Lecture 4, January 12, 2015 Bonding in H2

Lecture 4, January 12, 2015 Bonding in H2 Elements of Quantum Chemistry with Applications to Chemical Bonding and Properties of Molecules and Solids Course number: Ch125a; Room 147 Noyes Hours: 11-11:50am Monday, Wednesday, Friday William A. Goddard, III, wag@wag.caltech.edu 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Special Instructor: Julius Su <jsu@caltech.edu> Teaching Assistants: Hai Xiao <xiao@caltech.edu> Mark Fornace <mfornace@caltech.edu> Ch125a-Goddard-L02 copyright 2015 William A. Goddard III, all rights reserved 1

H 2 molecule, independent atoms Start with non interacting H atoms, electron 1 on H on earth, χ Ε (1) the other electron 2 on the moon, χ Μ (2) What is the total wavefunction, Ψ(1,2)? Ch125a-Goddard-L01 copyright 2009 2015 William A. A. Goddard III, III, all all rights reserved 2

H 2 molecule, independent atoms Start with non interacting H atoms, electron 1 on H on earth, χ Ε (1) the other electron 2 on the moon, χ Μ (2) What is the total wavefunction, Ψ(1,2)? Maybe Ψ(1,2) = χ Ε (1) + χ Μ (2)? Since the motions of the two electrons are completely independent, we expect that the probability of finding electron 1 at some location on earth to be independent of the probability of finding electron 2 at some location on the moon. Thus Ch125a-Goddard-L01 copyright 2009 2015 William A. A. Goddard III, III, all all rights reserved 4

Ch125a-Goddard-L01 H 2 molecule, independent atoms Start with non interacting H atoms, electron 1 on H on earth, χ Ε (1) the other electron 2 on the moon, χ Μ (2) What is the total wavefunction, Ψ(1,2)? Maybe Ψ(1,2) = χ Ε (1) + χ Μ (2)? Since the motions of the two electrons are completely independent, we expect that the probability of finding electron 1 at some location on earth to be independent of the probability of finding electron 2 at some location on the moon. Thus P(1,2) = P E (1)*P M (2) This is analogous to the joint probability, say of rolling snake eyes (two ones) in dice P(snake eyes)=p(1 for die 1)*P(1 for die 2)=(1/6)*(1/6) = 1/36 Question what wavefunction Ψ(1,2) leads to P(1,2) = P E (1)*P M (2)? copyright 2009 2015 William A. A. Goddard III, III, all all rights reserved 5

Answer: product of amplitudes Ψ(1,2) = χ Ε (1)χ Μ (2) leads to P(1,2) = Ψ(1,2) 2 = Ψ(1,2) * Ψ(1,2) = = [χ Ε (1)χ Μ (2)]* [χ Ε (1)χ Μ (2)] = = [χ Ε (1)* χ Ε (1)] [χ Μ (2)* χ Μ (2)] = = P E (1) P M (2) Conclusion the wavefunction for independent electrons is the product of the independent orbitals for each electron Ch125a-Goddard-L01 copyright 2015 William A. Goddard III, all rights reserved 6

Answer: product of amplitudes Ψ(1,2) = χ Ε (1)χ Μ (2) leads to P(1,2) = Ψ(1,2) 2 = Ψ(1,2) * Ψ(1,2) = = [χ Ε (1)χ Μ (2)]* [χ Ε (1)χ Μ (2)] = = [χ Ε (1)* χ Ε (1)] [χ Μ (2)* χ Μ (2)] = = P E (1) P M (2) Conclusion the wavefunction for independent electrons is the product of the independent orbitals for each electron Back to H 2, Ψ EM (1,2) = χ Ε (1)χ Μ (2) But Ψ ME (1,2) = χ Μ (1)χ Ε (2) is equally good since the electrons are identical Ch125a-Goddard-L01 copyright 2015 William A. Goddard III, all rights reserved 7

Answer: product of amplitudes Ψ(1,2) = χ Ε (1)χ Μ (2) leads to P(1,2) = Ψ(1,2) 2 = Ψ(1,2) * Ψ(1,2) = Ch125a-Goddard-L01 = [χ Ε (1)χ Μ (2)]* [χ Ε (1)χ Μ (2)] = = [χ Ε (1)* χ Ε (1)] [χ Μ (2)* χ Μ (2)] = = P E (1) P M (2) Conclusion the wavefunction for independent electrons is the product of the independent orbitals for each electron Back to H 2, Ψ EM (1,2) = χ Ε (1)χ Μ (2) But Ψ ME (1,2) = χ Μ (1)χ Ε (2) is equally good since the electrons are identical Also we could combine these wavefunctions χ Ε (1)χ Μ (2) ± χ Ε (1)χ Μ (2) Which is best? copyright 2015 William A. Goddard III, all rights reserved 8

