A Hartree-Fock Example Using Helium

Transcription

1 Univesity of Connecticut Chemisty Education Mateials Depatment of Chemisty June 6 A Hatee-Fock Example Using Helium Cal W. David Univesity of Connecticut, Cal.David@uconn.edu Follow this and additional woks at: Recommended Citation David, Cal W., "A Hatee-Fock Example Using Helium" 6. Chemisty Education Mateials..

2 A Hatee-Fock Example Using Helium C. W. David Depatment of Chemisty Univesity of Connecticut Stos, Connecticut Dated: June, 6 I. SYNOPSIS This mateial sets up a Hatee Fock computation fo the two electons of Helium so that the fomalism can be undestood in a simple context. II. INTRODUCTION The equations used in teating the Hatee-Fock methodology ae abstact enough that it is not at all clea whethe one undestands what they mean when applied to a eal wold case. Since the method is incopoated into a multitude of ab initio compute pogams, students have little oppotunity to actually wok with an example and see what is going on, why assumptions need to be made, and whee in the pocedue these assumptions ae used. This tact attempts to give students an example which can be used to help in the undestanding of this impotant methodology. III. THE MODEL Conside Helium s electons. They ae govened by a Schödinge Equation of the fom Hψ Eψ whee H us the Hamiltonian fo these electons. We wite this Hamiltonian in standad fom Ĥ op Ĥ + Ĥ + 3. whee H is the hydogenic Hamiltonian fo electon one, and H is obviously, the same fo electon, i.e., Ĥ i i Z i in atomic units whee Z fo helium. Fo the gound state, we wite the spatial pat of the wave function as ψ χ[ ]χ[ ] 3. whee we indicate the functional dependence using squae backets i.e., spatially symmetic, since we know that the spin pat αβ αβ is going to be antisymmetic. Notice that the poduct natue of ou Ansatz influences eveything else which follows. This emains an assumption, one which is not absolutely necessay! IV. THE HARTREE-FOCK SCHEME APPLIED TO HELIUM We seek a solution of the equation Ĥ op ψ Eψ using Equation 3. and Equation 3. ou Ansatz, which becomes Ĥ + Ĥ + χ[ ]χ[ ] Eχ[ ]χ[ ] Left multiplying by χ [ ] and integating ove dx dy dz we have space Typeset by REVTEX dx dy dz χ [ Ĥ + Ĥ + χ[ ]χ[ ] E dx dy dz χ [ ]χ[ ]χ[ ] 4. space

3 which is a function of, with a simila tem when using χ [ ], and integating ove space, i.e., dx dy dz χ [ ] Ĥ + Ĥ + χ[ ]χ[ ] E dx dy dz χ [ ]χ[ ]χ[ ] 4. space space which is a function of. Fo the fist of these Equation 4., assuming pe-nomalized obitals, χ we have: o + space space } dx dy dz {χ [ ] Ĥ χ[ ]χ[ ] } dx dy dz {χ [ ] Ĥ χ[ ]χ[ ] + dx dy dz {χ [ ] space } χ[ ]χ[ ] Eχ[ ] 4.3 < Ĥ > χ[ ] +Ĥχ[ ] } + dx dy dz {χ [ ] χ[ ] χ[ ] space Eχ[ ] 4.4 whee < Ĥ > } dx dy dz {χ [ ]Ĥχ[ ] ove it s own space with a simila tem fo electon. V. GENERATING THE SIMULTANEOUS EQUATIONS The tem < V > space } dx dy dz {χ [ ] χ[ ] is the key to this and vitually all othe self-consistent field methods scheme. Then ou fist SCF equation Equation 4. becomes [ ] < Ĥ > +Ĥ+ < V > χ[ ] Eχ[ ] 5. which is an equation fo χ[ ] based on one numbe < Ĥ > and two opeatos Ĥ and < V >, i.e., e-odeing tems and listing both equations 4.4 and 4.: { }] [Ĥ + < Ĥ > + < V > χ[ ] Eχ[ ] { }] [Ĥ + < Ĥ > + < V > χ[ ] Eχ[ ]5. symmetically, we have < V > dx dy dz {χ [ ] space } χ[ ] o, e-aanging, [Ĥ + {< V >}] χ[ ] E < Ĥ >χ[ ] [Ĥ + {< V >}] χ[ ] E < Ĥ >χ[ ]5.3 What we have hee is the essence of the method, i.e., two coupled equations one fo each electon, which explicitly includes the othe electon! VI. SIMULTANEOUS SOLVING OF THE TWO EQUATIONS The tick now is to solve each of these equations using stating assumptions concening the othe function,

