Physical Chemistry
8. Variational Method

Ruifeng ZHOU

Faculty of Science &Institute for the Advancement of Higher Education
Hokkaido University

Table of Content

1. Variational principle
2. Computational quantum chemistry12

1. Variational principle• The ground state (gs) is the stationary statecorresponding to the lowest energy.
• The wavefunction and energy of the ground state isusually the most important (at room temperature).
• If 𝜓𝜓𝑖𝑖 is a stationary state with energy 𝐸𝐸𝑖𝑖 ,� 𝑖𝑖 = 𝐸𝐸𝑖𝑖 𝜓𝜓𝑖𝑖𝐻𝐻𝜓𝜓
𝐸𝐸𝑖𝑖 ≥ 𝐸𝐸gs
• Any state, 𝜓𝜓, can be expressed as a linearcombination of the stationary states (basis).
𝜓𝜓 = � 𝑐𝑐𝑖𝑖 𝜓𝜓𝑖𝑖𝑖𝑖

31. Variational principle• The stationary states are orthonormal.
𝜓𝜓𝑖𝑖 𝜓𝜓𝑗𝑗 = 𝛿𝛿𝑖𝑖,𝑗𝑗• Thus, for the state, 𝜓𝜓,� = ∑𝑖𝑖 𝑐𝑐𝑖𝑖 𝜓𝜓𝑖𝑖 𝐻𝐻� ∑𝑗𝑗 𝑐𝑐𝑗𝑗 𝜓𝜓𝑗𝑗
𝐸𝐸 = 𝜓𝜓 𝐻𝐻𝜓𝜓

� 𝑗𝑗 𝜓𝜓𝑗𝑗 = � � 𝑐𝑐𝑖𝑖∗ 𝑐𝑐𝑗𝑗 𝐸𝐸𝑗𝑗 𝜓𝜓𝑖𝑖 𝜓𝜓𝑗𝑗
= � � 𝑐𝑐𝑖𝑖 𝜓𝜓𝑖𝑖 𝐻𝐻𝑐𝑐
𝑖𝑖

𝑗𝑗𝑖𝑖

𝑗𝑗

= � 𝑐𝑐𝑖𝑖 2 𝐸𝐸𝑖𝑖 ≥ 𝐸𝐸gs � 𝑐𝑐𝑖𝑖 2 = 𝐸𝐸gs𝑖𝑖𝑖𝑖

 𝐸𝐸 is the upper bound of 𝐸𝐸gs .

41. Variational principle• The lowest upper boundof 𝐸𝐸gs can be found using
variational method.
• Suppose there is a trialfunction, 𝜓𝜓𝑎𝑎 𝑥𝑥 , which is
the wavefunction with aparameter, 𝑎𝑎.
� 𝑎𝑎
𝐸𝐸 = 𝜓𝜓𝑎𝑎 𝐻𝐻𝜓𝜓•

𝐸𝐸 is a function of 𝑎𝑎.

𝐸𝐸gs

𝐸𝐸𝑎𝑎

1. Variational principle• The lowest upper bound of𝐸𝐸𝜓𝜓𝑎𝑎 is at the minimum of 𝐸𝐸 .
𝑑𝑑 𝐸𝐸=0𝑑𝑑𝑎𝑎• An even lower upper boundcan be found by a better
trial function.
• Multiple parameters can be𝐸𝐸gs
used, such as 𝜓𝜓𝑎𝑎,𝑏𝑏,𝑐𝑐,⋯ .
𝜕𝜕 𝐸𝐸𝜕𝜕 𝐸𝐸
𝜕𝜕 𝐸𝐸===⋯=0𝜕𝜕𝑎𝑎𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕5𝑎𝑎1. Variational principle• For example, find the upper bound of ground stateenergy of H atom using a Gaussian function as thetrial function.
• The Hamiltonian of hydrogen atom isℏ2 2
�=−∇ + 𝑉𝑉
𝐻𝐻2𝑚𝑚ℏ2 1 𝜕𝜕
𝜕𝜕
1
𝜕𝜕𝜕𝜕
2
� 2+ 2=−𝑟𝑟sin 𝜃𝜃
2𝑚𝑚 𝑟𝑟 𝜕𝜕𝑟𝑟
𝜕𝜕𝑟𝑟
𝑟𝑟 sin 𝜃𝜃 𝜕𝜕𝜃𝜃𝜕𝜕𝜃𝜃1𝜕𝜕 2e2�−+ 2 224𝜋𝜋𝜀𝜀0 𝑟𝑟
𝑟𝑟 sin 𝜃𝜃 𝜕𝜕𝜙𝜙6

