introduction to atomic and molecular physics -...

Ghent UniversityFaculty of Sciences

Introduction to Atomic and

Molecular Physics

Academic year 2015-2016

Lecturer: prof. dr. Philippe SmetCo-lecturer: prof. dr. Henk Vrielinck

Contents

1 Introduction and Recapitulation of Basic Concepts 1

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Quantum Mechanical Description of Hydrogen-Like Atoms . . . . . . . . . . . . 3

1.2.1 The Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2.2 Angular Momentum Operators: Definition and Commutation Relations . 51.2.3 Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2.4 Angular Part . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2.5 Radial Part . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.2.6 The Total Wave Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.2.7 Expectation Values of rk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.3 Coupling of Several Angular Momenta . . . . . . . . . . . . . . . . . . . . . . . . 311.3.1 Specifying Coupled States . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311.3.2 Allowed Values of j and m . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331.3.3 Calculating Clebsch-Gordon Coefficients . . . . . . . . . . . . . . . . . . . 341.3.4 Coupling of More Than Two Angular Momenta . . . . . . . . . . . . . . . 391.3.5 Vector Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2 The Fine Structure of Hydrogenic Atoms 41

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.2 Orbital and Spin Magnetic Moments . . . . . . . . . . . . . . . . . . . . . . . . . . 422.3 Spin-Orbit Coupling Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2.3.1 The Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.3.2 Influence on the Energy Levels of Hydrogenic Atoms . . . . . . . . . . . . 462.3.3 Illustration: Evaluation of the Spin-Orbit Interaction for a p Electron . . . 49

2.4 The Fine Structure of Hydrogenic Atoms . . . . . . . . . . . . . . . . . . . . . . . 52

3 Hydrogen-Like Atoms in a Magnetic Field 58

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583.2 The Zeeman Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603.3 The Paschen-Back Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 643.4 Intermediate Regime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

i

4 Techniques Of Approximation 72

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 724.2 Time-independent perturbation theory . . . . . . . . . . . . . . . . . . . . . . . . . 72

4.2.1 Two-level system: exact solutions and approximations . . . . . . . . . . . 734.2.2 General principle of perturbation theory . . . . . . . . . . . . . . . . . . . 784.2.3 Time-independent perturbation theory for non-degenerate levels . . . . . 794.2.4 Time-independent perturbation theory for degenerate levels . . . . . . . . 83

4.3 Time-dependent perturbation theory . . . . . . . . . . . . . . . . . . . . . . . . . . 864.3.1 Variation of constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 864.3.2 Harmonically oscillating perturbation . . . . . . . . . . . . . . . . . . . . . 88

4.4 Variation theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 914.4.1 The Rayleigh method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 914.4.2 The Rayleigh-Ritz method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 954.4.3 Some comments on the variation principle . . . . . . . . . . . . . . . . . . 99

4.5 The Hellmann-Feynman theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

5 The helium atom 102

5.1 The hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1025.1.1 The Coulomb integral J. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

5.2 Excited states of He . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1065.2.1 Exchange integral K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

5.3 Founding the Russell-Saunders notation . . . . . . . . . . . . . . . . . . . . . . . . 1085.3.1 [Lk, 1

r12] = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

5.3.2 〈α′ JM| O |αJM〉 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1105.3.3 Independency of M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1115.3.4 Spin-Orbit Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

5.4 The spectrum of helium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1135.5 Pauli principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

5.5.1 Permutations of (1 2 3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1175.5.2 Multiplication table for the group . . . . . . . . . . . . . . . . . . . . . . . 1175.5.3 Parity of the permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1175.5.4 Example for Slater determinants . . . . . . . . . . . . . . . . . . . . . . . . 118

6 Many electron atoms: Slater determinants and matrix elements 120

6.1 Many-electron atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1206.1.1 Evaluation of the matrix of the operator F = ∑N

i=1 f (i) within themanifold spanned by Slater determinants . . . . . . . . . . . . . . . . . . . 123

6.1.2 Evaluation of the matrix of the operator G = ∑Ni<j g(i, j) within the

manifold spanned by Slater determinants . . . . . . . . . . . . . . . . . . . 1276.2 Periodicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

ii

6.3 Slater atomic orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1286.4 Self-consistent fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1296.5 Term symbols and transitions of many-electron atoms . . . . . . . . . . . . . . . . 132

6.5.1 The diagonal sum rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1346.6 Hund’s rules and the relative energies of terms . . . . . . . . . . . . . . . . . . . . 1446.7 Alternative coupling schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

6.7.1 The jj coupling scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

7 Molecular Physics 156

7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1567.2 Born-Oppenheimer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

7.2.1 Formulation of the Born-Oppenheimer approximation . . . . . . . . . . . 1587.2.2 Application to the H+

2 molecular ion . . . . . . . . . . . . . . . . . . . . . . 1607.3 Molecular orbital theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

7.3.1 Linear Combinations of Atomic Orbitals (LCAO) . . . . . . . . . . . . . . 1637.3.2 The H2 molecule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1687.3.3 Example : benzene, a non-linear polyatomic conjugated π-system . . . . . 169

8 Group Theory 171

8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1718.2 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

8.2.1 Symmetry operations and elements . . . . . . . . . . . . . . . . . . . . . . 1718.2.2 Classification of molecules: point groups . . . . . . . . . . . . . . . . . . . 172

8.3 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1748.3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1748.3.2 Multiplication tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1748.3.3 Matrix representation (Nederlands: voorstelling, Deutsch: Darstellung) . . . 1758.3.4 Transformation of a matrix representation . . . . . . . . . . . . . . . . . . . 1788.3.5 Invariance of characters under basis transformations . . . . . . . . . . . . 1798.3.6 Characters and classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1808.3.7 Irreducible representations . . . . . . . . . . . . . . . . . . . . . . . . . . . 1808.3.8 The great and little orthogonality theorems . . . . . . . . . . . . . . . . . . 183

8.4 Reduced representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1868.4.1 The reduction of representations . . . . . . . . . . . . . . . . . . . . . . . . 1868.4.2 Symmetry-adapted bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

8.5 Symmetry and degeneracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1918.5.1 Transformation of an operator . . . . . . . . . . . . . . . . . . . . . . . . . 1918.5.2 Essential and accidental degeneracy . . . . . . . . . . . . . . . . . . . . . . 1928.5.3 Restricted representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1938.5.4 Symmetry reduction and lifting of degeneracy . . . . . . . . . . . . . . . . 193

8.6 Symmetry properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

iii

8.6.1 Transformation of p orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . 1958.6.2 Decomposition of direct-product bases . . . . . . . . . . . . . . . . . . . . 1978.6.3 Calculation of integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197

8.7 Character tables for molecular symmetry groups . . . . . . . . . . . . . . . . . . . 200

iv

Chapter 1

Introduction and Recapitulation ofBasic Concepts

1.1 Introduction

The spectrum of the Sun observed on the Earth’s surface essentially has three characteristicfeatures:

1. an overall spectrum approximating that of a black body at 5523 K (Fig. 1.1), as given byPlanck’s law:

I(ν, T) =2hν3

c21

ehνkT − 1

(1.1)

where I(ν, T) represents the energy emitted by a black body at temperature T per unitarea and per unit solid angle, in the frequency interval [ν, ν + dν].

2. absorption lines from chemical elements (such as H, Na, Fe and Ca) in the photosphereof the Sun (Fig. 1.2).1

3. absorption lines due to molecules such as O3, O2 and H2O in the Earth’s atmosphere.2

The absorption lines are caused by electronic transitions between the discrete energy levels ofatoms and molecules. This course aims to provide the theoretical framework for understandingthe electronic structure of atoms and molecules and explain, among others, the presence andcharacteristics of these absorption lines based on the fundamental laws of quantum mechanics.The absorption lines due to hydrogen atoms, for instance, occur at wavelengths of 410, 434,486 and 656 nm (lines h, f, F and C in Fig. 1.2). Hydrogen is a one-electron system, for whichthe quantum mechanical energy levels can be calculated exactly analytically - if relativistic

1http://www.harmsy.freeuk.com/fraunhofer.html2http://www.globalwarmingart.com/wiki/Image:Solar_Spectrum_png

1

1.1. INTRODUCTION

Figure 1.1: Solar spectrum, before and after passing through the Earth’s atmosphere. The spectralirrradiance is given per unit wavelength.

Figure 1.2: Visible part of the solar spectrum, with typical absorption lines (A, B: O2; C, F, f, h: H; D: Na;G: Fe & Ca).

effects are not taken into account, at least. The strong absorption lines at 590 nm, on theother hand, are due to Na atoms. The energy level scheme of such a polyelectronic system(one positive core, two or more electrons) cannot be calculated analytically, and one has toresort to certain approximative schemes and/or numerical methods. (Very) good results canbe obtained in this way, for both qualitative and quantitative analysis. One step further downthe line are molecules (multiple cores, multiple electrons), the simplest example being H+

2 . Animportant mathematical “toolbox” for treating molecules is group theory. All these subjects willbe discussed in this course.

Reference worksThis course is in part based on, and contains references to, following textbooks:Atkins & Friedman (A & F) - Molecular Quantum Mechanics.Bransden & Joachain (B & J) - Introduction to Quantum Mechanics.Arfken & Weber (A & W) - Mathematical Methods for Physicists.

2

1.2. QUANTUM MECHANICAL DESCRIPTION OF HYDROGEN-LIKEATOMS

1.2 Quantum Mechanical Description of Hydrogen-Like Atoms

1.2.1 The Hamiltonian

For a system in equilibrium consisting of a positive core and a (negative) electron, followingtime-independent Schrödinger equation applies:

[Hkin,e + Hkin,c + V

]Ψ (~re,~rc) = EΨ (~re,~rc) (1.2)

where ~re and ~rc are the coordinates of the electron and the core. The terms on the left-handside represent the kinetic energy of the electron, the kinetic energy of the core and the potentialenergy due to the Coulombic repulsion, respectively. The classical expressions for these termsare

Hkin,e =~p2

e2me

(1.3)

Hkin,c =~p2

c2mc

(1.4)

V =−Ze2

4πε0r= V(r) (1.5)

with me (mc) and ~pe (~pc) the mass and the impulse of the electron (core), r = |~r| with~r =~re −~rc

and Z the charge number of the core.3 Note that the Coulombic repulsion term only dependson the electron-core distance r. ’Solving the hydrogen atom’ means finding the eigenvectors Ψ(possible wave functions of the system) and eigenvalues E (possible discrete energy values ofthe system) for eigenvalue equation (1.2).By switching to the coordinate system (~R,~r) with

~R =me~re + mc~rc

me + mc(1.6)

~r = ~re −~rc (1.7)

Eq. (1.2) can be separated into two equations: one for the motion of the centre of mass and onefor the relative motion of the particles. The former is that of a free particle and is not of interesthere, while the latter is [

~p2

2µ+ V(r)

]Ψ (~r) = EΨ (~r) (1.8)

where µ is the reduced massµ =

memc

me + mc(1.9)

and~p = µ~v = µ

d~rdt

(1.10)

3Z = 1 for the hydrogen atom, but the generalisation to a charge Z will be convenient when treating otherone-electron systems such as He+ later on in this course.

3


Thus, we have reduced the original 6-dimensional problem to a 3-dimensional one.For a quantum mechanical treatment of the problem, the impulse ~p has to be replaced by itscorresponding operator −ih~∇ and Eq. (1.8) becomes

HΨ (~r) = EΨ (~r) (1.11)

where

H = − h2

2µ~∇2 + V(r) (1.12)

The Laplacian operator in this equation is composed of derivatives with respect to the relativecoordinates of the particle pair (~∇ = ~∇~r).4

Since V(r) is a central (spherically symmetric) potential, it is advantageous to switch fromCartesian to spherical coordinates, defined as follows (see Fig. 1.3):

x = r sin θ cos φ

y = r sin θ sin φ (1.13)

z = r cos θ

We will now first derive an expression for the operator ~∇2 in spherical coordinates.

Figure 1.3: Definition of the spherical coordinates (r, θ, φ).

The general expression for the Laplacian operator in orthogonal curvilinear coordinates ~q =

(q1, q2, q3) is5

~∇2 =1

h1h2h3

[∂

∂q1

(h2h3

h1

∂

∂q1

)+

∂

∂q2

(h3h1

h2

∂

∂q2

)+

∂

∂q3

(h1h2

h3

∂

∂q3

)](1.14)

4For a more rigorous quantum mechanical derivation of Eq. (1.12), see §3.8 (p. 81) and ’Further Information 4:Reduced mass’ (p. 488) in [A & F].

5See e.g. §2.2 in [A & W].

4


where

hi =

√√√√ 3

∑j=1

(∂rj

∂qi

)2

~r = (x, y, z) (1.15)

In spherical coordinates, i.e. ~q = (q1, q2, q3) = (r, θ, φ), this yields

h1 = 1 h2 = r h3 = r sin θ (1.16)

so that~∇2 =

1r2

∂

∂r(r2 ∂

∂r) +

1r2 sin θ

∂

∂θ(sin θ

∂

∂θ) +

1r2 sin2 θ

∂2

∂φ2 (1.17)

Substituting expression (1.17) in Eq. (1.12), we obtain

H =−h2

2µ

[1r2

∂

∂r(r2 ∂

∂r) +

1r2 sin θ

∂

∂θ(sin θ

∂

∂θ) +

1r2 sin2 θ

∂2

∂φ2

]+ V(r) (1.18)

Note that in Eq. (1.18) a separation of variables is “starting to show”: the first and last termdepend only on the distance r, while the second and third term have a prefactor r−2 but onlycontain derivatives in the angular coordinates θ and φ. We will see that a similar separation ofvariables is possible for the wave functions, i.e.

Ψnlm(~r) = Rnl(r)Ylm(θ, φ) (1.19)

where the indices n, l, m are quantum numbers that label the different eigenstates of the system.6

In order to solve Schrödinger equation (1.11) with Hamiltonian (1.18), we need to introduceangular momentum operators.

1.2.2 Angular Momentum Operators: Definition and Commutation Relations

In classical mechanics the angular momentum is defined as

~L =~r× ~p (1.20)

The corresponding quantum mechanical operator is obtained by replacing ~p by −ih~∇:

~L = −ih(~r× ~∇) (1.21)

or in Cartesian coordinates:

Lx = −ih(

y∂

∂z− z

∂

∂y

), Ly = −ih

(z

∂

∂x− x

∂

∂z

), Lz = −ih

(x

∂

∂y− y

∂

∂x

)(1.22)

6It will turn out that the radial part of the wave function is independent of the quantum number m, while theangular part is independent of n . We acknowledge this a priori with the notations Rnl(r) and Ylm(θ, φ).

5


Using Eqs. (1.22), following important commutation relations can be derived:

[Lx, Ly

]= ihLz,

[Ly, Lz

]= ihLx,

[Lz, Lx

]= ihLy (1.23)

which are the characteristic fingerprint of angular momentum operators. It follows from Eqs.(1.23) that [

~L2, Lx

]=[~L2, Ly

]=[~L2, Lz

]= 0 (1.24)

Their importance will become clear in the next section.The expressions for the angular momentum operators in spherical coordinates (r, θ, φ) are

Lx = −ih(− sin φ

∂

∂θ− cot θ cos φ

∂

∂φ

)(1.25)

Ly = −ih(

cos φ∂

∂θ− cot θ sin φ

∂

∂φ

)(1.26)

Lz = −ih∂

∂φ(1.27)

Note the independency of r.

Calculation of Li in spherical coordinates.Consider for example Ly . Using the chain rule we get

∂

∂x=

∂r∂x

∂

∂r+

∂θ

∂x∂

∂θ+

∂φ

∂x∂

∂φ(1.28)

∂

∂z=

∂r∂z

∂

∂r+

∂θ

∂z∂

∂θ+

∂φ

∂z∂

∂φ(1.29)

Using

r =√

x2 + y2 + z2 (1.30)

θ = arccoszr

(1.31)

φ = arctanyx

(1.32)

one finds

∂r∂x

= sin θ cos φ∂r∂z

= cos θ (1.33)

(1.34)∂θ

∂x=

cos θ cos φ

r∂θ

∂z= −sin θ

r(1.35)

(1.36)∂φ

∂x= − sin φ

r sin θ

∂φ

∂z= 0 (1.37)

Substituting these equations into Eq. (1.22) for Ly, we obtain Eq. (1.26).

6


Combining Eqs. (1.25), (1.26) and (1.27), we arrive at following expression:

~L2 = −h2[

1sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1sin2 θ

∂2

∂φ2

](1.38)

Because the operators Li (i = x, y, z) and ~L2 do not operate on the radial coordinate r, followingcommutation relations also hold:

[Lx, f (r)

]=[Ly, f (r)

]=[Lz, f (r)

]=[~L2, f (r)

]= 0 (1.39)

where f (r) is any spherically symmetric function.

1.2.3 Separation of Variables

Using Eq. (1.38), Hamiltonian (1.18) can be recast in the form

H =−h2

2µ

[1r2

∂

∂r(r2 ∂

∂r)−

~L2

h2r2

]+ V(r) (1.40)

We now propose a solution for the wave function of the form (1.19), so that the Schrödingerequation becomes7:(

−h2

2µ

[1r2

∂

∂r(r2 ∂

∂r)−

~L2

h2r2

]+ V(r)

)Rnl(r)Ylm(θ, φ) = EnRnl(r)Ylm(θ, φ) (1.41)

Using commutation relation (1.39), this can be rewritten as

−h2

2µ

[Ylm(θ, φ)

1r2

∂

∂r(r2 ∂

∂r)Rnl(r)−

Rnl(r)r2

~L2

h2 Ylm(θ, φ)

]

+ Ylm(θ, φ)V(r)Rnl(r) = EnRnl(r)Ylm(θ, φ) (1.42)

In this expression the separation of the angular and the radial part is explicit. If the Ylm(θ, φ)

are chosen to be eigenfunctions of ~L2 i.e. if8

~L2Ylm(θ, φ) = l(l + 1)h2Ylm(θ, φ) (1.43)

then Ylm(θ, φ) would cancel out and a purely radial differential equation would result. We willfirst determine the eigenfunctions of ~L2 (Section 1.2.4) and then solve the equation for the radialpart (Section 1.2.5).

7By labelling the energy level En, we anticipate the result that the energy of the wave function Rnl(r)Ylm(θ, φ)is independent of the quantum numbers l and m, i.e. independent of the angular part of the wave function. SeeSection 1.2.5.2

8The reason for chosing the particular form l(l + 1)h2 for the eigenvalue will become apparent later.

7


1.2.4 Angular Part

Note that, because of (1.39) and (1.40), following commutation relations hold:[H,~L2

]= 0 =

[H, Lz

](1.44)

The first equation reflects the conservation of total angular momentum, the second oneinvariance of the system under rotations about the z axis. Because ~L2 commutes with ~L(this follows directly from Eqs. (1.24)), but the components ~Li (i = x,y,z) do not mutuallycommute (Eqs. (1.23)), functions can only be eigenfunctions of ~L2 and a single ~Li componentsimultaneously. In the spherical coordinate system we defined, the expression for Lz isconsiderably simpler than that for Lx or Ly (see Eqs. (1.25) - (1.27)). We will therefore construct

eigenfunctions of ~L2 and Lz.

1.2.4.1 Eigenfunctions of Lz

Given the explicit expression (1.27), we are looking for solutions of following eigenvalueequation:

− ih∂

∂φΦm(φ) = mhΦm(φ) (1.45)

where we have called the eigenvalues mh, and the corresponding eigenfunctions Φm(φ). Thesolutions are

Φm(φ) =1√2π

eimφ (1.46)

where the prefactor is a normalisation constant. At this point, there is no restriction on thepossible values of m. However, we need to impose the boundary condition Φm(2π) = Φm(0)to ensure that the wave function is physically meaningful. This yields

eim2π = 1 (1.47)

which implies that m can only take on integer values (positive and negative)9, a first example ofquantisation. A measurement of the z-component of the angular momentum can therefore onlyyield values 0,±h,±2h, . . . Note that, if the z component is measured, the x and y componentsare undetermined since ~Lx and ~Ly do not commute with ~Lz (Eqs. (1.23)). Also note that thefunctions Φm(φ) are orthogonal:

∫ 2π

0Φ∗m(φ)Φm′(φ)dφ = δm,m′ (1.48)

m is called the magnetic quantum number.

9Later, when the spin moment will be introduced, this condition will be relaxed and m will be allowed to takeon half-integer values (+1/2,-1/2,+3/2,-3/2, . . . ) as well.

8


1.2.4.2 Eigenfunctions of ~L2

The results in the previous section suggest a further separation of variables in the angular partof the wave function:

Ylm(θ, φ) = Θlm(θ)Φm(φ) =1√2π

Θlm(θ)eimφ (1.49)

Functions Ylm(θ, φ) of this form obviously also are eigenfunctions of Lz. The explicit expressionfor Θlm(θ) can be found by demanding that the Ylm(θ, φ) are also eigenfunctions of ~L2 (cf. Eq.(1.43)):

~L2Ylm(θ, φ) = l(l + 1)h2Ylm(θ, φ) (1.50)

where we have written the eigenvalues as l(l + 1)h2. Substituting Eqs. (1.38) and (1.49) into theabove expression, we arrive at following equation for Θlm(θ):[

1sin θ

ddθ

(sin θ

ddθ

)+ l(l + 1)− m2

sin2 θ

]Θlm(θ) = 0 (1.51)

This is the associated Legendre equation, for which the solutions are the associated Legendrefunctions Pm

l (cos θ) which are given by (w = cos θ):

Pml (w) ≡ 1

2l l!(1− w2) m

2 dl+m

dwl+m

(w2 − 1

)l(−l ≤ m ≤ l) (1.52)

The associated Legendre functions satisfy the orthogonality condition

∫ π

0(Pm

l (cos θ))∗ Pml′ (cos θ) sin θdθ =

2(l + m)!(2l + 1)(l −m)!

δl,l′ (1.53)

and can alternatively be obtained from the following generating function (m ≥ 0):

Tm(w, s) =(2m)!(1− w2)m/2sm

2mm!(1− 2ws + s2)m+1/2 (1.54)

=∞

∑l=m

Pml (w)sl (|s| < 1) (1.55)

The total angular wave functions Ylm(θ, φ) are called the spherical harmonics and are given by

Ylm(θ, φ) = (−1)m

√(2l + 1)(l −m)!

4π(l + m)!Pm

l (cos θ)eimφ for m ≥ 0 (1.56)

Ylm(θ, φ) = (−1)mY∗l,−m(θ, φ) for m < 0 (1.57)

9


where the prefactor ensures normalisation and a phase factor (−1)m has been added.10 Thesefunctions are orthonormal, i.e.∫ π

0

∫ 2π

0Y∗1m(θ, φ)Y1′m′(θ, φ) sin θdθdφ = δl,l′δm,m′ (1.58)

l is called the azimuthal quantum number and can only be a non-negative integer:

l = 0, 1, 2, . . . (1.59)

while for every value of l, there are 2l + 1 allowed values for m,

m = −l,−l + 1, . . . , l − 1, l (1.60)

1.2.4.3 Spherical Harmonics for l = 0, . . . , 3

In Table 1.1 the explicit expressions for the spherical harmonics Ylm for l = 0, 1, 2 and 3 aregiven. Because Ylm(θ, φ) is the angular part of the wave function of an electron in a quantumstate with quantum numbers l and m, |Ylm(θ, φ)|2 corresponds to the probability density offinding such an electron at angular coordinates (θ,φ), irrespective of the distance r. Note thatthis quantity is independent of φ,

|Ylm(θ, φ)|2 =1

2π|Θlm(θ)|2 (1.61)

which again reflects the invariance of the system for rotation about the z axis. Since |Ylm(θ, φ)|2

depends only on θ, it is possible to visualise the probability distribution in a polar plot.11 Suchpolar plots for the lower-order spherical harmonics are given in Fig. 1.4 . Note the typical,well-known shapes for s, p, d and f orbitals (corresponding to l = 0, 1, 2 and 3 respectively).Also note that

|Ylm(θ, φ)|2 = |Yl,−m(θ, φ)|2 (1.62)

10The phase factor of a wave function can be chosen arbitrarily. Here we adopt the so-called Condon-Shortleyphase convention.

11The coordinates of a point in such a plot are (r, θ) ≡ (|Ylm(θ, φ)|2, θ)

10


l m Ylm(θ, φ)

0 0 Y0,0 = 1√4π

1 0 Y1,0 =√

34π cos θ

±1 Y1,±1 = ∓√

38π sin θe±iφ

2 0 Y2,0 =√

516π (3 cos2 θ − 1)

±1 Y2,±1 = ∓√

158π sin θ cos θe±iφ

±2 Y2,±2 =√

1532π sin2 θe±2iφ

3 0 Y3,0 =√

716π (5 cos3 θ − 3 cos θ)

±1 Y3,±1 = ∓√

2164π sin θ(5 cos2 θ − 1)e±iφ

±2 Y3,±2 =√

10532π sin2 θ cos θe±2iφ

±3 Y3,±3 = ∓√

3564π sin3 θe±3iφ

Table 1.1: Lower-order spherical harmonics.

11


Figure 1.4: Polar plots of the probability distribution |Ylm(θ, φ)|2 for the lower-order spherical harmonics[B & J].

12


1.2.4.4 Exercise: Alternative Determination of Angular Wave Functions Using LadderOperators

In the previous sections, we have derived explicit expressions for the spherical harmonicsfor general quantum numbers l and m. For a particular value of l, however, the Ylm(θ, φ)

eigenfunctions (m = −l, . . . ,+l) can be derived in an alternative, relatively simple way, usingladder operators.

Ladder operators, or shift operators, are defined as:

L+ = Lx + iLy L− = Lx − iLy (1.63)

Following important commutation relations can be derived using Eqs. (1.63):[~L2, L+

]= 0

[~L2, L−

]= 0 (1.64)

[Lz, L+

]= +hL+

[Lz, L−

]= −hL− (1.65)

This means that the function L±Ylm is an eigenfunction of

• ~L2 with eigenvalue h2l(l + 1):[~L2, L±

]Ylm = 0 ⇒ ~L2 [L±Ylm

]= L±

[~L2Ylm

]= l(l + 1)h2 [L±Ylm

](1.66)

• Lz with eigenvalue h(m± 1):

[Lz, L±

]Ylm = ±hL±Ylm ⇒ Lz

[L±Ylm

]− L±

[LzYlm

]= ±h

[L±Ylm

]⇒ Lz

[L±Ylm

]= h(m± 1)

[L±Ylm

](1.67)

In other words, the ladder operator L± generates (a multiplicative constant C aside) thespherical harmonic Yl,m±1:

L±Yl,m(θ, φ) = CYl,m±1(θ, φ) (1.68)

Fig. 1.5 illustrates this. C is in fact a Clebsch-Gordon coefficient and its value depends on thevalues of l and m (see Section 1.3).Combining Eqs. (1.63), (1.25) and (1.26), following expressions are obtained for the ladderoperators in spherical coordinates:

L+ = heiφ(

∂

∂θ+ i cot θ

∂

∂φ

)(1.69)

L− = he−iφ(− ∂

∂θ+ i cot θ

∂

∂φ

)(1.70)

13


Figure 1.5: The effect of ladder operators L+ and L− [A & F].

We will demonstrate the use of ladder operators for the construction of angular wave functionsfor the p orbitals (l = 1) below.

As discussed in Section 1.2.4.1, we already have (cf. Eq. (1.49))

Ylm(θ, φ) =1√2π

Θlm(θ)eimφ (1.71)

and we make use of the knowledge that the quantum number m can only take on the values−l,−l + 1, . . . , l − 1, l . This means we only need to determine Θ1,−1(θ), Θ1,0(θ) and Θ1,1(θ).

We will start by determining Y1,1(θ, φ) . L+ operating on Y1,1(θ, φ) (highest allowed value form) should yield zero12:

L+Y1,1(θ, φ) = L+Θ1,1(θ)1√2π

eiφ = 0 (1.72)

Using Eq. (1.69), one obtains

heiφ(

∂

∂θ+ i cot θ

∂

∂φ

)Θ1,1(θ)

1√2π

eiφ = 0

(dΘ1,1(θ)

dθ+ i cot θΘ1,1(θ)i

)e2iφ = 0

dΘ1,1(θ)

Θ1,1(θ)= cot θdθ

dΘ1,1(θ)

Θ1,1(θ)=

cos θdθ

sin θ=

d sin θ

sin θ

Θ1,1(θ) = C sin θ (1.73)

12In other words, the multiplicative constant C in Eq. (1.68) must be zero.

14


The integration constant C can be determined by normalising Θ1,1(θ):

1 =∫ π

0Θ∗1,1(θ)Θ1,1(θ) sin θdθ = C2

∫ π

0sin3 θdθ =

43

C2 ⇒ C =

√3

2(1.74)

Combining Eqs. (1.71), (1.73) and (1.74), we finally obtain

Y1,1(θ, φ) =

√3

8πsin θeiφ (1.75)

which corresponds to the function listed in Table 1.1 .

The function Y1,0 (m = 0) can now be determined by letting L− operate on Y1,1 (cf. Eq. (1.68)):

L−Y1,1(θ, φ) = CY1,0(θ, φ) (1.76)

We find, using Eq. (1.70)

L−Y1,1(θ, φ) = he−iφ(− ∂

∂θ+ i cot θ

∂

∂φ)

√3

8πsin θeiφ = −2h

√3

8πcos θ = CY1,0(θ, φ)

Normalisation then yields

∫ π

0

∫ 2π

0

3h2

2πC2 cos2 θ sin θdθdφ = 1

C =√

2h

and we arrive at13

Y1,0(θ, φ) = −√

34π

cos θ (1.77)

Y1,−1 can also be found by letting L− operate on Y1,0 (exercise):

Y1,−1(θ, φ) = −√

38π

sin θe−iφ (1.78)

ExerciseUse the same technique to determine the angular wave functions for the s orbital (l = 0, m =0) and the m = 2 eigenstate of the d orbital (l = 2). The first exercise is very short, the secondone is more time consuming.

13In Table 1.1 the minus sign is lacking for Y1,0. However, if Y1,0 is a ’good’ basis function, then −Y1,0 also is sincethe phase factor can be chosen arbitrarily.

15


1.2.4.5 Vector Model

The role of the quantum numbers l and m can be visualised to some extent by a (semi-)classicalvector model for the angular momentum. Since

~L2Ylm(θ, φ) = l(l + 1)h2Ylm(θ, φ) (1.79)

LzY1m(θ, φ) = mhY1m(θ, φ) (1.80)

l can be linked to the size of the angular momentum and m to the z component of~L :∣∣∣~L∣∣∣ =√

l(l + 1)h (1.81)

Lz = mh (1.82)

Furthermore, since the operators Lx, Ly and Lz do not mutually commute (Eqs. (1.23)), theorientation of the projection of ~L in the xy plane is undetermined (when the z componentís known). These three properties suggest the visualisation of ~L as a vector with length√

l(l + 1)h, describing a cone about the z axis with a (constant) projection on the z axis ofmh. This is illustrated in Fig. 1.6. The vector model can be useful as an intuitive image, but

Figure 1.6: Vector representation of the angular momentum for a quantum state with l = 3, in units h.

one should be cautious. For instance, following equation holds for the quantum mechanicaloperator ~L :

~L×~L = ih~L (1.83)

16


while for a classical vector~L, we have

~L×~L =~0 (1.84)

One should therefore not forget that the angular momentum is an operator and not a vector.

1.2.5 Radial Part

1.2.5.1 The Eigenvalue Equation

We now want to find an expression for the radial part Rnl(r) of the wave function. CombiningEqs. (1.42) and (1.79) yields

Ylm(θ, φ)

(−h2

2µ

[1r2

∂

∂r(r2 ∂

∂r)− l(l + 1)

r2

]Rnl(r) + V(r)Rnl(r)

)= EnRnl(r)Ylm(θ, φ) (1.85)

so that we get following eigenvalue equation for Rnl(r):(−h2

2µ

[1r2

ddr

(r2 ddr

)− l(l + 1)r2

]+ V(r)

)Rnl(r) = EnRnl(r) (1.86)

Defining an effective potential Veff(r) as

Veff(r) = V(r) +l(l + 1)h2

2µr2 = − Ze2

4πε0r+

l(l + 1)h2

2µr2 (1.87)

Eq. (1.86) yields [1r2

ddr

(r2 ddr

) +2µ

h2 (En −Veff(r))]

Rnl(r) = 0 (1.88)

Connection classical mechanics - quantum mechanicsThe shape and origin of Eq. (1.88) can be understood by considering the classical motion ofa particle with mass m in a central force field ~F(~r) = F(r)~er , with associated potential U(r).The total energy of such a particle is

W =12

mv2 + U(r) (1.89)

The velocity can be split up in its radial and tangential parts:

v2 = v2r + v2

θ =

(drdt

)2

+ r2(

dθ

dt

)2

(1.90)

17


On the other hand, the angular momentum is a conserved quantity:

d~Ldt

=~r× ~F =~0 (1.91)

and is given by~L =~r× ~p = m~r×~v (1.92)

so thatL =

∣∣∣~L∣∣∣ = mrvθ = mr2 dθ

dt(1.93)

Using this expression, Eq. (1.90) can be written as

v2 =

(drdt

)2

+L2

(mr)2 (1.94)

and the total energy, Eq. (1.89), as

W =12

m(drdt)2 +

L2

2mr2 + U(r) =12

m(drdt)2 + Ueff(r) (1.95)

with the effective potential Ueff defined as

Ueff(r) = U(r) +L2

2mr2 (1.96)

Notice the striking resemblance with Eq. (1.87), the main difference being the quantisation inthe latter expression because of the discrete values of l.The kinetic energy associated with angular momentum thus induces an extra potential termfor the radial motion of a particle in a central force field. It can intuitively be understood interms of a centrifugal force arising from the circular motion. Because of the r−2 dependence(i.e. r−3 for the force), it overpowers the (centripetal) Coulombic force at shorter distances andpushes the electron more outwards.

Substituting

Rnl(r) =unl(r)

r(1.97)

into Eq. (1.88) yieldsd2unl(r)

dr2 +2µ

h2 (En −Veff(r)) unl(r) = 0 (1.98)

Introducing the dimensionless variables ρ and n,

ρ =

√−8µEn

h2 r (1.99)

n =Ze2

4πε0h

√− µ

2En= Zα

√− µc2

2En(1.100)

18


with α the fine structure constant,

α =e2

4πε0hc≈ 1

137(1.101)

into Eq. (1.98) yields [d2

dρ2 −l(l + 1)

ρ2 +nρ− 1

4

]unl(ρ) = 0 (1.102)

In the asymptotic region ρ→ ∞, this equation simplifies to

d2

dρ2 unl(ρ) =14

unl(ρ) (1.103)

which leads us to propose that unl(ρ) contains a factor e+ρ2 or e−

ρ2 . Since

e+ρ2 → +∞ for ρ→ +∞ (1.104)

which is an unacceptable property for a physically sound wave function (not normalisable),the solution must be of the form

unl(ρ) = e−ρ/2 f (ρ) (1.105)

Substitution of Eq. (1.105) in Eq. (1.102) yields following equation for f (ρ) :[d2

dρ2 −d

dρ− l(l + 1)

ρ2 +nρ

]f (ρ) = 0 (1.106)

It is possible to show14, on the basis of the equation above, that f (ρ) (and consequently alsounl(ρ)) behaves as ρl+1 near the origin (ρ → 0). Therefore it makes sense to propose a solutionof the form

f (ρ) = ρl+1g(ρ) = ρl+1∞

∑k=0

ckρk (c0 6= 0) (1.107)

Substitution of Eq. (1.107) into Eq. (1.106) yields

∞

∑k=0

ρk+l−1 [(k + l + 1)(k + l)ck − (k + l + 1)ckρ− l(l + 1)ck + nckρ] = 0 (1.108)

For this equation to hold for all ρ , the coefficients have to be identically zero for all powers ofρ. In general, the coefficient for the term in ρk+l−1 contains two terms from the k term in thesummation,

(k + l + 1)(k + l)ck and − l(l + 1)ck (1.109)

and two terms from the (k− 1) term in the summation,

− ((k− 1) + l + 1)ck−1 and nck−1 (1.110)

14See § 7.2 and § 7.5 in [B & J].

19


Equalising the sum of these four terms to zero yields

ck =k + l − n

k2 + 2kl + kck−1 (1.111)

This implies thatck

ck−1→ 1

kfor k→ +∞ (1.112)

This ratio is the same as that in the power expansion of eρ or, consequently, ρpeρ for any finitevalue of p. However, it is physically inacceptable that g(ρ) ∝ eρ because this would lead to

unl(ρ) = e−ρ/2 f (ρ) ∝ eρ/2 (1.113)

which would make unl(ρ) (and consequently Rnl(ρ)) not normalisable. Therefore, thesummation in Eq. (1.107) must be finite, making g(r) a regular polynomial in ρ. Because ofthe recursion relation (1.111), ck = 0 implies that cl = 0 for all l > k. Assume the highest powerof ρ in the polynomial is nr, so that cnr+1 = 0 . Using Eq. (1.111), this leads to

n = nr + l + 1 (1.114)

We see that n ≡ n(E) (cf. Eq. (1.100)) is quantised: it can only be a positive natural number (1,2, 3, . . . ) because nr ≥ 0 and l ≥ 0. Furthermore, the allowed values of l are 0, 1, . . ., n− 1.15 nis called the principal quantum number and nr the radial quantum number.

1.2.5.2 The Energy Levels

The quantisation of n implies that also the energy levels are quantised. Using Eq. (1.100) wefind

En = − µ

2h2

(Ze2

4πε0

)2 1n2 (1.115)

= − e2

(4πε0)aµ

Z2

2n2 (1.116)

(1.117)

= −µc2

2(Zα)2

n2 , (n = 1, 2, 3, ...) (1.118)

where the so-called modified Bohr radius aµ has been introduced:

aµ =4πε0h2

µe2 =me

µa0 (1.119)

Note that15This is why, e.g., 1p or 2d orbitals do not exist.

20


• the energy levels depend only on the principal quantum number n, which justifies thenotation En .16

• En ∝ n−2 : energy differences between subsequent levels quickly drop with growing n .

• En < 0 : all associated states are bound states.

The energy levels of the hydrogen atom are shown in Fig. 1.7 . The spectral lines of hydrogenresult from transitions between different levels. This will be discussed in detail later.

Figure 1.7: Energy level scheme of atomic hydrogen [B & J].

1.2.5.3 The Radial Wave Functions

We still have to determine the expression for the radial wave functions Rnl(r) . Substituting Eq.(1.107) into Eq. (1.106), we obtain[

ρd2

dρ2 + (2l + 2− ρ)d

dρ+ (n− l − 1)

]g(ρ) = 0 (1.120)

This differential equation is identical to that of the associated Laguerre polynomials17:[ρ

d2

dρ2 + (p + 1− ρ)d

dρ+ (q− p)

]Lp

q (ρ) = 0 (1.121)

16We will later see that electron-electron interactions in polyelectronic systems as well as spin-orbit interactions(partially) lift this degeneracy.

17For definitions, properties and recursion relations, see [A & W].

21


for p = 2l + 1 and q = n + l. The explicit expression for g(ρ) = L2l+1n+l (ρ) is

L2l+1n+l (ρ) =

n−l−1

∑k=0

(−1)k+1 [(n + l)!]2

(n− l − 1− k)!(2l + 1 + k)!ρk

k!(1.122)

Alternatively, the associated Laguerre polynomials can be obtained using the generatingfunction

Up(ρ, s) =(−s)pe

−ρs1−s

(1− s)p+1 =∞

∑q=p

Lpq (ρ)

q!sq (|s| < 1) (1.123)

Associated Laguerre polynomials satisfy following orthogonality relation:∫ ∞

0e−ρρpLp

q (ρ)Lpq′(ρ)dρ ∝ δq,q′ (1.124)

Combining Eqs. (1.97), (1.105), (1.107), (1.122) and (1.99) we finally obtain following expressionfor the radial wave functions:

Rnl(r) = Ne−ρ/2ρl L2l+1n+l (ρ) (1.125)

with

ρ =2Znaµ

r (1.126)

N =

√√√√( 2Znaµ

)3 (n− l − 1)!

2n [(n + l)!]3(1.127)

The constant N ensures that Rnl is normalised.The radial wave functions satisfy the following orthonormality constraint:∫ ∞

0R∗nl(r)Rn′ l(r)r2dr = δn,n′ (1.128)

The lower-order wave functions (n = 1, 2 and 3) are listed in Table 1.2 and depicted in Fig. 1.8 .

The radial distribution function Dnl(r) is defined as the probability of finding an electron withwave function Ψnlm(r, θ, φ) in the interval [r, r + dr] from the core, irrespective of the direction.To obtain an expression for this function, we have to integrate out the angular coordinates fromthe total probability density:

∫ π

0dθ sin θ

∫ 2π

0dφΨ∗(r, θ, φ)Ψ(r, θ φ)r2dr = r2R2

nl(r)dr∫ π

0sin θdθ

∫ 2π

0Y∗lmYlmdφ

= r2R2nl(r)dr

(1.129)

22


n l Rnl(r)

1 0 R10 = 2(

Zaµ

) 32exp(−Zr

aµ)

2 0 R20 = 2(

Z2aµ

) 32(

1− Zr2aµ

)exp(− Zr

2aµ)

1 R21 = 1√3

(Z

2aµ

) 32(Zr

aµ)exp(− Zr

2aµ)

3 0 R30 = 2(

Z3aµ

) 32(

1− 2Zr3aµ

+ 2Z2r2

27a2µ

)exp(− Zr

3aµ)

1 R31 = 4√

29

(Z

3aµ

) 32(

1− Zr6aµ

) (Zraµ

)exp(− Zr

3aµ)

2 R32 = 427√

10

(Z

3aµ

) 32(

Zraµ

)2exp(− Zr

3aµ)

Table 1.2: Lower-order radial wave functions.

which leads toDnl(r) = r2R2

nl(r) (1.130)

The radial distribution functions for n = 1, 2 and 3 are also shown in Fig. 1.8 .

ExerciseProve that the distribution function Dn,n−1(r) (i.e. the maximum l for a certain n), has a maximumat

r =n2aµ

Z(1.131)

Note that this distance of maximum probability equals the (modified) Bohr radius for the 1s orbitalin hydrogen (i.e. the ground-state orbital). This is equal to the radius of the electron orbital in theBohr model, but in the current, fully quantum mechanical description, we are dealing with a diffuseelectron cloud for and we can only speak of a ’most probable distance’.

