Hartree-Fock Theory
The orbitalar approach Slater spin Mulliken Combination of Slater determinants Y(1,….N) = Pi fi(i) Pauli principle Solutions for S2
MO approach A MO is a wavefunction associated with a single electron. The use of the term "orbital" was first used by Mulliken in 1925. We are looking for wavefunctions for a system that contains several particules. We assume that H can be written as a sum of one-particle Hi operator acting on one particle each. H = Si Hi Y(1,….N) = Pi fi(i) E = Si Ei Robert Sanderson Mulliken 1996-1986 Nobel 1966
MO approach Friedrich Hund 1896-1997 Charles Alfred Coulson 1910-1974 MO theory was developed, in the years after valence bond theory (1927) had been established, primarily through the efforts of Friedrich Hund, Robert Mulliken, John C. Slater, and John Lennard-Jones. The word orbital was introduced by Mulliken in 1932. According to Hückel, the first quantitative use of MO theory was the 1929 paper of Lennard-Jones. The first accurate calculation of a molecular orbital wavefunction was that made by Charles Coulson in 1938 on the hydrogen molecule. By 1950, MO were completely defined as eigenfunctions of the self-consistent field
Generalizing the LCAO approach: A linear combination of atomic orbitals or LCAO Linus Carl Pauling 1901-1994 Nobel 1962 Sir John Lennard-Jones 1894-1954 It was introduced in 1929 by Lennard-Jones with the description of bonding in the diatomic molecules of the first main row of the periodic table, but had been used earlier by Pauling for H2+.
Schrödinger equation for LCAO HY = EY This is the set of linear equation solved for | Hij-E S ij |= 0 Secular determinant In an ab-initio calculation, all the Hij and Sij Integrals are rigorously calculated. To do that, one need defining a set of AOs (basis set) and the geometry. There is no parameterization. A subjective choice concerns that the basis set (always incomplete). The best calculation (SCF level or after IC) is that providing the lowest energy. Erwin Rudolf Josef Alexander Schrödinger Austrian 1887 –1961
Particles are electrons Born-Oppenheimer approximation 1927 } Operator acting on e Julius Robert Oppenheimer Berkeley- Los alamos 1904 –1967 Max Born German (1882-1970) H = [Te +VN-e + Vee] + TN + VN-N He = [Te +VN-e + Vee] Y = PY(e) PY(N) VN-e also acts on N; when N is considered as fixed: VN-e then becomes an operator acting on e. Should be a coupling term e N Separation of Y(e) and Y(N)
Born-Oppenheimer approximation The forces on both electrons and nuclei due to their electric charge are of the same order of magnitude; since the nuclei are so much more massive than the electrons, they must accordingly have much smaller velocities. . Nucleon mass / Electron mass = 1835 For the same kinetic energy, when an electron travels meter a nucleus of hydrogen travels 2.3 cm, that of carbon 6.7 millimeters and that of gold 1.7 millimeters. It is then considered as the movement of nuclei and electrons are independent, that electrons adapt instantaneously to the movement of nuclei. It is said that the electrons follow adiabatically the motion of the nuclei.
The adiabatic theorem 1928 A physical system remains in its instantaneous eigenstate if a given perturbation is acting on it slowly enough and if there is a gap between the eigenvalue and the rest of the Hamiltonian's spectrum. Vladimir Aleksandrovich Fock russian 1898–1974 Max Born German (1882-1970) This adiabatic principle is crucial because it allows us to separate the nuclear and electronic motion, leaving a residual electron-phonon interaction. From this point on it is assumed that the electrons respond instantaneously to the nuclear motion and always occupy the ground-state of that nuclear configuration.
Vibrations, nuclear motion Developing U around the equilibrium position (r = req) gives U(Dr) = U° + U’’(req)Dreq2: The angular frequency is w = (req/m) ½ depending on the reduced mass m=m1m2(m1+m2) 3N-6 degree of freedom + 3 translations and 3 rotations for the general case.
The surface energy obtained under the Born-Oppenheimer approximation: adiabatic surfaces. These are surfaces that can be achieved step by step from calculations developed. They may be significant changes in the characters of orbital obtained. According to an internuclear distance ionic or covalent change significantly. In the case of rapidly changing systems (collision dissociations from excited states), the nature of the orbital can not change and the Born-Oppenheimer approximation is no longer valid. The state surface is therefore a surface potential where the same dominant character is always found. Such surface, called diabatic surface, is in some region the ground state and for some region an excited state. Then, there is a jump from an adiabatic surface to another! The diabatic surfaces are inherently difficult to obtain since calculations lead to the system that is most stable.
