The Hamiltonian and Schrodinger Equation for Helium's Electrons

University of Connecticut OpenCommons@UConn

Chemistry Education Materials Department of Chemistry

6-15-2006 The aH miltonian and Schrodinger Equation for Helium's Electrons (Hylleraas) Carl W. David University of Connecticut, [email protected]

Follow this and additional works at: https://opencommons.uconn.edu/chem_educ Part of the Chemistry Commons

Recommended Citation David, Carl W., "The aH miltonian and Schrodinger Equation for Helium's Electrons (Hylleraas)" (2006). Chemistry Education Materials. 8. https://opencommons.uconn.edu/chem_educ/8 The Hamiltonian and Schrodinger Equation for Helium’s Electrons (Hylleraas)

C. W. David Department of Chemistry University of Connecticut Storrs, Connecticut 06269-3060 (Dated: June 15, 2006)

I. INTRODUCTION (remember, we are holding all the other {xi} constant) for electron 1 and electron 2, with equivalent terms for y For this two electron problem, the Schr¨odinger Equa- and z (two each) as a function or r1, r2 and ϑ, the angle tion has the form between the location vectors of the two electrons, ~r1 and ~r2. 2 2 2 ¯h 2 2 Ze Ze − ∇1 + ∇2 ψ − ψ − ψ = Eψ (1.1) 2m r1 r2 III. THE COORDINATE TRANSFORMATION where m is the mass of an electron, and the subscripts refer to electron 1 and 2 respectively. Here . Remember that 2 2 2 ∂ ∂ ∂ q ∇1 = + + 2 2 2 2 2 2 r1 = x + y + z ∂x1 ∂y1 ∂z1 1 1 1 where the subscript “1” means we are refering to electron and 1 (we have a similar expression for electron 2). − q It is traditional to set Z=2, since Z=1 is H , Z=3 2 2 2 r2 = x + y + z would be Li+, etc., i.e., to specialize to Helium itself. 2 2 2 Then, cross multiplying one has while, of course, 2m Ze2 2m Ze2 2m 2 2 p 2 2 2 − ∇1 + ∇2 ψ − 2 ψ − 2 ψ = E 2 ψ (1.2) r12 = (x1 − x2) + (y1 − y2) + (z1 − z2) ¯h r1 ¯h r2 ¯h which is the form most people start with. A. Preliminary Partial Derivatives

II. THE HAMILTONIAN We need, according to the chain rule, the following terms: Assuming inﬁnite nuclear masses, (m = melectron) one ∂ ∂r ∂ ∂y ∂ has = 1 + (3.1) ∂x1 ∂x1 ∂r1 ∂x1 ∂µ 2 2 2 2 ¯h 2 2 Ze Ze e Hop = − ∇ + ∇ − − + (2.1) 2m 1 2 r r r so, focussing on the ﬁrst of these, we ask, what is ∂r1 ? 1 2 12 ∂x1 We have We start with the idea of expressing the kinetic energy part of the Hamiltonian in a form appropriate for this p 2 2 2 2 2 2 ∂r1 ∂ x1 + y1 + z1 1 −1 ∂ x1 + y1 + z1 x1 problem. That operator surely has the form = = r1 = ∂x1 ∂x1 2 ∂x1 r1 2 (3.2) ¯h 2 2 − ∇1 + ∇2 Parenthetically, 2me −1 where ∇ has its traditional functional meaning: ∂r1 −1 ∂r1 x1 = 2 = − 3 (3.3) ∂x1 r1 ∂x1 r1 ∂2 ∂2 ∂2 ∇1 = 2 + 2 + 2 Then, we have ∂x1 ∂y1 ∂z1 −2 with a second almost identical term for electron 2’s ki- ∂r1 −3 ∂r1 = −2r1 netic energy operator. ∂x1 ∂x1 That means that we need to obtain ∂ 2 x = − 1 ∂x1 3 y1,z1,x2,y2,z2 r1 r1

