Quantum Mechanics

School of Physical Sciences

PH 502Wavemechanics & Quantum Physics

COURSE SCRIPT

Dr. Peter Blümler

Peter Blümler

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Peter Blümler

Peter Blümler

Peter Blümler

Dr. P. Blümler: PH 502 Wave Mechanics Syllabus page: 1

© Dr. P. Blümler, School of Physical Sciences, University of Canterbury, UK

Syllabus for Part II (Dr. Blümler)

7. The Foundations and Postulates of Quantum Mechanics:7.1 Philosophy and Concept of Axiomatic Physics

7.2 Operators and Eigenfunctions:7.2.1 Properties of Operators7.2.2 Eigenfunctions and Eigenvalues7.2.3 Linear Combinations7.2.4 Orthogonality, Normalisation and Orthonormality7.2.5 Hermiticity7.2.6 Commutators7.2.7 List of Important Operators

7.3 Postulates (Axioms) of Quantum Mechanics

7.3.1 Wavefunction7.3.2 Physical Systems7.3.3 Schrödinger’s Equation7.3.4 Probability and Expectation Value

7.4 Uncertainty Principle (principle of indeterminacy)

7.4 Dirac Notation

8. Confined Particles8.1 Recapitulation: 1D-Problem ‘particle in a box’

E.1 Excursus: Separation of variables

8.2 2D-Problem: Particle in a Square Well8.2.1 Example Quantum Corrals

8.3 3D-Problem ‘particle in a real box’, Degeneracy

9. The Harmonic Oscillator9.1 Classical Description

9.2 1D Harmonic Oscillator in QM9.2.1 Solution of Hermite’s Differential Equation9.2.2 Correspondence Principle

9.3 The 2D Harmonic Oscillator


10. Rotational Motion10.1 Classical Rotation in a 2D Plane

10.2 Some Quantum-Mechanical Considerations

E2: Excursus: Co-ordinate Transforms of the Laplacian OperatorE2.1 Method 1: Explicit Solution for 2D Circular Co-ordinatesE2.2 Method 2: Curvilinear Co-ordinate Transforms

10.3 Exact Solution for the Rigid Rotator

10.4 Particle Rotating on a Sphere

10.5 Angular Momentum

11. The Hydrogenic Atom11.1 Motion in a Coulomb field

11.2 Solution of the Radial Differential Equation

11.3 The Complete Solution: Atomic Orbitals

11.4 Linear Combinations: Hybrid Orbitals

11.5 Quantum numbers, Orbital Shapes and Degeneracy11.5.1 The Periodic Table of Elements

11.6 Electron Spin11.5.2 Selection Rules and Atomic Spectra

Dr. P. Blümler: PH 502 Wave Mechanics Useful physical constants page: 2


Useful constants

Planck constant: h 6.626076⋅10-34 Js = 4.135669⋅10-15 eV s

π=

2hh 1.054573⋅10-34 Js = 6.582122⋅10-16 eV s

Speed of light (in vac.): c 2.99792458⋅108 m/s

Electron mass: me 9.109390⋅10-31 kg510.9991 keV/c2

Proton mass: mp 1.672623⋅10-27 kg938.2723 MeV/c2

Neutron mass: mn 1.674929⋅10-27 kg939.5656 MeV/c2

Fundamental charge: e 1.602177⋅10-19 C

Compton wavelength:cm

h

e=λ C 2.42631058⋅10-12 m

Boltzmann constant: kB 1.380658⋅10-23 J/K

Vacuum permittivity 0ε ( )Jm/C10854188.8 212−⋅

04πε ( )Jm/C10112650.1 210−⋅

combinations:

hc = 1.9864⋅10-25 Jm = 1239.8 eV nmch = 3.1615⋅10-26 Jm = 197.33 eV nm

Dr. P. Blümler: PH 502 Wave Mechanics Greek Alphabet/Units page: 3


Units

The Greek alphabet

Α α alpha Ν ν nu

Β β beta Ξ ξ xi

∆ δ delta Ο ο omicron

Γ γ gamma Π π pi

Ε ε epsilon Ρ ρ rho

Ζ ζ zeta Σ σ, ς sigma

Η η eta Τ τ tau

Θ θ, ϑ theta Υ υ upsilon

Ι ι iota Φ φ, ϕ phi

Κ κ kappa Χ χ chi

Λ λ lambda Ψ ψ psi

Μ µ mu Ω ω omega

Prefix Exponent Symbol Prefix Exponent Symbol

deci -1 d deca 1 dacenti -2 c hecto 2 hmilli -3 m kilo 3 k

micro -6 µ mega 6 Mnano -9 n giga 9 Gpico -12 p tera 12 T

femto -15 f peta 15 Patto -18 a exa 18 E

zepto -21 z zetta 21 Zyocto -24 y yotta 24 Y

1Å = 10-10 m

Dr. P. Blümler: PH 502 Wave Mechanics Course Summary / Spine page: 4


COURSE SPINE

What is that?

This section is designed to give you a coarse outline of the learning outcomes of each section. Youmight find this useful for exam preparations and revision.

From the PDF-File you can always link to

Blue colour provides links to the particular section!

The following symbols mean:

Link to detailed discussion

Recap of material from previous courses

Mathematical detail, method or solution

Illustration in a MAPLE program

Workshop or exercise

Recap pervious knowledge!

Classical wave mechanics (PH 301)

Qualitative description of quantum mechanics (PH 301)

Partial differential equations (PH 501)



7. The Foundations and Postulatesof Quantum Mechanics:

7.2 Operators and Eigenfunctions

Definitions and examples

To know:1. Properties of operators, notation2. Eigenfunctions and eigenvalues, degeneracy, physical meaning3. Linear combination4. Orthogonality, normalisation and orthonormality5. Hermiticity, real eigenvalues6. Commutators

7. Important operators: UKHLLLpppzyx ˆ,ˆ,ˆ,ˆ,ˆ,ˆ,ˆ,ˆ,ˆˆˆˆ zyxzyx,,,

Exercises and workshops

7.3/7.4 The Postulates of Quantum Mechanics:

Definitions and examples

To know:1. Wavefunction, probability interpretation, normalisation2. Hermitian operators and physical observables3. Expectation values and deductions4. Schödinger equations5. Uncertainty principle (as in (7.4.1))


Dr. P. Blümler: PH 502 Wave Mechanics 7.2 Operators page: 6


7.2. Operators and Eigenfunctions:

What is an operator?

An operator is a symbol that tells you to do something to whatever follows the symbol.

More mathematically it can be understood as a symbol that represents a rule to transform a followingfunction into another function.

The notation in this script for operators is A and an expression like fA will be referred to as „A

operating on f “. For rigorousity we will very often use symbols like A representing a vector-operator. The underline shall simply remind you that this operator may act differently on individualco-ordinates (or vector components).

Comparison: Scalars, Functions, Vectors, Matrices, Operators

Scalars and Functions:

A linear function converts a scalar into another:

b = f(a)

Vectors and Matrices:

A matrix is a linear vector function, thus it converts a vector into another:

aMb =

example:

++++

++

=

===

=

333232131

323222121

313212111

3

2

1

333231

232221

131211

3

2

1

amamam

amamam

amamam

a

a

a

mmm

mmm

mmm

am

b

b

b

kikaMb

Functions and Operators:

An operator is a function of higher order, thus it generally converts a function into another function:

In quantum mechanics, physical observables (e. g. energy, position, linear and angular momentum)are represented mathematically by operators. For example the total energy (kinetic energy K plus

potential energy U) is represented by an operator UKH ˆˆˆ += , where H is called Hamilton-operator or simply Hamiltonian. It is defined as:

)(ˆ)( xaxb M=



Um

+∇−= 22

ˆ2

ˆ hH

We will later derive this àformula. The link to experimentally assessable values of the energy (its

average value) of a system described by H and in a state described by a wavefunction ψ(r) is given

by its àexpectation value H .

rrr d)(ˆ)(*ˆ ψψ= ∫ HH

7.2.1 Properties of Operators:

Most of the properties of operators are obvious. Nevertheless, they are summarised forcompleteness:

Sum and Difference: of two operators A and B :

( )( ) fff

fff

BABA

BABAˆˆˆˆ

ˆˆˆˆ

−=−

+=+

(7.2.1)

The product of two operators is defined by:

[ ]ff BABA ˆˆˆˆ = (7.2.2)

which means: First operate with Bon f then operate with A on the result of fB .

Equality: Two operators are equal if ff BA ˆˆ = holds for all functions f. (7.2.3)

Identity: The identity operator I does nothing (multiplies with 1 or adds 0) (7.2.4)

The nth power of an operator nA is defined as n successive applications of A , e.g.

[ ]fff AAAAA ˆˆˆˆˆ 2 == (7.2.5)

The exponential of an operator )êxp(A is defined by a Taylor expansion:

…++++==!3

ˆ

!2

ˆˆˆ)êxp(

32ˆ AAAIA Ae (7.2.6)

The associative law holds for operators:



( ) ( ) ff CBACBA ˆˆˆˆˆˆ = (7.2.7)

The commutative law does not generally hold for operators!

ff ABBA ˆˆˆˆ ≠ (7.2.8)

see àcommutators!

Examples:

1) Addition of a constant: cxfxf += )()(A

2) Derivative (wrt x): )()(dd

)(ˆ xfxfx

xf ′==D (could be named ‘prime operator’ or ‘dot

operator’ for derivatives wrt time)

3) For D operating on f(x) =exp(ax):

( ) ( ) ( ) f(x)aaxaaxx

axf(x) ==== expexpdd

expˆˆ DD

4) For 2D operating on f(x) =exp(ax2):

( )[ ] ( ) ( )[ ]( )( ) ( ) ( )( ) ( )f(x)axaaxaxaxaaxx

xa

axxaaxx

axf(x)

22222

2222

12expexp2expdd

2

exp2êxpddêxpˆˆˆ

+=+==

=

== DDDDD

2D corresponds to the second derivative 2

22

d

dˆx

=D

HINT: To calculate pure operator expressions it is good advice to include a ‘virtual function’(e.g. f) to remind you that you have to operate!

Example: For the two operators xddˆ ≡D and x≡x (meaning just multiply with x) calculate

Dx-xD ˆˆˆˆ :

[ ] [ ][ ] [ ]( ) 1ˆˆˆândˆˆˆˆ:hence

ˆˆˆˆˆ

ˆˆˆˆˆ

==

′=′=

′+==

Dx-xDDx-xD

xDx:Dx

DxD:xD

ff

fxff

fxfxff

and not 0! This illustrates that operators do not generally commute.

Do not forget to divide by f after your operations!!!



Linear Operators:

Almost all operators encountered in quantum mechanics are linear operators.

Definition: A linear operator has to satisfy the following two conditions:

( )( ) faaf

gfgf

AA

AAAˆˆ

ˆˆˆ

=

+=+(7.2.9)

where a is a constant and f and g are functions.

Examples:

1) D is a linear operator:

( ) ( )

( ) ( ) )(ˆd

)(d)(

dd

)(ând

)(ˆ)(ˆd

)(dd

)(d)()(

dd

)()(ˆ

xfaxxf

axafx

xaf

xgxfxxg

xxf

x g xfx

x g xf

DD

DDD

===

+=+=+=+

2) ( )2ˆ ≡S -an operator that squares- is not linear:

( ) ( )( ) ( ) )(ˆ)()()(ând

)()()()()(2)()()()()(ˆ

222

22222

xfaxfaxafxaf

xgxfxgxgxfxfx g xfx g xf

SS

S

≠==

+≠++=+=+

7.2.2 Eigenfunctions and Eigenvalues:

(German: ‘eigen’ = own, characteristic, individual, belonging to, specific, inherent!)

For aff =A the functions f are called eigenfunctions of the operator A , which scale the functionsby the eigenvalues a.

aff =A (7.2.10)

In other words: An eigenfunction of an operator A is a function f such that the application of A on fgives f again, times a constant, the eigenvalue, a.

If A is a linear operator with an eigenfunction g (and eigenvalue b) , then any multiple of g is also an

eigenfunction of A . (7.2.11)

Proof: ( ) dggcbgccgbgg ==== AAA ˆˆthenˆ)9.2.7(

Hence cg is an eigenfunction with the eigenvalue d = cb

operator eigenfunction eigenfunctioneigenvalue=



Example: axeax =)exp( is eigenfunction of D because axax eaex

=dd

(2axe is not an eigenfunction, because

222

dd axax eaxex

= ).

Degeneracy: If there is more than one (n = 1,2,..m) eigenfunction, nψ , to an eigenvalue, a,

they are called m-fold degenerate.

Deduction: (7.2.12)

Any linear combination of a set of degenerate eigenfunctions is also an eigenfunction.

Proof:

mψψψψ ,...,, 321 are all eigenfunctions of A having the same eigenvalue a.

mna nn ...,2,1forˆ =ψ=ψA . Hence, this state is m-fold degenerate.

The linear combination of a set of degenerate eigenfunctions is given by

qed!ˆˆˆ

111

1

agcaccg

cg

m

nnn

m

n ann

m

nnn

m

nnn

n

=ψ=ψ=ψ=

ψ≡

∑∑∑

∑

== ψ=

=

AAA

Physical meaning:

If a system is in an eigenstate of observable A (i.e., when the wavefunction is an eigenfunction of the

operator A ) then the àexpectation value of A is the eigenvalue of the wavefunction.

7.2.3 Linear Combinations:

A function g (which is no eigenfunction) can be expanded in terms of eigenfunctions nψ of an

operator A ( nnn a ψ=ψA ):

∑ ψ=

n

nncg . (7.2.13)

The new function g is a linear combination of eigenfunctions with the combination coefficients nc

(which might be complex!). (in mathematical terms: The eigenfunctions of an operator form acomplete set, from which other functions can be build up)



From this we can deduce the effect of an operator A on a (general) function g, which is not one ofthe eigenfunctions nψ :

gbacccg

n

nn

n

nb

nn

n

nn

n

nn

n

~ˆˆˆ)13.2.7(

=ψ=ψ=ψ=ψ= ∑∑∑∑≡

AAA

Hence this generates another function g~ , which also is a linear combination of the eigenfunctions;however with different coefficients.

A special case is given we use linear combinations of degenerate eigenfunctions (see àchapter11.4):

7.2.4 Orthogonality, Normalisation and Orthonormality:

Definitions:

1) Orthogonality:

Two functions nψ and mψ are orthogonal if the integral over all space (V) is zero.

0d* =ψψ∫ Vnm (7.2.14)

2) Normalisation:

A function nψ is normalised if the integral over all space (V) is one.

1d* =ψψ∫ Vnn (7.2.15)

3) Orthonormality:

If both properties are fulfilled, the functions are called orthonormal.

=≠

=δ=ψψ∫ mn

mnV nmnm if1

if0d* (7.2.16)

where nmδ is called Kronecker-delta.



7.2.5 Hermiticity:

Definition of Hermiticity:

An operator A is called hermitian, if it fulfils

( )∫∫∫∫ ψψ=ψψ=

ψψ=ψψ VVVV mnmnmnnm dˆdˆdˆdˆ **

*** AAAA * (7.2.17)

Deduction:

A hermitian operator has real eigenvalues (7.2.18)

Proof of Deduction (7.2.18):

A is assumed to be hermitian. 1) ψ=ψ aA

2) ***ˆ ψ=ψ a*A

On 1) we operate with *ψ from the left side and on 2) with ψ from the left,too:

1) ψψ=ψψ=ψψ **A* aaˆ

2) ψψ=ψψ=ψψ=ψψ *******ˆ aaa*A

we can integrate over all space or simply recognise, that the left sides of 1) and 2) are identical

(definition of hermiticity) because A was assumed to be hermitian. Hence the right sides must beidentical, which they are only when aa =* , which is only fulfilled when a is real (qed!).

Physical importance:

All observables (real numbers) correspond to hermitian operators, but -of course- not all hermitiansto observables. (see also à expectation values)

Deduction: (7.2.19)

Eigenfunctions corresponding to different eigenvalues of hermitian operators are orthogonal.