23


Figure 1.8: The radial wave functions Rnl(r) and the radial distribution functions Dnl(r) = r2R2nl(r) for

atomic hydrogen. The length scale is given in units aµ. [B & J]

24


1.2.6 The Total Wave Function

The total wave functions Ψnlm(r, θ, φ) of hydrogen-like atoms are finally obtained by combiningthe radial wave functions Rnl(r) (Eq. (1.125)) with the angular wave functions Ylm(θ, φ) (Eqs.(1.56) and (1.57)). The lower-order wave functions are listed in Table 1.3 .

Table 1.3: The normalised wave functions for hydrogen-like atoms for n = 1, 2 and 3 [B & J].

25


1.2.7 Expectation Values of rk

1.2.7.1 Definition

The expectation value of rk for a quantum state with quantum numbers n, l and m is defined as⟨rk⟩

nlm=∫

Ψ∗nlm(~r)rkΨnlm(~r)d~r (1.132)

or, after integrating out the angular part:⟨rk⟩

nlm≡⟨

rk⟩

nl=∫ ∞

0R2

nl(r)rk+2dr (1.133)

Note that⟨rk⟩ is independent of m.

These expectation values are frequently encountered in atomic physics, and general solutionsin terms of n and l are useful in a broad range of applications, including dipole transitions(k = 1), magnetic susceptibility (k = −2), spin-orbit coupling interactions (k = −3) and vander Waals interactions (k = −6).These are the results for |k| ≤ 3:

〈r〉nl =aµ

2Z[3n2 − l(l + 1)

](1.134)

⟨r2⟩

nl =a2

µn2

2Z2

[5n2 + 1− 3l(l + 1)

](1.135)

⟨r3⟩

nl =a3

µn2

8Z3

[35n(n2 − 1)− 30n2(l + 2)(l − 1) + 3(l + 2)(l + 1)l(l − 1)

](1.136)

⟨r−1⟩

nl=

Zaµn2 (1.137)

⟨r−2⟩

nl =Z2

a2µn3(l + 1

2 )(1.138)

⟨r−3⟩

nl =Z3

a3µn3l(l + 1

2 )(l + 1)(1.139)

The derivations for general n and l are quite tedious and difficult. We will consider two(simpler) examples here: the expression for

⟨r−3⟩ in the case of a 2p orbital (Section 1.2.7.2)

and for⟨r−1⟩ for arbitrary n and l (Section 1.2.7.3).

26


1.2.7.2 Exercise: Calculation of⟨r−3⟩ for a 2p orbital.

The radial part of the 2p orbital is (cf. Table 1.2)

R2p = R21 =1√3

(Z

2aµ

) 32(

Zraµ

)exp

(−Zr2aµ

)(1.140)

so that

⟨r−3⟩

2p =∫ ∞

0R∗2pR2pr−3r2dr

=13

(Z

2aµ

)3 ( Zaµ

)2 ∫ ∞

0exp

(−Zr

aµ

)rdr

=13

(Z

2aµ

)3 ∫ ∞

0exp(−u)udu (1.141)

Using the technique of partial integration, we obtain∫ ∞

0exp(−u)udu = −

∫ ∞

0ud(exp(−u))

= −u exp(−u)∣∣∣∞0+∫ ∞

0exp(−u)du

= − exp(−u)∣∣∣∞0

= 1

This leads to ⟨r−3⟩

2p =13

(Z

2aµ

)3

(1.142)

which is indeed the same result as Eq. (1.139) yields for n = 2 and l = 1.

1.2.7.3 Exercise: Calculation of⟨r−1⟩

nl

Substituting Eqs. (1.125) and (1.126) into Eq. (1.133) yields

⟨rk⟩

nl= N2

(naµ

2Z

)k+3 ∫ ∞

0e−ρρ2l

[L2l+1

n+l (ρ)]2

ρk+2dρ (1.143)

or, for k = -1, ⟨r−1⟩

nl= N2

(naµ

2Z

)2 ∫ ∞

0e−ρρ2l+1

[L2l+1

n+l (ρ)]2

dρ (1.144)

27


To evaluate the integral in the expression above,

∫ ∞

0e−ρρ2l+1

[L2l+1

n+l (ρ)]2

dρ (1.145)

we will make use of the generating function for the Laguerre polynomials (cf. Eq. (1.123)),

Uv(ρ, s) =(−s)ve

−ρs1−s

(1− s)v+1 =∞

∑q=v

Lvq(ρ)

q!sq (|s| < 1) (1.146)

Uw(ρ, t) =(−t)we

−ρt1−t

(1− t)w+1 =∞

∑q′=w

Lwq′(ρ)

q′!tq′ (|t| < 1) (1.147)

and calculate the expression

I =∫ ∞

0e−ρρ2l+1U2l+1(ρ, s)U2l+1(ρ, t)dρ (1.148)

using the two different expressions (analytical and summation) for the generating functions.Equating the two end results will allow evaluation of integral (1.145).

Calculation using the summation over Laguerre polynomials

Using Eqs. (1.146) and (1.147) and putting v = w = 2l + 1, we find

I =∞

∑q=2l+1

∞

∑q′=2l+1

sq

q!tq′

q′!

∫ ∞

0e−ρρ2l+1L2l+1

q (ρ)L2l+1q′ (ρ)dρ (1.149)

Because associated Laguerre polynomials satisfy (1.124), Eq. (1.149) reduces to

I =∞

∑q=2l+1

(st)q

(q!)2

∫ ∞

0e−ρρ2l+1

[L2l+1

q (ρ)]2

dρ (1.150)

Note that the right-hand side contains the integral of interest.

Calculation using the analytical expression

Inserting the analytical expressions for the generating functions in (1.148), again with v = w =

2l + 1 , we get

I =(−s)2l+1 (−t)2l+1

(1− s)2l+2 (1− t)2l+2

∫ ∞

0e−ρ[ s

1−s+t

1−t+1]ρ2l+1dρ (1.151)

28


The integral in the expression above is of the type∫ ∞

0e−aρρmdρ (1.152)

witha =

s1− s

+t

1− t+ 1 and m = 2l + 1 (1.153)

Using the technique of partial integration we get

Im =∫ ∞

0e−aρρmdρ

= −1a

[ρme−aρ

∣∣∣∞0−∫ ∞

0mρm−1e−aρdρ

]

=ma

Im−1

=ma

m− 1a

. . .1a

I0 (1.154)

Because I0 = 1a , we finally obtain

∫ ∞

0e−aρρmdρ =

m!am+1 (1.155)

or, in the current case:∫ ∞

0e−ρ[ s

1−s+t

1−t+1]ρ2l+1dρ =(2l + 1)!( s

1−s +t

1−t + 1)2l+2

=(2l + 1)! [(1− s)(1− t)]2l+2

(1− st)2l+2 (1.156)

Inserting Eq. (1.156) into Eq. (1.151) yields

I =(st)2l+1 (2l + 1)!

(1− st)2l+2 (1.157)

Using the Taylor expansion

(1 + x)n = 1 +n1!

x +n(n− 1)

2!x2 + . . . (1.158)

29


the right-hand side of Eq. (1.157) can be cast in the form

(2l + 1)! (st)2l+1[

1 +−2l − 2

1!(−st) +

(−2l − 2)(−2l − 3)2!

(−st)2 + . . .]

=∞

∑f=0

(st)2l+1+ f (2l + 1 + f )!f !

=∞

∑q=2l+1

(st)q q!(q− 2l − 1)!

(1.159)

and we get

I =∞

∑q=2l+1

(st)q q!(q− 2l − 1)!

(1.160)

Comparing Eqs. (1.150) and (1.160), we find

∫ ∞

0e−ρρ2l+1

[L2l+1

q (ρ)]2

dρ =(q!)3

(q− 2l − 1)!(1.161)

Plugging this result into Eq. (1.144), and using Eq. (1.127) we finally obtain

⟨r−1⟩

nl= N2

(naµ

2Z

)2 [(n + l)!]3

(n− l − 1)!

=

(2Znaµ

)3 (n− l − 1)!

2n [(n + l)!]3(naµ

2Z

)2 [(n + l)!]3

(n− l − 1)!

=Z

n2aµ(1.162)

which is identical to Eq. (1.137).

30

1.3. COUPLING OF SEVERAL ANGULAR MOMENTA

1.3 Coupling of Several Angular Momenta

1.3.1 Specifying Coupled States

Any vector~j is an angular momentum operator if following commutation relations are satisfied(cf. Eqs. (1.23)): [

jx, jy]= ihjz,

[jy, jz

]= ihjx,

[jz, jx

]= ihjy (1.163)

We already know that in the case of a single angular momentum~j , we can label the quantumstates as

∣∣jmj⟩

with

j 2 ∣∣jmj⟩

= j(j + 1)h2 ∣∣jmj⟩

(1.164)

jz∣∣jmj

⟩= mjh

∣∣jmj⟩

(1.165)

Assume now that there are two sources of angular momentum in a system, which we label~j1and~j2 . Because operators that refer to independent components of a system always mutuallycommute, we have [

j1q, j2q′]= 0 q, q′ = x, y, z (1.166)

and consequently [j 21 , j 2

2]=[j1z, j2z

]=[j 21 , j2z

]=[j 22 , j1z

]= 0 (1.167)

This means that j1, j2, j1z and j2z can be determined simultaneously and that the quantum statesof the composite system can be labelled as

∣∣j1mj1 j2mj2⟩=∣∣j1mj1

⟩ ∣∣j2mj2⟩

. This is the uncoupledscheme, in which we simply specify the individual components but do not provide informationon the way the two momenta couple.However, we can also specify the quantum states in a coupled scheme as

∣∣j1 j2 jmj⟩

where j andmj are the quantum numbers of the total angular momentum operator

~j =~j1 +~j2 (1.168)

This is possible because

1. ~j indeed is an angular momentum as it satisfies Eqs. (1.163), e.g.

[jx, jy

]=

[j1x + j2x, j1y + j2y

]=

[j1x, j1y

]+[j2x, j2y

]+[j1x, j2y

]+[j2x, j1y

]= ihj1z + ihj2z + 0 + 0

= ihjz (1.169)

31


This implies among others that Eqs. (1.164) and (1.165) apply and that j 2 and jz mutuallycommute.

2. j 2 and jz commute with j 21 and j 2

2 :

[j 2, j 2

1]=[j 2, j 2

2]

= 0 (1.170)

[jz, j 2

1]=[jz, j 2

2]

= 0 (1.171)

These identities follow directly from the fact that j 21 and j 2

2 commute with all componentsof both j1 and j2 .

In the coupled scheme, the orientation of the individual components are not specified but therelative orientation of the two coupling momenta ís.

Note that it is not possible to specify the value of mj1 or mj2 if j is already specified since thecorresponding operators do not commute, e.g.:

[j1z, j 2] = 2ih

(j1y j2x − j1x j2y

)6= 0 (1.172)

In other words, it not possible to label a quantum state as∣∣j1mj1 j2mj2 jmj

⟩and one has to choose

between the uncoupled and the coupled representation. Both pictures are valid but, dependingon the Hamiltonian, one can be more natural (mathematically more straightforward) to workin than the other. They correspond to two different basis sets which are connected by a unitarytransformation: ∣∣j1 j2 jmj

⟩= ∑

mj1,mj2

C(

j1 j2mj1mj2; j1 j2 jmj) ∣∣j1mj1 j2mj2

⟩(1.173)

The coefficients C(

j1 j2mj1mj2; j1 j2 jmj)

are called vector coupling coefficients or Clebsch-Gordancoefficients.For convenience of notation, we will drop j1 and j2 from now on, as well as the j subscripts, sothat the above expression turns into

|jm〉 = ∑m1,m2

C (m1m2; jm) |m1m2〉 (1.174)

Another, equivalent way of understanding this: the |m1m2〉 vectors are not, in general,eigenfunctions of jz or j 2. However, eigenfunctions of jz and j 2 cán be constructed fromappropriate linear combinations of |m1m2〉 vectors.

32


1.3.2 Allowed Values of j and m

A coupled |jm〉 state can be built only from uncoupled |m1m2〉 states for which:

m1 + m2 = m (1.175)

Intuitively this is immediately obvious: the component of the total orbital momentum aboutan axis has to be independent of the picture (coupled or uncoupled) and in the uncoupledrepresentation it is the sum of the individual components about that axis. Mathematically thiscan be seen as follows:

jz |jm〉 =(

jz1 + jz2)

∑m1,m2

C (m1m2; jm) |m1m2〉

hm |jm〉 = ∑m1,m2

h (m1 + m2)C (m1m2; jm) |m1m2〉

hm ∑m1,m2

C (m1m2; jm) |m1m2〉 = ∑m1,m2

h (m1 + m2)C (m1m2; jm) |m1m2〉 (1.176)

0 = ∑m1,m2

(m1 + m2 −m)C (m1m2; jm) |m1m2〉 (1.177)

This can only hold true in general if all basis set coefficients are identically zero, which leads to

C (m1m2; jm) = 0 if m1 + m2 6= m (1.178)

Making this restriction explicit in Eq. (1.174), we write

|jm〉 = ∑m1+m2=m

C (m1m2; jm) |m1m2〉 (1.179)

Note that this already guarantees that the |jm〉 are eigenfunctions of jz .Eq. (1.175) implies that the maximum value for m, and thus also the maximum value for j, is j1+ j2. Determining all other possible values for j requires some more thinking. It can be shown18

that the permitted states are

j = j1 + j2, j1 + j2 − 1, . . . , |j1 − j2| (1.180)

18See §4.10 in [A & F] or §4.4 in [A & W].

33


As always for angular momenta, there are 2j + 1 allowed values for m for a certain j:19

m = −j,−j + 1, . . . , j− 1, j (1.181)

1.3.3 Calculating Clebsch-Gordon Coefficients

The general derivation of an explicit expression for the Clebsch-Gordon coefficients is difficult,but for a particular case they can be computed relatively easily by considering/using thefollowing facts/rules:

• the coupled states with “extremal” quantum numbers can only arise from the uncoupledstates with “extremal” quantum numbers:

|j = j1 + j2, m = j1 + j2〉 = |m1 = j1; m2 = j2〉 (1.182)

|j = j1 + j2, m = −j1 − j2〉 = |m1 = −j1; m2 = −j2〉 (1.183)

Here we have introduced the notation convention of using a semicolumn “;” for theuncoupled states and a comma “,” for the coupled states. This convention will be appliedwhenever confusion is possible.

• from the definition of the ladder operator j± (cf. Eq. (1.63)), it follows that

j± = j1± + j2± (1.184)

• following identities apply for ladder operators:

j± |j, m〉 = [j(j + 1)−m(m± 1)]12 h |j(m± 1)〉 (1.185)

• the |j, m〉 states must be orthonormal, i.e.

〈j, m∣∣j′, m′

⟩= δj,j′δm,m′ (1.186)

(1.187)

∑m1+m2=m

|C (m1m2; jm)|2 = 1 (1.188)

19Note that the number of states in the coupled scheme is indeed equal to that in the uncoupled scheme (assumingj1 ≥ j2):

j1+j2

∑j=j1−j2

(2j + 1) =12[(j1 + j2)− (j1 − j2) + 1] [2(j1 − j2) + 1 + 2(j1 + j2) + 1]

= (2j2 + 1) (2j1 + 1)

where we have used the fact that the sum of all elements in an arithmetical row (a1, a2, . . . , an) is 12 n(a1 + an).

34


1.3.3.1 Example 1: j1 = j2 = 12

In the uncoupled picture, the |m1; m2〉 states are∣∣∣∣12;12

⟩ ∣∣∣∣12;−12

⟩ ∣∣∣∣−12

;12

⟩ ∣∣∣∣−12

;−12

⟩(1.189)

In the coupled picture, the allowed values for j are 1 and 0 (Eq. (1.180)), and the coupled |j, m〉states are

|0, 0〉 |1,−1〉 |1, 0〉 |1, 1〉 (1.190)

To determine the connection between these two sets of states, we start from Eq. (1.182):

|1, 1〉 =∣∣∣∣12;

12

⟩(1.191)

Now we can apply the ladder operator j− to find the other states with j = 1:20

j− |1, 1〉 =(

j1− + j2−) ∣∣∣∣12;

12

⟩

(1(1 + 1)− 1(1− 1))12 |1, 0〉 =

(12(

12+ 1)− 1

2(

12− 1)

) 12[ ∣∣∣∣−1

2;

12

⟩+

∣∣∣∣12;−12

⟩]

|1, 0〉 =1√2

[ ∣∣∣∣−12

;12

⟩+

∣∣∣∣12;−12

⟩](1.192)

Again applying j− we find |1,−1〉:

j− |1, 0〉 =(

j1− + j2−) 1√

2

[ ∣∣∣∣−12

;12

⟩+

∣∣∣∣12;−12

⟩]

(1(1 + 1)− 0(0− 1))12 |1,−1〉 =

1√2

[0 +

(12(

12+ 1)− 1

2(

12− 1)

) 12

+0 +(

12(

12+ 1)− 1

2(

12− 1)

) 12] ∣∣∣∣−1

2;−1

2

⟩

|1,−1〉 =

∣∣∣∣−12

;−12

⟩(1.193)

This result is in correspondence with Eq. (1.183).The only coupled state left to determine is |0, 0〉. It can only be composed of uncoupled stateswith m1 + m2 = m = 0:

|0, 0〉 = a∣∣∣∣12;−1

2

⟩+ b

∣∣∣∣−12

;12

⟩(1.194)

20We switch to units h for convenience.

35


Moreover, it has to be perpendicular to |1, 0〉 (cf. Eq. (1.186)):

〈1, 0| 0, 0〉 = 0 (1.195)

which yields

1√2

[a⟨−1

2;

12

∣∣∣∣ 12

;−12

⟩+ a

⟨12

;−12

∣∣∣∣ 12

;−12

⟩

+b⟨−1

2;

12

∣∣∣∣ −12

;12

⟩+ b

⟨12

;−12

∣∣∣∣ −12

;12

⟩]= 0

1√2[a.0 + a.1 + b.1 + b.0] = 0

1√2(a + b) = 0

b = −a (1.196)

Combined with the normalisation condition

a2 + b2 = 1 (1.197)

we obtain|0, 0〉 = 1√

2

[ ∣∣∣∣12;−12

⟩−∣∣∣∣−1

2;

12

⟩](1.198)

Exercise: verify that |1, 0〉 is a state with j = 1 and m = 0.This is equivalent to proving following identities

jz |1, 0〉 = 0. |1, 0〉 = 0 (1.199)

j 2 |1, 0〉 = 1(1 + 1) |1, 0〉 = 2. |1, 0〉 (1.200)

Eq. (1.199) is readily verified using Eq. (1.192). To verify Eq. (1.200), first show that

~j2 =~j 21 +~j 2

2 + 2~j1 ·~j2 =~j 21 +~j 2

2 + j1+ j2− + j1− j2+ + 2j1z j2z (1.201)

36


1.3.3.2 Example 2: j1 = 1, j2 = 32

In the uncoupled picture, the |m1; m2〉 states are∣∣∣∣1;32

⟩ ∣∣∣∣1;12

⟩ ∣∣∣∣1;−12

⟩ ∣∣∣∣1;−32

⟩∣∣∣∣0;

32

⟩ ∣∣∣∣0;12

⟩ ∣∣∣∣0;−12

⟩ ∣∣∣∣0;−32

⟩∣∣∣∣−1;

32

⟩ ∣∣∣∣−1;12

⟩ ∣∣∣∣−1;−12

⟩ ∣∣∣∣−1;−32

⟩In the coupled picture, the allowed values for j are 1

2 , 32 and 5

2 , and the coupled |j, m〉 states are∣∣∣∣52,52

⟩ ∣∣∣∣52,32

⟩ ∣∣∣∣52,12

⟩ ∣∣∣∣52,−12

⟩ ∣∣∣∣52,−32

⟩ ∣∣∣∣52,−52

⟩∣∣∣∣32,

32

⟩ ∣∣∣∣32,12

⟩ ∣∣∣∣32,−12

⟩ ∣∣∣∣32,−32

⟩∣∣∣∣12,

12

⟩ ∣∣∣∣12,−12

⟩We have ∣∣∣∣52,

52

⟩=

∣∣∣∣1;32

⟩(1.202)

Applying the ladder operator j−, we find

j−

∣∣∣∣52,52

⟩=

(j1− + j2−

) ∣∣∣∣1;32

⟩√

5∣∣∣∣52,

32

⟩=√

2∣∣∣∣0;

32

⟩+√

3∣∣∣∣1;

12

⟩∣∣∣∣52,

32

⟩=

√25

∣∣∣∣0;32

⟩+

√35

∣∣∣∣1;12

⟩(1.203)

37


and

j−

∣∣∣∣52,32

⟩=

(j1− + j2−

) (√25

∣∣∣∣0;32

⟩+

√35

∣∣∣∣1;12

⟩)

2√

2∣∣∣∣52,

12

⟩=

√25

(√2∣∣∣∣−1;

32

⟩+√

3∣∣∣∣0;

12

⟩)+

√35

(√2∣∣∣∣0;

12

⟩+√

4∣∣∣∣1;−1

2

⟩)∣∣∣∣52,

12

⟩=

1√10

∣∣∣∣−1;32

⟩+

√35

∣∣∣∣0;12

⟩+

√3

10

∣∣∣∣1;−12

⟩(1.204)

In a similar fashion one obtains∣∣∣∣52,−12

⟩=

√3

10

∣∣∣∣−1;12

⟩+

√35

∣∣∣∣0;−12

⟩+

1√10

∣∣∣∣1;−32

⟩(1.205)

∣∣∣∣52,−32

⟩=

√35

∣∣∣∣−1;−12

⟩+

√25

∣∣∣∣0;−32

⟩(1.206)

∣∣∣∣52,−52

⟩=

∣∣∣∣−1;−32

⟩(1.207)

Note that the latter result is again in correspondence with Eq. (1.183).

State∣∣ 3

2 , 32

⟩must be built of uncoupled states with m1 + m2 = m = 3

2 :∣∣∣∣32,32

⟩= a

∣∣∣∣0;32

⟩+ b

∣∣∣∣1;12

⟩(1.208)

Demanding it to be perpendicular to∣∣ 5

2 , 32

⟩yields√

25

a +

√35

b = 0 (1.209)

Combined with the normalisation condition a2 + b2 = 1, we get∣∣∣∣32,32

⟩= −

√35

∣∣∣∣0;32

⟩+

√25

∣∣∣∣1;12

⟩(1.210)

38


Applying the ladder operator j− to Eq. (1.210), we get the other j = 32 states:∣∣∣∣32,

12

⟩= −

√25

∣∣∣∣−1;32

⟩− 1√

15

∣∣∣∣0;12

⟩+ 2

√215

∣∣∣∣1;−12

⟩(1.211)

∣∣∣∣32,−12

⟩= −2

√215

∣∣∣∣−1;12

⟩+

1√15

∣∣∣∣0;−12

⟩+

√25

∣∣∣∣1;−32

⟩(1.212)

∣∣∣∣32,−32

⟩= −

√25

∣∣∣∣−1;−12

⟩+

√35

∣∣∣∣0;−32

⟩(1.213)

It is easy to verify that the∣∣ 3

2 , m⟩

states are perpendicular to the∣∣ 5

2 , m′⟩

states for m = m′,21 asrequired.

State∣∣ 1

2 , 12

⟩must be of the form∣∣∣∣12,

12

⟩= a

∣∣∣∣−1;32

⟩+ b

∣∣∣∣0;12

⟩+ c

∣∣∣∣1;−12

⟩(1.214)

Orthogonality to states∣∣ 5

2 , 12

⟩and

∣∣ 32 , 1

2

⟩and normalisation yields following set of equations:

1√10

a +

√35

b +

√310

c = 0

−√

25

a− 1√15

b + 2

√215

c = 0

a2 + b2 + c2 = 1 (1.215)

This leads to a = 1√2, b = − 1√

3and c = 1√

6, so that

∣∣∣∣12,12

⟩=

1√2

∣∣∣∣−1;32

⟩− 1√

3

∣∣∣∣0;12

⟩+

1√6

∣∣∣∣1;−12

⟩(1.216)

Again letting j− operate on the equation above, finally yields∣∣∣∣12,−12

⟩=

1√6

∣∣∣∣−1;12

⟩− 1√

3

∣∣∣∣0;−12

⟩+

1√2

∣∣∣∣1;−32

⟩(1.217)

1.3.4 Coupling of More Than Two Angular Momenta

Three angular momenta ~j1, ~j2 and ~j3 can be coupled consecutively: first couple two angularmomenta, e.g.

~j1 +~j2 =~j12 (1.218)

21For m 6= m′, this is trivial since the two states cannot have any |m1; m2〉 in common (because m1 + m2 = m) andthe |m1; m2〉 states are orthogonal.

39


as explained in the previous section, and then proceed by coupling the resulting total angularmomentum with the third angular momentum:

~j12 +~j3 =~j123 (1.219)

This algorithm can obviously be generalised for coupling an arbitrary number of angularmomenta. The end result is independent of the order in which the momenta are coupled.

1.3.5 Vector Model

The coupling of angular momenta can be interpreted from a pictorial point of view using thevector model introduced in Section 1.2.4.5. This approach may provide some more intuitiveunderstanding of certain aspects, e.g.

• that the absolute orientation of the individual momenta is not known in the coupledscheme.

• that in the coupling of two angular momenta with j1 = j2 = 12 , the z component of

the total angular momentum in the so-called coupled triplet state (j = 1) can be eitherpositive, zero or negative.

• . . .

One should realise, however, that the vector model also has strong limitations and caution isadvised in applying this pictorial image in calculations. For a more in-depth discussion, seee.g. §4.11 in [A & F].

40

Chapter 2

The Fine Structure of HydrogenicAtoms

2.1 Introduction

In the famous Stern and Gerlach experiment, a beam of silver atoms sent through aninhomogeneous magnetic field, with the field gradient perpendicular to the incoming beam,was observed to split up in two discrete components. The interaction with an inhomogoneous

Figure 2.1: The Stern-Gerlach experiment.

magnetic field is consistent with the presence of an angular moment, but since silver atomshave a zero orbital angular moment, this result points to the existence of an additional,intrinsic, electronic angular moment~s. This angular moment was termed spin.1 Furthermore,the presence of two discrete components, as opposed to the classically expected continuousspectrum, showed that this spin is quantised: it can only take on two “orientations” withrespect to an external field (spin up and spin down). Later, this spin was shown to automatically

1The term spin originates from the classical image of the electron as a charged sphere rotating around its ownaxis, thus generating a magnetic moment, even though later it turned out that the electron is in fact a point particleand that the spin does not have a classical analogue.

41

2.2. ORBITAL AND SPIN MAGNETIC MOMENTS

emerge from the Dirac equation, a special-relativistic, quantum-mechanical wave equation thatdescribes the electron in the presence of an external electromagnetic field: the electron possessesa spin~s with quantum numbers s = 1

2 and ms = ± 12 .

In Chapter 1 exact solutions for the Schrödinger equation for hydrogen-like atoms were found.However, the spin of the electron was completely ignored. The magnetic moments generatedby the orbital angular moment and the spin interact (spin-orbit coupling interaction), which canshift and split up the energy levels. The resulting changes in the atomic spectrum are referredto as the fine structure of the spectrum. We will derive expressions for the magnetic momentsassociated with orbital angular momenta and spins (Section 2.2) and for the spin-orbit couplinginteraction (Section 2.3) in hydrogen-like atoms, and then study the resulting fine structure(Section 2.4). We will see that the effect of spin-orbit interaction is quite limited for light atomssuch as hydrogen, but it can be a dominant feature in the spectra of heavy atoms.To avoid confusion, subscripts l and s will be added to the quantum number m when it refersto the orbital angular moment or the spin respectively.

2.2 Orbital and Spin Magnetic Moments

We will first calculate the magnetic moment arising from an orbital angular moment, basedon the classical picture of a particle with mass m and charge q moving in a circular orbit withradius r at a constant speed v around the z axis (Fig. 2.2). A current I is associated with this

Figure 2.2

moving charge, given byI =

qT

= qν (2.1)

with T = 1ν the period of the orbit. This current gives rise to a magnetic dipole moment ~m:

~m = IA~ez (2.2)

42

2.2. ORBITAL AND SPIN MAGNETIC MOMENTS

with A = πr2 the area enclosed by the orbit. Since

ν =ω

2π=

v2πr

(2.3)

we obtain~m =

12

qvr~ez (2.4)

On the other hand, the orbital angular moment~l of the particle is

~l =~r× ~p = mvr~ez (2.5)

Comparison of Eqs. (2.4) and (2.5) yields

~m = γ~l (2.6)

with the magnetogyric ratio γ defined as

γ =q

2m(2.7)

For an electron we have~m = γe~l γe = −

e2me

(2.8)

A more general (arbitrary orientation) and quantum-relativistic derivation yields the sameresult.Since the spin ~s is a purely quantum-relativistic property, a reasoning analogous to the oneabove cannot be applied. Experimentally, it is found that the associated magnetic moment foran electron is given by

~m = geγe~s (2.9)

with ge = 2.002319 the so-called g factor of the electron. The theory of quantum electrodynamicsyields exactly this value, while the Dirac theory yields ge = 2 .2 Note that the spin magneticmoment is roughly twice the value expected on the basis of a classical analogy.

The quantisation of angular moment (cf. Chapter 1) implies an analogous quantisation of theassociated magnetic moments:

• for the orbital angular moment:

mz = −µBml µB =eh

2me(2.10)

where we have introduced the Bohr magneton µB.

2The small deviation of ge from 2 is explained in quantum electrodynamics in terms of interactions of the chargedparticle with vacuum fluctuations (resulting from the Heisenberg uncertainty principle) of the electromagnetic field.

43

2.3. SPIN-ORBIT COUPLING INTERACTION

• for the spin:

mz = −geµBms µB =eh

2me(2.11)

2.3 Spin-Orbit Coupling Interaction

2.3.1 The Hamiltonian

The spin-orbit coupling energy can be calculated as the potential energy U of the spin magneticmoment ~m (Eq. (2.9)) in the magnetic field ~B induced by the orbital angular moment of theelectron:

U = −~m · ~B (2.12)

To get a suitable expression for ~B, recall that a particle with charge q and velocity ~v gives riseto a magnetic field

~B =µ0q

4πr2~v×~er (2.13)

Recapitulation: derivation of Eq. (2.13)We start from the experimental Bio-Savart’s law, which gives an expression for the magneticfield ~B induced by the current I running through a conductor (cf. Fig 2.3),

~B =µ0 I4π

∫l

~et ×~er

r2 dl (2.14)

and we will deduce the magnetic field generated by a single moving charged particle fromthis.Eq. (2.14) suggests that each infinitisemal piece of conductor dl gives rise to a magnetic fieldd~B:

d~B =µ0 I

4πr2~et ×~erdl (2.15)

=µ0

4πr2dqdt~et ×~erdl (2.16)

=µ0

4πr2 dqdl~et

dt×~er (2.17)

=µ0dq4πr2~v×~er (2.18)

The latter equation can be interpreted as stating that every particle with charge q induces amagnetic field

~B =µ0q

4πr2~v×~er (2.19)

which is identical to Eq. (2.13).Note that, since the electric field ~E is given by

44


~E =q

4πε0r2~er (2.20)

Eq. (2.19) can be rewritten as

~B =1c2~v× ~E (2.21)

where we have usedc =

1√

µ0ε0(2.22)

Biot-Savart’s law (Eq. (2.14)) and Eqs. (2.15), (2.18) and (2.21) remain valid in a relativistictheory, although Eq. (2.19) (Eq. 2.13) is not relativistically correct.

Figure 2.3

Figure 2.4

45


In the electron reference frame, the core with charge qc = Ze is orbiting around the electronwith velocity ~vc = −~ve (Fig. 2.4), so that

~B =µ0Ze4πr2 (−~ve)×~er

~B = −µ0Ze4π

~ve ×~rr3

~B =µ0Ze4πme

~lr3 (2.23)

with~l the orbital angular moment of the electron:

~l = me~r×~ve = −me~ve ×~r (2.24)

Substituting Eqs. (2.9) and (2.23) into Eq. (2.12), and switching to operators, we get

HSO = −(

geγe~s)·(

µ0Ze4πme

~lr3

)= ξ(r)~l ·~s (2.25)

ξ(r) = geµ0Ze2

8πm2e r3 (2.26)

The Dirac theory yieldsHSO = ξ(r)~l ·~s (2.27)

with

ξ(r) =µ0Ze2

8πm2e r3 (2.28)

so that our classical derivation yields a coupling strength ξ(r) about twice too large.3 Note that,using the expression for the electric potential

V =−Ze2

4πε0r(2.29)

the following, more general expression is found

ξ(r) =12

1m2

e c21r

dVdr

(2.30)

2.3.2 Influence on the Energy Levels of Hydrogenic Atoms

When the spin-orbit interaction is added to the Hamiltonian of the hydrogen atom, the corre-sponding Schrödinger equation no longer yields the same eigenvalues En and eigenfunctions

3The error originates from the non-relativistic treatment of the relative movement of core and electron.

46


|nlml〉 ≡ Rnl(r)Ylm(θ, φ) of Chapter 1. The 2n2 degeneracy of each En level4 will be partiallylifted and, in particular, the energy levels will depend on the quantum number l as well.The spin-orbit interaction term is very small compared to the energy difference betweensubsequent En (for relevant values of n) for the hydrogen atom: using Eq. (1.113) with Z =1 and aµ ≈ a0, we get

∆En ≈ E2 − E1 ≈38

e2

4πε0a0(2.31)

Since l and s are of the order h and taking r ≈ a0 and Z = 1, Eqs. (2.27) and (2.28) yield followingestimation for the spin-orbit coupling energy:

ESO ≈e2

8πε0m2e c2a3

0h2 (2.32)

It follows thatESO

∆En≈ 4

3h2

m2e c2a2

0≈ 7.10−5 (2.33)

The spin-orbit coupling can therefore be taken into account using perturbation theory:

H = H(0) + H(1) (2.34)

H(0) = HBohr = −h2

2µ~∇2 − Ze2

4πε0r(2.35)

H(1) = HSO = ξ(r)~l ·~s (2.36)

Since the perturbed level is degenerate in l, ml and ms (and s, but this can only have the value12 h in a one-electron system), the spin-orbit Hamiltonian operator has to be diagonalised withinthe |lmlsms〉 subsets.5 This means we need to calculate matrix elements of the type

⟨nl′m′lsm′s

∣∣ ξ(r)~l ·~s |nlmlsms〉 (2.37)

The operator~l ·~s is not diagonal in the uncoupled |lmlsms〉 basis subset, but using

~j 2 =(~l +~s

)2= ~l 2 +~s 2 + 2~l ·~s (2.38)

we can write~l ·~s = 1

2

(~j 2 −~l 2 −~s 2

)(2.39)

4There is an n2 degeneracy in l and ml and the factor two arises from the degeneracy in ms .5Two quantum labels s and ms should be added to the |nlml〉 functions (the eigenfunctions in the absence of the

spin-orbit coupling) because of the presence of the spin.

47


which shows that ~l · ~s ís diagonal in the coupled representation∣∣lsjmj

⟩with ~j = ~l +~s : the

matrix elements are of the type⟨nl′sj′m′j

∣∣∣ ξ(r)~l ·~s∣∣nlsjmj

⟩(2.40)

(2.41)

=⟨

nl′sj′m′j∣∣∣ ξ(r)

12

(~j 2 −~l 2 −~s 2

) ∣∣nlsjmj⟩

(2.42)

=12

h2 [j(j + 1)− l(l + 1)− s(s + 1)]⟨


∣∣nlsjmj⟩

(2.43)

=12

h2 [j(j + 1)− l(l + 1)− s(s + 1)]⟨


∣∣nlsjmj⟩

δll′δjj′δmjm′j(2.44)

where we have used the fact that ξ(r) does not have any angular dependency and theorthogonality of the coupled basis functions |lsjmj〉.Using Eq. (2.28) we find

⟨nlsjmj

∣∣ ξ(r)∣∣nlsjmj

⟩=

µ0Ze2

8πm2e〈nl| 1

r3 |nl〉 = µ0Ze2

8πm2e

⟨r−3⟩

nl (2.45)

where it has been made explicit that the expectation value⟨r−3⟩ ≡ ⟨r−3⟩

nl depends only onthe quantum numbers n and l . Plugging in Eq. (1.134) yields

⟨nlsjmj

∣∣ ξ(r)∣∣nlsjmj

⟩= 〈nl| ξ(r) |nl〉 =

µ0Ze2

8πm2e

Z3

a3µn3l(l + 1

2 )(l + 1)(2.46)

=Z4e2

8πε0c2m2e a3

µn3l(l + 12 )(l + 1)

(2.47)

Combining Eqs. (2.44) and (2.47), we obtain

ESO =12

h2 [j(j + 1)− l(l + 1)− s(s + 1)]Z4e2

8πε0c2m2e a3

µn3l(l + 12 )(l + 1)

(2.48)

which can be rewritten as (using Eq. (1.114)6

ESO =Z4e8me

512π4ε40c2h4

j(j + 1)− l(l + 1)− s(s + 1)2n3l(l + 1

2 )(l + 1)(2.49)

The spin-orbit coupling constant ζnl is defined via the equation

hcζnl = 〈nl| ξ(r) |nl〉 h2 (2.50)

6We take µ = me or, equivalently: aµ = a0 .

48


so thatESO =

12

hcζnl [j(j + 1)− l(l + 1)− s(s + 1)] (2.51)

ζnl has the unit of a wavenumber (m−1) and hcζnl is the corresponding energy. Introducing theRydberg constant RH

RH =mee4

8ε20h3c

(2.52)

and using the definition of the fine structure constant α (Eq. (1.99)), we find

ζnl =α2RHZ4

n3l(l + 12 )(l + 1)

(2.53)

Note that

• taking Z = 1, ESO ≈ 12 hcα2RH ≈ 2.10−5 hcRH, while the energy level separations are

of the order of hcRH, which again confirms that the spin-orbit interaction is but a smallperturbation for the hydrogen atom.

• ESO ∝ Z4, while the energy separations are ∝ Z2, so that the spin-orbit coupling willbecome relatively much more important for heavy atoms.

• due to the dependency on l, s and j, the spin-orbit interaction indeed (partially) lifts the2n2 degeneracy of the energy En of the unperturbed system. This will be demonstratedin Section 2.3.3 .

2.3.3 Illustration: Evaluation of the Spin-Orbit Interaction for a p Electron

We will calculate the spin-orbit coupling for a single p electron (l = 1, s = 12 ) in both the

coupled and uncoupled representations.

Uncoupled representationThe matrix elements to be calculated are of the form

〈nlmlsms| ξ(r)~l ·~s∣∣nlm′lsm′s

⟩=

hcζn,1

h2 〈nlmlsms|~l ·~s∣∣nlm′lsm′s

⟩(2.54)

Using~l ·~s = lx sx + ly sy + lz sz =

12

(l+ s− + l− s+

)+ lz sz (2.55)

and abbreviating |nlmlsms〉 as |ml ; ms〉, we find (in units h)

~l ·~s∣∣∣∣1;

12

⟩=

12

∣∣∣∣1;12

⟩(2.56)

~l ·~s∣∣∣∣1;−1

2

⟩=

1√2

∣∣∣∣0;12

⟩− 1

2

∣∣∣∣1;−12

⟩(2.57)

49


~l ·~s∣∣∣∣0;

12

⟩=

1√2

∣∣∣∣1;−12

⟩(2.58)

~l ·~s∣∣∣∣0;−1

2

⟩=

1√2

∣∣∣∣−1;12

⟩(2.59)

~l ·~s∣∣∣∣−1;

12

⟩=

1√2

∣∣∣∣0;−12

⟩− 1

2

∣∣∣∣−1;12

⟩(2.60)

~l ·~s∣∣∣∣−1;−1

2

⟩=

12

∣∣∣∣−1;−12

⟩(2.61)

and obtain following matrix representation for the operator ~l ·~s in the uncoupled representa-tion:

12 0 0 0 0 00 − 1

21√2

0 0 0

0 1√2

0 0 0 0

0 0 0 0 1√2

0

0 0 0 1√2− 1

2 0

0 0 0 0 0 12

(2.62)

Diagonalisation of this matrix7 yields following eigenfunctions and eigenvalues

φ1 =

∣∣∣∣1;12

⟩λ1 =

12

(2.64)

φ2 =1√3

∣∣∣∣1;−12

⟩+

√23

∣∣∣∣0;12

⟩λ2 =

12

(2.65)

φ3 = −√

23

∣∣∣∣1;−12

⟩+

1√3

∣∣∣∣0;12

⟩λ3 = −1 (2.66)

φ4 =

√23

∣∣∣∣0;−12

⟩+

1√3

∣∣∣∣−1;12

⟩λ4 =

12

(2.67)

φ5 = − 1√3

∣∣∣∣0;−12

⟩+

√23

∣∣∣∣−1;12

⟩λ5 = −1 (2.68)

φ6 =

∣∣∣∣−1;−12

⟩λ6 =

12

(2.69)

7Which is equivalent to solving the eigenvalue equation

~l ·~s |φ〉 = λ |φ〉 (2.63)

50


In conclusion, we get 4 times the eigenvalue

+12

hcζn,1 (2.70)

and twice the eigenvalue− hcζn,1 (2.71)

so that the 6-fold degeneracy in ml and ms of the p electron energy for the unperturbedHamiltonian is partially lifted by the spin-orbit interaction, as shown in Fig. 2.5.

Figure 2.5

Note that the “centre of gravity” of the energy levels is still the unperturbed energy En .