U(d) dNa-Cl PES: Potential Energy Surfaces. Covalent Na+ + Cl- Diabatic Adiabatic Na + Cl Diabatic Ionic dNa-Cl An ionic curve “naturally” correlates with ions whereas a covalent one dissociates to generate radicals. This leads to an avoided crossing.
Finding diabatic surfaces The diabatic surface should in principle be obtained by calculating the coupling terms <f1|TN|f2> allowing possible jumps from one adiabatic surface to another. This is uneasy since the basis set gives an avoided crossing without the coupling. The usual procedure is generally the reverse: impose “natural orbitals” and minimize the coupling terms to allow the surfaces crossing.
Diabatic- Adiabatic surfaces. Process velocity slow fast Born-Oppenheimer valid Not valid electrons adaptation to the nuclei motion Adapt (follow it) Do not (delay to follow nuclei) Natural MO character May switch Remains constant
Approximation of independent particles Douglas Rayner Hartree English 1897 1958 Vladimir Aleksandrovich Fock russian 1898–1974 HHartree= Si Fi The hamiltonian, HHartree , is the sum of Fock operators only operating on a single electron i. The wave function is the determinant of orbitals, f(i) to satisfy the Pauli principle. Fi f(i)=εi f(i) 1930 EHartree= Si εi
Approximation of independent particles Hartree Fock 1927 Fi = Ti + Sk Zik/dik+ Sj Rij Each one electron operator is the sum of one electron terms + bielectronic repulsions Rij is the average repulsion of the electron j upon i. The self consistency consists in iterating up to convergence to find agreement between the postulated repulsion and that calculated from the electron density. This ignores the real position of j vs. i at any given time.
Self-consistency Given a set of orbitals Y°i, we calculate the electronic distribution of j and its repulsion with i. This allows expressing and solving the equation to find new Y1i allowing to recalculate Rij. The process is iterated up to convergence. Since we get closer to a real solution, the energy decreases. Fi = Ti + Sk Zik/dik+ Sj Rij
Jij, Coulombic integral for 2 e Assuming Y= Y1Y2 Jij, Coulombic integral for 2 e Let consider 2 electrons, one in orbital Y1, the other in orbital Y2, and calculate the repulsion <1/r12>. Assuming Y= Y1Y2 This may be written Jij = = < Y1Y2IY1Y2> = (Y1Y1IY2Y2) Dirac notation used in physics Notation for Quantum chemists
Jij, Coulombic integral for 2 e Assuming Y= Y1Y2 Jij, Coulombic integral for 2 e Jij = < Y1Y2IY1Y2> = (Y1Y1IY2Y2) means the product of two electronic density Coulombic integral. This integral is positive (it is a repulsion). It is large when dij is small. When Y1 are developed on atomic orbitals f1, bilectronic integrals appear involving 4 AOs (pqIrs)
Jij, Coulombic integral involved in two electron pairs Assuming Y= Y1Y2 Jij, Coulombic integral involved in two electron pairs electron 1: Y1 or Y1 interacting with electron 2: Y2 or Y2 When electrons 1 have different spins (or electrons 2 have different spins) the integral = 0. <aIb>=dij. Only 4 terms are 0 and equal to Jij. The total repulsion between two electron pairs is equal to 4Jij.
Particles are electrons! Pauli Principle electrons are indistinguishable: |Y(1,2,...)|2 does not depend on the ordering of particles 1,2...: | Y (1,2,...)|2 = | Y (2,1,...)|2 Wolfgang Ernst Pauli Austrian 1900 1950 Thus either Y (1,2,...)= Y (2,1,...) S bosons or Y (1,2,...)= -Y (2,1,...) A fermions The Pauli principle states that electrons are fermions.
Particles are fermions! Pauli Principle The antisymmetric function is: YA = Y (1,2,3,...)-Y (2,1,3...)-Y (3,2,1,...)+Y (2,3,1,...)+... which is the determinant Such expression does not allow two electrons to be in the same state the determinant is nil when two lines (or columns) are equal; "No two electrons can have the same set of quantum numbers". One electron per spinorbital; two electrons per orbital. “Exclusion principle”
Two-particle case The simplest way to approximate the wave function of a many-particle system is to take the product of properly chosen wave functions of the individual particles. For the two-particle case, we have This expression is a Hartree product. This function is not antisymmetric. An antisymmetric wave function can be mathematically described as follows: Therefore the Hartree product does not satisfy the Pauli principle. This problem can be overcome by taking a linear combination of both Hartree products where the coefficient is the normalization factor. This wave function is antisymmetric and no longer distinguishes between fermions. Moreover, it also goes to zero if any two wave functions or two fermions are the same. This is equivalent to satisfying the Pauli exclusion principle.