Given: IR∈≠ψ=ψψ=ψ mnmnmmmnnn aaaaaa ,andwithândˆ AA

)1 ∫∫∫ ψψ=ψψ=ψψ VaVaV nmnnnmnm dddˆ *** A

2) ∫∫∫ ψψ=ψψ=ψψ VaVaV mnmmmnmn dddˆ *** A



Subtract the complex conjugate of 2) from 1):

∫∫∫∫ ψψ−ψψ=

ψψ−ψψ VaVaVV mnmnmnmnnm dddˆdˆ ***

*** AA

Because we have assumed A to be hermitian, the two terms on the left side are identical, hence:

( ) 0ddd ***** =ψψ−=ψψ−ψψ ∫∫∫ VaaVaVa nmmnmnmnmn

Because A is hermitian, the eigenvalues an and am are real hence nmm aaa ≠=* , hence the

difference of the eigenvalues is always different from zero and it follows 0d* =ψψ∫ Vnm , which

was to prove (qed!).

7.2.6 Commutators:

The key feature of operators -and the reason why they are used in quantum mechanics- is the factthat they àdo not generally commute (as we will later see with the uncertainty principle). Anotherapproach -originally proposed by Heisenberg- used matrices, which do not commute either.

Because of this it is useful to introduce a quantity which expresses this essential property. It is calledthe commutator operator [ , ] and defined as follows:

Definition of commutator operator: ABBABA ˆˆˆˆ]ˆˆ[ −≡, (7.2.20)

Notation:

If 0]ˆˆ[ =BA, it is said that ‘ A and B commute!’

If 0]ˆˆ[ ≠BA, it is said that ‘ A and B do not commute!’

(à example above)

Commutator algebra:

0]ˆˆ[ =A,A (7.2.21)

]ˆˆ[]ˆˆ[ A,BB,A −= (7.2.22)

for linear operators (àsee 7.2.9): ]ˆˆ[]ˆˆ[ B,AB,A bb = (7.2.23)

( )

]ˆˆ[ˆˆ]ˆˆ[

ˆˆˆˆˆˆˆˆˆˆˆˆ

ˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆ]ˆˆˆ[

C,ABCB,A

ACBCABCABCBA

CABCABACBCBAACBCBACB,A

+=

−+−=

+−−=−=

(7.2.24)

]ˆˆ[ˆˆ]ˆˆ[]ˆˆˆ[ C,BABC,AC,BA += (7.2.25)



( ) ( ) ]ˆˆ[]ˆˆ[ˆˆˆˆˆˆˆˆˆˆˆˆˆˆ]ˆˆˆ[ C,AB,AACABCABAACBCBACB,A +=−−+=+−+=+ (7.2.26)

]ˆˆ[]ˆˆ[]ˆˆˆ[ B,AC,AC,BA +=+ (7.2.27)

7.2.7 List of important operators:

The following table gives a list of typical observables in their classical definition and thecorresponding operators

observable classical QM-operator

position ( )zyx ,,=r ( )zyx ,,=r ( ) ( )zyx ,,ˆ,ˆ,ˆˆ == zyxr

linear momentum

( )zyx ppp ,,=p

==

=

∂∂∂∂∂∂

tztytx

z

y

x

mm

p

p

p

rp &

=∇=∇−=

=

∂∂∂∂∂∂

z

y

x

z

y

x

ii ˆˆ

ˆ

ˆ

ˆ

ˆ hhp

p

p

p

angular momentum( )zyx LLL ,,=L

see also below

prL ×=

=

z

y

x

L

L

L

−

−

−

=

−−−

=

=

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

xy

zx

yz

ixy

zx

yz

z

y

x

yx

xz

zy

h

pypx

pxpz

pzpy

L

L

L

Lˆˆˆˆ

ˆˆˆˆ

ˆˆˆˆ

ˆ

ˆ

ˆ

ˆ

kinetic energy, K ( )22221

zyx pppm

K ++=

∂

∂+

∂

∂+

∂

∂−=∇−=

2

2

2

2

2

222

2

2ˆ

2ˆ

zyxmmhh

K

potential energy, U U(x,y,z), e.g. 22

)( xk

xU = )(ˆ rU=U e.g. 2ˆ2

ˆ rUk

=

total energy, E E = K + UUH ˆˆ

2ˆ 2

2+∇−=

mh

In section à10.5 we will derive

φθ−φ

φθ−φ−

=

=

φ∂∂

φ∂∂

θ∂∂

φ∂∂

θ∂∂

sincotcos

coscotsin

ˆ

ˆ

ˆ

î

z

y

xh

L

L

L

L where φ∂∂

=izh

L is to remember!

à continue to next section (7.3)

Dr. P. Blümler: PH 502 Wave Mechanics 7.3 Postulates page: 15


7.3 Postulates (Axioms) of Quantum Mechanics:The postulates cannot be proven. They are axioms of quantum mechanics. However, they areplausible in the context of defining a fixed framework (mathematical rigorous and withoutcontradiction) for the description of quantum systems.Note that different authors very often use different formulations and ‘orders’ of these postulates!

7.3.1 The wavefunction

POSTULATE 1:

1a) Wavefunction:The state of a physical system is fully described by the wavefunction (=state function):

),...;,( 21 trrψwith r1, r2,... as the co-ordinates of particle 1, 2, ... and time t.

In this lecture we will deal exclusively with single particles and time-independent problems,hence the wavefunction simplifies to:

)(rψ (7.3.1)

The wavefunction must be single-valued, continuous (differentiable) and finite.

1b) There is a probabilistic interpretation (Born-Hypothesis) to the wavefunction:

The amplitude of ψ(x) is related to the probability of finding the particle at x. The amplitude isgiven by ψ*ψ, thus the probability P(r)dr of finding the particle in the volume dr is

probability density function: rrrrr d)()(*d)( ψψ≡P (7.3.2)

and is consequently normalised:

Normalisation condition: ∫

∞

∞−

≡ψψ 1d)()(* rrr (7.3.3)

This can be interpreted as finding the particle somewhere is one.

Comment: However it is customary to also normalise many-particle wavefunctions to 1.



7.3.2 Physical Systems

POSTULATE 2:

2a) A specific state of a system, in which a physical observable A has one specific value a,

must be described by (at least) one eigenfunction ψn to the operator A , of which thenan is an eigenvalue.

nn a ψ=ψA (7.3.4)

Hence, in any physical measurement of the observable A (associated with operator A ) theonly values that will ever be observed are the eigenvalues a, which satisfy eq. (7.3.4).

A particular and extremely important case is the Schrödinger equation, whichcorresponds to the total energy as the observable. (cf. àheuristic derivation)

2b) Every (classical) physical observable corresponds to a linear hermitian operator A inquantum mechanics, because it must yield real eigenvalues! (7.3.5)(see à7.2.5 and à7.2.7)

This postulate captures the central point of quantum mechanics: The values of dynamical variablescan be quantized!

If the system is a state, which is an eigenstate of A with eigenvalue a, then any measurement of thequantity A will result in value a.Although measurements must always yield an eigenvalue, the state doesn’t have to be an eigenstate

of A .

An arbitrary state g can be expanded in a complete set of eigenfunctions (or eigenvectors) of A (seeà linear combinations in 7.2.3).

Hence, iiii a ψ=ψψ Awith

and with eq. (7.2.13) ∑ ψ=n

i

iicg (7.3.6)

where n may go to infinity. In this case we only know that the measurement of A will yield one of thevalues ai, but we do not know which one in particular. However, we do know the probability that

eigenvalue am will occur. It is the squared magnitude of the coefficient 2mc (àprove see next part)



7.3.3 Expectation Values:

POSTULATE 3:

When a system is described by a wavefunction ψ , the mean value of the observable A is equal to

the expectation value A of the corresponding operator A . The expectation value is defined as

∫∫

ψψ

ψψ≡

V

V

d

dˆˆ

*

*AA (7.3.7)

or for normalised wavefunctions: ∫ ψψ≡ Vdˆˆ *AA (7.3.8)

Deductions:

• When mψ is an eigenfunction of the operator A (corresponding to the observable of interest

A), the determination of A always yields one result, the eigenvalue am. (7.3.9)

• When ψ is not an eigenfunction of A , a single measurement of A yields a single result, which is

one of the eigenvalues am of A . The probability that a particular eigenvalue am is measured is

equal to 2mc , where cm is the coefficient of the eigenfunction mψ in the expansion of ψ .(7.3.10)


If ψ is eigenfunction to A ( ψ=ψ aA ) and ψ is normalised then

aVaVaV ∫∫∫ =ψψ=ψψ=ψψ= dddˆˆ ***AA

Hence a series of experiments to determine A will have the average value a.

Proof of Deduction (7.3.10):Now the system is considered to be in no eigenstate, but rather in a state expressed by a linear

combination of eigenfunctions: ∑ ψ=ψn

nnc with nnn a ψ=ψA

∫∑∑∫∑∑

∫∑∑∫ ∑∑

ψψ=ψψ=

ψψ=

ψ

ψ=

VaccVacc

VccVcc

nmm

nn

nmnnmm n

nm

nmm n

nmn

nnm

mm

dd

dˆdˆˆ

****

** AAA

*

In a way it only makes sense to consider A as hermitian, because we want to calculate theexpectation value of a (physical, real) observable. Then the eigenfunctions mψ and nψ are



orthogonal (deduction 2b) and cancel for all mn ≠ . In other words, only the terms with n = mremain. If we further assume normalised wavefunctions, we yield

mm

mmm

mmmmmm

mm acaccVacc ∑∑∫∑ ==ψψ= 2*** dA

This is a sum of eigenvalues for that state, weighted by the square magnitude of the combinationcoefficients of the ‘probing’ wavefunction.

Example: Assume a wavefunction ψ which is a linear combination of two eigenstates

(eigenfunctions) of the linear, hermitian operator A with the (real) eigenvalues a and b. Hence,

bbaa cc ψ+ψ=ψ where aa aψ=ψA and bb bψ=ψA . Calculate the expectation value of A:

[ ] [ ]

[ ] [ ] [ ] [ ]

22

(normal.)1

**

)19.2.7(0

**

)19.2.7(0

**

(normal.)1

**

*

dddd

ddˆˆ

dˆdˆˆ

ba

bbbbababbabaaaaa

bbaabbaabbaabbaa

bbaabbaa

cbca

VcbcVcacVcbcVcac

VbcacccVcccc

VccccV

+=

ψψ+ψψ+ψψ+ψψ=

ψ+ψψ+ψ=ψ+ψψ+ψ=

ψ+ψψ+ψ=ψψ=

====

∫∫∫∫∫∫

∫∫

4342143421443442143421

**

*

AA

AAA

The average (or expectation) value of the observable A is a weighted average of eigenvalues, wherethe weights correspond to the square of the coefficients of the eigenfunctions the wavefunction iscomposed from.

Tools:

If the mean value of A is defined by ∫ ψψ= Vdˆˆ *AA

Consequently the mean value of A2 is defined by ∫ ψψ= Vdˆˆ 2*2 AA

It is also useful to define a mean square deviation: (7.3.11)

( ) ( ) ( ) ( )( )( )

( )222222

*2*2*

22*

22**

*2*2

ˆˆˆˆˆˆ2ˆ

dˆdˆˆ2dˆ

dˆˆˆ2ˆ

dˆˆˆˆˆˆdˆˆˆˆ

dˆˆˆˆdˆˆˆˆ

AAAAAAA

AAAA

AAAA

AAAAAAAAAA

AAAAAAAA

δ≡∆≡−=+−=

ψψ+ψψ−ψψ=

ψ+ψ−ψψ=

ψ+ψ−ψ−ψψ=ψ−ψ−ψ=

ψ−−ψ=ψ−ψ=−

∫∫∫∫

∫∫∫∫

VVV

V

VV

VV

The last definitions are common and useful ! A∆ is also called deviation operator and Aδ root-mean-square-deviation operator or rms-operator.



7.3.4 The Schrödinger-Equation:

The central equation of quantum mechanics has to be accepted as a postulate as well! However, youcan find some plausible and heuristic arguments in an àattachment

POSTULATE 4:

The wavefunction or state function of a system evolves in time according to the

time-dependent Schrödinger-equation:

ttit

∂ψ∂=ψ ),(),(ˆ rr hH (7.3.12)

However, in the following chapters we will exclusively work with time-independent systems, forwhich the time-independent Schrödinger-equation has been already introduced in postulate 2 àeq.(7.3.4):

time-independent Schrödinger-equation:

),(),(ˆ),(ˆ2

),(ˆ 22

tEttm

t rrrr ψ=ψ+ψ∇−=ψ UHh

(7.3.13)

Comment:These are the postulates to an extend necessary to understand the course. Other postulates e.g.concerning the symmetry of the wavefunction or alternatively the Pauli-exclusion principle may bedefined.


Dr. P. Blümler: PH 502 Wave Mechanics Heuristics: Schrödinger’s equation page: 20


Excursus: The Time-Independent Schrödinger Equation

We start with the one-dimensional classical wave equation,

2

2

22

2 1

t

u

vx

u

∂

∂=∂

∂ (1)

By introducing the separation of variables )()(),( tfxtxu ψ=

we obtain 2

2

22

2 )()()()(t

tf

v

x

x

xtf∂

∂ψ=∂

ψ∂ (2)

If we introduce one of the standard wave equation solutions for )(tf such as )exp( tiω (the constantcan be taken care of later in the normalisation), we obtain

)(d

)(d2

2

2

2x

vx

x ψω−=ψ (3)

Now we have an ordinary differential equation describing the spatial amplitude of the matter wave asa function of position. The energy of a particle is the sum of kinetic and potential parts

)(2

2xU

mp

E += (4)

which can be solved for the momentum, p, to obtain

( ))(2 xUEmp −= (5)

Now we can use de Broglie’s formula to get an expression for the wavelength

( ))(2 xUEm

hph

−==λ (6)

The term 22 /vω in equation (3) can be rewritten in terms of λ if we recall that πν=ω 2 andv=νλ :

( )22

2

2

22

2

2 )(244

hxUEm

vv

−=λ

π=νπ=ω (7)



When this result is substituted into equation (3) we obtain the famous time-independentSchrödinger equation

( ) 0)()(2

d

)(d22

2=ψ−+ψ xxUEm

x

x

h(8)

which is almost always written in the form

)()()(d

)(d2 2

22xExxU

x

xm

ψ=ψ+ψ− h (9)

This single-particle one-dimensional equation can easily be extended to the case of three dimensions,where it becomes

)()()()(ˆ2

22

rrrr ψ=ψ+ψ∇− EUm

h(10)

A two-body problem can also be treated by this equation if the mass m is replaced with a àreducedmass µ.

It is important to point out that this analogy with the classical wave equation only goes so far. Wecannot, for instance, derive the àtime-dependent Schrödinger equation in an analogous fashion(for instance, that equation involves the partial first derivative with respect to time instead of thepartial second derivative). In fact, Schrödinger presented his time-independent equation first, andthen went back and àpostulated the more general time-dependent equation.



The Time-Dependent Schrödinger Equation

We are now ready to consider the time-dependent Schrödinger equation. Although we were able toderive the single-particle time-independent Schrödinger equation starting from the classical waveequation and the de Broglie relation, the time-dependent Schrödinger equation cannot be derivedusing elementary methods and is generally given as a àpostulate of quantum mechanics. It is possibleto show that the time-dependent equation is at least reasonable if not derivable, but the argumentsare rather involved

The single-particle three-dimensional time-dependent Schrödinger equation is

),()(),(ˆ2

),( 22

2

2tUt

mt

ti rrrr ψ+ψ∇−=∂

ψ∂ hh (11)

where U is assumed to be a real function and represents the potential energy of the system. WaveMechanics is the branch of quantum mechanics with eq. (11) as its dynamical law. Note thatequation (11) does not yet account for spin or relativistic effects.