Coupled representationDeterminination of the Clebsch-Gordan coefficients for the coupled representation of a system

51

2.4. THE FINE STRUCTURE OF HYDROGENIC ATOMS

with~j1 =~l (l = 1) and~j2 =~s (s = 12 ) yields (exercise)8

∣∣∣∣32,32

⟩=

∣∣∣∣1;12

⟩≡ φ1 (2.72)

∣∣∣∣32,12

⟩=

1√3

∣∣∣∣1;−12

⟩+

√23

∣∣∣∣0;12

⟩≡ φ2 (2.73)

∣∣∣∣32,−12

⟩=

√23

∣∣∣∣0;−12

⟩+

1√3

∣∣∣∣−1;12

⟩≡ φ4 (2.74)

∣∣∣∣32,−32

⟩=

∣∣∣∣−1;−12

⟩≡ φ6 (2.75)

∣∣∣∣12,12

⟩= −

√23

∣∣∣∣1;−12

⟩+

1√3

∣∣∣∣0;12

⟩≡ φ3 (2.76)

∣∣∣∣12,−12

⟩= − 1√

3

∣∣∣∣0;−12

⟩+

√23

∣∣∣∣−1;12

⟩≡ φ5 (2.77)

The spin-obit interaction is diagonal in this basis and using Eq. (2.39) it can be verified that theeigenvectors indeed have the eigenvalues + 1

2 hcζn,1 (for the j = 32 quartet φ1, φ2, φ4 and φ6) and

− 12 hcζn,1 (for the j = 1

2 doublet φ3 and φ5):

~l ·~s∣∣∣∣32, mj

⟩=

12

(32

.52− 1.2− 1

232

) ∣∣∣∣32, mj

⟩=

12

∣∣∣∣32, mj

⟩(2.78)

~l ·~s∣∣∣∣12, mj

⟩=

12

(12

.32− 1.2− 1

232

) ∣∣∣∣12, mj

⟩= −

∣∣∣∣12, mj

⟩(2.79)

The outcome is not affected by the choice of basis set (coupled or uncoupled), but thecoupled representation clearly is more natural for the spin-orbit interaction. The remainingdegeneracies (one 2-fold, one 4-fold) are readily explained in the coupled representation interms of the independency of the energy levels on mj .

2.4 The Fine Structure of Hydrogenic Atoms

We have seen in the previous section that the degeneracy in l, ml and ms of the hydrogen energylevels is partially lifted by the spin-orbit interaction. More precisely: it splits each energy levelEn into one singlet term for l = 0 (j = s = 1

2 ) and n− 1 doublet terms for l = 1, 1, . . . , n− 1, eachof the latter splitting up into two levels, one for j = l + s = l + 1

2 and one for j = l − s = l − 12 .

8We again use the notation convention of labelling the coupled vectors as∣∣∣j, mj

⟩.

52


A level is labelled with a term symbol 2s+1lj =2lj and has a remaining 2j + 1 degeneracy in mj.9

E.g.: the n = 2 energy level splits up into 3 levels: 2s1/2 , 2 p1/2 and 2 p3/2 .Let us examine how the spectrum of the hydrogen atom looks like when the spin-orbit couplingis included. Using10

En = − Z2mee4

32π2ε20h2

1n2 (2.80)

and the expression for the fine structure constant α (Eq. (1.99), Eq. (2.49) can be rewritten as

ESO = −EnZ2α2 j(j + 1)− l(l + 1)− s(s + 1)nl(2l + 1)(l + 1)

(2.81)

Putting s = 12 , this simplifies to

ESO,j=l+ 12

= −En(Zα)2

n(2l + 1)(l + 1)(2.82)

ESO,j=l− 12

= +En(Zα)2

nl(2l + 1)(2.83)

for j = l − 12 . Note that Eq. (2.82) in principle does not apply for l = 0, since there is no spin-

orbit coupling interaction in this case. However, as can be derived from the Dirac equation,one should also take into account the Darwin term11, which only contributes for l = 0 orbitalsand yields the same value as the spin-orbit energy correction (2.82) for l = 0 .Based on the formulas above, one would not expect any degeneracy among the different 2lj

levels. However, in the Dirac theory, an exact degeneracy is found between levels within acertain n multiplet with different l but same j : 2 2s1/2 and 2 2 p1/2, 3 2s1/2 and 3 2 p1/2, 3 2 p 3

2and

3 2d 32, . . . The cause of this degeneracy is a relativistic correction to the kinetic energy in the

Hamiltonian which will be discussed below.

The relativistic expression for the kinetic energy T of a particle with rest mass m0 is

T =√

m20c4 + p2c2 −m0c2 (2.84)

where the impulse p is given byp =

m0v√1− v2

c2

(2.85)

9Each mj corresponds to a state and we will see later that this degeneracy in states can be lifted by an externalmagnetic field.

10We again put µ = me .11This term can be interpreted as originating from rapid quantum oscillations of the electron at the nucleus.

53


for velocities v c, we can approximate Eq. (2.84) with a Taylor expansion in(

pm0c

)2 1:

T = m0c2

[1 +(

pm0c

)2] 1

2

− 1

(2.86)

≈ m0c2([

1 +12

p2

m20c2− 1

8p4

m40c4

]− 1)

(2.87)

≈ 12

p2

m0− 1

8p4

m30c2

(2.88)

The first term on the right-hand side is the classical expression for the kinetic energy. Thesecond term is a first-order relativistic correction. To evaluate the influence of this smallperturbation on the energy level En, we need to calculate matrix elements of the type

⟨nl′m′ls

′m′s∣∣− ~p 4

8µ3c2 |nlmlsms〉 ≡⟨nl′m′l

∣∣− ~p 4

8µ3c2 |nlml〉 (2.89)

where we have made use of the fact that the operator is spin-independent. Because(~p 2

2µ+ V(r)

)|nlml〉 = En |nlml〉 (2.90)

we have~p 2 |nlml〉 = 2µ (En −V(r)) |nlml〉 (2.91)

and

~p4 |nlml〉 = ~p

2[2µ (En −V(r))] |nlml〉 (2.92)

= [2µ (En −V(r))] ~p2 |nlml〉 (2.93)

= [2µ (En −V(r))]2 |nlml〉 (2.94)

where we have used12 [~p 2, V(r)

]= 0 (2.95)

12V(r) ∝ r−1, ~p2

∝ ~∇ 2 and

~∇ 2(

1r

f (r))=

(~∇ 2 1

r

)f (r) +

1r

(~∇ 2 f (r)

)=

1r

(~∇ 2 f (r)

)so that [

~∇ 2,1r

]= 0

54


Since En and V(r) are independent of the angular coordinates, the perturbation operator isdiagonal in l and ml :

⟨nl′m′l

∣∣ ~p 4 |nlml〉 = 4µ2 ⟨nl′m′l∣∣ (En −V(r))2 |nlml〉 δll′δmlm′l

(2.96)

Putting

V(r) = − Ze2

4πε0r(2.97)

and using Eqs. (2.80), (1.132) and (1.133), we get following first-order energy correction:

Ekin,rel = 〈nlml | −~p 4

8µ3c2 |nlml〉 (2.98)

= − 12µc2

[E2

n +2EnZe2

4πε0

⟨r−1⟩

nl+

Z2e4

16π2ε20

⟨r−2⟩

nl

](2.99)

= − µe8Z4

512π4ε40h4c2n3

(− 3

4n+

22l + 1

)(2.100)

= −En(Zα)2

n

(3

4n− 2

2l + 1

)(2.101)

The total first-order energy shift due to the spin-orbit interaction and the relativistic correctionto the kinetic energy is:

• for j = l + 12 (using Eq. (2.82)):

ESO,j=l+1/2 + Ekin,rel = −En(Zα)2

n

[1

(2l + 1)(l + 1)+

34n− 2

2l + 1

](2.102)

= −En(Zα)2

n

[3

4n− 1

l + 1

](2.103)

• for j = l − 12 (using Eq. (2.83)):

ESO,j=l−1/2 + Ekin,rel = −En(Zα)2

n

[− 1

l(2l + 1)+

34n− 2

2l + 1

](2.104)

= −En(Zα)2

n

[3

4n− 1

l

](2.105)

Comparing Eqs. (2.103) and (2.105), it is clear that an n 2l(l+1/2) level will shift with the sameamount (relative to En) as an n 2l′(l′−1/2) with l′ = l + 1 . This explains the observed degeneracybetween e.g. 3 2s1/2 and 3 2 p1/2 .Note that the first-order corrected energy can for both cases (j = l + 1

2 and j = l− 12 ) be written

55


as

E(1)n,l,j = En

[1 +

(Zα)2

n

(− 3

4n+

1j + 1

2

)](2.106)

which makes the dependency on j (and independency on l) explicit. The combined influence ofthe spin-orbit interaction, the Darwin term and the relativistic correction to the kinetic energyare depicted in Figure 2.6A perturbational treatment of the more rigourous Dirac theory as well as the so-calledSommerfeld formula yield exactly the same result (Figure 2.7). Nevertheless, experimentally a(very) small splitting is observed between 2s1/2 and 2 p1/2 levels due to a shift of the 2s1/2 level(the Lamb shift). The Lamb shift is exactly predicted by quantum electrodynamics.

Figure 2.6

56


Figure 2.7

57

Chapter 3

Hydrogen-Like Atoms in a MagneticField

3.1 Introduction

In Chapter 1 we considered the following Hamiltonian for hydrogen-like atoms:

HBohr = −h2

2µ~∇2 − Ze2

4πε0r(3.1)

and found following energy eigenvalues and eigenfunctions:

EBohr = En = −µc2

2(Zα)2

n2 (n = 1, 2, 3, ...) (3.2)

ψnlml (~r) = Rnl(r)Ylml (θ, φ) (l = 0, 1, ..., n− 1 ; ml = −l,−l + 1, ..., l − 1, l) (3.3)

where Rnl(r) is given by Eq. (1.125) and Ylml (θ, φ) by Eqs. (1.56) and (1.57). In Chapter 2, weadded the spin-orbit interaction1 and a relativistic correction to the kinetic energy (which areof the same order of magnitude) to the Hamiltonian :

H = HBohr +

[ξ(r)~l ·~s− ~p4

8m30c2

](3.4)

= HBohr + HSO + Hkin,rel (3.5)

= HBohr + Hrelativity (3.6)

1or, in the case of l = 0, the Darwin term

58

3.1. INTRODUCTION

These additional terms are small perturbations to the Bohr Hamiltonian and perturbationtheory yielded energy eigenvalues

En + Erelativity = En + ESO + Ekin,rel (3.7)

with (Eq. (2.51))

ESO =12

hcζnl [j(j + 1)− l(l + 1)− s(s + 1)] (3.8)

and (Eq. (2.100))

Ekin,rel = −µe8Z4

512π4ε40h4c2n3

(− 3

4n+

22l + 1

)(3.9)

and eigenfunctions ψnlsjmj ≡ |nlsjmj〉 (grouped in n 2lj levels), which are linear combinationsof the |nlmlsms〉 eigenfunctions (coupled basis with ~j = ~l +~s). Note that for s = 1

2 , Eq. (3.8)reduces to

ESO,j=l+1/2 =12

hcζnl l (3.10)

ESO,j=l−1/2 = −12

hcζnl(l + 1) (3.11)

In this chapter, we will investigate the influence of an external magnetic field ~B on theenergy levels and eigenstates of hydrogen-like atoms. This magnetic field interacts withthe magnetic moments associated with the orbital angular moment and spin of the electron(Zeeman interactions), which gives rise to two additional terms in the Hamiltonian:

H = HBohr + Hrelativity + HZeeman (3.12)

withHZeeman = −~morbital · ~B− ~mspin · ~B = −γe

(~l + ge~s

)· ~B (3.13)

In the case ge = 1, the Zeeman Hamiltonian could be written as −γe~j · ~B, which is diagonal inthe |nlsjmj〉 basis (provided we choose the quantisation axis along the magnetic field direction).The fact that ge ≈ 2.0023 complicates matters.

The size of the Zeeman interaction depends on the magnetic field strength. Taking −γehB =

µBB as measure for the Zeeman energy shifts/splittings, EZ, and 12 α2hcRH as measure for the

energy shifts/splittings caused by the spin-orbit and kinetic-relativistic terms, Erelativity,2 wecan distinguish between three different regimes:

1. EZ Erelativity (Zeeman effect).Typical magnetic field strengths in the Sun’s photosphere or at the Earth’s surface are of

2As noted in Section 2.3.2, this is a measure for the spin-orbit contribution, but the kinetic-relativistic energyterm is of a similar magnitude.

59

3.2. THE ZEEMAN EFFECT

the order of 10−4 T, in which case EZ ≈ 10−5Erelativity .

2. EZ Erelativity (Paschen-Back effect).In certain man-made magnets, e.g., pulses of 1000 T (EZ ≈ 102Erelativity) can be generated.

3. EZ ≈ Erelativity (intermediate regime).For B in the order 1 - 10 T, we find EZ ≈ Erelativity . Conventional electromagnets have anupper limit of about 1.6 T.

Depending on the regime, a different approach is required for a satisfactory simultaneousdescription of the relativistic and magnetic field interactions. This is the subject of the nextsections. In particular, we will focus on the effect of these interactions on a certain n 2l level(fixed n and l). Since the relativistic kinetic energy contribution depends only on n and l (Eq.(3.9)), it shifts all states in an n 2l level equally. To study the relative energy differences betweenstates of such a level, we can therefore ignore this contribution and consider the Hamiltonian

HSO + HZeeman (3.14)

Note that for very strong magnetic fields (B ≥ 105 T), the Zeeman energy is of the same orderof magnitude (or larger) than the Bohr energy splittings, so that a perturbational treatment ofthe magnetic field is not valid. In this chapter, we always assume ∆EBohr EZ .

3.2 The Zeeman Effect

In this case, the Zeeman interaction lifts the remaining degeneracy in mj of the eigenfunctions|nlsjmj〉 of an n 2lj level, and the Zeeman energy contributions can be calculated usingfollowing perturbational scheme:

H = H(0) + H(1) (3.15)

H(0) = HBohr + HSO (3.16)

H(1) = HZeeman (3.17)

This means the Zeeman Hamiltonian (3.13) has to be diagonalised in the |mj〉 subsets (fixed n,l, j) of the |nlsjmj〉 basis, so that we need to calculate matrix elements of the type

〈nlsjm′j| HZeeman |nlsjmj〉 ≡ 〈nlsjm′j| − γe

(~l + ge~s

)· ~B |nlsjmj〉 (3.18)

60


Following operator equivalences exist for this particular type of matrix elements3,4:

~l = gl~j (3.19)

~s = gs~j (3.20)

with

gl =j (j + 1)− s (s + 1) + l (l + 1)

2j (j + 1)(3.21)

gs =j (j + 1) + s (s + 1)− l (l + 1)

2j (j + 1)(3.22)

Illustration of operator equivalencies (3.19) and (3.20)Consider the z components of the operators ~l, ~s and ~j in the case of a single p electron (l = 1,s = 1

2 ). The coupled |lsjmj〉 ≡ |j, mj〉 basis in terms of the uncoupled |lmlsms〉 ≡ |ml ; ms〉basis functions is given by following equations (cf. Section 2.3.3):∣∣∣∣32,

32

⟩=

∣∣∣∣1;12

⟩(3.23)

∣∣∣∣32,12

⟩=

√13

∣∣∣∣1;−12

⟩+

√23

∣∣∣∣0;12

⟩(3.24)

∣∣∣∣32,−12

⟩=

√23

∣∣∣∣0;−12

⟩+

√13

∣∣∣∣−1;12

⟩(3.25)

∣∣∣∣32,−32

⟩=

∣∣∣∣−1;−12

⟩(3.26)

for j = l + s = 32 , and ∣∣∣∣12,

12

⟩= −

√23

∣∣∣∣1;−12

⟩+

√13

∣∣∣∣0;12

⟩(3.27)

∣∣∣∣12,−12

⟩= −

√13

∣∣∣∣0;−12

⟩+

√23

∣∣∣∣−1;12

⟩(3.28)

for j = l − s = 12 . The matrix representations of the z components of the operators within the

j = 32 multiplet are (in units h)

3This can be proven using the Wigner-Eckart theorem, see e.g. the course ’Symmetriegroepen’.4Be ware: this equivalency is valid only within the |mj〉 subset.

61


[lz

]j= 3

2

=

1 0 0 00 1

3 0 00 0 − 1

3 00 0 0 −1

(3.29)

[sz]j= 32

=

12 0 0 00 1

6 0 00 0 − 1

6 00 0 0 − 1

2

(3.30)

[jz]

j= 32

=

32 0 0 00 1

2 0 00 0 − 1

2 00 0 0 − 3

2

(3.31)

and we see that, within this basis subset, the operators are indeed proportional:

lz =23

jz sz =13

jz (3.32)

The proportionality constants 23 and 1

3 are also retrieved from Eqs. (3.21) and (3.22)respectively, for l = 1 and s = 1

2 .Analogously, the operator representations in the j = 1

2 multiplet are

[lz

]j= 1

2

=

[23 00 − 2

3

][sz]j= 1

2=

[− 1

6 00 1

6

] [jz]

j= 12=

[12 00 − 1

2

](3.33)

so that nowlz =

43

jz sz = −13

jz (3.34)

which again is in correspondence with Eqs. (3.21) and (3.22) respectively.

Moreover, we can choose the z quantisation axis along the magnetic field direction, i.e. ~B = B~ez,so that the equivalent operators are diagonal in the |mj〉 subset:

〈nlsjm′j| − γeB (gl + gegs) jz |nlsjmj〉 = −γeB (gl + gegs) hmjδmjm′j(3.35)

The diagonal elements yield the Zeeman energy corrections EZ:

EZ = gjbmj (3.36)

withgj =

j (j + 1)− s (s + 1) + l (l + 1)2j (j + 1)

+ gej (j + 1) + s (s + 1)− l (l + 1)

2j (j + 1)(3.37)

62


andb = −γehB = µBB (3.38)

Using s = 12 and putting ge = 2, Eq. (3.37) simplifies to:

EZ,j=l+ 12

= b2l + 22l + 1

mj (3.39)

EZ,j=l− 12

= b2l

2l + 1mj (3.40)

The Zeeman effect is illustrated for 2s1/2, 2 p1/2 and 2 p3/2 levels in Figure 3.1 .The energies of Hamiltonian (3.14) in this regime are (combining Eqs. (3.10), (3.11), (3.39) and(3.40))

ESO + EZ =12

hcζnpl + b2l + 22l + 1

mj (j = l +12) (3.41)

ESO + EZ = −12

hcζnp(l + 1) + b2l

2l + 1mj (j = l − 1

2) (3.42)

Figure 3.1

63

3.3. THE PASCHEN-BACK EFFECT

3.3 The Paschen-Back Effect

In this case, the Zeeman Hamiltonian should be taken into account first:

H = H(0) + H(1) (3.43)

H(0) = HBohr + HZeeman (3.44)

H(1) = HSO (3.45)

The Zeeman interaction partially lifts the 2n2 degeneracy of the Bohr Hamiltonian in l, ml andms, so that we have to diagonalise the Zeeman Hamiltonian in the |lmlsms〉 subsets of an nlevel. The uncoupled basis is indeed most suitable since the Zeeman Hamiltonian is diagonalin it: taking the quantisation axis z along the magnetic field (~B = B~ez), we get

〈nl′m′lsm′s| HZeeman |nlmlsms〉 = 〈nl′m′lsm′s| b(

lz + ge sz

)|nlmlsms〉 (3.46)

= b (ml + gems) δll′δmlm′lδmsm′s (3.47)

These diagonal elements are the Zeeman energy corrections EZ and depend only on ml and ms.Note the independency of l. This leads to following table:

ml ms EZ

l +1/2 (l + ge/2) b ≈ (l + 1) bl −1/2 (l − ge/2) b ≈ (l − 1) b

l − 1 +1/2 (l − 1 + ge/2) b ≈ lbl − 1 −1/2 (l − 1− ge/2) b ≈ (l − 2) bl − 2 +1/2 (l − 2 + ge/2) b ≈ (l − 1) bl − 2 −1/2 (l − 2− ge/2) b ≈ (l − 3) bl − 3 +1/2 (l − 3 + ge/2) b ≈ (l − 2) bl − 3 −1/2 (l − 3− ge/2) b ≈ (l − 4) b

......

......

−l + 3 +1/2 (−l + 3 + ge/2) b ≈ (−l + 4) b−l + 3 −1/2 (−l + 3− ge/2) b ≈ (−l + 2) b−l + 2 +1/2 (−l + 2 + ge/2) b ≈ (−l + 3) b−l + 2 −1/2 (−l + 2− ge/2) b ≈ (−l + 1) b−l + 1 +1/2 (−l + 1 + ge/2) b ≈ (−l + 2) b−l + 1 −1/2 (−l + 1− ge/2) b ≈ −lb−l +1/2 (−l + ge/2) b ≈ (−l + 1) b−l −1/2 (−l − ge/2) b ≈ (−l − 1) b

64

3.3. THE PASCHEN-BACK EFFECT

We see that, in addition to the degeneracy in l, the(ml ; ms = − 1

2

)and

(m′l = ml − 2; ms = + 1

2

)states within a 2l level are also nearly degenerate.5 Thus, a 2(2l + 1) degenerate 2l level splitsup into 4 singlets

|l; 1/2〉 |l − 1; 1/2〉 | − l + 1;−1/2〉 | − l;−1/2〉 (3.48)

and 2(2l − 1) (nearly-degenerate) doublets. This is illustrated for a 2 p level in Figure 3.2 . The

Figure 3.2

spin-orbit interactionHSO = ξ(r)~l ·~s (3.49)

induces additional shifts and (partially) lifts the remaining (near-)degeneracies. However, theexpression

~l ·~s = 12

[l+ s− + l− s+

]+ lz sz (3.50)

shows that this operator does not couple between the degenerate states6: l is not shifted up ordown, and ml can shift with only one unit. In this case, perturbation theory for non-degeneratestates can be applied, even if there are (nearly) degenerate states. This means the first-order

5The energy difference between such levels is (ge − 2) b ≈ 0.0023b , which can easily be of the same order ofmagnitude or smaller than the spin-orbit contribution. This means we should in principle treat them as degeneratelevels when applying perturbation theory to include that contribution.

6I.e.: in the matrix representation of this operator, there are no off-diagonal elements between such states.

65

3.4. INTERMEDIATE REGIME

spin-orbit energy correction for an |nlmlsms〉 state is given by the expectation value of HSO

(the perturbation operator) in that state (the unperturbed basis function):

ESO = 〈nlmlsms| HSO |nlmlsms〉 = 〈nlmlsms| ξ(r)lz sz |nlmlsms〉 = hcζnlmlms (3.51)

The effect of the spin-orbit coupling is illustrated in Figure 3.2 for a p electron. The (near)degeneracy of the

(ml = 1, ms = − 1

2

)and

(ml = −1, ms = + 1

2

)is not lifted (in first order) by

the spin-orbit interaction in this particular case, but this result should not be generalised.We arrive at following total energy correction due to Hamiltonian (3.14) (combining Eqs. (3.47)and (3.51)):

EZ + ESO = b (ml + gems) + hcζnlmlms (3.52)

3.4 Intermediate Regime

In this case, the Zeeman and spin-orbit Hamiltonians should be treated on equal footing andwe apply following perturbational scheme:

H = H(0) + H(1) (3.53)

H(0) = HBohr (3.54)

H(1) = HZeeman + HSO (3.55)

H(1) lifts the degeneracy in ml and ms of an n 2l level, so that H(1) has to be diagonalised in the|ml ; ms〉 basis subset. Taking the quantisation axis along the magnetic field direction, we get7

H(1) = b(

lz + 2sz

)+ ξ(r)

[12

(l+ s− + l− s+

)+ lz sz

](3.56)

Note that neglecting the terms containing ladder operators - which yield off-diagonal elements- results in the Hamiltonian considered in the Paschen-Back regime. We find8

H(1) |n, l, ml ,+12〉 = b (ml + 1) h |n, l, ml ,+

12〉

+ξ(r)12

√l(l + 1)−ml(ml + 1)h2 |n, l, ml + 1,−1

2〉+ 0

+ξ(r)mlmsh2 |n, l, ml ,+12〉 (3.57)

7We again assume ge = 2 to simplify the calculations.8To avoid confusion we do not use the short-hand notation |ml ; ms〉 but |n, l, ml , ms〉, because the spin-orbit

coupling interaction depends on n and l.

66


H(1) |n, l, ml ,−12〉 = b (ml − 1) h |n, l, ml ,−

12〉

+0 + ξ(r)12

√l(l + 1)−ml(ml − 1)h2 |n, l, ml − 1,+

12〉

+ξ(r)mlmsh2 |n, l, ml ,−12〉 (3.58)

There are off-diagonal elements only between states |n, l, ml + 1,− 12 〉 and |n, l, ml ,+ 1

2 〉 , so thatthe matrix representation of H(1)

• is diagonal for the |n, l, ml , ms〉 states with extremal values for ml and ms, which aretherefore eigenfunctions of H(1):

|n, l, ml = +l, ms = +12〉 ESO + EZ = (l + 1)b +

12

lhcζnp (3.59)

|n, l, ml = −l, ms = −12〉 ESO + EZ = −(l + 1)b +

12

lhcζnp (3.60)

• otherwise consists of 2× 2 matrices in the(|n, l, ml + 1,− 1

2 〉, |n, l, ml ,+ 12 〉)

subsets: bml − 12 hcζnl(ml + 1) 1

2 hcζnl√

l(l + 1)−ml(ml + 1)

12 hcζnl

√l(l + 1)−ml(ml + 1) b(ml + 1) + 1

2 hcζnlml

(3.61)

Diagonalisation of these matrices yields the other eigenfunctions and correspondingenergies. The secular equation is

λ2 − 2b(2ml + 1)− hcζnl

2λ

+

[b2ml(ml + 1)− bhcζnl(ml +

12)− 1

4h2c2ζ2

nl l(l + 1)]

= 0 (3.62)

with solutions

λ =14[2b(2ml + 1)− hcζnl ]

(3.63)

± 14

√[2b(2ml + 1)− hcζnl ]

2 − 16b2ml(ml + 1) + 8hcζnlb(2ml + 1) + 4h2c2ζ2nl l(l + 1)

67


Example: a p levelWith l = 1 and s = 1

2 , the |ml ; ms〉 basis consists of six eigenfunctions: |1; 1/2〉, |1;−1/2〉,|0; 1/2〉, |0;−1/2〉, | − 1; 1/2〉 and | − 1;−1/2〉 . The matrix representation of H(1) (cf. (3.56))in this basis is

2b + 12 ζnp 0 0 0 0 0

0 − 12 ζnp

√2

2 ζnp 0 0 0

0√

22 ζnp b 0 0 0

0 0 0 −b√

22 ζnp 0

0 0 0√

22 ζnp − 1

2 ζnp 0

0 0 0 0 0 −2b + 12 ζnp

(3.64)

Figures 3.3, 3.4 and 3.59 shows the energy levels due to spin-orbit and Zeeman interactionsfor a 2p electron in function of the relative strength of the two interactions

• according to the intermediate-regime formulas (this section, orange lines). These are the’exact’ solutions in the sense that they can be applied for any value of b and ζnp .

• according to the Zeeman-regime formulas (Eqs. (3.39) and (3.40), dotted green line). Thisis valid only for b hcζnp .

• according to the Paschen-Back regime (Eq. (3.52), dashed black line). This is valid only forb hcζnp .

A value of ζ2p = 24.25 m−1 is used (Eq. (2.53) with Z = 1 and l = 1 ). Note that

• the basis functions with the ’extremal’ quantum numbers ( |1; 12 〉 = | 32 , 3

2 〉 and | −1;− 1

2 〉 = |32 ,− 3

2 〉) are exact eigenfunctions and consequently are valid in any regime.

• the other |j, mj〉 functions are good eigenfunctions in the Zeeman regime (cf. Figure3.3), but not valid at all in the Paschen-Back regime (cf. Figure 3.5). Likewise, the other|ml ; ms〉 functions are good eigenfunctions in the Paschen-Back regime, but not valid atall in the Zeeman regime.

9The only difference between the figures is the scale of the x and y axes!

68


Figure

3.3

69


Figure

3.4

70


Figure

3.5

71

Chapter 4

Techniques Of Approximation

4.1 Introduction

The central problem to be solved in quantum mechanics for stationary systems is the time-independent Schrodinger equation: we wish to determine the eigenvalues Ei (possible values ofE, i.e. possible energy values of the system) and corresponding eigenvectors ψi (possible wavefunctions for ψ, i.e. possible states of the system) of the eigenvalue equation

Hψ = Eψ (4.1)

However, in all but the simplest of systems, this equation cannot be solved analytically andone has to resort to approximations. In this chapter, a number of techniques will be discussedthat allow constructing such approximative solutions or are useful tools in that context.Section 4.2 deals with time-independent perturbation theory, a powerful and often-used techniquewhich can be employed whenever one wants to account for the effect of a small perturbationto a well-known system. Time-dependent perturbation theory, briefly discussed in Section 4.3,allows to calcultate probabilties of transitions between different energy states of a Hamiltonian.In Section 4.4, the basic principles of variation theory will be discussed and illustrated, whichallow systematic improvement of an initial guess for a wave function. Finally, we will brieflydiscuss the Hellmann-Feynman theorem, which can greatly facilitate calculation of the variationaldependence of a system’s energy on a particular parameter (Section 4.5).

4.2 Time-independent perturbation theory

In this section the key concepts and results of time-independent perturbation theory arepresented. Not all derivations are given in detail here, they can be found in any standard text

72

4.2. TIME-INDEPENDENT PERTURBATION THEORY

book on quantum mechanics. The exact solutions for a two-level system will first be discussed,as they will serve as reference point later on.

4.2.1 Two-level system: exact solutions and approximations

Consider a system governed by a Hamiltonian H(0) that has only two, orthonormal eigenstatesψ(0)1 and ψ

(0)2 with corresponding energies E(0)

1 and E(0)2 :

H(0)ψ(0)1 = E(0)

1 ψ(0)1 (4.2)

H(0)ψ(0)2 = E(0)

2 ψ(0)2 (4.3)

These form a complete orthonormal basis set. Now suppose there is an additional interactionwith Hamiltonian H(1), so that the total Hamiltonian becomes

H = H(0) + H(1) (4.4)

We assume here and in the following sections that the complete basis set for the undisturbedHamiltonian H(0) is also a complete basis set for the perturbation H(1). In fact, we choose (orbetter: approximate) the perturbation in such way that the basis set at hand is complete for it.We now want to find the eigenstates ψ+ and ψ− and corresponding energies E+ and E− of thissystem. These are the possible solutions for E and ψ in

Hψ = Eψ (4.5)

The new eigenstates can be expressed as linear combinations of the old eigenstates,1 i.e.

ψ = a1ψ(0)1 + a2ψ

(0)2 (4.6)

Determining the eigenstates thus means determining the coefficients ai .The energy eigenvalues and wave function coefficients are found by diagonalisation of theHamiltonian matrix representation in the

(ψ(0)1 , ψ

(0)2

)basis set:

[H11 H12

H21 H22

](4.7)

withHij = 〈ψ(0)

i | H |ψ(0)j 〉 (4.8)

1This condition is fulfilled because we assumed that the ψ(0)i states form a complete basis set.

73


As H is a Hermitian operator, H21 = H∗12. The energy eigenvalues are the solutions for E of∣∣∣∣∣ H11 − E H12

H21 H22 − E

∣∣∣∣∣ = 0 (4.9)

This yields

E± =12(H11 + H22)±

12

√(H11 − H22)

2 + 4H12H21 (4.10)

The secular equations for the coefficients a1 and a2 of the eigenstate corresponding to aneigenvalue E are [

H11 − E H12

H21 H22 − E

] [a1

a2

]=

[00

](4.11)

The first equation yields

a2 = −H11 − EH12

a1 (4.12)

Combining this with the normalisation condition of ψ,2

|a1|2 + |a2|2 = 1 (4.13)

we get

a1 =H12√

H212 + (H11 − E)2

(4.14)

a2 = − H11 − E√H2

12 + (H11 − E)2(4.15)

To avoid unnecessary complications, the results are given here in terms of matrix elementsof the complete Hamiltonian H = H(0) + H(1). In many (but not all) practical cases, theperturbation Hamiltonian H(1) produces no diagonal matrix elements, only off-diagonal matrixelements. Then H11 = E(0)

1 and H22 = E(0)2 .

Three limiting cases of this general problem are discussed in more detail in the boxes below.

Situation 1: initial degeneracy H11 = H22 = HIn the first situation the two energy levels of H0 are degenerate. Inserting this in (4.10), (4.14)and (4.15) we get

E+ = H + |H12| ψ+ =1√2(ψ

(0)1 +

|H12|H12

ψ(0)2 ) (4.16)

E− = H − |H12| ψ− =1√2(ψ

(0)1 −

|H12|H12

ψ(0)2 ) (4.17)

2The second secular equation is not independent from the first one and does not contain additional information.The normalisation condition eliminates the remaining degree of freedom (apart from a phase factor).

74


Note that the energies are linear in the perturbation and that the coefficients in thewavefunctions are independent of the perturbation. In the more general treatment we willsee that this is a characteristic of first-order degenerate perturbation theory. Also note that inthis case no approximation was made.As an example of this situation, one may picture the energy levels of a free electron in rest(double degeneracy in the electron spin state) in the presence of an external magnetic fieldalong the x-axis (H(1) = geµBBSx).

Situation 2: near degeneracyWe consider here the case where the eigenvalues of H0 are nearly degenerate. H11 is slightlylarger than H22, and in particular (H11 − H22)2 H12H21. An approximate solution ofthe problem will be obtained by expanding the eigenvalues (4.10) and coefficients of theeigenvectors (4.14) and (4.15) to lowest order in H11−H22

|H12| .

E+ =H11 + H22

2+ |H12|+

(H11 − H22)2

8 |H12|(4.18)

ψ+ =1√2((1 +

H11 − H22

4 |H12|)ψ

(0)1 + (1− H11 − H22

4 |H12|)ψ

(0)2 ) (4.19)

and

E− =H11 + H22

2− |H12| −

(H11 − H22)2

8 |H12|(4.20)

ψ− =1√2((1− H11 − H22

4 |H12|)ψ

(0)1 − (1 +

H11 − H22

4 |H12|)ψ

(0)2 ) (4.21)

Note that the wavefunctions are still orthogonal, but only approximately normalized. Thesolutions are very similar to those in Situation 1, but show small corrections to that case. Theeigenvalues exhibit corrections proportional to (H11−H22)

2

|H12| , the wavefunctions show corrections

of the order H11−H22|H12| . The highest energy level is pushed up a little further, the lowest level is

shifted downward by the correction. Because of the choice in ordering of the energy levels,E+ is closest to H11, and the corresponding wavefunction has a slightly larger coefficient a1

than a2.A practical example of this perturbation situation is encountered in the calcuation of energybands of semiconductors in the approximation of nearly free electrons (see Vastestoffysica).

Situation 3: far from degeneracyWe now assume that H11 and H22 differ largely: H12H21 (H11 − H22)2. We will nowexpand the eigenvalues 4.10 and coefficients of the eigenvectors 4.14 and 4.15 to lowest orderin |H12|

H11−H22.

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We obtain

E+ =H11 + H22

2+

H11 − H22

2+|H12|2

H11 − H22= H11 +

|H12|2

H11 − H22(4.22)

ψ+ = ψ(0)1 +

|H12|H11 − H22

ψ(0)2 (4.23)

E− =H11 + H22

2− H11 − H22

2− |H12|2

H11 − H22= H22 −

|H12|2

H11 − H22(4.24)

ψ− = − |H12|H11 − H22

ψ(0)1 + ψ

(0)2 (4.25)

Like in Situation 2 the wavefunctions are orthogonal, but only approximately normalized.The solutions resemble those of the unperturbed case very well.

The eigenvalues exhibit corrections proportional to |H12|2(H11−H22)

, the wavefunctions show

corrections of the order |H12|H11−H22

.The energy levels are pushed (slightly) farther apart from one another by the perturbation.

We will recognize all these features further on as the characteristics of second orderperturbation corrections.

Figure 4.1 shows the exact solutions for the energy levels and wavefunctions of a two-levelsystem.

Situation 1 is represented by the black-line solutions H11 = H22 = 0.

Situation 3 applies to all cases with H11 = −H22 6= 0, for small H12: in all cases the initialdependence on the non-diagonal element is quadratic. The range in which the quadraticapproximation works well is larger for larger |H11 − H22|.

For large H12 the solutions tend to those of the degenerate case. This is the range whereSituation 2 is valid.

76


Figure 4.1: Exact solutions for the energy eigenvalues E+ and E− (a en b) and for the coefficients ofthe eigenstates corresponding to E+ (c) and E− (d), as a function of the magnitude of the off-diagonalelement |H12|, for various magnitudes for the differences between the diagonal elements |H11 − H22|.

77


4.2.2 General principle of perturbation theory

Consider a system described by a Hamiltonian

H = H(0) + εH(1) (4.26)

for which

• Hamiltonian H(0) has a (possibly infinite) number of eigenstates ψ(0)i with corresponding

energies E(0)i , which are solutions of

H(0) |ψ(0)i 〉 = E(0)

i |ψ(0)i 〉 (4.27)

and are known. This is called the unperturbed system.

• εH(1) can be considered as a small perturbation: the energy shifts it causes must besmall compared to the energy differences between unperturbed levels with differentenergies.3,4 Furthermore, we assume again that the complete set of eigenfunctions for theunperturbed Hamiltonian also forms a complete set for the perturbation. Or alternatively,we approximate the perturbation by a Hamiltionian for which the wavefunctions |ψ(0)

i 〉form a complete set.

As an example, consider a hydrogen atom in the center-of-mass reference frame, with H(0)

accounting for the kinetic energy and the electron-core Coulomb interaction (Chapter 1). H(1)

could then be the interaction between the electronic spin and orbital momentum (Chapter 2) ora weak electromagnetic field (Chapter 3).

The central idea in perturbation theory then is to construct approximate solutions for the perturbedsystem in terms of the solutions for the unperturbed system. A good analogy to keep in mind isthe Taylor expansion of a (nicely behaving) function around a certain point (a), which allowsapproximating the value of the function at another point (b) without actually evaluating thefunction at b. And like in a Taylor expansion, the corrections to the unperturbed system canbe calculated up to an arbitrary order. We will consider corrections up to second order for theenergy and up to first order for the wave functions, which suffices in many practical cases inatomic, molecular and solid-state physics.

ε in the perturbation Hamiltonian may be considered as the knob that tunes the magnitudeof the perturbation. In practice, the considered perturbation is indeed very often dependent onan external (electric or magnetic) field whose magnitude determines how large the correctionsto the energies and wavefunctions are. In other cases, like in the example of spin orbit coupling,the magnitude of the perturbation is fixed by the physics of the problem. Nonetheless, ε

3This will become mathematically obvious later.4Degenerate energy levels require a different approach, as will become clear further on.

78


will turn out to be a very useful parameter that allows us to separate the various orders ofcorrections.

4.2.3 Time-independent perturbation theory for non-degenerate levels

4.2.3.1 Central equations

We first assume that all energy states are non-degenerate: with each energy eigenvalue justone eigenstate corresponds. An example of such a system is the linear harmonic oscillator, asdiscussed in Kwantummechanica 1. The expansion of the eigenvalues and eigenstates of the totalHamiltonian (4.26) reads

Ei = E(0)i + εE(1)

i + ε2E(2)i + . . . (4.28)

|ψi〉 = |ψ(0)i 〉+ ε |ψ(1)

i 〉+ ε2 |ψ(2)i 〉+ . . . (4.29)

The Schrodinger equation then becomes

(H(0) + εH(1))( |ψ(0)i 〉+ ε |ψ(1)

i 〉+ ε2 |ψ(2)i 〉+ . . .)

= (E(0)i + εE(1)

i + ε2E(2)i + . . .)( |ψ(0)

i 〉+ ε |ψ(1)i 〉+ ε2 |ψ(2)

i 〉+ . . .) (4.30)

This equation should hold for any value of ε (although we will only pay attention to smallvalues), so that solutions of this Schrodinger equation are found by solving the equation foreach power of ε separately. As we are only interested in the corrections to the energies up tosecond order (proportional to ε2) and to the wavefunctions up to first order (proportional to ε),the equations corresponding to the first three powers of ε give us all necessary information:

H(0) |ψ(0)i 〉 = E(0)

i |ψ(0)i 〉 (4.31)

H(0) |ψ(1)i 〉+ H(1) |ψ(0)

i 〉 = E(0)i |ψ

(1)i 〉+ E(1)

i |ψ(0)i 〉 (4.32)

H(0) |ψ(2)i 〉+ H(1) |ψ(1)

i 〉 = E(0)i |ψ

(2)i 〉+ E(1)

i |ψ(1)i 〉+ E(2)

i |ψ(0)i 〉 (4.33)

Equation (4.31) does not give new information: we have started the discussion with the knowneigenvalues and states of H(0) (see Equation (4.27)). We find the first-order corrections tothe eigenvalues and states by solving Equation (4.32) and the second-order correction to theeigenvalues from Equation (4.33). This is explained step by step in the following subsections.

4.2.3.2 First-order correction to the eigenvalues

We first address Equation 4.32, which we rewrite as

(H(0) − E(0)i ) |ψ(1)

i 〉 = (E(1)i − H(1)) |ψ(0)

i 〉 (4.34)

79


We have assumed that |ψ(0)i 〉 form a complete basis set of the problem including the

perturbation. Hence |ψ(1)i 〉 can be expanded as a linear combination of the functions |ψ(0)

i 〉

|ψ(1)i 〉 = ∑

jc(1)ij |ψ

(0)j 〉 (4.35)

This leads to(H(0) − E(0)

i )∑j

c(1)ij |ψ(0)j 〉 = (E(1)

i − H(1)) |ψ(0)i 〉 (4.36)

We find the first-order energy corrections by multiplying this equation with 〈ψ(0)i | , and noting

that the basis states |ψ(0)i 〉 are orthonormal

0 = E(1)i − 〈ψ

(0)i | H

(1) |ψ(0)i 〉 ⇒ E(1)

i = 〈ψ(0)i | H

(1) |ψ(0)i 〉 (4.37)

and multiplying the result with ε. Hence,

• the first-order corrections to the eigenvalues (energies) are linear in ε

• the corrections are found by computing the diagonal matrixelements of the perturbationHamiltonian H(1) in the basis of eigenfunctions of the unperturbed Hamiltonian |ψ(0)

i 〉.

In conclusion, to first order the energy eigenvalues read

Ei = E(0)i + ε〈ψ(0)

i | H(1) |ψ(0)

i 〉 (4.38)

4.2.3.3 First-order correction to the wavefunction

In order to find the coefficients c(1)ik we multiply Equation (4.36) with 〈ψ(0)k | , where k 6= i

〈ψ(0)k | (H(0) − E(0)

i )∑j

c(1)ij |ψ(0)j 〉 = 〈ψ

(0)k | (E(1)

i − H(1)) |ψ(0)i 〉

(E(0)k − E(0)

i )c(1)ik = −〈ψ(0)k | H

(1) |ψ(0)i 〉

c(1)ik =〈ψ(0)

k | H(1) |ψ(0)i 〉

(E(0)i − E(0)

k )(4.39)

Hence, we find as eigenstates of the perturbed Hamiltonian, correct to first order of perturba-tion

|ψi〉 = |ψ(0)i 〉+ ε ∑

j 6=i

〈ψ(0)j | H(1) |ψ(0)

i 〉

(E(0)i − E(0)

j )|ψ(0)

j 〉 (4.40)

We note that

• the perturbation Hamiltonian mixes all states j for which the non-diagonal element〈ψ(0)

j | H(1) |ψ(0)i 〉 6= 0 into the state i

80


• as the mixture coefficients c(1)ij are inversely proportional to E(0)i − E(0)

j , the largestcorrections are expected to come from states close in energy to the state that is beingcorrected

• the first-order correction to |ψ(0)i 〉 is perpendicular to |ψ(0)

i 〉

• we have silently made use of the fact that none of the energy levels is degenerate, becausefor all j 6= i, E(0)

i − E(0)j 6= 0 is a requirement for the formulas to be correct.

