Lecture 12 January 31, 2011 Symmetry, Homonuclear diatomics

1. Lecture 12 January 31, 2011 Symmetry, Homonuclear diatomics Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy 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. Last time

3. The role of symmetry in QM In this course we are concerned with the solutions of the Schrodinger equation, HΨ=EΨ, but we do actually want to solve this equation. Instead we want to extract the maximum information about the solutions without solving it. Symmetry provides a powerful tool for doing this. Some transformation R1is called a symmetry transformation if it has the property that R1(HΨ)=H(R1Ψ) The set of all possible symmetries transformations of H are collected into what is called a Group.

4. The definition of a Group 1). Closure: If R1,R2 e G (both are symmetry transformations) then R2 R1is also a symmetry transformation, R2 R1e G 2. Identity. The do-nothing operator or identity, R1 = ee G isclearly is a symmetry transformation 3. Associativity. If (R1R2)R3 =R1(R2R3). 4. Inverse. If R1e G then the inverse, (R1)-1e G,where the inverse is defined as (R1)-1R1 = e.

5. The degenerate eigenfunctions of H form a representation If HΨ=EΨ then H(R1Ψ)= E(R1Ψ) for all symmetry transformations of H. Thus the transformations amount the n denegerate functions, {S=(RiΨ), where RiΨieG} lead to a set of matrices that multiply in the same way at the group operators. The Mathematicians say that these functions form a basis for a representation of G. Of course the functions in S may not all be different, so that this representation can be reduced. The mathematicians went on to show that one could derive a set of irreducible representations that give all possible symmetries for the H. reorientations from which one can construct any possible.

6. Example, an atom. For an atom any rotations about any axis passing through the nucleus is a symmetry transformation. This leads to the group denoted as SO(3) by the mathematicians [O(3) indicates 3 three-dimensional real space, S because the inversion is not included). The irreducible representations of O(3) are labeled as S (non degenerate) and referred to as L=0 P (3 fold degenerate) and referred to as L=1 D (5 fold degenerate) and referred to as L=2 F (7 fold degenerate) and referred to as L=3 G (9 fold degenerate) and referred to as L=4

7. H2O, an example of C2v consider the nonlinear H2A molecule, with equal bond lengths, e.g. H2O, CH2, NH2 • The symmetry transformations are • e for einheit (unity) xx, yy, zz • C2z, rotation about the z axis by 2p/2=180º, x-x, y-y, zz • sxz, reflection in the xz plane, xx, y-y, zz • syz, reflection in the yz plane, x-x, yy, zz • Which is denoted as the C2v group.

8. Stereographic projections e e y y x x C2z C2z syz sxz sxz syz syz C2z C2z syz sxz sxz Consider the stereographic projection of the points on the surface of a sphere onto a plane, where positive x are circles and negative x are squares. Start with a general point, denoted as e and follow where it goes on various symmetry operations. This make relations between the symmetry elements transparent. e.g. C2zsxz= syz Combine these as below to show the relationships C2v

9. The character table for C2v Since (C2z)2 = e, (sxz)2 = e, (syz)2 = e We expect wavefunctions to be ±1 under each operation name 1-e N-e a1 a2 b1 b2 Since C2zsxz= syz the symmetries for syz are already implied by C2zsxz. Thus there are only 4 possible symmetries.

10. Symmetries for NH2 Ψ1 = A{(N2pya)[(Npx )(Hx)+(Hx)(Npx)](ab-ba)[(Npz)(Hz)+(Hz)(Npz)](ab-ba)} NHx bond NHz bond Applying sxz to the wavefunction leads to Ψ2 = A{(N2py)0[(NpR)(HR)+(HR)(NpR)](ab-ba)[(NpL)(HL)+(HL)(NpL)](ab-ba)} Each term involves transposing two pairs of electrons, e.g., 13 and 24 as interchanging electrons. Since each interchange leads to a sign change we find that sxzΨ1 = Ψ2 = Ψ1 Thus interchanging a bond pair leaves Ψ invariant Also the N(1s)2 and (2s)2 pairs are invariant under all operations