Typeset by REVTEX 2

2x1 ∂µ z1 r1r2µ z2 = − 4 = − 3 (3.10) r1 ∂z2 r1r2 r1 r2 Now, using the chain rule, we have For the second term in Equation 3.1 We start with the equation for the angle between the two radii ~r1 and ~r2. ∂ ∂r1 ∂ ∂µ ∂ From the law of cosines, we have = + ∂x1 ∂x1 ∂r1 ∂x1 ∂y q 2 2 r12 = r1 + r2 − 2r1r2 cos ϑ and the two multiplicative partial derivatives are known. Using Equation 3.5and Equation 3.4 we have while from vector algebra we know ∂ x1 ∂ x2 ~r1 · ~r2x1 ∂ ~r1 · ~r2 = + − 3 cos ϑ = ∂x1 r1 ∂r1 r1r2 r r2 ∂y r1r2 1 (An alternative formulation for this vector algebraic which is statement is: ∂ x1 ∂ x2 x1µ ∂ ~r1 · ~r2 x1x2 + y1y2 + z1z2 = + − 2 cos ϑ = = ≡ µ (3.4) ∂x1 r1 ∂r1 r1r2 r1 ∂y r1r2 r1r2 which deﬁnes µ = cosϑ. Since B. Second Partial Derivatives ∂ x1x2+y1y2+z1z2 ∂µ r1r2 = The second derivative would be ∂x1 ∂x1 x1 ∂ x2 x1µ ∂ gives us (using Equations 3.2 and 3.3) 2 ∂ + − 2 ∂ r1 ∂r1 r1r2 r1 ∂µ 2 = x x x + y y + z z ∂r−1 ∂x1 ∂x1 = 2 + 1 2 1 2 1 2 1 r r r ∂x 1 2 2 1 which would be

x ∂ x x µ ∂ x2 x1x2 + y1y2 + z1z2 x1 2 1 2 1 ∂ r ∂r ∂ r r − r2 ∂µ = − 3 ∂ 1 1 1 2 1 r1r2 r2 r1 2 = + ∂x1 ∂x1 ∂x1

x ~r · ~r x We have achieved a mixed representation of the second = 2 − 1 2 1 3 partial derivative. r1r2 r2 r1 Now we take the derivative with respect to x1 where appropriate, before converting ∂ to ∂ and ∂ . We ∂µ x r r µ x ∂x1 ∂r1 ∂µ 2 1 2 1 obtain = − 3 (3.5) ∂x1 r1r2 r2 r1 1 ∂ 2 ∂ Clearly, the other ﬁve groups of terms are equivalent. ∂ 1 ∂ r1 ∂r1 2 = + x1 (3.11) ∂µ y r r µ y ∂x1 r1 ∂r1 ∂x1 = 2 − 1 2 1 (3.6) 3 x2 ∂ ∂y1 r1r2 r2 r1 ∂ r r ∂µ + 1 2 (3.12) ∂x1 ∂µ z2 r1r2µ z1 µ ∂ = − 3 (3.7) − 2 (3.13) ∂z1 r1r2 r2 r1 r1 ∂µ µ ∂ ∂ 2 r1 ∂µ ∂µ x1 r1r2µ x2 −x1 (3.14) ∂x1 = − 3 (3.8) ∂x2 r1r2 r1 r2 Now we expand the partial derivatives with respect to x1 to their replacements. We are going to get a devil of a ∂µ y1 r1r2µ y2 = − 3 (3.9) lot of terms. We obtain for the ﬁrst term ∂y2 r1r2 r1 r2 3

∂2 2 = ∂x1 −1 2 2 1 ∂ ∂r1 ∂ x1 ∂ 1 ∂ −x1 ∂ x1 ∂ Equation − 3.11 → + x1 + → + x1 3 + (3.15) r1 ∂r1 ∂x1 ∂r1 r1 ∂x1∂r1 r1 ∂r1 r1 ∂r1 r1 ∂x1∂r1 x ∂ ∂ 2 −1 2 r1r2 ∂µ x ∂r ∂ x ∂ Equation − 3.12 → + → + 2 1 + 2 (3.16) ∂x1 r2 ∂x1 ∂µ r1r2 ∂x1∂µ µ ∂ Equation − 3.13 → − 2 (3.17) r1 ∂µ µ ∂ −2 ∂ 2 ∂µ r1 ∂r1 ∂ x1 ∂µ ∂ Equation − 3.14 − x1 → −x1µ − 2 ∂x1 ∂x1 ∂µ r1 ∂x1 ∂µ 2 x1µ ∂ − 2 (3.18) r1 ∂x1∂µ which is ∂2 2 = ∂x1 2 1 ∂ −x1 ∂ Equation − 3.15 → + 3 + r1 ∂r1 r1 ∂r1 x1 x1 ∂ x2 x1µ ∂ ∂ + − 2 (3.19) r1 r1 ∂r1 r1r2 r1 ∂µ ∂r1 x2 x1 ∂ x2 x1 ∂ x2 x1µ ∂ ∂ Equation − 3.16 → − 3 + + − 2 (3.20) r2 r1 ∂µ r1r2 r1 ∂r1 r1r2 r1 ∂µ ∂µ µ ∂ Equation − 3.17 → − 2 (3.21) r1 ∂µ −2 ∂r1 ∂ x1 x2 x1(~r1 · ~r2) ∂ Equation − 3.18 → −x1µ − 2 − 3 (3.22) ∂x1 ∂µ r1 r1r2 r1r2 ∂µ x1µ x1 ∂ x2 x1µ ∂ ∂ Equation − 3.18 → − 2 + − 2 (3.23) r1 r1 ∂r1 r1r2 r1 ∂µ ∂µ which is, upon cleaning up the expressions

