Lecture 3, January 14, 2011 energetics

1. Lecture 3, January 14, 2011 energetics Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy Course number: Ch120a Hours: 2-3pm 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 Teaching Assistants: Wei-Guang Liu <wgliu@wag.caltech.edu> Caitlin Scott <cescott@caltech.edu>

2. Course schedule Friday January 14: L3 and L4 Monday January 17: caltech holiday (MLKing) Wednesday January 19: wag L5 and L6 Friday January 21: wag L7 and L8, caught up Monday January 24: wag L9 Wednesday January 26: wag L10 and L11 Friday January 28: wag participates in a retreat for our nanotechnology project with UCLA Back on schedule Monday January 31: wag L12

3. Last time

4. Molecular Orbitals: Alternative way to view states of H2 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=Re and add electrons u: antibonding g: bonding Put 2 electrons in 2 orbitals, get 4 two-electron states 1-electron molecular orbitals

5. Analyze gu and ug states in 2 electron space ug gu 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/r12| Φ(1,2)> Worst case is for z1=z2, along diagonal Never have z1=z2 great EE Maximum at z1=z2 terrible EE

6. Ground and excited states from MO analysis

7. 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:

8. compare ground state from VB and MO Ground state MO wavefunction is half covalent and half ionic OK for R=Requilbrium = Re, but not for R = ∞

9. Compare bond dissociation for VB and MO MO limit half covalent and half ionic H- + H+ Covalent limit H + H

10. VB and MO for u excited states Pure covalent, but antibonding Pure ionic, but bonding

11. Energies for H2 states based on atomic orbitals (z=1)

12. Summary 2nd Postulate QM EQM = KEQM + PEQM where for a system with a potential energy function, V(x,y,z,t) PEQM= < Φ| V|Φ>=∫Φ(x,y,z,t)*V(x,y,z,t)Φ(x,y,z,t)dxdydz Just like Classical mechanics except weight V with P=|Φ|2 KEQM = (Ћ2/2me) <(Φ·Φ> where <(Φ·Φ> ≡ ∫[(dΦ/dx)2 + (dΦ/dx)2 + (dΦ/dx)2] dxdydz We have assumed a normalized wavefunction, <Φ|Φ> = 1 The stability of the H atom was explained by this KE (proportional to the average square of the gradient of the wavefunction). Also we used the preference of KE for smooth wavefunctions to explain the bonding in H2+ and H2. So far we have been able to use the above expressions to reason qualitatively. However to actually solve for the wavefunctions requires the Schrodinger Eqn., which we derive next.

13. Alternative form for QM KE One dimensional KEQM = (Ћ2/2me)<(dΦ/dx)|(dΦ/dx)> integrate by parts: ∫(du/dx) (v) dx = -∫(u)(dv/dx)dx if u,v  0 at boundaries Let du/dx = (dΦ*/dx) and v = dΦ/dx then KEQM = - (Ћ2/2me) ∫(Φ*)(d2Φ/dx2) dx = - (Ћ2/2me) <Φ|(d2/dx2)| Φ> KEQM = <Φ| - (Ћ2/2me) (d2/dx2)| Φ> = <Φ| | Φ> Where KE operator is = - (Ћ2/2me) (d2/dx2) ^ ^ ^ ^ KE KE KE KE Goddard form standard form Three dimensions KEQM = (Ћ2/2me) <(Φ·Φ> where <(Φ·Φ> ≡ ∫[(dΦ/dx)2 + (dΦ/dx)2 + (dΦ/dx)2] dxdydz integrating by parts: KEQM = <Φ| - (Ћ2/2me)2| Φ> = <Φ| | Φ> Where KE operator is = - (Ћ2/2me) 2 2 = [(dΦ/dx)2 + (dΦ/dx)2 + (dΦ/dx)2] Goddard form standard form

14. 3rd Postulate of QM, the variational principle E Eap Eex The ground state wavefunction is the system, Φ, has the lowest possible energy out of all possible wavefunctions. Consider that Φex is the exact wavefunction with energy Eex = <Φ’|Ĥ|Φ’>/<Φ’|Φ’> and that Φap = Φex + dΦ is some other approximate wavefunction. Then Eap = <Φap|Ĥ|Φap>/<Φap|Φap> ≥ Eex This means that for sufficiently small dΦ, dE = 0 for all possible changes, dΦ We write dE/dΦ = 0 for all dΦ This is called the variational principle. For the ground state, d2E/dΦ ≥ 0 for all possible changes

