Hylleraas wavefunction for He. dv 2. ,! r 2. )dv 1. in the trial function. A simple trial function that does include r 12. is ) f (r 12.

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1 Hylleraa wavefunction for He The reaon why the Hartree method cannot reproduce the exact olution i due to the inability of the Hartree wave-function to account for electron correlation. We know that the two electron in He repel one another and abent any other factor they would tay a far apart a poible o that the probability of finding them cloe to one another would be mall. Given the wave-function ψ the probability of finding electron 1 at the terminu of and in the volume element dv 1 while electron i at r in the volume element dv i ψ (, r )dv 1 dv which for the Hartree function i ϕ ( )ϕ ( r )dv 1 dv and the probability i uncorrelated in that the location of electron 1 i independent of the location of electron. Another way of looking at thi i to realize that the wave-function for the ground (pherically ymmetric) tate i a function of three variable: & r the ditance of the two electron from the nucleu and the ditance between the electron and the Hartree or orbital model doen t include in the trial function. A imple trial function that doe include i ϕ( )ϕ(r ) f ( ) and if we expand f ( ) around =, the region of interet, our trial function becomeϕ( )ϕ(r ) f ( ) = ϕ( )ϕ(r )( f () + f (1) () + f () () / +) and if we factor out f () we have an un-normalized trial function ϕ( )ϕ(r )(1+ b + c +) where b,c,are variational parameter. The firt to propoe a trial function of thi form wa Hylleraa and uing the function ψ = e α e αr (1+ b ) = e α ( +r ) (1+ b ) he determined the optimum value α = & b =.364 reulting in an energy au that i greater than the exact energy by.4%, a ignificant improvement over the imple product reult. Since thi calculation i omewhat tediou we reerve the detail to the appendix. Hylleraa explored everal trial function containing what are called the Hylleraa coordinate = + r,t = r & u = and laid the groundwork for ubequent calculation uing trial function of the form ψ = e α c ijk i t j u k ijk Pekeri (vide upra) ued thi functional form with 178 term and found E = and after correcting for the ize of the nucleu and relativitic effect predict the ionization energy 198,31.69 cm -1 compared to the experimental energy 198,31.8 cm -1. While trial function involving can give excellent reult they are not eaily interpreted wherea a function coniting of optimized orbital give a reaonable energy and i more in keeping with qualitative Jame F Harrion Michigan State Univerity 1

2 chemical idea. Alo wavefunction coniting of orbital product are almot alway the tarting point for more accurate calculation. Preliminarie Appendix. He like atom and a Hylleraa wavefunction. Our goal i to ue the variation principle to calculate the energy of a two-electron atom in an S tate uing the implet Hylleraa wavefuntion ψ = e α ( +r ) (1+ b ) where α & b are variational parameter and i the eparation between electron. The energy i given by E = ψ * Ĥψ dτ ψ * ψ dτ with the Hamiltonian in atomic unit Ĥ = Z Z + 1 r Becaue of in the wavefunction it convenient to ue what are called Hylleraa coordinate,t,u ( ) intead of,θ 1,ϕ 1 and r,θ,ϕ. Thee are defined a = + r,t = r &u =. We need to have the integrand ψ * Ĥψ and the volume element dτ in thee coordinate o let begin with the volume element. Firt write dτ a (figure 1) Figure 1 Jame F Harrion Michigan State Univerity

3 z θ 1 r y nucleu x dτ = 8π d r dr inθ 1 dθ 1 The factor 8π come about becaue the orientation of the triangle defined by,r &θ 1 i arbitrary and we have integrated over ϕ 1 &ϕ each contributing a factor of π and then one θ contributing a factor of. Uing = + r r coθ 1 we have d = r inθ 1 dθ 1 = udu giving dτ = 8π r ud dr du from the definition of &t we have r = t 4 and uing the Jacobian (,r ) (,t) = 1 we have d dr = 1 ddt with the final reult dτ = π u( t )ddudt with the limit, u, & u t u Now for the integrand. When we ue Hylleraa coordinate it will be impler to rewrite the kinetic energy term a follow. Let conider the integral Jame F Harrion Michigan State Univerity 3