∂2 2 = ∂x1 2 2 2 1 ∂ x1 ∂ x1 ∂ Equation − 3.19 → − 3 + 2 2 r1 ∂r1 r1 ∂r1 r1 ∂r1 2 x1 x2 x1µ ∂ + − 2 (3.24) r1 r1r2 r1 ∂r1∂µ 2 x2 x1 ∂ x2x1 ∂ Equation − 3.20 → − 3 + 2 + r2 r1 ∂µ r1r2 ∂r1∂µ x2 x2 x1µ ∂ ∂ − 2 (3.25) r1r2 r1r2 r1 ∂µ ∂µ µ ∂ Equation3.21 → − 2 (3.26) r1 ∂µ −2x1 ∂ x1x2 2 ~r1 · ~r2 ∂ Equation − 3.22 → −x1µ 4 − 3 − x1 5 (3.27) r1 ∂µ r1r2 r1r2 ∂µ 2 2 x1µ x1 ∂ x1µ x2 x1µ ∂ Equation3.23 → − 2 − 2 − 2 2 (3.28) r1 r1 ∂r1∂µ r1 r1r2 r1 ∂µ 4 and, cleaning up again, we have ∂2 2 = ∂x1 2 2 2 2 1 ∂ x1 ∂ x1 ∂ x1 x2 x1µ ∂ Equation − 3.24 → − 3 + 2 2 + − 2 (3.29) r1 ∂r1 r1 ∂r1 r1 ∂r1 r1 r1r2 r1 ∂r1∂µ 2 2 x2 x1 ∂ x2x1 ∂ x2 x2 x1µ ∂ Equation − 3.25 → − 3 + 2 + − 2 2 (3.30) r2 r1 ∂µ r1r2 ∂r1∂µ r1r2 r1r2 r1 ∂µ µ ∂ Equation3.26 → − 2 (3.31) r1 ∂µ 2 2x1µ ∂ x1x2 ∂ Equation − 3.27 → + 4 − 3 (3.32) r1 ∂µ r1r2 ∂µ 2 ~r1 · ~r2 ∂ Equation − 3.27 → +x1 5 (3.33) r1r2 ∂µ 2 2 2 2 2 2 x1µ ∂ x1µx2 ∂ x1µ ∂ Equation3.28 → − 3 − 3 2 + 4 2 (3.34) r1 ∂r1∂µ r1r2 ∂µ r1 ∂µ which is, penultimately, when all six term are added together, ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 2 + 2 + 2 + 2 + 2 + 2 = ∂x1 ∂y1 ∂z1 ∂x2 ∂y2 ∂z2 2 2 2 3 ∂ r1 ∂ r1 ∂ Equation − 3.29 → − 3 + 2 2 + r1 ∂r1 r1 ∂r1 r1 ∂r1 2 2 2 2 2 x1x2 + y1y2 + z1z2 ∂ (x1 + y1 + z1 )µ ∂ 2 − 3 r1r2 ∂r1∂µ r1 ∂r1∂µ 2 2 2 3 ∂ r2 ∂ r2 ∂ Equation − 3.29 → − 3 + 2 2 + r2 ∂r2 r2 ∂r2 r2 ∂r2 2 2 2 2 2 x1x2 + y1y2 + z1z2 ∂ (x2 + y2 + z2 )µ ∂ 2 − 2 r2r1 ∂r2∂µ r2r1 ∂r2∂µ 2 x1x2 + y1y2 + z1z2 ∂ r2r1µ ∂ Equation − 3.30 → − 3 + 2 + r1r2 ∂µ r1r2 ∂r1∂µ 2 2 r2 (x1x2 + y1y2 + z1z2)µ ∂ 2 2 − 3 2 r1r2 r1r2 ∂µ 2 2 2 2 x2x1 + y1y2 + z1z2 ∂ (x2x1 + y1y2 + z1z2) ∂ r1 r1r2y ∂ Equation − 3.30 → − 3 + 2 + 2 2 − 3 2 r2r1 ∂µ r2r1 ∂r2∂µ r2r1 r2r1 ∂µ 3µ ∂ 3µ ∂ 3µ ∂ 3µ ∂ Equation − 3.31 → − 2 − 2 − 2 − 2 r2 ∂µ r1 ∂µ r2 ∂µ r1 ∂µ 2 2 2r1µ ∂ 2r2µ ∂ Equation − 3.32 → + 3 + 3 r1 ∂µ r2 ∂µ ~r1 · ~r2 ∂ ~r1 · ~r2 ∂ Equation − 3.32 → − 3 + 3 r1r2 ∂µ r2r1 ∂µ ~r2 · ~r1 ∂ ~r2 · ~r1 ∂ Equation − 3.33 → − 3 + 3 r2r1 ∂µ r2r1 ∂µ 2 2 2 2 2 r1µ ∂ ~r1 · ~r2µ ∂ µ ∂ Equation − 3.34 → − 3 − 3 2 + 2 2 r1 ∂r1∂µ r1r2 ∂µ r1 ∂µ 2 2 2 2 2 r2µ ∂ ~r1 · ~r2µ ∂ µ ∂ Equation − 3.34 → − 3 − 3 2 + 2 2 r2 ∂r2∂µ r1r2 ∂µ r2 ∂µ