7

1. Variational principle𝜓𝜓𝑎𝑎,𝑏𝑏2−𝑎𝑎𝑟𝑟= 𝑏𝑏𝑒𝑒• The trial function must be normalized.
𝜓𝜓𝑎𝑎,𝑏𝑏 𝜓𝜓𝑎𝑎,𝑏𝑏 = 12𝜋𝜋�0𝜋𝜋∞

� �0032∞𝜓𝜓 2 𝑟𝑟 2 sin 𝜃𝜃 𝑑𝑑𝑟𝑟𝑑𝑑𝜃𝜃𝑑𝑑𝜙𝜙 = 4𝜋𝜋 �02 22−2𝑎𝑎𝑟𝑟
𝑏𝑏 𝑒𝑒𝑟𝑟 𝑑𝑑𝑟𝑟

𝜋𝜋=12𝑎𝑎• The trial function has one parameter, 𝑎𝑎.
= 𝑏𝑏 22𝑎𝑎
𝜓𝜓𝑎𝑎 𝑟𝑟 =𝜋𝜋342−𝑎𝑎𝑟𝑟𝑒𝑒1. Variational principle

� 𝑎𝑎
𝐸𝐸 = 𝜓𝜓𝑎𝑎 𝐻𝐻𝜓𝜓
∞2 1 𝜕𝜕ℏ𝜕𝜕∗
22 𝑑𝑑𝑑𝑑= 4𝜋𝜋 � 𝜓𝜓𝑎𝑎 −𝑟𝑟+𝑉𝑉𝜓𝜓
𝑟𝑟𝑎𝑎2 𝜕𝜕𝜕𝜕2𝑚𝑚𝑟𝑟
𝜕𝜕𝜕𝜕013ℏ2e2−=𝑎𝑎 −2𝜋𝜋 3 2 𝑎𝑎2𝑚𝑚𝜀𝜀0• At the minimum of 𝐸𝐸 ,3𝑑𝑑 𝐸𝐸
3ℏ2 e2−2 1=−2𝜋𝜋=02𝑚𝑚 𝜀𝜀0𝑑𝑑𝑎𝑎𝑎𝑎𝑚𝑚2 e48𝑎𝑎 ==32218𝜋𝜋 𝜀𝜀0 ℏ
9𝜋𝜋𝑎𝑎0 281. Variational principle

𝑚𝑚e4
𝐸𝐸min = −12𝜋𝜋 3 𝜀𝜀0 2 ℏ2
• The exact ground state energy is known as𝑚𝑚e4
𝐸𝐸1 = −32𝜋𝜋 2 𝜀𝜀0 2 ℏ2
• So,8𝐸𝐸1 ≈ 0.8488𝐸𝐸1 > 𝐸𝐸1
𝐸𝐸min =3𝜋𝜋
• It is “not too bad”!

910

1. Variational principle• The trial function with optimized parameter isψ(a0-3/2)

0.5
0.480.2829𝑟𝑟 223−𝑟𝑟−22−𝑎𝑎0𝑒𝑒 9𝜋𝜋𝑎𝑎0≈ 0.2765𝑒𝑒
𝑎𝑎0 2
1|1,0,0⟩ =

0.3
0.2-8

Exact H1sBest trial
Other trials

𝜋𝜋𝑎𝑎0 3𝑟𝑟
−𝑎𝑎𝑒𝑒 0<E> (eV)0.64𝜓𝜓 =3𝜋𝜋𝑎𝑎032-10-12

0.1
0.00

1

23r(a0)45-14E1
123
45a (2/9πa02)67

1. Variational principle• Usually, the variational method is only used for thecalculation of the ground state (|0⟩).
• However, it can be used for the calculation ofexcited states (|𝑛𝑛⟩).
• All the excited states are orthogonal to the groundstate, i.e., 0 𝑛𝑛 = 0.
• Thus, the first excited state (|1⟩) is the lowest
energy state in the subspace of Hilbert space,which is orthogonal to |0⟩.
• |1⟩ can be calculated by variational method in thatsubspace.

