Using a Spreadsheet to Solve the Schrödinger Equations for Small
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Using a Spreadsheet to Solve the Schrödinger Equations for Small Molecules: An Undergraduate Quantum Chemistry Lab Jacob Buchanan, Benjamin Livingston, Yingbin Ge* Department of Chemistry Central Washington University “I think you should be more explicit here in Step Two.” By Sidney Harris, Copyright 2007, The New Yorker Goals Better understanding of quantum chemistry • by visualizing quantum phenomena on a spreadsheet • by solving Schrödinger equation for molecules by hand • by breaking the total energy into pieces – Potential energy: N-N repulsion, e-e repulsion, e-N attraction. – Kinetic energy of each electron. • by hand calculating exchange and correlation energy on a spreadsheet. 3 Prior Knowledge • Multivariable differential and integral calculus • Quantum mechanical postulates: πψ = oψ ψπψπτ <o> = (when ψ is a real function.) ψψπτ • Linear algebra and differential equations are not required • Experience with Excel or Google spreadsheet 4 Quantum chemistry exercises on a spreadsheet Ex. 1. Visualization of tunneling effect in a H atom Ex. 2. Visualization of particle-like properties of waves Ex. 1-2 help students better understand abstract QM concepts and gain experience with spreadsheet calculations. Ex. 3. Solving the Schrödinger equation for H2+ and H2 Ex. 4. Exchange energy in H2*: first singlet vs triplet excited states Ex. 5. Correlation energy in H2: a 2-determinant wave function 5 Ex. 1. Visualization of tunneling effect in a H atom −1 2 1 π» π − π = πΈπ 2 π 1 2 Students can easily verify that π = π −π is an eigenfunction with πΈ = − . Radius (r) … Total E Potential E Kinetic E = -1/2 = -1/r =1/r – 1/2 0.5 -0.5 -2.0 1.5 0.6 -0.5 -1.7 1.2 0.7 -0.5 -1.4 0.9 … … … 1.5 Total E Potential E Kinetic E 1.0 0.5 0.0 -0.5 -1.0 -1.5 -2.0 0.5 1.5 2.5 3.5 Radius 6 Ex. 2. Visualization of particle-like properties of waves Double-slit diffraction patterns of electrons visually illustrate the wave-like properties of particles. Photoelectric effects illustrate (less visually) the particle-like property of light. Visualization of the particle-like property of waves on a spreadsheet: Ψ =[ cos θ + cos 2θ + cos 3θ + cos(4θ) ] / 4 θ cos(θ) cos(2θ) cos(3θ) cos(4θ) Ψ -3 -0.99 0.96 -0.91 0.84 -0.02 -2 -0.42 -0.65 0.96 -0.15 -0.06 -1 0.54 -0.42 -0.99 -0.65 -0.38 0 1.00 1.00 1.00 1.00 1.00 1 0.54 -0.42 -0.99 -0.65 -0.38 2 -0.42 -0.65 0.96 -0.15 -0.06 3 -0.99 0.96 -0.91 0.84 -0.02 Ψ2 0.00 0.00 0.14 1.00 0.14 0.00 0.00 1 Ψ2 0 -3 0 3 7 Combination of many waves exhibit particle-like property π Ψ= cos πθ /π π=1 π=1 π = 100 π = 10 Ψ2 -4 0 4 -4 0 wave # of values the momentum (p = h/λ) may have: 1 10 4 -4 0 4 particle 100 Uncertainty of momentum Uncertainty of position 8 Ex. 3. Solving the Schrödinger equation for H2+ and H2 H2 + • Introduction to solving problems without analytical solutions. • Nucleus positions fixed with a bond length of 1.5 a.u. • Calculate every potential and kinetic energy component approximately. • Need to evaluate integrals on a spreadsheet. 9 Integral calculation on a spreadsheet • Using a spreadsheet to estimate the value of an integral: πΏ πΏ π(π₯)ππ₯ ≅ −πΏ π π π=−πΏ • For a fast decaying function f(x,y,z), the function value at several thousand randomly selected data points are calculated within a given volume. • The product of the volume and the average function value is approximately the integral. 