The Hartree-Fock Method Applied to Helium's Electrons

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]

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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 ψ = 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 ˆ WAVE FUNCTION APPROXIMATIONS HψHopψ = Eψ

We start with the Hamiltonian (in symbolic form): which becomes

ˆ ˆ ˆ 1 Hop = H1 + H2 + r12 1 Hˆ1 + Hˆ2 + 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 ˆ 1 2 Z Hi = − ∇i − dx1dy1dz1 we have 2 ri

Z Z ∗ ˆ ˆ 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 ∗ ˆ ˆ 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) ∗ ˆ dx1dy1dz1φ1(r1)H1φ1(r1)φ1(r2) space 1 Z ∗ ˆ + dx1dy1dz1φ1(r1)H2φ1(r1)φ1(r2) or space 1

Z ˆ ˆ ∗ 1 < 1|H1|1 > 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 < 1|V |1 >≡ dx1dy1dz1φ1(r1) φ1(r1) 1 2 1 1 space 1 r12 then cycle around again. < 1|V |1 > (and its equivalent, < 2|V |2 >) are func- is the key to this (and virtually all other “self-consistent tions of coördinates,i.e., field” methods) scheme, and we define it as < 1|V |1 > for future reference. Symmetrically, we have < 1|V |1 >= f(~r2) Z ∗ 1 < 2|V |2 >≡ dx2dy2dz2φ1(r2) φ1(r2) and space 2 r12 is the key to this (and virtually all other “self-consistent < 2|V |2 >= f(~r1) field” methods) Then our SCF equation becomes h i is the key to this (and virtually all other “self-consistent < 1|Hˆ1|1 > +Hˆ2+ < 1|V |1 > 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 < 1|V |1 >), i.e., re-ordering terms, Z ∗ 1 < 1|V |1 >≡ dx1dy1dz1φ1(r1) φ1(r1) h n oi space 1 r12 Hˆ2 + < 1|Hˆ1|1 > + < 1|V |1 > φ1(r2) = Eφ1(r2) h n oi Hˆ + < 1|Hˆ |1 > + < 2|V |2 > 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 < 1|V |1 > (r , ϑ , ϕ ) ≡ r2dr sin ϑ dϑ 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 ϑ2 cos ϑ1 + sin ϑ1 sin ϑ2 cos(ϕ2 − ϕ1) 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 ˆ Hop = − 2 2 → − 2 2 (3.1) on the ground states 1s2, we would have, as an example, 2µR dϑ 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 ψ (for “real” centered on the nucleus. Since the distance from the helium’s electrons) is 3

  2 ∂ψ ∂ψ  2 ∂ r ∂ sin ϑ1 2 ¯h  1 1 ∂r1 1 ∂ϑ1 1 ∂ ψ − 2  + + 2 2  + 2me r1 ∂r1 sin ϑ1 ∂ϑ1 sin ϑ1 ∂ϕ1

 2 ∂ψ ∂ψ  ∂ r ∂ sin ϑ2 2 1 2 ∂r2 1 ∂ϑ2 1 ∂ ψ  2  + + 2 2  (4.1) r2 ∂r2 sin ϑ2 ∂ϑ2 sin ϑ2 ∂ϕ2 

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 = −ı¯h ∂x , etc.) induced form

