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Chapter 9 Many-Electron Atoms Chapter 9: Slide 1 Outline • The Hamiltonian for Multielectron Atoms • The Hartree Method: Helium • Koopman’s Theorem • Extension to Multielectron Atoms • Antisymmetrized Wavefunctions: Slater Determinants • The Hartree-Fock Method • Hartree-Fock Orbital Energies for Argon • Electron Correlation Chapter 9: Slide 2 The Hamiltonian for Multielectron Atoms Helium 2 2 2 2 2e 2 2e 2 e2 SI Units: H 1 (r1 ) 2 (r2 ) Z=2 2m 2m 40 r1 40 r2 40 r12 1 2 1 2 2 1 Atomic Units: H 1 (r1 ) 2 (r2 ) 2 2 2 r1 r2 r12 Multielectron Atoms N 1 1 N 2 N Z 1 H i r i r 2 i 1 i 1 i i 1 j ij Elect Elect- Elect- KE Nuc Elect PE PE N 1 1 1 1 1 1 1 1 i r r r i 1 j ij r23 r24 r34 r35 12 13 Chapter 9: Slide 3 Atomic Orbitals In performing quantum mechanical calculations on multielectron atoms, it is usually assumed that each electron is in an atomic orbital, , which can be described as a Linear Combination of Hydrogen-like orbitals, which are called Slater Type Orbitals (STOs). These STOs are usually denoted as i (although the text uses the notation, bi) . Thus: ci i The goal of quantum mechanical calculations is to find the values of the ci which minimize the energy (via the Variational Principle). These STOs are also used to characterize the Molecular Orbitals occupied by electrons in molecules. We will discuss these STOs in significantly greater detail in Chapter 11, when we describe quantum mechanical calculations on polyatomic molecules. Chapter 9: Slide 4 Outline • The Hamiltonian for Multielectron Atoms • The Hartree Method: Helium • Koopman’s Theorem • Extension to Multielectron Atoms • Antisymmetrized Wavefunctions: Slater Determinants • The Hartree-Fock Method • Hartree-Fock Orbital Energies for Argon • Electron Correlation Chapter 9: Slide 5 The Hartree Method: Helium Hartree first developed the theory, but did not consider that electron wavefunctions must be antisymmetric with respect to exchange. Fock then extended the theory to include antisymmetric wavefunctions. We will proceed as follows: 1. Outline Hartree method as applied to Helium 2. Show the results for atoms with >2 electrons 3. Discuss antisymmetric wavefunctions for multielectron atoms (Slater determinants) 4. Show how the Hartree equations are modified to get the the “Hartree-Fock” equations. Chapter 9: Slide 6 Basic Assumption Each electron is in an orbital, i (e.g. a sum of STOs). The total “variational” wavefunction is the product of one electron wavefunctions: (r1 , r2 ) 1 (r1 ) 2 (r2 ) Procedure init “Guess” initial values the individual atomic orbitals: (r1 ) and 2 (r2 ) init 1 (This would be an initial set of coefficients in the linear combination of STOs) Let’s first look at electron #1. Assume that it’s interaction with the second electron (or with electrons #2, #3, #4, ... in multielectron atoms) is with the average “smeared” out electron density of the second electron. SI Units Atomic Units 2 2 eff e (r2 ) init (r2 ) init V1 (r1 ) e eff 2 or V1 (r1 ) 2 dr2 dr2 40 r12 r12 1 init 2init (2) 2 ( 2) r12 Chapter 9: Slide 7 It can be shown (using the Variational Principle and a significant amount of algebra) that the “effective” Schrödinger equation for electron #1 is: 1 2 2 1 init H 1eff 1 11 where H 1eff 1 V1eff V1eff 2init (2) 2 ( 2) 2 r1 r12 elect elect- “Effective” KE Nuc elect-elect PE PE This equation can be solved exactly to get a new estimate for the function, 1new (e.g. a new set of coefficients of the STOs). There is an analogous equation for 2: 1 2 1 init H 2 2 22 eff where H 2 2 eff 2 V2eff V2eff 1init (1) 1 (1) 2 r2 r12 This equation can be solved exactly to get a new estimate for the function, 2new (e.g. a new set of coefficients of the STOs). Chapter 9: Slide 8 A Problem of Consistency init We used initial guesses for the atomic orbitals, (r2 ) and 2 (r2 ) , init 1 eff eff to compute H1 and H 2 . 