Of course the time-dependent equation can be used to derive the time-independent equation. If wewrite the wavefunction as a product of spatial and temporal terms, )()(),( tft rr ψ=ψ , then eq.(11) becomes

)()(ˆ2

)()(

)( 22

2

2rrr ψ

+∇−=

∂

∂ψ U

mtf

t

tfi

hh (12)

or )()(ˆ2)(

1)()(

22

2

2rr

rψ

+∇−

ψ=

∂

∂U

mt

tftf

i hh(13)

Since the left-hand side is a function of f(t) only and the right hand side is a function of r only, thetwo sides must equal a constant. If we tentatively designate this constant E (since the right-hand sideclearly must have the dimensions of energy), then we extract two ordinary differential equations,namely

hiE

t

tftf

−=2

2

d

)(d)(

1(14)

and )()()()(ˆ2

22

rrrr ψ=ψ+ψ∇− EUm

h(15)

The latter equation is once again the àtime-independent Schrödinger equation. The formerequation is easily solved to yield



= h

Etitf exp)( (16)

The Hamiltonian in eq. (15) is a àHermitian operator, and the eigenvalues of a Hermitian operatormust be real, so E is real. This means that the solutions f(t) are purely oscillatory, since f(t) never

changes in magnitude (recall Euler's formula ϑ±ϑ=ϑ± sincos ie i ). Thus if

−ψ=ψ h

Etit exp)(),( rr (17)

then the total wave function ),( trψ differs from )(rψ only by a phase factor of constant magnitude.

There are some interesting consequences of this. First of all, the quantity 2),( trψ is time

independent, as we can easily show:

)()(exp)(exp)(),(),(),( ***2 rrrrrrr ψψ=

−ψ

ψ=ψψ=ψ hh

EtiEtittt (18)

Secondly, the àexpectation value for any time-independent operator is also time-independent, if),( trψ satisfies eq. (17). By the same reasoning applied above,

VVtt d)(ˆ)(d),(ˆ),(ˆ ** rrrr ψψ=ψψ= ∫∫ AAA (19)

For these reasons, wave functions of the form (17) are called stationary states. The state ),( trψ is

„stationary“, but the particle it describes is not!

Of course eq. (17) represents a particular solution to eq. (11). The general solution to eq. (11) willbe a àlinear combination of these particular solutions, i.e.

−ψ=ψ ∑ h

tEict j

jjj exp)(),( rr (20)

Dr. P. Blümler: PH 502 Wave Mechanics 7.4 Uncertainty Principle page: 24


7.4 The Uncertainty Principle:The uncertainty principle (or better: principle of indeterminacy) is NOT strictly a postulate!

A typical (but weak) definition is:

Uncertainty principle (principle of indeterminacy):Denies the ability to determine simultaneously and with arbitrarily high precision, the values ofparticular pairs of observables, named complementary variables. (e.g. position and linearmomentum):

2ˆˆ h

≥δδ xpx (7.4.1)

where 22 ˆˆˆ AAA −=δ defines the uncertainty of the observables (specifically the root mean

square deviations from the mean.

Why is this a weak definition? We still do not know what observables are complementary and whytheir uncertainty is 2/h≥ .

Can one measure an eigenvalue/observable precisely? What are the conditions for this?state a: when A is measured, one determines precisely a.state b: when B is measured, one determines precisely b.

At the moment we claim:

Deduction: (7.4.2)

If two eigenstates are to have simultaneously precisely defined values, their corresponding operators

must commute. That is ABBA ˆˆˆˆ = or using definition à (7.2.20): 0ˆˆ[ =]B,A

Proof of deduction (7.4.2):

eigenstates: ψ=ψ aA and ψ=ψ bB or ψ is simultaneously eigenfunction to A and B .

hence, ψ=ψ=ψ=ψ=ψ=ψ=ψ=ψ ABBBAABA ˆˆˆˆˆˆˆˆ aaabbabb

and ABBA ˆˆˆˆ = or 0ˆˆˆˆ =− ABBA

But what we have deduced so far is only the necessary condition. We have not proven the

sufficient part. This is 0ˆˆˆˆ =− ABBA implies that A and B have the same eigenfunction. Or if

ψ=ψ aA is ψ also eigenfunction to B for 0ˆˆˆˆ =− ABBA ?

Given: ψ=ψ aA , then ψ=ψ=ψ BBAB ˆˆˆˆ aa



because of ABBA ˆˆˆˆ = we can rewrite:

( ) ( )ψ=ψ

ψ=ψ=ψ

BBA

BBAABˆˆˆ

ˆˆˆˆˆ

a

a

when compared with ψ=ψ aA it directly follows, that ψ∝ψB because both equations must be

fulfilled. If we name the proportionality constant b or ψ=ψ bB , we have proven the claim.

Since we have not made any specific reference in our proof, this is true for all linear operators!

Note: For simplification we have been not really mathematically rigorous. The exact proof it is necessary that ψ is member

of a complete set and non-degenerate. Otherwise linear combinations of B are also eigenfunctions that fulfil thecondition.

Now we can appreciate the importance of operators, the fact that they do not generally commuteand why a commutator was defined!!

For a more general description of the uncertainty principle, we want to investigate now the case,that the two operators do not commute and suppose this property in the most general form:

CB,A ˆ]ˆˆ[ i=Furthermore, the system is in an arbitrary state described by ψ , which is not necessarily an

eigenfunction ofA or B .When we measure the system, we get

∫ ψψ= Vdˆˆ *AA and ∫ ψψ= Vdˆˆ *BB

but we are more interested in the spread of values around each of these mean values (or in otherwords, how broad the values are distributed, or how certain the measurement is). Hence, we useanother deviation operator

AAa ˆˆˆ −≡∆ and BBb ˆˆˆ −=∆

and calculate its commutator:

( )( ) ( )( )( ) ( )

CB,AABBA

BAABBAABBABAABBA

AABBBBAABBAAb,a

ˆ]ˆˆ[ˆˆˆˆ

ˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆ

ˆˆˆˆˆˆˆˆ]ˆˆ,ˆˆ[]ˆˆ[

i==−=

+−−−+−−=

−−−−−=−−=∆∆

Because we are only interested in real outcomes, we let A and B be hermitian, hence a∆ and b∆are also hermitian and we define an integral

( ) ( ) ( ) ( )ψ∆−∆αψ∆+∆α=ψ∆−∆αψ∆−∆α= ∫ ba*b*ababa ˆˆ*ˆˆˆˆwheredˆˆ22

iiiViI

with α as a real but arbitrary parameter. This integral relates the two deviations and measures how

much a∆ is different from b∆ (via α) and it is always greater then zero (because of the square).

( ) ( )( ) ( )( ) 0dˆˆ*ˆdˆˆ*ˆdˆˆ2

≥ψ∆−∆αψ∆+ψ∆−∆αψ∆α=ψ∆−∆α= ∫∫∫ ViiViViI ba*bba*aba

now we use the hermiticity of a∆ and b∆ :



( )( ) ( )( )( )( ) [ ]

0ˆˆˆ]ˆˆ[ˆˆ

dˆˆˆˆˆˆ*dˆˆˆˆ*

dˆˆˆ*dˆˆˆ*

222222

222

≥α+∆+∆α=∆∆α−∆+∆α=

ψ∆∆α+∆∆α−∆+∆αψ=ψ∆−∆α∆+∆αψ=

ψ∆−∆α∆ψ+ψ∆−∆α∆ψα=

∫∫∫∫

Cbab,aba

abbabababa

babbaa

i

ViiVii

ViiViI

This can be rearranged to

0ˆ4

ˆˆ

ˆ2

ˆˆ

2

2

2

2

22 ≥

∆−∆+

∆+α∆=

a

Cb

a

CaI

This statement is true for all real α, but we are searching for the minimum deviation (minimaluncertainty). Because α only scales the first term, the minimum is for this term to vanish or

2ˆ2

ˆ

a

C

∆−=α .

The minimum deviation is hence found for

222

2

2

2

ˆ41ˆˆ

0ˆ4

ˆˆ

Cba

a

Cb

≥∆∆

≥∆

−∆

Because AAa ˆˆˆ −≡∆ , ( ) ( )22222 ˆˆˆˆˆˆˆ AAAAAAa δ≡∆≡−=−=∆

we can rewrite the equation using rms-operators:

( ) ( )CBA

CBA

ˆ21ˆˆ

ˆ41ˆˆ 222

≥δδ

≥δδ

This is a more general form of the uncertainty principle!

Uncertainty principle (principle of indeterminacy):When the commutator of two operators A and B is different from zero,

CB,A ˆ]ˆˆ[ i= (7.4.3)

the observables to these operators are complementary. and hence cannot be determined witharbitrary precision. The highest precision that can be achieved is then given by:

CBA ˆ21ˆˆ ≥δδ (7.4.4)

where 22 ˆˆˆ AAA −=δ defines the uncertainty of the observables.



Examples:1) Calculate the uncertainty of space and associated linear momentum (e.g. in x-direction):

hhhhhhhi

iixix

xixx

xixixx =−=−

∂∂

−∂∂

=∂∂

−∂∂

=]ˆˆ[ p,x

hence h=C and we directly get eq. (7.4.1): 2

ˆˆ h≥δδ xpx

2) Now calculate the uncertainty of spatial observable y and linear momentum in

x-direction: 0]ˆˆ[ =−∂∂

−∂∂

=∂∂

−∂∂

=ixi

yxi

yyxixi

yxhhhhh

p,y

Hence, we can determine both simultaneously and precise. The two operators commute and aretherefore not complementary!


Dr. P. Blümler: PH 502 Wave Mechanics 7.5 Dirac Notation page: 28


7.5 Dirac Bracket Notation:In this optional chapter a very elegant and simple notation is introduced. However, it will be NOTUSED in following chapters to avoid confusion. This part should simply provide you with thenecessary definitions, so that you are enabled to read quantum mechanical literature which often usesthis type of notation.

Dirac’s bracket notation is an abbreviation for the typical integrals in quantum mechanics and we willsee that may postulates can be expressed in a more handy and short style.The basic definition is:

Vnm nm dˆˆ *∫ ψψ≡ AA (7.5.1)

where the symbol n is called a ket and denotes the state expressed by wavefunction nψ . The

complex conjugate of the wavefunction *nψ is denoted by the bra n .

When bras and kets are strung together with an operator between them, like as in the bracket

nm A the integral of eq. (7.5.1) is to be understood. When the operator is simply multiplication

by 1 (or I ) the operator is omitted from the bracket.

Hence the following conditions from the previous postulates become in bracket notation:

ànormalisation condition: 1=nn (7.5.2)

àorthonormality condition: nmmn δ= (7.5.3)

àhermiticity: *ˆˆ mnnm AA = (7.5.4)




8 Confined particles:

Solutions Excursus 1: Separation of variables

To know:1. Solutions for particle in 1D box

…,3,2,1,8

,sin2

222 ==

π=ψ n

ma

hnE

axn

nan

2. Zero-point energy, uncertainty principle3. Separation of variables (approach to 2D/3D problems)4. Degeneracy


To know:

General recipe for solving QM-problems1. Determine potential energy2. Determine boundary conditions3. Find suitable co-ordinate system4. Try to separate Schrödinger equation5. Solve differential equation6. Determine constants/quantum numbers via boundary

conditions, normalisation, zero-point energy.

Dr. P. Blümler: PH 502 Wave Mechanics 8 Confined Particles page: 30


8. Confined Particles

8.1 Recapitulation: 1D-Problem ‘particle in a box’Mechanical analogy: Vibrating string, àsee [1] page 554ff.

Fig. 8.1: Co-ordinate system for the 1D problem.

The Hamiltonian for a particle in the infinitely deep well is:

)(d

d2

ˆ2

22xU

xm- +=

hH (8.1.1a)

with:

><∞≤≤

=axx

axxU

and0for

0for0)( (8.1.1b)

The particle cannot penetrate the infinitely high potentials at x = 0 and x = a. Therefore, thewavefunction is only non-zero for the region defined by U(x) = 0, resulting in the Hamiltonian wealready know for free translational motion. The àSchrödinger equation is then

ψ−=ψ22

2 2

d

d

hmE

x(8.1.2)

with the general solution: 22),sin()cos()( −≡+=ψ hmEkkxBkxAx (8.1.3)

Considering the boundary conditions ψ(0) = ψ(a) = 0 one gets ψ(0) = A = 0 andψ(a) = B sin(ka) = 0. The only non-trivial (particle actually exists) solution is ka being an integermultiple of π .

…,3,2,1, ≡π

= nan

k (8.1.4)

This implies a quantized energy of the particle:

…h,3,2,1,

82 2

22

2

222==

π= n

ma

hn

ma

nEn (8.1.5)

To complete the problem, only B remains to be determined. This is done by ànormalising theproblem. Because the probability to find the particle in the box must be unity, one simply has to solvethe following equation:

0x

a

U(x)



( ) ( )[ ]2

sindsind)()(12

02

4222

0

2

0

* aBBxBxxx

a

axn

nax

axn

aa

nn =−==ψψ≡ ππ

π∫∫ (8.1.6)

with aB 2= the complete solution of (8.1.1) is:

Normalised solution for a particle confined in a 1D potential well:

…,3,2,1,8

,sin2

222 ==

π=ψ n

ma

hnE

axn

nan (8.1.7)

n = 1

n = 2

n = 3

n = 4

ψ ( )xn

Em

ah

[8

/ ]

22

n

ψ ψ ( ) ( )x xn n*

x x0 0a a

Fig. 8.2: 1D-confined particle: Graphical representation of the wavefunctions (left) and the probabilitydensities (right) versus the energy of the state for n = 1, 2, 3 and 4.



8.2 2D-Problem: Particle in a Square WellThe particle is now confined in a rectangular well with linear dimensions a1 and a2 as depicted in Fig.8.3. The mechanical analogy to this problem would be a vibrating rectangular drum or sheet, whoseedges are fixed.

Fig. 8.3: Geometry of the rectangular well. Note the graphical necessity to give infinity a value!

The Hamiltonian for this problem is:

),(2

ˆ2

2

2

22yxU

yxm+

∂

∂+

∂

∂−=

hH (8.2.1a)

with:∞

≤≤≤≤=

otherallfor

0and0for0),( 21 ayax

yxU (8.2.1b)

Hence the àSchrödinger equation is (cf. argument above à(8.1.4)):

),(2),(),(

22

2

2

2yx

mE

y

yx

x

yxψ

−=

∂

ψ∂+

∂

ψ∂

h(8.2.2)

This partial differential equation can only be solved if we can separate the variables ( seeàexcursus E1). We now substitute (8.2.2) with ψ(x,y) = X(x)Y(y) and divide by XY as shown inàexcursus E1:

22 hmEYYXX −=′′+′′ (8.2.3)

0

a1

0

a2

xyU(x,y)∞∞

∞∞

∞

0



Each term must be a constant summing up to the energy term, which we therefore separate as wellE = EX + EY. So (8.2.3) can be separated into two similar parts:

EEEyYmE

yYxXmE

xX YXYX

≡+−=′′−=′′ with)(2

)(and)(2

)(22 hh

(8.2.4)

Each of these equations has the form of à(8.1.2) for the 1D-problem and even the boundaryconditions are the same! So we can readily write the solution as:


+=+=

π

π==ψ

22

22

21

21

2

,

2

2

1

1

21,

8

sinsin2

)()(

2121

2121

a

n

a

nm

hEEE

ayn

axn

aayYxX

Yn

Xnnn

nnnn

(8.2.5)

nn1

21

1

2

2

3

3

4

4

x

y

a1

a2

a) b)

c)

d)

Fig. 8.4: Wavefunctions for a 2D well with a2 = 2 a1: a) ψ11, b) ψ12, c) ψ11, d) matrix with greyshaded pictures for n1,n2 = 1,2,3,4. Dark colours represent negative values, bright colourspositive values.



A typical feature of 2D/3D systems becomes apparent when the box is geometrically square (cubefor 3D, àsee Fig. 8.6), that is a1 = a2 = a. Then the energies are

( )22

212

2

,821

nnam

hE nn += (8.2.6)

That means that ψab and ψba have exactly the same energy, although the wavefunctions are different.The property of different states having the same energy is called degeneracy, and the correspondingstates to this energy are said to be degenerate.

The wavefunctions of degenerate states typically can be transformed by simple geometricaltransformation (like 90° rotations). Thus, degeneracy can always be traced to an aspect of thesystem’s symmetry.