15. Summary deriviation of Schrödinger Equation ^ ^ ^ KE KE KE EQM = <Φ| | Φ> + < Φ| V|Φ> = <Φ| Ĥ | Φ> where the Hamiltonian is Ĥ ≡ + V and = - (Ћ2/2me)2 And we assume a normalized wavefunction, <Φ|Φ> = 1 V(x,y,z,t) is the (classical) potential energy for the system Consider arbitraryΦap = Φex + dΦ and require that dE= Eap – Eex = 0 Get <dΦ|Ĥ-Eex|Φex>] = ∫ dΦ*[(Ĥ-Eex)Φex] = 0 for all possible dΦ This [(Ĥ-Eex)Φex] = 0 or the Schrödinger equation Ĥ Φex = EexΦex The exact ground state wavefunction is a solution of this equation

16. Excited states The Schrödinger equation Ĥ Φk = EkΦk Has an infinite number of solutions or eigenstates (German for characteristic states), each corresponding to a possible exact wavefunction for an excited state For example H atom: 1s, 2s, 2px, 3dxy etc Also the g and u states of H2+ and H2. These states are orthogonal: <Φj|Φk> = djk= 1 if j=k = 0 if j≠k Note < Φj| Ĥ|Φk> = Ek< Φj|Φk> = Ekdjk We will denote the ground state as k=0 The set of all eigenstates of Ĥ is complete, thus any arbitrary function Ө can be expanded as Ө = Sk Ck Φk where <Φj| Ө>=Cj or Ө = SkΦk <Φk| Ө>

17. The Hamiltonian for H2+ Coordinates of H atom r 1 1 1 For H atom the Hamiltonian is Ĥ = - (Ћ2/2me)2– e2/r or Ĥ = - ½ 2– 1/r (in atomic units) For H2+ molecule the Hamiltonian (in atomic units) is Ĥ = - ½ 2+ V(r) where Coordinates of H2+ We will rewrite this as

18. The Schrödinger Equation for H2+ The exact (electronic wavefunction of H2+ is obtained by solving Here we can ignore the 1/R term (not depend on electron coordinates) to write where e is the electronic energy Then the total energy E becomes R E= e + 1/R Since v(r) depends on R, the wavefunction φ depends on R. E(R) Thus for each R we solve for φ and e and add to 1/R to get the total energy E(R) Born-Oppenheimer approximation

19. Symmetry Considerations ^ ^ I I After inversion the electron is remains a distance of ra from one of the nuclei and rb from the other, but their identities are transposed. Thus the potential energy, v(r) is unchanged by inversion, v(-x,-y,-z) ≡ v( r) = v(r) where r is considered as the 3D vector with components x,y,z The operation of inversion (denoted as ) through the origin of a coordinate system changes the coordinates as x  -x y  -y z  -z Taking the origin of the coordinate system as the bond midpoint, inversion changes the electronic coordinates as illustrated. Thus we say that v(r) is invariant under inversion

20. Considering symmetry ^ ^ ^ I I I Under inversion the kinetic energy terms in the Hamiltonian are also unchanged Hence the full Hamiltonian is invariant under inversion h(-x,-y,-z) ≡ h( r) = h(x,y,z) = h(r) Now consider that we had solved hφ=eφ for the exact wavefunction φ and apply the inversion to both sides hφ= e φ Which we can rewrite as h(-r)φ(-r)=eφ(-r) But h(-r) = h(r) because of inversion symmetry Thus h(r)φ(-r)=eφ(-r)

21. 4th postulate of QM Consider the exact eigenstate of a system HΦ = EΦ and multiply the Schrödinger equation by some CONSTANT phase factor (independent of position and time) exp(ia) = eia eiaHΦ = H (eiaΦ) = E (eiaΦ) Thus Φ and (eiaΦ) lead to identical properties and we consider them to describe exactly the same state. 4th Postulate: wavefunctions differing only by a constant phase factor describe the same state

22. Continue with symmetry discussion We just derived that because h(-r) = h(r) Then for any eigenfunction φ(r)of h h(r)φ(r) = eφ(r) It must be that φ(-r) also is an eigenfunction of the same h with the same energy, e. h(r)φ(-r) = eφ(-r) φg(r) Thus for a system with inversion, each nondegenerate eigenstate is of either g or u inversion symmetry. φg(r) = + φg(r) g for gerade or even φu(r) = - φu(r) u for ungerade or odd φu(r)

23. Now consider symmetry for H2 molecule For H2 we use the coordinate system at the right. Using the same conventions and assumptions as for H2+ leads to where 1/r12 is the interaction between the two electrons and Contains all terms depending only on the coordinates of electron 1 inverting the coordinates of electron 1 leaves h(1) invariant but not H(1,2)

24. Invert coordinates of electron 1 As before h(1) is invariant and so is h(2), but clearly something is different for H2. The problem is 1/r12. Inverting electron 1 or 2 separately does not preserve the r12 distance.