4.2.3.4 Second-order correction to the energy

For calculating the second-order energy corrections, we multiply Equation (4.33) with 〈ψ(0)i | ,

and make use of the fact that |ψ(0)i 〉 and |ψ(1)

i 〉 are orthogonal

〈ψ(0)i | (E(0)

i − H(0)) |ψ(2)i 〉+ 〈ψ

(0)i | H

(1) |ψ(1)i 〉 = E(2)

i

⇒ E(2)i = 〈ψ(0)

i | H(1) |ψ(1)

i 〉 (4.41)

Now inserting in this equation the expression (4.39) we found for |ψ(1)i 〉, we obtain

E(2)i = ∑

j 6=i

〈ψ(0)i | H(1) |ψ(0)

j 〉〈ψ(0)j | H(1) |ψ(0)

i 〉

E(0)i − E(0)

j

or

E(2)i = ∑

j 6=i

∣∣∣〈ψ(0)j | H(1) |ψ(0)

i 〉∣∣∣2

E(0)i − E(0)

j

(4.42)

Correct to the second order of perturbation, the energies thus read

Ei = E(0)i + ε〈ψ(0)

i | H(1) |ψ(0)

i 〉+ ε2 ∑j 6=i

∣∣∣〈ψ(0)j | H(1) |ψ(0)

i 〉∣∣∣2

E(0)i − E(0)

j

(4.43)

We note that

• second order corrections to Ei come from all states j for which the non-diagonal element〈ψ(0)

j | H(1) |ψ(0)i 〉 6= 0

• the second order energy corrections are proportional to ε2. Very often a perturbation dueto an external field produces no diagonal matrix elements in the basis in which H(0) isdiagonal. The lowest order energy corrections are then quadratic in the magnitude of theapplied field.

• as the mixture coefficients c(1)ij are inversely proportional to E(0)i − E(0)

j , the largestcorrections are expected to come from states close in energy to the energy that is beingcorrected

81


• second-order corrections from levels lying above the level that is being corrected arenegative, those coming from levels below the corrected level are positive. In particular,second-order corrections push down the lowest energy level and push up the highestenergy level. This is in agreement with what we have seen for the two-levels-system.Thus, second-order corrections always increase the total energy span of the system

• non-degeneracy of the energy states is again a requirement for the validity of theformulas.

In the following subsection the analysis is adapted to take into account the possibility ofdegenerate energy levels.

Example - exerciseAs an example of second order perturbation theory for non-degenerate energy levels, weconsider the system of a harmonic oscillator, perturbed by non-harmonic terms. This exampleis particularly useful, as it applies to vibrations of diatomic molecules (or molecules consistingof more than two atoms, if one considers the vibrations corresponding to only one normalmode). Molecular vibrations will be briefly introduced in the final chapter of this course.As introduced in Kwantummechanica 1, the Hamiltonian of a linear harmonic oscillator, whichwe here consider as the unperturbed system, is given by

H(0) =p2

2M+

Kx2

2(4.44)

In the case of a diatomic molecule, x represents the deviation from the equilibrium bondlength and M is the reduced mass of the two-body system. The characteristic frequency(eigenmode) of this oscillator is

ω =

√KM

(4.45)

It is convenient to introduce the ladder operators

a =

√Mω

2h(x +

i pMω

) a† =

√Mω

2h(x− i p

Mω) (4.46)

and the number operatorN = a† a (4.47)

The Hamiltonian can then be rewritten as

H(0) = hω(N +12) (4.48)

The eigenstates and energies for this unperturbed harmonic oscillator are

|ψ(0)n 〉 = |n〉 E(0)

n = hω(n +12) (4.49)

82


It is further interesting to note that

a |n〉 =√

n |n− 1〉 a† |n〉 =√

n + 1 |n + 1〉 (4.50)

and[a, a†] = aa† − a† a = 1 (4.51)

Finally, the expressions for x and p as linear combinations of the ladder operators are alsogiven

x =

√h

2Mω(a + a†) p = i

√hMω

2(a† − a) (4.52)

All energy levels of this system are indeed non-degenerate.

Exercise

• Calculate the corrections to the wavefunctions up to first and to the energies to secondorder, if H(1) = εx. This perturbation represents the situation of placing the harmonicoscillator (heteronuclear diatomic molecule) in a constant electric field (magnitude ε)parallel to the molecular axis.

• Calculate the corrections to the wavefunctions up to first and to the energies to secondorder, if H(1) = εx3. This perturbation models the anharmonic part of the interatomicinteraction (large vibrations, ε fixed).

4.2.4 Time-independent perturbation theory for degenerate levels

If two or more levels of the unperturbed system are degenerate, Eqs. (4.43) and (4.40) are notvalid.5 Such a situation, which we already encountered in the general treatment of the two-level system (Situation 1), requires a different approach.

Let us assume that H(0) has n distinct energy eigenvalues E(0)i , each with a degeneracy νi.

n and νi may be finite or infinite numbers. |ψ(0)iα 〉 is an orthonormal basis set in which H(0) is

diagonal. H(1) is, however, not necessarily diagonal in this basis and may, in particular, producenon-diagonal matrix elements between |ψ(0)

iα 〉 and |ψ(0)iβ 〉, which share the same zero-order

energy eigenvalue E(0)i . This set of eigenstates is not unique, though: any linear combination of

eigenstates sharing the same eigenvalue is a new eigenstate with the same eigenvalue. Hence,for each eigenvalue E(0)

i we try and find νi linear combinations of the original eigenstates

|φ(0)iα 〉 = ∑

β

ciα,iβ |ψ(0)iβ 〉 (4.53)

5Note that the denominators in the final terms in these equations would be zero in this case, and lead to aninfinite result, if the perturbation Hamiltonian produces a non-diagonal element between these states.

83


such that all non-diagonal elements of H(1) between states with the same zero-order energy E(0)i

vanish. Then, non-diagonal elements of H(1) only exist between states with different zero-orderenergies, and Eqs. (4.43) and (4.40) can be applied again.

Finding the linear combinations (4.53) mathematically corresponds to diagonalizing H(1)

within the subbasis |ψ(0)iα 〉 of degenerate states with energy E(0)

i . The problem to be solvedthus is ∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

H(1)i1,i1 − E(1)

i H(1)i1,i2 . . . H(1)

i1,iνi

H(1)i2,i1 H(1)

i2,i2 − E(1)i . . . H(1)

i2,i

......

. . ....

H(1)iνi ,i1

H(1)iνi ,i2

. . . H(1)iνi ,iνi− E(1)

i

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣= 0 (4.54)

whereH(1)

iα,iβ = 〈ψ(0)iα | H

(1) |ψ(0)iβ 〉 (4.55)

This yields νi energy eigenvalues E(1)iα , which may all be different when the perturbation lifts the

degeneracy completely, or some of which may still coincide (incomplete lifting of degeneracy).The energy eigenvalues of the complete Hamiltonian, correct to first order, now are

Eiα = E(0)i + εE(1)

iα (4.56)

The coefficients ciα,iβ of the new eigenstate |φ(0)iα 〉 corresponding with the eigenvalue E(1)

iα arefound by solving

H(1)i1,i1 − E(1)

iα H(1)i1,i2 . . . H(1)

i1,iνi

H(1)i2,i1 H(1)

i2,i2 − E(1)iα . . . H(1)

i2,iνi

......

. . ....

H(1)iνi ,i1

H(1)iνi ,i2

. . . H(1)iνi ,iνi− E(1)

iα

ciα,i1

ciα,i2

...

ciα,iνi

=

0

0

...

0

(4.57)

After applying this procedure for all n degenerate energy levels of H(0), we have found a basis|φiα〉 in which H(1) can only have non-diagonal matrix elements between states with differentzero-order energy

H(1)iαi ,jβ j

= 〈φiαi |ˆH(1) |φjβ j〉 6= 0 ⇒ i 6= j (4.58)

84


and for which we can apply the formulas derived above for the wavefunctions, corrected tofirst order

|φiαi〉 = |φ(0)iαi〉+ ε ∑

j 6=i∑β j

〈φ(0)jβ j| H(1) |φ(0)

iαi〉

(E(0)i − E(0)

j )|φ(0)

jβ j〉 (4.59)

and for the energies, corrected to second order

Eiαi = E(0)i + εE(1)

iαi+ ε2 ∑

j 6=i∑β j

∣∣∣〈φ(0)jβ j| H(1) |φ(0)

iαi〉∣∣∣2

E(0)i − E(0)

j

(4.60)

We end this subsection with a few remarks

• note that second-order corrections lower the energy of the ground state of the system andraise the energy of the highest excited state. This conclusion has already been drawn fromEq. (4.43)

• in many cases where H(0) has degenerate energy levels, the degeneracies are notcompletely lifted by the first-order corrections. The remaining degeneracies may be liftedby the second-order corrections. However, in some cases, even after applying the second-order corrections the degeneracy is not completely lifted. This is the case when H(0) andH(1) share symmetry elements, as we shall see in Chapter 8

• we have started the analysis here from an orthonormal basis of eigenstates of H(0).Strictly speaking the eigenstates |ψ(0)

iα 〉 with the same eigenvalue E(0)i do not have to

be orthonormal. In that case the secular equation for finding the first-order energycorrections (4.61) is generalized to det|H(1) − E(1)

i S| = 0∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

H(1)i1,i1 − E(1)

i Si1,i1 H(1)i1,i2 − E(1)

i Si1,i2 . . . H(1)i1,iνi− E(1)

i Si1,iνi

H(1)i2,i1 − E(1)

i Si2,i1 H(1)i2,i2 − E(1)

i Si2,i2 . . . H(1)i2,i − E(1)

i Si2,iνi

......

. . ....

H(1)iνi ,i1− E(1)

i Siνi ,i1 H(1)iνi ,i2− E(1)

i Siνi ,i2 . . . H(1)iνi ,iνi− E(1)

i Siνi ,iνi

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣= 0 (4.61)

with Siα,iβ =⟨

ψ(0)iα

∣∣∣ ψ(0)iβ

⟩. In the Linear Combinations of Atomic Orbitals technique for

calculating molecular energy levels (see Chapter 7) we will, indeed, start off from a non-orthogonal basis of orbitals centered on different atoms, and we shall use this equation tofind molecular orbitals for the system

• more generally, in the context of perturbation theory (also for time-dependent perturba-tion theory) it is important to know which matrix elements of H(1) are zero and which arenot. In Chapter 8 we will see that symmetry helps us a lot in answering this question.

85

4.3. TIME-DEPENDENT PERTURBATION THEORY

Example - the hydrogen atom in a uniform electric field : Stark effectExercise: calculate the effect of applying a uniform electric field ~E = ε~ez on the energy levelsof the ground state and the first excited state of the hydrogen atom, neglecting fine-structureeffects.

As unperturbed Hamiltonian H(0) take the spinless Hamiltonian of Chapter 1, Equation (1.8),and the unperturbed wavefunctions are those in Table 1.3.A uniform electric field introduces an extra potential energy term in the Hamiltonian

− e V = e ε z = e ε r cos θ = H(1) (4.62)

• First consider the effect on the first excited state with n = 2. Demonstrate that in firstorder this level splits in three and calculate the magnitude of the splitting. Two levelsare to first order not affected by the applied electric field. Where does the lowest ordercorrection to these levels come from (the magnitude of the correction is not asked for)?What is the sign of this correction?

• Demonstrate that the ground state (n = 1) is only affected in second order. Calculatethe contribution from the states n = 2 and n = 3 to this second-order correction.

• How does the electric field affect the n = 2→ n = 1 spectral line of the hydrogen atom?

4.3 Time-dependent perturbation theory

Time-dependent perturbation theory allows us to calculate the probability that a quantumme-chanical system in a given state makes a transition to another state under stimulation by atime-dependent field. In atomic and molecular physics, this time-dependent field is usually anoscillating electromagnetic field, that via Fourier analysis can be decomposed into harmoniccomponents. For this reason, we will confine the theory here to the particular case of aperturbation with a harmonic time-dependence. We will make use of Dirac’s method of variationof constants.

4.3.1 Variation of constants

We consider a time-independent unperturbed system with known energy eigenvalues and acorresponding complete set of wavefunctions

H(0)ψ(0)n = E(0)

n ψ(0)n (4.63)

The time-dependent wavefunctions

Ψ(0)n = ψ

(0)n e−iE(0)

n t/h (4.64)

86


are solutions of the time-dependent, unperturbed Schrodinger equation

H(0)Ψ(0) = ih∂Ψ(0)

∂t(4.65)

We now consider a small time-dependent perturbation εH(1)(t). The exact form of the time-dependence is not specified at this moment. Later on, we will only consider harmonicperturbations.Like before we assume that any eigenfunction of the problem including the perturbation

(H(0) + εH(1)(t)

)Ψ = ih

∂Ψ∂t

(4.66)

can be expanded in eigenfunctions of the unperturbed system

Ψ(t) = ∑n

an(t)Ψ(0)n (t) = ∑

nan(t)ψ

(0)n e−iE(0)

n t/h (4.67)

Switching to bracket notation, we find

ε ∑n

an(t)e−iE(0)n t/hH(1)(t) |ψ(0)

n 〉 = ih ∑n

dan

dte−iE(0)

n t/h |ψ(0)n 〉 (4.68)

By multiplying with ψ(0)∗m and integrating over space, we find an expression for the time-

derivative of the coefficient am, which we further on write as am

am =ε

ih ∑n

an〈ψ(0)m | H(1)(t) |ψ(0)

n 〉ei(

E(0)m −E(0)

n

)t/h (4.69)

If we now assume that the perturbation is switched on at t = 0, we find by integrating

am(t) = am(0) +ε

ih

∫ t

0∑n

an(t)〈ψ(0)m | H(1)(t) |ψ(0)

n 〉ei(

E(0)m −E(0)

n

)t/hdt (4.70)

So far we have not yet made use of approximations. However, equation (4.70) only yieldsthe time-dependence of the coefficients we are looking for, if we already know these time-dependences, which is of course an unwanted situation. We can find an approximate solutionfor the problem by explicitly making use of the fact that the perturbation is small and that weapply it for a relatively short time. We further assume that at t = 0 the system is in a well-defined eigenstate ψ

(0)n , so that an(0) = 1 and for all m 6= n, am(0) = 0. Since the perturbation

is small, for relatively short times we may assume that an remains close to 1 and all othercoefficients am remain much smaller than 1. Inserting this in equation (4.69) we find

am =ε

ih〈ψ(0)

m | H(1)(t) |ψ(0)n 〉e

i(

E(0)m −E(0)

n

)t/h (4.71)

andam(t) =

ε

ih

∫ t

0〈ψ(0)

m | H(1)(t) |ψ(0)n 〉e

i(

E(0)m −E(0)

n

)t/hdt (4.72)

87


The probability Pn→m of finding the system that was originally in |ψ(0)n 〉 after switching on the

perturbation for a time t in the state |ψ(0)m 〉, which we call the transition probability between

the initial state |ψ(0)n 〉 and the final state |ψ(0)

m 〉, is given by

Pn→m = |am(t)|2 (4.73)

If (initially) the transition probability is found proportional with time, it makes sense to definea transition rate Wn→m

Wn→m = Pn→m/t (4.74)

From equation (4.72) we see that both these quantities are proportional to∣∣∣〈ψ(0)

m | H(1)(t) |ψ(0)n 〉∣∣∣2.

4.3.2 Harmonically oscillating perturbation

If we consider the interaction of an atom or a molecule with an oscillating electromagnetic fieldderived from a vector potential ~A(t), the perturbation for electric dipole transitions is given by(or better, approximated by)

H(1)(t) =em~A(t) · ~p = − ieh

m~A(t) · ~∇ (4.75)

where m represents the electron mass and−e its charge. An electromagnetic field with arbitrarytime-dependence can be decomposed into harmonic components by Fourier analysis. For thisreason, it is sufficient to consider monochromatic waves with angular frequency ω0 > 0

~A(t) = ~A0e−iω0t + ~A+0 eiω0t (4.76)

The magnitude of the vector potential∣∣∣~A0

∣∣∣ takes over the role of ε in the analysis in section4.3.1. We now introduce the following notations for the transition matrix element

H(1)mn = − ieh

m~A0 · 〈ψ(0)

m | ~∇ |ψ(0)n 〉 = −H(1)+

mn (4.77)

and for the Bohr angular frequency

ωmn =E(0)

m − E(0)n

h(4.78)

By inserting these in equation (4.72) we find

Pn→m =

∣∣∣H(1)mn

∣∣∣2h2

∣∣∣∣∫ t

0

(ei(ωmn−ω0)t + ei(ωmn+ω0)t

)dt∣∣∣∣2 (4.79)

88


After executing the integration over t this becomes

Pn→m =

∣∣∣H(1)mn

∣∣∣2h2

∣∣∣∣∣ ei(ωmn−ω0)t − 1ωmn −ω0

+ei(ωmn+ω0)t − 1

ωmn + ω0

∣∣∣∣∣2

(4.80)

which can be transformed into

Pn→m =

∣∣∣H(1)mn

∣∣∣2 t2

h2

∣∣∣∣ sin ((ωmn −ω0)t/2)(ωmn −ω0)t/2

ei(ωmn−ω0)t/2 +sin ((ωmn + ω0)t/2)

(ωmn + ω0)t/2ei(ωmn+ω0)t/2

∣∣∣∣2(4.81)

The function sin(u)/u only takes appreciable values if its argument is close to zero. SincePn→m → 0 for t→ 0, transition probabilties considerably different from zero are only obtainedfor ω0 ≈ ±ωmn or hω0 ≈ ±

(E(0)

m − E(0)n

). The physical interpretation is clear: in transition

processes energy should be (approximately) conserved. The energy of the photon, indeed,matches the energy difference between the two states of the quantum mechanical system.For E(0)

m 6= E(0)n only one of these conditions can be fulfilled at a time. The two terms in equation

(4.81) then have largely different sizes and only one needs to be retained. For E(0)m > E(0)

n thecondition for resonance absorption can be met hω0 = E(0)

m − E(0)n , in which case

Pn→m =

∣∣∣H(1)mn

∣∣∣2 t2

h2 (4.82)

An equal transition probability is found for resonance (stimulated) emission when hω0 =

E(0)n − E(0)

m (when E(0)m < E(0)

n ). Finally, inserting the value of the transition matrix element(4.77) back into equation (4.82), we find for the transition probability between two infinitelysharp energy levels

Pn→m =e2∣∣∣~A0

∣∣∣2 t2

m2

∣∣∣〈ψ(0)m | ~∇ |ψ(0)

n 〉∣∣∣2 δ

(E(0)

m − E(0)n ± hω0

)(4.83)

The appearance of the matrix element 〈ψ(0)m | ~∇ |ψ(0)

n 〉, or more generally 〈ψ(0)m | H(1) |ψ(0)

n 〉, in theformula for transition probability leads to the notion of selection rules. If 〈ψ(0)

m | H(1) |ψ(0)n 〉 = 0

the transition probability is 0 and the transition is called forbidden. For 〈ψ(0)m | H(1) |ψ(0)

n 〉 6= 0 thetransition is called allowed. In Chapter 8 we shall see that symmetry is of great help in derivingselection rules. E.g. for eigenfunctions with a well-defined parity (for which the Hamiltonian isinvariant under inversion of the cooridates), there are parity selection rules. As the ~∇ operatorhas odd parity, electric dipole transitions are only possible between states with different parity.

In a final step we examine the probability for transitions to a final state that is part of acontinuous distribution between E f ,min and E f ,max characterized by a density of states ρ(E f ).For simplicity we restrict the discussion to absorption transitions, but for emission the same

89


result is obtained. We further denote

|ψ(0)n 〉 = |i〉 E(0)

n = Ei |ψ(0)m 〉 = | f 〉 E(0)

m = E f (4.84)

Fon any single energy level within the distribution of final states the transition probability isstill given by equation (4.81), in which we here only retain the term for optical absorption (firstterm). We find the transition probability to the distribution of by multiplying with the densityof final states and integrating over the final energy. We thus find

Pi→ f =t2

h2

∫ E f ,max

E f ,min

sin2 ((E f − Ei − hω0)t/2h)(

(E f − Ei − hω0)t/2h)2

∣∣∣H(1)i f

∣∣∣2 ρ(E f )dE f (4.85)

The function (sin(x)/x)2 is very sharply peaked around x = 0. The transition matrix elementand density of states are assumed to be smooth functions of E f and approximately take constantvalues in the range where (sin(x)/x)2 significantly differs from zero. Furthermore, since theintegrand quickly becomes zero as

∣∣E f − Ei − hω0∣∣ becomes larger, the integration range may

safely be expanded over the full range from −∞ to +∞. Hence, we find

Pi→ f =t2

h2

∣∣∣H(1)i f

∣∣∣2 ρ(Ei + hω0)2ht

∫ +∞

−∞

sin2 (x)x2 dx =

2π

h

∣∣∣H(1)i f

∣∣∣2 ρ(Ei + hω0) t (4.86)

The resulting transition probability is proportional with time, yielding a time-independent(initial) transition rate

Wi→ f =2π

h

∣∣∣H(1)i f

∣∣∣2 ρ(Ei + hω0) (4.87)

Equation (4.87) is known as Fermi’s Golden Rule.

90

4.4. VARIATION THEORY

4.4 Variation theory

Variation theory allows optimisation of wave functions of a particular functional form: it givesa procedure to determine the best representative (i.e. the one with the lowest energy) of aspecific class of functions. Two frequently used variational treatments are the Rayleigh method(Section 4.4.1) and the Rayleigh-Ritz method (Section 4.4.2).

4.4.1 The Rayleigh method

Consider a system with Hamiltonian H and a trial wave function ψtrial, which can have any(physically acceptable) functional form. The Rayleigh ratio E then is defined as

E =

∫ψ∗trialHψtrialdτ∫ψ∗trialψtrialdτ

(4.88)

where τ comprises all spatial and spin coordinates. E is an energy functional, as its valuedepends on the function ψtrial. The importance of this quantity lies in the variation theorem,which states that following relation holds for any ψtrial:

E ≥ E0 (4.89)

where E0 is the true ground-state energy of the system. The equality holds if and only if thetrial wave function is the true ground-state wave function of the system.

Proof of the variation theoremWe can expand the trial wave function in the eigenfunctions ψi of the Hamiltonian, which weassume to be orthonormal:

ψtrial = ∑i

ciψi

∫ψ∗i ψjdτ = δij (4.90)

We find ∫ψ∗trialHψtrialdτ = ∑

i∑

jc∗i cj

∫ψ∗i Hψjdτ (4.91)

= ∑i

∑j

c∗i cj

∫ψ∗i Ejψjdτ (4.92)

= ∑i

∑j

c∗i cjEjδij (4.93)

= ∑i|ci|2 Ei (4.94)

91


≥ E0 ∑i|ci|2 (4.95)

where the latter inequality follows from the fact that Ei ≥ E0 and |ci|2 ≥ 0 . Consequently

E ≥E0 ∑

i|ci|2

∑i|ci|2

= E0 (4.96)

The Rayleigh method is minimisation of the Rayleigh ratio for a particular functional form,which yields the optimal function of that form. Moreover, since the Rayleigh ratio cannot beless than the true ground-state energy, it provides an upper bound for the latter.

4.4.1.1 Example 1

We propose the functional form e−kr for the ground-state wave function of hydrogenic atoms.We can then use the Rayleigh method to find the optimal function of this form, i.e. the value ofk which minimises the energy. The Hamiltonian is

H = − h2

2µ~∇2 − Ze2

4πε0r(4.97)

but since the trial wave function has no angular dependence, it suffices to consider the radialderivatives in the Laplacian (cf. Eq. (1.17)):

H = − h2

2µ

1r2

ddr

(r2 d

dr

)− Ze2

4πε0r(4.98)

H = − h2

2µ

(2r

ddr

+d2

dr2

)− Ze2

4πε0r(4.99)

Consequently, we have

∫ψ∗trialHψtrialdτ =

∫ π

0sin θdθ

∫ 2π

0dφ∫ +∞

0ψ∗trialHψtrialr2dr (4.100)

= 4π∫ +∞

0e−kr

(He−kr

)r2dr (4.101)

Since (2r

ddr

+d2

dr2

)e−kr =

(−2k

r+ k2

)e−kr (4.102)

92


we get

∫ψ∗trialHψtrialdτ = 4π

[h2kµ

∫ +∞

0e−2krrdr− h2k2

2µ

∫ +∞

0e−2krr2dr

− Ze2

4πε0

∫ +∞

0e−2krrdr

](4.103)

The integrals in the above expressions reduce to (substituting u = −2kr2)

∫ +∞

0e−2krrdr =

1(2k)2

∫ +∞

0e−uudu =

14k2 (4.104)

∫ +∞

0e−2krr2dr =

1(2k)3

∫ +∞

0e−uu2du =

14k3 (4.105)

where we have used∫ +∞

0e−uuz−1du = Γ (z) (4.106)

Γ (n) = (n− 1)! (n = 0, 1, . . .) (4.107)

Eq. (4.103) simplifies to ∫ψ∗trialHψtrialdτ =

πh2

2µk− Ze2

4ε0k2 (4.108)

Furthermore, we have∫ψ∗trialψtrialdτ =

∫ π

0sin θdθ

∫ 2π

0dφ∫ +∞

0e−2krr2dr = 4π

14k3 =

π

k3 (4.109)

so that

E =h2k2

2µ− Ze2k

4πε0(4.110)

Considered as a function of k, this is a parabola and the minimum is found by differentiatingwith respect to k:

dEdk

= 0 (4.111)

k =Ze2µ

4πε0h2 (4.112)

so that the best ground-state wave function of this functional form is

ψ0 = e− Ze2µ

4πε0 h2 r= e−

raµ (4.113)

93


with energy (plugging Eq.(4.112) into Eq. (4.110))

Emin = − Z2e4µ

32π2ε20h2 = −13.6 eV (4.114)

Note that Eqs. (4.113) and (4.114) correspond to the true 1s wave function (Table 1.2) and energy(Eq. (1.115) with n = 1). This is due to the fact that the exact wave function is of the same formas the trial function.

4.4.1.2 Example 2

Now consider a trial wave function of the form e−kr2for hydrogenic atoms.

The relevant integrals are

∫ψ∗trialψtrialdτ = 4π

∫ +∞

0e−2kr2

r2dr (4.115)

∫ψ∗trialHψtrialdτ = 4π

∫ +∞

0e−kr2

(He−kr2

)r2dr (4.116)

Since (2r

ddr

+d2

dr2

)e−kr2

=(−6ke−kr2

+ 4k2r2)

e−kr2(4.117)

the right-hand part of Eq. (4.116) equals

4π

[− h2

2µ

[−6k

∫ +∞

0e−2kr2

r2dr + 4k2∫ +∞

0e−2kr2

r4dr]− Ze2

4πε0

∫ +∞

0e−2kr2

rdr

](4.118)

Using Eqs. (4.106) and (4.107), together with

Γ (z + 1) = zΓ (z) (4.119)

Γ(

12

)=

∫ +∞

0e−uu−1/2du = 2

∫ +∞

0e−t2

dt =∫ +∞

−∞e−t2

dt =√

π (4.120)

the radial integrals in Eqs. (4.115) and (4.118) can be reduced to (putting u = 2kr2)

∫ +∞

0e−2kr2

r2dr =1

4√

2k3/2

∫ +∞

0e−uu1/2du =

14√

2k3/2

12√

π (4.121)

∫ +∞

0e−2kr2

r4dr =1

8√

2k5/2

∫ +∞

0e−uu3/2du =

18√

2k5/2

34√

π (4.122)

∫ +∞

0e−2kr2

rdr =14k

∫ +∞

0e−udu =

14k

(4.123)

94


and we obtain ∫ψ∗trialψtrialdτ =

( π

2k

) 32

(4.124)

∫ψ∗trialHψtrialdτ =

3π3/2h2

4√

2µk−

12 − Ze2

4ε0k−1 (4.125)

so that

E =3h2

2µk− Ze2√

2ε0π3/2

√k (4.126)

and the optimal value for k is again found by differentiation:

dEdk

= 0 (4.127)

k =Z2e4µ2

18π3ε20h4 (4.128)

The corresponding ground-state energy is (plugging Eq. (4.128) into Eq. (4.126))

Emin = − Z2e4µ

12π3ε20h2 = −11.5 eV (4.129)

Note that this is indeed higher than the true ground-state energy (cf. Eq. (4.114)) by 2.1 eV.

4.4.2 The Rayleigh-Ritz method

The Rayleigh-Ritz method is a particular case of the Rayleigh method, in which the trialwave function is a linear combination of fixed basis functions ψi (which in general are noteigenfunctions of the Hamiltonian):

ψtrial = ∑i

ciψi (4.130)

and the coefficients ci in this expansion are the parameters to be optimised. To facilitate thecalculations, we will assume all basis functions and coefficients are real. The Rayleigh ratio is

E =

∑i,j

c∗i cj

∫ψ∗i Hψjdτ

∑i,j

c∗i cj

∫ψ∗i ψjdτ

(4.131)

=

∑i,j

cicjHij

∑i,j

cicjSij(4.132)

95


where we have defined

Hij =∫

ψ∗i Hψjdτ =∫

ψi Hψjdτ (4.133)

Sij =∫

ψ∗i ψjdτ =∫

ψiψjdτ (4.134)

Note that Sij = δij if the basis functions are orthonormal. Differentiation with respect tocoefficient ck yields

dEdck

=

∑i,j

[(∂ci

∂ck

)cj + ci

(∂cj

∂ck

)]Hij

∑i,j

cicjSij(4.135)

−

[∑i,j

[(∂ci

∂ck

)cj + ci

(∂cj

∂ck

)]Sij

] [∑i,j

cicjHij

](

∑i,j

cicjSij

)2 (4.136)

Because∂ci

∂ck= δik (4.137)

we get

dEdck

=

∑j

cjHkj + ∑i

ci Hik

∑i,j

cicjSij−

[∑

jcjSkj + ∑

iciSik

] [∑i,j

cicjHij

](

∑i,j

cicjSij

)2 (4.138)

=

∑j

cj(

Hkj − ESkj)+ ∑

ici (Hik − ESik)

∑i,j

cicjSij(4.139)

Equating these expressions (there is one for every coefficient ck) to zero will yield a set ofequations for the optimal values of the coefficients. The expression is zero only if the numeratorvanishes. Because6

Hki = Hik Ski = Sik (4.140)

the numerator can be rewritten as (dummy summation indices)

∑i

ci (Hki − ESki) + ∑i

ci (Hik − ESik) = 2 ∑i

ci (Hik − ESik) (4.141)

6The matrix represenation of a quantum mechanical operator is Hermitian and all values are assumed real here,so that it is symmetric.

96


Thus, the set of equations to be solved is

∑i

ci (Hik − ESik) = 0 (k = 1, 2, . . .) (4.142)

The existence of a non-trivial solution requires that the secular determinant is zero:

|Hik − ESik| = 0 (4.143)

The roots of this equation are the possible values of E , the lowest of which is the optimalapproximation for the ground-state energy of the system for the chosen set of basis functions.The coefficients of the corresponding optimal approximation for the ground-state wavefunction are found by solving the secular equations (4.142) for that value of E .

4.4.2.1 Example

Suppose we want to investigate the dependency of the ground-state wave function and energyon the core mass mc for a hydrogen atom (isotope mass effect). The Hamiltonian will only beaffected via the reduced mass in the kinetic energy term:

H = − h2

2µ~∇2 − e2

4πεor(4.144)

µ =memc

me + mc(4.145)

and the true eigenfunctions and energies of this Hamiltonian are given by the Ψnlml of Table 1.3and Eq. (1.115) respectively, the only difference being the value of µ and aµ . Although theseexact solutions are known, the Rayleigh-Ritz method will be applied here as a demonstration.Consider following trial function:

ψtrial = c1ψ1 + c2ψ2 (4.146)

withψ1 =

(πa3

0)− 1

2 e−r

a0 ψ2 =(32πa3

0)− 1

2

(2− r

a0

)e−

r2a0 (4.147)

i.e. a linear combination of the 1s and 2s hydrogen orbitals (Table 1.3) in the case of an infinitelyheavy nucleus.7 Note that ψ1 and ψ2 are not eigenfunctions of Hamiltonian (4.144), e.g.:

Hψ1 =(πa3

0)− 1

2

[− h2

2µ

(2r

ddr

+d2

dr2

)− e2

4πεor

]e−

ra0 (4.148)

=(πa3

0)− 1

2

[(2h2

2µa0− e2

4πε0

)1r

e−r

a0 − h2

2µa20

e−r

a0

](4.149)

7If mc = +∞, then µ = me and aµ = a0.

97


Using the definitions of the Bohr radius a0 and the Rydberg constant R,

a0 =4πε0h2

mee2 R =mee4

8ε20h3c

(4.150)

and definingγ =

me

mc=

me

µ− 1 (4.151)

this can be simplified to

Hψ1 =(πa3

0)− 1

2

[e2

4πε0

(me

µ− 1)

1r

e−r

a0 − e2

8πε0a0

me

µe−

ra0

](4.152)

=e2

8πε0a0

[2γa0

1r− (γ + 1)

]ψ1 (4.153)

= hcR[

2γa01r− (γ + 1)

]ψ1 (4.154)

Eq. (4.154) shows that ψ1 is an eigenfunction of H (with energy −hcR) only if γ = 0, i.e.only if µ = me (mc = +∞), as expected. Calculation of the Hamiltonian matrix elements isstraightforward. E.g.:

H11 =∫

ψ∗1(r)Hψ1(r)d~r (4.155)

= hcR[

2γa04π∫ +∞

0ψ∗1(r)

1r

ψ1(r)r2dr− (γ + 1)∫

ψ∗1(r)ψ1(r)d~r]

(4.156)

= hcR[

2γa04π1

πa30

∫ +∞

0e−

2ra0 rdr− (γ + 1)

](4.157)

= hcR[

2γa04π1

πa30

a20

4− (γ + 1)

](4.158)

= hcR (γ− 1) (4.159)

In an analogous fashion, it can be shown that

H22 =14

hcR (γ− 1) (4.160)

H12 = H21 =8√

227

hcRγ (4.161)

98


Because ψ1 and ψ2 are orthonormal, we have Sij = δij and the secular determinant simplifies to∣∣∣∣∣∣∣H11 − E H12

H21 H22 − E

∣∣∣∣∣∣∣ = E2 − (H11 + H22) E +(

H11H22 − H212)

(4.162)

and the roots of this equation are

E =18

hcR (γ− 1)

5± 3

√1 +

213

38

(γ

γ− 1

)2 (4.163)

For the real hydrogen atom, we have γ = 0.000545, and the lowest of the two roots is:

E =18

hcR (γ− 1)

5 + 3

√1 +

213

38

(γ

γ− 1

)2 ≈ hcR (γ− 1) = −0.999455hcR (4.164)

which is slightly higher than the true ground-state energy −hcR, in accordance with Eq. (4.89).The corresponding secular equations are

c1 (H11 − E) + c2H12 = 0 (4.165)

c1H21 + c2 (H22 − E) = 0 (4.166)

but these are not linearly independent. Combination with the normalisation condition c21 + c2

2 =

1 yieldsψ = 1.00000ψ1 − 0.00054ψ2 (4.167)

which shows that, within the fixed basis set (4.147), the optimal ground-state wave functionhas a very limited mixing in of the 2s orbital.

4.4.3 Some comments on the variation principle

As illustrated in the previous sections, a variational approach can yield an upper bound for thetrue ground-state energy by virtue of the variation theorem. However, variational techniquescan also be applied to find a lower bound, so that the true value can be determined moreprecisely. An upper bound for the first excited state can also be calculated using a variationalprinciple. However, such calculations are often very hard in practice.Finally, one should keep in mind that the wave function is restricted from the start to aparticular functional form, so that the optimal ground-state wave function and energy resultingfrom a variational treatment may still be poor approximations for the true ones. Moreover,there is no guarantee that a wave function which is optimised with respect to the energy -even one reproducing the ground-state energy well - is suited for an accurate determination ofanother property of the system, such as its dipole moment .

99

4.5. THE HELLMANN-FEYNMAN THEOREM

4.5 The Hellmann-Feynman theorem

Consider a system with a Hamiltonian H which depends on a parameter λ. An eigenfunctionψ and its energy eigenvalue E will in general also depend on λ. The question now is how theenergy varies with changing λ, i.e. we want to calculate the derivative of E with respect to λ.The Hellmann-Feynman theorem states that

dEdλ

=

⟨δHδλ

⟩ψ

(4.168)

where the right-hand side is an expectation value for wave function ψ :⟨δHδλ

⟩ψ

≡ 〈ψ| δHδλ|ψ〉 ≡

∫ψ∗(λ)

δH(λ)

δλψ(λ)dτ (4.169)

The significance of this theorem lies in the fact that the dependence of the wave function on theparameter need not be known to calculate the term on the right-hand side and that the partialderivative may be very simple. This theorem can be employed in quite diverse situations, but isparticularly useful for studying the effect of an external electric or magnetic field on a quantumsystem.

Proof of the Hellmann-Feynman theoremAssuming the wave function is normalised, i.e.∫

ψ∗(λ)ψ(λ)dτ = 1 (4.170)

the energy is

E ≡ E(λ) =∫

ψ∗(λ)H(λ)ψ(λ)dτ (4.171)

so that

dEdλ

=∫

δψ∗(λ)

δλH(λ)ψ(λ)dτ +

∫ψ∗(λ)

δH(λ)

δλψ(λ)dτ (4.172)

+∫

ψ∗(λ)H(λ)δψ(λ)

δλdτ (4.173)

We have ∫δψ∗(λ)

δλH(λ)ψ(λ)dτ = E(λ)

∫δψ∗(λ)

δλψ(λ)dτ (4.174)

and, because the Hamiltonian operator is Hermitian and E is real, also

100

4.5. THE HELLMANN-FEYNMAN THEOREM

∫ψ∗(λ)H(λ)

δψ(λ)

δλdτ =

∫ [H(λ)ψ(λ)

]∗ δψ(λ)

δλdτ (4.175)

= E(λ)∫

ψ∗(λ)δψ(λ)

δλdτ (4.176)

so that

dEdλ

= E(λ)∫ [

δψ∗(λ)

δλψ(λ) + ψ∗(λ)

δψ(λ)

δλ

]dτ +

∫ψ∗(λ)

δH(λ)

δλψ(λ)dτ (4.177)

= E(λ)δ

δλ

[∫ψ∗(λ)ψ(λ)dτ

]+∫

ψ∗(λ)δH(λ)

δλψ(λ)dτ (4.178)

=∫

ψ∗(λ)δH(λ)

δλψ(λ)dτ (4.179)

where we have used Eq. (4.170).

As an example, consider the standard hydrogen Hamiltonian:

H = − h2

2µ~∇2 − Ze2

4πε0r(4.180)

which has eigenfunctions ψnlml and energy eigenvalues (cf. Eq. (1.116))

En = −12

e2Z2

(4πε0)aµ

1n2 (4.181)

We can apply the Hellmann-Feynman theorem with λ = Z:

dEn

dZ=

⟨δ

δZ

[− h2

2µ~∇2 − Ze2

4πε0r

]⟩ψnlml

(4.182)

− e2Z(4πε0)aµ

1n2 = − e2

4πε0

⟨r−1⟩

ψnlml

(4.183)

so that ⟨r−1⟩

ψnlml

=Z

aµn2 (4.184)

in accordance with Eq. (1.137). Note how simple the calculation is compared to that in Section1.2.7.3.

101

Chapter 5

The helium atom

5.1 The hamiltonian

For the He atom (Z = 2) we have two electrons, each containing kinetic and potential energywith respect to the nucleus, which can be described by H1 and H2. We use r1 and r2 forthe distance between electron and nucleus. Futhermore, there is an additional potentialenergy term (H12) describing the repulsion between both electrons, which only depends onthe distance r12 = |~r1 −~r2| between both electrons. Consequently, the total Hamiltonian can bewritten as

H = H1 + H2 + H12 (5.1)

= − h2

2me~∇2

1 −Ze2

4πε0r1− h2

2me~∇2

2 −Ze2

4πε0r2+

e2

4πε0r12(5.2)

= − h2

2me(~∇2

1 + ~∇22)−

2e2

4πε0r1− 2e2

4πε0r2+

e2

4πε0r12(5.3)

Note that we should use the reduced mass, instead of me. However, this error is not importantin the remainder of the discussion.This Hamiltonian, although fairly simple at first sight, cannot be solved analytically. Oneoption to approximate the solution is treating the contribution of H12 as a perturbation. We willsee that this is actually not the case - due to the strong contribution of the repulsion betweenboth electrons - and so this approach should be further refined. Nevertheless, we consider herethe unperturbed system as

H(0) = H1 + H2 (5.4)

= − h2

2me~∇2

1 −Ze2

4πε0r1− h2

2me~∇2

2 −Ze2

4πε0r2(5.5)

102

This Hamiltonian is a sum of two independent terms, which leads to the eigenfunction beingthe product of the individual eigenfunctions. Suppose that H = H1 + H2 and that ψi is aneigenfunction for Hi with energy Ei. If we take ψ = ψ1ψ2, then we can write

Hψ = (H1 + H2)ψ = (H1 + H2)ψ1ψ2 (5.6)

= (H1ψ1)ψ2 + ψ1(H2ψ2) (5.7)

= E1ψ1ψ2 + E2ψ1ψ2 (5.8)

= (E1 + E2)ψ (5.9)

because Hiψj = 0 if i 6= j. This can easily be generalized for an Hamiltonian with n independentterms.In the case of helium, we can write for the wavefunction for H(0)

ψ(~r1,~r2) = ψn1l1ml1(~r1)ψn2l2ml2(~r2) (5.10)

The energy levels for hydrogenic atoms are given by ((1.115) and (2.52)):

En = −hcRZ2

n2 (5.11)

For the helium (Z = 2) atom we find

E = E1,n1 + E2,n2 = −4hcR(

1n2

1+

1n2

2

)(5.12)

5.1.1 The Coulomb integral J.

If we now take the influence of the electron-electron repulsion term H12 into account, we findfor the first-order correction to the energy

E(1) = 〈n1l1ml1; n2l2ml2|e2

4πε0r12|n1l1ml1; n2l2ml2〉 = J (5.13)

The energy correction J is called the Coulomb integral, and can be written as 1

J =e2

4πε0

∫|ψn1l1ml1(~r1)|2

(1

r12

)|ψn2l2ml2(~r2)|2dτ1τ2 (5.14)

This positive integral can be interpreted in the following way. The probability of finding thefirst electron in the volume element dτ1 is given by |ψn1l1ml1(~r1)|2dτ1. The corresponding chargeis obtained by multiplying with −e. Similarly, −e|ψn2l2ml2(~r2)|2dτ2 is the charge for the volumeelement dτ2. The product then yields the Coulombic interaction, and the total contribution tothe potential energy is obtained by integration over the entire volume.