8.2.1 Example Quantum CorralsDon Eigler and co-workers à[2] used the tip of a scanning tunnelling microscope (STM) to placeiron atoms on a copper surface (cf. Fig. 8.5a). After finishing the construction of such quantumcorrals they used the same instrument in a different mode to scan the surface, which exhibitedprobability distributions of a trapped electron.

a) b)

c) Fig. 8.5: STM measurements of quantum corrals: a) Different stages in the assembly of a round

corral, b) the finished round quantum corral with a diameter of 143 Å, c) a rectangular corral.



8.3 3D-problem ‘particle in a real box’We have now all tools to solve the 3D Schrödinger equation

),,(),,(),,(),,(ˆ2

22

zyxEzyxzyxUzyxm

ψ=ψ+ψ∇−h

(8.3.1)

for a particle in a 3D box with the following boundary conditions:

∞

≤≤≤≤≤≤=

otherallfor

0and0and0for0),,( 321 azayax

zyxU (8.3.2)

Derive the following solution of the problem:


…,3,2,1,,with8

sinsinsin8

),,(

32123

23

22

22

21

21

2

,,

3

3

2

2

1

1

321,,

321

321

=

++=

π

π

π=ψ

nnna

n

a

n

a

nm

hE

a

zn

ayn

axn

aaazyx

nnn

nnn

(8.3.3)

Problem 8.1: Find degenerate states in Fig. 8.4d. Can you tell by looking at the displayedprobability density? Find other states for n1,n2 > 4!

Problem 8.2: Assuming the rectangular corral in Fig. 8.5.c was scanned at 4.2K and contains asingle electron, can you estimate the distance of the iron atoms using eq. (8.2.5)?Discuss how realistic your result is and what assumptions might be wrong!

Thermal energy E = kBT.

Comparing the thermal and QM energy one can work out the distance of the STMtip, assuming a particle in a 3D box.

Problem 8.3: How degenerate is (5,6,7) for a particle in a cubic box? (It is not 6)

References:1. Mary L. Boas: „Mathematical Methods in the Physical Sciences“ 2nd ed., Wiley,

New York 1983.

2. M. F. Crommie, C. P. Lutz, D. M. Eigler: „Confinement of electrons to quantumcorrals on a metal surface“, Science 262(5131) (1993) 218-220; M. F. Crommie, C.P. Lutz, D. M. Eigler and E. J. Heller, Physica D 83 (1995) 98-108.



Fig. 8.6: The first energy levels of a partic le in a cubical box with a1 = a2 = a3 = a together with thequantum numbers n1, n2, n3 of the corresponding wavefunctions and the degeneracy of eachenergy level. The ground state energy is E0 = E111 = 3h2/(8ma2).


Dr. P. Blümler: PH 502 Wave Mechanics Excursus 1 page: 38


E1 Excursus: Separation of variables:

The problem satisfies Laplace’s equation

∂

∂+

∂

∂+

∂

∂≡∇=ψ∇

2

2

2

2

2

222 ÔperatorLaplacianthewith0),,(ˆ

zyxzyx (E1.1)

However, we want to restrict it to two dimensions for the moment and try a solution of independentvariables, that is:

)()(),( yYxXyx =ψ (E1.2)

Substituting (E1.2) into (E1.1), we get

0d

)(d)(

d

)(d)(

2

2

2

2=+

y

yYxX

x

xXyY (E1.3)

Note that we can write this now in ordinary rather than partial derivatives, because X depends onlyon x and Y on y!We now divide (E1.3) by XY to separate the terms:

0d

)(d)(

1

d

)(d)(

1

onlyondepends

2

2

onlyondepends

2

2=+

44 344 2144 344 21yx

y

yYyYx

xXxX

(E1.4)

Non-trivial solutions must therefore fulfil the following condition:

0withd

)(d)(

1

d

)(d)(

1 22

2

2

2≥−=−= kk

y

yYyYx

xXxX

(E1.5)

or YkYXkX 22 and =′′−=′′ (E1.6)

The constant k2 is called the separation constant and (E1.6) has the following solutions:

−=

=)exp(

)exp()(and

)cos(

)sin()(

ky

kyyY

kx

kxxX (E1.7)

The solutions (or linear combinations) of the initial problem are therefore:

==ψ

−

−

kxe

kxe

kxe

kxe

yYxXyx

ky

ky

ky

ky

cos

cos

sin

sin

)()(),( (E1.8)



9. The Harmonic Oscillator

9.1 Classical description

Detail Classical Description

To know:

1. Potential energy: 22

xU k= with 2222 2 xmmk νπ=ω=

2. Classical forbidden regime

9.2 1D Harmonic Oscillator in QM

Detail (solution of differential equation)

Detail (discussion of solution, graphs) MAPLE exercise

To know:

1. Schrödinger-Equation: )()(2

)(d

d2

22

2

22xExx

mx

xmψ=ψ

ω+ψ−

h

2. Solutions: Energy and quantum numbers

( ) ( ) …h ,2,1,0with21

21 =ω+=ν+= sshsE

Wavefunction = Gaussian times Hermite-polynomial

xmxhmeN sss hω=νπ=ξξ=ξψ ξ− 2with)()( 22H

3. Zero-point energy 200 ν=⇒= hEs

4. Potential tunneling (QM ↔ Classical)

5. Correspondence principle




9.3 The 2D and 3D Harmonic Oscillator

Detail (separation of variables)


To know:

1. Schrödinger-Equation: )()(2)(ˆ2

22222

rrrr ψ=ψνπ+ψ∇− Emm

h

with ),,( zyxr =

2. Solutions: Energy and quantum numbers*

( )…,3,2,1,0,,with

23

=

ν+++=

zyx

zyxsss

sss

hsssEzyx

Wavefunction = Gaussian times Hermite-polynomial

xmxhmeN sss hω=νπ=ξξ=ξψ ξ− 2with)()( 22H

3. Degeneracy

*for identical oscillation frequencies in all 3 dimensions.

Dr. P. Blümler: PH 502 Wave Mechanics 9 Harmonic Oscillator page: 41


9. The Harmonic Oscillator

9.1 Classical description

Fig. 9.1 The one-dimensional harmonic oscillator. Displacement from equilibrium (x = 0) is denoted byx.

The classical equation of motion of a particle with mass m is given by àHooke’s law (cf. Fig. 9.1):

0or 20 =ω+

−=

xx

kxxm

&&

&&(9.1.1)

with 20ω= mk as the spring constant and ω0 the natural frequency. The energy of the system is then UKE =+=

In the turning points ±x0 the particle comes to a rest (cf. Fig. 9.2) and thus the kinetic energy K = 0

and the energy is entirely potential 202

xUE k== (9.1.3)

For 20

2 xx > , the kinetic energy is negative, which is classically forbidden.

Fig. 9.2: The potential U(x) of the harmonic oscillator. The turning points of the oscillation are atx = ±x0, where U(x0) = E.

For the classical case. the probability PCL(x) of finding the particle in the interval dx around the pointx at any time is



tT

txxP d

2d

d)( 0

0

CL

πω

== (9.1.4)

with T0 as the period. Combining eqn. (9.1.2) and (9.1.3) gives ( ) ( )220

20

2dd xx

tx −ω= and thus

220

0CL

2

1dd

2)(

xxxt

xP−π

=π

ω= (9.1.5)

The probability must be ànormalised (N as a normalisation constant) according to

( ) 12

arcsin2

d2

d)(0

0

0

0

210

00

220

CL ≡=

π

=−π

=−−

−

−∫∫

NxxN

xxxN

xxPx

x

x

x

x

x

(9.1.6)

Thus N = 2 and 22

0

CL 1)(

xxxP

−π= (9.1.7)



9.2 1D Harmonic Oscillator in QM

Since the àpotential energy is: 22222212

21 2)( xmxmkxxU νπ=ω== (9.2.1)

with ω as the angular frequency or ω = 2πν, the Schrödinger equation is derived straight-forwardly:

[ ] 0)(28

)(d

d

)()()()(d

d2

2222

2

2

2

2

22

=ψνπ−π

+ψ

ψ=ψ+ψ−

xxmEh

mx

x

xExxUxxm

h

(9.2.2)

We substitute and simplify with xh

mhE ν

π≡ξν

≡λ 2and2

(9.2.3)

thus (9.2.2) becomes: )()(d

d 22

2ξψλ−=ξψ

ξ−

ξ(9.2.4)

Note: The presented solution uses operators. Brehm and Mullin provide you with analternative way using series expansion (check their pages 244-250).

The left side of equation (9.2.4) ‘naively’ reminds you of ‘equation a2-b2 = (a+b) (a-b)’, but becareful because we are dealing with operators here. So we try:

( )

ψ

−ξ−

ξ=

ψξ−ψ′ξ+ψ−ψ′ξ−ψ ′′=ξψ−ψ′

ξ+

ξ=ψ

ξ−

ξ

ξ+

ξ

1d

d

dd

dd

dd

22

2

2

(9.2.5a)

and analogously

ψ

+ξ−

ξ=ψ

ξ+

ξ

ξ−

ξ1

d

ddd

dd 2

2

2(9.2.5b)

To simplify further we introduce the following operators as abbreviations:

ξ−

ξ≡

ξ+

ξ≡ +

ddând

ddˆ aa (9.2.6)

So eqn. (9.2.5) now look like this:

( ) )(1ˆˆ)(d

d 22

2ξψ+=ξψ

ξ−

ξ+aa (9.2.7a)



( ) )(1ˆˆ)(d

d 22

2ξψ−=ξψ

ξ−

ξ+aa (9.2.7b)

From this equation or the corresponding formulation of the Schrödinger equation in (9.2.4), which isthen (where ψλ(ξ) denotes the wavefunction to the energy term λ)

)()1()(ˆˆ ξψ+λ−=ξψ λλ+aa (9.2.8a)

)()1()(ˆˆ ξψ−λ−=ξψ λλ+aa (9.2.8b)

we see that that for all wavefunctions 2ˆˆˆˆ =− ++ aaaa (9.2.9)

Now we take the Schrödinger equation of (9.2.8b) and operate on both sides with a

)(ˆ)1()(ˆˆˆ ξψ−λ−=ξψ λλ+ aaaa (9.2.10)

and combine it with eq. (9.2.9):

( )

)(ˆ)3()(ˆˆˆ

)(ˆ2)(ˆ)1()(ˆˆˆ

)(ˆ)1()(ˆ2ˆˆ

ξψ−λ−=ξψ

ξψ+ξψ−λ−=ξψ

ξψ−λ−=ξψ−

λλ+

λλλ+

λλ+

aaaa

aaaaa

aaaa

(9.2.11)

This corresponds to the Schrödinger equation (9.2.8) for the energy (eigenvalue) λ-2, or

)()3()(ˆˆ )2()2( ξψ−λ−=ξψ −λ−λ+aa (9.2.12)

Therefore, )(ˆ λψa must be proportional to )2( −λψ . If we repeat the procedure of (9.2.10 -11), thefunctions corresponding to energy levels (λ-4), (λ-6), .... can be created. For this reason theoperator a is called annihilation operator, because operated on ψ it generates the wavefunction

that corresponds to the next lower energy level. Analogously +a is called creation operator,

because it increases the energy level. The pair a and +a are also called ladder operators.

However, annihilation is more useful at the moment, because we know that the decrease of energycannot be continued indefinitely, since kinetic and mean potential energy are positive it cannotbecome negative. Therefore, there must be a minimum energy assigned by λmin. and a applied to

)( minλψ consequently must give 0: 0ˆ )( min =ψ λa . (9.2.13)

We operate on both sides with +a and use eq. (9.2.8b) to set it in relation to the Schrödingerequation:

)(min

)( minmin )1(ˆˆ0 λλ+ ψ−λ−=ψ= aa (9.2.14)

thus, λmin.= 1. As demonstrated above, λ can take integer values differing by 2, so we conclude thatλ = 1, 3, 5, 7, ... or with λ ≡ 2s + 1, where s = 0, 1, 2, 3,... Now were-substitute using the definition in à(9.2.3):



( ) ( ) …h ,2,1,0with2 2

121 =ω+=ν+=

νλ= sshs

hE (9.2.15)

What is left is the determination of the wavefunction, which is now rather simple. Therefore, we takethe wavefunction corresponding to the lowest energy level and all we have to do is to operate on itwith a+ to get the next higher wavefunction and continuation of this procedure gives us any ψs welike. (Note that now s rather then λ is used as a label.)

( ))(

)(ˆ)(

0times)(d

ddd

dd

dd

0

ξψ

ξ−

ξ−

ξ−

ξ−=

ξψ=ξψ

ξξξξ

+

ss

sss

N

N

…

a

(9.2.16)

with Ns as a ànormalisation constant. So all left to determine is ψ0, from which we know it mustsatisfy à(9.2.13) or

ξξ−=ψ=ψψ

=ψ

ξ+

ξ=ψ dlnd

dor0

ddˆ 0

0

000a (9.2.17)

which can readily be solved to be a Gaußian: 200

2)( ξ−=ξψ eN (9.2.18)

ψ1, ψ2, ψ3, etc. can then be found using eq. (9.2.16) and are

[ ]

[ ]

[ ]L

ξ−ξ−ξ=ξψ

ξ−−ξ=ξψ

ξ−ξ=

ξ−ξ−=

ξ−

ξ−

ξ∝ξψ

2exp128)(

2exp24)(

2exp2

2exp2

2exp

dd

)(

23

33

22

22

2

1

22

1

N

N

N

(9.2.19)

the polynomials in squared brackets are known as the Hermite polynomials H1, H2, H3,... whichare summarised in àTable 9.1 together with the normalisation constants Ns, which are obtained by

the requirement 1d)()(* =ξξψξψ∫∞

∞−

.

Normalised solution for the 1D harmonic oscillator:

( ) ( ) …h ,2,1,0with21

21 =ω+=ν+= sshsE (9.2.15)

xmxhmeN sss hω=νπ=ξξ=ξψ ξ− 2with)()( 22H (9.2.20)



Table 9.1: Energy levels Es, quadratic normalisation constants 2sN and Hermite polynomials H s

for various s

s Es [J] 2sN H s(ξ)

02νh

hmπν4 1

12

3 νh

hmπν4

21 ξ2

22

5 νh

hmπν4

81 24 2 −ξ

32

7 νh

hmπν4

481 ξ−ξ 128 3

42

9 νh

hmπν4

3841 124816 24 +ξ−ξ

52

11 νh

hmπν4

38401 ξ+ξ−ξ 12016032 35

....

s ( ) ν+ hs21

hm

ssπν4

!2

1 22

d

d)1( ξ−ξ

ξ− ee

s

ss

It can be shown that:

0)(22)( =+− ′′′ ys(y)yy sss HHH (9.2.21)

0)(2)(2)( 11 =−+ −+ yyysy sss HHH (9.2.22)

)(2)( 1 ysy ss −=′ HH (9.2.23)

Exercise 9.1: Show that the normalisation constants in Tab. 9.1 are correct! (Use:

axxa

2d)exp( 2

0

2 π=−∫

∞

for a > 0)



1

2

0

3

4

5

Eh

[]ν

3/2

5/2

7/2

9/2

11/2

1/2

ψ ξ( ) ψ ξ ψ ξ*( ) ( )s

ξ ξ00

Fig. 9.3: Graphs of the wavefunctions (left) and probability density functions (right) of the 1Dharmonic oscillator for the first 5 energy levels versus dislocation y. The red curves on theright represent the probability density functions of the classical harmonic oscillator (cf. àeq.(9.1.7)).

On the right side of Fig. 9.3 we observe, that with increasing energy or s the quantum mechanical

probability density )(QM ξsP is becoming more and more similar to the classical probability density

)(ξCLP (red curves in Fig. 9.3). This observation is further illustrated in Fig. 9.4, where the twofunctions are compared for E20 = 21hν/2.



P(

)ξ

ξ0

Fig. 9.4: Illustration of the correspondence principle: Black curve: Quantum mechanical probabilitydensity function of a high energy state (ψ20(ξ))2. Red curve: Classical probability densityfunction from à(9.1.7).