25. For multielectron systems, the inversion symmetry applies only if we invert 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.

26. Symmetry for H2 ^ I Just as for H2+, inversion symmetry of H2 leads to the result that every eigenstate of H2 is either g or u Consider the VB wavefunctions = does NOT have symmetry But ] is I [ = + I ] is = - Which is why we labeled them as such

27. Now consider inversion symmetry for MO wavefunctions This is easy since each MO has inversion symmetry Thus

28. 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/r12 + 1/R = H(1,2) We will denote this transposition as t where tΦ(1,2) = Φ(2,1) Note that t2 = e, the einheit or identity operator t2Φ(1,2) = Φ(1,2) Thus t is of order 2 and the previous arguments on inversion apply equally to transposition

29. 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

30. Permutational symmetry for the u MO states of H2 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

31. permutational symmetry for H2 wavefunctions symmetric antisymmetric

32. Electron spin, 5th postulate QM b B=0 Increasing B a 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 This leads to the 5th 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 a for spin up and b for spin down. Thus the ground states of H atom are φ(xyz)a(spin) and φ(xyz)b(spin)

33. Electron spin b B=0 Increasing B a So far we have considered the electron as a point particle with mass, me, and charge, -e. In fact the electron has internal coordinates, that we refer to as spin, with two possible angular momenta. +½ or a or up-spin and -½ or b 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. DE = -gBzsz Now the wavefunction of an atom is written as a spinorbital ψ(r,s) where r refers to the vector of 3 spatial coordinates, x,y,z ands refers to the internal spin coordinates

34. 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)a, with up-spin, ms = +1/2 φ(r)b, with down-spin, ms = -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 H2+ the ground state becomes φg(r)a and φg(r)b While the excited state becomes φu(r)a and φu(r)b

35. Spin states for 2-electron systems Since each electron can have up or down spin, any two-electron system, such as H2 molecule will lead to 4 possible spin states each with the same energy Φ(1,2) a(1) a(2) Φ(1,2) a(1) b(2) Φ(1,2) b(1) a(2) Φ(1,2) b(1) b(2) This immediately raises an issue with permutational symmetry Since the Hamiltonian is invariant under interchange of the spin for electron 1 and the spin for electron 2, the two-electron spin functions must be symmetric or antisymmetric with respect to interchange of the spin coordinates, s1and s2 Symmetric spin Neither symmetric nor antisymmetric Symmetric spin

36. Spin states for 2 electron systems Combining the two-electron spin functions to form symmetric and antisymmetric combinations leads to Φ(1,2) a(1) a(2) Φ(1,2) [a(1) b(2) + b(1) a(2)] Φ(1,2) b(1) b(2) Φ(1,2) [a(1) b(2) - b(1) a(2)] Adding the spin quantum numbers, ms, to obtain the total spin projection, MS = ms1 + ms2 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 component, 0. It is called a singlet state. MS +1 0 -1 0 Symmetric spin Antisymmetric spin

37. Spinorbitals The Hamiltonian does not depend on spin the spatial and spin coordinates are independent. Hence the total wavefunction can be written as a product of a spatial wavefunction, φ(s), called an orbital, and a spin function, х(s) = a or b. We refer to the composite as a spinorbital ψ(r,s) = φ(s) х(s) where r refers to the vector of 3 spatial coordinates, x,y,z whiles to the internal spin coordinates.

