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
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
4
Quantum chemistry exercises on a spreadsheet




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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
5
Ex. 1.Visualization of tunneling effect in a H atom
−1 2
1
𝛻 𝜓 − 𝜓 = 𝐸𝜓
2
𝑟
1
2
Students can verify that 𝜓 = 𝑒 −𝑟 is an eigenfunction with 𝐸 = − au.
Radius Total E Potential E Kinetic E
(r)
= -1/2 = -1/r
=1/r – 1/2
1
-0.5
-1.0
0.5
4
-0.5
-0.25
-0.25
…
…
…
…
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.
7
Ex. 2. Visualization of particle-like properties of waves
Photoelectric effects illustrate (less visually) the particle-like property of light.
8
Ex. 2. Visualization of particle-like properties of waves
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)
-3 -0.99
0.96 -0.91
0.84
-2 -0.42 -0.65
0.96 -0.15
-1
0.54 -0.42 -0.99 -0.65
0
1.00
1.00
1.00
1.00
1
0.54 -0.42 -0.99 -0.65
2 -0.42 -0.65
0.96 -0.15
3 -0.99
0.96 -0.91
0.84
Ψ
-0.02
-0.06
-0.38
1.00
-0.38
-0.06
-0.02
Ψ2
0.00
0.00
0.14
1.00
0.14
0.00
0.00
1
Ψ2
0
-3
0
3
9
Combination of many waves exhibit particle-like
property
𝑁
Ψ=
cos 𝑛θ /𝑁
𝑛=1
𝑁=1
𝑁 = 100
𝑁 = 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
10
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.

11
Integral calculation on a spreadsheet
12
H2 +
rA
HA
e- (x, y, z)
6
rB
HB
RAB = 1.5
6
7.5
∞
Ψ𝑂Ψ𝑑τ ≅ 𝑉
−∞
𝑁
𝑖=1 Ψ
𝜏𝑖 𝑂Ψ 𝜏𝑖
𝑁
13
1
2
1
𝑟
1
2
− 𝛻 2 (𝑒 −𝑟 ) = ( − ) 𝑒 −𝑟
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
…
-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
…
𝑒 −𝑟𝐴 𝑒 −𝑟𝐵 φA+φB
rA
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
…
rB
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
e 1-
HA
r12
e2-
6
HB
6
7.5
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.
15
There are more columns but calculations can still be done
conveniently on a spreadsheet.
x1
…
y1
…
z1
…
x2
…
y2
…
φ1A φ1B
… …
φ2A φ2B σ1
… … …
-ψ2/r1A
-ψ2/r1B
Attraction
z2
…
-ψ2/r2A
σ2
…
r1A
…
r1B
…
r2A
…
r2B
…
r12
…
Ψ
Ψ2
σ1σ2 …
-ψ2/r2B
1
-2ψ𝛻 2 ψ
Kinetic
Repulsion Energy
ψ2/r12
16
Average of 10 runs (each contains 8000 sampling)
2
H2+
H2
Kinetic E
a. u.
0
Kinetic E
e-e repulsion
Total E
Total E
Potential E
Attraction
-2
Spreadsheet
HF/STO-3G
Potential E
Attraction
-4
17
Range and standard deviation of 10 runs
2
H2 +
H2
Kinetic E
Kinetic E
0
e-e repulsion
Total E
a. u.
Potential E
Total E
Attraction
-2
Potential E
Attraction
-4
18
Comparison of H2 energy calculated using
spreadsheet and using HF theory with
various basis sets
-0.6
-0.7
HF/STO-1G
HF/STO-2G
HF/STO-3G
HF/cc-pV6Z
-0.8
E (au)
Spreadsheet
-0.9
-1.0
-1.1
-1.2
0.5
1
1.5
2
2.5
3
3.5
Bond distance (au)
19
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:
1
< >
𝑟12
=
ψ𝑟1 ψd𝜏
ψ2/r12
12
≅
ψψd𝜏
ψ2
ψ𝑆1 = (σ1σ∗2 + σ2σ∗1)
ψ 𝑇1 = (σ1σ∗2- σ2σ∗1)
20
Ex. 5. Correlation energy in H2: ψ𝑔𝑠 + C𝑒𝑥 ψ𝑒𝑥
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.
21
Ex. 5. Correlation energy in H2: ψ𝑔𝑠 + C𝑒𝑥 ψ𝑒𝑥
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:
ψ𝑔𝑠 = σ1σ2
w/o correlation
ψ𝑐𝑜𝑟𝑟 = σ1σ2 + Cexσ∗1σ∗2
w/ correlation
Ecorr ≅
(ψ𝑐𝑜𝑟𝑟 2/r12)
ψ𝑐𝑜𝑟𝑟2
−
(ψ𝑔𝑠 2/r12)
ψ𝑔𝑠 2
22
Acknowledgements

