Lecture 5,6 January 15, 2010 - PowerPoint PPT Presentation

Lecture 5,6 January 15, 2010

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Lecture 5,6 January 15, 2010

Lecture 5,6 January 15, 2010

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1. Lecture 5,6 January 15, 2010 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> Ted Yu <tedhyu@wag.caltech.edu>

2. Course schedule Wednesday January 13, wag in Florida for review of Chevron catalysis program Friday January 15, 2pm L5 (regular time), make up for Jan. 13 Friday January 15, 3pm L6 (regular time) Monday Jan. 18 holiday, remember MLK, no lecture Wednesday Jan. 20, 2pm L7 regular time Friday Jan. 22, 2pm L8 (regular time)

3. Last time

4. For multielectron systems, inversion inverts all electron coordinates simultaneously We define I to be inversion symmetry of H2 leads to the result that every eigenstate of H2 is either g or u

5. Permutation Symmetry transposing the two electrons in H(1,2) must leave the Hamiltonian invariant since the electrons are identical H(2,1) = h(2) + h(1) + 1/r12 + 1/R = H(1,2) We will denote this 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 the previous arguments on inversion apply equally to transposition 1 2 Every exact two electron wavefunction must be either symmetric, s, or antisymmetric, a

6. permutational symmetry for H2 wavefunctions symmetric antisymmetric

7. Electron spin, 5th postulate QM b B=0 Increasing B a 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. 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 DE = -gBzsz In a magnetic field the ground state of H atom splits into two states

8. Spinorbitals The Hamiltonian does not depend on spin the spatial and spin coordinates are independent. Hence the 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. 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.

9. 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

10. 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) Thus for any eigenstate of H(1,2) Ψ(1,2) = E Ψ(1,2) We can write H(2,1) Ψ(2,1) = E Ψ(2,1) And hence H(1,2) Ψ(2,1) = E Ψ(2,1) Thus Ψ(2,1) is also an eigenfunction of H(1,2) with the same E as Ψ(1,2) If the state is nondegenerate, Ψ(2,1) = tΨ(1,2) = l Ψ(1,2) Where t is the transposition operator interchange all coordiantes (space and spin) of electrons 1 and 2 Since interchanging the electrons twice gets us back where we started, t2 = e, the einheit or unity or do-nothing operator. But this means that t2Ψ(1,2) = tΨ(2,1) = l tΨ(2,1) = l2Ψ(1,2) And hence l = ±1

11. Permutational symmetry continued • 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 But if we interchange only the spatial coordinates, we get the same results and also for the spin coordinates Thus factoring the wave function as spatial and spin coordinates Ψ(1,2) = (1,2)(1,2) We know that either (2,1) = +(1,2) or (2,1) = -(1,2) and (2,1) = +(1,2) or (2,1) = -(1,2)

12. 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

13. 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

14. 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 permutaions 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

15. 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

16. The 6th postulate of QM: the Pauli Principle For every state 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

17. 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

18. 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

19. 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

20. 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

21. 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)

22. 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.

23. 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

24. 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

25. 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

26. 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

27. Details of deriving energy: one electron termss 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

28. 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

29. 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

30. 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

31. 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.

32. Energy for 2 electrons with same 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

33. 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

34. 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

35. 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)] We will analyze the energy for this wavefunction next.

36. H atom, excited states r +Ze In atomic units, the Hamiltonian h = - (Ћ2/2me)2– Ze2/r Becomes h = - ½ 2– Z/r Thus we want to solve hφk = ekφk for all excited states k φnlm = Rnl(r) Zlm(θ,φ) product of radial and angular functions Ground state: R1s = exp(-Zr), Zs = 1 (constant)

37. The excited angular states of H atom, 1 nodal plane z z pz + px - + x - x φnlm = Rnl(r) Zlm(θ,φ) the excited angular functions, Zlm must have nodal planes to be orthogonal to Zs 3 cases with one nodal plane Z10=pz= r cosθ (zero in the xy plane) Z11=px= rsinθcosφ (zero in the yz plane) Z1-1=py= rsinθsinφ (zero in the xz plane) We find it useful to keep track of how often the wavefunction changes sign as the φ coordinate is increased by 2p = 360º If m=0, denote it as s: pz = ps If m=1, denote it as p: px, py = pp If m=2 we call it a d function If m=3 we call it a f function