Consider H 2 at R= The wavefunction for H2 at long R Two equivalent wavefunctions Φ LR Φ RL Φ LR Φ RL Φ LR Φ RL At R= these are all the same, what is best for finite R Ch125a-Goddard-L01 copyright 2015 William A. Goddard III, all rights reserved 9

Plot two-electron wavefunctions along molecular axis R LR+RL R LR-RL L L x L R L R z Lower KE (good) Higher PE (bad) EE poor higher KE (bad) Lower PE (good) EE good Ch125a-Goddard-L01 copyright 2015 William A. Goddard III, all rights reserved 12

Details E(H 2+ ) 1 = KE1 + PE1a + PE1b + 1/Rab for electron 1 E(H 2+ ) 2 = KE2 + PE2a + PE2b + 1/Rab for electron 2 E(H 2 ) = (KE1 + PE1a + PE1b) + (KE2 + PE2a+PE2b) + 1/Rab+EE EE = <Φ(1,2) 1/r 12 Φ(1,2)> assuming <Φ(1,2) Φ(1,2)>=1 Note that EE = dx 1 dy 1 dz 1 dx 2 dy 2 dz 2 [ Φ(1,2) 2 /r 12 ] > 0 since all terms in integrand > 0 Ch125a-Goddard-L01 copyright 2015 William A. Goddard III, all rights reserved 13

Valence Bond wavefuntion of H2 LR-RL Nodal theorem says ground state is nodeless, thus Φ g is better than Φ u LR+RL Valence bond: start with ground state at R= and build molecule by combining best state of atoms Ch125a-Goddard-L01 copyright 2015 William A. Goddard III, all rights reserved 14

Molecular Orbitals: Alternative way to view states of H 2 Valence bond: start with ground state at R= and build molecule by bonding atoms Molecular orbitals (MO): start with optimum orbitals of one electron molecule at R=R e and add electrons H2: two MO s u: antibonding g: bonding Ch125a-Goddard-L01 copyright 2015 William A. Goddard III, all rights reserved 15

Compare ground state MO and VB near Re KE: best possible EE: too much ionic character KE: pretty good EE: excellent Now consider large R: Ch125a-Goddard-L01 copyright 2015 William A. Goddard III, all rights reserved 17

compare ground state from VB and MO MO wavefunction is half covalent and half ionic OK for R=R equilbrium = R e, but not for R = Ch125a-Goddard-L01 copyright 2015 William A. Goddard III, all rights reserved 18

Analyze gu and ug states in 2 electron space gu ug Both have exactly the same KE and same PE Now combine into gu+ug and gu ug Which is better? Ch125a-Goddard-L01 copyright 2015 William A. Goddard III, all rights reserved 23

Analyze gu and ug states in 2 electron space gu ug All four have one nodal plane and lead to same KE and same PE except for the electron-electron repulsion term gu-ug gu+ug <Φ(1,2) 1/r 12 Φ(1,2)> Worst case is for z 1 =z 2, along diagonal Never have z 1 =z 2 great EE Ch125a-Goddard-L01 Maximum at z 1 =z 2 terrible EE copyright 2015 William A. Goddard III, all rights reserved 24

Ground and excited states from MO analysis Thus gu-ug is best excited state because it has very low electron repulsion. The electrons have zero chance of both being at the same spot Ch125a-Goddard-L01 copyright 2015 William A. Goddard III, all rights reserved 25

The uu or 2 nd g state In the MO description while the VB description leads to The connections between these states are where Ch125a-Goddard-L01 For R=, S=0 and λ = 1, pure VB For R=0, S=1 and λ = 0, pure MO For R e =1.4a 0, S=.7 and λ = 0.176 copyright 2015 William A. Goddard III, all rights reserved 27

Now consider symmetry for H 2 molecule For H 2 we use the coordinate system at the right. Using the same conventions and assumptions as for H 2 + leads to where 1/r 12 is the interaction between the two electrons and Contains all terms depending only on the coordinates of electron 1 Next consider inverting the coordinates of electron 1 Ch120a-Goddard-L02 29