2 2 2 2 2 2 2 2 ¯h 2 2 ¯h ∂ ∂ ∂ ∂ ∂ ∂ − ∇1 + ∇2 = − 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 ∂ψ  2 2 2 2 2 ∂ r 2meR sin ϑ1 ∂ϕ1 sin ϑ2 ∂ϕ2 1 1 ∂r1 0 r2  ∂r  1 1 with sin ϑi = 1 for i=1 and 2. Lastly, defining η = ϕ2 − ϕ1, and using the chain rule, (with a similar expression for r2) i.e., ψ 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  ∂ψ  2 ∂ sin ϑ1 2 ¯h  1 ∂ϑ1 1 ∂ ψ which completes this derivation of part of Equation 3.1. − + + 2  2 2  We differ by a factor of two from Equation 3.1 and have 2meR  sin ϑ1 ∂ϑ1 sin ϑ1 ∂ϕ1 retained me rather than using µ (which is inappropriate  ∂ψ  here). ∂ sin ϑ2 2 1 ∂ϑ2 1 ∂ ψ   + 2 2  sin ϑ2 ∂ϑ2 sin ϑ2 ∂ϕ2  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 ϑ 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 ϑ1 cos ϑ2 + sin ϑ1 sin ϑ2 cos(ϕ2 − ϕ1) a two-dimensional form, using ϕ1 and ϕ2 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 ϑ differentiation do not appear in the final result, so ¯h ∂ ψ ∂ ψ Z Z 1 π − 2 2 + 2 + − − + ψ = Eψ it behooves us to fix ϑ1 = ϑ2 = 2 for both electrons, 2meR ∂ϕ1 ∂ϕ2 R R r12 which is a value that make sin ϑ1 and sin ϑ2 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. “work”. In our notation this becomes The repulsive part of the potential energy is 1 1 1 √ = 2Rp1 − cos(ϕ − ϕ ) r p 2 2 2 1 12 r1 + r2 − 2r1r2 cos η (law of cosines) which in this specialized case is which is again two-dimensional in the same sense as Equation 5.1. 1 1 = √ √ (5.2) We thus have pR2 + R2 − 2R2 cos η 2R 1 − cos η

¯h2 ∂2 ∂2 2Z 1 Hˆ = − + − + √ (5.3) op 2 2 2 p 2meR ∂ϕ1 ∂ϕ2 R 2R 1 − cos(ϕ2 − ϕ1)

(where Z/R is the potential energy of attraction of an wave function as electron to the nucleus) as the Hamiltonian operator, complete with (constant) nuclear-electron attraction and ψ = χ[~r ]χ[~r ] = χ[ϕ ]χ[ϕ ] (5.5) electron-electron repulsion. 1 2 1 2 We write this Hamiltonian in “standard” form (we indicate the functional dependence using square ˆ ˆ ˆ 1 brackets) i.e., spatially symmetric, since we know that Hop = H1 + H2 + √ p (5.4) 2R 1 − cos(ϕ2 − ϕ1) the spin part (α(1)β(2) − α(2)β(1)) is going to be antisymmetric. ˆ where H1 is the hydrogenic Hamiltonian for electron one, We seek a “solution” of the equation and Hˆ2 is obviously, the same for electron 2, i.e.,

2 2 2 Hˆopψ = Eψ ˆ ¯h 2 Z ¯h ∂ Z Hi = − 2 ∇i − = − 2 − 2meR R 2meR ∂φi R and using Equation 5.3 as decomposed using Equation For the ground state, we write the spatial part of the 5.4, which becomes (using the Ansatz Equation 5.5)

! ˆ ˆ 1 H1 + H2 + √ p χ[ϕ1]χ[ϕ2] = Eχ[ϕ1]χ[ϕ2] 2R 1 − cos(ϕ2 − ϕ1)

∗ Left multiplying by χ [ϕ1] and integrating over dϕ1 we have

Z ! ! ∗ ˆ ˆ 1 dϕ1 χ [ϕ1) H1 + H2 + √ p χ[ϕ1]χ[ϕ2] = space 1 2R 1 − cos(ϕ2 − ϕ1) Z ∗ E dϕ1 (χ [ϕ1)χ[ϕ1]χ[ϕ2]) (5.6) space 1

∗ with a similar term when using χ [ϕ2], and integrating over space 2, i.e.,

Z ! ! ∗ ˆ ˆ 1 dϕ2 χ [ϕ2] H1 + H2 + √ p χ[ϕ1]χ[ϕ2] = space 2 2R 1 − cos(ϕ2 − ϕ1) Z ∗ E dϕ2 (χ [ϕ2]χ[ϕ1]χ[ϕ2]) space 2 5

For the ﬁrst of these (Equation 5.6), assuming pre-normalized orbitals we have: Z n ∗ o dϕ1 χ [ϕ1] Hˆ1χ[ϕ1]χ[ϕ2] space 1 Z n ∗ o + dϕ1 χ [ϕ1] Hˆ2χ[ϕ1]χ[ϕ2] space 1 Z ( ! ) ∗ 1 + dϕ1 χ [ϕ1] √ p χ[ϕ1]χ[ϕ2] space 1 2R 1 − cos(ϕ2 − ϕ1)