The term x1x2 + y1y2 + z1z2 is just r1r2µ, yields the gorgeous cancellations (see above) leading to which becomes ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 2 + 2 + 2 + 2 + 2 + 2 = ∂x1 ∂y1 ∂z1 ∂x2 ∂y2 ∂z2 5

2 ∂ ∂2 Equation − 3.35 → + 2 (3.35) r1 ∂r1 ∂r1 2 2 r1r2µ ∂ µ ∂ Equation − 3.35 → + 2 − = 0 (3.36) r1r2 ∂r1∂µ r1 ∂r1∂µ 2 ∂ ∂2 Equation − 3.35 → + 2 (3.37) r2 ∂r2 ∂r2 2 2 r1r2µ ∂ µ ∂ Equation − 3.35 → + 2 − = 0 (3.38) r2r1 ∂r2∂µ r2 ∂r2∂µ 2 2 2 r1r2µ ∂ µ ∂ 1 r1r2µ ∂ Equation − 3.35 → − 3 + + 2 − 3 2 (3.39) r1r2 ∂µ r1 ∂r1∂µ r1 r1r2 ∂µ 2 2 2 r2r1µ ∂ r2r1µ ∂ 1 r1r2µ ∂ Equation − 3.35 → − 3 + 2 + 2 − 3 2 (3.40) r2r1 ∂µ r2r1 ∂r2∂µ r2 r2r1 ∂µ 3µ ∂ 3µ ∂ Equation − 3.35 → − 2 − 2 (3.41) r2 ∂µ r1 ∂µ 2µ ∂ 2µ ∂ Equation − 3.35 → + 2 + 2 (3.42) r1 ∂µ r2 ∂µ µ ∂ µ ∂ µ ∂ µ ∂ Equation − 3.35 → − 2 + 2 − 2 + 2 = 0 (3.43) r1 ∂µ r1 ∂µ r2 ∂µ r2 ∂µ µ ∂2 Equation − 3.35 → − (3.44) r1 ∂r1∂µ µ ∂2 Equation − 3.35 → − (3.45) r2 ∂r1∂µ We obtain ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 2 + 2 + 2 + 2 + 2 + 2 = ∂x1 ∂y1 ∂z1 ∂x2 ∂y2 ∂z2 2 ∂ ∂2 Equation − 3.35 → + 2 (3.46) r1 ∂r1 ∂r1 2 ∂ ∂2 Equation − 3.37 → + 2 (3.47) r2 ∂r2 ∂r2 µ ∂ µ ∂2 1 µ2 ∂2 Equation − 3.39 → − 2 + + 2 − 2 2 (3.48) r1 ∂µ r1 ∂r1∂µ r1 r1 ∂µ | {z } µ ∂ µ ∂2 1 µ2 ∂2 Equation − 3.40 → − 2 + + 2 − 2 2 = 0 (3.49) r2 ∂µ r2 ∂r2∂µ r2 r2 ∂µ | {z } 3µ ∂ 3µ ∂ Equation − 3.41 → − 2 − 2 (3.50) r2 ∂µ r1 ∂µ 2µ ∂ 2µ ∂ Equation − 3.42 → + 2 + 2 (3.51) r1 ∂µ r2 ∂µ µ ∂2 Equation − 3.44 → − (3.52) r1 ∂r1∂µ | {z } µ ∂2 Equation − 3.45 → − (3.53) r2 ∂r1∂µ | {z } (where we notice that the underbraced material (above) cancels) which (almost) ﬁnally becomes