1[B&J]: p.465 ’Two electron atoms’.

103

Figure 5.1: Interpretation of J [A&F].

To evaluate the interaction integral J, we have to find an expression for r12 in terms of thevariables for ri, θi, φi.

Figure 5.2: Defining ri, θi, φi. ω is the angle between~r1 and~r2.

The distance between both electrons can be written as, where we consider r1 > r2:

r12 =√

r21 + r2

2 − 2r1r2 cos ω (5.15)

withcos ω = cos θ1 cos θ2 + sin θ1 sin θ2 cos(φ1 − φ2) (5.16)

which leads to1

r12=

1r1

[1 +

(r2

r1

)2

− 2(

r2

r1

)cos ω

]−1/2

(5.17)

104

If we put s = r2r1

and w = cos ω, then

1r12

=1r1[1 + s2 − 2ws]−1/2 =

1r1

∞

∑k=0

Pk(cos ω)

(r2

r1

)k

(5.18)

This is a consequence of the equation 2

T(w, s) = (1− 2ws + s2)−1/2 =∞

∑l=0

Pl(w)sl , with |s| < 1 (5.19)

Given our condition r1 > r2, the condition |s| < 1 is fulfilled.Making use of the addition theorem for spherical harmonics

Pk(cos ω) =4π

2k + 1

k

∑m=−k

Yk,m(θ1, φ1)Y∗k,m(θ2, φ2) (5.20)

so that finally1

r12= 4π

∞

∑k=0

12k + 1

k

∑m=−k

Yk,m(θ1, φ1)Y∗k,m(θ2, φ2)rk

2

rk+11

(5.21)

In the case of r1 < r2, r1 and r2 have to be switched in (5.21).We consider the situation where both electrons are in the 1s orbital, as we can expect the 1s2

configuration to have the lowest energy. The wavefunction for each of the individual electronsthen comes down to ψi =

√Z3

πa30e−Zri/a0 . In this case, the wavefunctions ψi do not contribute to

the angular part of the integration (note that this will not be the case for orbitals other than s).The integration over the spherical harmonics in 1

r12is only different from zero for k = m = 0.

The summation in (5.21) therefore reduces to

1r12

= 4π1√4π

1√4π

1r1

(5.22)

which is true for r1 > r2. In the other case, 1r12

= 1r2

.With this, the integration of J becomes

J =e2

4πε0

(Z3

πa30

)2 ∫ ∞

0

∫ ∞

0

e−2Z(r1+r2)/a0

r12r2

1r22dr1dr2

∫ π

0sin θ1dθ1

∫ π

0sin θ2dθ2

∫ 2π

0dφ1

∫ 2π

0dφ2

(5.23)

2[B&J]: p.260, equation [6.82].

105

5.2. EXCITED STATES OF HE

and by replacing 1/r12:

J =e2

4πε0

(Z3

πa30

)2 ∫ ∞

0

[∫ r2

0

e−2Zr1/a0

r2r2

1dr1 +∫ ∞

r2

e−2Zr1/a0

r1r2

1dr1

]e−2Zr2/a0r2

2dr2 · 4 · 4π2

=e2

4πε0

(Z3

πa30

)2

(4π)2 527

a50

Z5

=58

Za0

e2

4πε0(5.24)

For He (Z = 2), we find J = 5.510−18J = 34eV. To calculate the ground state energy, we have toadd J to the energies following from H1 and H2. For the case of a He core and one electron (ina 1s orbital), the binding energy (5.11) equals Ei = 4 · 13.6eV = 54.4eV.The total energy for the He ground state is then

E = (−54.4− 54.4 + 34)eV = −74.8eV (5.25)

This value can experimentally be obtained by determining the energy to fully ionize a neutralHe atom (from He to He+, and from He+ to He2+). It gives 78.9eV, from which thecalculated result deviates by 4.1eV. The agreement is not perfect, but reasonable, given thatthe perturbation (which yielded J) is not small.

5.2 Excited states of He

5.2.1 Exchange integral K

We will now look at the excited states of He, where one or both electrons occupy a differentorbital than 1s, e.g. 1s and 2s. The wavefunctions are then described by, depending on whichelectron is in what state:

ψn1l1ml1(~r1)ψn2l2ml2(~r2) or ψn2l2ml2(~r1)ψn1l1ml1(~r2) (5.26)

This can be written as a(~r1)b(~r2) or b(~r1)a(~r2), which we will denote by state 1 and state 2,respectively. The energy for the unperturbed system is given by Ea + Eb. For the perturbedsystem, we use the perturbation theory for degenerate states, as both states 1 and 2 havethe same energy. Therefore, we use the secular determinant, for which the following matrix

106

5.2. EXCITED STATES OF HE

elements have to be calculated:

H11 = 〈a(~r1)b(~r2)|H1 + H2 +e2

4πε0r12|a(~r1)b(~r2)〉 = Ea + Eb + J (5.27)

H22 = Ea + Eb + J (5.28)

H12 = 〈a(~r1)b(~r2)|H1 + H2 +e2

4πε0r12|b(~r1)a(~r2)〉 (5.29)

= (Ea + Eb)〈a(~r1)b(~r2)| |b(~r1)a(~r2)〉+ 〈a(~r1)b(~r2)|e2

4πε0r12|b(~r1)a(~r2)〉 (5.30)

= 〈a(~r1)b(~r2)|e2

4πε0r12|b(~r1)a(~r2)〉 = H21 (5.31)

We made use of the orthogonality of the states ψn1l1ml1 and ψn2l2ml2 , which yields e.g.

〈a(~r1)b(~r2)| b(~r1)a(~r2)〉 = 0. (5.32)

The integral for H12 is called the exchange integral K, which is positive:

K =e2

4πε0〈a(~r1)b(~r2)|

1r12|b(~r1)a(~r2)〉 > 0 (5.33)

The secular determinant can thus be written as

H11 − ES11 H12 − ES12

H21 − ES21 H22 − ES22=

H11 − E H12

H21 H22 − E

=Ea + Eb + J − E K

K Ea + Eb + J − E= 0 (5.34)

In the first line we used that S11 =∫(a(1)b(2))∗a(1)b(2)d~r1d~r2 = 1 = S22 and S12 = S21 = 0

because of states 1 and 2 being orthonormal. The solutions of (5.34) are

E = Ea + Eb + J ± K (5.35)

The eigenfunctions are

ψ±(~r1,~r2) =1√2[a(~r1)b(~r2)± b(~r1)a(~r2)] (5.36)

or in terms of the different orbitals and their quantum numbers

ψ±(~r1,~r2) =1√2[ψn1l1ml1(~r1)ψn2l2ml2(~r2)± ψn2l2ml2(~r1)ψn1l1ml1(~r2)] (5.37)

107

5.3. FOUNDING THE RUSSELL-SAUNDERS NOTATION

The degeneracy of the functions a(1)b(2) and b(1)a(2) is now removed by the electronrepulsion. Indeed, the energy of the two wavefunctions ψ+ and ψ− differs by a value 2K.This is a pure quantum mechanical effect, without classical counterpart, leading to a correctionto the Coulomb integral J. Fig. 5.3 illustrates the probability to find both electrons in one of thewavefunctions. For ψ− we find that the probability is zero for r12 = 0, or~r1 = ~r2. This meansthat both electrons tend to avoid each other. This dip in the probability density |ψ−|2 is calledthe Fermi hole. For ψ+, there is a probability to find both electrons at the same place, whichleads to a larger energy, J + K, compared to the energy for the ψ− wavefunction (= J − K).

Figure 5.3: Probability density for ψ− and ψ+ [A&F].

The key difference between ψ+ and ψ− can be seen by looking at the influence of interchangingboth electrons. ψ− is antisymmetric, and changes sign:

ψ−(2, 1) =1√2[a(2)b(1)− b(2)a(1)] = −ψ−(1, 2) (5.38)

On the other hand, ψ+ is symmetric when changing electrons:

ψ+(2, 1) =1√2[a(2)b(1) + b(2)a(1)] = ψ+(1, 2) (5.39)

5.3 Founding the Russell-Saunders notation

5.3.1 [Lk, 1r12] = 0

We first show that the components of ~L commutate with 1r12

, or more specifically that

[Lk,1

r12] = 0, for k = x, y, z,+,− (5.40)

108


For the two electrons:Lx = l1x + l2x (5.41)

We will first calculate

[l1x,1

r12] f = (l1x

1r12− 1

r12l1x) f (5.42)

= l1xf

r12− 1

r12l1x f (5.43)

=1i

(y1

∂

∂z1− z1

∂

∂y1

)f

r12− 1

r12

1i

(y1

∂

∂z1− z1

∂

∂y1

)f (5.44)

=fi

(y1

∂

∂z1− z1

∂

∂y1

)1

r12(5.45)

Where we used that (uv)′ = u′v + uv′. Using the explicit form of r12 (in cartesian coordinates!)

r12 =√(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2 (5.46)

We find for the derivative along z1:

∂

∂z1

1r12

= −12[(x1 − x2)

2 + (y1 − y2)2 + (z1 − z2)

2]−3/22 (z1 − z2) (5.47)

= − 1r3

12(z1 − z2) (5.48)

and similarly∂

∂y1

1r12

= − 1r3

12(y1 − y2) (5.49)

For (5.45) we thus find

fi

(y1

∂

∂z1− z1

∂

∂y1

)1

r12= − f

ir−3

12 [y1(z1 − z2)− z1(y1 − y2)] (5.50)

= if

r312(z1y2 − y1z2) (5.51)

In complete analogy one can prove that

[l2x,1

r12] f = i

fr3

12(z2y1 − y2z1) (5.52)

This can also be understood by applying the permutation 1 2 and noting that r12 = r21. Byaddition of (5.51) and (5.52), one obtains[

Lx,1

r12

]f =

[l1x + l2x,

1r12

]f = 0 (5.53)

109


By the same method, one easily obtains[Ly,

1r12

]f =

[Lz,

1r12

]f = 0 (5.54)

By noting that (see Chapter 1)

~L2 = L2x + L2

y + L2z , and L± = Lx ± iLy (5.55)

we find[~L2,

1r12

] = [L+,1

r12] = [L−,

1r12

] = 0 (5.56)

5.3.2 〈α′ JM| O |αJM〉

If O is an hermitian operator, and ~J an angular momentum operator, satisfying the relation

[ Jk, O] = [~J2, O] = 0 (5.57)

with k = x, y, z,+,−, then we prove that

〈α′ J′M′| O |αJM〉 = δJ J′δMM′〈α′ JM| O |αJM〉 (5.58)

in which 〈α′ JM| O |αJM〉 is independent from M. Indeed, from (5.57) we get JzO = O Jz and itfollows that on the one hand

〈α′ J′M′| O Jz |αJM〉 = M〈α′ J′M′| O |αJM〉 (5.59)

and on the other hand

〈α′ J′M′| JzO |αJM〉 = 〈αJM| ( JzO)† |α′ J′M′〉∗ (5.60)

= 〈αJM| (O† J†z ) |α′ J′M′〉∗ (5.61)

Here we used the definition of the adjoint, or hermitian conjugate operator 3, and also (AB)† =

B† A†4. Finally, because O and Jz are hermitian operators, they are auto-adjoint (O† = O andJ†z = Jz) 5.

〈α′ J′M′| JzO |αJM〉 = 〈αJM| (O† Jz) |α′ J′M′〉∗ (5.62)

= M′〈αJM| O† |α′ J′M′〉∗ (5.63)

= M′〈α′ J′M′| O |αJM〉 (5.64)

3[B&J] p.196, eq. [5.30].4[B&J] p.197, eq. [5.33].5[B&J] p.196, eq. [5.31].

110


We finally obtain by combining (5.59) and (5.64)

M′〈α′ J′M′| O |αJM〉 = M〈α′ J′M′| O |αJM〉 (5.65)

or(M′ −M)〈α′ J′M′| O |αJM〉 = 0 (5.66)

so 〈α′ J′M′| O |αJM〉 = 0 if M′ 6= M.Furthermore, ~J2O = O~J2 and in complete analogy one finds

[J′(J′ + 1)− J(J + 1)]〈α′ J′M′| O |αJM〉 = 0 (5.67)

leading to〈α′ J′M′| O |αJM〉 = 0, except if J′(J′ + 1) = J(J + 1) (5.68)

Consequently J′ = J or J′ = −J − 1. The latter is physically meaningless, so that J′ = J.

5.3.3 Independency of M

Finally, we start from〈α′ J′M′| O J+ |αJM〉 = 〈α′ J′M′| J+O |αJM〉 (5.69)

The left side yields

〈α′ J′M′| O J+ |αJM〉 =√

J(J + 1)−M(M + 1)〈α′ J′M′| O |αJM + 1〉 (5.70)

The right side gives

〈α′ J′M′| J+O |αJM〉 = 〈αJM| ( J+O)† |α′ J′M′〉∗ (5.71)

= 〈αJM| O† J†+ |α′ J′M′〉∗ (5.72)

= 〈αJM| O† J− |α′ J′M′〉∗ (5.73)

=√

J(J + 1)−M′(M′ − 1)〈αJM| O† |α′ J′M′ − 1〉∗ (5.74)

=√

J(J + 1)−M′(M′ − 1)〈α′ J′M′ − 1| O |αJM〉 (5.75)

(5.76)

Both matrix elements are zero unless M′ = M + 1 and J′ = J, and one then has

〈α′ JM + 1| O |αJM + 1〉 = 〈α′ JM| O |αJM〉 (5.77)

111


5.3.4 Spin-Orbit Hamiltonian

The spin-orbit hamiltonian is

HSO =N

∑i=1

ξ(ri)~li ~si (5.78)

In general, we have〈i| AB |k〉 = ∑

m〈i| A |m〉〈m| B |k〉 (5.79)

Using the Wigner-Eckart theorem, one has the following operator equivalencies:

〈LM′LSM′S| ξ(ri)~li ~si |LMLSMS〉 (5.80)

= ∑L′′M′′L S′′M′′S

〈LM′LSM′S| ξ(ri)~li |L′′M′′LS′′M′′S〉〈L′′M′′LS′′M′′S |~si |LMLSMS〉 (5.81)

= 〈LM′LSM′S| ξ(ri)~li |LMLSM′S〉〈LMLSM′S|~si |LMLSMS〉 (5.82)

= Ki(L, S)Mi(L, S)〈LM′LSM′S|~L |LMLSM′S〉〈LMLSM′S| ~S |LMLSMS〉 (5.83)

= Ki(L, S)Mi(L, S)〈LM′LSM′S|~L ~S |LMLSMS〉 (5.84)

Here Ki and Mi do not depend upon ML, M′L, MS or M′S. Taking

ξ(LS) =N

∑i=1

Ki(L, S)Mi(L, S) (5.85)

one obtains by summation

〈LM′LSM′S| HSO |LMLSMS〉 = ξ(LS)〈LM′LSM′S|~L ~S |LMLSMS〉 (5.86)

Furthermore~J = ~L + ~S (5.87)

so that~L ~S =

12(~J2 −~L2 − ~S2) (5.88)

and

〈LM′LSM′S| HSO |LMLSMS〉 =ζ(LS)

2[J(J + 1)− L(L + 1)− S(S + 1)]δM′J MJ

δJ J′ (5.89)

with J = L + S, L + S− 1, ..., |L− S|. The first order energy-correction is

E(LSJ) = E(LS) +ζ(LS)

2[J(J + 1)− L(L + 1)− S(S + 1)] (5.90)

Consequently, the configurations are split up by the Coulomb repulsion in terms characterizedby L and S. The terms themselves are split up in multiplets according to the rule of Landé:

E(L, S, J)− E(L, S, J − 1) =ζ(LS)

2[J(J + 1)− (J − 1)J] = ζ(LS)J (5.91)

112

5.4. THE SPECTRUM OF HELIUM

Therefore, it is necessary to represent the wave function by the quantum numbers L, S and J,or by the Russell-Saunders notation 2S+1LJ .

5.4 The spectrum of helium

For the ground state, 1s2, we have a single term with energy 2Ea + J (5.25). Because we havel1 = l2 = 0, we find an S term for the ground state (L = 0).For the excited states where the electrons occupy different orbitals we find two states withenergy Ea + Eb + J + K and Ea + Eb + J − K. This energy difference should become clear fromabsorption spectroscopy on He.We can limit ourselves here to the case of 1s1nl1, as configurations nl1n’l’1 with n, n′ > 1would lead to an excitation energy higher than the (first) ionization energy of He. This 1s1nl1

configuration leads to terms with L = l2 (l1 = 0), so for 1s12s1 we have an S term, for 1s12p1 a Pterm, etc. Based on the selection rule ∆l = ±1, transitions can occur between S and P, betweenP and D, but not between S and S or S and D. The selection rules between the ψ− and ψ+

wavefunctions are

(symmetrical↔ symmetrical) and (anti-symmetrical↔ anti-symmetrical).

This means that transitions between symmetrical and antisymmetrical wavefunction combi-nations are not allowed. For instance consider the dipole moment for such a transition. Theelectrical dipole operator (see also exercises) is given by −e~r1 − e~r2, so that the dipole momentfor this transition is

µ = −e∫

ψ∗+(~r1,~r2)(~r1 +~r2)ψ−(~r1,~r2)d~r1d~r2 (5.92)

When we now interchange both electrons (i.e. change the labels 1 and 2), the integrand shouldremain the same, as otherwise the integral will vanish. Because ψ− changes sign (and ψ+

does not), this integrand changes sign, which makes that transitions between symmetric andantisymmetric wavefunctions are not allowed.Regarding the multiplicities, we have to consider the spin as well. As s1 = s2 = 1

2 , we expectsinglet (S = 0) and triplet (S = 1) states to occur. For the singlet terms we then find thatJ = L, for the triplet terms we get J = L + 1, L, L − 1 provided that L > 0. For the 1s1np1

configuration, we can then expect levels such as 1P1, 3P0, 3P1 and 3P2, which are then furthersplit by the spin-orbit coupling. Naively - but see below! - we can expect that each of them willoccur as a symmetric and an asymmetric combination.

Fig. 5.4 shows the energy levels based on the observed spectrum of helium. This isto some extent consistent with the abovementioned analysis. Each 1s1nl1 configurationleads to two terms (symmetric and antisymmetric), which differ in energy. As the groundstate is (orbitally) symmetric, we know that the allowed transitions from this level go to

113

5.5. PAULI PRINCIPLE

Figure 5.4: Energy level scheme of He, classified into symmetric and antisymmetric orbitalwavefunctions. Selected allowed and forbidden transitions are indicated. [A&F].

other symmetric states. We can also see that the selection rules (§5.4) apply. When bothsymmetrical and antisymmetrical combinations for the same configuration can be found, thenthe antisymmetrical one has the lowest energy.A key observation is that all the symmetric states appear to be singlets, while all theanti-symmetric states are triplets, wich is in contrast to our naive considerations on themultiplicities we made above. For instance, for the 1s1np1 configuration there are only fourterms, instead of eight. This is to be explained in the next section.

5.5 Pauli principle

In the above discussion we did not take the spin state of both electrons into account. We alreadysaw (see exercises) that a combination of two spins s1 = 1

2 and s2 = 12 on the one hand leads to

S = 0 (with α for spin up (ms =12 ) and β for spin down (ms = − 1

2 )

σ−(1, 2) =1√2[α(1)β(2)− β(1)α(2)] (5.93)

Note that this state is antisymmetric:

σ−(2, 1) = −σ−(1, 2) (5.94)

114


On the other hand, three states belong to S = 1, which are all three symmetric:

σMS=+1+ (1, 2) = α(1)α(2) (5.95)

σMS=0+ (1, 2) =

1√2[α(1)β(2) + β(1)α(2)] (5.96)

σMS=−1+ (1, 2) = β(1)β(2) (5.97)

(5.98)

These four spin states can be combined with both orbital states ψ+ and ψ−, in total eight states.Experimentally it turns out that the following combinations are effectively observed:

ψ+(1, 2)σ−(1, 2), ψ−(1, 2)σMS=+1+ (1, 2), ψ−(1, 2)σMS=0

+ (1, 2), ψ−(1, 2)σMS=−1+ (1, 2) (5.99)

while the following ones are not observed:

ψ−(1, 2)σ−(1, 2), ψ+(1, 2)σMS=+1+ (1, 2), ψ+(1, 2)σMS=0

+ (1, 2), ψ+(1, 2)σMS=−1+ (1, 2) (5.100)

Observing the effect of particle interchange, we see that the allowed states are antisymmetric,while the forbidden states are symmetric. This is extended for multiple electrons into the Pauliprinciple:

Pauli principleThe total wavefunction, including the spin, must be antisymmetricwith respect to the interchange of any pair of electrons.

This principle has a fundamental aspect, as it can be extended for fermions (particles with halfintegral spin, such as electrons) and bosons (particles with integral spin, such as photons): Thetotal wavefunction must be antisymmetric (symmetric) with respect to the interchange of anypair of fermions (bosons). An important consequence of the Pauli principle is that two electronscannot have identical quantum numbers, or

Pauli exclusion principleNo pair of electrons can occupy the same state.

As an illustration, consider two electrons with identical quantum numbers, so also with thesame spin. If we put the quantization axis along their spin, we have spin up. The totalspin state is then symmetric, α(1)α(2). Based on the Pauli principle, the orbital state has tobe antisymmetric, e.g. ∝ a(1)b(2) − a(2)b(1). If both electron have identical wavefunctions,then a = b and the wavefunction vanishes, showing that two electrons cannot have the samequantum numbers. If both electrons are in the same orbital, then the spin state must beantisymmetric, e.g. ∝ α(1)β(2) − α(2)β(1). This is similar to the spins pairing up (note that

115


there also exist a spin-up/spin-down combination for S = 1).Wavefunctions satisfying the Pauli principle can be written as Slater determinants, for instance

ψ(1, 2) = ψ1s(~r1)ψ1s(~r2)σ−(1, 2) (5.101)

=1√2

ψ1s(~r1)ψ1s(~r2)[α(1)β(2)− β(1)α(2)] (5.102)

=1√2

ψ1s(~r1)α(1) ψ1s(~r1)β(1)ψ1s(~r2)α(2) ψ1s(~r2)β(2)

(5.103)

This can be written somewhat more concise, by introduction of the spinorbital:

ψα1s(1) = ψ1s(~r1)α(1), ψ

β1s(1) = ψ1s(~r1)β(1) (5.104)

so that the ground state of the He atom can be expressed as

ψ(1, 2) =1√2

ψα1s(1) ψ

β1s(1)

ψα1s(2) ψ

β1s(2)

(5.105)

This state is anti-symmetric (changing two columns leads to a sign change) and if both electronswould occupy the same state (orbital and spin!), then there are two identical columns, leadingto a vanishing wavefunction.The Slater determinants can be written in a general form, for N electrons and φ1,..., φN

spinorbitals:

ψ(1, 2, ..., N) =1√N!

φ1(1) φ2(1) ... φN(1)φ1(2) φ2(2) ... φN(2)| | ... |

φ1(N) φ2(N) ... φN(N)

(5.106)

In Fig. 5.4 we saw that, based on experimental observations, the triplet states (with S = 1) arelower in energy than the singlet states (with S = 0). For the triplet states the spins are parallel,characterized by symmetric spin functions ((5.95) to (5.97)). Consequently, we have that theorbital wavefunction must be antisymmetrical, which means that both electrons tend to avoideach other (e.g. the probability to find them at r12 = 0 is zero). This effect of electrons withparallel spins avoiding each other is called the spin correlation. A consequence of the need foran antisymmetric orbital wavefunction is that the Coulomb repulsion between both electronswill be lower, compared to the case where we have the symmetric orbital wavefunction incombination with the antisymmetric spin function σ−(1, 2). Hence, the triplet states will havea lower energy than the singlet states 6.

6It reality we have based our reasoning solely on the first order perturbation theory. In addition, there will alsobe a perturbation of the wavefunctions, which also has to be taken into account when determining the relative orderof the triplet and singlet states.

116


5.5.1 Permutations of (1 2 3)

Let’s now illustrate the construction of Slater determinants for three electrons. First, we haveto consider that there are six possible permutations of (1 2 3), which can be labeled as follows:

E =

(1 2 31 2 3

); K =

(1 2 31 3 2

)(5.107)

A =

(1 2 32 3 1

); L =

(1 2 33 2 1

)(5.108)

B =

(1 2 33 1 2

); M =

(1 2 32 1 3

)(5.109)

5.5.2 Multiplication table for the group

These permutations form a group. One of the key properties of a group is that the combinationof any two elements in the group also belongs to the group. We will deal with group theory inmore detail later. As a consequence, it is useful to construct a multiplication table, which showsthe result of combining two elements in the group (RS = T, with R in the left column and S inthe top row).

G(P) E A B K L ME E A B K L MA A B E M K LB B E A L M KK K L M E A BL L M K B E AM M K L A B E

5.5.3 Parity of the permutations

P(E) = P(A) = P(B) = +1 (5.110)

P(K) = P(L) = P(M) = −1 (5.111)

The ’parity’ is obtained by composition of the generating permutations, in which just twoelements are permuted. There are C2

n such generating permutations, in this case C23 = 3!

2!1! = 3.In this example, these are K, L and M. One takes P(E) = 1 and P(G) = −1 for all thegenerating permutations.P is a homomorphism, implying

P(X ·Y) = P(X)⊗P(Y) (5.112)

117


and this is to be verified directly on the multiplication table. In this table, it is also to be seenthat K, L and M generate the complete group. Indeed, combinations of K, L and M generate E,A and B, making up the entire group.

5.5.4 Example for Slater determinants

Take, for instance:ψa1(1)ψa2(2)ψa3(3) (5.113)

in which

ψa1(1) ≡ ψ(n1,l1,s1=12 ,ml1,ms1)

(~r1) (5.114)

ψa2(2) ≡ ψ(n2,l2,s2=12 ,ml2,ms2)

(~r2) (5.115)

ψa3(3) ≡ ψ(n3,l3,s3=12 ,ml3,ms3)

(~r3) (5.116)

The Slater determinant is an anti-symmetrical linear combination:

|a1a2a3〉 =1√3!

∑PP(P) · P[ψa1(1)ψa2(2)ψa3(3)] (5.117)

We will now prove that this form is equivalent to the following form:

|a1a2a3〉 =1√3!·

ψa1(1) ψa2(1) ψa3(1)ψa1(2) ψa2(2) ψa3(2)ψa1(3) ψa2(3) ψa3(3)

(5.118)

=1√3![ψa1(1)ψa2(2)ψa3(3)− ψa1(1)ψa3(2)ψa2(3) (5.119)

−ψa2(1)ψa1(2)ψa3(3) + ψa3(1)ψa1(2)ψa2(3) (5.120)

+ψa2(1)ψa3(2)ψa1(3)− ψa3(1)ψa2(2)ψa1(3)] (5.121)

This can readily be written in terms of the permutations as follows:

|a1a2a3〉 =1√3![+E− K−M + A + B− L](ψa1(1)ψa2(2)ψa3(3)) (5.122)

=1√3![P(E) · E + P(K) · K + P(M) ·M (5.123)

+P(A) · A + P(B) · B + P(L) · L](ψa1(1)ψa2(2)ψa3(3)) (5.124)

=1√3!

∑PP(P) · P[ψa1(1)ψa2(2)ψa3(3)] (5.125)

Remark.In the previous discussion, we used, for example

ψa2(1)ψa3(2)ψa1(3) = A[ψa1(1)ψa2(2)ψa3(3)] (5.126)

118


The operator A acts on the quantum numbers, not the electrons in the following way

A[ψa1(1)ψa2(2)ψa3(3)] = ψaA(1)(1)ψaA(2)(2)ψaA(3)(3) (5.127)

= ψa2(1)ψa3(2)ψa1(3) (5.128)

and analogous results for the other permutations.

ExerciseShow that, according to the multiplication table, AK = M and KA = L.

119

Chapter 6

Many electron atoms: Slaterdeterminants and matrix elements

6.1 Many-electron atoms

In the previous chapter, the wavefunction of the ground state of the He atom was described bythe product of two hydrogenic wavefunctions. By taking the repulsion between the electronsinto account, the description could be improved. However, since this contribution is not small,first order perturbation theory resulted in a overestimation of the electron-electron term. A wayto improve the description is by taking into account the expansion of the atom as a consequenceof the repulsion. This swelling can to some extent be included in the Hamiltonian by replacingthe actual nuclear charge, Ze, by a so called effective nuclear charge, Zeffe.The inter-electronic contribution for electron 1 is given by

V = ∑i 6=1

e2

4πε0r1i(6.1)

From classical electrostatics we know that if the i-th electron is located completely outside theelectronic cloud formed by the other electrons, then Gauss’ law yields∮

S(ri)

~E~eNdS =qincluded

ε0(6.2)

=(Z− N + 1)e

ε0(6.3)

so thatE4πr2

i =(Z− N + 1)e

ε0(6.4)

120

6.1. MANY-ELECTRON ATOMS

Figure 6.1: Classical electrostatics state that a spherically symmetric charge distribution can be replacedby a point charge at the center.

orE =

(Z− N + 1)e4πε0r2

i(6.5)

SinceF = −eE = −dV

dri(6.6)

one has

V(ri) =∫

(Z− N + 1)e2

4πε0r2i

dri = −(Z− N + 1)e2

4πε0ri+ c1 (6.7)

If the i-th electron is located entirely within the electronic cloud formed by the other electrons,Gauss’ law yields ∮

S(ri)

~E~eNdS =qincluded

ε0=

Zeε0

(6.8)

so thatE4πr2

i =Zeε0

(6.9)

orE =

Ze4πε0r2

i(6.10)

SinceF = −eE = −dV

dri(6.11)

one has

V(ri) =∫ Ze2

4πε0r2i

dri = −Ze2

4πε0ri+ c1 (6.12)

In general, one thus has

H′0 =N

∑i=1

[(− h2

2m~∇2 − Ze2

4πε0ri

)+ V(ri)

](6.13)

121


and the perturbation hamiltonian becomes

H′1 =

[N

∑i<j

e2

4πε0rij

]−[

N

∑i=1

V(ri)

](6.14)

in these equations Atkins and Friedman (see A&F eq. 7.44) take

V(ri) ≈σe2

4πε0ri(6.15)

where −σe is an effective charge that repels the charge -e of the i-th electron. This results ina reduction of the nuclear charge Ze to (Z − σ)e, in which σ is the nuclear screening constant(see Table 6.1). This decrease in nuclear charge is called shielding. The diagonalisation of H′1will be obtained a priori by choosing the Russell-Saunders coupling scheme, for which theeigenfunctions are

|αLSML MS〉 (6.16)

Our task is to find the unitary transformation of these eigenfunctions to the Slater-determinants

|α; l1; s1 = 1/2, ml1 , ms1 ; l2; s2 = 1/2, ml2 , ms2 ; . . . ; lN , sN = 1/2, mlN , msN 〉 (6.17)

in which α stands for all other possible quantum numbers and to compute matrix elementssuch as

〈αLSML MS| H′1 |αLSML MS〉 (6.18)

and〈αLSML MS| ˆH′SO |αLSML MS〉 (6.19)

in the perturbation sequence

H′0 H′1 H′SO =N

∑i=1

ξ(ri)~li ~si (6.20)

H He1s 1 1.6875

Li Be B C N O F Ne1s 2.6906 3.6848 4.6795 5.6727 6.6651 7.6579 8.6501 9.64212s 1.2792 1.9120 2.5762 3.2166 3.8474 4.4916 5.1276 5.75842p 2.4214 3.1358 3.8340 4.4532 5.1000 5.7584

Na Mg Al Si P S Cl Ar1s 10.6259 11.6089 12.5910 13.5754 14.5578 15.5409 16.5239 17.50752s 6.5714 7.3920 8.2136 9.0200 9.8250 10.6288 11.4304 12.23042p 6.8018 7.8258 8.9634 9.9450 10.9612 11.9770 12.9932 14.00823s 2.5074 3.3075 4.1172 4.9032 5.6418 6.3669 7.0683 7.75683p 4.0656 4.2852 4.8864 5.4819 6.1161 6.7641

Table 6.1: Values of Zeff = Z - σ for neutral ground-state atoms. (From [A & F], p233)

122


It is readily seen that the matrix elements of operators of type F and G have to be computed forwhich:

F =N

∑i=1

f (i) and G =N

∑i<j

g(i, j) (6.21)

e.g.

F =N

∑i=1

V(ri) or F =N

∑i=1

ξ(ri)~li ~si and G =N

∑i<j

e2

4πε0rij(6.22)

We now show how these matrix elements can be written in terms of one and two electronmatrix elements.

6.1.1 Evaluation of the matrix of the operator F = ∑Ni=1 f (i) within the manifold

spanned by Slater determinants

〈a1a2a3|3

∑i=1

f (i) |b1b2b3〉 (6.23)

|a1a2a3〉 =1√3!

∑PP(P) · P (Ψa1(1)Ψa2(2)Ψa3(3)) (6.24)

|b1b2b3〉 =1√3!

∑P′P(P′) · P′ (Ψb1(1)Ψb2(2)Ψb3(3)) (6.25)

〈a1a2a3|3

∑i=1

f (i) |b1b2b3〉 =13! ∑

P∑P′P(P)P(P′)×

∫Ψ∗aP(1)

(1) f (1)ΨbP′(1)(1)dτ1 ×

∫Ψ∗aP(2)

(2)ΨbP′(2)(2)dτ2 ×

∫Ψ∗aP(3)

(3)ΨbP′(3)(3)dτ3

+∫

Ψ∗aP(1)(1)ΨbP′(1)

(1)dτ1 ×∫

Ψ∗aP(2)(2) f (2)ΨbP′(2)

(2)dτ2 ×∫

Ψ∗aP(3)(3)ΨbP′(3)

(3)dτ3

+∫

Ψ∗aP(1)(1)ΨbP′(1)

(1)dτ1 ×∫

Ψ∗aP(2)(2)ΨbP′(2)

(2)dτ2 ×∫

Ψ∗aP(3)(3) f (3)ΨbP′(3)

(3)dτ3

(6.26)

Remarks:

1) There is no set ai = aj, ∀i, j ∈ 1, 2, 3; in general ∀i, j ∈ 1, 2, . . . N(i 6= j)

2) Take D = a1, a2, a3⋂b1, b2b3; if #D < 2 = 3− 1, in general #D < N − 1, the matrix

element 〈a1a2a3| ∑3i=1 f (i) |b1b2b3〉, will be zero due to the fact that at least one of the

integrals∫

Ψ∗aP(i)(i)ΨbP′(i)

(i)dτi in each of the three terms within square brackets in (6.26)is zero. The matrix element consequently will only be different from zero if:#D ≥ N − 1, in this case #D ≥ 2.

123


iii) The third factor is always zero but if

|aP(3)〉 = |bB·P”(3)〉 (6.35)

and the latter implies P(3) = P”(3)

P(3) = P”(3) 6= 2(6.36)

But since P(2) = P"(2) and P(3) = P"(3) one must also have P(1) = P"(1), so P = P". Furthermore,one must have P(1) = P"(1) = 2; in this case, the second and the third term in (6.30) must be zero;consequently the sum in (6.30) reduces to

13! ∑

P||P(1)=2P2(P)P(B)〈aP(1)| f |bB·P(1)〉 (6.37)

In general, there are (N-1)! permutations with one fixed element, P(i) = j; so in our case thereare two. In this example, these permutations are A and M. (6.37) can then further be simplifiedto

13! ∑

P||P(1)=2P(B)〈a2| f |b1〉 =

13!× 2!×P(B)× 〈a2| f |b1〉 (6.38)

Herein, one has〈a2| f |b1〉 =

∫Ψ∗a2

(~r1) f (~r1)Ψb1(~r1)d~r1 (6.39)

which does not depend on the integration variable~r1, (i.e. a dummy index). So (6.38) becomes

13×P(B)× 〈a2| f |b1〉 (6.40)

One may retrace the reasoning of the previous pages leading to (6.40) for the second term in(6.30), which yields

13! ∑

P||P(2)=2P(B)〈a2(2)| f (2) |b1(2)〉 =

13×P(B)× 〈a2| f |b1〉 (6.41)

and, in analogy, for the third term in (6.30), which is

13! ∑

P||P(3)=2P(B)〈a2(3)| f (3) |b1(3)〉 =

13×P(B)× 〈a2| f |b1〉 (6.42)

So the three terms in (6.30) yield the same result. Summing these, one finally obtains

〈a1a2a3|3

∑i=1

f (i) |b1b2b3〉 = P(B)× 〈a2| f |b1〉 (6.43)

This result may be generalised in an obvious way for N dimensions and in the case ak 6= bP0(k)

and aj = bP0(j) for all j = 1, 2, . . . , k− 1, k + 1, . . . , N, wherein P0 is the ordering permutation,

125


mapping equal quantum sets on equal ranks. So one has

〈a1a2 . . . aN |N

∑i=1

f (i) |b1b2 . . . bN〉 = P(P0)× 〈ak| f |bP0(k)〉 (6.44)

ii) case #D = 3Let us take

a1 = b3 = bL(1)

a2 = b2 = bL(2)

a3 = b1 = bL(3)

(6.45)

consequentlyaP(j) = bL·P(j) ∀j ∈ 1, 2, 3 ∀P ∈ G(P) (6.46)

We again examine the term within the brackets in (6.30). In this second example we have toreplace B by L. The first term consequently is

〈aP(1)| f |bL·P”(1)〉 × 〈aP(2)| bL·P”(2)〉 × 〈aP(3)| bL·P”(3)〉 (6.47)

i) The second factor is zero but ifP(2) = P”(2)

ii) The third factor is zero but if:P(3) = P”(3)

Since P(2) = P”(2) and P(3) = P”(3) one must also have P(1) = P”(1), so P = P”. In contradictionto the previous case, there is no limitation for the value of P(1) = P”(1). The first term in (6.30)here becomes

13! ∑

PP2(P)P(L)〈aP(1)| f |bL·P(1)〉 =

13! ∑

PP(L)〈aP(1)| f |aP(1)〉 (6.48)

=P(L)

3!· 2! ·

N=3

∑j=1〈aj| f |aj〉 (6.49)

=P(L)

3·

N=3

∑j=1〈aj(1)| ˆf (1) |aj(1)〉 (6.50)

(6.51)

For the second and third term in (6.30) one finds similar results, so (6.30) becomes

〈a1a2a3|3

∑i=1

f (i) |b1b2b3〉 =P(L)

3×

∑3j=1〈aj(1)| f (1) |aj(1)〉

+∑3j=1〈aj(2)| f (2) |aj(2)〉

+∑3j=1〈aj(3)| f (3) |aj(3)〉

(6.52)

126


In (6.52) the three terms within brackets are equal, they do not depend upon the integrationvariable, noted (i) between round brackets, so (6.52) reduces to

〈a1a2a3|3

∑i=1

f (i) |b1b2b3〉 = P(L)×3

∑j=1〈aj| f |aj〉 (6.53)

and it seems straightforward to generalise this to N dimensions as follows

〈a1a2 . . . aN |N

∑i=1

f (i) |b1b2 . . . bN〉 = P(P0)×N

∑j=1〈aj| f |aj〉 (6.54)

wherein P0 is the ordering permutation for which

aj = bP0(j) ∀j ∈ 1, 2, . . . , N (6.55)

Finally, for the unit permutation P0 = E, one obtains the following simplified result

〈a1a2 . . . aN |N

∑i=1

f (i) |a1a2 . . . aN〉 =N

∑j=1〈aj| f |aj〉 (6.56)

6.1.2 Evaluation of the matrix of the operator G = ∑Ni<j g(i, j) within the manifold

spanned by Slater determinants

In complete analogy to (6.52) and (6.44) one can prove after lengthy calculations

i. If A and B differ in more than two quantum sets

〈A| G |B〉 = 0 (6.57)

ii. If A and B differ in the sets (ak, al) and (bm, bn)

〈A| G |B〉 = ± [〈akal | g |bmbn〉 − 〈akal | g |bnbm〉] (6.58)

Wherein the ± sign corresponds to the parity of the permutation placing ak above bm andal above bn and the common quantum sets on the same rank.

iii. If A and B differ but in the sets ak and bl

〈A| G |B〉 = ±∑t 6=k

[〈akat| g |blat〉 − 〈akat| g |atbl〉] (6.59)

iv. For diagonal matrix elements one has

〈A| G |A〉 = ∑k<t

[〈akat| g |akat〉 − 〈akat| g |atak〉] (6.60)

127

6.2. PERIODICITY

Wherein the sum runs over all pairs (akat) in A in the same order as the quantum sets in A.

6.2 Periodicity

The ground state electronic configuration is often constructed with the use of the building-upor aufbau principle. This procedure reflects the structure of the periodic table and the Pauliexclusion principle. It gives rise to an order of energy levels as determined by the penetrationand shielding (see previous sections). However, with such a simple procedure it is not possibleto obtain the fully correct configuration as predicted by the periodic table. Thus, as we arenot interested in the sum of one-electron energies, but rather in the lowest total energy, it issometimes necessary to redistribute the electrons over the available orbitals. For example,relocating electrons so that a half-filled d5 or completely filled d10 configuration is obtained,can lower the total energy of the atom. Apart from this, in most cases the building up principlegives a fairly good result if one keeps in mind the following two remarks:

• electrons always tend to occupy separate orbitals before filling up half-occupied orbitalsin order to minimize their energy

• Hund’s rule of maximum multiplicity. Electrons first occupy separate orbitals withparallel spin.

Once the electronic configuration is determined, it is possible to extract several properties. Oneof these properties is the ionization energy I, defined as the energy needed to extract an electronfrom the atom at 0K (Eg → E+

g + e−g ). This energy increases when going from left to right inthe periodic table, as the increase in nuclear attraction dominates over the electron-electronrepulsion. Although one should keep in mind that this ionization energy depends on thetype of orbital from which the electron is removed, e.g. it requires more energy to removean electron from a half-filled orbital, which is reflected by the dip in ionization energy betweenN (1s22s22p1

x2p1y2p1

z) and O (1s22s22p2x2p1

y2p1z).