10 H2+ e- (x, y, z) rA HA Y rB HB RAB = 1.5 X Z ∞ ΨπΨπτ ≅ π −∞ π π=1 Ψ ππ πΨ ππ π 11 1 2 1 π 1 2 − π» 2 (π −π ) = ( − ) π −π Random coordinates x y 0.3 -1.6 0.7 -0.5 -0.5 -0.4 1.5 -0.6 1.2 -1.6 1.8 -1.7 0.0 … z -0.5 1.7 1.0 1.2 -1.6 -0.8 1.2 1.2 -1.8 0.2 1.4 -0.8 -1.2 … π −ππ΄ π −ππ΅ φA+φB rA -2.4 0.5 1.4 -1.8 0.8 3.0 -0.7 0.8 -2.2 0.4 2.9 0.5 -1.9 … rB 1.5 2.4 1.9 2.2 1.9 3.1 2.0 1.5 3.1 1.7 3.7 2.0 2.3 … φA φB σ 2.4 0.226 0.088 2.4 0.089 0.089 1.9 0.152 0.152 2.2 0.115 0.115 1.9 0.154 0.154 3.1 0.045 0.045 2.0 0.139 0.139 1.5 0.213 0.213 3.1 0.046 0.046 1.7 0.192 0.192 3.7 0.026 0.026 2.0 0.142 0.142 2.3 0.104 0.104 … … … Attraction Kinetic E σ2 -σ2/rA -σ2/rB σ[(1/rA-1/2)φA+(1/rB-1/2)φB] 0.314 0.099 0.067 0.041 0.010 0.178 0.032 0.013 0.013 -0.003 0.305 0.093 0.049 0.049 0.003 0.229 0.053 0.024 0.024 -0.002 0.309 0.095 0.051 0.051 0.003 0.091 0.008 0.003 0.003 -0.001 0.278 0.077 0.039 0.039 0.001 0.425 0.181 0.117 0.117 0.026 0.093 0.009 0.003 0.003 -0.002 0.384 0.147 0.089 0.089 0.016 0.052 0.003 0.001 0.001 -0.001 0.285 0.081 0.042 0.042 0.001 0.208 0.043 0.019 0.019 -0.003 … … … … … H2 e1- HA r12 e2- Y HB X Z Nucleus positions fixed with a bond length of 1.5 a.u. No exchange energy between electrons with opposite spin. Hartree product wave function ψ = σ(1)σ(2) is used. 13 There are more columns but calculations can still be done conveniently on a spreadsheet. x1 … y1 … z1 … φ1A φ1B … … -ψ2/r … 1A y2 … φ2A φ2B … … -ψ2/r … x2 … 1B -ψ2/r … z2 … r1A … σ1 … σ2 … 2A -ψ2/r … r1B … r2A … r2B … r12 … Ψ Ψ2 σ1σ2 … 2B ψ2/r … 1 12 -2ψπ» 2 ψ … 14 Average of 10 runs (each contains 8000 sampling) 2 H2+ H2 Kinetic E Kinetic E e-e repulsion 0 a. u. Total E Total E Potential E Attraction -2 Spreadsheet HF/STO-3G Potential E Attraction -4 15 Range and standard deviation of 10 runs 2 H2+ H2 Kinetic E Kinetic E e-e repulsion 0 Total E a. u. Potential E Total E Attraction -2 Potential E Attraction -4 16 Ex. 4. Exchange energy in H2*: first singlet vs. triplet excited states No 2 electrons can have the same set of quantum numbers. Electrons with same spin can better avoid each other than electrons with opposite spin, which reduces the e-e repulsion. Hartree product and determinantal wave functions can be used to estimate the e-e repulsion for the S1 (w/o exchange) state and T1 state (w/exchange). 1 < > π12 = ψπ1 ψdπ ψ2/r12 12 ≅ ψψdπ ψ2 ψπ»π = σ1σ∗2 (S1) ψπππ‘ = (σ1σ∗2- σ2σ∗1) πΌ1 πΌ2 (T1) 17 Difference between S1 and T1 excited states: σ* σ σα1σ*β1 (S1) w/o exchange Use Hartree product ψπ»π = σ1σ*2 σα1σ*α1 (T1) w/ exchange Use determinantal ψπππ‘ 18 Exchange energy is the essentially the difference between repulsion of electrons with same spin vs. repulsion of electrons with opposite spin. ψπ»π σ1σ*2 ψπππ‘ ψπ»π 2 σ1σ∗2 - σ2σ∗1 … Exchange E ≅ ψπππ‘ 2 ψπ»π 2/r12 ψπππ‘ 2/r12 … … … ψπππ‘ 2/r12 − 2 ψπππ‘ ψπ»π 2/r12 ψπ»π 2 19 Ex. 5. Correlation energy in H2: ψππ + Cππ₯ ψππ₯ Hartree-Fock energy (1 determinant) E Correlation energy Exact energy (infinite # of determinants) • 2-determinant wave function that includes a parametric constant Cex for the excited state determinant. • Variational method to minimize Ecorr. • Minimum energy using the 2-determinant wave function compared to the HF energy with Cex = 0. 20 Ex. 5. Correlation energy in H2: ψππ + Cππ₯ ψππ₯ Hartree-Fock energy (1 determinant) E Correlation energy from double excitation CID energy (2 determinants) Using Hartree product functions to calculate correlation energy approximately for a H2 molecule: ψππ = σ1σ2 w/o correlation ψππππ = σ1σ2 + Cexσ∗1σ∗2 w/ correlation Ecorr ≅ (ψππππ 2/r12) ψππππ2 − (ψππ 2/r12) ψππ 2 21 Acknowledgements • Central Washington University (CWU) College of the Sciences Summer Writing Grant & Faculty Development Fund • CWU Office of Graduate Studies & Research Faculty Travel Fund • CWU Department of Chemistry • Dr. Robert Rittenhouse for helpful discussion 22 Questions, comments, & suggestions? 23 −1 π π 1 π π 1 π2 2 π + π πππ + 2π 2 ππ ππ π πππ ππ ππ π ππ2 π ππ 2 −1 π 2π 2 ππ π π2 ππ + = −1 π 2π 2 ππ π2 = 1 π 2π 2 ππ π 2 π −π = 1 (2ππ −π 2 2π = 1 ( π − 1 ) 2 π ππ 1 π π πππ ππ π π πππ ππ + 1 π − π = πΈπ π 1 π2 π ππ2 π ππ2 π −π π −π − π 2 π −π ) π −π 24 H atom calculations • Familiarize students with spreadsheet format. • Calculation of kinetic energy and potential energy. • Comparison to analytical solution. 25 Compilation • Results from each student compiled using Google spreadsheet • Results averaged and compared to HF/STO-3G calculations and experimental data. 26