Invert coordinates of electron 1 As before h(1) is invariant and so is h(2), but clearly something is different for H 2. The problem is 1/r 12. Inverting electron 1 or 2 separately does not preserve the r 12 distance. Ch120a-Goddard-L02 30

For multielectron systems, the inversion symmetry applies only if we inverts all electron coordinates simultaneously Thus we define I to be Which we write as This leads to Now we see that H(1,2) is invariant under inversion. Ch120a-Goddard-L02 31

Symmetry for H 2 Just as for H 2+, inversion symmetry of H 2 leads to the result that every eigenstate of H 2 is either g or u Consider the VB wavefunctions ^ I But = does NOT have symmetry I [ ] = + is I ] = - is Which is why we labeled them as such Ch120a-Goddard-L02 32

Now consider inversion symmetry for MO wavefunctions This is easy since each MO has inversion symmetry Thus Ch120a-Goddard-L02 33

Permutation Symmetry transposing the two electrons in H(1,2) must leave the Hamiltonian invariant since the electrons are identical 1 2 H(2,1) = h(2) + h(1) + 1/r 12 + 1/R = H(1,2) We will denote this transposition as τ where τφ(1,2) = Φ(2,1) Note that τ 2 = e, the einheit or identity operator τ 2 Φ(1,2) = Φ(1,2) Thus τ is of order 2 and the previous arguments on inversion apply equally to transposition Ch120a-Goddard-L02 34

Permutational symmetry of wavefunctions Every exact two-electron wavefunction must be either symmetric, s, or antisymmetric, a with respect to transposition Applying this to the product MO wavefunctions leads to No symmetry Thus symmetry alone, would tell us that the ug and gu product wavefunctions are wrong Ch120a-Goddard-L02 35

Permutational symmetry for the u MO states of H 2 We saw previously that combining these wavefunctions leads to a low lying covalent state and a high energy ionic state Which lead properly to symmetric and antisymmetric permutation states Ch120a-Goddard-L02 36

permutational symmetry for H2 wavefunctions symmetric antisymmetric Ch120a-Goddard-L02 37

He atom describe using He+ orbitals With 2 electrons, we can form the ground state of He by putting both electrons in the He + 1s orbital, just like the MO state of H 2 Ψ He (1,2) = A[(Φ 1s α)(φ 1s β)]= Φ 1s (1)Φ 1s (2) (αβ-βα) Ε He = 2<1s h 1s> + J 1s,1s First lets review the energy for He +. Writing Φ 1s = exp(-ζr) we determine the optimum ζ for He + as follows <1s KE 1s> = + ½ ζ 2 (goes as the square of 1/size) <1s PE 1s> = - Zζ (linear in 1/size) Applying the variational principle, the optimum ζ must satisfy de/ dζ = ζ - Z = 0 leading to ζ = Z, This value of ζ leads to KE = ½ Z 2, PE = -Z 2, E=-Z 2 /2 = -2 h 0. write PE=-Z/R 0, so that the average radius is R 0 =1/ζ 40

J 1s,1s = e-e energy of He atom He + orbitals Now consider He atom: E He = 2(½ ζ 2 ) 2Zζ + J 1s,1s e 1 How can we estimate J 1s,1s Assume that each electron moves on a sphere With the average radius R 0 = 1/ζ And assume that e 1 at along the z axis (θ=0) Neglecting correlation in the electron motions, e 2 will on the average have θ=90º so that the average e1-e2 distance is ~sqrt(2)*r 0 Thus J 1s,1s ~ 1/[sqrt(2)*R0] = 0.71 ζ A rigorous calculation (notes chapter 3, appendix 3-C page 6) Gives J 1s,1s = (5/8) ζ R 0 e 2 41

The optimum atomic orbital for He atom He atom: E He = 2(½ ζ 2 ) 2Zζ + (5/8)ζ Applying the variational principle, the optimum ζ must satisfy de/ dζ = 0 leading to 2ζ - 2Z + (5/8) = 0 Thus ζ = (Z 5/16) = 1.6875 KE = 2(½ ζ 2 ) = ζ 2 PE = - 2Zζ + (5/8)ζ = -2 ζ 2 E= - ζ 2 = -2.8477 h 0 Ignoring e-e interactions the energy would have been E = -4 h 0 The exact energy is E = -2.9037 h 0 (from memory, TA please check). Thus this simple approximation accounts for 98.1% of the exact result. 42