The principle that for large quantum numbers (e.g. energies) the quantum mechanical probabilitydistribution function PQM resembles the classical probability distribution function PCL is called‘correspondence principle’ (see Fig. 9.4), it describes the limiting regime where quantum mechanicsmeets classical mechanics (c.f. àde Broglie relation).

correspondence principle: r

rrdd

)()(lim CLQM tPPs

s∝=

∞→(9.2.24)

In other words, the system behaves only classically, when there is sufficient energy.

Exercise 9.2: Demonstrate the correspondence principle of a particle in a 1D-box.




Fig. 9.5: Classical picture of a 2D harmonic oscillator.

The solution of the 2D harmonic oscillator is omitted (see àeq. (8.2.3) and the following treatmentof the 3D harmonic oscillator). The results are:


( ) ( ) ( ) ( )…

hh

,2,1,0,with

21

21

21

21

=

ω++ω+=ν++ν+=

yx

yyxxyyxxss

ss

sshshsEyx

(9.3.1)

22 22)()(),( υ−ξ− υξ=υξψ eeN

yxyxyx ssssss HH (9.3.2)

ymyhm

xmxhm

yy

xx

h

h

ω=νπ=υ

ω=νπ=ξ

2and

2with

yxyx

ssss hm

ssN

yxyxν+ν

π=

+4

!!2

1)(

2 (9.3.3)



E/hν

0

1

2

3

4

5

energylevels

(00)

(01) (10)

(02) (11) (20)

(03) (12) (21) (30)

(04) (13) (22) (31) (40)

quantum numbers( , )s sx y

degeneracy

1

2

3

4

5

Fig. 9.6: Energy diagram for a symmetrical (Ux = Uy = U) two-dimensional harmonic oscillator.Compare with Fig. 9.7.

s1s2

2

2

3

30

0

1

1

Fig. 9.7: Wavefunctions for a symmetrical (Ux = Uy = U) two-dimensional harmonic oscillator. Blackfor negative, white for positive values. Compare the pattern for the degenerate states fromFig. 9.6.



Fig. 9.8: The first four wavefunctions with identical quantum numbers (ψ00, ψ11, ψ22, ψ33) for asymmetric potential. (Note: The states with mixed quantum numbers are not shown and notboth potentials are drawn. The red line is an diagonal average).




The potential shall be given by 2222 22

)( rrr νπ== mk

U (9.4.1)

with 2222 zyxr ++= (9.4.2)

The Schrödinger equation is then:

)()(2)(ˆ2

22222

rrrr ψ=ψνπ+ψ∇− Emm

h(9.4.3)

Completely analogously we substitute as in the 1D case (cf. àeq. (9.2.3)):

( ) ZYXZYX EEEhh

Eλ+λ+λ≡++

ν=

ν≡λ

22(9.4.4)

and: zh

my

hm

xh

m νπ≡ζ

νπ≡υ

νπ≡ξ 2,2,2 (9.4.5)

which yields the Schrödinger equation in the following form:

( ) ),,(),,(),,( 2222

2

2

2

2

2ζυξλψ−=ζυξψζ+υ+ξ−ζυξψ

ζ∂

∂+

υ∂

∂+

ξ∂

∂(9.4.6)

As in the 2D and 3D treatment of a confined particle in zero potential (cf. àchapter 8.2 and 8.3) weassume the solution to be a product of functions X,Y,Z, which depend on one spatial co-ordinateonly:

)()()(),,( ζυξ=ζυξψ ZYX (9.4.7)

we substitute into (9.4.6) and divide by XYZ:

0d

)(d)(

1

d

)(d)(

1

d

)(d)(

1 22

22

22

=

λ+ζ−

ζ

ζζ

+

λ+υ−

υ

υυ

+

λ+ξ−

ξ

ξξ

ZYX ZZ

YY

XX

(9.4.8)

When each term becomes zero the equation is fulfilled, for instance for the first term:

)()(d

)(d

0)()(d

)(d

0d

)(d)(

1

22

22

22

ξλ−=ξξ−ξ

ξ

=ξλ+ξξ−ξ

ξ

=λ+ξ−ξ

ξξ

XXX

XXX

XX

X

X

X

(9.4.9)

This the same expression as eq. (9.2.4), of which we know the results (eq. (9.2.15) and (9.2.16)).Hence, the solution of the Schrödinger equation for the 3D harmonic oscillator is given by:




(for a symmetrical potential*)

( ) ( ) ( ) ( )…,3,2,1,0,,with

23

21

21

21

=

ν+++=ν++ν++ν+=

zyx

zyxzyxsss

sss

hssshshshsEzyx (9.4.10)

222 222)()()(),,( ζ−υ−ξ− ζυξ=ζυξψ eeeN

zyxzyxzyx sssssssss HHH (9.4.11)

zhmyhmxhm νπ=ζνπ=υνπ=ξ 2and,2,2with (9.4.12)

23

)(2 4

!!!2

1

νπ

= ++ hm

sssN

zyxssssss zyxzyx

(9.4.13)

* in eq. (9.4.1) we have assumed a symmetrical potential, represented by a single k or ν.

Fig. 9.9: Energy diagram for the first 4 states of a symmetrical 3D harmonic oscillator.




10. Rotational Motion

10.1/2 Classical Rotation / QM Considerations

Detail Classical Description

To know:1. Linear / Rotational Motion:2. Reduced Mass3. Boundary Conditions

10.3 Exact Solution for the Rigid Rotator

Detail (Co-ordinate transform) Detail (discussion of solution, graphs)

To know:1. Schrödinger-Equation in circular co-ordinates2. Solutions and quantum numbers

…,3,2,1,0with2

1)( ±±±=

π=φψ φ meim

m

hhmL

ImE m

zm == and2

22

3. Uncertainty and m = 0

4. Shape of wavefunctions, auxiliary functions, Lz




10.4 Particle Rotating on a Sphere:

Detail (Co-ordinate transform) Detail (Solution of differential eq.)


To know:1. Schrödinger-Equation in spherical co-ordinates.2. Integration limits for spherical co-ordinates.3. Solutions (spherical harmonics)*

Schrödinger-eq. ),()1(),(ˆ 2 φθ+−=φθΛ lm

lm ll YY

wavefunctions φθ=φθ imlmlm

lm eN )(cos),( PY

energy I

llEl 2)1(

2h+=

4. Quantization conditions lmll ≤≤−= and,2,1,0 …5. Orbital Shapes (simple ones)*, dependence on l, m6. Probability density, (θφ)-dependence7. Degeneracy

* not in detail, but their composition and dependence (the most simple you could remember!)




10.5 Angular Momentum

Detail Detail (discussion of solution, graphs)

To know:

1. Linear definitions, eg. .

−=−= ∂

∂∂∂

yziyzx zyhpzpyL ˆˆˆˆˆ

2. φ∂∂=

izhL and 222 ˆˆ Λ−= hL applications to spherical harmonics

3. Spatial quantization: )1(ˆ += llhL


Dr. P. Blümler: PH 502 Wave Mechanics 10 Rotational Motion page: 57


10. Rotational Motion

10.1 Classical Rotation in a 2D PlaneWe remember the classical expressions for linear and rotational motion, as listed in Table 10.1:

Tab.10.1: Classical expressions for linear and rotational motion in a plane

linear motion rotational motion

co-ordinates Cartesian: r = x,y circular: r, φ

mass m moment of inertia:

∑=i

ii rmI 2

first derivative wrt t linear velocity:

yvxv yx && == ,

rv &=

angular velocity:

πν=ω×

=ω 2,2

vvr

vr

momentum linear momentum:

p = mvangular momentum:

ω=×= ILprL ,

force: linear force:

apF m== &torque:

ω=×= &IFrT

kinetic energy:

mp

mvK22

1 22 ==

IL

IK22

1 22 =ω=

The typical case, where we wish to apply the following treatment, is the rotation of a (at least)diatomic molecule around its centre of mass (Fig. 10.1).

Fig. 10.1: a) Rotation of a diatomic molecule around its centre of mass (CM). b) Without loss ofgenerality the system can be described by the rotation of a reduced mass µ.

µx

y

z

r0

L

φ

a) b)



First we want to reduce the problem to a ‘single mass’ problem. Since the rotation is around thecentre of mass. In good agreement with nature we can assume ‘point-masses’:

2102211 and rrrrmrm +== (10.1.1)

hence: 021

120

21

211211 andor)( r

mmm

rrmm

mrrrmrm

+=

+=−= (10.1.2)

The moment of inertia of the molecule is: 222

211 rmrmI += (10.1.3)

Substitution of (10.1.2) in (10.1.3) gives:

( ) ( )( )

( ) ( )

( )''theaswith

21

2120

20

21

21202

21

1221202

21

21

2202

21

22

1

massreducedmm

mmrI

rmm

mmr

mm

mmmmr

mm

mmr

mm

mmI

+=µµ=

+=

+

+=

++

+=

(10.1.4)

Analogously we can reduce the masses of a many-body problem, without loss of generality!

10.2 Some Quantum-Mechanical ConsiderationsFor a simple start, the particle shall be confined on a circular path (e.g. a bead on a circular wire).Thus moving in zero-potential, and the total energy is kinetic only.

IL

r

Lm

pKE

222

2

20

22=

µ=== (10.2.1)

We can relate this equation to àde Broglie’s law:

λ=

λ=

hrL

hp or (10.2.2)

The consequence of (10.2.2) is that the angular momentum is inversely proportional to thewavelength of the particle. Thus, the shorter the wavelength the higher the angular momentum, thehigher the energy. If λ can take arbitrary values, after one round the particle-wave will have adifferent amplitude at the same position. Therefore, the corresponding wavefunction will havedifferent values at the same position, which is a violation of the first QM-postulate. Hence, only suchλ are quantum mechanically allowed, which form standing waves on the orbit (see Fig. 10.2). Inother words λ has to be an integer multiple of the path length:

λ=π mr02 (10.2.3)



Otherwise the particle-waves could interfere destructivelyand finally vanish. We haven’t specified the quantisationnumber m yet. Because the particle can move clock- andcounter-clock wise we allow positive and negative integers.Like in the case of a particle in a box, we might not allowm = 0, because this would correspond to zero energy, butwe might reconsider this after a more complete quantummechanical treatment.

With the cyclic boundary conditions and principle thoughts,we have already found expressions for the quantized energyof the ‘particle on a ring’, the ‘rigid rotator’, orunrestricted motion on a circular path. All left to do is to

combine àeqs.(10.2.1)-(10.2.3):

…h,3,2,1with

2822

22

20

2

220

2

2

20

22±±±==

π=

λ== m

Im

rI

mrh

I

rh

IL

Em (10.2.4)

Note: Careful that you do not confuse m with mass. At the moment we might call it the azimuthalquantum number !!

10.3 Exact Solution for the Rigid RotatorWe do not have to consider a potential, so the àSchrödinger equation for the rigid rotator isformally the same as for the free particle in two-dimensional Cartesian co-ordinates with theboundary conditions of being constrained to a circular path:

Schrödinger equation for a free particle in 2D

),(2),(),(

),(ˆ22

2

2

22 yx

mE

y

yx

x

yxyx ψ

−=

∂

ψ∂+

∂

ψ∂=ψ∇

h(10.3.1)

However, Cartesian co-ordinates are not a suitable co-ordinate frame. 2D-polar or circular co-ordinates are better suited for the problem, because r = r0 is a constant. Therefore we transform eq.(10.3.1) into circular co-ordinates (r,φ) (see àExcursus 2: eq. (E2.16)) and replace the mass m bythe àreduced mass µ.

Fig. 10.2: a) Unacceptable and b) acceptable wavefunctions



The àSchrödinger equation for the rigid rotator in circular co-ordinates is therefore:

),(2

),(1

),(1

),(ˆ22

2

22 φψµ−=φψ

φ∂

∂+φψ

∂∂

∂∂=φψ∇ r

Er

rr

rr

rrr

h(10.3.2)

This looks more complicated then (10.3.1), but since r is not a variable but a constant r0 (suppliedby the rigidity of the rotator’s axis), it simplifies to (using àeq. (10.2.4) to express E as L2 and m,eq. (10.1.4) for I):

)()()(2

)(d

d

)(2

)(d

d1

22

2

2

20

2

2

22

2

20

φψ−=φψ−=φψµ−

=φψφ

φψµ−

=φψφ

mLEr

E

r

hh

h (10.3.3)

We acknowledge that the partial differentials became ordinary, and that this equation has beensolved already for the free one-dimensional particle (see àpart 5c of John Dore’s part). At themoment we still consider m as unrestricted in value despite our previous considerations.

The solutions to (10.3.3) are (for generality we allow complex solutions):

φ−φ +=φψ imim BeAe)( (10.3.4)

The boundary conditions may be expressed as:

)2()( φ+πψ=φψ (10.3.5)

φ−π−φπφ−φ

φ+π−φ+πφ−φ

+=+

+=+

imimimimimim

imimimim

eBeeAeBeAe

BeAeBeAe

22

)2()2(

(10.3.6)

This can only be fulfilled if m is an integer (because 12 =π±ime for m = 0, ±1, ±2, ...), which alsosatisfies our previous considerations, and the wavefunction reproduces itself on the next circuit.Energy and angular momentum are quantized.

Since we allow m to have negative and positive integer values, let us compare the cases m and -m ineq. (10.3.4). Obviously they correspond to the same state when A and B are swapped. The

corresponding energies ( 2mEm ∝ ) are the same, so the states are degenerate. However, the

angular momentum L is pointing in opposite directions (+z and -z). Hence, the two states correspondto rotation in opposite directions. Because the same considerations hold for the wavefunction witheither A = 0 or B = 0, we want to eliminate this unnecessary ambiguity and rewrite eq. (10.3.4) as:

φ=φψ imm Ae)( (10.3.7)



This also allows the determination of A by normalisation (Note the integration limits for φ):

12d1dd)()(* 22

0

22

0

22

0

=π=φ=φ=φφψφψ ∫∫∫ππ

φφ−π

AAeeA imim (10.3.8)

Hence, π

=21A and we could readily write down the solution. However, there is still the unsolved

case m = 0! For the discussion of a particle in a box we have used two arguments, as to why thecorresponding state doesn’t exist: 1) It wouldn’t agree with àHeisenberg’s uncertainty principle(since the particle was confined, it has had a finite position range, so the momentum/energy must notbecome zero). 2) The corresponding wavefunction (cf. àeq. (8.1.7)) would become zero, whichwould violate the àfirst postulate.

Let’s now consider the same arguments for the case of the rigid rotator. The second can be solved

directly and 0)(21

0 ≠=φψπ

. It is constant but not zero, so it exists! The first argument based on

the fact, that the particle is confined. Actually it is not, it moves free on its circular path in zeropotential. The idea that it is constrained to the ring originates from the viewpoint of its path in 2D-Cartesian co-ordinates, but we have solved the problem in circular co-ordinates, and we will findexpressions for its wavefunction for any φ. (The àSchrödinger equation is identical with that of anunconstrained particle in 1D Cartesian co-ordinates.)

Therefore, we allow m = 0 as an eigenvalue of the Schrödinger equation1.

Normalised solution of the Schrödinger equation for the rigid rotator

…,3,2,1,0with2

1)( ±±±=

π=φψ φ meim

m (10.3.9)

hhmL

ImE m

zm == and2

22 (10.3.10)

From the solutions we directly observe, that each state is doubly degenerate due to rotation inopposite directions (Lz pointing up- or downwards) but with the same speed (energy). Except form = 0 where Lz is zero and the question of direction doesn’t arise.

There are several ways to depict the wavefunctions. A problem arises, because they are complex

)sin()cos( φ+φ=φ mimeim . We get somewhat around the problem by plotting the real part only(since the imaginary is just 90° phase shifted) as done in Fig. 10.3.

1 The treatment in àeq. (10.2.4) might still be treated as correct, because we didn’t explicitly consider circular co-ordinates at that point. However, it has no particular meaning, because we didn’t solve the Schrödinger equationfor 2D-Cartesian co-ordinates.



However, this gives a wrong picture, because it suggests, that the particle is distributed non-uniformlyover the circle. As shown in àeq. (10.3.8) the probability distribution is independent of φ and henceuniform. Hence, when the particle is in a state of definite angular momentum it’s distribution iscompletely uniform!