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
Benjamin Livingston (CWU Alumnus pursuing
PhD in Math at Oregon State University)
Dr. Robert Rittenhouse
CWU COTS Summer Writing Grant & Faculty
Development Fund (YG)
CWU Office of Graduate Studies & Research
Faculty Travel Fund (YG)
CWU Department of Chemistry
23
Questions?
24
Difference between S1 and T1 excited states:
σ*
σ
σα1σ*β1 (S1)
w/o exchange
Use Hartree product ψ𝐻𝑃 = σ1σ*2
σα1σ*α1 (T1)
w/ exchange
Use determinantal ψ𝑑𝑒𝑡
25
Exchange energy is the essentially the difference
between repulsion of electrons with same spin vs.
repulsion of electrons with opposite spin.
ψ𝐻𝑃
σ1σ*2
ψ𝑑𝑒𝑡
ψ
𝐻𝑃
2
ψ
𝑑𝑒𝑡
2
σ1σ∗2 - σ2σ∗1 …
…
Exchange E ≅
ψ𝑑𝑒𝑡 2/r12
−
ψ𝑑𝑒𝑡 2
ψ
𝐻𝑃
2/r
12
…
ψ
𝑑𝑒𝑡
2/r
12
…
ψ𝐻𝑃 2/r12
ψ𝐻𝑃 2
26
−1 𝜕
𝜕
1 𝜕
𝜕
1
𝜕2
2
𝑟
+
𝑠𝑖𝑛𝜃
+
2𝑟 2 𝜕𝑟
𝜕𝑟
𝑠𝑖𝑛𝜃 𝜕𝜃
𝜕𝜃
𝑠𝑖𝑛2 𝜃 𝜕𝜙 2
−1 𝜕
2𝑟 2 𝜕𝑟
𝜕
𝑟2
𝜕𝑟
+
=
−1 𝜕
2𝑟 2 𝜕𝑟
𝑟2
=
1
𝜕
2𝑟 2 𝜕𝑟
𝑟 2 𝑒 −𝑟
=
1
(2𝑟𝑒 −𝑟
2
2𝑟
=
1
(
𝑟
−
1
)
2
𝜕
𝜕𝑟
1 𝜕
𝑠𝑖𝑛𝜃 𝜕𝜃
𝜕
𝑠𝑖𝑛𝜃
𝜕𝜃
+
1
𝜓 − 𝜓 = 𝐸𝜓
𝑟
1
𝜕2
𝑠𝑖𝑛2 𝜃 𝜕𝜙2
𝑒 −𝑟
𝑒 −𝑟
− 𝑟 2 𝑒 −𝑟 )
𝑒 −𝑟
27
H atom calculations
Familiarize students with spreadsheet
format.
 Calculation of kinetic energy and
potential energy.
 Comparison to analytical solution.

28
Compilation
Results from each student compiled using
Google spreadsheet
 Results averaged and compared to
HF/STO-3G calculations and experimental
data.

29