Interpretation: The optimum atomic orbital for He atom Ψ He (1,2) = Φ 1s (1)Φ 1s (2) with Φ 1s = exp(-ζr) We find that the optimum ζ = (Z 5/16) = 1.6875 With this value of ζ, the total energy is E= - ζ 2 = -2.8477 h 0 This wavefunction can be interpreted as two electrons moving independently in the orbital Φ 1s = exp(-ζr) which has been adjusted to account for the average shielding due to the other electron in this orbital. On the average this other electron is closer to the nucleus about 31% of the time so that the effective charge seen by each electron is 2.00-0.31=1.69 The total energy is just the sum of the individual energies. Ionizing the 2 nd electron, the 1 st electron readjusts to ζ = Z = 2 With E(He + ) = -Z 2 /2 = - 2 h 0. thus the ionization potential (IP) is 0.8477 h 0 = 23.1 ev (exact value = 24.6 ev) 43

Electron spin, 5 th postulate QM Consider application of a magnetic field Our Hamiltonian has no terms dependent on the magnetic field. Hence no effect. But experimentally there is a huge effect. Namely The ground state of H atom splits into two states β B=0 Increasing B α This leads to the 5 th postulate of QM In addition to the 3 spatial coordinates x,y,z each electron has internal or spin coordinates that lead to a magnetic dipole aligned either with the external magnetic field or opposite. We label these as α for spin up and β for spin down. Thus the ground states of H atom are φ(xyz)α(spin) and φ(xyz)β(spin) Ch120a-Goddard-L02 44

Ch120a-Goddard-L02 Electron spin So far we have considered the electron as a point particle with mass, m e, and charge, -e. In fact the electron has internal coordinates, that we refer to as spin, with two possible angular momenta. +½ or α or up-spin and -½ or β or down-spin But the only external manifestation is that this spin leads to a magnetic moment that interacts with an external magnetic field to splt into two states, one more stable and the other less stable by an equal amount. β ΔE = -γb z s z B=0 Increasing B α Now the wavefunction of an atom is written as a spinorbital ψ(r,σ) where r refers to the vector of 3 spatial coordinates, x,y,z and σ refers to the internal spin coordinates 45

Spin states for 1 electron systems Our Hamiltonian does not involve any terms dependent on the spin, so without a magnetic field we have 2 degenerate states for H atom. φ(r)α, with up-spin, m s = +1/2 φ(r)β, with down-spin, m s = -1/2 The electron is said to have a spin angular momentum of S=1/2 with projections along a polar axis (say the external magnetic field) of +1/2 (spin up) or -1/2 (down spin). This explains the observed splitting of the H atom into two states in a magnetic field Similarly for H 2 + the ground state becomes φ g (r)α and φ g (r)β While the excited state becomes φ u (r)α and φ u (r)β Ch120a-Goddard-L02 46

Spin states for 2 electron systems Combining the two-electron spin functions to form symmetric and antisymmetric combinations leads to Φ(1,2) α(1) α(2) Φ(1,2) [α(1) β(2) + β(1) α(2)] Φ(1,2) β(1) β(2) Symmetric spin S=1 0-1 Φ(1,2) [α(1) β(2) - β(1) α(2)] Antisymmetric spin 0 S=0 Adding the spin quantum numbers, m s, to obtain the total spin projection, M S = m s1 + m s2 leads to the numbers above. The three symmetric spin states are considered to have spin S=1 with components +1.0,-1, which are referred to as a triplet state (since it leads to 3 levels in a magnetic field) The antisymmetric state is considered to have spin S=0 with just one Ch120a-Goddard-L02 component, 0. It is copyright called 2011 a William singlet A. Goddard state. III, all rights reserved M S +1 47

Spinorbitals The Hamiltonian does not depend on spin, thus the spatial and spin coordinates are independent. Hence the total wavefunction can be written as a product of a spatial wavefunction, φ(σ), called an orbital, and a spin function, х(σ) = α or β. We refer to the composite as a spinorbital ψ(r,σ) = φ(σ) х(σ) where r refers to the vector of 3 spatial coordinates, x,y,z while σ to the internal spin coordinates. Ch120a-Goddard-L02 48

spinorbitals for two-electron systems Thus for a two-electron system with independent electrons, the wavefunction becomes Ψ(1,2) = Ψ(ρ 1,σ 1,ρ 2,σ 2 ) = ψ a (ρ 1,σ 1 ) ψ b (ρ 2,σ 2 ) = φ a (ρ 1 ) х a (σ 1 ) φ b (ρ 2 ) х b (σ 2 ) =[φ a (ρ 1 ) φ b (ρ 2 )][х a (σ1) х b (σ 2 )] Where the last term factors the total wavefunction into space and spin parts Ch120a-Goddard-L02 49