1112

1. Variational principle• For example, the ground state of harmonic
oscillation is an even function.
14𝑚𝑚𝑚𝑚𝑥𝑥 2𝑚𝑚𝜔𝜔−|0⟩ =𝑒𝑒 2ℏ𝜋𝜋𝜋• Any odd function is orthogonal to |0⟩.
• Use an odd trial function in the following form for|1⟩.
2−𝑎𝑎𝑥𝑥𝜓𝜓𝑎𝑎,𝑏𝑏 = 𝑏𝑏𝑏𝑏𝑒𝑒
+∞
2 𝜋𝜋
3𝑏𝑏
2−𝑎𝑎 2 = 1
𝜓𝜓𝑎𝑎,𝑏𝑏 𝜓𝜓𝑎𝑎,𝑏𝑏 = � 𝑏𝑏 2 𝑥𝑥 2 𝑒𝑒 −2𝑎𝑎𝑥𝑥 𝑑𝑑𝑑𝑑 =4 2−∞

13

1. Variational principle
32𝑎𝑎2𝑏𝑏 2 = 4𝜋𝜋• There is only one parameter, 𝑎𝑎.
• The mean kinetic energy is222ℏ𝑑𝑑𝜓𝜓1 𝑑𝑑𝜓𝜓1ℏ 𝑑𝑑
𝐸𝐸K = 𝜓𝜓𝑎𝑎 −𝜓𝜓 =2𝑚𝑚 𝑑𝑑𝑥𝑥 2 𝑎𝑎
2𝑚𝑚 𝑑𝑑𝑥𝑥 𝑑𝑑𝑥𝑥

22
ℏ2 2 3 +∞ −𝑎𝑎𝑥𝑥 23ℏ
22−𝑎𝑎𝑥𝑥
2𝑎𝑎𝑎𝑎 �𝑒𝑒− 2𝑎𝑎𝑥𝑥 𝑒𝑒𝑑𝑑𝑑𝑑 ==2𝑚𝑚 𝜋𝜋2𝑚𝑚−∞

1. Variational principle• The mean potential energy is
12𝑉𝑉 = 𝑏𝑏 𝜓𝜓𝑎𝑎 𝑚𝑚𝜔𝜔2 𝑥𝑥 2 𝜓𝜓𝑎𝑎
2+∞
22 33𝑚𝑚𝜔𝜔
2𝑎𝑎2 𝑚𝑚𝜔𝜔2 � 𝑥𝑥 4 𝑒𝑒 −2𝑎𝑎𝑥𝑥 𝑑𝑑𝑑𝑑 ==2𝜋𝜋8 𝑎𝑎−∞

• The mean total energy is3ℏ23 𝑚𝑚𝜔𝜔2𝐸𝐸 = 𝐸𝐸K + 𝑉𝑉 =𝑎𝑎 +2𝑚𝑚8 𝑎𝑎• At the minimum of 𝐸𝐸 ,14

151. Variational principle• Thus,𝑑𝑑 𝐸𝐸3ℏ2 3 𝑚𝑚𝜔𝜔2=−=022𝑚𝑚 8 𝑎𝑎𝑑𝑑𝑎𝑎𝑚𝑚𝜔𝜔
𝑎𝑎 =2ℏ|1⟩ = 23ℏ𝜔𝜔
𝐸𝐸 =21𝑚𝑚3 𝜔𝜔3 4𝜋𝜋ℏ3𝑥𝑥𝑒𝑒−𝑚𝑚𝜔𝜔 2𝑥𝑥2ℏ• This is the exact first excited state because a
perfect trial function is used.

1. Variational principle• Suppose the ground state (|0⟩) is already known.
• The trial function for |1⟩ can be created in thefollowing way (normalization needed).
𝜓𝜓 = 𝜑𝜑 − 0 𝜑𝜑 |0⟩• 𝜑𝜑 is an arbitrary wavefunction.
• 𝜓𝜓 and |0⟩ are orthogonal because
0 𝜓𝜓 = 0 𝜑𝜑 − 0 𝜑𝜑 0 0 = 0• |1⟩ can be found by varying 𝜑𝜑 until the minimum of
𝐸𝐸 is found.
• |2⟩ can be calculated by variational principle in thesubspace orthogonal to both |0⟩ and |1⟩.