2 C. Final Cancellations 2 ∂ ∂ Equation − 3.46 → + 2 (3.54) r1 ∂r1 ∂r1

∂2 ∂2 ∂2 ∂2 ∂2 ∂2 2 + 2 + 2 + 2 + 2 + 2 = ∂x1 ∂y1 ∂z1 ∂x2 ∂y2 ∂z2 6

2 ∂ ∂2 3µ ∂ 3µ ∂ Equation − 3.50 → − − (3.58) Equation − 3.47 → + 2 (3.55) 2 2 r2 ∂r2 ∂r2 r2 ∂µ r1 ∂µ µ ∂ 1 − µ2 ∂2 2µ ∂ 2µ ∂ Equation − 3.48 → − + (3.56) + + (3.59) 2 2 2 r2 ∂µ r2 ∂µ r1 ∂µ r1 ∂µ 1 2 µ ∂ 1 − µ2 ∂2 Equation − 3.49 → − 2 + 2 2 (3.57) which ﬁnally, and we mean that(!) is r2 ∂µ r2 ∂µ

∂2 ∂2 ∂2 ∂2 ∂2 ∂2 2 + 2 + 2 + 2 + 2 + 2 = ∂x1 ∂y1 ∂z1 ∂x2 ∂y2 ∂z2 2 2 2 2 ∂ ∂ 2 ∂ ∂ 1 1 ∂ 1 1 2 ∂ + 2 + + 2 − 2 + 2 2µ + 2 + 2 (1 − µ ) 2 (3.60) r1 ∂r1 ∂r1 r2 ∂r2 ∂r2 r1 r2 ∂µ r1 r2 ∂µ OK, we lied. There is a traditional form which we have to include:

∂2 ∂2 ∂2 ∂2 ∂2 ∂2 2 + 2 + 2 + 2 + 2 + 2 = ∂x1 ∂y1 ∂z1 ∂x2 ∂y2 ∂z2 ∂ r2 ∂ ∂ r2 ∂ ∂ (1 − µ2) ∂ 1 1 ∂r1 1 2 ∂r2 1 1 ∂µ 2 + 2 + 2 + 2 (3.61) r1 ∂r1 r2 ∂r2 r1 r2 ∂µ which is, one must believe, the most compact form possible.

IV. THE r1, r2, r12 FORM and use the chain rule to obtain ∂ ∂r ∂ ∂r ∂ We had the following expression for the Kinetic Energy = 1 + 12 Operator: ∂x1 ∂x1 ∂r1 ∂x1 ∂r12

∂2 ∂2 ∂2 ∂2 ∂2 ∂2 which is, by direct diﬀerentiation + + + + + = ∂x2 ∂y2 ∂z2 ∂x2 ∂y2 ∂z2 1 1 1 2 2 2 ∂ x1 ∂ x1 − x2 ∂ 2 2 = + ∇1 + ∇2 = ∂x1 r1 ∂r1 r12 ∂r12 r2∂ r2∂ 1 ∂ 1 1 ∂ 2 1 1 ∂(1 − µ2) ∂ ∂r1 ∂r2 ∂µ since 2r1dr1 = 2x1dx1. We have 2 + 2 − 2 + 2 (4.1) r1 ∂r1 r2 ∂r2 r1 r2 ∂µ 2 ∂ ∂ ∂ ∂x1 which we derived in r1, r2, ϑ space. Now we turn to a 2 = (4.2) ∂x1 ∂x1 diﬀerent spatial representation, r1, r2, r12. Again we seek x1 ∂ x1−x2 ∂ 1 2 2 2 ∂ + ∂ r1 ∂r1 r12 ∂r12 the Kinetic Energy Operator. 2µ (∇1 + ∇2) 2 = (4.3) We start with ∂x1 ∂x1