11. NH2 symmetry continued Consider now the singly occupied Npx orbital C2z changes the sign but, sxz does not. Thus Npx transforms as b1 and The total wavefunction transforms as B1. Include the S= 1/2 , leads to The 2B1 state 1-e N-e a1 a2 b1 b2

12. Symmetries for H2O and CH2 CH2 NH2 H2O A{(O2py)2[(Opx )(Hx)+(Hx)(Opx)](ab-ba)[(Opz)(Hz)+(Hz)(Opz)](ab-ba)} H2O OHx bond OHz bond CH2 A{(C2py)0[(Cpx )(Hx)+(Hx)(Cpx)](ab-ba)[(Cpz)(Hz)+(Hz)(Cpz)](ab-ba)} Since have enen number electrons in 2py, wavefunction is invarient under all symmetry transformations, thus must be 1A1.

13. Now do triplet state of CH2 A{(C2sa)1(2pxa)1[(CpL )(HL)+(HL)(CpL)](ab-ba)[(CpR)(HR)+(HR)(CpR)](ab-ba)} CHL bond CHR bond Soon we will consider the triplet state of CH2 in which one of the 2s nonbonding electrons (denoted as s to indicate symmetric with respect to the plane of the molecule) is excited to the 2px orbital (denoted as p to indicate antisymmetric with respect to the plane) Since we know that the two CH bonds are invariant under all symmetry operations, from now on we will write the wavefunction as y z A{[(CHL)2(CHR)2](Csa)1(Cpa)1} Here s is invariant (a1) while p transforms as b1. Since both s and p are unpaired the ground state is triplet or S=1 p=2px s=2s Thus the symmetry of triplet CH2 is 3B1

14. Second example, C3v, with NH3 as the prototype z x A{[(Npy )(Hy)+(Hy)(Npy)](ab-ba)[(Npx )(Hx)+(Hx)(Npx)](ab-ba)[(Npz)(Hz)+(Hz)(Npz)] (ab-ba)} NHx bond NHc bond NHb bond We will consider a system such as NH3, with three equal bond lengths. Here we will take the z axis as the symmetry axis and will have one H in the xz plane. The other two NH bonds will be denoted as b and c.

15. Symmetry elements for C3v b b b b b b x x x x x x z x c c c c c c Take the z axis out of the plane. The six symmetry operations are: C3 e C32= C3-1 sxzC32 sxz sxzC3

16. The C3v symmetry group b sxzC32 C3 e x sxz sxzC3 C32= C3-1 c The sxzC3 transformation corresponds to a reflection in the bz plane (which is rotated by C3 from the xz plane) and the sxzC32transformation corresponds to a reflection in the cz plane (which is rotated by C32 from the xz plane). Thus these 3 reflections are said to belong to the same class. Since {C3 and C3-1 do similar things and are converted into each other by sxz we say that they are in the same class.

17. The character table for C3v The E symmetry (irreducible representation) is of degree 2, which means that if φpx is an eigenfunction of the Hamiltonian, the so is φpy and they are degenerate. This set of degenerate functions would be denoted as {ex,ey} and said to belong to the E irreducible representation. The characters in this table are used to analyze the symmetries, but we will not make use of this until much later in the course. Thus an atom in a P state, say C(3P) at a site with C3v symmetry, would generally split into 2 levels, {3Px and 3Py} of 3E symmetry and 3Pz of 3A1 symmetry.

18. Application for C3v, NH3 z x sLP We will write the wavefunction for NH3 as x =A{(sLP)2[(NHb bond)2(NHc bond)2(NHx bond)2]} c b where we combined the 3 Valence bond wavefunctions in 3 pair functions and we denote what started as the 2s pair as sLP Consider first the effect of sxz. This leaves the NHx bond pair invariant but it interchanges the NHb and NHc bond pairs. Since the interchange two pairs of electrons the wavefunction does not change sign. Also the sLP orbital is invariant.