Permutational symmetry again For a two-electron system the Hamiltonian is invariant (unchanged) upon transposition of the electrons (changing both spatial and spin coordinates simultaneously) H(2,1) = H(1,2) Again the simultaneous transposition of space-spin coordinates of two electrons is of order 2 Thus for every eigenstate of the Hamiltonian we obtain either Ψs(1,2) = +1 Ψs(1,2) Ψa(1,2) = -1 Ψa(1,2) Here the transposition interchanges both spin and space components of the wavefunction simultaneously Ch120a-Goddard-L02 50

Permutational symmetry continued Factoring the wave function in terms of a spatial wavefunction depending only on spatial coordinates and a spin spatial wavefunction depending only on spin coordinates leads to Ψ(1,2) = Φ(1,2)χ(1,2) We know that the H(1,2) is separately unchanged by transposing either just the spatial coordinates or the spin coordinates Thus either Φ(2,1) = +Φ(1,2) or Φ(2,1) = -Φ(1,2) and either χ(2,1) = +χ(1,2) or χ(2,1) = -χ(1,2) Ch120a-Goddard-L02 51

Our Hamiltonian for H 2, Permutational symmetry, summary H(1,2) =h(1) + h(2) + 1/r12 + 1/R Does not involve spin This it is invariant under 3 kinds of permutations Space only: ρ 1 ρ 2 Spin only: σ 1 σ 2 Space and spin simultaneously: (ρ 1,σ 1 ) ç è (ρ 2,σ 2 ) Since doing any of these interchanges twice leads to the identity, we know from previous arguments that Ψ(2,1) = ± Ψ(1,2) symmetry for transposing spin and space coord Φ(2,1) = ± Φ(1,2) symmetry for transposing space coord Χ(2,1) = ± Χ(1,2) symmetry for transposing spin coord Ch120a-Goddard-L02 52

Permutational symmetries for H 2 and He H 2 Have 4 degenerate g ground states for H 2 He Have 4 degenerate u excited states for H 2 Have 4 degenerate ground state for He Ch120a-Goddard-L02 53

Permutational symmetries for H 2 and He H 2 BUT THIS IS ALL WRONG The g state of H 2 is nondegenerate The u excited state is 3 fold degenerate He The He ground state is nondegenerate How can we fix this obvious disagreement with experiment! Ch120a-Goddard-L02 Have 4 degenerate g ground states for H 2 Have 4 degenerate u excited states for H 2 Have 4 degenerate ground state for He 54

Permutational symmetries for H 2 and He H 2 He Ch120a-Goddard-L02 the only states observed are those for which the wavefunction changes sign upon transposing all coordinates of electron 1 and 2 Leads to the 6 th postulate of QM 55

The 6 th postulate of QM: the Pauli Principle For every eigenstate of an electronic system H(1,2, i j N)Ψ(1,2, i j N) = EΨ(1,2, i j N) The electronic wavefunction Ψ(1,2, i j N) changes sign upon transposing the total (space and spin) coordinates of any two electrons Ψ(1,2, j i N) = - Ψ(1,2, i j N) We can write this as τ ij Ψ = - Ψ for all i and j Ch120a-Goddard-L02 56

Implications of the Pauli Principle Consider two independent electrons, 1 on the earth described by ψ e (1) and 2 on the moon described by ψ m (2) Ψ(1,2)= ψ e (1) ψ m (2) And test whether this satisfies the Pauli Principle Ψ(2,1)= ψ m (1) ψ e (2) - ψ e (1) ψ m (2) Thus the Pauli Principle does NOT allow the simple product wavefunction for two independent electrons Ch120a-Goddard-L02 57

Quick fix to satisfy the Pauli Principle Combine the product wavefunctions to form a symmetric combination Ψ s (1,2)= ψ e (1) ψ m (2) + ψ m (1) ψ e (2) And an antisymmetric combination Ψ a (1,2)= ψ e (1) ψ m (2) - ψ m (1) ψ e (2) We see that τ 12 Ψ s (1,2) = Ψ s (2,1) = Ψ s (1,2) (boson symmetry) τ 12 Ψ a (1,2) = Ψ a (2,1) = -Ψ a (1,2) (Fermion symmetry) Thus for electrons, the Pauli Principle only allows the antisymmetric combination for two independent electrons Ch120a-Goddard-L02 58