6.3 Slater atomic orbitals

Deriving a definitive analytical form for the atomic orbitals of many-electrons atoms is notpossible, but it is in any case useful to have a set of approximate atomic orbitals that can replacethe actual wavefunctions. These slater type orbitals (STOs) are constructed as follows:

• an orbital with quantum numbers n, l and ml and atomic number Zψnlml (r, θ, φ) = Nrne f f−1e−Ze f f ρ/ne f f Ylml (θ, φ)

• The effective principal quantum number, neff, is defined as followsn→ neff : 1→ 1 2→ 2 3→ 3 4→ 3.7 5→ 4.0 6→ 4.2

128

6.4. SELF-CONSISTENT FIELDS

Figure 6.2: The variation of the first ionization energy through Period 2 of the periodic table [A& F].

• Zeff is taken from Table 6.1.

In this table the values for the effective nuclear charge are obtained by fitting STOs tonumerically computed wavefunctions. However, one should keep in mind that these orbitalswith different value of n but similar l and ml values are not orthonormal and that ns-orbitalswith n>1 completely vanish at the nucleus.

6.4 Self-consistent fields

A numerical solution of the Schrödinger equation would lead to the most appropriate atomicorbitals. The original method was developed by Hartree and is referred to as the self-consistentfield method. By introducing the effects of electronic exchange, Fock improved this procedureand the obtained orbitals are called Hartree-Fock orbitals. The starting point is that each ofthe electrons moves in a spherically symmetric potential created by the nucleus and the otherelectrons. The Schrödinger equation is then solved for that specific electron, but with the chargeinside the sphere depending on the position of the electron (see figure 6.1). As this approachstarts from the assumption of known wavefunctions (so that the spherical averaged potentialcan be calculated), these wavefunctions must be approximated by STOs. After solving theSchrödinger equation for all electrons, one ends up with improved wavefunctions. An iterativeprocedure is then applied until the set of new wavefunctions does not differ significantlyanymore from the previous set. At this point the wavefunctions are called self-consistent andare a good approximation of the true many-electron wavefunctions. Let us now look at theseHartree-Fock equations. We will not go into detail on the full derivation of these equations, butwill focus on their interpretation. The Hamiltonian is defined as:

∑i

hi +12 ∑

i,j(i 6=j)

e2

4πε0rij(6.61)

129


in which hi is the hydrogenic Hamiltonian for electron i in the field of a nuclear charge Ze andis often called the core Hamiltonian. The Hartree-Fock equation for a space orbital ψs occupiedby a single electron is then:

h1 + ∑r(2Jr − Kr)

ψs(1) = εsψs(1) (6.62)

and the sum runs over all occupied wavefunctions and εs is the one-electron orbital energy.The Coulomb operator is then given by:

Jrψs(1) =∫

ψ∗r (2)(

e2

4πε0r12

)ψr(2)dτ2

ψs(1) (6.63)

and reflects the interaction between electron 1 and 2 in the orbital ψr. The exchange operator,Kr is defined as:

Krψs(1) =∫

ψ∗r (2)(

e2

4πε0r12

)ψs(2)dτ2

ψr(1) (6.64)

and takes into account the spin correlation. From equations 6.63 and 6.64 it is clear that allwavefunctions are needed to set up J and K. The orbital energies can then be obtained in thefollowing way:

εs =∫

ψ∗s (1)h1ψs(1)dτ1 + ∑r(2Jsr − Ksr) (6.65)

withJsr =

∫ψ∗s (1)Jrψs(1)dτ1

= e2

4πε0

∫ψ∗s (1)ψ∗r (2)

(1

r12

)ψr(2)ψs(1)dτ1dτ2

(6.66)

which is nothing else than the Coulomb integral as introduced in Chapter 5 for the heliumatom, and:

Ksr =∫

ψ∗s (1)Krψs(1)dτ1

= e2

4πε0

∫ψ∗s (1)ψ∗r (2)

(1

r12

)ψr(1)ψs(2)dτ1dτ2

(6.67)

being the exchange integral. The total energy is then, after eliminating the double counted e-einteractions:

E =2N

∑s

εs −N

∑r,s

(2Jrs − Krs) (6.68)

In the case of the Helium atom in the 1s2 ground-state configuration, we have then for theenergy of one electron:

ε1s = E1s + (2J1s,1s − K1s,1s)

= E1s + J1s,1s(6.69)

and the total energy becomes:

E = 2ε1s − (2J1s,1s − K1s,1s)

= 2 (E1s + J1s,1s)− J1s,1s

= 2E1s + J1s,1s

(6.70)

130


which is exactly the same energy as the one obtained in Chapter 5. The one electron energy εr

is the energy needed to remove an electron from the atom if we consider the other electrons tobe in a fixed distribution. As a consequence, this energy is equal to the ionization energy of anelectron in that orbital. This is called the Koopmans’ theorem:

Ir ≈ εr (6.71)

and the approximation comes from the "fixed distribution" assumption, as the remaining N-1electrons do change their distribution and Hartree-Fock orbital energies as compared to the Nelectron atom. In general, solutions of the Hartree-Fock equations are available and the totalelectron density can then easily be calculated by summing the squares of the wavefunctionsfor each electron. These values display the expected shell structure as can be seen in Figure6.3. Note that the total electron density shows the shell structure as a series of inflections:it decreases monotonically without intermediate maxima and minima. However, one shouldkeep in mind that the suggested Hartree-Fock orbitals are not the only possible nor the bestorbitals that can be constructed. As they are generated starting from the orbital approximation,they are rooted in an approximate central-field form of the potential. In contrast, the truewavefunctions depend on the separation of the electrons and not on their distance to thenucleus. Taking this separation into account forms the basis of the correlation problem, which isnowadays extensively investigated. Another way to improve the generated wavefunctions isby solving the Dirac equations instead of the non-relativistic Schrödinger equations. Especiallyfor heavy atoms, these relativistic effects are of high importance and cannot be neglected ifwe want to consider several properties, such as the colour of gold, lanthanide contraction, theinert-pair effect and even the liquid character of mercury.

Figure 6.3: A representation of the electron density calculated for a many-electron atom [A& F].

131

6.5. TERM SYMBOLS AND TRANSITIONS OF MANY-ELECTRON ATOMS

6.5 Term symbols and transitions of many-electron atoms

The state of a many-electron atom is described by the total spin, S, the total orbital angularmomentum, L, and the total angular momentum J. In the LS coupling or Russell-Saunderscoupling scheme, these three quantities are defined as follows:

S = s1 + s2, s1 + s2 − 1, ..., |s1 − s2| (6.72a)

L = l1 + l2, l1 + l2 − 1, ..., |l1 − l2| (6.72b)

J = L + S, L + S− 1, ..., |L− S| (6.72c)

For more than two electrons in the valence shell, the above series need to be applied severaltimes. The core electrons on the other hand should never be taken into account as their angularmomenta add up to zero.

Example: construct the term symbols for the (2p)1(3p)1 configurationFor this configuration we have that l1 = l2 = 1 resulting in L = 2, 1, 0 corresponding with aD, P and S term. Two electrons always result in S = 1, a triplet state, and S = 0, a singlet statefor the spin. Combining these two quantities in expression 6.72c results in the following termsymbols:

3D3,3 D2,3 D1,1 D2,3 P2,3 P1,3 P0,1 P1,3 S1,1 S0

One should pay attention when deriving term symbols for equivalent electrons. Consider forexample a (2p)2 configuration. In this case L = 2,1,0 and S = 1,0. However, due to the Pauliprinciple, the combination of L = 2 and S = 1, which would result in a 3D term, is excluded. Aneasy way to determine the possible term symbols for a specific configuration, is by drawing a

table of microstates. An electron occupying the ml orbital is then referred to as (+ml), for spin α

and (−ml) for spin β. A possible microstate for a (2p)2 configuration can for instance be (

+1,−1)

and should then be linked to the appropriate MSML cell.For each microstate, one should determine the corresponding L and S value. For the p2

configuration, the microstate (+1,−1) must belong to L = 2, S = 0, and thus a 1D term. As there

are five different states for L = 2 (ML = 2,1,0,-1,-2), we can strike out five different microstates

in the table. In the next row, the state (+1,

+0) should correspond with L = 1 and S = 1 and thus

a 3P term. After striking out the corresponding nine states (ML = 1,0,-1 and MS = 1,0,-1), only

one state is left. And this (+0,−0) microstate then obviously corresponds with L = 0 and S = 0,

and thus the 1P state.

Note that the symmetry in MS and ML is such that the same terms are contained in each ofthe four ±MS ±ML cells. As a consequence the positive part of the table contains enoughinformation about the configuration and we will leave out the strictly negative part from nowon.

132


ML ↓ MS → +1 0 −1

+2 (+1,−1)

1D

+1 (+1,

+0) (

+1,−0) , (

−1,

+0) (

−1,−0)

3P 3P, 1D 3P

0 (+1,-

+1) (

+1,-−1) , (-

−1,

+1), (

+0,−0) (

−1,-−1)

3P 3P, 1D, 1S 3P

−1 (-+1,

+0) (-

+1,−0) , (-

−1,

+0) (-

−1,−0)

3P 3P, 1D 3P

−2 (-+1, -−1)

1D

Example: d2 configurationIn a d2 configuration there are 10 possible spinorbitals for the two electrons. Taking intoaccount the Pauli exclusion principle, this results in C2

10 = 10.91.2 = 45 possible combinations

or microstates (Note that this holds only for electrons in the same shell. For two d-electrons ina different shell, we obtain 100 different microstates as they already differ in principal quantumnumber)

ms1ml1

ms2ml2 ML MS+2

−2 4 0

+2

+1 3 1

−2

+1 3 0

+2

−1 3 0

−2−1 3 -1

+1−1 2 0

+2

+0 2 1

−2

+0 2 0

+1

+0 1 1

−1

+0 1 0

+2

−0 2 0

−2−0 2 -1

+1−0 1 0

−1−0 1 -1

+0−0 0 0

+2 -

+1 1 1

−2 -

+1 1 0

+1 -

+1 0 1

−1 -

+1 0 0

+0 -

+1 -1 1

−0 -

+1 -1 0

+2 -

−1 1 0

−2 -−1 1 -1

+1 -−1 0 0

−1 -−1 0 -1

+0 -−1 -1 0

−0 -−1 -1 -1 -

+1 -−1 -2 0

+2 -

+2 0 1

−2 -

+2 0 0

+1 -

+2 -1 1

−1 -

+2 -1 0

+0 -

+2 -2 1

−0 -

+2 -2 0 -

+1 -

+2 -3 1 -

−1 -

+2 -3 0

+2 -

−2 0 0

−2 -−2 0 -1

+1 -−2 -1 0

−1 -−2 -1 -1

+0 -−2 -2 0

−0 -−2 -2 -1 -

+1 -−2 -3 0 -

−1 -−2 -3 -1 -

+2 -−2 -4 0

These 45 different microstates can then be put in the so called table of microstates. Limitingourself to the positive part of this table (ML ≥ 0, MS ≥ 0), we get

MS ↓ ML → 0 +1 +2 +3 +4

1 (+2-

+2),(

+1-

+1) (

+2-

+1), (

+1

+0) (

+2+0) (

+2+1)

3F, 3P 3F, 3P 3F 3F

0 (+2-−2),(−2-

+2),(

+1-−1), (

+2-−1),(−2-

+1),(

+1−0), (

+2−0) , (

−2+0), (

+1−1) (

+2−1),(−2+1) (

+2−2)

(−1-

+1), (

+0−0) (

−1+0) 3F, 1G,1D 3F, 1G 1G

3F,3P, 1G,1D,1S 3F,3P, 1G,1D

133


Example: p3 configurationIn a p3 configuration there are 6 spinorbitals for 3 electrons, resulting in C3

6 = 6.5.41.2.3 = 20 different

microstates

ms1ml1

ms2ml2

ms3ml3 ML MS+1

−1

+0 2 1

2

−1

+0−0 1 - 1

2

+0−0 -

+1 -1 1

2

−0 -

+1 -−1 -2 - 1

2+1

−1

−0 2 - 1

2

−1

+0 -

+1 0 1

2

+0−0 -−1 -1 - 1

2+1

−1 -

+1 1 1

2

−1

+0 -−1 0 - 1

2

+0 -

+1 -−1 -2 1

2+1

−1 -

−1 1 - 1

2

−1−0 -

+1 0 - 1

2+1

+0

−0 1 1

2

−1−0 -−1 0 - 3

2+1

+0 -

+1 0 3

2

−1 -

+1 -−1 -1 - 1

2+1

+0 -

−1 0 1

2+1

−0 -

+1 0 1

2+1

−0 -

−1 0 - 1

2+1 -

+1 -

−1 -1 1

2

And the table of microstates is then given by

MS ↓ ML → 0 +1 +2

32 (

+1+0-

+1)

4S12 (

+1+0-−1), (

+1−0-

+1), (

−1+0-

+1) (

+1−1-

+1), (

+1+0−0) (

+1−1+0)

4S,2P, 2D 2P, 2D 2D

6.5.1 The diagonal sum rule

Evaluation of the energies of the terms

1) (np)2 configurationIf we look at the configuration table we have that

E(3P) = DM(+1+0) = 〈

+1+0 | H’

1 |+1+0〉 (6.73)

with

H’1 =

N

∑i<j

e2

4πε0rij−

N

∑i=1

σe2

4πε0ri(6.74)

134


one further has

E(1D) = DM(+1−1) (6.75)

E(3P) + E(1D) + E(1S) = DM(+1-−1) + DM(

−1-

+1)+ DM(

+0−0) (6.76)

2) (nd)2 configuration

E(3F) = DM(+2+1)

E(3F) + E(3P) = DM(+2-

+1) + DM(

+1+0)

E(1G) = DM(+2−2)

E(3F) + E(1D) + E(1G) = DM(+2−0) + DM(

−2+0) + DM(

+1−1)

E(3F) + E(3P) + E(1G)+ E(1S) + E(1D) = DM(+2-−2) + DM(

−2-

+2) + DM(

+1-−1) + DM(-

−1-

+1) + DM(

+0−0)

3) (np)3 configuration

E(4S) = DM(+1+0 -

+1)

E(2D) = DM(+1−1

+0)

E(2D) + E(2P) = DM(+1−1-

+1) + DM(

−1+0−0)

Making use of the reduction rule for

DM(+1−1 -

+1) = 〈

+1−1 -

+1| H’

1 |+1−1 -

+1〉

with

〈A| G |A〉 =N

∑i<j

[〈aiaj| g |aiaj〉 − 〈aiaj| g |ajai〉

]and using equation (5.14) and (5.33) we find

E(4S) = J(p1, p0)− K(p1, p0)

+ J(p1, p−1)− K(p1, p−1)

+ J(p0, p−1)− K(p0, p−1)

E(2D) = J(p1, p1)

+ J(p1, p0)− K(p1, p0)

+ J(p1, p0)

135


E(2D) + E(2P) = J(p1, p1) + 2J(p1, p−1)− K(p1, p−1)

+ 2J(p1, p0)− K(p1, p0) + J(p0, p0)

These Coulomb and Exchange integrals can then be evaluated according to the methodexplained in Example (7.3), page 2211, solving the problem of evaluating the energies ofthe terms on a quantitative basis. We now generalise this in the following pages.

Evaluation of the direct and exchange integrals

The two-electron integrals to be evaluated are

〈ab| e2

4πε0r12|cd〉 =

∫∫Ψ*

a(1)Ψ*b(2)

e2

4πε0r12Ψ*

c(1)Ψ*d(2)dτ1dτ2 (6.77)

in whichΨa(i) = Rna la(ri)Ylama

l(θi, ϕi)χma

s (σi) (6.78)

and (equation (5.21))1

r12= 4π

∞

∑k=0

12k + 1

+k

∑m=−k

Ykm(1)Y∗km(2)rk<

rk+1>

(6.79)

Substitution yields

〈ab| e2

4πε0r12|cd〉 =4π

∞

∑k=0

12k + 1

Rk(nala, nblb; nclc, ndld)×

+k

∑m=−k

π∫θ=0

2π∫ϕ=0

Y*lama

l(1)Y*

lbmbl(2)Ykm(1)Y*

km(2)Ylcmcl(1)Yldmd

l(2) sin θ1 sin θ2dθ1dθ2dϕ1dϕ2

×[χma

s (1), χmcs(1)

]×[χmb

s(2), χmd

s(2)]

(6.80)

in which we used

Rk(nala, nblb; nclc, ndld) =e2

4πε0

∫ ∞

0

∫ ∞

0Rna la(1)Rnb lb(2)

rk<

rk+1>

Rnc lc(1)Rnd ld(2)r21r2

2dr1dr2 (6.81)

We first examine the factors containing θ and ϕ. These reduce to four factors∫ π

0Θlama

l(θ1)Θkm(θ1)Θlcmc

l(θ1) sin θ1dθ1 ×

∫ π

0Θlbmb

l(θ2)Θkm(θ2)Θldmd

l(θ2) sin θ2dθ2

× 1(√2π)3

∫ 2π

0ei(−ma

l +m+mcl )ϕ1dϕ1 ×

1(√2π)3

∫ 2π

0ei(−mb

l−m+mdl )ϕ2dϕ2

(6.82)

1A&F

136


The latter two are equal to (2π)−1/2 when−mal + m + mc

l = 0

−mbl −m + md

l = 0(6.83)

and zero if this is not fulfilled, so

m = mal −mc

l = mdl −mb

l

and:

mal + mb

l = mcl + md

l

(6.84)

Defining the coefficients ck as follows

ck(lml , l′m′l) =√

22k + 1

∫ π

0Θk,ml−m′l

(θ)Θl,ml (θ)Θl′,m′l(θ) sin θdθ (6.85)

we have that

〈ab| e2

4πε0r12|cd〉 = δma

s mcs δmb

s mdsδma

l +mbl ,mc

l +mdl

×∞

∑k=0

Rk(nala, nblb; nclc, ndld)ck(lamal , lcmc

l )ck(ldmd

l , lbmbl )

(6.86)

The ck are infinite numbers and were evaluated by Gaunt (see page 138). Taking then c = aand d = b in (6.86), one obtains the Coulomb - integral

J(a, b) = 〈ab| e2

4πε0r12|ab〉

= ∑k

Fk(nala, nblb)ak(lamal , lbmb

l )(6.87)

and with c = b and d = a we get the exchange integral

K(a, b) = 〈ab| e2

4πε0r12|ba〉

= δmas mb

s ∑k

Gk(na, la, nblb)bk(lamal , lbmb

l )(6.88)

137


138


in which we defined Fk(nala, nblb) = Rk(nala, nblb; nala, nblb)

Gk(nala, nblb) = Rk(nala, nblb; nblb, nala)(6.89a)

ak(kamal , lbmb

l ) = ck(lamal , lama

l )ck(lbmb

l , lbmbl )

bk(kamal , lbmb

l ) =[ck(lama

l , lbmbl

]2 (6.89b)

The coefficients Fk and Gk can be calculated if the radial functions Rnl(r) are known explicitly. Ifnot, they are treated as parameters. The ak and bk are tabulated (see Appendix). For equivalentelectrons we have

Fk(nl , nl) = Gk(nl , nl) (6.90)

From the ak, bk tables it appears that these coefficients have a common denominator which weabsorb in Fk, Gk by the notation-conventionFk = Fk

common denom ak

Gk = Gk

common denom bk

(6.91)

Finally, one should be aware that for ak the sign of m and m’ may be permuted. This followsfrom definition and

Θl,−m = (−1)mΘl,m (6.92)

Whereas for bk only the upper or the lower signs may be combined.As an example, we apply this on the first-order perturbaton for the (np)2 configuration

139


Example: (np)2 configurationCalculation of the energies of the terms

E(1D) = DM(+1−1) = J(p1, p1) (6.93)

= 1 · F0 +1

25F2 = F0 + F2 (6.94)

E(3P) = DM(+1+0) = J(p1, p0)− K(p1, p0) (6.95)

= F0 − 2F2 − 3G2 = F0 − 5F2 (6.96)

in which we used that for equivalent electrons F2 = G2.

E(1S) = DM(+1 -−1) + DM(

−1 -

+1) + DM(

+0−0)−

[E(3P) + E(1D)

](6.97)

= J(p1, p−1) + J(p1, p−1 + J(p0, p0)− [(F0 + F2) + (F0 − 5F2)] (6.98)

= 2(F0 + F2) + (F0 + 4F2)− [2F0 − 4F2] (6.99)

= F0 + 10F2 (6.100)

Even without evaluating the radial integrals, this leads to a useful check

E(1S)− E(1D)

E(1D)− E(3P)=

9F2

6F2= 1.50 (6.101)

and typical experimental results are

CI (2p)2 1.13NII (2p)2 1.14

(2p)(3p) configuration interaction;see fig 7.17 page 240 [A & F]OIII (2p)2 1.14

SiI (3p)2 1.48FeI (4p)2 1.50SnI (5p)2 1.39LaII (6p)2 18.43

LS-coupling breaks downPbI (6p)2 0.62

140


Example: (np)3 configuration

Calculation of the energies of the terms

E(4S) = DM(+1+0 -

+1)

= J(p1, p0)− K(p1, p0)

+ J(p1, p−1)− K(p1, p−1)

+ J(p0, p−1)− K(p0, p−1)

=

[(F0 − 2

25F2)−(

325

G2)]

+

[(F0 +

125

F2)−(

625

G2)]

+

[(F0 − 2

25F2)−(

325

G2)]

= (3F0 − 3F2)− 12G2

= 3F0 − 15F2

(6.102)

E(2D) = DM(+1−1+0)

=J(p1, p1)

+ J(p1, p0)− K(p1, p0)

+ J(p1, p0)

=

[(F0 +

125

F2)]

+

[(F0 − 2

25F2)−(

325

G2)]

+

[(F0 − 2

25F2)]

= (3F0 − 3F2)− 3G2

= 3F0 − 6F2

(6.103)

141


Similar calculations for E(2P) lead to

E(2P) = 3F0 (6.104)

Combining the above expressions for the term-energies leads to

E(2P)− E(2D)

E(2D)− E(4S)=

6F2

8F2= 0.75 (6.105)

and typical experimental results are

NI (2p)3 0.667OII (2p)3 0.500

(2p)(3p) configuration interaction;see fig 7.17 page 240 [A & F]SII (3p)3 0.509

AsI (4p)3 0.715SbI (5p)3 0.908SnI (5p)2 1.39

LS-coupling breaks downBiI (6p)3 1.12

6.5.1.1 Evaluation of the multiplet splittings

In the Russell-Saunders case, one has the following sequence of perturbations

H′0 H′1 HSO (6.106)

with

H′0 =N

∑i=1

− h2

2me~∇2

i −Ze2

4πε0ri+ V(ri)

(6.107a)

H′1 =N

∑i<j

e2

4πε0rij−

N

∑i=1

V(ri) (6.107b)

HSO =N

∑i=1

ξ(ri)~li ·~si = ξ(LS)~L · ~S (6.107c)

Applying then the diagonal sum rule, we relate ζ(LS), the spin-orbit coupling constant, to theone-electron ζnl .

142


Example: (np)2 configuration

∑terms in ML, MS cell

〈LMLSMS| ξ(LS)~L · ~S |LMLSMS〉 =r

∑k=1〈Ak|

N

∑i=1

ξ(ri)~li ·~si |Ak〉 (6.108)

in which the |LMLSMS〉 are the term eigenfunctions and the |Ak〉 the Slater determinants.

• ML = 2, MS = 0 cell: no splitting

• ML = 1, MS = 1 cell:

1 · 1 · ζ(3P) = 1 · 12· ζnp + 0 · 1

2ζnp (6.109)

=⇒ ζ(3P) =12

ζnp (6.110)

with

ζnl =α2RZ4

e f f

n3l(l + 12 )(l + 1)

(6.111)

143

6.6. HUND’S RULES AND THE RELATIVE ENERGIES OF TERMS

6.6 Hund’s rules and the relative energies of terms

In order to identify the lowest energy term for a specific configuration, Friedrich Hundproposed the following three rules:

1. The term with maximum multiplicity lies lowest in energy.This rule is a consequence of the spin correlation: due to the Fermi hole, electronswith parallel spin can contract towards the nucleus without excessive electron-electronrepulsion.

2. For a given multiplicity, the term with the highest value of L lies lowest in energy.From a classical point of view, this rule can be understand as follows: a high value of Lmeans that the electrons are orbiting in the same direction, so they will be less often ineach others proximity, leading to a lower average repulsion.

3. For atoms with less than half-filled shells, the level with the lowest value of J lies lowestin energy, and for more than half-filled shells, the highest J lies lowest in energy.

One should keep in mind that the above rules can be used to predict the lowest energy term,but not necessarily to order all the terms, because the structure of an atom is in most casesmore accurately described by a configuration interaction, and not by a single configuration. Ifwe consider for instance the 3s13d1 excited state configuration of magnesium, we expect the 3Dstate to be lower in energy than the 1D state, but experimentally the opposite is found. Thiscan be explained by the fact that the 1D term is in fact a combination of 75% 1D from the 3s13d1

configuration and 25% 1D from the 3p2 configuration. Suppose that these two have a similarenergy, the electron-electron repulsion removes the degeneracy and the lowest term may endup below the 3D term, because this level remains unchanged as their is no 3p2 3D term.

Figure 6.4: Configuration interaction between the 3s13d1 and 3p2 configuration of Mg, resulting in alowering of the 1D term.(Fig. 7.17 from [A&F])

144

6.7. ALTERNATIVE COUPLING SCHEMES

6.7 Alternative coupling schemes

In the Russell-Saunders scheme, a term symbol consists of a defined value for L and S. Howeverthis holds no longer if the spin-orbit coupling starts to dominate. Within the vector model, thiscan be interpreted in the following way. In the Russell-Saunders scheme, the individual orbitaland spin momenta precess rapidly around their resultant, but slowly, due to the weak spin-orbit coupling, around the combination of both, being J.When the spin-orbit coupling is strong, which is in general the case for heavy atoms, theindividual li and si precess fast around their resultant ji, but these ji’s precess slowly aroundtheir resultant J. In other words we can say that the individual orbital and spin momentumcouple first to a combined angular momentum ji, before adding up to a total angularmomentum J. This scheme is called jj-coupling.

Figure 6.5: Vector model for both the LS coupling scheme (left) and the jj coupling scheme (right) (from[A&F] p 240-241).

6.7.1 The jj coupling scheme

Recall that

H′0 =N

∑i=1

− h2

2me~∇2

i −Ze2

4πε0ri+ V(ri)

(6.112a)

H′1 =N

∑i<j

e2

4πε0rij−

N

∑i=1

V(ri) (6.112b)

HSO =N

∑i=1

ξ(ri)~li ~si = ξ(LS)~L ~S (6.112c)

→ H′0 H′1 HSO : LS coupling→ H′0 HSO H′1 : jj coupling

145


We will now show that the matrix of HSO is diagonal in the basis |a1a2 . . . aN〉, with ai =

(ni, li, si = 1/2, ji, mji).

Remark that ξ(ri)~li ~si is not diagonal in the basis |ni, li, si = 1/2, mli , msi〉, so this does notprovide an adequate representation for the the first-order perturbation scheme.HSO is an operator of the type F = ∑N

i fi ; hence it is sufficient to check whether

〈n l s = 1/2 j mj| ξ(LS)~l ~s |n’ l’ s’=1/2 j’ mj’〉 (6.113)

is diagonal. In the following we leave out the redundant s, s’. Since

~j2 = (~l +~s)2 = ~l2 +~s2 + 2~l ·~s (6.114)

we have that~l ·~s = 1

2(~j2 −~l2 −~s2) (6.115)

so that~l ·~s |n’ l’ j’ m’j〉 =

12(~j2 −~l2 −~s2) |n’ l’ j’ mj’〉 (6.116)

In the first-order perturbation scheme, all matrix elements pertain to the same configuration,so n = n’ and l = l’. We evaluate

〈n l j mj| ξ(r)~l ~s |n l j’ mj’〉 = ∑()”〈n l j mj| ξ(r) |n" l" j" mj"〉〈n" l" j" mj"|~l ~s |n l j’ m’j〉

= 〈n l j mj| ξ(r) |n l j’ m’j〉〈n l j’ m’j|~l ~s |n l j’ m’j〉 (6.117)

Since [ξ(r), lk

]= 0 (6.118)

and[ξ(r), sk] = 0 (6.119)

for k = x, y, z, +, - and hence [ξ(r), jk

]= 0 (6.120)

We have that [ξ(r),~j2

]= 0 (6.121)

and the matrix of ξ(r) within the basis |α j mj〉 is diagonal and independent on mj. We thendefine

〈n l j mj| ξ(r) |n l j’ mj’〉 = 〈〈n l j| ξ(r)|n l j〉〉δj j’δmj mj’

= 〈〈n l j| ξ(r)|n l j〉〉 (6.122)

146


(6.117) then becomes

〈n l j mj| ξ(r)~l ~s |n l j’ mj’〉 = 〈〈n l j| ξ(r)|n l j〉〉 · 〈n l j mj|~l ~s |n l j mj〉 × δj j’ × δmj mj’

=12〈〈n l j|ξ(r)|n l j〉〉

[j(j + 1)− l(l + 1)− 3

4

]×δj j’ × δmj mj’ (6.123)

We define 〈〈n l j| ξ(r)|n l j〉〉 = ζnlj. On the basis of the commutation relations (6.118) and(6.119), the matrix of ξ(r) is not only independent upon mj but also on ml and ms, and thus onj. Hence one may write ζnlj = ζnl . Equation (6.117) finally yields

〈n l j mj| ξ(r)~l ~s |n l j’ mj’〉 = ζnl ·12

[j(j + 1)− l(l + 1)− 3

4

]δj j’δmj mj’ (6.124)

147


Example 1: (npn’p configuration)For each electron j = 3

2 or j = 12 , which yields 4 possible combinations (and their four

respective energy levels). We again list the nonnegative MJ values1. j = 3

2 and j′ = 32 (16 wavefunctions)

MJ = mj + m′j Slater determinants J

3 ( 32 , 3

2 ) 32 ( 3

2 , 12 ), ( 1

2 , 32 ) 3, 2

1 ( 32 ,− 1

2 ), ( 12 , 1

2 ), (− 12 , 3

2 ) 3, 2, 10 ( 3

2 ,− 32 ), ( 1

2 ,− 12 ),(− 1

2 , 12 ), (− 3

2 , 32 ) 3, 2, 1, 0

With a slaterdeterminant defined as

(mj, m′j) =12

Ψnp 32 mj

(1) Ψn′p 32 m′j

(1)

Ψnp 32 mj

(2) Ψn′p 32 m′j

(2)

(6.125)

2. j = 32 and j′ = 1

2 (8 wavefunctions)


2 ( 32 , 1

2 ) 21 ( 3

2 ,− 12 ), ( 1

2 , 12 ) 2, 1

0 (− 12 , 1

2 ), ( 12 ,− 1

2 ) 2, 1

3. j = 12 and j′ = 3

2 (8 wavefunctions)


2 ( 12 , 3

2 ) 21 ( 1

2 , 12 ), (− 1

2 , 32 ) 2, 1

0 ( 12 ,− 1

2 ), (− 12 , 1

2 ) 2, 1

4. j = 12 and j′ = 1

2 (4 wavefunctions)


1 ( 12 , 1

2 ) 10 ( 1

2 ,− 12 ), (− 1

2 , 12 ) 1, 0

We now evaluate the first-order energy splittings induced by the spin-orbit coupling

〈A| HSO |A〉 =N

∑i=1〈ni li ji mji | ζ(ri)~li ~si |ni li ji mji〉 (6.126)

148


For table 1, we have that

(6.126) = ζnp ·12

[32· 5

2− 1 · 2− 3

4

]+ ζn′p ·

12·[

32· 5

2− 1 · 2− 3

4

]=

12(ζnp + ζn′p

)(6.127)

For table 2:

〈A| HSO |A〉 = ζnp ·12

[32· 5

2− 1 · 2− 3

4

]+ ζn′p ·

12·[

12· 3

2− 1 · 2− 3

4

]=

12

ζnp − ζn′p

For table 3:

= ζnp ·12

[12· 3

2− 1 · 2− 3

4

]+ ζn′p ·

12·[

32· 5

2− 1 · 2− 3

4

]= −ζnp +

12

ζn′p

For table 4:

= ζnp ·12

[12· 3

2− 1 · 2− 3

4

]+ ζn′p ·

12·[

12· 3

2− 1 · 2− 3

4

]= −ζnp − ζn′p

149


Example 2: (np2 configuration) (15 states)1. j = 3

2 and j′ = 32 (6 wavefunctions)


2 ( 32 , 1

2 ) 21 ( 3

2 ,− 12 ) 2

0 ( 32 ,− 3

2 ), ( 12 ,− 1

2 ) 2, 0

2. j = 32 and j′ = 1

2 (8 wavefunctions)


2 ( 32 , 1

2 ) 21 ( 3

2 ,− 12 ), ( 1

2 , 12 ) 2, 1

0 (− 12 , 1

2 ), ( 12 ,− 1

2 ) 2, 1

3. j = 12 and j′ = 1

2 (1 wavefunction)


0 ( 12 ,− 1

2 ) 0

For table 1:

(6.126) = ζnp

For table 2:

(6.126) = −12

ζnp

For table 3:

(6.126) = −2ζnp

6.7.1.1 Coulomb repulsion in jj coupling

Recall that H′0 HSO H′1 in jj coupling. We start with wavefunctions A[∏N

i=1 |ni, li, si = 1/2, ji, mji〉]

in which A is the antisymmetrisation-operator turning the ket-products into slater determi-nants satisfying the Pauli principle. Since, for ~J = ~j1 + ~j2 in the example npn’p and np2, onehas [

Jk,1

r12

]= 0 , k = x, y, z,+,−

it will be useful to transform to a representation |n1l1s1 j1; n2l2s2 j2; JMJ〉 diagonalizing theperturbation Hamiltonian. Furthermore, the matrix of H′1 will be independent upon MJ. Thesplitting energies can then be derived by the diagonal sum rule. For n p 3

2 , n’ p 32 and J = 3, the

150


configuration table yields

E(1)(J = 3) = 〈32

32| H′1 |

32

32〉 = T(np

32

32

; n′p32

32) (6.128)

in which

T(a; b) =∫∫

Ψ*a(1)Ψ

*b(2)

e2

4πε0r12Ψa(1)Ψb(2)dτ1dτ2−

∫∫Ψ*

a(1)Ψ*b(2)

e2

4πε0r12Ψb(1)Ψa(2)dτ1dτ2

(6.129)containing a direct and an exchange integral. To compute these integrals using the appendices,we use the relation

Ψn,l,j= 32 ,mj=

32= Ψn,l,ml=1,ms=

12

(6.130a)

or|n, l, j =

32

, mj =32〉 = |n, l, ml = 1, ms =

12〉 (6.130b)

so:

T(np32

32

; n′p32

32) =

∫∫Ψ∗np1, 1

2(1)Ψ∗n′p1, 1

2(2)

e2

4πε0r12Ψnp1, 1

2(1)Ψn′p1, 1

2(2)dτ1dτ2−∫∫

Ψ∗np1, 12(1)Ψ∗n′p1, 1

2(2)

e2

4πε0r12Ψn′p1, 1

2(1)Ψnp1, 1

2(2)dτ1dτ2 (6.131a)

or using the ak and bk values

T(np32

32

; n′p32

32) = J(np1,

12

; n′p1,12)− K(np1,

12

; n′p1,12) (6.131b)

= F0 + F2 − G0 − G2 (6.131c)

The next cell in the configuration table yields

E(1)(J = 3) + E(1)(J = 2) = 〈32

12| H′1 |

32

12〉+ 〈1

232| H′1 |

12

32〉

= T(np32

32

; n′p32

12) + T(np

32

12

; n′p32

32) (6.132)

To reduce this problem, we first evaluate the Clebsch Gordan coefficients for

〈nlsmlms| ←→ |nlsjmj〉 (6.133)

Starting from

|32

,32〉 = |1;

12〉 (6.134)

151


and applying the ladder operator j− = l− + s−, we find

j− |32

,32〉 = (l− + s−) |1;

12〉√

32(

32+ 1)− 3

2(

32− 1) |3

2,

12〉 =

√1(1 + 1)− 1(1− 1) |0;

12〉+

√12(

12+ 1)− 1

2(

12− 1) |1;−1

2〉

√3 |3

2,

12〉 =√

2 |0;12〉+ |1;−1

2〉

so we have that

|32

,12〉 =

√23|0;

12〉+

√13|1;−1

2〉 (6.135)

When applying j− again we find

j− |32

,12〉 = (l− + s−)

[√23|0;

12〉+

√13|1;−1

2〉]

√32(

32+ 1)− 1

2(

12− 1) |3

2,−1

2〉 =

√23

[√1(1 + 1)− 0 | − 1;

12〉+

√12(

12+ 1)− 1

2(

12− 1) |0;−1

2〉]

+

√13

[√1(1 + 1)− 1(1− 1) |0;−1

2〉]

so we have that

|32

,−12〉 =

√13| − 1;

12〉+

√23|0;−1

2〉 (6.136)

Again applying j−

j− |32

,−12〉 = (l− + s−)

[√13| − 1;

12〉+

√23|0;−1

2〉]

√3 |3

2,−1

2〉 =√

3 | − 1;−12〉

and thus finally

|32

,−32〉 = | − 1;−1

2〉 (6.137)

152


Using now 〈 32 , 1

2 |12 , 1

2 〉 = 0 we find that

|12

,12〉 = α |0;

12〉+ β |1;−1

2〉

⇒√

23· α +

√13· β = 0

⇒ α ∼√

13

and β ∼ −√

23

(6.138)

Combined with the normalisation condition

α2 + β2 = 1 (6.139)

we obtain

α =

√13

and β = −√

23

(6.140)

and thus

|12

,12〉 =

√13|0;

12〉 −

√23|1;−1

2〉 (6.141)

Again applying the ladder operators

j− |12

,12〉 = (l− + s−)

[√13|0;

12〉 −

√23|1;−1

2〉]

leads to

|12

,−12〉 =

√23| − 1;

12〉 −

√13|0;−1

2〉 (6.142)

Recalling equation (6.132)

E(1)(J = 3) + E(1)(J = 2) = T(np32

32

; n′p32

12) + T(np

32

12

; n′p32

32) (6.143)

and evaluating the first term

T(np32

32

; n′p32

12) =

∫∫Ψ∗np 3

232(1)Ψ∗n′p 3

212(2)

e2

4πε0r12Ψnp 3

232(1)Ψn′p 3

212(2)dτ1dτ2

−∫∫

Ψ∗np 32

32(1)Ψ∗n′p 3

212(2)

e2

4πε0r12Ψn′p 3

212(1)Ψnp 3

232(2)dτ1dτ2

(6.144)

153


With the obtained Clebsch Gordan coefficients we have that

Ψnp 32

32

= |1; 12 〉

Ψ∗np 3

232

= 〈1; 12 |

Ψ∗n′p 3

212

=√

23 〈0; 1

2 | +√

13 〈1;− 1

2 |

Ψn′p 32

12

=√

23 |0; 1

2 〉+√

13 |1;− 1

2 〉

(6.145)

Combining equations (6.144) and (6.145) we find that

∫∫Ψ∗np 3

232(1)Ψ∗n′p 3

212(2)

e2

4πε0r12Ψnp 3

232(1)Ψn′p 3

212(2)dτ1dτ2 =

23

J(np1, n′p0) +13

J(np1, n′p1)

(6.146)and

∫∫Ψ∗np 3

232(1)Ψ∗n′p 3

212(2)

e2

4πε0r12Ψn′p 3

212(1)Ψnp 3

232(2)dτ1dτ2 =

[√23

]2

K(np1, n′p0) (6.147)

and thus

T(np32

32

; n′p32

12) =

23

J(np1, n′p0) +13

J(np1, n′p1)−23

K(np1, n′p0)

=23(F0 − 2

25F2) +

13(F0 +

125

F2)− 23(

325

G2)

=23(F0 − 2F2) +

13(F0 + F2)−

23(3G2)

T(np32

32

; n′p32

12) = F0 − F2 − 2G2 (6.148)

The other T term in equation (6.132) yields the same result. Consequently one gets

E(1)(J = 2) = 2(F0 − F2 − 2G2)− E(1)(J = 3)

= F0 − 3F2 + G0 − 3G2 (6.149)

After further calculation, one finally obtains the results listed in the following table

npn’p J E(1)

j = 32 j′ = 3

2 3 F0 + F2 − G0 − G2

2 F0 − 3F2 + G0 − 3G2

1 F0 + F2 − G0 − G2

0 F0 + 5F2 + G0 + 5G2

j = 32 j′ = 1

2 2 F0 − G2

j = 12 j′ = 3

2 1 F0 − 5G2

j = 12 j′ = 1

2 1 F0 − G0

0 F0 + G0

154


For the np2 configuration similar results are found, which are listed in the following table andcompared to the LS-coupling results

np2

J J jj'

1S 10

1D 1

0

2

3P -5

0

0

1/2

1/2

-1

2S+1LJ

2

1

0

0

2

21

0

5

-3

-1

-5

-1/2

-2

1 3/2 3/2

3/2 1/2

1/2 1/2

(A) (B) (D) (C)LS coupling JJ coupling

with the correlation - diagram2

(A): no spin-orbit coupling (only H′1)

(B): weak spin-orbit coupling

(C): no Coulomb repulsion

(D): weak Coulomb repulsion

2[A&F] fig 7.20, p244

155

Chapter 7

Molecular Physics

7.1 Introduction

In this chapter, an introduction is given to the quantum mechanical basis of molecular bondformation. A complete understanding of the electronic structure of polyatomic moleculesand their vibrational and rotational substructure is beyond the scope of this course. In thisintroduction, we will confine ourselves to a qualitative description of diatomic molecules, withthe homonuclear diatomic molecular ion H+

2 as example. A qualitative discription of the exactsolution will be given for this ion, and it will be used as generic example for the MolecularOrbital (MO) theory, based on Linear Combinations of Atomic Orbitals (LCAO). In the final sectionof this chapter, the extension of LCAO-MO theory to polyatomic molecules will briefly bediscussed, based on a group theoretical representation of molecular orbitals (see Chapter 8).LCAO-MO also forms the basis for periodic calculations of electronic band structure in solids(see Vastestoffysica and Vastestof- en nanofysica).