1 2 2 1 init H 1eff 1 11 where H 1eff 1 V1eff V1eff 2init (2) 2 ( 2) 2 r1 r12 1 2 1 init H 2 2 2 2 eff where H 2 2 V2eff eff 2 V2eff 1init (1) 1 (1) 2 r2 r12 new new We then solved the equations to get new orbitals, 1 (r2 ) and 2 (r2 ) eff eff If these new orbitals had been used to calculate V1 and V2 , we would have gotten different effective potentials. Oy Vey!!! What a mess!!! What can we do to fix the problem that the orbitals resulting from solving the effective Schrödinger equations are not the same as the orbitals that we used to construct the equations?? Chapter 9: Slide 9 The Solution: Iterate to Self-Consistency new new Repeat the procedure. This time, use 1 (r1 ) and 2 (r2 ) to construct V1eff and V2eff and solve the equations again. newer Now, you’ll get an even newer pair of orbitals, (r1 ) and 2 (r2 ) newer 1 BUT: You have the same problem again. The effective Hamiltonians that were used to compute this newest pair of orbitals were constructed from the older set of orbitals. Well, I suppose you could repeat the procedure again, and again, and again, and again, until you either: (1) go insane (2) quit Chemistry and establish a multibillion dollar international trucking conglomerate (please remember me in your will). Chapter 9: Slide 10 Fortunately, the problem is not so dire. Usually, you will find that the new orbitals predicted by solving the equations get closer and closer to the orbitals used to construct the effective Hamiltonians. When they are sufficiently close, you stop, declare victory, and go out and celebrate with a dozen Krispy Kreme donuts (or pastrami sandwiches on rye, if that’s your preference). When the output orbitals are consistent with the input orbitals, you have achieved a “Self-Consistent Field” (SCF). Often, you will reach the SCF criterion within 10-20 iterations, although it may take 50-60 iterations or more in difficult cases. While the procedure appears very tedious and time consuming, it’s actually quite fast on modern computers. A single SCF calculation on a moderate sized molecule (with 50-100 electrons) can take well under 1 second. Chapter 9: Slide 11 The Energy A. The total energy 1 2 1 2 2 2 1 1 2 2 1 2 2 1 H 1 (r1 ) 2 (r2 ) 1 (r1 ) 2 (r2 ) 2 2 r1 r2 r12 2 r1 2 r2 r12 He He 1 1 2 2 He 1 2 H H 1 (1) H 2 ( 2) where H 1He 1 and H 2 2 2 r12 2 r1 2 r2 H1 and H2 are just each the Hamiltonian for the electron in a He+ ion. E * H d 1 (1)2 (2) H 1 (1)2 (2) He He 1 E 1 (1) 2 (2) H 1 (1) H 2 ( 2) 1 (1)2 (2) r12 We’re assuming that 1 and 2 have both been normalized. Chapter 9: Slide 12 He He 1 E 1 (1) 2 (2) H 1 (1) H 2 ( 2) 1 (1)2 (2) r12 E I1 I 2 J12 Remember, this is the total energy of the two electrons. I1 1 (1) 2 (2) H 1He (1) 1 (1) 2 (2) 2 (2) 2 (2) 1 (1) H 1He (1) 1 (1) 1 (1) H 1He 1 (1) I1 is the energy of an electron in a He+ ion. He He He I 2 1 (1) 2 (2) H 2 (2) 1 (1) 2 ( 2) 1 (1) 1 (1) 2 (2) H 2 ( 2) 2 ( 2) 2 ( 2) H 2 1 (2) I2 is the energy of an electron