16

17

1. Variational principle• Analogue in 3D Euclidean space.
|2⟩𝜓𝜓 = 𝜑𝜑 − 0 𝜑𝜑 |0⟩0 𝜑𝜑 |0⟩

𝜑𝜑|0⟩

|1⟩

1. Variational principle• The electronic Hamiltonian of Helium atom
(suppose the nucleus is stationary) is�
𝐻𝐻2ℏ2e2 2122=−∇1 + ∇2 −+ −2𝑚𝑚
4𝜋𝜋𝜀𝜀0 𝑟𝑟1 𝑟𝑟2𝑟𝑟⃑1 − 𝑟𝑟⃑2

• The corresponding Schrödinger equation2
does not have analytical solutions.
• According to the Aufbau principle, bothelectrons in ground state He atom occupy the1s AO with singlet spin.

18

19

1. Variational principle𝜓𝜓 𝑟𝑟⃑1 , 𝑟𝑟⃑2 = 𝜓𝜓1s 𝑟𝑟⃑1 𝜓𝜓1s 𝑟𝑟⃑2
• The 1s AO can be approximated by the 1shydrogen like orbital (HLO).
𝜓𝜓1s 𝑟𝑟⃑ → 𝜓𝜓𝜁𝜁 𝜌𝜌 =𝜁𝜁 3 −𝜁𝜁𝜌𝜌
𝑒𝑒
3
Zeff
𝜋𝜋𝑎𝑎0

• 𝜌𝜌 (dimensionless) is the length in atomic unit.
𝑟𝑟𝜌𝜌 ≡𝑎𝑎0• 𝜁𝜁 = 𝑍𝑍eff is the effective atomic number(effective nuclear charge in unit of e).

Shielding1. Variational principle

20• The trial function has 𝜁𝜁 as the parameter.
𝜁𝜁 3𝜓𝜓 𝜌𝜌1 , 𝜌𝜌2 = 3 3 𝑒𝑒 −𝜁𝜁 𝜌𝜌1+𝜌𝜌2
𝜋𝜋𝑛𝑛 𝑎𝑎0• 𝐸𝐸 is a function of 𝜁𝜁. The detailed calculation isskipped.
5𝜁𝜁2� 𝜁𝜁 𝜌𝜌1 , 𝜌𝜌2 = −2𝜁𝜁 + 8𝜁𝜁 −𝐸𝐸1
𝐸𝐸 = 𝜓𝜓𝜁𝜁 𝜌𝜌1 , 𝜌𝜌2 𝐻𝐻𝜓𝜓4••

−𝜁𝜁 2 + 4𝜁𝜁 𝐸𝐸1 is the mean energy of HLO witheffective atomic number 𝜁𝜁 in He+.
5𝜁𝜁
− 𝐸𝐸1 is the mean e-e potential energy (𝑉𝑉ee ).
4

211. Variational principle• If 𝜁𝜁 = 1, the trial function is the 1s orbital of H.
• As 𝜁𝜁 increases, the trial function becomes moreconcentrated to the nucleus and 𝑉𝑉ee increases.
+.
• If 𝜁𝜁 = 2, the trial function is the 1s orbitalofHe
1.6875
1.4He+

1.2

ψ(a0-3/2)

1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2

1.0
0.8
0.6
0.4
0.2
0.0

0

H

12r(a0)3-4-2

E (E1)

1.6

02

Ek+VenVee
Total

46
8
0.5

H

1.0

1.5

He+

2.0
Zeff

2.5

3.0

22

1. Variational principle•𝐸𝐸𝑑𝑑 𝐸𝐸
27= −4𝜁𝜁 +𝐸𝐸1𝑑𝑑𝜁𝜁
4𝑑𝑑 𝐸𝐸
reaches its minimum when𝑑𝑑𝜁𝜁= 0, where27𝜁𝜁 == 1.6875
16729
𝐸𝐸1 ≈ −77.49eV
𝐸𝐸min =128
• Experiments show that the total ionization energyof ground state He is 79.01eV.
• This result is “close”.