Generalization: Slater determinant . The expression can be generalized to any number of fermions by writing it as a determinant. John Clarke Slater 1900-1976
Kij, Exchange integral for 2 e Obeying Pauli principleY= 1/√ IY1Y2I Slater determinant Kij, Exchange integral for 2 e Let consider 2 electrons, one in orbital Y1, the other in orbital Y2, and calculate the repulsion <1/r12>. Assuming Y= 1/√ IY1Y2I Kij is a direct consequence of the Pauli principle Kij = = < Y1Y2IY2Y1A> = (Y1Y2IY1Y2) Dirac notation used in physics Notation for Quantum chemists
Kij, Exchange integral two electron pairs Obeying Pauli principleY= 1/√ IY1Y2I Slater determinant Kij, Exchange integral It is also a positive integral (-K12 is negative and corresponds to a decrease of repulsion overestimated when exchange is ignored). two electron pairs
Interactions of 2 electrons Obeying Pauli principleY= 1/√ IY1Y2I Slater determinant Interactions of 2 electrons
Unpaired electrons: sgsu Singlet and triplet states bielectronic terms Jij+Kij │YiYj│+ │YiYj│ Jij │YiYj│ │YiYj│ Kij Jij-Kij │YiYj│ │YiYj│ │YiYj│- │YiYj│
Obeying Pauli principleY= 1/√ IY1Y2I Slater determinant Jj and Kj operators We need to express the ij bielectronic repulsion as one electron operator Rj acting on Yj and representing the repulsion with an average distribution of j. Rj is the sum of 2 operators Jj and Kj.
Slater rules Hartree-Fock Operators are 1 or 2 electron operators ( i or ij). Developing the polyelectronic wavefunction leads only to the following contributions <c1Ic1> <c2Ic2> ….. <ciIOIci> ….. <cNIcN> or <c1Ic1> <c2Ic2> ….. <cicjIOIcicj> ….. <cNIcN> Only the terms <c1Ic1>=1 remain, the others being nil: <c1Ic1>=0
Slater rules Non zero terms arise from 3 possibilities: Dl = DR <DlIOIDR> = S [<kikjIOIlilj> - <kikjIOIlilj>] Dl differ from DR by one orbital i <DlIOIDR> = <kiIOIlj> <DlIOIDR> = S [<kikjIOIlilj> - <kikjIOIlilj>] Dl differ from DR by two orbitals i and j <DlIOIDR> = <kikjIOIlilj> - <kikjIOIlilj> N i<j N ij
Electronic energy i is the index for an orbital: EE = Sεi εi = <fiIhIfi> + S (2Jij-Kij) occ occ j Each i electron interacts with 2 j electrons with opposite spins Jij-Kij with that of the same spin for and Jij for that with different spin.
Electronic energy – total energy The sum of the energies of all the electrons contains the ij repulsion twice: EE = Sεi εi = <fiIhIfi> + S (2Jij-Kij) occ occ i j occ occ = S [hii occ EE = S[hii occ ] ] +2 S Repij + S repij i i j ESCF is defined as the energy of YSCF applying Slater rules Pauli principle occ occ occ ESCF = <YSCFIHIYSCF> = Shii + S S(2Jij-Kij) i j<i i occ ESCF = EE - Repij = Shii + Repij ETOTAL = ESCF +EN i
Restricted Hartree-Fock Closed-shell system: an MO is doubly occupied or vacant. The spatial functions are independent from the spin. Assuming k and l with spin a, the expression of a Fock matrix-element, Fkl, is occ occ Fkl = hkl + S [(klIjj)-(klIjj)] - S (klIjj) J of spin a J of spin b Fkl = h + 2J – K and εi = hii + S (2Jij-Kij) ESCF = 2S εi – SS(2Jij-Kij) There is no exchange term for electrons with different spin (<aIb>=0) Altogether there are 2 J for one K. occ j occ occ occ i j<i i
Restricted Hartree-Fock Why the HOMO-LUMO electron gap is overestimated? εi = hii + S (2Jij-Kij) occ j For a virtual i orbital the repulsion concerns all the electrons i For an occupied i orbital the repulsion concerns all the electrons minus 1 (itself) i The repulsion is larger for an unoccupied level than for an occupied one!
