Using a Spreadsheet to Solve the Schrödinger Equations for H 2 in the Ground and Excited States Jacob Buchanan Department of Chemistry Central Washington

Using a Spreadsheet to Solve the Schrödinger Equations for H2 in the Ground and Excited States

Jacob Buchanan

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

3

GoalsBetter understanding of quantum chemistry by visualizing quantum phenomena on a

spreadsheetby solving Schrödinger equation for

molecules by handby 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.

4

Prior Knowledge

5

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. 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

6

Ex. 1. Visualization of tunneling effect in a H atom

−12𝛻2𝜓−

1𝑟𝜓=𝐸𝜓

Students can verify that is an eigenfunction with .

0.5 1.0 1.5 2.0 2.5 3.0 3.5-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5 Total EPotential E

Radius

Radius

Total E

Potential E

Kinetic E

(r) = -1/2 = -1/r =1/r – 1/21 -0.5 -1.0 0.54 -0.5 -0.25 -0.25

… … … …

7


Double-slit diffraction patterns of electrons visually illustrate the wave-like properties of particles.

8


Photoelectric effects illustrate (less visually) the particle-like property of light.


9

Visualization of the particle-like property of waves on a spreadsheet:

θ cos(θ)cos(2

θ)cos(3

θ)cos(4

θ) Ψ Ψ2

-3 -0.99 0.96 -0.91 0.84 -0.02 0.00-2 -0.42 -0.65 0.96 -0.15 -0.06 0.00-1 0.54 -0.42 -0.99 -0.65 -0.38 0.140 1.00 1.00 1.00 1.00 1.00 1.001 0.54 -0.42 -0.99 -0.65 -0.38 0.142 -0.42 -0.65 0.96 -0.15 -0.06 0.003 -0.99 0.96 -0.91 0.84 -0.02 0.00

-3 0 30

1

] / 4

Ψ 2

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Combination of many waves exhibit particle-like property

Ψ =∑𝑛=1

𝑁

cos (𝑛θ )/𝑁

𝑁=1 𝑁=100𝑁=10

-4 0 4 -4 0 4-4 0 4

Ψ 2

wave particle

# of values the momentum (p = h/λ) may have:1 10 100

Uncertainty of momentum

Uncertainty of position

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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.

Ex. 3. Solving the Schrödinger equation for H2

+ and H2

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Integral calculation on a spreadsheet

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HA HB

e- (x, y, z)

7.5

6

6rA rB

RAB = 1.5

∫−∞

∞

Ψ �̂�Ψ 𝑑 τ≅ 𝑉∑𝑖=1

𝑁

Ψ (𝜏 𝑖 ) �̂�Ψ (𝜏 𝑖 )

𝑁

H2+

x y z rA rB φA φB σ σ2 -σ2/rA -σ2/rB

σ[(1/rA-1/2)φA+(1/rB-1/2)φB]

0.3 -0.5 -2.4 1.5 2.4 0.226 0.088 0.314 0.099 0.067 0.041 0.010-1.6 1.7 0.5 2.4 2.4 0.089 0.089 0.178 0.032 0.013 0.013 -0.0030.7 1.0 1.4 1.9 1.9 0.152 0.152 0.305 0.093 0.049 0.049 0.003

-0.5 1.2 -1.8 2.2 2.2 0.115 0.115 0.229 0.053 0.024 0.024 -0.002-0.5 -1.6 0.8 1.9 1.9 0.154 0.154 0.309 0.095 0.051 0.051 0.003-0.4 -0.8 3.0 3.1 3.1 0.045 0.045 0.091 0.008 0.003 0.003 -0.0011.5 1.2 -0.7 2.0 2.0 0.139 0.139 0.278 0.077 0.039 0.039 0.001

-0.6 1.2 0.8 1.5 1.5 0.213 0.213 0.425 0.181 0.117 0.117 0.0261.2 -1.8 -2.2 3.1 3.1 0.046 0.046 0.093 0.009 0.003 0.003 -0.002

-1.6 0.2 0.4 1.7 1.7 0.192 0.192 0.384 0.147 0.089 0.089 0.0161.8 1.4 2.9 3.7 3.7 0.026 0.026 0.052 0.003 0.001 0.001 -0.001

-1.7 -0.8 0.5 2.0 2.0 0.142 0.142 0.285 0.081 0.042 0.042 0.0010.0 -1.2 -1.9 2.3 2.3 0.104 0.104 0.208 0.043 0.019 0.019 -0.003… … … … … … … … … … … …

Kinetic EAttractionRandomcoordinates 𝑒−𝑟 𝐴 φA+φB

() = ()

𝑒−𝑟 𝐵

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HA HB

e1-

7.5

6

6

e2-

r12

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.

