University of Connecticut OpenCommons@UConn Chemistry Education Materials Department of Chemistry 2009 The aH rtree-Fock Method Applied to Helium's Electrons Carl W. David University of Connecticut, [email protected] Follow this and additional works at: https://opencommons.uconn.edu/chem_educ Part of the Physical Chemistry Commons Recommended Citation David, Carl W., "The aH rtree-Fock Method Applied to Helium's Electrons" (2009). Chemistry Education Materials. 72. https://opencommons.uconn.edu/chem_educ/72 The Hartree Fock Method Applied to Helium's Electrons C. W. David (Dated: March 5, 2009) I. SYNOPSIS where Z = 2 for Helium. For the ground state, we write the spatial part of the The difficulties of applying the Hartree-Fock method to wave function as many body problems is illustrated by treating Helium's electrons up to the point where tractability vanishes. Second, the problem of applying Hartree-Fock methods = φ1(r1)φ1(r2) to the helium atom's electrons, when they are constrained to remain on a sphere, is revisited. The 6-dimensional total energy operator is reduced to a 2-dimensional one, i.e., spatially symmetric, since we know that the spin part and the application of that 2-dimensional operator in the (α(1); β(2) − α(2)β(1))is going to be antisymmetric. Hartree-Fock mode is discussed. We seek a \solution" of the equation II. HELIUM HAMILTONIAN AND STARTING ^ WAVE FUNCTION APPROXIMATIONS H Hop = E We start with the Hamiltonian (in symbolic form): which becomes ^ ^ ^ 1 Hop = H1 + H2 + r12 1 H^1 + H^2 + φ1(r1)φ1(r2) = Eφ1(r1)φ1(r2) r12 where H1 is the hydrogenic hamiltonian for electron one, and H2 is obviously, the same for electron 2, i.e., Left multiplying by φ1(r1) and integrating over ^ 1 2 Z Hi = − ri − dx1dy1dz1 we have 2 ri Z Z ∗ ^ ^ 1 ∗ dx1dy1dz1 φ1(r1) H1 + H2 + φ1(r1)φ1(r2) = E dx1dy1dz1φ1(r1)φ1(r1)φ1(r2) (2.1) space 1 r12 space 1 with a similar term when using φ1(r2), and integrating over space 2, i.e., Z Z ∗ ^ ^ 1 ∗ dx2dy2dz2 φ1(r2) H1 + H2 + φ1(r1)φ1(r2) = E dx2dy2dz2φ1(r2)φ1(r1)φ1(r2) space 2 r12 space 2 Z We then obtain, for the first of these (Equation 5.6), ∗ 1 + dx1dy1dz1φ1(r1) φ1(r1)φ1(r2) assuming pre-normalized orbitals: space 1 r12 Z = Eφ1(r2) (2.2) ∗ ^ dx1dy1dz1φ1(r1)H1φ1(r1)φ1(r2) space 1 Z ∗ ^ + dx1dy1dz1φ1(r1)H2φ1(r1)φ1(r2) or space 1 Z ^ ^ ∗ 1 < 1jH1j1 > φ1(r2) + H2φ1(r2) + φ1(r2) dx1dy1dz1φ1(r1) φ1(r1) = Eφ1(r2) (2.3) space 1 r12 Typeset by REVTEX 2 R ∗ where H1;1 = φ1(r1)(H1) φ1(r1) over it's own space. The \trick" now is to solve each of these equations for The term starting assumptions concerning the other function, i.e., Z assume a form for φ1(r1) and solve for φ1(r2), then use ∗ 1 this new form for φ (r ) to solve for φ (r ), which you < 1jV j1 >≡ dx1dy1dz1φ1(r1) φ1(r1) 1 2 1 1 space 1 r12 then cycle around again. < 1jV j1 > (and its equivalent, < 2jV j2 >) are func- is the key to this (and virtually all other \self-consistent tions of co¨ordinates,i.e., field” methods) scheme, and we define it as < 1jV j1 > for future reference. Symmetrically, we have < 1jV j1 >= f(~r2) Z ∗ 1 < 2jV j2 >≡ dx2dy2dz2φ1(r2) φ1(r2) and space 2 r12 is the key to this (and virtually all other \self-consistent < 2jV j2 >= f(~r1) field” methods) Then our SCF equation becomes h i is the key to this (and virtually all other \self-consistent < 1jH^1j1 > +H^2+ < 1jV j1 > φ1(r2) = Eφ1(r2) (2.4) field” methods) To see what these integrals entail (in terms of actually which is an equation for φ1(r2) based on one \number" carrying them out), we have (from above) and two operators (H2 and < 1jV j1 >), i.e., re-ordering terms, Z ∗ 1 < 1jV j1 >≡ dx1dy1dz1φ1(r1) φ1(r1) h n oi space 1 r12 H^2 + < 1jH^1j1 > + < 1jV j1 > φ1(r2) = Eφ1(r2) h n oi H^ + < 1jH^ j1 > + < 2jV j2 > φ (r ) = Eφ (r ) which, assuming we are looking solely for the ground 1 2 1 1 1 1 state of the Helium atom's electrons, based on the ground (2.5) states 1s2, we would have, as an example, ! Z 1 < 1jV j1 > (r ;# ;' ) ≡ r2dr sin # d# dφ e−2αr1 2 2 2 1 1 1 1 1 p 2 2 space 1 r1 + r2 − 2r1r2 cos γ with electrons to the nucleus was constant in this model, the Hamiltonian simplified to such an extent as to make the cos γ = cos #2 cos #1 + sin #1 sin #2 cos('2 − '1) problem