= Eχ[ϕ2] (5.7) or

< 1|Hˆ1|1 > χ[ϕ2] + Hˆ2χ[ϕ2] + Z ( ! ) ∗ 1 dϕ1 χ [ϕ1] √ p χ[ϕ1] χ[ϕ2] = Eχ[ϕ2] (5.8) space 1 2R 1 − cos(ϕ2 − ϕ1)

R ∗ where < 1|Hˆ1|1 >= dϕ1χ [ϕ1]H1χ[ϕ1] over it’s own The last term on the l.h.s. of Equation 5.8 allows us space (and a similar term for electron 2). to deﬁne

Z ( ! ) ∗ 1 < 1|V |1 >≡ dϕ1 χ [ϕ1] √ p χ[ϕ1] space 1 2R 1 − cos(ϕ2 − ϕ1) which is the key to this (and virtually all other “self-consistent ﬁeld” methods) scheme. Symmetrically, we have

Z ( ! ) ∗ 1 < 2|V |2 >≡ dϕ2 χ [ϕ2] √ p χ[ϕ2] space 2 2R 1 − cos(ϕ2 − ϕ1)

Then our SCF equation becomes i.e., assume a form for χ[ϕ1] and solve for χ[ϕ2], then use this new form for χ[ϕ ] to solve for χ[ϕ ], which one then h ˆ ˆ i 2 1 < 1|H1|1 > +H2+ < 1|V |1 > χ[ϕ2] = Eχ[ϕ2] (5.9) cycles around again and again, until convergence of some kind is achieved. which is an equation for χ[ϕ2] based on one “number” < 1|V |1 > (and its equivalent, < 2|V |2 >) are func- ˆ ˆ (< 1|H1|1 >) and two operators (H2 and < 1|V |1 >), tions of co¨ordinates,i.e., i.e., re-ordering terms, h n oi < 1|V |1 >= f[ϕ2] Hˆ + < 1|Hˆ |1 > + < 1|V |1 > χ[ϕ ] = Eχ[ϕ ] 2 1 2 2 and h n oi Hˆ + < 2|Hˆ |2 > + < 2|V |2 > χ[ϕ ] = Eχ[ϕ ] 1 2 1 1 < 2|V |2 >= f[ϕ1] (5.10) To see what these integrals entail (in terms of actually or, re-arranging, carrying them out), we have for a real electron-electron interaction term h i Hˆ2 + {< 1|V |1 >} χ[ϕ2] = (E− < 1|Hˆ1|1 >)χ[ϕ2] Z ∗ 1 h i < 1|V |1 > [ϕ2] ≡ dϕ1 χ [ϕ1] χ[ϕ1] space 1 r12 Hˆ1 + {< 2|V |2 >} χ[ϕ1] = (E− < 2|Hˆ2|2 >)χ[ϕ1] (5.11) which would become Z 2π ( !) The “trick” now is to solve each of these equations 2 1 < 1|V |1 > [ϕ2] ≡ dϕ1 χ [ϕ1] √ p for starting assumptions concerning the other function, 0 2R 1 − cos(φ2 − φ1) 6

This integral evaluates to a function of ϕ2 (see Appendix which are, in spherical polar coördinates: II)! The opposite obtains for < 2|V |2 >, i.e., it evaluates ˆ ~r1 = r1 sin ϑ1 cos ϕ1î + r1 sin ϑ1 sin ϕ1ˆj + r1 cos ϑ1k to a function of ϕ1. It is this double interdependence which makes the Hartree-Fock scheme “entangled” in the sense that the and equation for χ[ϕ2] depends on ϕ1 and vice versa and ˆ the coupled Equations 5.10 commingle the two variables ~r2 = r2 sin ϑ2 cos ϕ2î + r2 sin ϑ2 sin ϕ2ˆj + r2 cos ϑ2k again. Now Appendix I, The angle between two radius vectors 2 2 2 r12 = r1 + r2 − 2r1r2 cos η Let and ˆ ~r1 = x1î + y1ˆj + z1k r2 = (x − x )2 + (y − y )2 + (z − z )2 and 12 1 2 1 2 1 2 ˆ ~r2 = x2î + y2ˆj + z2k which, in double spherical polar coordinates becomes