19. Consider the effect of C3 Next consider the C3 symmetry operator. It does not change sLP. It moves the NHx bond pair into the NHb pair, moves the NHb pair into the NHc pair and moves the NHc pair into the NHx pair. A cyclic permutation on three electrons can be written as (135) = (13)(35). For example. φ(1)φ(3)φ(5) φ(3)φ(5)φ(1) (say this as e1 is replaced by e3 is replaced by e5 is replace by 1) This is the same as φ(1)φ(3)φ(5) φ(1)φ(5)φ(3)  φ(3)φ(5)φ(1) The point is that this is equivalent to two transpostions. Hence by the PP, the wavefunction will not change sign. Since C3 does this cyclic permutation on 6 electrons,eg (135)(246)= (13)(35)(24)(46), we still get no sign change. (135) (13) (35) Thus the wavefunction for NH3 has 1A1 symmetry

20. x Linear molecules, C∞v symmetry z O=C=O Consider the wavefunction for one electron in a linear molecule. Here we use polar coordinates, r = sqrt(x2+y2), a, z (axis along z) Since the wavefunction has period of 2p in a, the a dependence of any wavefunction can be expanded as a Fourier series, φ(r,a,z)=f(r,z){a0 +S m=1m=∞ [am cos ma + bm sin ma]

21. x Linear molecules, C∞v symmetry z O=C=O Consider the wavefunction for one electron in a linear molecule. Here we use polar coordinates, r = sqrt(x2+y2), a, z (axis along z) Since the wavefunction has period of 2p in a, the a dependence of any wavefunction can be expanded as a Fourier series, φ(r,a,z)=f(r,z){a0 +S m=1m=∞ [am cos ma + bm sin ma] Clearly the kinetic energy will increase with m, so that for the samef(r,z), we expect m=0 lowest, then m=±1, then m=±2, m=±3, etc

22. x Linear molecules, C∞v symmetry z O=C=O Consider the wavefunction for one electron in a linear molecule. Here we use polar coordinates, r = sqrt(x2+y2), a, z (axis along z) Since the wavefunction has period of 2p in a, the a dependence of any wavefunction can be expanded as a Fourier series, φ(r,a,z)=f(r,z){a0 +S m=1m=∞ [am cos ma + bm sin ma] Clearly the kinetic energy will increase with m, so that for the samef(r,z), we expect m=0 lowest, then m=±1, then m=±2, m=±3, etc Also if we rotate the molecule about the z axis by some angle b, the states with the same m get recombined [cos m(a+b)] = (cos ma)(cosmb) – (sin ma)(sin mb) [sin m(a+b)] = (sin ma)(cosmb) + (cos ma)(sin mb) Which means that the wavefunctions with the same m are degenerate

23. x C∞v symmetry group z O=C=O The symmetry operators are: Rz(a): counterclockwise rotation by an angle a about the z axis sxz: reflection in the xz plane (this takes +a into –a)

24. x C∞v symmetry group z O=C=O The symmetry operators are: Rz(a): counterclockwise rotation by an angle a about the z axis sxz: reflection in the xz plane (this takes +a into –a) s’ = Rz(a) sxz Rz(-a); reflection in a plane rotated by an angle a from the xz plane (there are an infinite number of these)

25. s’ = Rz(a) sxz Rz(-a) is a reflection in the plane rotated by an angle a from the xz plane y y y y e e e e x x x x Take the z axis out of the plane. The six symmetry operations are: y e a a x -a C-a Ca sxz y a e x s’ s’

26. x C∞v symmetry group name 1-e N-e s p d f g z O=C=O The symmetry operators are: Rz(a): counterclockwise rotation by an angle a about the z axis sxz: reflection in the xz plane (this takes +a into –a) s’ = Rz(a) sxz Rz(-a); reflection in a plane rotated by an angle a from the xz plane (there are an infinite number of these) e: einheit (unity) This group is denoted as C∞v,The character table (symmetries) are