2 2 2 2 r12 = (x1 − x2) + (y1 − y2) + (z1 − z2) which is

2 ∂ x1 ∂ ∂ x1−x2 ∂ ∂ r1 ∂r1 r12 ∂r12 2 = + (4.4) ∂x1 ∂x1 ∂x1 which becomes

−1 2 1 ∂ ∂r1 ∂ x1 ∂ + x1 + (4.5) r1 ∂r1 ∂x1 ∂r1 r1 ∂x1∂r1 −1 2 1 ∂ ∂r12 ∂ x1 − x2 ∂ + + (x1 − x2) + (4.6) r12 ∂r12 ∂x1 ∂r12 r12 ∂x1∂r12 7 which is seen to be

∂ ∂ ∂ ∂ ! 1 ∂ x1 ∂ x1 x1 ∂r1 x1 − x2 ∂r1 Equation 4.5 → − x1 3 + + (4.7) r1 ∂r1 r1 ∂r1 r1 r1 ∂r1 r12 ∂r12 ∂ ∂ ∂ ∂ ! 1 ∂ (x1 − x2) ∂ x1 − x2 x1 ∂r12 x1 − x2 ∂r12 Equation 4.6 → + − (x1 − x2) 3 + + (4.8) r12 ∂r12 r12 ∂r12 r12 r1 ∂r1 r12 ∂r12 and this becomes

2 1 ∂ x1 ∂ 3 ∂ 1 ∂ 2 ∂ Equation IV → − 3 → − → (4.9) r1 ∂r1 r1 ∂r1 r1 ∂r1 r1 ∂r1 r1 ∂r1 2 2 2 x1 ∂ ∂ +Equation IV → 2 2 → 2 (4.10) r1 ∂r1 ∂r1 ∂ ∂ 2 x1(x1 − x2) ∂r1 ~r1 · ~r12 ∂ Equation IV → + → 2 (4.11) r1r12 ∂r12 r1r12 ∂r1∂r12 2 1 ∂ (x1 − x2) ∂ 3 1 ∂ 2 ∂ Equation 4.8 → + − 3 → − → + (4.12) r12 ∂r12 r12 ∂r12 r12 r12 ∂r12 r12 ∂r12 x (x − x ) ∂2 ~r · ~r ∂2 Equation 4.8 → + 1 1 2 → 1 12 (4.13) r1r12 ∂r12∂r1 r1r12 ∂r1∂r12 2 2 2 r12 ∂ ∂ +Equation 4.8 → 2 2 → 2 (4.14) r12 ∂r12 ∂r12 which becomes

2 2 2 ∂ 2 ∂ ~r1 · ~r12 ∂ 2 ∂ ∂ 2 + + 2 + + 2 ∂r1 r1 ∂r1 r1r12 ∂r1∂r12 r12 ∂r12 ∂r12 which is for the 3 r1 terms, i.e., we get a second set from the r2 terms, leading to

2 2 ∇1 + ∇2 = ∂2 2 ∂ ∂2 2 + + 2r ˆ1 · rˆ12 ∂r1 r1 ∂r1 ∂r1∂r12 ∂2 2 ∂ ∂2 + 2 + + 2r ˆ2 · rˆ12 ∂r2 r2 ∂r2 ∂r2∂r12 4 ∂ ∂2 + + 2 2 (4.15) r12 ∂r12 ∂r12

This is our result! When combined with the potential energy operator, one can easily form the Hamiltonian of this system in this representation.