2. Computational quantum chemistry23• In computational quantum chemistry, a basis set isa set of finite number of functions, 𝜓𝜓1 , ⋯ , 𝜓𝜓𝑛𝑛 .
• The trial function, 𝜓𝜓, is a linear combination of𝜓𝜓1 , ⋯ , 𝜓𝜓𝑛𝑛 , which depends on 𝑐𝑐𝑖𝑖 and parameters of𝜓𝜓𝑖𝑖 .
𝑛𝑛𝜓𝜓 = � 𝑐𝑐𝑖𝑖 𝜓𝜓𝑖𝑖
𝑖𝑖=1

• Calculations using the basis set expanded towards
a complete (infinite) set of functions are said
to approach the complete basis set (CBS) limit.

2. Computational quantum chemistry• Slater type orbitals (STOs) are based on HLOs butwith no nodes.
ℛ𝜁𝜁,𝑛𝑛 𝜌𝜌 = 𝑁𝑁𝜌𝜌𝑛𝑛−1 𝑒𝑒 −𝜁𝜁𝜁𝜁24• 𝑛𝑛 is the principal quantum number.
• 𝜁𝜁 is Slater exponent, corresponding to the effective
atomic number.
𝑍𝑍eff𝜁𝜁 =𝑛𝑛• 𝑁𝑁 is the constant for normalization.
• The integrals of STOs are difficult (especially themulticentre ones in molecules).

2. Computational quantum chemistry25• Gaussian type orbitals (GTOs) are widely usedbecause their efficient implementations of HartreeFock methods and density functional theory (DFT).
2𝑛𝑛−1−𝑎𝑎𝑟𝑟
𝑒𝑒ℛ𝑎𝑎,𝑛𝑛 = 𝑁𝑁𝑟𝑟

• An individual primitive GTO gives a rather poor
description for the wavefunction near the nucleus.
• In the basis set STO-nG, 𝑛𝑛 (usually from 2 to 6,most frequently 3) GTOs are fitted to a single STO.
• For example, the H 1s AO with STO-2G is𝑟𝑟 2𝑟𝑟 23−0.151623 2−0.851819 2𝑎𝑎𝑎𝑎00 + 0.27181𝑒𝑒
𝜓𝜓STO−2G 𝑎𝑎0 2 = 0.11757𝑒𝑒

26

2. Computational quantum chemistry• The energies and shapes of STO-nG basis sets forH1s orbitals using variational principle are shown
0.6below.
Exact H1s
𝐸𝐸 𝐸𝐸10.84880.9716
0.98980.99700.99900.9996

STO-1G H1s
STO-2G H1s
STO-3G H1s

0.5

ψ(a0-3/2)

H1senergySTO-1G
STO-2G
STO-3G
STO-4G
STO-5G
STO-6G

0.4
0.3
0.2
0.1
0.00123r(a0)452. Computational quantum chemistry27• For variational calculation by computers, a more
efficient algorithm to optimize the trial functions isneeded.
• Hartree and Fock developed the Hartree-Fock (HF)
method.
• It is assumed that the electronic state can beapproximated by a single Slater determinant.
• For example, the Slater determinant of He is1 𝜓𝜓1s 𝑟𝑟⃑1 ↑ 𝜓𝜓1s 𝑟𝑟⃑1 ↓𝜓𝜓 𝑟𝑟⃑1 , 𝑟𝑟⃑2 =2 𝜓𝜓1s 𝑟𝑟⃑2 ↑ 𝜓𝜓1s 𝑟𝑟⃑2 ↓28

2. Computational quantum chemistry• When looking for one of the one-electron orbitals(e.g., AOs), the other electrons are regarded as asteady charge density distribution (cloud), which
provides an electrostatic field.
Repulsion
Repulsion−−𝑍𝑍

Attraction

𝑍𝑍−

Attraction

Electron cloud29

2. Computational quantum chemistry• The effective Hamiltonian of electron 1 in He is22ℏ
2e1
�1eff = −∇1 2 −+ 𝑉𝑉ee 𝑟𝑟⃑1 ≡ 𝐹𝐹� 1𝐻𝐻2𝑚𝑚
4𝜋𝜋𝜀𝜀0 𝑟𝑟1•1𝑉𝑉ee𝑟𝑟⃑1 is the potential energy of electron 1 in themean-field of electron 2, called Coulomb operator.
• HF method is based on mean-field theory.
• Suppose the wavefunction of electron 2 is 𝜓𝜓 0 𝑟𝑟⃑2 .
1𝑉𝑉ee𝑟𝑟⃑12e�
=4𝜋𝜋𝜀𝜀0𝜓𝜓 0𝑟𝑟⃑2𝑟𝑟⃑1 − 𝑟𝑟⃑22𝑑𝑑3 𝑟𝑟⃑22. Computational quantum chemistry30• 𝜓𝜓 0 𝑟𝑟⃑2 is an initial guessed wavefunction calledbase.
• 𝐹𝐹� 1 is the 1st Fock operator.