27. The symmetry functions for C∞v Lower case letters are used to denote one-electron orbitals

28. Application to FH The ground state wavefunction of HF is A{(F2px)2(F2py)2[(Fpz)(H)+(H)(Fpz)](ab-ba)} In C∞v symmetry, the bond pair is s (m=0),while the px and py form a set of p orbitals (m=+1 and m=-1). Consider the case of up spin for both px and py Ψ(1,2) = A{φxaφya}=(φxφy- φyφx) aa Rotating by an angle g about the z axis leads to φa =cosgφx +singφy and φb =cosgφy -singφx This leads to (φaφb- φbφa) = [(cosg)2 +(sing)2] }=(φxφy- φyφx) Thus (φxφy- φyφx) transforms as S.

29. Continuing with FH Thus the (px)2(py)2 part of the HF wavefunction A{(F2px)2(F2py)2[(Fpz)(H)+(H)(Fpz)](ab-ba)} Since both aa and bb transform like S; the total wavefunction transforms as S The symmetry table, demands that we also consider the symmetry with respect to reflection in the xz plane. Here px is unchanged while py changes sign. Since there are two electrons in py the wavefunction is invariant. Thus the ground state of FH has 1S+ symmetry

30. x z Next consider the ground state of OH We write the two wavefunctions for OH as 2Px Ψx=A{(sOHbond)2[(pxa)(pya)(pxb)]} Ψy=A{(sOHbond)2[(pxa)(pya)(pyb)]} We saw above that A{(pxa)(pya)} transforms like S. thus we need examine only the transformations of the downspin orbital. But this transforms like p. 2Py Thus the total wavefunction is 2P. Another way of describing this is to note that A{(px)2(py)2} transforms like S and hence one hole in a (p)4shell, (p)3 transforms the same way as a single electron, (p)1

31. x z Now consider the ground state of NH A{(NH bond)2(N2pxa)(N2pya)} We saw earlier that up-spin in both x and y leads to S symmetry. Considering now the reflection, sxz, we see that with just one electron in py, we now get S-. Thus the ground state of NH is 3S-.

32. Now consider Bonding H atom to all 3 states of C x z Bring H1s along z axis to C and consider all 3 spatial states. (2px)(2pz) O 2pz singly occupied. H1s can get bonding Get S= ½ state, Two degenerate states, denote as 2P (2py)(2pz) (2px)(2py) No singly occupied orbital for H to bond with

33. x z Ground state of CH (2P) The full wavefunction for the bonding state 2Px A{(2s)2(OHs bond)2(O2pxa)1} 2Py A{(2s)2(OHs bond)2(O2pya)1}

34. x z Bond a 2nd H atom to the ground state of CH Starting with the ground state of CH, we bring a 2nd H along the x axis. Get a second covalent bond This leads to a 1A1 state. No unpaired orbtial for a second covalent bond.

35. x z θe Re Analyze Bond in the ground state of CH2 Ground state has 1A1 symmetry. For optimum bonding, the pz orbital should point at the Hz while the px orbital should point at the Hx. Thus the bond angle should be 90º. As NH2 (103.2º) and OH2 (104.5º), we expect CH2 to have bond angle of ~ 102º

36. But, the Bending potential surface for CH2 1B1 1Dg 1A1 3B1 3Sg- 9.3 kcal/mol The ground state of CH2 is the 3B1 state not 1A1. Thus something is terribly wrong in our analysis of CH2

37. Short range Attractive interaction sz with H Compare bonding in BeH+ and BeH Long range Repulsive interaction with H BeH TA’s check numbers, all from memory 2 eV BeH+ has long range attraction no short range repulsion 3 eV 1 eV BeH+ 1eV Repulsive orthogonalization of zs with sz H

38. Compare bonding in BeH and BeH2 BeH+ MgH+ 1S+ 3.1 eV R=1.31A 2.1 eV R=1.65 A 1.34 eV R=1.73A 2.03 eV R=1.34A 2S+ ~3.1 eV 1S+ linear ~2.1 e Expect linear bond in H-Be-H and much stronger than the 1st bond Expect bond energy similar to BeH+, maybe stronger, because the zs orbital is already decoupled from the sz. TA’s check numbers, all from memory Cannot bind 3rd H because no singly occupied orbitals left.