The central approximation for calulating molecular energy levels is the Born-Oppenheimerapproximation (see Section 7.2). It is based on the great difference in mass between electronsand nuclei. As a result, electrons can respond almost instantaneously to displacements of thenuclei. This allows to separate the complex many-body Schrodinger equation of the moleculeinto an equation for the nuclear wavefunction, and one for the electronic wavefunction thatdepends on the nuclear coordinates. For each position of the nuclear coordinates, the electronicwavefunction, electronic energy and total energy of the molecule can thus be calculated. Thisprocedure allows to calculate a potential energy curve as a function of interatomic distance fordiatomic molecules, as shown in Figure 7.1. For polyatomic molecules, a potential energy surfacecan be calculated, which depends on several relative nuclear coordinates, labeled as bonddistances and bond angles. A necessary condition for molecular bonding is that a minimumoccurs (at least one) in this potential energy surface. This is the case in Figure 7.1a: in its

156

7.1. INTRODUCTION

equilibrium geometry the nuclei in this diatomic molecule will be at a distance R0 . In Figure7.1b no minimum is found as a function of the nuclear coordinate. The equilibrium interatomicdistance is ∞ in this case: no molecular bonding occurs.

Figure 7.1: Total electronic and nuclear molecular energy as a function of the interatomic distance ofa diatomic molecule. (a) A minimum occurs in the energy curve : molecular bonding occurs at theequilibrium interatomic distance R0. (b) No minimum occurs in the energy curve: no mulecular bondingoccurs.

Even in the Born-Oppenheimer approximation, only for the simplest molecular structure, theH+

2 ion, an analytical solution can be found. For all other cases, numerical modeling based onvariational techniques is required. Computational molecular modeling is a very active field ofcontemporary research with applications in physics, chemistry, biology, pharmacy, etc. Popularmethods include the Hartree-Fock method, Density Functional Theory and Molecular Dynamics.These methods will not be discussed in this course. Their principles and applications are topicsin the courses of Computationele fysica and of many elective master courses (e.g. Atomic andmolecular physics, Simulations and modeling for the nanoscale, Computationele materiaalfysica, Capitaselecta vastestoffysica).

Around the equilibrium molecular geometry, the potential energy evolves approximatelyquadratically in the nuclear coordinates. This suggests that the nuclei oscillate harmonicallyaround this equilibrium geometry with characteristic vibrational frequencies. Molecules alsorotate around their center of gravity with characteristic angular frequencies. These motionaleffects give rise to vibrational and rotational spectra, which can be detected in infraredabsoption spectroscopy and Raman scattering experiments. They also result in vibrationalstructure in optical absorption and luminescence spectra. This introduction will also notdeal with these aspects. They are discussed from a phenomenological and experimentalviewpoint in the courses Vastestoffysica, Vastestof- en nanofysica, Optical spectroscopy of materialsand Luminescence. Their computational treatment is given in Atomic and molecular physics and

157

7.2. BORN-OPPENHEIMER

Simulations and modeling for the nanoscale.

7.2 Born-Oppenheimer

7.2.1 Formulation of the Born-Oppenheimer approximation

Consider a molecule with n electrons, labeled i, and ν nuclei, labeled α. Ignoring spin-effects,the energy eigenvalues of the molecule may be found by solving the Schrodinger equation

H Ψ = E Ψ

H =[Hel]

+[Hnucl

]=

[−∑n

i=1h2

2me~∇2

i + V(~ri, ~Rα)]+[−∑ν

α=1h2

2Mα~∇2

α

](7.1)

in which V(~ri, ~Rα) comprises the Coulomb repulsion between electrons, the Coulomb attractionbetween electrons and nuclei, and the Coulomb repulsion between nuclei

V(~ri, ~Rα) = ∑i<1

Vij + ∑α,i

Vαi + ∑α<β

Vαβ (7.2)

Although Vαβ is only dependent on nuclear coordinates, it is customary to include it in Hel .We will try and find a way of separating this equation into two equations, one of which willprovide the electronic wavefunction and energy, and one for the nuclear wavefunction andenergy. It is important to realize that the wavefunction for the complete problem cannot besimply written as the product of a wavefunction that only depends on the electron coordinatesand one that only depends on the nuclear coordinates

Ψ(~ri, ~Rα) 6= φel(~ri)χnucl(~Rα) (7.3)

A more subtle treatment is required. In the following we will see that, within the Born-Oppenheimer approximation a separation of the wavefunction

Ψ(~ri, ~Rα) = φel(~ri, ~Rα)χnucl(~Rα) (7.4)

is possible.

Because Mα me we may assume that the motion of the electrons instantaneously adapts tochanges in the positions of the nuclei. Hence we may assume that the electron wavefunctionφel is a solution of the Schrodinger equation

Hel φel = Eel φel (7.5)

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Both Eel and φel depend on the nuclear coordinates ~Rα, which are treated as external parametersof the Schrodinger equation for the electronic energy and wavefunction. In this way, wecalculate Eel as a function of the nuclear coordinates ~Rα.

We now try to also treat the nuclear motion in a quantum mechanical way. To this end, weassume

Ψ = φel χnucl (7.6)

where χnucl does not explicitly depend on the coordinates of the electrons. Inserting thiswavefunction in Equation (7.1) we find

(Hel + Hnucl)(φel χnucl) = E (φel χnucl)

(Hel φel) χnucl + Hnucl (φel χnucl) = E (φel χnucl)

Hnucl (φel χnucl) = (E− Eel) (φel χnucl) (7.7)

Taking the form of Hnucl into account (see Equation (7.1), the left-hand term in this equationcan be worked out further

Hnucl (φel χnucl) = φel (Hnucl χnucl)−ν

∑α=1

h2

Mα(~∇α φel) · (~∇α χnucl)−

ν

∑α=1

h2

2Mα(~∇2

α φel) χnucl

(7.8)

The nuclear motion is very slow compared with the electronic motion. As a result, φel variesonly very slowly with the nuclear coordinates. For this reason, last two terms on the right-handside of Equation (7.8) are much smaller than the first. In the Born-Oppenheimer approximation,these two terms are completely neglected. Then φel can be divided out from Equation (7.7),which reduces to a Schrodinger equation for χnucl

Hnucl χnucl = (E− Eel) χnucl

(−ν

∑α=1

h2

2Mα

~∇2α + Eel) χnucl = E χnucl (7.9)

We may summarize the Born-Oppenheimer procedure for calculating molecular energy levelsas follows :

• Equation (7.5) allows us to calculate φel and Eel as a function of the nuclear coordinates~Rα. Eel is the sum of the electronic contribution to the total energy of the molecule andthe Coulomb repulsion between the nuclei.

• Eel(~Rα) is the potential energy surface in the Schrodinger equation for χnucl . Theminimum of this potential energy surface yields the equilibrium geometry of the

159


molecule.1

• Equation (7.9) is the equation of motion for the nuclei and describes the vibrational androtational state of the molecule.

The neglected terms in (7.8) may afterwards be treated as a perturbation and will mix theproduct wavefunctions φel χnucl . For most molecules this perturbation is very smal and veryoften they are not taken into account.

In the remainder of this chapter we will concentrate on Equation (7.5) for calculating theelectronic wave function and energy levels. For simplicity we will drop the subscript "el"

H φ = E φ (7.10)

Equation (7.9) may then be used to calculate the vibrational and rotational substructure of theseelectronic levels. This lies beyond the scope of this introductory chapter.

7.2.2 Application to the H+2 molecular ion

Even with the Born-Oppenheimer approximation, the only molecular system that can be solvedanalytically is the H+

2 molecular ion. It consists of just one electron, moving in the Coulombpotential energy of the two H+ nuclei, which are labeled A and B, and which are at aninternuclear distance R. The (electronic) Hamiltonian of H+

2 thus is

H = − h2

2me~∇2 − e2

4πε0rA− e2

4πε0rB+

e2

4πε0R(7.11)

rA and rB represent the distances of the electron from the nuclei A and B, respectively, as shownin Figure 7.2. The corresponding Schrodinger equation can be solved as a function of R byseparation of variables, after a transition from Cartesian to elliptical coordinates, with the nucleiin the focal points

µ =rA + rB

Rµ ≥ 1

ν =rA − rB

R− 1 ≤ ν ≤ 1

tan ϕ =yx

(7.12)

The wavefunction can be written as

φ(µ, ν, ϕ) = M(µ) N(ν) Φ(ϕ) (7.13)

1In general, more than one (local) minimum may be found, which implies that several conformations of themolecule may exist.

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Figure 7.2: H+2 in Cartesian coordinates, defining the polar angle ϕ.

and the Schrodinger equation can be separated into equations for M, N, and

d2

dϕ2 Φ = −λ2Φ (7.14)

The latter equation has as solutions

Φ(ϕ) = e±iλϕ (7.15)

Solving the equations for M and N is mathematically quite tedious and therefore beyond thescope of this introductory chapter. For the current qualitative analysis it is sufficient to note thatthe resulting Eel is a function of λ2, so that the energy levels corresponding with the molecularorbitals MNΦ, are non-degenerate for λ = 0, and doubly degenerate for λ 6= 0. These molecularorbitals are labeled σ for λ = 0, π for λ = 1, δ for λ = 2, etc. These one-electron molecularorbitals (for linear molecules) remain eigenfunctions of lz, the projection of the orbital angularmomentum on the molecular axis: lzΦ = λΦ.

Figure 7.3 shows the numerical results for the potential energy curves of H+2 for the various

molecular orbitals expressed in atomic units: R as multiples of a0 and E as multiples of 2Ry(Ry = Rydberg energy = 13.6 eV). All orbital energies steeply rise as R → 0. This is a resultof the repulsion energy between the nuclei. For R → 0 the molecular orbitals should resemblethose of 2He+, with an axial distortion (Stark-like effect), because there are still two separatenuclei instead of one (for He). On the other hand, for R → ∞ the Coulomb repulsion betweenthe nuclei vanishes and the system consists of a hydrogen atom and a H+ ion. The orbitalsand energies are in that case the same as those for the hydrogen atom. In Figure 7.3 themolecular orbitals are additionally labeled according to their asymptotic behavior for R → 0.The subscripts g (from the German word gerade, English: even) and u (ungerade, English: odd)refer to the parity of the function under inversion of the electonic coordinate~r → −~r. On therighthand side of the figure, the atomic orbitals of hydrogen of the corresponding energy are

161


Figure 7.3: Molecular orbital energies as a function of the the internuclear distance for H+2 .

given. In the R → ∞ limit, we may consider the molecular orbitals as linear combinations ofhydrogen atomic orbitals (LCAO) as follows

1sσg = 1sA + 1sB 2pσu = 1sA − 1sB 3dσg = 2pzA − 2pzB . . . (7.16)

Certain potential energy curves exhibit a minimum, these correspond to bonding molecularorbital states. Ohters are non-bonding. It is customary to make a futher distinction betweenanti-bonding states, which are linear combinations of atomic orbitals on different atoms, andtruly non-bonding states, which are simply orbitals centered on just one atom.

We now focus attention to the lowest two molecular orbital states of H+2 . The fact that one of

them, the ground state 1sσg, corresponds to a bonding state, whereas the other does not, mayalso be understood by looking at the contour diagrams of these orbitals for a given R. In thebonding state the electron density is concentrated in the space between the nuclei, whereas inthe anti-bonding state the electron density in the region between the two nuclei is depleted.Figure 7.3 shows that the calculations predict an equilibrium interactomic distance Req ≈ 2a0

and an extra binding energy ∆E ≈ 0.2 Ry ≈ 2.7 eV (with respect to a hydrogen atom and a1H+ ion). This is very close to the results found in experiments: Req = 106 pm and ∆E = 2.648eV.

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7.3. MOLECULAR ORBITAL THEORY

Figure 7.4: Contour plot in the z− x-plane of the 1sσg and the 2pσu molecular orbitals of H+2 .

7.3 Molecular orbital theory

7.3.1 Linear Combinations of Atomic Orbitals (LCAO)

Arriving at the exact solutions for H+2 already requires highly advanced mathematical tech-

niques. Moreover, this is the only molecular problem that is, within the Born-Oppenheimerapproximation, analytically solvable. We will thus need suitable methods to approximate themolecular orbitals for the more general problems. Equation (7.16) and Figure 7.4 indicate thata reasonable approximation can be found by expanding the molecular orbitals in one-electronorbitals on the atoms in the molecule. This is known as the Linear Combination of AtomicOrbitals (LCAO) method. Such an expansion produces in the case of H+

2 for R→ ∞ indeed thecorrect result. In principle, in order to arrive at the exact solution, a complete basis set shouldbe considered for the expansion, implying that all atomic orbitals should be included in it. Inpractice, however, and certainly for a qualitative analysis, we restrict the expansion to a limitedset of atomic orbitals. We illustrate the method here for H+

2 , but it may be readily extended toother diatomic and polyatomic molecules.

For reconstructing the ground state of H+2 we try linear combinations of just two atomic

orbitals, the 1s orbital centered on atom A (φ1s,A) and the 1s orbital centered on atom B (φ1s,B).For simplicity of notations, in the following the bra and ket notations for the wavefunctions willbe used again. An arbitrary linear combination of these atomic orbitals is

|ψ〉 = cA |φ1s,A〉+ cB |φ1s,B〉 (7.17)

with|φ1s,A〉 =

1√πa3

0

e−rAa0 |φ1s,B〉 =

1√πa3

0

e−rBa0 (7.18)

DefiningS = 〈φ1s,A| φ1s,B〉 (7.19)

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we find that the requirement that |ψ〉 be normalized, yields

〈(cA〈φ1s,A| + cB〈φ1s,B| )| (cA |φ1s,A〉+ cB |φ1s,B〉)〉 = c2A + c2

B + 2cAcBS = 1 (7.20)

because S is real in this case. For the considered wavefunctions, S is also positive.

Next, we impose that |ψ〉 be an eigenfunction of H for H+2 (Equation (7.11))

H(cA |φ1s,A〉+ cB |φ1s,B〉) = E(cA |φ1s,A〉+ cB |φ1s,B〉) (7.21)

Multiplying these equations with 〈φ1s,A| and 〈φ1s,B| , we find a system of linear equations forthe coefficients cA and cB. We define further

HAA = 〈φ1s,A| H |φ1s,A〉 HBB = 〈φ1s,B| H |φ1s,B〉

HAB = 〈φ1s,A| H |φ1s,B〉 HBA = 〈φ1s,B| H |φ1s,A〉 (7.22)

For the considered linear combinations of 1s states of H+2 , a homonuclear diatomic molecule,

HAA = HBB and HAB = HBA. This further implies that all these matrix elements are real. Thesystem of linear equations then becomes

cAHAA + cBHAB = E(cA + cBS)

cAHAB + cBHAA = E(cAS + cB) (7.23)

The requirement that this system has non-trivial solutions leads to the secular equation∣∣∣∣∣ HAA − E HAB − ESHAB − ES HAA − E

∣∣∣∣∣ = 0

⇒ (HAA − E)2 − (HAB − ES)2 = 0 (7.24)

leading to two solutions

E+ =HAA + HAB

1 + Sc+1 = c+2 =

1√1 + S

E− =HAA − HAB

1− Sc−1 = −c−2 =

1√1− S

(7.25)

We may already anticipate that HAB < 0 (see further) and see that E+, corresponding with thesymmetric bonding linear combination, is the solution with lowest energy. The anti-symmetricanti-bonding linear combination has higher energy.

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In this simple case we may evaluate the integral HAA

HAA = 〈φ1s,A| (−h2

2me~∇2 − e2

4πε0rA− e2

4πε0rB+

e2

4πε0R) |φ1s,A〉 = E1s − C +

e2

4πε0R(7.26)

with

C = 〈φ1s,A|e2

4πε0rB|φ1s,A〉 =

e2

4πε0

∫φ2

1s,A1rB

dτ (7.27)

In a similar way we find

HAB = 〈φ1s,A| (−h2

2me~∇2 − e2

4πε0rA− e2

4πε0rB+

e2

4πε0R) |φ1s,B〉 = E1sS− J +

e2S4πε0R

(7.28)

with

J = 〈φ1s,A|e2

4πε0rA|φ1s,B〉 =

e2

4πε0

∫φ1s,Aφ1s,B

1rA

dτ (7.29)

Thus, we find as energy of the bonding linear combination

E+ =E1s − C + e2

4πε0R + E1sS− J + e2S4πε0R

1 + S= E1s +

e2

4πε0R− J + C

1 + S(7.30)

and for the anti-bonding linear combination

E− =E1s − C + e2

4πε0R − E1sS + J − e2S4πε0R

1− S= E1s +

e2

4πε0R+

J − C1− S

(7.31)

where S, C and J are functions of the distance between the nuclei R. These may be calculatedexactly making use of the elliptical coordinates defined in Equation (7.12). The results are

S = (1 +Ra0

+R2

3a20)e−

Ra0 (7.32)

C =e2

4πε0R1− (1 +

Ra0)e−

2Ra0 (7.33)

J =e2

4πε0a0(1 +

Ra0)e−

Ra0 (7.34)

Minimizing Equation (7.30) by variation of R yields the equilibrium bond length and minimumenergy for H+

2 within this approximation. One finds ∆E = 1.76 eV and Req = 2.5 a0, inreasonable agreement with experiment and with the exact calculation. A better approximationmay be found by including more atomic orbitals in the basis set, by allowing also a0 to vary, byallowing different forms for the atomic orbitals (Gaussian-type orbitals, e.g.), etc.

The LCAO results for the energy levels are shown in Figure 7.5. Note that the molecular orbitalenergy levels are not displaced symmetrically around the E1s level. They only would be if theeffects of Coulomb repulsion between the nuclei, C and S are zero, or would accidentally cancel

165


each other. In general, the center of gravity of the two molecular energy levels lies higher thanthe atomic energy levels.

Figure 7.5: LCAO result for energy levels for linear combinations of 1s states for H+2 .

For homonuclear diatomic molecules following real one-electron LCAOs can be constructed(apart from a normalization constant)

σg1s = 1sA + 1sB

σu1s = 1sA − 1sB

σg2s = 2sA + 2sB

σu2s = 2sA − 2sB

σg2p = 2pzA − 2pzB

σu2p = 2pzA + 2pzB

πg2p = 2pxA − 2pxB ; 2pyA − 2pyB

πg2u = 2pxA + 2pxB ; 2pyA + 2pyB

and so on. σ, π, δ . . . refers to the quantum number |λ|: the σ levels are non-degenerate, allothers are two-fold degenerate. g and u refer to the parity of the LCAO. Especially for large Rthese LCAOs are good approximations for the true molecular orbitals. On the other hand,we have seen that for R → 0 the molecular orbitals of H+

2 approach those of 2He+. Theapproximations in these two limiting cases should thus evolve into one another as R varies.This is represented in Figure 7.6 in a so-called correlation diagram. On the left-hand side of thefigure we find the joint-nucleus limit: the atomic energy levels for an atom, with a mass andcharge that is the sum of those of the two atoms in the molecule, are split in levels with equal|m| accounting for the fact that the two nuclei do not perfectly coincide (Stark-like effect). Onthe right-hand side, the formation of symmetric and anti-symmetric linear combinations oforbitals on the individual atoms in the molecule leads to a splitting of the atomic levels of theseparated atoms. On the figure, the situation for selected homonuclear diatomic molecules isindicated. Note that R in this figure gives an indication of the relative interatomic distances forthese molecules, but should not be interpreted as a linear scale.

The procedure outlined here for H+2 can also be applied to construct one-electron molecular or-

166


Figure 7.6: Correlation of molecular orbitals in a two-center system for equal nuclear charges. [HerzbergG, Molecular Spectra and Molecular Structure: I. Diatomic Molecules]

bitals for other homonuclear or heteronuclear diatomic molecules, and polyatomic molecules,each time using an appropriate (finite, incomplete) basis set of atomic orbital functions. Grouptheory greatly helps in determining which Hamiltonian matrix elements and overlap integralsshould be calculated and which are zero (see Section 8.6.3). In some cases, group theory allowsus to determine the LCAOs corresponding with the different molecular energy levels withoutneed for calculating any matrix element. For finding approximate solutions of problems,degenerate and non-degenerate time-independent perturbation theory (Chapter 4) can be used.The ground state configuration of a many-electron molecule may then be found by filling up themolecular energy level system with electrons, taking into account the spin degeneracy of thelevels, the Pauli exclusion principle, and the Coulomb repulsion and exchange energy betweenthe electrons, in a similar way as for atoms (see Chapters 5 and 6). This is illustrated below forthe H2 molecule.

167


7.3.2 The H2 molecule

Determining the electronic configurations and many-electron energy terms for polyelectronicmolecules basically follows the same route as for atoms. We describe the principles here for H2.Its lowest energy term is found by placing a second electron in the σg1s molecular orbital. Theelectron configuration is (σg1s)2 and the orbital two-electron wavefunction is σg1s(1)σg1s(2)(labeling the electrons 1 and 2). The orbital wavefunction is symmetric under exchange of thetwo electrons, so in order to make the total wavefunction antisymmetric, the spin part shouldread 1√

2[α(1)β(2)− β(1)α(2))]. The total spin of the molecule thus is S = 0.

The ground state energy term of H2 is labeled 1Σg, by analogy with the notations forpolyelectronic atoms. More generally terms for linear molecules are labeled 2S+1Λ(±)

Ω(p), with

• S the total electron spin of the molecule

• Λ the total projection of the orbital angular momentum on the molecular axis : Λ = ∑i λi,in spectroscopic notation (Σ, Π, ∆, Φ, . . .)

• Ω the projection of the total angular momentum along the molecular axis. It is onlyincluded in the notation when the level splitting due to spin-orbit interaction becomesimportant

• p the parity of the spatial product wave function: g for even and u for odd. For the parityof product functions, g× g = g = u× u and g× u = u = u× g

• ± refers to the mirror symmetry (+ for symmetric and - for anti-symmetric) through anarbitrary plane containing the molecular axis

The symbols referring to parity and reflection symmetry are only added when applicable andare therefore placed between brackets in the general notation.

The total electronic and nuclear Coulomb repulsion energy of the ground state for H2 is givenby

E = 2 E1s +e2

4πε0R− 2

J + C1 + S

+ Ece (7.35)

with Ece the energy arising from the direct and exchange contributions to the Coulombrepulsion between the electrons. The energy is again a function of the internuclear distance(or bond length) R, and by minimizing it one finds within the LCAO approximation Req = 74pm and ∆E = 3.7 eV. The binding energy now refers to the energy of two separated H atomsin the 1s state. The experimental bond length (74.1 pm) agrees very well with this result, butfor the binding energy (∆Eexp = 4.5 eV) there is still considerable room for improvement. Suchimprovement may be found by allowing the ground state molecular configuration to interactwith one or several excited state configurations of low energy. This method is referred to as

168


configuration interaction.

For other diatomic molecules (homo- or heteronuclear) and polyatomic molecules, a similarroute is followed to find the electronic ground state. First, one-electron molecular orbitals areconstructed from the valence electrons of the atoms (the inner shell core electrons on each atomare considered not to take part in bonding). These molecular orbitals are then occupied withthe valence electrons, considering the (orbital and spin) degeneracies of the levels, in such waythat the total energy is minimal.

As a final remark to this section, we note that for non-linear poyatomic molecules lz does nolonger commute with the Hamiltonian and Λ is no longer a good quantum number, implyingthat it cannot be used in the notation for the energy terms. Instead of Λ the symmetry species(irrep) of the product spatial wavefunction is then used (see Chapter 8).

7.3.3 Example : benzene, a non-linear polyatomic conjugated π-system

Conjugated π-systems are (mostly) planar molecules existing of carbon and hydrogen atoms.Their orbitals with local σ and π symmetry have largely different energies and belong todifferent symmetry species, so they can be treated separately. The complete molecules arenon-linear, and as λ is not a good quantum number the LCAO orbitals of the completemolecule cannot be attributed σ or π symmetry. Considering the bonds between pairs of atomsseparately, the labels σ and π, however, still make sense. Considering now planar moleculesonly consisting of C and H atoms, the in-plane σ orbitals are linear combinations of the 1sorbitals of H and the three 2sp2 hybridized orbitals of C. The bonding σ molecular orbitalshave low energy and are completely filled. Each C atom has one valence electron left in a 2pz

orbital, perpendicular to the plane of the molecule. The π-bonding between these electronsstabilizes the planarity of the molecule.

For such molecules the Huckel approximation allows to obtain a qualitative picture of the valenceenergy levels and the resulting ground state properties of the molecule. This approximationassumes that

• all overlap integrals between atoms in the molecule are zero : Sij = δij

• all diagonal matrix elements of the Hamiltonian have the same value Hii = α, with α < 0

• all off-diagonal matrix elements of the Hamiltonian are zero, except those betweennearest neighbor atoms, which are all set equal to β < 0.

As an example of the Huckel approximation for conjugated π-systems, we calculate the π energylevels for benzene (= cyclohexene) C6H6. This molecule with D6h symmetry, is shown in Figure7.7.

169


Figure 7.7: Benzene molecule, defining the labeling of the C atoms

Following the rules of the Huckel approximation, we find as secular equation for the π energylevels ∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

α− E β 0 0 0 β

β α− E β 0 0 00 β α− E β 0 00 0 β α− E β 00 0 0 β α− E β

β 0 0 0 β α− E

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣= 0

With mathematical program packages (e.g. Maple or Matlab) one easily finds the eigenvaluesand eigenvectors to be

α + 2β 1√6(pz1 + pz2 + pz3 + pz4 + pz5 + pz6)

α + β 1√12(2pz1 + pz2 − pz3 − 2pz4 − pz5 + pz6)

12 (pz2 + pz3 − pz5 − pz6)

α− β 1√12(2pz1 − pz2 − pz3 + 2pz4 − pz5 − pz6)

12 (pz2 − pz3 + pz5 − pz6)

α− 2β 1√6(pz1 − pz2 + pz3 − pz4 + pz5 − pz6)

(7.36)

In the benzene molecule, the six 2pz valence electrons occupy the two lowest-lying energylevels (three molecular orbitals which all can be doubly occupied). These are the bondingπ-orbitals. The higher-energy anti-bonding orbitals remain unoccupied. The many-electronground state term is 1A1g.

Exercise: find the result (7.36) without explicitly solving the secular equation, by means ofgroup theory. Find also the irreps corresponding with the eigenvalues and eigenstates.

170

Chapter 8

Group Theory

8.1 Introduction

Symmetry is a very important aspect of a physical system. Group theory provides themathematical framework for dealing with symmetry and is of fundamental importance inmany branches of science, including elementary-particle physics, nuclear physics, solid-statephysics, spectroscopy, and quantum chemistry. It enables predicting, e.g., the form molecularorbitals (see Section 8.4.2), or how a certain interaction will split up a degenerate energy level(qualitatively, see Section 8.5), and if an integral is (or is not) necessarily zero (see Section8.6.3). It can do so without detailed knowledge of the physical system, the precise form ofthe interaction or the wavefunctions, etc. - only their symmetry properties need to be known.Group theory is quite mathematical in nature, but this chapter is meant to be an introductionillustrating its power when applied to physically relevant problems. Although the field ofapplications is much broader, the theory and examples will be restricted to molecular systems,here. Instead of adopting a rigorous mathematical approach, the key properties and equationswill be illustrated and example results will be generalised. For a completer and more generaldiscussion, we refer to the elective Master course Symmetriegroepen and various excellenttextbooks on this subject (e.g. Tinkham M, Group Theory and Quantum Mechanics).

8.2 Symmetry

8.2.1 Symmetry operations and elements

A symmetry operation of an object is an operation (rotation, reflection,. . . ) that leaves that objectapparently unchanged. As objects we will consider molecules, here. For every symmetryoperation, there is a corresponding symmetry element: the point, line or plane with respectto which the symmetry operation is carried out. If translational symmetry is not taken into

171

8.2. SYMMETRY

account (which would be necessary for, e.g., the classification of crystal structures), five typesof symmetry operations/elements can be distinguished:

• E: the unity operation, i.e. doing nothing.

• Cn: an n-fold rotation, i.e. a rotation over an angle 2π/n about an axis of symmetry(which is the corresponding symmetry element). Note that one should take the largestpossible value of n for a given rotational axis. If there is more than one axis of symmetry,the one with the highest value of n is called the principal axis (Nederlands: hoofdas) -provided it is unique.

• σ: a reflection with respect to a mirror plane (which is the symmetry element). A mirrorplane is called vertical (σv) if it contains the principal axis and horizontal (σh) if it isperpendicular to the principal axis. A vertical mirror plane that bisects two C2 axes thatare perpendicular to the principal axis, is called dihedral (σd).

• i: an inversion around a centre of symmetry (which is the symmetry element).

• Sn: an n-fold improper (Nederlands: oneigenlijke) rotation, also called rotation-reflection,which consists of an n-fold rotation, followed by a reflection in a (horizontal) planeperpendicular to the n-fold axis. Note that this definition differs from that given inMateriaalfysica ! We adopt this definition here because it is common in the literature ofgroup theory and it is consistent with the multiplication and character tables listed furtheron. When choosing the other definition (n-fold rotation, followed by inversion) these tablesneed to be adapted.

8.2.2 Classification of molecules: point groups

The symmetry of an object can be classified according to the collection of all symmetry elementsof that object, which is called the point group. The fact that this collection indeed constitutes agroup in the mathematical sense of the word, will be discussed below. Point group refers to thefact that only operations are considered whose corresponding symmetry elements all intersectin at least one point.1 Figure 8.1 allows determining the point group of a given molecule.

ExerciseDetermine the point group of (a) a benzene molecule C6H6, (b) a water moleculeand (c) an ammonia molecule (NH3).

(a) Benzene is a non-linear molecule with a (unique) 6-fold rotational axis (C6) -the principal axis -, with six 2-fold rotational axes (C2) perpendicular to it. Themolecular plane is a mirror plane, which is perpendicular to C6 (σh). We arrive atpoint group D6h.

1If translational symmetry is also considered, the symmetry classification needs to be done according to the spacegroup, as is the case in crystallography (see Materiaalfysica).

172

8.2. SYMMETRY

(b) Water is a non-linear molecule with a single C2 axis - the principal axis. Thereis no mirror plane perpendicular to C2, but there are two vertical mirror planescontaining the principal axis (2σv). The point group consequently is C2v .

(c) Ammonia is a non-linear molecule with a single 3-fold rotational axis - theprincipal axis -, and three vertical mirror planes. The point group is C3v .

Figure 8.1: Scheme for the determination of a point group [Atkins PW and Friedman RS, MolecularQuantum Mechanics].

173

8.3. GROUPS

8.3 Groups

8.3.1 Definition

The mathematical definition of a group is a set of elements (G) together with an operation (oftensomewhat misleadingly called multiplication and denoted · here) with the following properties:

1. The set contains a unity element E, so that R · E = E · R = R for all elements R of the set.

2. Associativity: R · (S · T) = (R · S) · T ≡ R · S · T .

3. Closure (Nederlands: geslotenheid): if R and S are elements of G, than so is R · S.

4. The unique inverse of each element of the set is also part of the set. S is the inverse of R ifS · R = R · S = E .

The symmetry operations of a point group form a group: the group elements2 are the symmetryoperations themselves, and the group operation simply is the consecutive application of thesesymmetry operations. We will denote this as RS . Note that

• the definition of a group does not require commutativity of the operation. Indeed,consecutive rotation about two different axes, e.g., in general is not commutative. Theorder in which we apply symmetry operations therefore is important: with RS we meanthat S is applied first, and after that R.

• property 2 implies that one can combine more than two symmetry operations in whateverway, as long as the relative order is respected.

• property 3 implies that the result of two consecutive symmetry operations is equivalentto a single symmetry operation of the same group: if R and S are symmetry operations ofa point group, then the result obtained by letting RS operate on the object is the same asthat by letting T operate on the object, with T an element of the group.

• the inverse element of R is denoted R−1 .

• The number of elements in the set is called the order of the group and denoted h . Exceptfor D∞h and C∞v, all molecular symmetry groups in the scheme (Fig. 8.1) have finiteorder. Most of the examples given in this chapter will deal with finite-order symmetrygroups.

8.3.2 Multiplication tables

The closure condition (property 3) allows one to set up a multiplication table in which theeffect of all possible combinations of two symmetry operations is given. Consider again

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8.3. GROUPS

Figure 8.2: Symmetry operations for the NH3 molecule.

the NH3 molecule (Figure 8.2) with symmetry group C3v (see exercise above). There are sixsymmetry operations (group of order 6, h = 6): the unity operation E, rotation about the 3-foldaxis counterclockwise (C+

3 ) or clockwise (C−3 ) and reflections with respect to the three mirrorplanes (σv, σ′v and σ′′v ). We get following multiplication table for all combinations RS, with theconvention of putting operation S at the top and operation R at the left-hand side:

E C+3 C−3 σv σ′v σ′′v

E E C+3 C−3 σv σ′v σ′′v

C+3 C+

3 C−3 E σ′′v σv σ′vC−3 C−3 E C+

3 σ′v σ′′v σvσv σv σ′v σ′′v E C+

3 C−3σ′v σ′v σ′′v σv C−3 E C+

3σ′′v σ′′v σv σ′v C+

3 C−3 E

Table 8.1: Multiplication table for C3v

Note that each row and column contains every element exactly once.

8.3.3 Matrix representation (Nederlands: voorstelling, Deutsch: Darstellung)

Up to now, we have represented the symmetry operations in a visual fashion. It is also possibleto use matrix representations, which allows a more mathematical approach. To determine thematrix representation for a specific symmetry operation, we need to choose a basis, i.e. a setof functions on which the effect of the symmetry operations can be mapped. In the case of theNH3 molecule, we can use, e.g., the s orbitals of the different atoms, which we label sA, sB andsC for the hydrogen atoms and sN for the nitrogen atoms. The dimension d of this particularbasis is 4. It is convenient to write this basis as a vector with four components (sN ,sA,sB,sC). In

2Note the unfortunate double use of the word element: (i) symmetry element (rotational axis, mirror plane,. . . )and (ii) element of a set in the context of a group.

175

8.3. GROUPS

general, we can denote a basis as ~f = ( f1, f2,..., fd), but usually the vector superscript is dropped( f ).

Consider, e.g., the effect of σv on the NH3 molecule:

σv

sN

sA

sB

sC

=

sN

sA

sC

sB

(8.1)

We can generalise this result to the effect of an operation R on a basis f being given by

R fi = ∑j

Dij(R) f j (8.2)

The Dij(R) ≡ [D(R)]ij coefficients constitute a matrix, denoted D(R), which is a representationof operation R in basis f . The matrix representation for σv in the case of the NH3 molecule, e.g.,is

σv f = D(σv)

sN

sA

sB

sC

=

1 0 0 00 1 0 00 0 0 10 0 1 0

sN

sA

sB

sC

(8.3)

and those of the other symmetry operations can be derived in an analogous fashion:

D(E) =

1 0 0 00 1 0 00 0 1 00 0 0 1

D(C+3 ) =

1 0 0 00 0 0 10 1 0 00 0 1 0

D(C−3 ) =

1 0 0 00 0 1 00 0 0 10 1 0 0

D(σv) =

1 0 0 00 1 0 00 0 0 10 0 1 0

D(σ′v) =

1 0 0 00 0 0 10 0 1 00 1 0 0

D(σ′′v ) =

1 0 0 00 0 1 00 1 0 00 0 0 1

Note that if3

RS = T (8.4)

thenD(R)D(S) = D(T) (8.5)

3The operation between R and S in Eq. (8.4) is the consecutive application of the symmetry operations, while inEq. (8.5), the operation between D(R) and D(S) is matrix multiplication.

176

8.3. GROUPS

In a more rigorous approach, this homomorphism between the group of symmetry operationsand the group of square matrices is taken as the definition of a representation (see Symme-triegroepen). We state that the basis f generates the representation D.

Proof of Eq. (8.5)Assume that RS = T, then RS f = T f = h with

hi = ∑j

Dij(T) f j (8.6)

but also, assuming S f = g, so that Rg = R(S f ) = RS f = h

hi = ∑k

Dik(R)gk = ∑k

Dik(R)

(∑

jDkj(S) f j

)= ∑

j

(∑

kDik(R)Dkj(S)

)f j (8.7)

Comparison of the two equations above shows that

Dij(T) = ∑k

Dik(R)Dkj(S) (8.8)

orD(T) = D(R)D(S) (8.9)

Example of Eq. (8.5)We know from the multiplication table that σvC+

3 = σ′v for the C3v group. We can verify thispurely visually: since

C+3 (sN , sA, sB, sC) = (sN , sC, sA, sB) (8.10)

we getσvC+

3 (sN , sA, sB, sC) = σv(sN , sC, sA, sB) = (sN , sC, sB, sA) (8.11)

and indeed, we also have

σ′v(sN , sA, sB, sC) = (sN , sC, sB, sA) (8.12)

It is however also possible to establish that σvC+3 = σ′v with the matrix representations:

D(σv)D(C+3 ) =

1 0 0 00 1 0 00 0 0 10 0 1 0

1 0 0 00 0 0 10 1 0 00 0 1 0

=

1 0 0 00 0 0 10 0 1 00 1 0 0

= D(σ′v)

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8.3. GROUPS

Note that the character (Nederlands: karakter) χ - defined as the sum of the diagonal elements(trace, Nederlands: spoor) - of the matrix representations are different for the different ’types’ ofsymmetry operations.

χ(E) = 4 χ(C+3 ) = 1 χ(C−3 ) = 1

χ(σv) = 2 χ(σ′v) = 2 χ(σ′′v ) = 2

As we will see, the character is a quantity of key importance in group theory.

8.3.4 Transformation of a matrix representation

The choice of a basis is of course arbitrary: since (sN , sA, sB, sC) is a suitable basis set, any set offour linearly independent combinations of these basis functions also forms a basis. 4 E.g.:

f ′ =

s1

s2

s3

s4

=

sN

1√3(sA + sB + sC)

1√6(2sA − sB − sC)

1√2(sB − sC)

(8.13)

Basis f can be transformed into basis f ′ by means of a unitary matrix C: f ′ = C f , or, for thedifferent components:

f ′i = ∑j

Cij f j (8.14)

In this example

C =

1 0 0 00 1√

31√3

1√3

0 2√6− 1√

6− 1√

6

0 0 1√2− 1√

2

(8.15)

If operation R has a matrix representation D(R) in basis f , it will in general have another matrixrepresentation D′(R) in basis f ′. The connection between the two matrix representations isgiven by

D′(R) = CD(R)C−1 (8.16)

4Although strictly speaking all linear combinations are suitable, we will only consider normalized linearcombinations here that preserve the orthogonality of the original basis (if it was orthogonal). The basistransformations we use are thus unitary (C−1 = C†), or in the case of real transformations orthogonal (C−1 = CT).

178

8.3. GROUPS

Example: determination of the matrix representation of C+3 in basis f ′

Using Eqs. 8.3.4 and 8.15 we find

D′(C+3 ) = CD(C+

3 )C−1

=

1 0 0 00 1√

31√3

1√3

0 2√6− 1√

6− 1√

6

0 0 1√2− 1√

2

·

1 0 0 00 0 0 10 1 0 00 0 1 0

·

1 0 0 00 1√

32√6

0

0 1√3− 1√

61√2

0 1√3− 1√

6− 1√

2

=

1 0 0 00 1 0 0

0 0 − 12 −

√3

2

0 0√

32 − 1

2

Similar calculations yield the other matrix representations in the basis f ′:

D′(E) =

1 0 0 00 1 0 00 0 1 00 0 0 1

D′(C+3 ) =

1 0 0 00 1 0 0

0 0 − 12 −

√3

2

0 0√

32 − 1

2

D′(C−3 ) =

1 0 0 00 1 0 0

0 0 − 12

√3

2

0 0 −√

32 − 1

2

D′(σv) =

1 0 0 00 1 0 00 0 1 00 0 0 −1

D′(σ′v) =

1 0 0 00 1 0 0

0 0 − 12 −

√3

2

0 0 −√

32

12

D′(σ′′v ) =

1 0 0 00 1 0 0

0 0 − 12

√3

2

0 0√

32

12

Although the representations D and D′ are not identical, they are also not fundamentallydifferent, because they are connected by a similarity transformation of the basis that generatesthem. We will call them equivalent representations, or simply equal and will state that D = D′.Within the context of this course, D = D′ thus means that a similarity transformation can befound that transforms the one representation into the other.

8.3.5 Invariance of characters under basis transformations

Note that the characters χ of the matrix representations in the new basis f ′ are the same as thosein the old basis f . This is in general true and this property is a direct result of the invariance ofthe trace of a product of matrices under cyclic permutation5:

χ′(R) = tr(

D′(R))= tr

(CD(R)C−1

)= tr

(C−1CD(R)

)= tr (D(R)) = χ(R) (8.17)

5I.e. tr(ABCD) = tr(BCDA) = tr(CDAB) = tr(DABC). Beware: in general tr(ABCD) 6= tr(BACD).

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8.3. GROUPS

8.3.6 Characters and classes

Consider a certain element of the group (i.e. a symmetry operation), R . The set of elementsthat is generated by calculating X−1RX with X running over all elements of the group is calleda class. In other words, two symmetry operations R and S belong to the same class if there is asymmetry operation X in the group so that X−1RX = S . Members of the same class are said tobe conjugate (Nederlands: toegevoegd). Note that

• a class can be generated completely from one of its elements.

• from the definition, it is immediately clear that the unity element E forms a class by itself.

• an element is always conjugate to itself.

Exercise: determine the different classes of C3v

Take e.g. C+3 and consider all elements generated by X−1RX with X an element of the group.

The inverse elements and the products are readily found using multiplication table 8.1. E.g.:

σ−1v C+

3 σv = σv(C+3 σv) = σvσ′′v = C−3 (8.18)

All other elements also generate either C+3 or C−3 . These two symmetry operations therefore

form a class (2C3). Similarly, one finds that σv, σ′v and σ′′v form a class (3σv). Together with E,there are three classes: E, 2C3 and 3σv.A word of caution: it is not in general true that all rotations form one class and all reflections another.

An important feature of a class is that all its members have equal characters: assume R and Sbelong to the same class, then

χ(S) = trD(S) = tr(

D−1(X)D(R)D(X))= tr

(D(X)D−1(X)D(R)

)= trD(R) = χ(R)

(8.19)We can therefore speak of the character of a class. Note that two different classes can havedifferent characters, but may also have the same character.