• The problem becomes the 1st HF equation.
𝐹𝐹� 1 𝜓𝜓 1 = 𝜖𝜖 1 𝜓𝜓 1

• 𝜓𝜓 1 and 𝜖𝜖 1 can calculate by computer (see later).
• With calculated 𝜓𝜓 1 , the 2nd Fock operator of He
can be created.
22
ℏ2e
222
�
∇2 −
𝐹𝐹 = −+ 𝑉𝑉ee 𝑟𝑟⃑22𝑚𝑚4𝜋𝜋𝜀𝜀0 𝑟𝑟2

31

2. Computational quantum chemistry•2𝑉𝑉ee𝑟𝑟⃑2 is the potential energy of electron 2 in themean-field of electron 1.
2𝑉𝑉ee

𝑟𝑟⃑22

𝜓𝜓 12

𝜓𝜓 𝑘𝑘−12e�=4𝜋𝜋𝜀𝜀0𝑟𝑟⃑1𝑟𝑟⃑1 − 𝑟𝑟⃑2𝑑𝑑3 𝑟𝑟⃑1

• 𝜓𝜓 2 can be calculated by the 2nd HF equation.
𝐹𝐹� 2 𝜓𝜓 2 = 𝜖𝜖 2 𝜓𝜓 2
• The iteration process can be repeated.
𝑘𝑘𝑉𝑉ee2e�𝑟𝑟⃑ =4𝜋𝜋𝜀𝜀0𝑟𝑟′⃑𝑟𝑟⃑ − 𝑟𝑟′⃑𝑑𝑑3 𝑟𝑟′⃑2. Computational quantum chemistry3222ℏ2e𝑘𝑘𝑘𝑘2�∇ −+ 𝑉𝑉ee 𝑟𝑟⃑
𝐹𝐹 = −
2𝑚𝑚4𝜋𝜋𝜀𝜀0 𝑟𝑟
𝐹𝐹� 𝑘𝑘 𝜓𝜓 𝑘𝑘 = 𝜖𝜖 𝑘𝑘 𝜓𝜓 𝑘𝑘• The process can be truncated if the iteration shows
the sign of convergence.
• If the iteration converges,
lim 𝜓𝜓 𝑘𝑘 = 𝜓𝜓𝑘𝑘→∞lim 𝐹𝐹� 𝑘𝑘 = 𝐹𝐹�𝑘𝑘→∞lim 𝜖𝜖 𝑘𝑘 = 𝜖𝜖𝑘𝑘→∞

33

2. Computational quantum chemistryCalculate 𝜓𝜓 1 𝑟𝑟⃑1−Base𝜓𝜓 0 𝑟𝑟⃑2

Calculate 𝜓𝜓 2 𝑟𝑟⃑2−

Calculate 𝜓𝜓 3 𝑟𝑟⃑1−

𝜓𝜓 1 𝑟𝑟⃑1𝜓𝜓 2 𝑟𝑟⃑2Calculate 𝜓𝜓 4 𝑟𝑟⃑2−

Calculate 𝜓𝜓 5 𝑟𝑟⃑1−

𝜓𝜓 3 𝑟𝑟⃑1

𝜓𝜓 4 𝑟𝑟⃑2⋯−

Self-consistent𝜓𝜓 𝑟𝑟⃑

2. Computational quantum chemistry• The converged 𝜓𝜓 𝑟𝑟⃑ is a self-consistent fieldbecausee2𝜓𝜓 𝑟𝑟′⃑ 2 3
�
𝑑𝑑 𝑟𝑟′
𝑉𝑉ee 𝑟𝑟⃑ =⃑4𝜋𝜋𝜀𝜀0𝑟𝑟⃑ − 𝑟𝑟′⃑ℏ2 22e2∇ −+ 𝑉𝑉ee 𝑟𝑟⃑
𝐹𝐹� = −2𝑚𝑚4𝜋𝜋𝜀𝜀0 𝑟𝑟
� 𝑟𝑟⃑ = 𝜖𝜖𝜓𝜓 𝑟𝑟⃑𝐹𝐹𝜓𝜓

• HF method is also a called self-consistent field(SCF) method.