39. Energetics

40. Energy for 2 electron product wavefunction Consider the product wavefunction Ψ(1,2) = ψa(1) ψb(2) And the Hamiltonian H(1,2) = h(1) + h(2) +1/r12 + 1/R In the details slides next, we derive E = < Ψ(1,2)| H(1,2)|Ψ(1,2)>/ <Ψ(1,2)|Ψ(1,2)> E = haa + hbb + Jab + 1/R where haa =<a|h|a>, hbb =<b|h|b> Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)>=ʃ [ψa(1)]2 [ψb(1)]2/r12 Represent the total Coulomb interaction between the electron density ra(1)=| ψa(1)|2 and rb(2)=| ψb(2)|2 Since the integrand ra(1) rb(2)/r12 is positive for all positions of 1 and 2, the integral is positive, Jab > 0

41. Details in deriving energy: normalization First, the normalization term is <Ψ(1,2)|Ψ(1,2)>=<ψa(1)|ψa(1)><ψb(2) ψb(2)> Which from now on we will write as <Ψ|Ψ> = <ψa|ψa><ψb|ψb> = 1 since the ψi are normalized Here our convention is that a two-electron function such as <Ψ(1,2)|Ψ(1,2)> is always over both electrons so we need not put in the (1,2) while one-electron functions such as <ψa(1)|ψa(1)> or <ψb(2) ψb(2)> are assumed to be over just one electron and we ignore the labels 1 or 2

42. Details of deriving energy: one electron terms Using H(1,2) = h(1) + h(2) +1/r12 + 1/R We partition the energy E = <Ψ| H|Ψ> as E = <Ψ|h(1)|Ψ> + <Ψ|h(2)|Ψ> + <Ψ|1/R|Ψ> + <Ψ|1/r12|Ψ> Here <Ψ|1/R|Ψ> = <Ψ|Ψ>/R = 1/R since R is a constant <Ψ|h(1)|Ψ> = <ψa(1)ψb(2) |h(1)|ψa(1)ψb(2)> = = <ψa(1)|h(1)|ψa(1)><ψb(2)|ψb(2)> = <a|h|a><b|b> = ≡ haa Where haa≡ <a|h|a> ≡ <ψa|h|ψa> Similarly <Ψ|h(2)|Ψ> = <ψa(1)ψb(2) |h(2)|ψa(1)ψb(2)> = = <ψa(1)|ψa(1)><ψb(2)|h(2)|ψb(2)> = <a|a><b|h|b> = ≡ hbb The remaining term we denote as Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)> so that the total energy is E = haa + hbb + Jab + 1/R

43. The energy for an antisymmetrized product, Aψaψb The total energy is that of the product plus the exchange term which is negative with 4 parts Eex=-< ψaψb|h(1)|ψb ψa >-< ψaψb|h(2)|ψb ψa >-< ψaψb|1/R|ψb ψa > - < ψaψb|1/r12|ψbψa > The first 3 terms lead to < ψa|h(1)|ψb><ψbψa >+ <ψa|ψb><ψb|h(2)|ψa >+ <ψa|ψb><ψb|ψa>/R But <ψb|ψa>=0 Thus all are zero Thus the only nonzero term is the 4th term: -Kab=- < ψaψb|1/r12|ψbψa >which is called the exchange energy (or the 2-electron exchange) since it arises from the exchange term due to the antisymmetrizer. Summarizing, the energy of the Aψaψb wavefunction for H2 is E = haa + hbb + (Jab –Kab) + 1/R