εi = - IP Koopmans theorem εi = hii + S (2Jij-Kij) εi IP = 2S εj - occ j This theorem applies for a “vertical transition”: sudden without time for electronic reorganization. The orbitals do not relax! Tjalling C. Koopmans Dutch Nobel Prize in Economic Sciences in 1975 This theorem is obvious for Hückel-type approaches: The AOs are fixed. The total energy is the sum of the orbital energies: This theorem is obvious for Hückel-type approaches: The AOs are fixed. The total energy is the sum of the orbital energies: It is less obvious in HF since ESCF S εi j occi j occ j occi j occi IP = 2S εj - 2S εj = 2S εj - 2S εj εi - j j j (Physica, 1, 104 (1933) For DFT, the Janak theorem is a generalization of Koopman’s theorem. J. F. Janak, Phys. Rev. B 18, 7165; J.P. Perdew, R. G. Parr, M. Levy et J. L. Balduz Jr., Phys. Rev. Lett. 23 (1982) 1691.
Koopmans theorem εk = - IP Tjalling C. Koopmans
What is wrong with Koopmans theorem? The AOs are not “fixed” but relax (“breath”); the Slater exponents depend on the bielectronic repulsion. The subtraction in the previous slide suppose no variation. The relaxation is important for polarisable systems (metals, the work function is generally weak). Calculating with AOs for the neutral species underestimates the energy for the ionized species. IP is overestimated by Koopmans theorem. How to proceed more rigorously? Calculate the IP as the difference of the total energy for two states.
Cu+ (Koop.) (exp.) Na 1s 658.4 662 79.4 81.2 2s 82.3 81.3 5.2 6.0 2p 71.83 69.6 2.8 3.66 3s 10.65 9.6 .378 .372 3p 7.28 6.1 3d 1.61 0.79
Open-Shell – spin contamination Let assume that here is one a electron more than b. For them Fock matrix elements, there is one more exchange term for the a electrons than for the b’s. The matrix elements are not the same and solving the secular determinants for a and b independently leads to different solutions. fi fi . The UHF (unrestricted Hartree-Fock method) solves the secular equation for a and b independently and obtains different spatial functions for a and b. This does not allow to have eigenfunctions of the spin operators. k k j j i i
ROHF The same spatial function is taken whatever the spin is. Robert K. Nesbet This allows eigenfunctions of spin operators. This allows separating spatial and spin functions. It is not consistent with variational principle (that does better without this constraint). 2e occ 1e occ F = h + S [( Ijj)-( jI j)] - S ( Ijj)-1/2 ( jI j) j j
Spin operator The spinorbitals should be solution of SZ and S2. This is not the case for UHF.
Correspondence principle 1913/1920 For every physical quantity one can define an operator. The definition uses formulae from classical physics replacing quantities involved by the corresponding operators For an angular momentum Niels Henrik David Bohr Danish 1885-1962 Classical expression Quantum expression lZ= xpy-ypx
Angular momentum In quantum mechanics angular momentum is defined by an operator where r and p are the position and momentum operators respectively. In particular, for a single particle with no electric charge and no spin, the angular momentum operator can be written in the position basis as This orbital angular momentum operator is the most commonly encountered form of the angular momentum operator, though not the only one. It satisfies the following canonical commutation relations: , It follows, for example,
Relations in Angular momentum L2 is the norm and Lx, Ly, Lz are the projections. [Jx, Jy]= i Jz [Jx, J2]= 0 [Jy, Jz]= i Jx eigenfunctions of J2 are j(j+1) [Jz, Jx]= i Jy eigenfunctions of Jz are m J2Ij,m> = j(j+1) Ij,m> and Jz Ij,m> = m Ij,m> In atomic units, h =1
Introducing J+ = Jx+iJy and J- = Jx-iJy J2 = Jx2 + Jy2 + Jz2 = Jz2 + ½ (J+J- + J-J+ ) J+J- and J-J+ operators have |j,m> as eigenfunction: J+J- |j,m> = (j+m)(j-m+1) |j,m> J-J+ |j,m> = (j+m+1)(j-m) |j,m>
spin For an electron: s =1/2 → S2=s (s +1) =3/4 and ms=±1/2. A spin vector is either a |s,ms> = |1/2,1/2> or b: |s,ms> = |1/2,-1/2>
spin For the projection Sz=Ms = Ss ms For the norm S2 = S12+S22+2S1S2 For several electrons: the total spin is the vector sum of the individual spins. For the projection Sz=Ms = Ss ms For the norm S2 = S12+S22+2S1S2 S2=[S1z2+S1z+S1-S1+]+[S2z2+S2z+S2-S2+]+[(S1+S2- ++S1-S2+)+2S1zS2z] For 2 electrons, there are 4 spin functions: a(1) a(2), b(1) b(2), a(1) b(2) and a(2) b(1) that are solutions of Sz. Are they solution of S2?