34

35

2. Computational quantum chemistryCalculate
properties

Nuclearpositions
Initial guessof base
Fock
operator
Iteration
Solve HFequation
YesConverge?

No• If the iteration does notconverge, change thebase until it converges.
• The self consistenteigenenergy (𝜖𝜖) of selfconsistent Fork operator
is the HF energy.
• The summary of HF
algorithm is shown on
the right.

36

2. Computational quantum chemistry• In HF method, if an approximately complete basis
set, 𝜓𝜓1 , ⋯ , 𝜓𝜓𝑛𝑛 , is used,
𝑛𝑛𝑛𝑛𝑘𝑘𝜓𝜓 𝑘𝑘 = � 𝑐𝑐𝑗𝑗 𝜓𝜓𝑗𝑗
𝑗𝑗=1

𝑛𝑛𝑘𝑘
𝑘𝑘
𝐹𝐹� 𝑘𝑘 � 𝑐𝑐𝑗𝑗 𝜓𝜓𝑗𝑗 = 𝜖𝜖 𝑘𝑘 � 𝑐𝑐𝑗𝑗 𝜓𝜓𝑗𝑗
𝑗𝑗=1

𝑗𝑗=1

• Take inner product with 𝜓𝜓𝑖𝑖 from left.
𝑛𝑛� 𝑐𝑐𝑗𝑗
𝑗𝑗=1𝑘𝑘𝑛𝑛𝑘𝑘
𝜓𝜓𝑖𝑖 𝐹𝐹� 𝑘𝑘 𝜓𝜓𝑗𝑗 = 𝜖𝜖 𝑘𝑘 � 𝑐𝑐𝑗𝑗 𝜓𝜓𝑖𝑖 𝜓𝜓𝑗𝑗
𝑗𝑗=1

37

2. Computational quantum chemistry• Define𝑘𝑘𝜓𝜓𝑖𝑖 𝐹𝐹� 𝑘𝑘 𝜓𝜓𝑗𝑗 = 𝐹𝐹𝑖𝑖,𝑗𝑗𝜓𝜓𝑖𝑖 𝜓𝜓𝑗𝑗 = 𝑆𝑆𝑖𝑖,𝑗𝑗

𝑘𝑘𝐹𝐹𝑖𝑖,𝑗𝑗

form the entries of Fork matrix, 𝑭𝑭 𝑘𝑘 , whose
calculation (many integrals) can be speedup by 4-5
orders of magnitude with GTOs compared to STOs.
• Rearrange the equation.
•𝑛𝑛�

𝑗𝑗=1

𝑘𝑘𝐹𝐹𝑖𝑖,𝑗𝑗− 𝜖𝜖 𝑘𝑘 𝑆𝑆𝑖𝑖,𝑗𝑗

𝑐𝑐𝑗𝑗

𝑘𝑘=0382. Linear combination of atomic orbitals• The linear equations are equivalent to𝑘𝑘𝑘𝑘
𝜖𝜖 𝑆𝑆1,1 − 𝐹𝐹1,1𝜖𝜖𝑘𝑘⋮𝑘𝑘𝑆𝑆𝑛𝑛,1 − 𝐹𝐹𝑛𝑛,1⋯⋱⋯𝑘𝑘
𝑘𝑘
𝜖𝜖 𝑆𝑆1,𝑛𝑛 − 𝐹𝐹1,𝑛𝑛𝜖𝜖𝑘𝑘⋮𝑘𝑘𝑆𝑆𝑛𝑛,𝑛𝑛 − 𝐹𝐹𝑛𝑛,𝑛𝑛𝑐𝑐1⋮ =0𝑐𝑐𝑛𝑛• To get non-zero solution of 𝑐𝑐1 , ⋯ , 𝑐𝑐𝑛𝑛 , the coefficient
determinant must be 0.
• This condition is called the secular equation.
𝑘𝑘
𝑘𝑘
𝑘𝑘
𝑘𝑘
𝜖𝜖 𝑆𝑆1,1 − 𝐹𝐹1,1 ⋯ 𝜖𝜖 𝑆𝑆1,𝑛𝑛 − 𝐹𝐹1,𝑛𝑛=0⋮
⋱
⋮𝑘𝑘
𝑘𝑘
𝑘𝑘
𝑘𝑘
𝜖𝜖 𝑆𝑆𝑛𝑛,1 − 𝐹𝐹𝑛𝑛,1 ⋯ 𝜖𝜖 𝑆𝑆𝑛𝑛,𝑛𝑛 − 𝐹𝐹𝑛𝑛,𝑛𝑛