in a He+ ion. J12 1 (1) 2 (2) 1 1 (1)2 (2) (r ) (r ) (r ) (r ) dr dr * 1 1 1 1 * 2 2 2 2 1 2 r12 r1 r2 r12 J12 is the Coulomb Integral and represents the coulombic repulsion energy of the two electrons Chapter 9: Slide 13 The Energy B. The Individual Orbital Energies, 1 and 2 1 2 2 1 2 2 1 H1eff 1 11 where H1eff 1 V1eff 1 2 (2) 2 ( 2) 2 r1 2 r1 r12 1 2 2 1 1 1 (1) H eff 1 1 (1) 1 (1) 1 2 (2) 2 (2) 1 (1) 2 r1 r12 1 2 2 1 1 1 (1) 1 1 (1) 1 (1) 2 (2) 2 (2) 1 (1) 2 r1 r12 1 1 1 (1) H1He 2 (1) 1 (1)2 (2) 1 (1)2 (2) r12 1 I1 J12 where J12 1 (1)2 (2) 1 1 (1)2 (2) 1*1 2*2 dr1dr2 r12 r12 Chapter 9: Slide 14 The Energy B. The Individual Orbital Energies, 1 and 2 (Cont’d.) 1 2 1 2 1 H 2 2 2 2 eff where H 2 2 V2eff 2 1 (1) eff 2 2 1 (1) 2 r2 2 r2 r12 1 2 2 1 2 2 (2) H eff 2 2 (2) 2 (2) 2 1 (1) 1 (1) 2 (2) 2 r2 r12 1 2 2 1 2 2 (2) 2 2 (2) 2 (2) 1 (1) 1 (1) 2 (2) 2 r2 r12 1 2 2 (2) H 2He 2 (2) 1 (1)2 (2) 1 (1)2 (2) r12 2 I 2 J12 where J12 1 (1)2 (2) 1 1 (1)2 (2) 1*1 2*2 dr1dr2 r12 r12 Chapter 9: Slide 15 1 I1 J12 2 I 2 J12 The sum of orbital energies: 1 2 I1 J12 I 2 J12 I1 I 2 2J12 C. Total Energy versus sum of orbital energies The sum of orbital energies: 1 2 I1 I 2 2J12 The total energy: E I1 I 2 J12 The sum of the orbital energies has one too many Coulomb integrals, J12. The reason is that each orbital energy has the full electron-electron repulsion – You’re counting it one time too many!!! Chapter 9: Slide 16 1 2 I1 I 2 2J12 E I1 I 2 J12 Therefore: E ( I1 J12 ) ( I 2 J12 ) J12 E 1 2 J12 Chapter 9: Slide 17 Outline • The Hamiltonian for Multielectron Atoms • The Hartree Method: Helium • Koopman’s Theorem • Extension to Multielectron Atoms • Antisymmetrized Wavefunctions: Slater Determinants • The Hartree-Fock Method • Hartree-Fock Orbital Energies for Argon • Electron Correlation Chapter 9: Slide 18 Koopman’s Theorem Estimation of Atomic (or Molecular) Ionization Energies Ionization Energy (IE): M M+ + e- M is a neutral atom or molecule E IE E ( M ) E ( M ) 1 I1 J12 I1 I 2 J12 I 2 E I 2 1 I 2 E E ( He ) E ( He ) IE (He ) I2 is the energy of the He+ ion E is the energy of the He atom Koopman’s Theorem: The ionization energy of an atom or molecule can be estimated as -H, which is the orbital energy of the highest occupied orbital. Chapter 9: Slide 19 M M+ + e- M is a neutral atom or molecule E IE E ( M ) E ( M ) Koopman’s Theorem: The ionization energy of an atom or molecule can be estimated as -H, which is the orbital energy of the highest occupied orbital. There are two approximations in using Koopman’s theorem to estimate ionization energies which limit the accuracy: 1. Electron “relaxation” of the remaining N-1 electrons is neglected. 2. Differences in the “correlation energy” [to be discussed later] of the electrons in the ion and neutral atom are ignored. To obtain an accurate estimate of the ionization energy, one should perform quantum mechanical energy calculations on the neutral atom and ion to get E(M) and E(M+), from which the IE can be computed by the definition. Chapter 9: Slide 20 Electron Affinity Electron Affinity (EA): M + e- M- M is a neutral atom or molecule E EA E ( M ) E ( M ) With this “new” definition of Electron Affinity, a negative value of EA means that adding an electron to the atom is an exothermic process. Note: The text comments that Koopman’s theorem can also be used to calculate the electron affinity, EA = -L (energy of the lowest unoccupied