V. YET ANOTHER FORMULATION

In yet another coordinate scheme (double spherical polar coordinates) we have:

2 2 ∂ ∂ 2 ! ∂r ∂ sin ϑ1 ¯h 1 1 ∂r1 1 ∂ϑ1 ∂ = − 2 + 2 sin ϑ1 + 2 2m r1 ∂r1 sin ϑ1 ∂ϑ1 ∂φ1

2 ∂ ∂ 2 !! ∂r ∂ sin ϑ2 1 2 ∂r2 1 ∂ϑ2 ∂ + 2 + 2 sin ϑ2 + 2 r2 ∂r2 sin ϑ2 ∂ϑ2 ∂φ2 Ze2 Ze2 e2 − − + (5.1) r1 r2 r12 8 and again, in another set of mixed coordinates,

2 2 ∂ ∂ 2 ! ∂r ∂ sin ϑ1 ¯h 1 1 ∂r1 1 ∂ϑ1 ∂ = − 2 + 2 sin ϑ1 + 2 2m r1 ∂r1 sin ϑ1 ∂ϑ1 ∂φ1

2 ∂ ∂ 2 !! ∂r ∂ sin ϑ2 1 2 ∂r2 1 ∂ϑ2 ∂ + 2 + 2 sin ϑ2 + 2 r2 ∂r2 sin ϑ2 ∂ϑ2 ∂φ2 Ze2 Ze2 e2 − − + (5.2) r r p 2 2 1 2 r1 + r2 − 2r1r2 cos ϑ12

which can be transformed into an equation without “12” so, we have subscripts, since

~r1 · ~r2 = r1r2 cos ϑ12

r1 sin ϑ1 cos φ1r2 sin ϑ2 cos φ2 + r1 sin ϑ1 sin φ1r2 sin ϑ2 sin φ2 + r1r2 cos ϑ1 cos ϑ2 = r1r2 cos ϑ12 which can be be solved for cos ϑ12. We obtain, now in a single consistent coordinate system,

2 2 ∂ ∂ 2 ! ∂r ∂ sin ϑ1 ¯h 1 1 ∂r1 1 ∂ϑ1 ∂ = − 2 + 2 sin ϑ1 + 2 2m r1 ∂r1 sin ϑ1 ∂ϑ1 ∂φ1

2 ∂ ∂ 2 !! ∂r ∂ sin ϑ2 1 2 ∂r2 1 ∂ϑ2 ∂ + 2 + 2 sin ϑ2 + 2 r2 ∂r2 sin ϑ2 ∂ϑ2 ∂φ2 Ze2 Ze2 e2 − − + (5.3) r r p 2 2 1 2 r1 + r2 − 2r1r2 (cos ϑ1 cos ϑ2 + sin ϑ1 sin ϑ2 cos(φ1 − φ2))

This last form shows explicitly not only the non- [4] showed that there is no Frobenius solution to the separability of the Schrödinger Equation for Helium’s Schrödinger equation (see also Coolidge and James [5] electrons, but how horribly intertwined the coördinates ). actually are due to the r12 term. The analytical situation was clarified by Fock [6] (for the English translation, see Fock [7]) who found that in- troducing hyperspherical coördinates required that loga- VI. DISCUSSION (I) rithmic terms exist in the expansion of the wave function. This result overshadowed Bartlett’s similar [8] indepen- dent discovery. A review of the current situation in this We have seen that the Schrödinger Equation for the field may be found in the work of Abbot and Maslen [9] 2-electron atom/ion has a 6-dimensional representation as well as in the recent work of Myers et al [10]. The in double-3-space {x1, y1, z1, x2, y2, z2}. We assert else- convergence of the Fock expansion has been investigated 1 where that this is reducible to a {r1, r2, r12} set for S by [11]. states. As of the date of writing, the best computation of the The first major attack on the solution to this prob- energy of the ground electronic state of Helium is due to lem was due to Hylleraas, vide infra[1, 2], who obtained Schwartz [12, 13]. spectacular (for the time) energies for the Helium atom’s electrons. Bartlett [3] vide supra showed that the series solution to the Schrödinger equation using the Hylleraas’ expan- VII. DISCUSSION (II) sion gave rise to equations which yielded different values for the same coëfficients depending on which equa- If one substitutes a series (Ansatz) into the appro- tions were used to determine them. Further, Withers priate Schrödinger equation, one expects that one can 9 sequentially obtain recurrence relations between linked and we will evaluate one term of this set to see what is coëfficients with only boundary conditions effecting the going on. We have resolution of these linked recurrence relations. Then, using these recurrence relations to determine as many coëfficients as possible relative to arbitrary ones, one ex- ∂2ψ ∂2e−α(r1+r2)+βr12 pects that this truncated and partially evaluated Ansatz, 2 = 2 (7.3) when used in a variational calculation, will lead to the ∂x1 ∂x1 fastest possible convergence to the exact answers (and coëfficients) as the truncation of the series is altered. One expects the variationally determined coëfficients to substituting Equation (7.1) into Equation (7.2). monatonically approach their limiting “exact” values as the series is extended. First, one has An alternative approach might be to ask, what is the potential energy function which gives rise to the simplest correlated wave function? Consider the function ∂e−α(r1+r2)+βr12 x1 x1−x2 −α(r1+r2)+βr12 ∂ ∂ −α r + β r e ∂x1 1 12 ψ = e−α(r1+r2)+βr12 (7.1) = (7.4) ∂x1 ∂x1 ∂e−α(r1+r2)+βr12 x2 x1−x2 −α(r1+r2)+βr12 What, we ask, is the potential energy function which has ∂ ∂ −α r − β r e ∂x2 2 12 this function as an eigenfunction? = (7.5) ∂x2 ∂x2 We will work in the full six dimensional coördinate system. Then, we have