2. Computational quantum chemistry• This is a 𝑛𝑛th degree equation about 𝜖𝜖 𝑘𝑘 with 𝑛𝑛solutions (counting multiplicities/degeneracies).
• Note the coefficient matrix is a Hermitian matrix.
• The spectral theorem says that any Hermitianmatrix can be diagonalized by a unitary matrix.
• The calculation of secular equation is fast bycomputer.
• The ground state corresponds to the lowest 𝜖𝜖 𝑘𝑘 .
• When 𝜖𝜖 𝑘𝑘 is found, 𝑐𝑐1 , ⋯ , 𝑐𝑐𝑛𝑛 , i.e., 𝜓𝜓 𝑘𝑘 can becalculated.

392. Computational quantum chemistry40• For example, the wavefunction of ground state He
atom is calculated using the basis set of 5 STOs
(𝑛𝑛 = 1).
• The convergence is reached when
3𝑎𝑎0 2 𝜓𝜓= 0.075738𝑒𝑒 −1.43𝜌𝜌 + 0.043658𝑒𝑒 −2.4415𝜌𝜌+ 0.017295𝑒𝑒 −4.096𝜌𝜌 − 0.0273𝑒𝑒 −6.4843𝜌𝜌
+ 0.06675𝑒𝑒 −7.978𝜌𝜌𝐸𝐸 𝑍𝑍eff = −77.49eV
𝜖𝜖 = −77.87eV
𝐸𝐸 = −79.01eV• This result is more accurate than using one 𝑍𝑍eff .

41

2. Computational quantum chemistryψ(a0-3/2)

1.5Zeff=1.6875
HF-5STO

1.0

0.5

0.00

12r (a0)3

2. Computational quantum chemistry42• As a type of variational method, 𝜖𝜖 > 𝐸𝐸gs .

• Obviously, 𝜖𝜖 depends on the basis set.
• The lowest possible by HF method 𝜖𝜖 is calculatedwith CBS, which is called the HF limit (𝐸𝐸HF ).
• However, 𝐸𝐸HF > 𝐸𝐸gs .

• 𝐸𝐸HF even does NOT approach 𝐸𝐸gs (error of ~1% forHe).
 The self-consistent wavefunction is the solution tothe HF equation, but NOT to Schrödinger equation.
It does NOT approach the exact solution.

43

2. Computational quantum chemistry• The correlation energy isCE ≡ 𝐸𝐸HF − 𝐸𝐸gs > 0• CE is caused by electronic correlation.
 The distributions of electrons depend on each other(mean-field approximation is inaccurate).
Independentdistributions

Mean-field approximation

More likely

Less likely

Electronic correlation

2. Computational quantum chemistry

44• The variational approaches to correct the result
from HF method are post-HF methods, such as fullconfiguration interaction (full CI).
• Typically, the overall orbital is no longer a Slater
determinant, and the mean-field is abandoned.
• A very simple example of the trial function of He is= −77.49eV
𝜓𝜓 = 𝑁𝑁𝑒𝑒 −𝑎𝑎 𝑟𝑟1 +𝑟𝑟2 1 + 𝑏𝑏 𝑟𝑟⃑1 − 𝑟𝑟⃑2 𝐸𝐸 𝑍𝑍𝜖𝜖eff= −77.87eV
𝐸𝐸min = −78.68eV
𝐸𝐸 = −79.01eV• It is even better than 5 STOs basis set, because it
includes 𝑟𝑟⃑1 − 𝑟𝑟⃑2 in the trial function, which reflects
electronic correlation.