orbital). However, this is not commonly used and is very inaccurate. Note: The “old” definition of Electron Affinity is the energy “released” when an electron is added to a neutral atom. EA(old) = - EA(new) Chapter 9: Slide 21 Outline • The Hamiltonian for Multielectron Atoms • The Hartree Method: Helium • Koopman’s Theorem • Extension to Multielectron Atoms • Antisymmetrized Wavefunctions: Slater Determinants • The Hartree-Fock Method • Hartree-Fock Orbital Energies for Argon • Electron Correlation Chapter 9: Slide 22 The Hartree Method for Multielectron Atoms The Hartree method for the more general N electron atom is a straightforward extension of the method outlined for the two electrons in Helium Each of the N electrons has an effective Hamiltonian. For electron #1, for example: 1 2 Z H 1eff 1 11 where H 1eff 1 V1eff 2 r1 elect elect- “Effective” KE Nuc elect-elect PE PE 2 init 2 init 2 N 1 init (r2 ) init 3 (r3 ) N (rN ) V1eff j ( j) j ( j) dr2 dr3 init 2 drN j 2 r1 j r12 r13 r1N As before, we are assuming that electron #1 is interacting with the “smeared out” electron density of electrons #2 to N. Chapter 9: Slide 23 There are equivalent equations for each electron, i, of the N electrons: 1 Z H ieff i ii where H ieff i2 Vi eff 2 ri 2 1 init init (rj ) j eff init j Vi j ( j) dr j j i rij j i rij As in the two electron case, one assumes that the total wavefunction is the product of one electron wavefunctions: N (r1 , r2 , , rN ) i (ri ) 1 (r1 ) 2 (r2 ) N (rN ) i 1 Initial guesses are made for each of the atomic functions, iinit, which are used to compute the effective potentials, Vieff, and the N equations are solved to get a new set of ’s. The procedure is repeated (iterated) until the guess wavefunctions are the same as the ones which are computed; i.e. until you reach a Self-Consistent Field (SCF) Chapter 9: Slide 24 The Energy N N 1 E i J ij i 1 i 1 j i E 1 2 J12 J13 J14 J 23 J 24 i is the orbital energy of the i’th. electron. This is the eigenvalue of the effective Hamiltonian for the i’th. electron Jij is the Coulomb Integral describing the repulsion between an electron in orbital i and an electron in orbital j. 2 i (r1 ) j (r2 ) 2 1 J ij i j i j dr1dr2 r12 r1 r2 r12 Note: If N=2 (i.e. He), the above expression for E reduces to E 1 2 J12 Chapter 9: Slide 25 Math. Preliminary: Determinants A determinant of order N is an NxN array a11 a12 a13 ... a1N of numbers (elements). The total number of a21 a2 N elements is N2. a31 a3 N Second Order Determinant a11 a12 aN1 aN 2 aN 3 ... a NN a11a22 a21a12 a21 a22 Note: The expansion has 2 terms Third (and higher) Order Determinant: Expansion by Cofactors a11 a12 a13 a a23 a a a a a21 a22 a23 a11 22 a12 21 23 a13 21 22 a32 a33 a31 a33 a31 a32 a31 a32 a33 a11 (a22a33 a32a23 ) a12 (a21a33 a31a23 ) a13 (a21a32 a31a22 ) Note: The expansion has 6 terms Chapter 9: Slide 26 Fourth Order Determinant a11 a12 a13 a14 a21 a22 a23 a24 a11 3 x3 a12 3x3 a13 3x3 a14 3x3 a31 a32 a33 a34 a41 a42 a43 a44 Note: Each 3x3 determinant has 6 terms. Therefore, the 4x4 determinant has 4x6 = 24 terms. General Properties of Determinants Property #1: An NxN determinant has N! terms. Property #2: If two columns or rows of a determinant are exchanged, then the value of the determinant changes sign. Property #3: If two columns or rows of a determinant are the same, then the value of the determinant is 0. Chapter 9: Slide 27 Outline • The Hamiltonian for