38. spinorbitals for two-electron systems Thus for a two-electron system with independent electrons, the wavefunction becomes Ψ(1,2) = Ψ(r1,s1,r2,s2) = ψa(r1,s1) ψb(r2,s2) = φa(r1) хa(s1) φb(r2) хb(s2) =[φa(r1) φb(r2)][хa(s1) хb(s2)] Where the last term factors the total wavefunction into space and spin parts

39. Permutational symmetry again For a two-electron system the Hamiltonian is invariant (unchanged) upon transposition of the electrons (changing both spatial and spin coordinates) H(2,1) = H(1,2) Again the simultaneous transposition of space-spin coordinats 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

40. Permutational symmetry continued Factoring the wave function as spatial and spin coordinates Ψ(1,2) = (1,2)(1,2) We know that the H(1,2) is separately unchanged by transposing either just the spacial coordinates or the spin coordinates Thus (2,1) = +(1,2) or (2,1) = -(1,2) and either (2,1) = +(1,2) or (2,1) = -(1,2)

41. Permutational symmetry, summary Our Hamiltonian for H2, 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: r1  r2 Spin only: s1 s2 Space and spin simultaneously: (r1,s1)  (r2,s2) 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

42. Permutational symmetries for H2 and He H2 Have 4 degenerate g ground states for H2 Have 4 degenerate u excited states for H2 He Have 4 degenerate ground state for He

43. Permutational symmetries for H2 and He H2 Have 4 degenerate g ground states for H2 BUT THIS IS ALL WRONG The g state of H2 is nondegenerate The u excited state is 3 fold degenerate The He ground state is nondegenerate How can we fix this obvious disagreement with experiment! Have 4 degenerate u excited states for H2 He Have 4 degenerate ground state for He

44. Permutational symmetries for H2 and He H2 He the only states observed are those for which the wavefunction changes sign upon transposing all coordinates of electron 1 and 2 Leads to the 6th postulate of QM

45. The 6th 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 tijΨ = - Ψ for all I and j

46. Implications of the Pauli Principle Consider two independent electrons, 1 on the earth described by ψe(1) and 2 on the moon described by ψm(2) Ψ(1,2)= ψe(1) ψm(2) And test whether this satisfies the Pauli Principle Ψ(2,1)= ψm(1) ψe(2) ≠ - ψe(1) ψm(2) Thus the Pauli Principle does NOT allow the simple product wavefunction for two independent electrons

47. Quick fix to satisfy the Pauli Principle Combine the product wavefunctions to form a symmetric combination Ψs(1,2)= ψe(1) ψm(2) + ψm(1) ψe(2) And an antisymmetric combination Ψa(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2) We see that t12Ψs(1,2) = Ψs(2,1) = Ψs(1,2) (boson symmetry) t12Ψa(1,2) = Ψa(2,1) = -Ψa(1,2) (Fermion symmetry) Thus for electrons, the Pauli Principle only allows the antisymmetric combination for two independent electrons

48. Consider some simple cases: identical spinorbitals Ψ(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2) Identical spinorbitals: assume that ψm = ψe Then Ψ(1,2)= ψe(1) ψe(2) - ψe(1) ψe(2) = 0 Thus two electrons cannot be in identical spinorbitals Note that if ψm = eia ψe where a is a constant phase factor, we still get zero

49. Consider some simple cases: orthogonality Consider the wavefunction Ψold(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2) where the spinorbitals ψm and ψe areorthogonal hence <ψm|ψe> = 0 Define a new spinorbital θm = ψm + lψe (ignore normalization) That is NOT orthogonal to ψe. Then Ψnew(1,2)= ψe(1) θm(2) - θm(1) ψe(2) = ψe(1) θm(2) + l ψe(1) ψe(2) - θm(1) ψe(2) - l ψe(1) ψe(2) = ψe(1) ψm(2) - ψm(1) ψe(2) =Ψold(1,2) Thus the Pauli Principle leads to orthogonality of spinorbitals for different electrons, <ψi|ψj> = dij = 1 if i=j =0 if i≠j

50. Consider some simple cases: nonuniqueness Starting with the wavefunction Ψold(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2) Consider the new spinorbitals θm and θe where θm = (cosa) ψm + (sina) ψe θe = (cosa) ψe - (sina) ψm Note that <θi|θj> = dij Then Ψnew(1,2)= θe(1) θm(2) - θm(1) θe(2) = +(cosa)2ψe(1)ψm(2) +(cosa)(sina) ψe(1)ψe(2) -(sina)(cosa) ψm(1) ψm(2) - (sina)2ψm(1) ψe(2) -(cosa)2ψm(1) ψe(2) +(cosa)(sina) ψm(1) ψm(2) -(sina)(cosa) ψe(1) ψe(2) +(sina)2ψe(1) ψm(2) [(cosa)2+(sina)2] [ψe(1)ψm(2) - ψm(1) ψe(2)] =Ψold(1,2) ψm ψe a θm θe a a Thus linear combinations of the spinorbitals do not change Ψ(1,2)