2
2
between distributions ρab (r1 ) = ϕa (r1 ) * ϕb (r2 ) And ρba (r2 ) = ϕa * (r2 )ϕb (r2 ) It is called exchange interaction. It only arises for spin-­‐orbitals that have equal spin. In the energy it enters with a minus sign EHF = ∑ haσ +
aσ
1
1
σσ '
σσ
J ab
− ∑ kab
∑
2 aσ
2 a ,b,σ
bσ '
If one has a closed shell with equal α and β orbitals, the expression reduces to EHF = 2∑ ha + 2∑ (2 J ab − kab ) a
a ,b
The exchange interaction is the reason that if orbitals are degenerate, it is preferable to align spins parallel. It maximizes the number of “ -­‐ k ab ” like-­‐spin contributions. See Hund’s rules for many-­‐electron atoms. We now have an expression (did not derive) for the Hartree-­‐Fock energy. This expression can be !
minimized to find an optimal set of orbitals. If one expands the orbitals in a basis set X i ( r ) , one obtains a matrix eigenvalue equation. ∑F c
ij
j
jλ
= ε λ ciλ (orthonomal basis) The matrix F depends on the occupied orbitals. For this reason the equations have to be solved “self-­‐
consistently” Chapter 9 – Many Electron Atoms 126 Winter 2013 Chem 356: Introductory Quantum Mechanics Hartree-­‐Fock Iteration Scheme Guess starting orbitals ϕ a( old ) Construct F ⎡ϕaold ⎤ Diagonalize Fc = cε pick up the lowest energy orbitals → F ⎡⎣ϕ anew ⎤⎦ Check if F ⎡ϕanew ⎤ = F ⎡ϕaold ⎤ If not: update orbitals to get new set of improved orbitals If F ⎡ϕanew ⎤ ≈ F ⎡ϕaold ⎤ convergence reached ⎣
⎦
!
⎣
⎣
⎦
⎦
⎣
⎣
⎦
⎦
This is called the self consistent field (SCF) iteration scheme. It is a powerful numerical scheme to minimize the energy of a single determinant. Some Notes: -­‐ The orbitals in HF are not unique. The only thing that matters is the determinant. In a determinant I can make linear combinations of rows and columns. Same is true for orbitals -­‐
-­‐
-­‐
-­‐
-­‐
-­‐
1
2
(ϕ
a
The energy E HF is not equal to + ϕb )
∑ε
a
1
2
(ϕ
a
− ϕ b ) = − ϕ a ⋅ ϕ b One needs to use the full expression, given before. a
HF is surprisingly accurate in many cases. For molecules it often yields ~ 99% of the exact energy. For purposes in chemistry this is not accurate enough. 1% of energy > 100 kcal mol-­‐1 typically! HF SCF equations are tricky. It is easy to get trapped in a higher energy solution. Pitfall of non-­‐
linear optimization. In practice one uses a basis set. The accuracy of a HF calculation is limited by the size of the basis set. Hartree-­‐Fock results can be qualitatively wrong when you expect degeneracies between ground and excited states; or when certain orbitals in a degenerate manifold are occupied, and others not in the HF determinant. Usually one needs to go beyond HF to get qualitatively correct results then. One always needs to go beyond HF to get quantitatively accurate results Chapter 9 – Many Electron Atoms 127