Multielectron Atoms • The Hartree Method: Helium • Koopman’s Theorem • Extension to Multielectron Atoms • Antisymmetrized Wavefunctions: Slater Determinants • The Hartree-Fock Method • Hartree-Fock Orbital Energies for Argon • Electron Correlation Chapter 9: Slide 28 Slater Determinants Review: The Pauli Antisymmetry Principle ^ The permutation operator, Pij , exchanges the coordinates of two electrons in a wavefunction. Permuting two identical particles will not change the probability density: P (r , r ) p (r , r ) ˆ 2 ij i j 2 ij i j Therefore: pij 1 Pauli Principle: All elementary particles have an intrinsic angular momentum called spin. There are two types of particles, with different permutation properties: Bosons: Integral spin (0, 1, 2,…) Pij() = + Fermions: Half integral spin (1/2, 3/2,…) Pij() = - Chapter 9: Slide 29 Electrons (s = ½) are fermions. Therefore, wavefunctios are antisymmetric with respect to electron exchange (permutation). ˆ Pij (ri , rj ) (rj , ri ) (ri , rj ) Note that the permutation operator exchanges both the spatial and spin coordinates of the electrons. Review: Ground State Helium 1 1s (r1 )11s (r2 ) 2 1s (r1 ) 11s (r2 ) 2 2 1 or 1s(1)11s(2) 2 1s(1) 11s(2) 2 This wavefunction is antisymmetric 2 with respect to exchange of Shorthand electrons 1 and 2. 1s (1)1s (2)1 2 1 2 1 or 2 Factored Form Chapter 9: Slide 30 The electron configuration of ground state Lithium is 1s22s1. The wavefunction, (1, 2, 3) 1s (1)11s (2) 2 2 s (3) 3 , just won’t do. It’s not either symmetric or antisymmetric with respect to electron exchange. An appropriate antisymmetric wavefunction is: 1s (1)11s (2) 2 2s (3) 3 1s (1)1 2s (2) 21s (3) 3 1 (1, 2, 3) 1s(1) 11s (2) 2 2s (3) 3 1s (1) 1 2s (2) 21s (3) 3 6 2s (1)11s (2) 21s (3) 3 2s (1)11s (2) 21s (3) 3 Question: How do I know that this wavefunction is antisymmetric? Answer: Try it out. Exchange electrons 1 and 2. Terms 1 and 3 switch with each other, but each with opposite sign. Terms 2 and 5 switch with each other, but each with opposite sign. Terms 4 and 6 switch with each other, but each with opposite sign. Voila!! The wavefunction has changed sign. Chapter 9: Slide 31 Question: How did I figure out how to pick out the appropriate six terms? Answer: It was easy!! Mookie showed me how. Problem: The Mookster won’t be around to write out the antisymmetric wavefunctions for you on a test. Solution: I guess I should impart the magic of King Mookie, and show you how it’s done. Chapter 9: Slide 32 Slater Determinants The ground state Helium wavefunction is: (1,2) 1 1s(1)11s(2) 2 1s(1)11s(2) 2 2 It can be written as a 2x2 determinant, called a Slater determinant (named after J. C. Slater, who first came up with the idea). 1 1s(1)1 1s(1) 1 (1,2) 1 1s(1)11s(2) 2 1s(1) 11s(2) 2 2 1s(2) 2 1s(2) 2 2 Note that different “spinorbitals”** are put in different columns. Different electrons are put in different rows. The coefficient is to normalize the antisymmetrized wavefunction. **A spinorbital is just the combination of the spatial and spin part of an orbital taken together. Chapter 9: Slide 33 1 1s(1)1 1s(1) 1 (1,2) 1s(2) 2 1s(2) 2 1 1s(1)11s(2) 2 1s(1) 11s(2) 2 2 2 Two properties of determinants come in very handy. Property #2: If two columns or rows of a determinant are exchanged, then the value of the determinant changes sign. 1 1s(2) 2 1s(2) 2 P (1,2) (2,1) 2 1s(1)1 1s (1) 1 12 1 1s(1) 11s(2) 2 1s(1)11s(2) 2 (1,2) 2 Hey!! That’s nice!! A Slater Determinant is automatically antisymmetric with respect to the exchange of two electrons. Chapter 9: Slide 34 Property #3: If two columns or rows of a determinant are the same, then the value of the determinant is 0. Let’s put both electrons in the same spinorbital, say 1s, and see what happens. 