2 2 2 2 2 2 2 ∂ ψ ∂ ψ ∂ ψ ∂ ψ ∂ ψ ∂ ψ ∇6ψ = 2 + 2 + 2 + 2 + 2 + 2 (7.2) We obtain ∂x1 ∂y1 ∂z1 ∂x2 ∂y2 ∂z2

∂2ψ α αx2 β β(x − x )2 α2x2 β2(x − x )2 x (x − x ) 1 1 2 1 1 2 1 1 2 −α(r1+r2)+βr12 2 = − + 3 + − 3 + 2 + 3 − 2αβ e ∂x1 r1 r1 r12 r12 r1 r12 r1r12 ∂2ψ α αx2 β β(x − x )2 α2x2 β2(x − x )2 x (x − x ) 2 1 2 2 1 2 2 1 2 −α(r1+r2)+βr12 2 = − + 3 + − 3 + 2 + 3 − 2αβ e ∂x2 r2 r2 r12 r12 r2 r12 r2r12

When we add the ﬁve other sets of terms similar to these, we obtain 2 3α α 3α α 3β β 2 2 ~r1 · ~r12 ~r2 · ~r12 ∇6ψ = − + − + + − + α + β − 2αβ − ψ (7.6) r1 r1 r2 r2 r12 r12 r1r12 r2r12 which is 2 2α 2α 2β 2 2 ~r1 · ~r12 ~r2 · ~r12 ∇6ψ = − − + + α + β − 2αβ − ψ (7.7) r1 r2 r12 r1r12 r2r12 What this is saying is that ψ is not an eigenfunction, since the term proportional to αβ shouldn’t be there if it were. 1 2 Z Z 1 Z − α Z − α 1 − β 1 2 2 ~r1 · ~r12 ~r2 · ~r12 − ∇6ψ − − + = − − + − α + β + αβ − ψ (7.8) 2 r1 r2 r12 r1 r2 r12 2 r1r12 r2r12

[1] E. A. Hylleraas, Zeits. f. Physik 48, 469 (1928). [5] A. S. Coolidge and H. M. James, Phys. Rev. 51, 855 [2] E. A. Hylleraas, Zeits. f. Physik 54, 347 (1929). (1937). [3] J. H. Bartlett Jr., J. J. Gibbons Jr., and C. G. Dunn, [6] V. A. Fock, Izv. Akad. Nauk SSSR, Ser. Fiz. 18, 161 Phys. Rev. 47, 679 (1935). (1954). [4] C. S. Withers, Phys. Rev. A 30, 1506 (1984). [7] V. A. Fock, Det Koneglige Norske Videnskabers Selskabs 10

Forhandlinger 31, 138 (1958). [11] J. D. Morgan III, Theor. Chim. Acta 69, 181 (1986). [8] J. H. Bartlett, Phys. Rev. 51, 661 (1937). [12] C. Schwartz, Experiment and Theory in Com- [9] P. C. Abbott and E. N. Maslen, J. Phys. A 20, 2043 putations of the He Atom Ground State, (1987). http://lanl.arxiv.org/PS cache/physics/pdf/0208/0208005.pdf. [10] C. R. Myers, C. J. Umrigar, J. P. Sethna, and J. J. M. [13] C. Schwartz, Int. J. Mod. Phys. . III, Phys. Rev. A. 44, 5537 (1991).