8.3.7 Irreducible representations

In the original basis (sN , sA, sB, sC) (page 176), the matrix representations of all symmetryoperations have the following block-diagonal form:

180

8.3. GROUPS

This suggests a split-up of the basis in sN and (sA, sB, sC): the symmetry operations do not ‘mixup’ the sN basis function with the other basis functions. In fact, it implies that (i) sN forms abasis by itself, i.e. all symmetry operations have a matrix representation in the basis sN , namelythe 1-dimensional identity matrix [1] and (ii) the functions sA, sB and sC form a basis on theirown, i.e. all symmetry operations have a matrix representation in the basis (sA, sB, sC), namely

E :

1 0 00 1 00 0 1

C+3 :

0 0 11 0 00 1 0

C−3 :

0 1 00 0 11 0 0

σv :

1 0 00 0 10 1 0

σ′v :

0 0 10 1 01 0 0

σ′′v :

0 1 01 0 00 0 1

This breaking up of a higher-dimensional matrix representation in several lower-dimensionalmatrix representations is called the reduction of the representation. In this particular case, wehave following reduction:

D(4) = D(1) ⊕ D(3) (8.20)

where ⊕ is called the direct sum. sN is said to generate or to span (the) D(1) (matrixrepresentations). Analogously, D(3) is spanned by the (sA, sB, sC) basis.Note that the representations in both bases (1- and 3-dimensional) reproduce the multiplicationtable (which is a specific property of a point group - in that sense they represent the pointgroup), but the D(1) matrix representations clearly are not very ‘informative’ in this respect: allsymmetry transformations are represented by the same matrix (1). It is called an unfaithful(Nederlands: niet-getrouwe) representation and is also referred to as the totally symmetricrepresentation. Nevertheless, this representation is essential for representation theory.It is possible to in turn reduce D(3): in the basis f ′ = (s1, s2, s3, s4) (see page 178), all matrixrepresentations are of the form

which corresponds to the reduction

D(4) = D(1) ⊕ D(1) ⊕ D(2) (8.21)

181

8.3. GROUPS

so that in fact we have obtained the reduction

D(3) = D(1) ⊕ D(2) (8.22)

The two-dimensional matrix representations are

D(2)(E) =

[1 00 1

]D(2)(C+

3 ) =

[− 1

2 −√

32√

32 − 1

2

]D(2)(C−3 ) =

[− 1

2

√3

2

−√

32 − 1

2

]

(8.23)

D(2)(σv) =

[1 00 −1

]D(2)(σ′v) =

[− 1

2 −√

32

−√

32

12

]D(2)(σ′′v ) =

[− 1

2

√3

2√3

212

]

and in the second one-dimensional representation all matrices again are [1]. Thus, thebasis function s1 = sA + sB + sC spans the same representation D(1) as sN . The physicalinterpretation is clear: s1 has the same symmetry as sN . Indeed, both are invariant under all sixsymmetry operations. We arrive at a very important conclusion, which allows a more rigorousmathematical definition of the rather vague concept ‘the same symmetry’: two functions are ofthe same symmetry if they span a basis for the same matrix representation.One may wonder if it is possible to reduce D(2) further, i.e. if we can we find a basistransformation C′ so that all two-dimensional matrices in the basis f ′′ = C′ f ′ are block diagonal(which would imply, in this particular case, that they are diagonal). It turns out that this is notpossible. D(2) is therefore called an irreducible representation (irrep, Nederlands: niet-reduceerbarevoorstelling (NRV)). Obviously, D(1) is also irreducible. It turns out that each group of finiteorder has a finite number of irreps (with corresponding symmetry types or symmetry species). Allother representations can be written as direct sums of these irreps.To each irrep of a group, we can connect a label Γ(l) (where (l) only functions as an index- it is not the dimension!), called the symmetry species (Nederlands: symmetrietype). One-dimensional irreps are denoted A or B, two-dimensional irreps E and three-dimensional irrepsT. If a set of functions forms a basis for a certain irrep, those functions are said to span thatirrep.In summary, we have chosen a basis, which generated a matrix representation of the groupelements (symmetry operations). This representation is in general reducible into lower-dimensional representations (irreps). Equivalently: the basis can be split up in lower-dimensional bases, each of which spans a representation that is not further reducible. Allbasis functions (linear combinations of the originally chosen basis functions) that span the sameirrep, share the same symmetry properties.Symmetry species can be classified according to the characters of the corresponding irrep, as

the latter form a unique fingerprint of an irrep. Symmetry species Γ(1) of irrep D(1), e.g., hascharacters (1, 1, 1) for the three classes (E, 2C3, 3σv), and is labelled A1 (the totally symmetricrepresentation). For the 2-dimensional irrep E, we find (2, -1, 0). All irreps of a group and theircharacters can be summarised in a so-called character table, which is a condense ‘summary’ of

182

8.3. GROUPS

that group (see further).

We will now focus our attention to following, intimately related questions:

• which symmetry species are contained in a particular group?

• to which sum of irreps can an arbitrary representation be reduced?

• which basis functions span a certain irrep?

In order to answer them, some mathematical machinery will be introduced in the next section.This machinery is based on two theorems.

8.3.8 The great and little orthogonality theorems

The great orthogonality theorem (GOT) states that for a group of order h, with D(l)(R) the matrixrepresentation of a symmetry operation R in a dl-dimensional irrep of symmetry species Γ(l) ofthe group and D(l′)(R) the matrix representation of R in a dl′-dimensional irrep of symmetryspecies Γ(l′), following equation holds

∑R

[D(l)

ij (R)]∗

D(l′)i′ j′ (R) =

hdl

δll′δii′δjj′ (8.24)

The right-hand side yields

• zero for any matrix elements (arbitrary i, i′, j, j′) of different irreps (l 6= l′).

• zero for different matrix elements (i 6= i′ or j 6= j′) within the same irrep.

• hdl

if the same matrix element is chosen within the same irrep.

Illustration of the GOT for NH3

The matrix representations for the 1-dimensional irrep A1 (with basis sN) are the identitymatrices [1] for all six symmetry operations. For Γ(l) = Γ(l′) = A1, Eq. (8.24) translates to(i = i′ = j = j′ = 1)

∑R

D(A1)11 (R)∗D(A1)

11 (R) = 1 + 1 + 1 + 1 + 1 + 1 =61

(8.25)

On the other hand, consider matrix elements 11 and 12 of the two-dimensional irrep E (usingEqs. (8.23)):

∑R

D(E)11 (R)∗D(E)

12 (R) = 0−√

34

+

√3

4+ 0−

√3

4+

√3

4= 0 (8.26)

183

8.3. GROUPS

The ‘weaker’ little orthogonality theorem (LOT) suffices for most situations in practice. It states

∑R

[χ(l)(R)

]∗χ(l′)(R) = hδll′ (8.27)

This implies that

• the sum of the squares of the characters of a group for any irrep equals the order of thegroup.

• the sum of the products of the corresponding characters for two different irreps is zero.

The LOT can be derived from the GOT.

Proof of the LOTPutting i = j and i′ = j′ and summing over i and i′, the GOT yields

∑i,i′

∑R

[D(l)

ii (R)]∗

D(l′)i′i′ (R) = ∑

i,i′

hdl

δll′δii′δii′

∑R

(∑

i

[D(l)

ii (R)]∗)(

∑i′

D(l′)i′i′ (R)

)=

hdl

δll′ ∑i,i′

δii′

∑R

[χ(l)(R)

]∗χ(l′)(R) = hδll′

Since all symmetry operations within a class have the same character, it is possible to furthersimplify Eq. (8.27) by summing over the different classes instead of the symmetry operationsseparately:

∑c

g(c)[χ(l)(c)

]∗χ(l′)(c) = hδll′ (8.28)

where g(c) is the number of symmetry operations in class c. For l = l′ (i.e. considering thesame irrep), this yields

∑c

g(c)|χ(l)(c)|2 = h (8.29)

Two other important and very useful properties can be derived from the GOT - for which wewill not present a proof here:

• The number of symmetry species (i.e. the number of irreps) of a group is equal to thenumber of classes of that group.

• The sum of the squares of the dimensions of all irreps of a group equals the order of thegroup:

∑l

d2l = h. (8.30)

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8.3. GROUPS

C3v E 2C3 3σvA1 1 1 1A2 1 1 -1E 2 -1 0

Table 8.2: Character table of point group C3v

The LOT (Eq. (8.28)) and the above two properties provide all the necessary information toconstruct the character table of a group, containing all symmetry species of the group and theircharacters for all classes. As an example, the character table for C3v is given in Table 8.2 .

Illustration: construction of the character table of C3v

Since there are three classes, there must be three symmetry species (irreps), which welabel Γ(i) (i = 1,2,3). Every group contains the unfaithful representation of 1-dimensionalmatrices [1] . This means that Γ(1) = A1 with corresponding characters (1,1,1).Since there are six symmetry operations (h = 6), the dimensions of Γ(2) and Γ(3) must be 1and 2 to satisfy Eq. (8.30). We hence label them A2 and E, respectively. d2 = 1 and d3 = 2imply that

χ(2)(E) = 1 χ(3)(E) = 2 (8.31)

The remaining χ(A2)(2C3), χ(A2)(3σv), χ(E)(2C3) and χ(E)(3σv) can be obtained by applyingEq. (8.28) for different combinations of l and l′:1. For l = l′ = 2:

1 + 2(χ(2)(2C3))2 + 3(χ(2)(3σv))

2 = 6 (8.32)

2. For l = 1 and l′ = 2:1 + 2χ(2)(2C3) + 3χ(2)(3σv) = 0 (8.33)

3. For l = l′ = 3:4 + 2(χ(3)(2C3))

2 + 3(χ(3)(3σv))2 = 6 (8.34)

4. For l = 1 and l′ = 3:2 + 2χ(3)(2C3) + 3χ(3)(3σv) = 0 (8.35)

Combination of these equations yields

χ(2)(2C3) = 1 χ(2)(3σv) = −1 χ(3)(2C3) = −1 χ(3)(3σv) = 0 (8.36)

In a similar fashion, character tables can be constructed for all point groups. The charactertables of selected molecular point groups are tabulated at the end of this chapter (source: [A &F, Molecular Quantum Mechanics] and [Altman & Herzig Point Group Theory Tables]). For otherpoint groups, they can also be found, e.g. on http : //www.webqc.org/symmetry.php .

185

8.4. REDUCED REPRESENTATIONS

Exercises

• Construct the character table for C2v , the symmetry group of the water molecule

• Construct the character table for Td , the symmetry group of methane (CH4) and thepoint symmetry group of the diamond lattice

8.4 Reduced representations

8.4.1 The reduction of representations

Given a set of basis functions, which symmetry species (which irreps) of a certain group dothese span? We have seen that any matrix representation can be written as a direct sum ofirreps (block diagonalisation):

D(R) = D(Γ(1))(R)⊕ D(Γ(2))(R)⊕ ... (8.37)

We can express this more conveniently in terms of the symmetry species of the irreps:

Γ = ∑l

al Γ(l) (8.38)

Our goal is now to be able to determine the coefficients al in a general case.Note that the reduction of matrix representations is nothing but a similarity transformation(Nederlands: gelijkvormigheidstransformatie) 6 and that the trace of a matrix is invariant undersimilarity transformations. Consequently, the character of the original representation mustequal the sum of the characters of the irreps:

χ(R) = ∑l

al χ(l)(R) (8.39)

Multiplication with[χ(l′)(R)

]∗and summation over all symmetry operations of the group

yields

∑R

[χ(l′)(R)

]∗χ(R) = ∑

R∑

lal

[χ(l′)(R)

]∗χ(l)(R)

= h ∑l

alδll′

= hal′

where we have used the LOT (Eq. (8.27)). We obtain

al =1h ∑

R

[χ(l)(R)

]∗χ(R) (8.40)

6In the analysis given in this chapter, transformations are further restricted to unitary and orthogonaltransformations.

186


or, again switching to a summation over classes:

al =1h ∑

cg(c)

[χ(l)(c)

]∗χ(c) (8.41)

This equation allows us to calculate the coefficients al based on nothing but the classes andtheir characters.

Illustration of Eq. (8.41) in the case of the s-orbital basis for C3v

We have derived the four-dimensional representation this basis induces above (see Section8.3.3), and have shown that by a suitable orthogonal transformation, it can be reducedinto irreps (see Section 8.3.4) twice the one-deminsional totally symmetric representationA1 and the two-dimensional representation E. The success of this reduction depended onour knowledge of the suitable tranformation. Now we derive this result based on Eq. (8.41),without knowledge of the orthogonal transformation needed.Starting from the original four-dimensional representation in Section 8.3.3, the characters ofthe classes (E, 2C3, 3σv) are (4, 1, 2). Applying Eq. (8.41) yields

aA1 =16[1× 1× 4 + 2× 1× 1 + 3× 1× 2] = 2

aA2 =16[1× 1× 4 + 2× 1× 1 + 3× (−1)× 2] = 0

aE =16[1× 2× 4 + 2× (−1)× 1 + 3× 0× 2] = 1

and we indeed arrive at the reduction Γ = 2A1 ⊕ E.

8.4.2 Symmetry-adapted bases

The reduction of a representation is equivalent to finding the similarity transformation thatreduces the matrix representations of all symmetry operations to block-diagonal forms, witheach block corresponding to an irrep. There is a corresponding reduction of the original basisf = ( f1, . . . , fn) in basis functions f ′i which are linear combinations of the fi and are groupedper irrep. This basis f ′ is adapted to the symmetry of the system: every f ′i transforms accordingto a certain symmetry species (irrep) of the system, whereas the original basis functions fi ingeneral were mixtures of different symmetry species.In the previous section, we found a procedure to determine the different irreps contained in agiven representation Γ(l) . Now the objective is to determine the corresponding basis functions,i.e. the symmetry-adapted basis of that representation. The mathematical tools needed for thisare the generalized projection operators connected to the representation Γ(l):

P(l)ij =

dl

h ∑R

[D(l)

ij (R)]∗

R (8.42)

They are linear combinations of the different symmetry operations of the group with therepresentation matrix elements as weighing factors. The effect of a generalized projection

187


operator on a basis function f (l′)

i′ of a basis for an irrep Γ(l′) is

P(l)ij f (l

′)i′ = f (l)j δll′δii′ (8.43)

This means that the operator P(l)ij connected to an irrep Γ(l) will generate

• zero when it operates on a basis function belonging to any other irrep Γ(l) (l′ 6= l) or onany basis function belonging to the same irrep that is not f (l)i (i′ 6= i) .

• basis function f (l)j when it operates on the basis function f (l)i belonging to the same irrep.

The latter implies that projection operators allow determination of the complete basis f (l) ifonly one of the basis functions is known.

Proof of Eq. (8.43)Consider the set of functions f (l

′) that form a basis for the irrep D(l′) with dimension dl′ ofsymmetry species Γ(l′) in a group of order h. The effect of a symmetry operation R on a basisfunction can be written as

R f (l′)

i′ = ∑j′

D(l′)i′ j′ (R) f (l

′)j′ (8.44)

Multiplication with[

D(l)ij (R)

]∗and summation over all symmetry operations yields

∑R

[D(l)

ij (R)]∗

R f (l′)

i′ = ∑R

∑j′

[D(l)

ij (R)]∗

D(l′)i′ j′ (R) f (l

′)j′

= ∑j′

f (l′)

j′

(∑R

[D(l)

ij (R)]∗

D(l′)i′ j′ (R)

)

= ∑j′

f (l′)

j′h

dl′δll′δii′δjj′

= f (l′)

jhdl

δll′δii′

where we have used the GOT (Eq. (8.24)). Considering Eq. (8.42), we arrive at Eq. (8.43).

On the other hand, the effect of the projection operator P(l)ii (i = j) on an arbitrary basis f =

( f1, f2, ...) (i.e. it does not need to be a basis for an irrep), like (sN , sA, sB, sC) for NH3, is

P(l)ii f j = b(l)ji f (l)i (8.45)

This operator, which is a projection operator in the strict sense (i.e. an idempotent operatorP(l)

ii P(l)ii = P(l)

ii ) thus generates the ith basis function of irrep Γ(l) , unless this basis functionis not contained in f j, or, in other words, is orthogonal to it. The operators P(l)

ij with i 6= j are

188


so-called nilpotent operators (P(l)ij P(l)

ij = 0).

Proof of Eq. (8.45)The basis functions f j in general are linear combinations of symmetry-adapted basisfunctions:

f j = ∑j′,l′

b(l′)

jj′ f (l′)

j′ (8.46)

where the summation runs over all basis functions of all irreps and the b(l′)

j′ are the expansioncoefficients.Using the definition of the projection operator (Eq. (8.43)) with i = j, one obtains

P(l)ii f j = ∑

j′,l′b(l′)

jj′

(P(l)

ii f (l′)

j′

)= ∑

j′,l′b(l′)

jj′ f (l)i δll′δij′ = b(l)ji f (l)i (8.47)

The basis functions are determined apart from a multiplicative constant, b(l)i ), which can beeliminated by normalization. Hence, the projections operators P(l)

ii generate the symmetry-adapted basis functions f (l)i . Note, however, that b(l)i can be zero, which means basis functionf (l)i is not contained in f j , or in mathematical terms, the functions are orthogonal.

In summary: applying P(l)ii to an arbitrary basis function generates f (l)i and applying P(l)

ij to

f (l)i generates all other basis functions f (l)j of irrep Γ(l) .

The problem with this method is that all matrix representations need to be known explicitlyin order to determine the projection operators. We can avoid this problem by employing analternative projection operator p(l) for which only the characters need to be known (which arereadily available from the character table of the group):

p(l) = ∑i

P(l)ii =

dl

h ∑i,R

[D(l)

ii (R)]∗

R =dl

h ∑R

[χ(l)(R)

]∗R (8.48)

Its effect on an arbitrary basis function f j is

p(l) f j = ∑i

P(l)ii f j = ∑

ib(l)i f (l)i (8.49)

The equation above reveals the downside to this approach: in general a linear combinationof all basis functions of irrep Γ(l) is generated. For one-dimensional irreps this does not poseany problem since there is only one basis function. For higher-dimensional irreps, the differentbasis functions have to be extracted from the linear combinations. In certain practical cases(but not in all cases) this can easily be handled, as illustrated in the box below. If one wayor another the complete irreducible representations are known, it is preferred to use the actualprojection operators P(l)

ii to find the basis functions belonging to the different rows (or columns)of a representation separately.

189


Illustration: projection operators applied to the s-basis for C3v

Since Γ = 2A1 ⊕ E, we only need to consider A1 and E . Following procedure is convenient:

1. Construct a table in which the effect of all symmetry operations on the original basisfunctions (here: sN , sA, sB and sC) is summarized. (Determination of R f j.)

2. Multiply each result in the column with the character of the corresponding operationfor the different irreps. (Determination of χ(l)(R)R f j.)

3. Sum the results per column and multiply with dlh . (Determination of p(l) f j .)

The table issN sA sB sC

E sN sA sB sC

C+3 sN sC sA sB

C−3 sN sB sC sA

σv sN sA sC sB

σ′v sN sC sB sA

σ′′v sN sB sA sC

(8.50)

Symmetry species A1

We have χ(R) = 1 for all operations. For the first and second column we obtain

p(A1)sN =16(sN + sN + sN + sN + sN + sN) = sN

p(A1)sA =16(sA + sC + sB + sA + sC + sB) =

13(sA + sB + sC)

and the third and fourth columns yield the same function. Thus, the basis functions s1 = sN

and s2 = 13 (sA + sB + sC) both form a symmetry-adapted basis that spans the (1-dimensional)

irrep A1 . Since these are linearly independent functions, A1 indeed shows up twice in thereduction of the s-basis representation.

Symmetry species EThe characters of E are (2, -1, -1, 0, 0, 0) for the six symmetry operations. The four columnsyield

p(E)sN =26(2sN − sN − sN + 0 + 0 + 0) = 0

p(E)sA =26(2sA − sC − sB + 0 + 0 + 0) =

13(2sA − sB − sC) = sx

p(E)sB =13(2sB − sA − sC) = sy

p(E)sC =13(2sC − sB − sA) = sz

190

8.5. SYMMETRY AND DEGENERACY

The three generated functions are, however, not linearly independent. We can then extracttwo linearly independent functions which are orthogonal. One solution is s3 = sx =13 (2sA − sB − sC) and s4 = sy − sz = sB − sC . The basis (s3,s4) or - since basis functionsare determined apart from a multiplicative constant - the basis (2sA − sB − sC,sB − sC) thusspans the (2-dimensional) irrep E . It can be verified that projection operator p(A2) returnszero for every basis function, reflecting the fact that A2 is not contained in the (reducible)representation generated by this basis.

8.5 Symmetry and degeneracy

Noether’s theorem (See Inleiding tot de theoretische fysica) describes the close relation betweensymmetry and conservation laws. A similar relation exists between symmetry and degeneracyin quantum mechanical systems. In following subsections, it will be explained how theeigenvalues of an operator may be labeled with the symmetry species or irreps of the symmetrygroup of this operator and how the dimension of these irreps determine the degeneragy ofthe eigenvalues. As these are general properties of quantum mechanical operators, they areespecially true for Hamitonians and their eigenvalues, energy levels. We will also see howreduction in symmetry leads to a splitting of energy levels and how group theory allowsto qualitatively predict this splitting, without one having to calculate matrix elements of theHamiltonian. A more general and rigorous discussion is given in the elective master courseSymmetriegroepen.

8.5.1 Transformation of an operator

Consider an operator O and a basis set |ψi〉. The matrix representation of O in this basis is

[Oij]=[〈ψi| O |ψj〉

](8.51)

Now consider a base transformation R so that

|φk〉 = R |ψi〉 |ψi〉 = R−1 |φk〉 (8.52)

The matrix elements of O become

Oij = 〈ψi| O |ψj〉 = 〈φk| ROR−1 |φl〉 = (ROR−1)kl (8.53)

Hence, the effect of a transformation R on an operator O is ROR−1.The equation above holds true for any transformation. We may now look for symmetrytranformations S of the operator O such that

SOS−1 = O (8.54)

191


These form a group, the symmetry group of the operator. In the following, we wil consider thisoperator to be the total energy operator of the system, the Hamiltonian.

8.5.2 Essential and accidental degeneracy

Consider a Hamiltonian H with symmetry group G. Assume that an eigenvalue E of thisHamiltonian is found with corresponding eigenfunction |ψ〉

H |ψ〉 = E |ψ〉 (8.55)

We now apply a symmetry transformation to this equation

SH |ψ〉 = E S |ψ〉⇒ SHS−1S |ψ〉 = E S |ψ〉

⇒ HS |ψ〉 = E S |ψ〉 (8.56)

which holds for any S ∈ G. Hence, we find that S |ψ〉 is an eigenfunction of H with the sameeigenvalue E. This already indicates that degeneracy may exist. However, for degeneracy toactually occur, |ψ〉 and S |ψ〉 should be linearly independent functions.

As for all S ∈ G, S |ψ〉 is an eigenfuction of H with eigenvalue E, all linear combinations ofthese eigenfunctions are also eigenfunctions of H with the same eigenvalue E. Consequently,Equation (8.56) holds true for P(l)

ii |ψ〉, a particular linear combination of S |ψ〉. Now we mayapply the projection operators for all irreps of G on the known eigenfunction and determinethe linearly independent eigenfunctions with this eigenvalue, and thus the true degeneracy ofthe energy level. Hereunder, we discuss the various possibilities

• ∀i, P(l)ii |ψ〉 = 0: H has no eigenfunctions that are basis functions of irrep Γ(l) with the

eigenfunction E.

• P(l)ii |ψ〉 = b(l)ψi | f

(l)i 〉 and dl = 1: the basis function | f (l)i 〉 of Γ(l) is an eigenfunction of

H with eigenvalue E. As Γ(l) is one-dimensional, no other eigenfunctions of H with thesame eigenvalue E can be found in this way.

• P(l)ii |ψ〉 = b(l)ψi | f

(l)i 〉 and dl > 1: the basis function | f (l)i 〉 of Γ(l) is an eigenfunction of H.

Also for this eigenfunction, applying Equation (8.56), we find again that for all S ∈ GS | f (l)i 〉 is an eigenfunction of H with eigenvalue E. In particular, P(l)

ij | f(l)i 〉 = | f

(l)j 〉 is an

eigenfunction of H with eigenvalue E, that is linearly independent from | f (l)i 〉. In thisway we find dl linearly independent eigenfunctions of H with the same eigenvalue E.

In most cases, we will find only one l for which P(l)ii |ψ〉 6= 0. Thus it makes perfectly sense to

label the (energy) eigenvalues of H with the irreps of its symmetry group G. The degeneracyof energy levels is then determined by the dimension of the corresponding irrep. In such case,

192


all degeneracy of the system is imposed by symmetry, this is called essential degeneracy.

It is, however, also possible that P(l)ii |ψ〉 6= 0 for more than one l. In that case the degeneracy

of the eigenvalue E is the sum of the dimensions of all irreps for which P(l)ii |ψ〉 6= 0. The

coincidence of energy eigenvalues corresponding with different irreps is called accidentaldegeneracy. It is not imposed by symmetry and can very often be lifted by changing one(or more) parameter(s) of the quantum mechanical system, leaving its symmetry intact. Anaccidental degeneracy that occurs systematically (e.g. the degeneracy of the energy levels ofthe hydrogen atom with different l but the same n) indicates that the Hamiltonian has moresymmetry transformations (a larger symmetry group) than those that have been recognized sofar, and G is a subgroup of the true symmetry group of H.

8.5.3 Restricted representation

Consider a subgroup g ⊂ G, so that ∀Sg ∈ g : Sg ∈ G. We now consider a representation Γ(S)of G with S ∈ G with characters χ(S). This representation can be restricted to g by consideringonly Γ(Sg) with Sg ∈ g, and the characters of this restricted representation are χ(Sg).

8.5.4 Symmetry reduction and lifting of degeneracy

Consider an undisturbed Hamiltonian H(0) with symmetry group G. Now a perturbation εH(1)

is applied. The total Hamiltonian H = H(0) + εH(1) will in general have lower symmetry thanH(0), its symmetry group g is a subgroup of G. We now further assume that no accidental, butonly essential degeneracy occurs. Hence, each eigenvalue of H(0) may be labeled with an irrepΓ(l) of G and is dl-fold degenerate. The same holds true for H : each of its energy eigenvaluesmay be labeled with an irrep γ(k) of g and is dk-fold degenerate. As the dimensions of irreps inthe subgroup g are in general smaller than those in G, we may already anticipate that certainenergy levels will have to split up. This problem is discussed more thoroughly below.

Now consider a particular eigenvalue E(l)0 of H(0), corresponding with the irrep Γ(l). In order

to find out how this energy level splits up when the perturbation is applied, we restrict thisirrep of G to g. In general, the resulting representation is not an irrep of g, but can be reducedin irreps of g, following the recipe of Equation (8.41)

Γ(l)(Sg) = ∑k

akγ(k)(Sg) ∀Sg ∈ g (8.57)

This reduction determines in how many sublevels the considered energy level splits afterapplying the perturbation and what the remaining degeneracy of each resulting sublevel is. Wedo not have to calculate a single Hamiltonian matrix element to arrive at this result. It shouldbe noted, though, that the information we obtain in this way is purely qualitative: group theorydoes not allow to determine the magnitude of the splittings, nor the relative ordering in energy

193


of the various sublevels.Throughout this course, we have seen numerous examples of symmetry lowering that leads tosplitting of energy levels (and spectral lines). Two notable examples are the Stark effect (exercisein Chapter 4) and the Zeeman effect (Chapter 3) which both correspond to a symmetry loweringfrom the symmetry of a free atom (three-dimensional rotation group, an example of a groupwith h = ∞), to uniaxial symmetry (C∞v or D∞h , also a groups with h = ∞).

Illustration: Methane molecule with a basis of s functionsThe symmetry group of methane is Td, it has one carbon atom in the center and 4 hydrogenatoms at the corners of a regular tetrahedron. The character table of Td has been derived inan exercise above.

Td E 3C2 8C3 6S4 6σd

A1 1 1 1 1 1A2 1 1 1 −1 −1E 2 2 −1 0 0T1 3 −1 0 1 −1T2 3 −1 0 −1 1

We consider the basis spanned by the five s functions centered on the atoms of this moleculesC, sH1 , sH2 , sH3 , sH4 . The characters of the representation Γ generated by this basis, are

χ(E) = 5 χ(C2) = 1 χ(C3) = 2 χ(S4) = 1 χ(σd) = 3

The reduction of this representation in Td is

Γ = 2ATd1 ⊕ TTd

2 (8.58)

When only linear combinations of these five basis functions are considered, this molecule hasthree energy levels, two of them are non-degenerate, while one has a threefold degeneracy.

Consider now that this molecule is slightly distorted along one of its C − H bonds. Thisdistortion may be the result of applying a strain or electric field along this bond axis, or simplyby replacing one of the H atoms by another atom, or even another hydrogen isotope. By thisdistorsion, the symmetry of the molecule is lowered to C3v. We now have to reduce Γ in C3v

Γ = 3AC3v1 ⊕ EC3v (8.59)

and find 4 distinct energy levels, 3 non-degenerate and one twofold degenerate. By lookingat the reduction of the irreps of Td in C3v , we get further insight in the energy level structureof the distorted molecule. C3v retains following symmetry elements of Td : E, 2 out of the 8 C3

rotations, and 3 out of the 6 σd mirror planes, that are labeled σv in C3v.

194

8.6. SYMMETRY PROPERTIES

Hence we find

ATd1 = AC3v

1 ATd2 = AC3v

2 ETd = EC3v

TTd1 = AC3v

2 ⊕ EC3v TTd2 = AC3v

1 ⊕ EC3v

If a small distortion is applied, the two A1 energy levels remain essentially unchanged, whilethe T1 level splits up into a non-degenerate and a doubly degenerate level. As the perturbationis small, the energy splitting it causes will also be small.

8.6 Symmetry properties

8.6.1 Transformation of p orbitals

The hydrogenic p orbitals are of the form7

px = f (r)x, py = f (r)y, pz = f (r)z (8.60)

Since every symmetry operation in a point group preserves the distance between two points,the function f (r) is invariant with respect to these operations. We can therefore consider thebasis (x, y, z) instead. In other words: the basis (px,py,pz) is of the same symmetry as the basis(x,y,z).We now want to know how this basis transforms under the symmetry operations of the C3v

point group, i.e. we want to determine which symmetry species it spans. We define x and y asin Fig. 8.3 and z as being orthogonal to the plane, so that (~ex,~ey,~ez) forms a right-handed axissystem.

Figure 8.3: Definition of functions x and y [A & F]. One can imagine, e.g., NH3 with the H atoms at thecorners of the triangle and the nitrogen atom in the center, but lifted out of the plane of the paper, alongthe z axis (cf. Fig. 8.2 .)

7The pz orbital is the spherical harmonic Y1,0, the px and py orbitals are linear combinations of Y1,1 and Y1,−1:

px =1√2(Y1,−1 −Y1,1) py =

i√2(Y1,−1 + Y1,1)

195


There are three classes (E, 2C3 and 3σv ) and it suffices to determine the representation matrixfor one member of each class to determine the characters:

1. The identity operation E of course has the 3× 3 identity matrix as representation.

2. Consider C+3 (counterclockwise rotation about z axis over 2π

3 ) for the class 2C3:

C+3

xyz

=

−12 x +

√3

2 y−√

32 x− 1

2 yz

=

−12

√3

2 0

−√

32 − 1

2 00 0 1

x

yz

(8.61)

3. For the 3σv class, consider σv (reflection with respect to the mirror plane yz):

σv

xyz

=

−xyz

=

−1 0 00 1 00 0 1

x

yz

(8.62)

Thus we find the characters χ(E) = 3 , χ(2C3) = 0 and χ(3σv) = 1 . Note that for the reductionwe want to make, we only need the characters, so it is sufficient to determine the diagonalelements of the matrix representations of one symmetry operation in each class: we do notneed the full matrix representation.

We can now reduce the representation using Eq. (8.41):

aA1 =16[1× 1× 3 + 2× 1× 0 + 3× 1× 1] = 1

aA2 =16[1× 1× 3 + 2× 1× 0 + 3× (−1)× 1] = 0

aE =16[1× 2× 3 + 2× (−1)× 0 + 3× 0× 1] = 1

The reduction therefore is Γ = A1 ⊕ E . Since z is invariant under all symmetry operations, itmust span the irrep A1 . Consequently, x and y must span the irrep E. The formal procedure offinding a symmetry-adapted basis using projection operators yields the same result. This alsoimplies that the basis (px,py,pz) spans A1 ⊕ E, with pz spanning A1 and (px,py) spanning E.

Now assume we would like to determine the symmetry species spanned by the d orbitals,which have the form

dxy = xy f (r) dyz = yz f (r) dzx = zx f (r)

(8.63)

dx2−y2 = (x2 − y2) f (r) dz2 = (3z2 − r2) f (r)

One option would be to go through the same procedure all over again, that is to determine thecharacters of the classes, perform the reduction in irreps and determine the symmetry-adapted

196


bases by means of projection operators. However, the (not purely r-dependent parts of the) dorbitals are products of the (not purely r-dependent parts of the) p orbitals, and it turns out thatit is possible to use the information we have already obtained on the latter. Put more generally:if it is known which symmetry species are spanned by a basis ( f1, f2, f3,. . . ), it is possible todeduce which symmetry species are spanned by a basis built from their products. The nextsection deals with this matter.

8.6.2 Decomposition of direct-product bases

It can be proven that, if the functions f (l)i (i = 1,. . . ,dl) form a basis that spans irrep Γ(l) withdimension dl , and the functions f (l

′)j′ (j’ = 1,. . . ,d′l) form a basis that spans irrep Γ(l′) with

dimension d′l , then the product functions f (l)i f (l′)

j′ form a basis that spans a representation withdimension dld′l . This representation is called the direct-product representation. The character of asymmetry operation in this representation, χ(R), is the product of the corresponding charactersof the irreps:

χ(R) = χ(l)(R)χ(l′)(R) (8.64)

These (easily determinable) characters allow reduction of the direct-product representation intoirreps.

Example: reduction of the basis (xz, yz) for the C3v point groupThe (xz, yz) basis is the direct product of the bases (x, y) and z, which span the irreps E andA1 respectively. The characters of these irreps are (2, -1, 0) and (1, 1, 1) respectively, so thatthe characters of the direct-product representation are (2, -1, 0) × (1, 1, 1) ≡ (2×1,-1×1,0×1) =(2, -1, 0), which again are the characters of E. This is denoted as

A1 ⊗ E = E (8.65)

8.6.3 Calculation of integrals

It can be shown that, if f (l)i is a function of a basis spanning irrep Γ(l), and f (l′)

j is a function ofa basis spanning irrep Γ(l′), then, for a symmetric range of integration, it holds that∫ [

f (l)i

]∗f (l′)

j dτ ∝ δll′δij (8.66)

An integration over all space, e.g., is an example of a symmetric range of integration for theangular coordinates. Note that Eq. (8.66)

• implies that any two basis functions belonging to different irreps or any two differentrows (or columns) of the same irrep necessarily are orthogonal.

• does not necessarily yield a non-zero result in the case i = j and l = l′, as the integral canaccidentally be zero (that is: the proportionality constant due to the integration over theradial coordinate can be zero).

197


• does not yield the actual value of the integral if it is not necessarily zero.

A consequence of Eq. (8.66) is that an integral, for a symmetric range of integration, mustvanish unless the integrand is a basis function for the totally symmetric irrep (A1). Integrals ofthe type ∫ [

f (l)i

]∗f (l′)

j dτ (8.67)

will yield zero if the reduction of the direct-product Γ(l) ⊗ Γ(l′) does not contain A1 (or itsequivalent). Among these integrals are the direct overlap integrals between wavefunctions ondifferent atoms 〈ψa| φb〉.

For integrals of the type ∫ [f (l)i

]∗f (l′)

j f (l′′)

k dτ (8.68)

we first reduce the direct product of Γ(l) and Γ(l′) : Γ(l) × Γ(l′) = ∑k akΓ(k) . Next, we reduce thedirect products Γ(k) × Γ(l′′) for all k occuring in that expansion. If A1 does not occur in any ofthese expansions, the integral necessarily is zero. This is a very useful and important result asintegrals of the type (8.68) are simply operator matrix elements in quantum mechanics:

< a|Ω|b >=∫

ψ∗a Ω φb dτ (8.69)

with Ω some operator: Ω = H yields matrix elements of the Hamiltonian of the system, Ω = ~µ

yields dipole transition matrix elements,. . .

It is further interesting to note that the Hamiltonian of a quantum mechanical system remainsinvariant under all symmetry operations of the system (this actually defines the symmetryoperations of a quantum mechanical system). This means that the operator H spans the totallysymmetric representation of the symmetry group A1. On the other hand, Ω = 1 yields theoverlap integral between the two wavefunctions. Obviously, 1 also spans A1. Consequently,when calculating Hamiltonian and direct overlap matrices, zeros appear in the same positionsin the matrix (at the same matrix elements).

As a final remark, we note that if ψa and φb are basis function belonging to a particular row(column) of an irrep of the symmetry group, the stronger theorem (8.66) can be used fordetermining whether or not integrals of the type (8.68) are zero.

198


Example: overlap of orbitals in NH3

Determine which orbitals of nitrogen in NH3 can overlap with the symmetry-adapted linearcombinations of hydrogen s orbitals.

We have found a symmetry-adapted basis (s1,s2,s3,s4), with s1 = sN , s2 = sA + sB + sC, s3 =

2sA− sB− sC and s4 = sB− sC , and established that both s1 and s2 span A1 while (s3,s4) spansE. On the other hand, we found that a nitrogen pz orbital spans A1 and (px,py) spans E . Since

A1 × A1 = A1

E× A1 = A1 × E = E

E× E = A1 + A2 + E

we find that there can only be overlap between following combinations of nitrogen (left) andhydrogen (right) orbitals

• (s or pz) and s2.

• (px or py) and (s3 or s4).

Adopting the definitions of the x, y, z reference frame in Fig. 8.3 and considering the labelingof the H atoms in Fig. 8.2, we see that px overlaps with s4 and py with s3. This result canalso be obtained by determining to which row (column) of E the basis functions belong viaprojection.

199

8.7. CHARACTER TABLES FOR MOLECULAR SYMMETRY GROUPS

8.7 Character tables for molecular symmetry groups

200


201


202


203


204

Appendixak(lm,l’m’)

l l’ m m’ k = 0 k = 2s s 0 0 1s p 0 ±1 1

0 0 1s d 0 ±2 1

0 ±1 10 0 1

s f 0 ±3 10 ±2 10 ±1 10 0 1

p p ±1 ±1 1 1±1 0 1 -2 ×1/250 0 1 4

p d ±1 ±2 1 2±1 ±1 1 -1±1 0 1 -2 ×1/350 ±2 1 -40 ±1 1 20 0 1 4

p f ±1 ±3 1 5±1 ±2 1 0±1 ±1 1 -3±1 0 1 -4 ×1/750 ±3 1 -100 ±2 1 00 ±1 1 60 0 1 8

ak(lm,l’m’)

l l’ m m’ k = 0 k = 2 k = 4 k = 6d d ±2 ±2 1 4 1

±2 ±1 1 -2 -4±2 0 1 -4 ×1/49 6 ×1/441±1 ±1 1 1 16±1 0 1 2 -240 0 1 4 36

d f ±2 ±3 1 10 3±2 ±2 1 0 -7±2 ±1 1 -6 1±2 0 1 -8 6±1 ±3 1 -5 -12±1 ±2 1 0 28±1 ±1 1 3 ×1/105 -4 ×1/693±1 0 1 4 -240 ±3 1 -10 180 ±2 1 0 -420 ±1 1 6 60 0 1 8 36

f f ±3 ±3 1 25 9 1±3 ±2 1 0 -21 -6±3 ±1 1 -15 3 15±3 0 1 -20 18 -20±2 ±2 1 0 ×1/225 49 ×1/1089 36 ×1/7361.64±2 ±1 1 0 -7 -90±2 0 1 0 -42 120±1 ±1 1 9 1 225±1 0 1 12 6 -3000 0 1 16 36 400

bk(lm,l’m’)

l l’ m m’ k = 1 k = 3 k = 5s p 0 ±1 1 ×1/3

0 0 1s f 0 ±3 1

0 ±2 1 ×1/70 ±1 10 0 1

p d ±1 ±2 6 3±1 ±1 3 9±1 0 1 180 ±2 0 ×1/15 15 ×1/2450 ±1 3 240 0 4 27±1 ∓2 0 45±1 ∓1 0 30

d f ±2 ±3 15 10 1±2 ±2 5 20 5±2 ±1 1 24 15±2 0 0 20 35±1 ±3 0 25 7±1 ±2 10 15 24±1 ±1 8 2 50±1 0 3 ×1/35 2 ×1/315 800 ±3 0 25 280 ±2 0 0 630 ±1 6 9 900 ±0 9 16 100±2 ∓3 0 0 210±2 ∓2 0 0 126±2 ∓1 0 10 70±1 ∓3 0 0 84±1 ∓2 0 25 112±1 ∓1 0 15 105

bk(lm,l’m’)

l l’ m m’ k = 0 k = 2 k = 4 k = 6s s 0 0 1s d 0 ±2 1

0 ±1 1 ×1/50 0 1

p p ±1 ±1 1 1±1 0 0 3 ×1/250 0 1 4±1 ∓1 0 6

p f ±1 ±3 45 1±1 ±2 30 3±1 ±1 18 6±1 ±0 9 100 ±3 0 70 ±2 15 ×1/175 12 ×1/1890 ±1 24 150 0 27 16±1 ∓3 0 28±1 ∓2 0 21±1 ∓1 3 15

d d ±2 ±2 1 4 1±2 ±1 0 6 5±2 0 0 4 15±1 ±1 1 1 16±1 0 0 1 ×1/49 30 ×1/4410 0 1 4 36±2 ∓2 0 0 70±2 ∓1 0 0 35±1 ∓1 0 6 40

bk(lm,l’m’)

l l’ m m’ k = 0 k = 2 k = 4 k = 6f f ±3 ±3 1 25 9 1

±3 ±2 0 25 30 7±3 ±1 0 10 54 28±3 0 0 0 63 84±2 ±2 1 0 49 36±2 ±1 0 15 32 105±2 0 0 20 3 224±1 ±1 1 9 ×1/49 1 ×1/1089 225 ×1/7361.64±1 0 0 2 15 3500 0 1 16 36 400±3 ∓3 0 0 0 924±3 ∓2 0 0 0 462±3 ∓1 0 0 42 210±2 ∓2 0 0 70 504±2 ∓1 0 0 14 378±1 ∓1 0 24 40 420

introduction to atomic and molecular physics -...

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