A better representation is shown in àFig. 10.4, where an auxiliary function is constructed byconnecting each point of the perimeter to the centre and to mark on this radius a length proportionalto the amplitude of the wavefunction (see Fig. 10.4a). In other words, this is the plot of the amplitudein polar co-ordinates. Hence, for ψ0 we would obtain a circle.

This procedure allows us to project the wavefunction into a plane and the overlay of its real andimaginary part (see Fig. 10.4c and d). The sum of their magnitudes than somehow resembles theprobability distribution (note overlapping parts in Fig. 10.4c/d, they correspond to the gaps from acircle).

Fig. 10.3:

Graphical representation of thewavefunction (real part) for the rigidrotator (m = 0,1,2,3). Thewavefunctions for m = -1, -2, -3look identical, but the direction ofrotation alters (small arrow) and theangular momentum (bold arrow)points down.



+

_

_ +

+

+

__+

+

_

_

Fig. 10.4: a) Construction of the auxiliary function from the real part of ψ±1. b) Same for ψ±2. c)The auxiliary function from a). Dark colour denotes the real part, brighter grey theimaginary part. d) Same as c) for ψ±2 or b) respectively.

10.4 Particle Rotating on a Sphere:

The assumption, that we can describe rotational motion of moleculesby moving on a circular path is quite naive. Because how should themolecules in a macroscopic sample know which plane is designated.Typically there is no preferred direction. Therefore, a description byrotation on the surface of a sphere is a more likely approximation.

We still assume the molecule to be completely rigid along its bond (r= r0 = const.). From our experience with other two-dimensionalquantum systems, we expect 2 quantum numbers necessary todescribe the system, because there are two degrees of freedom inrotation (see Fig. 10.5).

Fig. 10.5:Schematic representation ofa particle moving on thesurface of a rigid sphere. Thearrows indicate the twodegrees of freedom inrotation.

a) b)

c) d)



Hence, we have to express the Schrödinger equation in 3D polar (spherical) co-ordinates with awavefunction that depends on both spherical angles (and the radius), ψ(r, θ, φ). Fig. 10.6 showstheir interdependence.

Fig. 10.6: Relation between Cartesian and spherical co-ordinates and their range.

In the excursus to this chapter it is shown, how the Laplacian operator is converted into spherical co-ordinates (see àeq. (E2.2.24))

2

2

2222

22

sin

1sin

sin

11ˆφ∂

∂

θ+

θ∂∂

θθ∂∂

θ+

∂∂

∂∂

=∇rrr

rrr

Following the same assumptions as in section à10.3, we can directly derive the àSchrödingerequation (r = r0 = const., mass: m → µ):

),(2

),(sin

1),(sin

sin

122

2

220

20

φθψµ

−=φθψφ∂

∂

θ+φθψ

θ∂∂

θθ∂∂

θ hE

rr(10.4.1)

After multiplication with θ220 sinr and setting Ir =µ 2

0 the equation becomes:

),(sin2

),(),(sinsin 222

2φθψθ−=φθψ

φ∂

∂+φθψ

θ∂∂

θθ∂∂

θhIE

(10.4.2)

We try a àseparation of variables by expressing the wavefunction by a product of functions thatdepend on one variable only:

)()(),( φΦθΘ=φθψ (10.4.3)



We substitute in (10.4.2) and divide by ΘΦ :

θ−=φΦφφΦ

+θΘ

θθ

θθΘθ 2

22

2sin

2)(

d

d)(

1)(

dd

sindd

)(sin

hIE

(10.4.4)

We rearrange the terms that depend on θ or φ on either side of the equation and substitute the

energy term by 22 −≡λ hIE :

)(d

d)(

1sin)(

dd

sindd

)(sin

2

22 φΦ

φφΦ−=θλ+θΘ

θθ

θθΘθ

(10.4.5)

To be solvable each side must correspond to a constant! We take a closer look at the right term andrecognise it as identical to àeq. (10.3.3) when choosing the separation constant to be m2. So wehave to solve two equations depending on θ and φ:

)()(d

d 22

2φΦ−=φΦ

φm (10.4.6)

and 22sin)(dd

sindd

)(sin

m=θλ+θΘ

θθ

θθΘθ

(10.4.7)

The first we have already solved in the previous section and can copy its solution from à(10.3.9):

…,2,1,0with2

1)( ±±=

π=φΦ φ meim

m (10.4.8)

Equation (10.4.7) however is new and more complicated. First we multiply with θθΘ 2sin/)( andrearrange:

0)(sin

)(dd

sindd

sin1

2

2=θΘ

θ−λ+θΘ

θθ

θθm

(10.4.9)

Now we want to get rid of all the sinθ terms. Because we have to take its derivative we choose forsubstitution a new function

ζ=ζ cos)(P (10.4.10)

and with θθ−=ζ dsind and the Pythagoras relation ζ−=ζ 22 sin1 equation (10.4.9) becomes:

( ) 0)(1d

)(d1

dd

2

22 =ζ

ζ−−λ+

ζζ

ζ−ζ

PmP

(10.4.11)



or ( ) 0)(1

)(2)(12

22 =ζ

ζ−−λ+ζ′ζ−ζ′′ζ− P

mPP (10.4.12)

This type of equation is called associated Legendre differential equation (see e.g. à[1] page 504)

with eigenvalues )1( +=λ ll and has solutions )(ζmlP for lml ≤≤− and l as an integer called

the associated Legendre functions. They have the following general form of (Legendre) polynomialsof degree l and order m:

( ) ( )l

ml

ml

lmm

l l

m

1d

d

!2

1)1()( 22 2

−ζζ

ζ−−=ζ+

+P (10.4.13)

and can be normalised by the following integral over the entire range (N as the normalisationconstant):

( )

( ))!(2

)!(12

1)!(

)!(

122

d)( 21

1

2

ml

mllN

ml

ml

lNN

ml

ml

+

−+=

=−+

+=ζζ∫

−

P

(10.4.14)

Note that ζ = cosθ has a range from -1 to 1!

(We might derive this solution either by series expansion or by expressing them by the mth derivativeof the Legendre functions Pl(ζ), which can be generated by ladder operators of the type

ζ−ζ−=ζ

+ ldd2 )1(b and ζ+ζ−=

ζl

dd2 )1(b , completely analogous to the previous treatment of

the harmonic oscillator. However the solution is quite lengthy due to the interdependence of m and l)

We have seen that the θ dependent part of the Laplacian operator in spherical co-ordinatesproduces differential equations of the Legendre type. It is common practice and useful to denote thepart with the angular derivatives of the Laplacian by a new operator, which we will call Legendreoperator (or Legendrian)2:

2

2

22

sin

1sin

sin1ˆ

φ∂

∂

θ+

θ∂∂

θθ∂∂

θ=Λ (10.4.15)

2 The Legendrian measures the curvature of a function relative to the surface of a sphere. Thus its application to awavefunction that represents a sphere is zero, because a sphere in spherical co-ordinates is flat.



With these tools we rewrite the Schrödinger equation in à(10.4.1):

)()(),(and2

with)()()()(ˆ2

2 φΦθΘ=φθψ=λφΦθΘλ−=φΦθΘΛhIE

(10.4.16)

We have seen that …,2,1,0with2

1)( ±±=

π=φΦ φ meim

m

Re-substituting ζ = cosθ the solutions of eq. (10.4.9) are:

lmllm ≤≤−θ=θΘ with)(cos)( P (10.4.17)

Because m is limited by l, it is often referred to as ml. As eigenvalues for the associated Legendreequation we found with the condition that l is an integer:

…h

,2,1,0with)1(2

2=+==λ lll

IE(10.4.18)

So the complete wavefunction is given by (using the ànormalisation condition in (10.4.14))

( )

lmll

eml

mll lm

imlm

≤≤−=

φθ≡θ+π

−+=φΦθΘ=φθψ φ

and,2,1,0with

),()(cos)!(4

)!(12)()(),(

…

YP(10.4.19)

This lengthy expression is typically abbreviated by (again) new functions ),( φθlmY called spherical

harmonics. We summarise the mathematical treatment:

Particle rotating on a sphere:

Schrödinger equation: ),()1(),(ˆ 2 φθ+−=φθΛ lm

lm ll YY (10.4.20)

Wavefunction (spherical harmonics):( ) φθ

+π

−+=φθ iml

ml

m eml

mll)(cos

)!(4

)!(12),( PY (10.4.21)

Energy: I

llEl 2)1(

2h+= (10.4.22)

Quantization conditions: lmll ≤≤−= and,2,1,0 … (10.4.23)



Table 10.2: Composition of the spherical harmonics ),( φθlmY for the quantum numbers

l ≤ 3 and -l ≤ m ≤ +l

l m El * )(ζlmP ** )(cosθl

mP lmN *** ),( φθl

mY

0 0 0 1 1π2

1π4

1

1 0 2 ζ θcosπ3

21 θπ cos

43

1 ±1 2 21 ζ−∓ θsin∓ π6

41 φ±

π θ iesin83∓

2 0 6 ( )13 221 −ζ ( )1cos3 2

21 −θ

π5

21 ( )1cos3 2

165 −θπ

2 ±1 6 213 ζ−ζ∓ θθsincos3∓ π30

121 φ±

π θθ iesincos815∓

2 ±2 6 ( )213 ζ− θ2sin3π30

241 φ±

π θ 223215 sin ie

3 0 12 ( )ζ−ζ 35 321 θθ− cos)sin52( 2

21

π7

21 θθ−π cos)sin52( 2

167

3 ±1 12 ( ) 2223 115 ζ−−ζ∓ ( ) θ−θ sin1cos5 2

23∓

π21

121 ( ) φ±

π θ−θ iesin1cos5 26421∓

3 ±2 12 ( )ζζ− 2115 θθcossin15 2π

210120

1 φ±π θθ 22

32105 cossin ie

3 ±3 12 ( ) 232115 ζ−∓ θ3sin15∓ π35

1201 φ±

π θ 336435 sin ie∓

* in multiples of ( )I22h = l(l+1)

** from àeq. (10.4.13)

***from àeq. (10.4.14)

So far for math, but what do the spherical harmonics look like? We have solved a problem for aparticle on a sphere, and the result is a (sometimes) complex wavefunction on the surface of thesphere. How can one visualise this? The problem is identical with projecting the globe of the earthonto a 2D-map! We are used to projections of the earth onto a cylinder, so that the poles becomeas large as the equator, although they are points (pseudo Mercator-projectons). In our case thelatitudes correspond to θ from 0 (south pole) to π (north pole) and the longitudes to φ from 0(Greenwich) to 2π (again Greenwich, so UK is at the left and right edge). Although this is a rathercontorted view of reality, you might find it instructive (see àFig. 10.7).

These wavefunctions projected on a sphere are more realistic, but you are restricted to the view on asingle hemisphere (see àFig. 10.8).

Another possibility is to plot auxiliary functions (like in the 2D case, cf. àFig. 10.4), but this time wewill plot the amplitude of the wavefunctions in spherical co-ordinates (see àFig. 10.9). Like in the2D-case this disables us to distinguish the sign of φ, but it is a more realistic representation of thewavefunction. (To understand the relation between Fig. 10.7, 10.8 and 10.9, find the nodes -nownodal planes- in àFig. 10.7 and mark them on the projections onto a globe, as in àFig. 10.8.Realise that φ is spanning the entire diameter, so that nodes separated by π -or half the image widths



in àFig. 10.7- correspond to the same nodal plane in àFig. 10.8 and à10.9. For a bettercomparison switch to àgraph 10.10, where all three representations are compared next to eachother.)

Finally we want to calculate the probability density function:

)()()()(

),(),(),(

**

*QM

φφθΘ⋅φφθΘ=

φθφθ=φθ

mlmm

lm

lm

lmlmP YY

(10.4.24)

Consulting àeq. (10.4.8) We directly see that 1)()(* =φφ⋅φφ mm and from àeq. (10.4.13) that

)()(* θΘ=θΘ lm

lm , because )(θΘ l

m is always real. Hence, (10.4.24) becomes:

( )2QM )()( θΘ=θ lmlmP (10.4.25)

We have just proven that the probability density function is (like in the 1D case) homogeneous with

respect to φ. In other words, )(QM θlmP is not homogeneous with respect to θ.

àFig. 10.10 shows )(QM θlmP in the three different visualisation modes we have used (map

corresponds to àFig. 10.7, globe to à10.8, aux. to à10.9). Note the similarity of the probabilitydensity functions along the diagonal of Fig. 10.10! The auxiliary functions look almost alike (e.g.

)()()( QM33

QM22

QM11 θ≈θ≈θ ±±± PPP ), however they are not identical! They have the same symmetry,

but since they correspond to different energy levels, their curvature along θ must increase with thequantum numbers, as it can be clearly seen form the map-graphs! Also note that the nodal planesnow correspond to black colour and since we have taken the square of the wavefunction, we mightno longer consider them as planes but nodal cones (planes are then cones for θ=π/2 and 0)!

The last question we are going to consider is the energy term scheme (see àFig. 10.11) for theparticle on a sphere and the degeneracy of its levels. As shown by àeq. (10.4.22), the energy of thesystem doesn’t depend on m, although m is always spanning a range -l ≤ m ≤ +l resulting in thesame energy. Hence the degeneracy of each energy level is 2 times l (each m is covering the range ofl twice) +1 (for the zero):

Degeneracy of El is 2l+1 for l = 0, 1, 2, .... (10.4.26)

Adjacent levels have a energy difference of :

[ ]I

lI

llllEEE ll

22

1 )1(2

)1()2)(1(hh

+=+−++=−=∆ + (10.4.27)

Note: Some authors also introduce a so-called rotational constant B. It is defined as

hcBllcBllEcI

B l )1(2)1(hence4

+=π+=π

= hh(10.4.28)

with c the speed of light.



),( φθmlY m = 0 m = +1 m = -1 m = +2 m = -2 m = +3 m = -3

Real Real Imag Real Imag Real Imag Real Imag Real Imag Real Imaginary

l = 0

l = 1

l = 2

l = 3

Fig. 10.7: Cylindrical projections of the wavefunctions ),( φθmlY of a particle on a sphere (pseudo Mercator-projections). The vertical axis is θ from 0 to

π , the horizontal axis is φ from 0 to 2π . White colours depict positive and black negative amplitudes. Nodes are in mid-grey! (Note that this is arather distorted view of the wavefunctions).



),( φθmlY m = 0 m = +1 m = -1 m = +2 m = -2 m = +3 m = -3

Real Real Imag Real Imag Real Imag Real Imag Real Imag Real Imaginary

l = 0φ = 0

θ = 0

l = 1

l = 2

l = 3

Fig. 10.8: Projections of the wavefunctions from Fig. 10.7 on a sphere. The view is from θ = 30° and φ = 270° from 0 to 2π (on the globe of the earth thiswould correspond to somewhere over the southern Pacific). White colours depict positive and black negative amplitudes. Nodes are in mid-grey!



Fig. 10.9: Projection of the magnitude of the wavefunctions ),( φθmlY in spherical co-ordinates, generating a 3D-version of the auxiliary functions as

described in àFig. 10.4.

),( φθmlY m = 0 m = ±1 m = ±2 m = ±3

Real Real Imaginary Real Imaginary Real Imaginary

l = 0

l = 1

l = 2

l = 3



)(QM θlmP m = 0 m = ±1 m = ±2 m = ±3

map globe aux. map globe aux. map globe aux. map globe aux.

l = 0 φ = 0

θ = 0

l = 1

l = 2

l = 3

Fig. 10.10: The probability density functions of the particle on a sphere. As shown in àeq. (10.4.25) it is only a function of θ. The three differentvisualisations are directly compared. Map corresponds to the pseudo Mercator-projection of àFig. 10.7. Globe is a plot of the wavefunction onthe sphere as in àFig. 10.8, and aux. is the plot of the amplitude in spherical co-ordinates (cf. àFig. 10.9). Note that the levels gray scale nowfrom 0 to max, so the nodal planes/cones are black!