2 2 2 2 r12 = (r1 sin ϑ1 cos φ1 − r2 sin ϑ2 cos φ2) + (r1 sin ϑ1 sin φ1 − r2 sin ϑ2 sin φ2) + (r1 cos ϑ1 − r2 cos ϑ2) which is

2 2 2 r12 = r1 − r2 − 2r1r2 (sin ϑ2 sin ϑ1 (cos φ2 cos φ1 + sin φ2 sin φ1) + cos ϑ1 cos ϑ2) which means that

2 2 2 2 r1 + r2 − 2r1r2 cos η = r1 + r2 − 2r1r2 (sin ϑ2 sin ϑ1 (cos φ2 cos φ1 + sin φ2 sin φ1) + cos ϑ1 cos ϑ2) so that

cos η = (sin ϑ2 sin ϑ1 (cos φ2 cos φ1 + sin φ2 sin φ1) + cos ϑ1 cos ϑ2)

or, Since r cos η = cos ϑ cos ϑ + sin ϑ sin ϑ cos(φ − φ ) α 1 − cos α 2 1 2 1 2 1 sin = ± 2 2

Appendix II, Evaluating the Hartree-Fock Equations we have Z 2π 2 2 ! A1 sin ϕ1 < 1|V |1 > [ϕ2] = dϕ1 √ The functions ϕ2−ϕ1 0 2 2R sin( 2 ) χ [ϕ ] = A sin nϕ + B cos nϕ (5.12) n 1 n 1 n 1 Although tractable, we see that we do not achieve the χn[ϕ2] = Cn sin nϕ2 + Dn cos nϕ2 (5.13) simplicity of the original Summerfield et al. paper. No- tice the explicit ϕ2 dependence! which are designed to give proper behavior at ϕ + 2π in standard manner are appropriate as eigenfunctions of the unperturbed Hamiltonians. Appendix III-Hyperspherical Polar Coördinates Arbitrarily choosing B1 = 0 and n=1 we have Consider transforming from the traditional Z 2π 2 2 ! A1 sin ϕ1 x , y , z , x , y , z not into double spherical coördinates, < 1|V |1 > [ϕ2] = dϕ1 √ p 1 1 1 2 2 2 0 2R 1 − cos(ϕ2 − ϕ1) but instead into a single 6-dimensional coördinate 7 scheme analogous to normal 3-dimensional spherical the rest of the transformation of the kinetic energy oper- polar coördinates. ator. Parenthetically, we note that the volume element We would have in this 6-dimensional space is

5 4 3 2 x1 = r cos ϑ dV = r sin ϑ sin ϕ1 sin ϕ2 sin ϕ3 sin ϕ4drdϑdϕ1dϕ2dϕ3dϕ4 y1 = r sin ϑ cos ϕ1 and z1 = r sin ϑ sin ϕ1 cos ϕ2 x2 = r sin ϑ sin ϕ1 sin ϕ2 cos ϕ3 ∂ r2 ∂χ 2 1 ∂r 1 y = r sin ϑ sin ϕ sin ϕ sin ϕ cos ϕ ∇ χ = + Λopχ 2 1 2 3 4 r5 ∂r r2 z2 = r sin ϑ sin ϕ1 sin ϕ2 sin ϕ3 sin ϕ4 (5.14) where Λop is the 5-dimensional angular part of the kinetic following Sommerfeld [2] where his p = 4. In this case, energy operator [3]. Since r1 and r2 do not separate 2 2 2 we have r = r1 + r2, when we sum the squares of the cleanly in this hyperspherical coördinatesystem, it is far coördinates. The ties between fully polar and double from clear whether or not this latter system is superior spherical polar coördinates might lead to some simplifi- to the double spherical polar system used above for our cation, but to see if that is true one needs to slog through purposes.

[1] J. H. Summerﬁeld, G. S. Beltrame, and J. G. Loeser, J. [3] R. J. White and F. H. Stillinger Jr., J. Chem. Phys., 52, Chem. Ed., 76, 1430 (1999) 5800 (1970). [2] A. Sommerfeld, ”Partial Diﬀerential Equations in Physics”, Academic Press, 1949, page 227