4 f f fdv = f x + f y + f z dxdydz and conider the term f f dxdydz = dy dz x f f x f x f x dx = dy dz f x dx where the term f f = becaue we aume f (±) = x Extending thi to the remaining coordinate reult in f f dxdydz = f x + f y + f z dxdydz = f i f dxdydz and o ψ Ĥ ψ dτ = 1 1 ψ i 1 ψ + ψ i ψ Z Z + 1 r dτ Our goal i to expre the integrand in term of the Hylleraa coordinate. Let firt conider 1 ψ. 1 ψ = x 1 î + y 1 ĵ + z 1 ˆk uing the chain rule x 1 = x 1 + and by ymmetry = y 1 y 1 y 1 ( y y ) 1 u u and + u = x 1 x 1 u x 1 x 1 (x x ) 1 u u Jame F Harrion Michigan State Univerity 4

5 = z 1 z 1 z 1 (z z ) 1 u u o after ome algebra 1 ψ i 1 ψ = in a imilar fahion + + u + u + u i r u r 1 u ψ i ψ = + + u Adding thee and recognizing that + u + u r 1 i r r u + r u i r = + t u 8 & r = 4 t & 1 1 = 4t r t we obtain + + u + (u t ) u( t ) u t(u ) u( t ) u 4Zψ t + ψ u In the text we calculated the energy of Helium uing the trial function ψ = α 3 π e α α 3 π e α = α 3 π e α ( + r ) = α 3 π e α and found α = 7 8 & E = Let redo thi calculation uing Hylleraa coordinate to ee how the integration work. Since thi trial function depend only on and ψ i normalized the energy i given by E = π 4Zψ t + ψ u( t )ddudt u There are three term in the integrand. Firt the kinetic energy. Jame F Harrion Michigan State Univerity 5

6 u( t )d du dt = α 8 π e α u( t )ddudt = α 8 e α u d u du ( t )dt π o e α d u du u u( t )d du dt = α now the nuclear attraction.. ( t )dt = e α d u(u u3 3 )du = 8 15 e α 5 d = 1 π 4Zψ u( t )ddudt = 4Z t π e α d u du and latly the electron electron repulion contribution E = π ψ ( t )ddudt = π 4Zψ t + ψ u( t )ddudt u e α u 5α d du ( t )dt = 8π u dt = Zα π E = α Zα α and o de dα = α Z = reult in α = Z 5 16 and for He α = with E = 16 a required. Now let evaluate the energy aociated with the Hylleraa function ψ = e α ( +r ) (1+ b ) = e α (1+ bu) When we evaluate the energy aociated with a trial function that depend only on & u uch a above we will encounter integral that have the form Jame F Harrion Michigan State Univerity 6

7 I(N, M ) = N e α d u M du ( t )dt = The firt few being Firt the overlap integral +u N M I (N,M) 5 8α α α M + 3( M + )( M + 4) ( M + N + 4) M +N +5 (α ) α 8 ψ ψ = +u e αs (1+ bu) u( t )ddudt = e αs d u(1+ bu + b u )du ( t )dt ψ ψ = I(,1) + bi(,) + b I(,3) = 1 + b 35 8α 7 + b 6 α 8 and now for the expectation value. Note that derivative wrt t vanih. ψ Ĥ ψ = + u + (u t ) u( t ) u ψ 4Z t + ψ u( t )ddudt u u( t )ddudt = α (I(,1) + bi(,) + b I(,3)) = 1 α 4 + b 35 8α 5 + b 6 Jame F Harrion Michigan State Univerity 7

8 u u( t )ddudt = b I(,1) = b (u t ) u ddudt = αb e α (1+ bu)du (u t )dt = b 5 4α 3 5 b +u 4Zψ uddudt = Z α + b 15 5 α b α 7 and latly ψ ( t )ddudt = I(,) + bi(,1) + b I(,) = 5 8α + b 5 α b 16α 7 The energy i given by the ratio α ( 5 5α Z)α + b Z + b α ( Z) E = 1+ b 35 8α + 6 b α and for He α 7 8 α + b 5α b α E = 1+ b 35 8α + 6 b α = N D minimizing E with repect to α & b reult in α = & b =.365& E =.8911 Jame F Harrion Michigan State Univerity 8