02

6

12

20

energylevels

quantum number

degeneracy

l = 0l = 1

l = 2

l = 3

l = 4

2 +1ll l(+1)

13

5

7

9

EI h [2 / ]2

Fig. 10.11: Energy diagram for a particle on a sphere.

10.5 Angular MomentumBefore we finish the chapter, we want to investigate the angular momentum of unrestricted motion ona sphere, or free 3D rotation. The angular momentum is defined as (see àTab. 10.1)

−−−

==

×

=×=

=

xy

zx

yz

zyx

zyx

z

y

x

z

y

x

ypxp

xpzp

zpyp

ppp

zyx

p

p

p

z

y

x

L

L

L eeeprL (10.5.1)

Also from Tab. 10.1 we see that the kinetic energy can be expressed in terms of angular momentum.Since we didn’t have to consider potential energy in àsection 10.4, we can express the total energyas kinetic energy.

IL

KE2

2== (10.5.2)

with 2222zyx LLLL ++=⋅= LL (10.5.3)

In the context of eigen-equations and the entire context of quantum-mechanics, it is more useful toexpress angular momentum Lx, Ly, Lz and L2as operators. The straightforward way is to rewrite(10.5.1) for each component of L by the operators of position and linear momentum:

−=−= ∂

∂∂∂

yziyzx zyhpzpyL ˆˆˆˆˆ (10.5.4)

( )zxizxy xz ∂

∂∂∂ −=−= hpxpzL ˆˆˆˆˆ (10.5.5)

−=−= ∂

∂∂∂

xyixyz yxhpypxL ˆˆˆˆˆ (10.5.6)



As before we rewrite these equations in polar notation, which is more suitable to the problem. FromàFig. 10.6 we get the expressions for the Cartesian co-ordinates:

θ=φθ=φθ=

cos

sinsincossin

rz

ryrx

(10.5.7)

Hence,

( )

φθ−φθ=

φθ−θ=

θ−φθ=

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

xyiz

zxiy

yzix

rr

rr

rr

sinsincossinˆ

cossincosˆ

cossinsinˆ

h

h

h

L

L

L

(10.5.8)

Now, we express the partial derivatives of the polar co-ordinates by the Cartesian (you might usethe recipe of àE2.2 - the other way around) or work it out:

yr

xr

zz

yy

xx

zr

yr

xr

zz

yy

xx

zyxzrz

yry

xrx

r

∂∂

+φθ+∂∂

φθ−=∂∂

φ∂∂

+∂∂

φ∂∂

+∂∂

φ∂∂

=φ∂∂

∂∂

θ−∂∂

φθ+∂∂

φθ=∂∂

θ∂∂

+∂∂

θ∂∂

+∂∂

θ∂∂

=θ∂∂

∂∂

θ+∂∂

φθ+∂∂

φθ=∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

=∂∂

cossinsinsin

sinsincoscoscos

cossinsincossin

(10.5.9)

After some reformulation (for which we do not need r∂∂ , because L must not depend on r), weget:

φ∂∂

φ∂∂

θ∂∂

φ∂∂

θ∂∂

=

φθ−φ=

φθ−φ−=

iz

iy

ix

h

h

h

L

L

L

ˆ

sincotcosˆ

coscotsinˆ

(10.5.10)

Check the result by substituting the expressions of (10.5.9) into (10.5.10) and try to derive (10.5.8).This is easier then the other way around ( θθ=θ sin/coscot ).

We are left with deriving an equation for )ˆ(ˆ)ˆ(ˆ)ˆ(ˆ)ˆ(ˆˆ2zzyyxx LLLLLLLLL ++== . When we do

by using the expressions in (10.5.10), this becomes quite lengthy, but try it. The approach here isdifferent by looking at àeq. (10.5.2). If the system has no potential energy (like in àsection 10.4),the total energy is purely kinetic and expressed by L2. In operator notation we conclude, that the

following eigen-equation must be valid for 2L :

),(2),(ˆ2 φθψ=φθψ IEL (10.5.11)



When we compare this with àeq. (10.4.16) -substituting the terms-

),(2

),(ˆ2

2 φθψ−=φθψΛhIE

(10.5.12)

we find (cf. à(10.4.15), the same result we would get using the explicit expressions in (10.5.10)) :

φ∂

∂

θ+

θ∂∂

θθ∂∂

θ

=Λ−=

2

2

2

2222

sin

1sin

sin1ˆˆ

ihhL (10.5.13)

Since we have solved eq. (10.5.12) in section 10.4 (àsee (10.4.20)), all we have to do is tosubstitute:

…hh ,2,1,0with),()1(),(ˆ),(ˆ 2222 =φθ+=φθΛ−=φθ lll lm

lm

lm YYYL (10.5.14)

So the eigenvalues of 2L with respect to the spherical-harmonics are simply )1(2 +llh .

A further look to eq. (10.5.10) reveals another term, we came across in solving the Schrödinger-

equation for the rotators. It is φ∂∂=

izhL and when applied to the Φ -part of the spherical-

harmonics (see eq. (10.4.8)) we directly get:

lmlmemN

ei

Ni

N

lm

imml

imml

ml

lmz

≤≤−φθ=π

θΘ=

πφ∂

∂θΘ=φΦ

φ∂∂

θΘ=φθ

φ

φ

with),(2

)(

21

)()()(),(ˆ

Y

Y

hh

hhL

(10.5.15)

We have just seen that the magnitude of the total angular momentum L is quantized having

eigenvalues )1( +llh and that additionally zL is quantized by eigenvalues mh .

As long as we do not assign the problem to specific Cartesian direction - for instance by applying anadditional (magnetic) field- there is nothing specific about the z co-ordinate. The simple dependenceon φ occurs because we had chosen it to be perpendicular to z. We will later see, that this choicemakes the x and y component of the angular momentum non-measurable, as their contribution to theprobability density function vanishes (see àeq. (10.4.25)). However, if we would have chosen x ory to lie perpendicular to φ the angular momentum relative to them would turn out to be quantized,and the remaining other would be non-determinable.

These results are really important, because we can now conclude, that any system with a sphericallysymmetric potential U(r) -hence a central force problem, which can be separated from the angularterms- has solutions for its angular terms as above.

Graphically we can summarise these considerations. As an example we consider l = 2 and hence

6)1(ˆ hh =+= llL . If a preferred direction exists (as in the presence of an external field) or

we simply chose it to be z, mz h=L with m = -2, -1, 0, 1, 2. Thus 5 different directions of theangular momentum vector exist relative to the z-axis, as visualised in Fig. 10.12. Note that a



completely parallel or anti-parallel alignment to the z-axis is not possible, since its maximumprojection on the preferred axis is hl , while having a length of lll hh >+ )1( . (Only for very large

values of l -as for macroscopic objects- is lll ≈+ )1( , so that the object is able to rotate solely

around a preferred axis).

As discussed above (and proven later) if we select z as a preferred axis, x- and y-component of theangular momentum are uncertain or have equal probability to lie on cones (see Fig. 10.13) with half-apex-angles ϑ determined by ( ))1(/arccos +=ϑ llm . (Do not think of the angular momentum

vector sweeping out one of the cones, but simply lying at some position on the cones).


h

6h

h−

h2−

h2

zm = +2

m = -2

m = +1

m = -1

m = 06h0

Fig. 10.12: Vector model of angular momentum:

Possible alignments of L with respect to the z-axis.

z

Fig. 10.13: Cones of location for the angularmomentum vector



Excursus 2: Co-ordinate Transforms of the LaplacianOperator:

E2.1 Method 1: Explicit solution for 2D circular co-ordinates.

The Laplacian is2

2

2

22ˆ

yx ∂

∂+∂

∂=∇ (E2.1)

We want to express it in circular co-ordinates (r,φ) (see àFig. 10.1b)

φ=φ= sinandcos ryrx (E2.2)

with 22222 or yxryxr +=+= (E2.3)

and xy

xy

arctanortancossin

=φφ=φφ

= (E2.4)

The total differential of an arbitrary function f is:

φφ∂

∂+

∂∂

= dddf

rrf

f (E2.5)

hence x

fxr

rf

xf

∂φ∂

φ∂∂

+∂∂

∂∂

=∂∂

(E2.6)

and y

fyr

rf

yf

∂φ∂

φ∂∂

+∂∂

∂∂

=∂∂

(E2.7)

so we have to find expressions for (using eq. (E2.2) to (E2.4))

( ) ( ) φ==+

=+=+∂∂

=∂∂ −

cos21

222

21222122

rx

yx

xyxxyx

xxr

(E2.8)

rr

y

yx

y

x

yxy

xxx

y

φ−=−=

+−=

+−=

∂∂

=∂φ∂ sin

1

1arctan

2222

2

2(E2.9)

φ==+

=∂∂

sin22 r

y

yx

yyr

(E2.10)

ryx

xxy

x

y

φ=

+=

+=

∂φ∂ cos

1

1122

2

2(E2.11)



Therefore, we can express the first partial derivatives of f by substituting (E2.8/9) into (E2.6) and(E2.10/11) into (E2.7):

φ∂∂φ

−∂∂

φ=φ

φ∂∂

−φ∂∂

=∂∂

≡f

rrf

rf

rf

xf

f x sincos

sincos (E2.12)

φ∂∂φ

+∂∂

φ=∂∂

≡f

rrf

yf

f y cossin (E2.13)

To find the second partial derivative we substitute f in (E2.6) by xf and in (E2.7) by yf :

xf

xr

rf

rf

x

f xxx

∂φ∂

φ∂∂

+∂∂

∂∂

=∂

∂=

∂

∂2

2(E2.14)

yf

yr

rf

yf

y

f yyy

∂φ∂

φ∂∂

+∂∂

∂∂

=∂

∂=

∂

∂2

2(E2.15)

We have to be careful, because xf and yf depend on r and φ! Substituting (E2.12) and (E2.8/9)into (E2.14):

2

2

2

2

2

222

2

22

2

22

2

22

2

2

2

sincossin2cossincossinsincos

sinsincossincossin

cossinsin

cos

sinsincoscos

sincos

φ∂

∂+

φ+

φ∂∂φφ

+∂φ∂

∂φφ−

φ∂∂∂φφ

−∂∂φ

+∂

∂φ=

φ

φ∂

∂φ−φ∂

∂φφ−φ∂∂

∂φ+∂∂φ−−

φ

∂φ∂∂φ−

φ∂∂φ+

∂

∂φ=

φ

φ∂

∂φ−∂∂

φφ∂

∂−φ

ϕ∂

∂φ−∂∂

φ∂∂=

∂

∂

f

r

f

rrf

rrf

rrf

rr

f

rf

rf

rrf

rf

rf

rf

rr

f

rf

rrff

rrf

rx

f

and analogously by substituting (E2.13) and (E2.10/11) into (E2.15):

2

2

2

2

2

222

2

22

2

2

coscossin2cossincossincossin

coscossinsin

cossin

φ∂

∂+φ+

φ∂∂φφ−

∂φ∂∂φφ+

φ∂∂∂φφ+

∂∂φ+

∂

∂φ=

φ

φ∂

∂φ+

∂∂

φφ∂

∂+φ

φ∂

∂φ+

∂∂

φ∂∂

=∂

∂

f

r

f

rrf

rrf

rrf

rr

f

rf

rrff

rrf

ry

f



The Laplacian is then given by the sum of the last two equations:

( )2

2

2

2222

2

222

2

2

2

22 cossincossin

cossinˆφ∂

∂+φ+φ+∂∂φ+φ+

∂

∂φ+φ=∂

∂+∂

∂=∇ f

rrf

rr

f

y

f

x

ff

and using Pythagoras: 1cossin 22 =φ+φ and getting rid of f

2

2

22

2

22

2

2

2

2

22 1111ˆ

φ∂

∂+

∂∂

∂∂=

φ∂

∂+∂∂+

∂

∂=∂

∂+∂

∂=∇rr

rrrrrrryx

(E2.16)

where we used a nice abbreviation for the two terms that depend on r.

E2.2 Method 2: Curvilinear Co-ordinate TransformsIt can be shown (see à[1] page 431- 433) that for orthogonal co-ordinates

( )( )( )

=

=∇

θφ∂∂

φθ∂∂

θφ∂∂

φθ

∂∂∂∂∂∂

hh

hh

hh

hhhr

r

r

r

z

y

x 1ˆ (E2.17)

with ( ) ( )2222rz

ry

rx

rh∂∂

∂∂

∂∂ +

+=

( ) ( )2222θ∂

∂θ∂

∂θ∂

∂θ +

+= zyxh (E2.18)

2222

+

+

=

φ∂∂

φ∂∂

φ∂∂

φzyxh

and

φ∂∂

φ∂∂

+θ∂∂

θ∂∂

+∂∂

φ∂∂

=++=∇φ

θ

θ

φθ

φθ∂∂

∂∂

∂∂

hhh

h

hh

rhhh

rhhhrr

rrzyx

1ˆ2

2

2

2

2

22 (E2.19)

The differentials can be derived by the Jacobian :

φθ= dddddd rJzyx (E2.20)

with the Jacobian (determinant) φθ

φ∂∂

θ∂∂

∂∂

φ∂∂

θ∂∂

∂∂

φ∂∂

θ∂∂

∂∂

==φθ∂

∂= hhh

rzyx

J r

zzrz

yyry

xxrx

),,(),,(

(E2.21)



Application to circular co-ordinates:

For the 2D example in àE2.1, we derive from eq. (E2.2):

( ) ( ) ( ) 1sincos 22222 =φ+φ=

+=

∂∂

∂∂

ry

rx

rh

( ) ( ) 22222

2 cossin rrrh yx =φ+φ−=

+

=

φ∂∂

φ∂∂

φ

and hence:

=∇

φ∂∂

∂∂ r

rr1ˆ and

2

2

22 11111ˆ

φ∂

∂+

∂∂

∂∂

=

φ∂∂

φ∂∂

+

∂∂

∂∂

=∇rr

rrrrrr

rrr

Application to spherical co-ordinates:

The relation between Cartesian and polar co-ordinates is (see àFig. 10.6):

θ=φθ=φθ=

cos

sinsincossin

rz

ryrx

(E2.22)

The scaling factors for curvilinear transforms are (see (E2.18))

( ) ( ) ( ) ( )

( ) 1cossincossin

cossinsincossin

2222

2222222

=θ+φ+φθ=

θ+φθ+φθ=+

+= ∂

∂∂∂

∂∂

rz

ry

rx

rh

( ) ( ) ( ) ( ) ( ) 22222222 sinsincoscoscos rrrrh zyx =φ−+φθ+φθ=+

+=

θ∂∂

θ∂∂

θ∂∂

θ

( ) ( ) θ=φθ+φθ−=

+

+

=

φ∂∂

φ∂∂

φ∂∂

φ2222

2222 sincossinsinsin rrrh zyx

and hence (see à(E2.17))

( )( )

( )

θ

θ

θ=

θ

θ

θ=∇

φ∂∂

θ∂∂

∂∂

φ∂∂

θ∂∂

∂∂

r

r

r

rr

r

r

r

rrsin

sin

sin

1sin

sin

sin

1ˆ

2

2

2

2(E2.23)

and (see à(E2.19)):

( ) ( )

2

2

2222

2

22

2

sin

1sin

sin

11

sin1

sinsinsin

1ˆ

φ∂

∂

θ+

θ∂∂

θθ∂∂

θ+

∂∂

∂∂

=

φ∂∂

θφ∂∂

+θ∂∂

θθ∂∂

+∂∂

θ∂∂

θ=∇

rrrr

rr

rr

rr(E2.24)

The differential expression is (see à(E2.21)):

φθθ= dddsinddd 2 rrzyx (E2.24)

Dr. P. Blümler: PH 502 Wave Mechanics Excursion 2 page: 82


Tab. E2.1: You might find the following table generally useful:

Cartesian 2D circular 3D-cylindrical 3D-sphericalco-ordinates: x,y,z

φ=

φ=

sin

cos

ry

rx

zz

ry

rx

′=

φ=

φ=

sin

cos

θ=

φθ=

φθ=

cos

sinsin

cossin

rz

ry

rx

rh - 1 1 1θh - - 1=′zh rφh - r r rsinθ

∇

∂∂∂∂∂∂

z

y

x

φ∂∂

∂∂ r

rr1

′∂∂φ∂

∂∂∂

z

rr

r1

θ

θ

θφ∂∂

θ∂∂

∂∂

r

r

r

r

rsin

sin

sin

1

2

2

2∇2

2

2

2

2

2

zyx ∂∂

∂∂

∂∂ ++

2

2

2

11

φ∂

∂+

∂∂

∂∂

rrr

rr 2

2

2

2

211

zrrr

rr ′∂

∂+φ∂

∂+

∂∂

∂∂

2

2

2222

2 sin

1sin

sin

11

φ∂

∂

θ+

θ∂∂

θθ∂

∂

θ+

∂∂

∂∂

rrrr

rr

dV dx dy dz φddrr zrr ′φddd φθθ dddsin2 rr



11. The Hydrogen Atom

Detail (Solution of differential eq.)


