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The Hamiltonian and Schrodinger Equation for Helium's Electrons

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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.
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