H2

16

There are more columns but calculations can still be done conveniently on a spreadsheet.

x1 y1 z1 x2 y2 z2 r1A r1B r2A r2B r12

… … … … … … … … … … …

φ1A φ1B φ2A φ2B σ1 σ2 Ψ Ψ2

… … … … … … σ1σ2 …

-ψ2/r1A -ψ2/r1B -ψ2/r2A -ψ2/r2B ψ2/r12 -ψψ

AttractionRepulsion

Kinetic Energy

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-4

-2

0

2

Spreadsheet

HF/STO-3G

a.

u.

Potential E

Kinetic E

Total E

Attraction

Potential E

Total E

Attraction

e-e repulsion

Kinetic EH2+ H2

Average of 10 runs (each contains 8000 sampling)

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-4

-2

0

2

a.

u.

Kinetic E

Attraction

Potential E

Total E

H2+ H2

e-e repulsionKinetic E

Total E

Potential E

Attraction

Range and standard deviation of 10 runs

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Comparison of H2 energy calculated using spreadsheet and using HF theory with various basis sets

0.5 1 1.5 2 2.5 3 3.5-1.2

-1.1

-1.0

-0.9

-0.8

-0.7

-0.6Spreadsheet

HF/STO-1G

HF/STO-2G

HF/STO-3G

HF/cc-pV6Z

Bond distance (au)

E (

au)

20

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. This reduction of e-e repulsion is called exchange energy.

The e-e repulsion for the S1 state and T1 state:<> =

(- )

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Ex. 5. Correlation energy in H2: +

Correlation energy

• 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.

Exact energy (infinite # of determinants)

Hartree-Fock energy (1 determinant)

E

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Ex. 5. Correlation energy in H2: +

Using Hartree product functions to calculate correlation energy approximately for a H2 molecule: w/o correlation w/ correlation

Hartree-Fock energy (1 determinant)

CID energy (2 determinants)

Correlation energy from double excitation

Ecorr

E

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Acknowledgements

Benjamin Livingston (CWU Alumnus pursuing PhD in Math at Oregon State University)

Dr. Robert RittenhouseCWU COTS Summer Writing Grant &

Faculty Development Fund (YG)CWU Office of Graduate Studies &

Research Faculty Travel Fund (YG)CWU Department of Chemistry

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Questions?

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σ

σ*

σα1σ*β1 (S1) σα1σ*α1

(T1)w/o exchange w/ exchangeUse Hartree product = σ1σ*2 Use determinantal

Difference between S1 and T1 excited states:

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2 2 2/r122/r12

σ1σ*2 - … … … …

Exchange E

Exchange energy is the essentially the difference between repulsion of electrons with same spin vs. repulsion of electrons with opposite spin.

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−12𝑟 2 [ 𝜕𝜕𝑟 (𝑟2 𝜕

𝜕𝑟 )+ 1𝑠𝑖𝑛𝜃

𝜕𝜕 𝜃 (𝑠𝑖𝑛𝜃 𝜕

𝜕𝜃 )+ 1𝑠𝑖𝑛2𝜃 ( 𝜕2

𝜕𝜙2 )]𝜓 − 1𝑟 𝜓=𝐸𝜓

)

( )

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H atom calculationsFamiliarize students with

spreadsheet format.Calculation of kinetic energy and

potential energy.Comparison to analytical

solution.

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CompilationResults from each student

compiled using Google spreadsheet

Results averaged and compared to HF/STO-3G calculations and experimental data.

Documents

Using a Spreadsheet to Solve the Schrödinger Equations for H 2 in the Ground and Excited States Jacob Buchanan Department of Chemistry Central Washington