1 1s(1)1 1s(1)1 (1,2) 1 1s(1)11s(2) 2 1s(1)11s(2) 2 0 2 1s(2) 2 1s(2) 2 2 This explains the more commonly stated form of the Pauli Principle: No two electrons can occupy the same orbital with the same spin. Chapter 9: Slide 35 The Lithium Ground State Wavefunction The electron configuration of ground state Lithium is 1s22s1. The antisymmetrized wavefunction is: 1s (1)1 1s (1) 1 2s (1)1 1 (1, 2, 3) 1s (2) 2 1s (2) 2 2s (2) 2 3! 1s (3) 3 1s (3) 3 2s (3) 3 1 The factor, ,is to normalize the wavefunction (which has 3! terms) 3! Expanding the wavefunction 1s (2) 2 2 s (2) 2 1s (2) 2 2 s (2) 2 1s (1)1 1s (1) 1 1 1s (3) 3 2s (3) 3 1s (3) 3 2 s (3) 3 (1, 2, 3) 6 1s (2) 2 1s (2) 2 2 s (1)1 1s (3) 3 1s (3) 3 Chapter 9: Slide 36 1s (1)1 1s (2) 2 2s (3) 3 2s (2) 21s (3) 3 1 (1, 2, 3) 1s(1) 1 1s (2) 2 2s (3) 3 2s (2) 21s (3) 3 6 2s (1)1 1s (2) 21s(3) 3 1s (2) 21s (3) 3 1s (1)11s (2) 2 2s (3) 3 1s (1)1 2s (2) 21s (3) 3 1 (1, 2, 3) 1s(1) 11s (2) 2 2s (3) 3 1s (1) 1 2s (2) 21s (3) 3 6 2s (1)11s (2) 21s (3) 3 2s (1)11s (2) 21s (3) 3 We discussed earlier that this expanded (6 term) wavefunction is antisymmetric with respect to electron exchange. Chapter 9: Slide 37 1s (1)1 1s (1) 1 2s (1)1 1 (1, 2, 3) 1s (2) 2 1s (2) 2 2s (2) 2 3! 1s (3) 3 1s (3) 3 2s (3) 3 The antisymmetry can also be shown by using the property of determinants. Exchanging two electrons: 1s (1)1 1s (1) 1 2s (1)1 1 (1, 3, 2) 1s (3) 3 1s (3) 3 2s (3) 3 (1, 2, 3) 3! 1s (2) 2 1s (2) 2 2s (2) 2 Let’s put all 3 electrons in the 1s orbital: 1s(1)1 1s(1) 1 1s(1)1 1 (1, 2, 3) 1s (2) 2 1s (2) 2 1s (2) 2 0 3! 1s(3) 3 1s(3) 3 1s (3) 3 Chapter 9: Slide 38 General (1)1 (1) 1 n (1) 1 1 (2) 2 (2) 2 (1, 2, 3,) N! ( N ) N ( N ) N n ( N ) N Shorthand Notations (Various types) 1s (1) 1s(1) 2s(1) 1 (1, 2, 3) 1s (2) 1s(2) 2s (2) 3! 1s (3) 1s(3) 2s(3) Use bars to indicate spin. Lack of a bar means the spin is Chapter 9: Slide 39 Show diagonal terms only. Lithium 2s3 3 1 1s (1)1 1s (2) 2 3! Beryllium 1 1s (1)1 1s (2) 2 2 s (3) 3 2 s ( 4) 4 4! Other shorthand notations include: Leaving out the normalization constant. Leaving out the normalization constant and electron numbering. Beryllium 1s 1s 2s 2s To avoid confusion, the only shorthand I might use is the diagonal form at the top of this page. Chapter 9: Slide 40 Outline • The Hamiltonian for Multielectron Atoms • The Hartree Method: Helium • Koopman’s Theorem • Extension to Multielectron Atoms • Antisymmetrized Wavefunctions: Slater Determinants • The Hartree-Fock Method • Hartree-Fock Orbital Energies for Argon • Electron Correlation Chapter 9: Slide 41 The Hartree-Fock Method Hartree’s original method neglected to consider that the wavefunction in a multielectron atom (or molecule) must be antisymmetric with respect to electron exchange. The Hartree-Fock is an extension, using antisymmetrized wavefunctions. It results in additional “Exchange” terms in the Effective Hamiltonians and “Exchange Integrals” in the expression for the energy. We actually encountered Exchange Integrals when we calculated the energy of excited state Helium in the 1s12s1 electron configuration. Chapter 9: Slide 42 Review: The Energy of Triplet State Helium (1s12s1) antisym(r1 , r2 ) spin 1M S 1 T1 1s (r1 )2 s (r2 ) 2 s (r1 )1s (r2 ) 1 2 2 antisym(r1 , r2 ) 1 1s (r1 )2 s (r2 ) 2 s (r1 )1s (r2 ) 1s(1)2s(2) 2s(1)1s(1) 1 2 2 Remember that spin does not contribute directly to the energy. 