44. The energy of the antisymmetrized wavefunction The total electron-electron repulsion part of the energy for any wavefunction Ψ(1,2) must be positive Eee =∫ (d3r1)((d3r2)|Ψ(1,2)|2/r12 > 0 This follows since the integrand is positive for all positions of r1 and r2 then We derived that the energy of the Aψa ψb wavefunction is E = haa + hbb + (Jab –Kab) + 1/R Where the Eee = (Jab –Kab) > 0 Since we have already established that Jab > 0 we can conclude that Jab > Kab > 0

45. Separate the spinorbital into orbital and spin parts Since the Hamiltonian does not contain spin the spinorbitals can be factored into spatial and spin terms. For 2 electrons there are two possibilities: Both electrons have the same spin ψa(1)ψb(2)=[Φa(1)a(1)][Φb(2)a(2)]= [Φa(1)Φb(2)][a(1)a(2)] So that the antisymmetrized wavefunction is Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)a(2)]= =[Φa(1)Φb(2)- Φb(1)Φa(2)][a(1)a(2)] Also, similar results for both spins down Aψa(1)ψb(2)= A[Φa(1)Φb(2)][b(1)b(2)]= =[Φa(1)Φb(2)- Φb(1)Φa(2)][b(1)b(2)] Since <ψa|ψb>= 0 = < Φa| Φb><a|a> = < Φa| Φb> We see that the spatial orbitals for same spin must be orthogonal

46. Energy for 2 electrons with same spin The total energy becomes E = haa + hbb + (Jab –Kab) + 1/R where haa≡ <Φa|h|Φa> and hbb ≡ <Φb|h|Φb> where Jab= <Φa(1)Φb(2) |1/r12 |Φa(1)Φb(2)> We derived the exchange term for spin orbitals with same spin as follows Kab ≡ <ψa(1)ψb(2) |1/r12 |ψb(1)ψa(2)> `````= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)><a(1)|a(1)><a(2)|a(2)> ≡ Kab where Kab≡ <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)> Involves only spatial coordinates.

47. Energy for 2 electrons with opposite spin Now consider the exchange term for spin orbitals with opposite spin Kab ≡ <ψa(1)ψb(2) |1/r12 |ψb(1)ψa(2)> `````= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)><a(1)|b(1)><b(2)|a(2)> = 0 Since <a(1)|b(1)> = 0. Thus the total energy is Eab = haa + hbb + Jab + 1/R With no exchange term unless the spins are the same Since <ψa|ψb>= 0 = < Φa| Φb><a|b> There is no orthogonality condition of the spatial orbitals for opposite spin electrons In general < Φa| Φb> =S, where the overlap S ≠ 0

48. Summarizing: Energy for 2 electrons When the spinorbitals have the same spin, Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)a(2)] The total energy is Eaa = haa + hbb + (Jab –Kab) + 1/R But when the spinorbitals have the opposite spin, Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)b(2)]= The total energy is Eab = haa + hbb + Jab + 1/R With no exchange term Thus exchange energies arise only for the case in which both electrons have the same spin

49. Consider further the case for spinorbtials with opposite spin Neither of these terms has the correct permutation symmetry separately for space or spin. But they can be combined [Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)]= A[Φa(1)Φb(2)][a(1)b(2)]-A[Φb(1)Φa(2)][a(1)b(2)] Which describes the Ms=0 component of the triplet state [Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)]= A[Φa(1)Φb(2)][a(1)b(2)]+A[Φb(1)Φa(2)][a(1)b(2)] Which describes the Ms=0 component of the singlet state Thus for the ab case, two Slater determinants must be combined to obtain the correct spin and space permutational symmetry

50. Consider further the case for spinorbtials with opposite spin The wavefunction [Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)] Leads directly to 3Eab = haa + hbb + (Jab –Kab) + 1/R Exactly the same as for [Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)a(2)] [Φa(1)Φb(2)-Φb(1)Φa(2)][b(1)b(2)] These three states are collectively referred to as the triplet state and denoted as having spin S=1 The other combination leads to one state, referred to as the singlet state and denoted as having spin S=0 [Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)] If <Φa|Φb> = 0 then 1Eab = haa + hbb + (Jab +Kab) + 1/R