38. The excited angular states of H atom, 2 nodal planes z dz2 + - - x + Z20 = dz2 = [3 z2 – r2 ] m=0, ds Z21 = dzx = zx =r2 (sinθ)(cosθ)cosφ Z2-1 = dyz = yz =r2 (sinθ)(cosθ)sinφ Z22 = dx2-y2 = x2 – y2 = r2 (sinθ)2 cos2φ Z22 = dxy = xy =r2 (sinθ)2 sin2φ m = 1, dp 57º m = 2, dd Where we used [(cosφ)2 – (sinφ)2]=cos2φ and 2cosφ sinφ=sin2φ Summarizing: one s angular function (no angular nodes) called ℓ=0 three p angular functions (one angular node) called ℓ=1 five d angular functions (two angular nodes) called ℓ=2 seven f angular functions (three angular nodes) called ℓ=3 nine g angular functions (four angular nodes) called ℓ=4 ℓ is referred to as the angular momentum quantum number There are (2ℓ+1) m values, Zℓm for each ℓ

39. Summarizing the angular functions • So far we have • one s angular function (no angular nodes) called ℓ=0 • three p angular functions (one angular node) called ℓ=1 • five d angular functions (two angular nodes) called ℓ=2 • Continuing we can form • seven f angular functions (three angular nodes) called ℓ=3 • nine g angular functions (four angular nodes) called ℓ=4 • where ℓ is referred to as the angular momentum quantum number • And there are (2ℓ+1) m values for each ℓ

40. The real (Zlm) and complex (Ylm) momentum functions Here the bar over m  negative

41. Excited radial functions Clearly the KE increases with the number of angular nodes so that s < p < d < f < g Now we must consider radial functions, Rnl(r) The lowest is R10 = 1s = exp(-Zr) All other radial functions must be orthogonal and hence must have one or more radial nodes, as shown here Zr = 7.1 Zr = 2 Zr = 1.9 Note that we are plotting the cross section along the z axis, but it would look exactly the same along any other axis. Here R20 = 2s = [Zr/2 – 1]exp(-Zr/2) and R30 = 3s = [2(Zr)2/27 – 2(Zr)/3 + 1]exp(-Zr/3)

42. Combination of radial and angular nodal planes ˉ ˉ ˉ R R R Combining radial and angular functions gives the various excited states of the H atom. They are named as shown where the n quantum number is the total number of nodal planes plus 1 The nodal theorem does not specify how 2s and 2p are related, but it turns out that the total energy depends only on n. Enlm = - Z2/2n2 The potential energy is given by PE = - Z2/2n2 ≡ -Z/ , where =n2/Z Thus Enlm = - Z/(2 ) angular nodal planes Size (a0) radial nodal planes total nodal planes name 1s 0 0 0 1.0 2s 1 1 0 4.0 2p 1 0 1 4.0 3s 2 2 0 9.0 3p 2 1 1 9.0 3d 2 0 2 9.0 4s 3 3 0 16.0 4p 3 2 1 16.0 4d 3 1 2 16.0 4f 3 0 3 16.0

43. Sizes hydrogen orbitals Hydrogen orbitals 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f Angstrom (0.1nm) 0.53, 2.12, 4.77, 8.48 1.7A 0.74A H 4.8 H C H--H H H H H H H H H

44. Hydrogen atom excited states -0.033 h0 = -0.9 eV 4s 4p 4d 4f -0.056 h0 = -1.5 eV 3s 3p 3d -0.125 h0 = -3.4 eV 2s 2p -0.5 h0 = -13.6 eV 1s Energy zero

45. Plotting of orbitals: line cross-section vs. contour

46. Contour plots of 1s, 2s, 2p hydrogenic orbitals

47. Contour plots of 3s, 3p, 3d hydrogenic orbitals

48. Contour plots of 4s, 4p, 4d hydrogenic orbtitals

49. Contour plots of hydrogenic 4f orbitals

50. He atom – using He+ orbitals With 2 electrons, we can form the ground state of He by putting both electrons in the He+ 1s orbital ΨHe(1,2) = A[(Φ1sa)(Φ1sb)]= Φ1s(1)Φ1s(2)(ab-ba) EHe= 2<1s|h|1s> + J1s,1s First lets review the energy for He+. Writing Φ1s= exp(-zr) we determine the optimum z for He+ as follows <1s|KE|1s> = + ½ z2 (goes as the square of 1/size) <1s|PE|1s> = - Zz (linear in 1/size) Applying the variational principle, the optimum z must satisfy dE/dz = z -Z = 0 leading to z =Z, KE = ½ Z2, PE = -Z2, E=-Z2/2 = -2 h0. writing PE=-Z/R0, we see that the average radius is R0=1/z Now consider He atom: EHe = 2(½ z2) – 2Zz + J1s,1s