Consider some simple cases: identical spinorbitals Ψ(1,2)= ψ e (1) ψ m (2) - ψ m (1) ψ e (2) Identical spinorbitals: assume that ψ m = ψ e Then Ψ(1,2)= ψ e (1) ψ e (2) - ψ e (1) ψ e (2) = 0 Thus two electrons cannot be in identical spinorbitals Note that if ψ m = e iα ψ e where α is a constant phase factor, we still get zero Ch120a-Goddard-L02 59

Consider some simple cases: orthogonality Consider the wavefunction Ψ old (1,2)= ψ e (1) ψ m (2) - ψ m (1) ψ e (2) where the spinorbitals ψ m and ψ e are orthogonal hence <ψ m ψ e > = 0 Define a new spinorbital θ m = ψ m + λ ψ e (ignore normalization) In which θ m is NOT orthogonal to ψ e. Then Ψ new (1,2)= ψ e (1) θ m (2) - θ m (1) ψ e (2) = ψ e (1) θ m (2) + λ ψ e (1) ψ e (2) - θ m (1) ψ e (2) - λ ψ e (1) ψ e (2) = ψ e (1) ψ m (2) - ψ m (1) ψ e (2) =Ψ old (1,2) Thus any overlap between 2 spinorbitals is eliminated by the Pauli Principle. We say that the PP leads to orthogonality of spinorbitals for different electrons, <ψ i ψ j > = δ ij = 1 if i=j Ch120a-Goddard-L02 =0 if i j 60

Consider some simple cases: nonuniqueness Starting with the wavefunction Ψ old (1,2)= ψ e (1) ψ m (2) - ψ m (1) ψ e (2) Consider the new spinorbitals θ m and θ e where ψ m ψ e θ m = (cosα) ψ m + (sinα) ψ e θ e = (cosα) ψ e - (sinα) ψ m Note that <θ i θ j > = δ ij Then Ψ new (1,2)= θ e (1) θ m (2) - θ m (1) θ e (2) = α +(cosα) 2 ψ e (1)ψ m (2) +(cosα)(sinα) ψ e (1)ψ e (2) θ m -(sinα)(cosα) ψ m (1) ψ m (2) - (sinα) 2 ψ m (1) ψ e (2) -(cosα) 2 ψ m (1) ψ e (2) +(cosα)(sinα) ψ m (1) ψ m (2) -(sinα)(cosα) ψ e (1) ψ e (2) +(sinα) 2 ψ e (1) ψ m (2) [(cosα) 2 +(sinα) 2 ] [ψ e (1)ψ m (2) - ψ m (1) ψ e (2)] =Ψ old (1,2) θ e α Thus linear combinations of the spinorbitals do not change Ψ(1,2) Ch120a-Goddard-L02 61

Determinants The determinant of a matrix is defined as The determinant is zero if any two columns (or rows) are identical Adding some amount of any one column to any other column leaves the determinant unchanged. Thus each column can be made orthogonal to all other columns. (and the same for rows) The above properties are just those of the Pauli Principle Thus we will take determinants of our wavefunctions. Ch120a-Goddard-L02 62

The antisymmetrized wavefunction Now put the spinorbitals into the matrix and take the deteminant Where the antisymmetrizer can be thought of as the determinant operator. Similarly starting with the 3!=6 product wavefunctions of the form The only combination satisfying the Pauil Principle is Ch120a-Goddard-L02 63

Example: Interchanging electrons 1 and 3 leads to From the properties of determinants we know that interchanging any two columns (or rows) that is interchanging any two spinorbitals, merely changes the sign of the wavefunction Guaranteeing that the Pauli Principle is always satisfied Ch120a-Goddard-L02 64

Historical Note The idea of combining functions to form say a determinant in the same way we would combine numbers algebraically goes back to a German Mathemetician, Stäckel in ~1905 Later Heisenberg introduced the concept of an antisymmetrized product of orbitals in discussing magnetic systems in ~1925 For some reason they are now called Slater determinants, I believe this is because Slater showed that the simplest allowed product wavefunction would have this form, in ~1928 Warning: these dates are my recollections from reading these papers when I was a graduate student, they may be a bit off Ch120a-Goddard-L02 65