a(1)a(2) is solution of S2 S2 a(1) a(2) = 2 a(1) a(2) a(1) (Sz2+Sz+S-S+) a(2) [1/2]2 + 1/2 + 0 0.75 a(2) (Sz2+Sz+S-S+) a(1) 2Sza(1)Sza(2) 2 ½ ½ 0.5 S-a(1)S+a(2) S+a(1)S-a(2) total 2 S2 a(1) a(2) = 2 a(1) a(2)
a(1)b(2) is not a solution of S2 a(1) (Sz2+Sz+S-S+) b(2) [1/2]2 - 1/2 + 1 0.75 b(2) (Sz2+Sz+S-S+) a(1) [1/2]2 + 1/2 + 0 2Sza(1)Szb(2) 2 ½ (-½) -.5 S-a(1)S+b(2) b(1) a(2) S+a(1)S-b(2) total 2 S2 a(1) b(2) = a(1) b(2) + b(1) a(2)
a(1)b(2)± b (1) a (2) are solutions of S2 S2 a(1) b(2) = a(1) b(2) + b(1) a(2) S2 b(1) a(2) = b(1) a(2) + a(1) b(2) S2 (a(1) b(2) + b(1) a(2)) = 2 (a(1) b(2) + b(1) a(2)) S2 (a(1) b(2) - b(1) a(2)) = 0 (a(1) b(2) - b(1) a(2)) The spinorbitals must then be a product of such expressions by a constant (a spatial term which is a constant for the spin operator). We should be able to separate the spatial and the spin contributions.
Use of indentation operators upon vector sums S2=Sz2+Sz+S-S+ S-S+ |1,1> = 0 S-S+ |1,0> = (1+0+1)(1-0) |1,0> = 2|1,0> S-S+ |1,-1> = (1-1+1)(1+1) |1,-1> = 2 |1,-1> Singlet S2|0,0> = [0+0+0] =0 |0,0> Triplet S2|1,1> = [1+1+0] |1,1> Triplet S2|1,0> = [0+0+2] |1,0> Triplet S2|1,-1> = [1-1+2] |1,-1>
S2 as an operator on determinant S2(D)= Pab(D) + [(na-nb)2 +2na +2nb ] D Pab is an operator exchanging the spins :a b na and nb are the # of electrons of each spin Application: S2(IaaI) = IaaI+IaaI = -IaaI+IaaI = 0 S2(IabI) = IabI+IabI S2(IabI-IbaI) = 2 (IabI+IabI)= 2 (IabI-IbaI) S2(IabI+IbaI) = IabI+IbaI+IabI+IbaI = -IbaI-IabI+IabI+IabI = 0
A single Slater determinant is not necessarily an eigenfunction of S2 Separation of spatial and spin functions IabI= a(1)a(1) b(2)b(2) - b(1)b(1) a(2)a(2) = a(1)b(2) a(1)b(2) - b(1)a(2) a(2)b(1) is not a eigenfunction of S2 if ab
A combination of Slater determinant then may be an eigenfunction of S2 IabI± IbaI = a(1)b(2) a(1)b(2) - b(1)a(2) a(2)b(1) ± b(1)a(2) a(1)b(2) - a(1)b(2) a(2)b(1) + IabI+ IbaI = [a(1)b(2) + b(1)a(2)] (a(1)b(2) - a(2)b(1)) IabI- IbaI = [a(1)b(2) - b(1)a(2)] (a(1)b(2) + a(2)b(1)) are a eigenfunction of S2 The separation of spatial and spin functions is possible If the spatial function (a or b) is associated with opposite spins
UHF: Variational solutions are not eigenfunctions of S2 Iab’I+ Iba’I = a(1)b’(2) a(1)b(2) - b’ (1)a (2) a(2)b(1) + b(1)a’(2) a(1)b(2) – a’(1)b(2) a(2)b(1) =[a(1)b’(2) + b(1)a’(2)](a(1)b(2) - [a’(1)b(2) + b’(1)a(2)] a(2)b(1)) Not equal The separation of spatial and spin functions is not possible If 2 spatial functions (a a’; b b’) are associated with opposite spins