4 3 i.e., assume a fom fo χ[ ] and solve fo χ[ ], then use this new fom fo χ[ ] to solve fo χ[ ], which you then cycle aound again. Ah, but life is not so kind. Thee ae majo poblems left which make the above pesciption faught with peil. Conside the integal using the law of cosines to eplace < V > space } dx dy dz {χ [ ] + χ[ ] cos γ 6. whee γ is the angle between and thee is an equivalent tem fo < V >. We see that < V > is a function of the coödinates of electon i.e., < V > [ ], and < V > is a function of the coödinates of electon i.e., < V > [ ]! This intemingling is the cause of ou gief! Paenthetically, we note that if thee was a simplification hee, quantum chemisty would be tactable, and chemisty might theefoe ossify. VII. EVALUATING THE REQUISITE INTEGRAL < V > space } dx dy dz {χ [ ] χ[ ] x x + y y + x z 7. Even if one wee to assume that the gound state hee has the fom of s obitals, i.e., χ[ ] e α eithe integal fom Equation 6. o 7. emains an intactable integal! We would have, fom Equation 6. < V > π { } π d sin ϑ dϑ dϕ e α + e α cos γ 7. which clealy emains a function of afte integation. γ as above efes to the angle between and while the latte efes to the spheical pola angle fom the z axis. This is a none tivial integal. + cos γ whee γ is the angle between and, it seem easonable to convet to double spheical pola coödinates in ode to isolate maximally the adial pats, i.e., Conside x x + y y + z z which equals, in double spheical pola coödinates which is sin ϑ cos φ sin ϑ cos φ + sin ϑ sin φ sin ϑ sin φ + cos ϑ cos ϑ sin ϑ sin ϑ cos φ cos φ + sin φ sin φ + cos ϑ cos ϑ

5 4 which means that + cos γ + sin ϑ sin ϑ cos φ cos φ + sin φ sin φ + cos ϑ cos ϑ so that cos γ sin ϑ sin ϑ cos φ cos φ + sin φ sin φ + cos ϑ cos ϑ o, cos ϑ cos ϑ + sin ϑ sin ϑ cosφ φ which would, on the suface, make the desied integal simple. But it doesn t. π { } π < V > d sin ϑ dϑ dϕ e α + e α cos ϑ cos ϑ + sin ϑ sin ϑ cosφ φ 7.3 VIII. AN ASIDE which we e-wite one moe time If + cos γ then choosing tempoaily we have and allowing x + cos γ we have + x x cos γ + x xµ whee µ cos γ. Then, geneating the Taylo expansion as a function of µ we have + + x +x xµ µ + µ! +x xµ µ µ But +x xµ x + x xµ 3 which, at µ is +x xµ +x xµ x + x 3/ x +x xµ 3 3x + x 5/

6 5 and etc., so 3 +x xµ 3 x +x xµ 3 + x + 3x +x xµ 5/ x + x 3/ µ 35x + x 7/ which we e-wite as 3x + x + x / + x 3/ µ+! + x 5/ µ + 53x3 3! + x 7/ µ x4 4! + x 9/ µ x5 5! + x / µ 5 8. which, we emembe, means n n n i n i i + n! x n µ n + x n+ i + x n cos ϑ cos ϑ + sin ϑ sin ϑ cosφ φ n n! + x n+ whee we have chosen to slightly simplify the tem cos φ cos φ + sin φ sin φ cosφ φ IX. RETURNING TO THE MAIN PROBLEM We need to substitute the expansion into the integal, and then integate tem by tem. We obtain e α n n i i + n! π π < V > d sin ϑ dϑ dϕ n cos ϑ cos ϑ + sin ϑ sin ϑ cosφ φ n + e α n+ It is obvious that thee ae no futhe simplifications, and we ae doomed to evaluate an infinite numbe of tems, doing non-tivial integals adding insult to injuy. One sees whee semi-empiical methods come fom now, since declaing the entie integal to have a value sidesteps actually finding that value. How shewd.