tractable from the Hartree-Fock point of view. The authors introduce the kinetic energy operator Clearly, this is a non-trivial integral, and is not suscepti- without derivation as ble to the treatments used when the integral is done over both r and r . which, assuming we are looking solely for 1 2 1 d2 1 d2 the ground state of the Helium atom's electrons, based ^ Hop = − 2 2 ! − 2 2 (3.1) on the ground states 1s2, we would have, as an example, 2µR d# 2µR dζ Solving the equation set Equation 2.5 is non-trivial, to say the least. We here derive a variant of Equation 3.1 and explain and explore how the Hartree Fock method works in this simple case. III. AN EXAMPLE ILLUSTRATING HOW EASY MISTAKES CAN BE MADE; ELECTRONS ON A SPHERE USING HF METHODOLOGY IV. 6-DIMENSIONAL HELIUM HAMILTONIAN A simplified model for helium's electrons, useful in studying the Hartree-Fock method, was introduced [1] in The \true" kinetic energy operator portion of the which helium's electrons moved on the surface of a sphere Hamiltonian operating on a wave function (for \real" centered on the nucleus. Since the distance from the helium's electrons) is 3 8 2 2 @ @ 3 2 @ r @ sin #1 2 ¯h < 1 1 @r1 1 @#1 1 @ − 2 4 + + 2 2 5 + 2me :r1 @r1 sin #1 @#1 sin #1 @'1 2 2 @ @ 39 @ r @ sin #2 2 1 2 @r2 1 @#2 1 @ = 2 4 + + 2 2 5 (4.1) r2 @r2 sin #2 @#2 sin #2 @'2 ; which is six-dimensional (see Appendix III). It is written spherical polar co¨ordinates. This is comparable to the @ in the infinite nuclear mass approximation using double algorithmic (from px = −{¯h @x , etc.) induced form 2 2 2 2 2 2 2 2 ¯h 2 2 ¯h @ @ @ @ @ @ − r1 + r2 = − 2 + 2 + 2 + 2 + 2 + 2 2me 2me @x1 @y1 @z1 @x2 @y2 @z2 The 1-subscript refers to electron one, and the 2-subscript and eliminates taking partial derivatives with respect to refers to the other electron. these angles. This means that we are working in the x- We endeavor to obtain Equation 3.1 from Equation 4.1. y plane of each electron, i.e., the electrons are not on a First, we note that the radius for both electrons in this sphere, they are on a circle in the x-y plane! model is fixed at r1 = r2 = R, so the radial kinetic energy We now have contributors vanish (there is no radial kinetic energy), ( ) ¯h2 1 @2 1 @2 − + (4.2) 2 2 @ 3 2 2 2 2 2 @ r 2meR sin #1 @'1 sin #2 @'2 1 1 @r1 0 r2 4 @r 5 1 1 with sin #i = 1 for i=1 and 2. Lastly, defining η = '2 − '1, and using the chain rule, (with a similar expression for r2) i.e., 6= f(r1; r2) where we have r is the magnitude of ~r and r is the magnitude of ~r . 1 1 2 2 ¯h2 @2 @2 ¯h2 @2 We assume then that ~r1 = ~r2 = R, and then are left with − 2 2 + 2 = − 2 2 (4.3) 2meR @η @η meR @η 82 @ 3 2 @ sin #1 2 ¯h < 1 @#1 1 @ which completes this derivation of part of Equation 3.1. − + + 2 4 2 2 5 We differ by a factor of two from Equation 3.1 and have 2meR : sin #1 @#1 sin #1 @'1 retained me rather than using µ (which is inappropriate 2 @ 39 here). @ sin #2 2 1 @#2 1 @ = 4 + 2 2 5 sin #2 @#2 sin #2 @'2 ; V. THE HARTREE-FOCK EQUATIONS We have achieved a reduction in dimensionality from 6 to 4. The reduction of the Hartree-Fock treatment of he- Next, we note that the angle between the two radius lium's electrons from a six-dimensional to a \one- vectors, denoted as # in the original paper, is given (here dimensional" (in η) problem is wonderful, in making it denoted by η) as tractable, but misses the sense of how the Hartree-Fock method is supposed to work, and might be better left in cos η = cos #1 cos #2 + sin #1 sin #2 cos('2 − '1) a two-dimensional form, using '1 and '2 as variables, so that students could see what is going on. (see Appendix I). Therefore, we continue the derivation using the two In order to continue along the path of deriving Equa- dimensional form: tion 3.1 from Equation 4.1 we note that the terms involv- 2 2 2 ing # differentiation do not appear in the final result, so ¯h @ @ Z Z 1 π − 2 2 + 2 + − − + = E it behooves us to fix #1 = #2 = 2 for both electrons, 2meR @'1 @'2 R R r12 which is a value that make sin #1 and sin #2 equal to one, (5.1) 4 (where Z=2 for helium) for the total energy operator, as which is the simplification which makes the original paper suggested by Equation 4.2.