1 2 2 1 2 2 1 1 H 1 2 H1 H 2 2 r1 2 r2 r12 r12 E trip 1 1s(1)2s(2) 2s(1)1s(2) H 1 1s(1)2s(2) 2s(1)1s(2) 2 2 Chapter 9: Slide 43 E trip 1s 2 s J 1s 2 s K 1s 2 s 1 2 3 4 1 1 where J1s 2 s 1s (1)2 s (2) 1s (1)2 s (2) K1s 2 s 1s (1)2 s (2) 2 s (1)1s (2) r12 r12 1s (1) 2 2 s (2) 2 [1s (1)2 s (1)] [1s (2)2 s (2)] J 1s 2 s dr1dr2 K1s 2 s dr1dr2 r12 r12 1. Energy of electron in 1s He+ orbital 3. Coulomb (repulsion) Integral 2. Energy of electron in 2s He+ orbital 4. Exchange Integral Chapter 9: Slide 44 E trip 1s 2 s J 1s 2 s K 1s 2 s 1 2 3 4 1s (1) 2 2 s (2) 2 J 1s 2 s dr1dr2 Always positive r12 3. Coulomb (repulsion) Integral The integrand of the Coulomb integral represents the repulsion of two infinitesimal electron densities, (1)=1s(1)2 and (2)=2s(2)2, separated by a distance, r12. The repulsion is summed over all infinitesimal electron densities. [1s (1)2 s (1)] [1s (2)2 s (2)] K1s 2 s dr1dr2 Usually positive r12 4. Exchange Integral Arises purely from the antisymmetry of the spatial function with respect to electron exchange. It has no classical analog. If the above calculation had been performed with a simple product wavefunction, spat = 1s(1)2s(2), there would be no exchange integral Chapter 9: Slide 45 The Hartree-Fock Energy N N Hartree Energy: E i J ij Use simple product wavefunction: i 1 i 1 j i not antisymmetric w.r.t. exchange 2 2 1 i (r1 ) j (r2 ) J ij i j i j dr1dr2 r12 r1 r2 r12 Jij is the Coulomb Integral describing the repulsion between an electron in orbital i and an electron in orbital j. Chapter 9: Slide 46 Hartree-Fock Energy: E i 2 J ij K ij N N i 1 i 1 j i Use antisymmetrized wavefunction: Slater Determinant J ij i j 1 i j i* (r1 )i (r1 ) * (r2 ) j (r2 ) j dr1dr2 Coulomb r12 r1 r2 r12 Integral K ij i j 1 ji (r ) (r ) (r ) (r ) dr dr i * 1 j 1 * j 2 i 2 Exchange 1 2 r12 r1 r2 r12 Integral The Exchange Integral arises from the antisymmetry of the wavefunction, and has no classical analog. Note: There is an error in the text’s HF energy. The way the authors are writing it, the top summation index should be N/2 Chapter 9: Slide 47 Outline • The Hamiltonian for Multielectron Atoms • The Hartree Method: Helium • Koopman’s Theorem • Extension to Multielectron Atoms • Antisymmetrized Wavefunctions: Slater Determinants • The Hartree-Fock Method • Hartree-Fock Orbital Energies for Argon • Electron Correlation Chapter 9: Slide 48 Hartree-Fock Orbital Energies for Ar 0 eV Separated particles -16.1 eV 3p Note that the ns and np orbitals have different -34.8 eV 3s energies. This is due to screening of the p electrons. Koopman’s Theorem IE - -260 eV 2p Electron Removed IE(exp) IE(Koop) -335 eV 2s 1s 3206 eV 3227 eV 2s -- 335 2p 249 260 3s 29.2 34.8 3p 15.8 16.1 -3227 eV 1s Chapter 9: Slide 49 Outline • The Hamiltonian for Multielectron Atoms • The Hartree Method: Helium • Koopman’s Theorem • Extension to Multielectron Atoms • Antisymmetrized Wavefunctions: Slater Determinants • The Hartree-Fock Method • Hartree-Fock Orbital Energies for Argon • Electron Correlation Chapter 9: Slide 50 Electron Correlation The principal approximation of the Hartree-Fock method is that a given electron interacts with the “smeared-out” electron density of the remaining N-1 electrons. Actually, the other N-1 electrons are point particles, just like the one we’re considering. Thus, the motion of the electrons are correlated. That is, they try to avoid each other. High Low Energy Energy Not favored Favored Chapter 9: Slide 51 Because the Hartree-Fock (HF) method does not consider the specific electron-electron repulsions, which tend to keep two electrons apart, the HF energy is invariably too high. The difference between the “exact” electronic energy and the HF energy is called the “Correlation Energy”, Ecorr. 0 Ecorr Eexact EHF Generally, the correlation energy is very small compared to the total energy (usually <1%) EExact However, in absolute terms, this can still represent EHF a rather large energy. The “exact” electronic energy can be measured as the negative of the sum of the Ionization Energies. N Eexact Eexp IEi EHF i 1 EExact Ecorr Eexact E HF Eexp E HF Chapter 9: Slide 52 EHF Helium Ecorr EHF = -2.862 au EExact EExact = -2.904 au Eexact EHF % Error x100 1.4% Eexact However, the correlation energy can still be very large in absolute terms. 2625 kJ / mol Ecorr Eexact EHF 0.042 au x 110 kJ / mol 1 au Chapter 9: Slide 53 EHF Argon Ecorr EHF = -526.807 au EExact EExact = -527.030 au Eexact EHF % Error x100 0.04% Eexact However, the correlation energy can still be very large in absolute terms. 2625 kJ / mol Ecorr Eexact EHF 0.223 au x 585 kJ / mol 1 au For many applications (e.g. geometries and frequencies), inclusion of the correlation energy is not that important. However, for applications involving bond breaking and bond making (e.g. reactions), inclusion of the correlation energy is critical in order to get good results. We will qualitatively discuss methods used to determine the correlation energy in a later chapter. Chapter 9: Slide 54 An Example: Calculated Ionization Energy and Electron Affinity of Fluorine Ionization Energy (IE): M M+ + e- M is a neutral atom or molecule E IE E ( M ) E ( M ) Electron Affinity (EA): M + e- M- M is a neutral atom or molecule E EA E ( M ) E ( M ) Methods: E(HF) = HF/6-311++G(3df,2pd) Hartree-Fock Energy E(QCI) = QCISD(T)/6-311++G(3df,2pd) Correlated Energy This is the HF energy with a correction for electron correlation calculated at the QCISD(T) level (later Gator). Chapter 9: Slide 55 Species E(HF) E(QCI) F -99.402 au -99.618 au F+ -98.825 -98.984 F- -99.446 -99.737 IE ( HF ) E ( F ) E ( F ) kJ / mol 98.825 ( 99.402) 0.577 au 2625 1515 kJ / mol au Similarly: IE (QCI ) 1664 kJ / mol EA( HF ) E ( F ) E ( F ) kJ / mol 99.446 ( 99.402) 0.044 au 2625 116 kJ / mol au Similarly: EA(QCI ) 312 kJ / mol Chapter 9: Slide 56 Quantity Expt. HF QCI IE 1681 kJ/mol 1514 kJ/mol 1664 kJ/mol EA -328 -115 -312 Koopman’s Theorem IE Energy of highest occupied orbital at HF/6-311++G(3df,2pd) level H = -0.733 au IE -H = +0.733 au • 2625 kJ/mol / au = 1924 kJ/mol Notes: (1) Koopman’s Theorem gives only rough approximation for Ionization Energy (2) Accurate calculations of the IE or EA require the use of energies corrected for electron correlation. Chapter 9: Slide 57 EHF Ecorr Eexact EHF 0 E HF ( F ) Eexact ( F ) EQCI ( F ) Eexact EQCI Ecorr F Ecorr F Ecorr F IE(HF)=1514 kJ/mol IE(QCI)=1664 kJ/mol IE E ( F ) E ( F ) 0 EHF (F ) EA E ( F ) E ( F ) 0 Eexact ( F ) EQCI ( F ) EA(HF)= -115 kJ/mol EHF ( F ) EA(QCI)= -312 kJ/mol Eexact ( F ) EQCI ( F ) Chapter 9: Slide 58