To know:1. Coulomb potential, effective potential, angular momentum barrier, energy

spectrum, Schrödinger-equation for central force problem.2. Bohr-model, -radius3. Radial waveform (Laguerre polynomials)*4. Solutions*

lmlnln

nh

eZE

rRNr

n

nlm

≤≤−−≤=ε

µ−=

φΦθΘ=φθψ

and1and,...3,2,1with

8

)()()(),,(

2220

42

5. Quantum numbers n,l,m and degeneracy.6. Radial distribution function (construction, spherical Jacobian)7. Orbitals*8. Basic idea of hybrid orbitals

* not in detail, but their dependence (the most simple you could remember!)


Dr. P. Blümler: PH 502 Wave Mechanics 11 Hydrogenic Atom page: 84


11. The Hydrogenic Atom

11.1 Motion in a Coulomb fieldThe motion of a single electron in the Coulomb-field of charged nucleus is to be solved. Weremember that the force varies as F(r) ∝ 1/r2 and hence the potential energy (F = -dU/dr) asU(r) ∝ 1/r. To be exact the Coulomb-potential between a nucleus of charge Ze and an electron ofcharge -e is:

rZe

rU0

2

4)(

πε−= (11.1.1)

with ( )Jm/C10854188.8 2120

−⋅=ε the permittivity in the vacuum. As introduced in àpart 10.1

we can simplify the two-particle problem to a ‘quasi’ one-particle problem using its reduced mass

nucleuse

nucleusemmmm

+⋅

=µ , where mnucleus is the mass of the nucleus.

For instance for 1H: Z = 1 and pe

pe

mm

mm

+

⋅=µ .

Hence the àSchrödinger-equation for this problem is:

ψ=ψπε

−ψ∇µ

− Er

Ze

0

22

2

4ˆ

2h

(11.1.2)

Because the problem is similar to the rotation of a particle on the surface on the sphere, but now witha radial potential, an expression of this equation in spherical co-ordinates, ),,( φθψ r , appears to beappropriate (using àeqn. (E2.24)) :

0),,(4

2

),,(sin

1sin

sin

11

0

2

2

2

2

2222

2

=φθψ

πε+

µ

+φθψ

φ∂

∂

θ+

θ∂∂

θθ∂∂

θ+

∂∂

∂∂

rr

ZeE

rrrr

rrr

h

(11.1.3)

As before, separation of variables is tried by :

)()()(),,( φΦθΘ=φθψ rRr (11.1.4)



Insertion into eq. (11.1.3) gives

04

2

d

d

sindd

sindd

sindd

dd

0

2

2

2

2

2222

2

=ΦΘ

πε+µ

+φ

Φ

θ

Θ+

θΘ

θθθ

Φ+

ΦΘ

Rr

ZeE

r

R

r

RrR

rrr

h

(11.1.5)

following the typical separation-procedure eq. (11.1.5) is multiplied by ΘΦRr2

:

04

2

d

d

sin

1dd

sindd

sin1

dd

dd1 2

0

2

22

2

22 =

πε+

µ+

φ

Φ

θΦ+

θΘ

θθθΘ

+

r

rZe

ErR

rrR h

(11.1.6)

and rearranged:

φ

Φ

θΦ+

θΘ

θθθΘ

−=

πε+

µ+

2

2

22

0

2

22

d

d

sin

1dd

sindd

sin1

42

dd

dd1

rr

ZeE

rR

rrR h

(11.1.7)

Hence, the problem is separated into a radial part (left) and an angular part (right). The right side(square brackets) we identify as the differential equation for the particle on a sphere (see àeq.(10.4.15) to (10.4.20)) and can readily insert the solutions (spherical harmonics):

)1(ˆ1

ˆ14

2dd

dd1

)20.4.10(2

)19.4.10(

2)15.4.10(

2

0

2

22

+=Λ−=

ΘΦΛΘΦ

−=

πε+

µ+

ll

rr

ZeE

rR

rrR

lml

m

YY

h(11.1.8)

0)()1(

42

d)(d2

d

)(d

0)1(4

2dd

2d

d1

0)1(4

2dd

dd1

20

2

22

2

22

0

2

22

22

2

0

2

22

=

+−

πε+

µ++

⋅=+−

πε+

µ+

+

=+−

πε+

µ+

rRr

llr

ZeE

rrR

rr

rR

r

Rllr

rZe

ErR

rr

Rr

R

llrr

ZeE

rR

rrR

h

h

h

(11.1.9)

This is not surprising, because the problem can be understood as a particle moving on a collection ofspherical shells. However, what is left to find is its distribution over these shells, the radial part.



Now we want to go -in a way- some steps backwards, to understand our findings. Therefore, weexpress equation (11.1.9) in the standard form of the Schrödinger-equation:

)()()()(ˆ2

)()()(2

)1()(

dd

dd1

2

eff22

2

22

2

2

rERrRrUrR

rERrRrUr

llrR

rr

rr

=+∇µ

−

=

+

µ

++

µ−

h

hh

(11.1.10)

where an effective potential )(2

)1()(2

2eff rU

r

llrU +µ

+= h is introduced with U(r) as the

Coulomb-potential from àeq. (11.1.1). The first term appears due to the rotation of the electron,which counter-acts the Coulomb-potential as a repulsive core for l > 0. It becomes infinitely high asr → 0, hence prevents the system to collapse, and is therefore called “angular momentumbarrier”. The nature of the two potentials and their sum is depicted in Fig. 11.1.

-1.0

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1.0

2 4 6 8 10

U [

10

J]

-17

r [Å]

U r( )

U r ( )e f f

a.m.b.

a) eff

U

[10

J

]-1

8

-2

-1

0

1

2

r [Å]2 4 6 8 10

l = 0

l = 1

l = 2

l = 3b)

Fig. 11.1: Potentials for 1H: a) green: Coulomb-potential, U(r); blue: „angular momentum barrier“(a.m.b.) and red curve: sum of a) and b), the effective potential Ueff(r). Calculations forl = 1. b) Ueff(r) for l = 0, 1, 2 and 3.

In Fig. 11.1 we recognise the attracting nature of the Coulomb-potential (negative) and the repellingcharacter of the “angular momentum barrier”. If the energy becomes higher then the Coulomb-termand the angular momentum term dominates, the particle is retracted from the nucleus and mustbehave like a free (unbound) particle, hence without quantized energy states (energies form acontinuum, see Fig. 11.2). This process is called “ionisation” and defined by Ueff. Due to the widerpotential range for higher energies, we can also conclude that the energy levels for the bound statesmust converge to that limit. Due to the finite value of the potential at a discrete bound energy state wecan finally expect the particle to tunnel beyond the classical limits (that is where the energy levelmeets the potential).



11.2 Solution of the Radial Differential Equation

The aim of this part is to solve àeq. (11.1.9). This will be done in elementary steps, which requiresmany substitutions to guide through the essential maths. We start with the last line of (11.1.9) andrewrite it in a shorter notation:

0)1(

422

0)()1(

42

d)(d2

d

)(d

20

2

2

20

2

22

2

=

+−

πε+

µ+′+′′

=

+−

πε+

µ++

Rr

llr

ZeER

rR

rRr

llr

ZeE

rrR

rr

rR

h

h(11.2.1)

the terms in the squared brackets can be simplified by suitable substitution:

2hEµ

≡η and 2

0

2

4 hπε

µ≡α Ze (11.2.2)

leading to:

0)1(2

22

2=

+−

α+η+′+′′ R

r

llR

rRR

rR (11.2.3)

We are now testing the equation for r →∞, meaning the electron leaves the nucleus to become a freeparticle. In eq, (11.2.3) that causes the terms that depend on 1/r to vanish:

02 =η+′′ ∞∞ RR (11.2.4)

eff

U

[10

J]

-18

-1

0

1

2

r [Å]2 4 6 8 10

continuum of unbound, free-particle states

discrete,boundstates

Fig. 11.2: Sketch of the quantum mechanical energy spectrum asto be expected from general considerations.



The solution is the same as for the free (non-restricted/non-confined) particle:

( ) ( )riBriArR η++η−=∞ 2exp2exp)( (11.2.5)

where η ∝ E represents the energy. This general (complex) solution can only be ànormalised (tothe real value 1), when we assume E or η < 0, so that η−−=η−−=η 2212 ii with a

positive radicand. (Try these normalisations as examples or alternatively substitute with a negative,see eq. (11.2.8)) This can also be suggested when looking at àFig. 11.2 (but now we havemathematical proof)! The normalisation condition then reads as:

( ) ( )( ) 1d2exp2exp

0

2=η−−+η−∫

∞

rrBrA (11.2.6)

The first term becomes infinite for r → ∞ and is hence unacceptable, because we know that thewavefunction must vanish, while the second (becoming zero) is acceptable for these conditions.Therefore, A = 0 and

( )rBrR η−−=∞ 2exp)( (11.2.7)

or with changing our substitution from (11.2.2) to

hEµ−

=η−≡β2

2 (11.2.8)

Now, we try a first approach for a general solution by incorporating the knowledge for r → ∞ andignore B, knowing that the final normalisation of the wavefunction must recover it.

)()()( rferfRrR rβ−∞ == (11.2.9)

with f(r) as a new unknown function! Hence,

( )( ) ( )

( ))()(2)(

)()()()()(

)()()()()(

)()(

2 rfrfrfe

rfrferfrferR

rfrferferferR

rferR

r

rr

rrr

r

β+′β−′′=

β−′β−′β−′′=′′

β−′=β−′=′

=

β−

β−β−

β−β−β−

β−

(11.2.10)

These results are used in the radial part of the Schrödinger-equation (àeq. (11.2.3)):

( ) ( ) 0)()1(2

2)()(2

)()(2)(2

2 =

+−

α+η+β−′+β+′β−′′ β−β−β− rfe

r

llr

rfrfer

rfrfrfe rrr

Multiplication with exp(βr) and rearranging yields



0)()1()(2

2)(1

2)(

0)()1(2

2)(2

)(2

)()(2)(

22

22

=

+−

β+α+β+η+′

β−+′′

=

+−

α+η+

β−′+β+′β−′′

rfr

llr

rfr

rf

rfr

llr

rfr

rfr

rfrfrf

(11.2.11)

and with η−=β 22 (see definition in à(11.2.8))

0)()1()(2

)(1

2)(2

=

+−

β+α+′

β−+′′ rf

r

llr

rfr

rf (11.2.12)

So far things do not really look simplified. We try another approach for the evaluation of theunknown part of this equation f(r) by expressing is as a power series (with coefficients b) and tryingto find an upper limit (quantum number) to meet the boundary conditions, which means that the seriesis transformed into a polynomial of a degree given by this upper limit:

q

qqrbrf ∑

∞

=

≡0

)( and hence q

qq

r rberR ∑∞

=

β−=0

)( (11.2.13)

for substitution into eq. (11.2.12), we calculate the first and second derivative:

∑

∑

∑

−

−

−=′′

=′

=

q

qq

q

qq

q

qq

rbqqrf

rbqrf

rbrf

2

1

)1()(

)(

)(

(11.2.14)

Used in eq. (11.2.12) this gives:

( ) ( ) ( )[ ][ ] ( )[ ] 0222)1(2)1(

0)1(2222)1(

12

21122

=β−β−α++−+−

=+−β−α+β−+−

−−

−−−−−

∑∑

∑q

qq

qq

q

q

qq

qq

qq

qq

qq

rbqrbllqqq

rbllrbrqbrqbrbqq

(11.2.15)

In order to become zero the coefficients for each power of r have to cancel. Hence, we match the

indices for qr and set them to be zero:

[ ] [ ][ ]

[ ]( )

)1()3)(2(22

)1()2(2)1)(2()()1(2

0)()1(2)1()2(2)1)(2(

1

2

12

+−++α−β+β

=+−++++

β−α−+β=

=β−α−+β−+−++++

+

+

++

llqqq

llqqqq

b

b

bqbllqqq

q

q

qq

(11.2.16)



We further simplify this by reducing the index, q, by 1, which results in the following recursion-formula:

( ))1()1)(2(

)1(2

)1()1)(2(2)1(2

1 +−++α−+β

=+−++α−β+−β

=+ llqqq

bllqq

qbb qqq (11.2.17)

The next step is to find a limiting case for q→∞ and hence dominating α and l2, hence assuming q >l:

2

2

)1)(2()1(

2lim 1 +

β=

+++β

≈+∞→ q

b

qqq

bb qqq

q(11.2.18)

The purpose of a recursion-formula is to address the next higher coefficient by a lower. So we wantto try to express bq by b0 (the lowest coefficient relating to a constant). We start by calculating b1, b2

etc... to end up with a general expression for large q:

( )

( ) ( )

( ) ( )

( ) 0

03

23

02

12

001

2!

2

...

2543

12

51

243

12

41

231

32

bq

b

bbb

bbb

bbb

qq β=

β⋅⋅

=β≈

β⋅

=β≈

β=β≈

(11.2.19)

This reminds us of the Taylor-expansion series for the exp-function, we recall:

( )

( )

rrr

q

qrq

qq

r

q

qar

ebeebrR

qr

berberR

qar

e

βββ−

∞

=

β−∞

=

β−

∞

=

==

β==

=

∑∑

∑

02

0

00

)19.2.4(

0

)13.2.4(

0

22)(

!2

2)(hence

!

(11.2.20)

Without loss of generality, we can deposit the 2 in the normalisation constant and get

rebrR β= 0)( (11.2.21)

However, this cannot be the final result, because we have to remember the boundary condition0)(lim =

∞→rR

r and from the definition in à(11.2.8) we reassure that 0>β ! Therefore an upper

limit p has to be introduced to stop the series and convert it into a polynomial of degree p, e.g.



0 ≤ q ≤ p., which means that the series stops after p or 01 =+pb . Hence 02 =+pb and all

following coefficients vanish due to eq. (11.2.17) and for 0≠pb this gives:

0)1()1)(2(

)1(21 =

+−++α−+β

=+ llppp

bb pp (11.2.22)

The equation is fulfilled for a) the nominator = 0 and b) the denominator ≠ 0, or:

np

p

α≡

+α

=β

α=+β

1

)1(a)

(11.2.23)

where a new (simplified) quantum number, n = p + 1, is introduced.

1

)1()1()1()1)(2(

0)1()1)(2(b)00

+≥>

+>++>++

≠+−++>>

ln

ln

llnnllpp

llpp 32144344 21

(11.2.24)

As at the start of the derivation of this formula we (c.f. àeq. (11.2.18)) we have to assume p+1 < l.As expected, we found another quantum number n, related to l, which is related to m (cf. àeq.(10.4.23)). From condition a) we can directly calculate the energies associated with the radial part ofthe wavefunction, by re-substituting à(11.2.2) and à(11.2.8):

…h

hh

,3,2,11for832

4

2

2220

42

2220

2

42

20

2

=+≥ε

µ−=

επ

µ−=

πε

µ=

µ−

lnnh

eZ

n

eZE

n

ZeE

n

(11.2.25)

This is exactly the result Bohr found in his treatment of the hydrogen atom and we also identify ourparameter

B

1r

=α (11.2.26)

where Br is the Bohr-radius.



The only thing left to do, is the exact determination of the wavefunction R(r). We summarise our‘equipment’ for this task:

α−+β+−++

=

+−++α−+β

=

=+≥

−==β=α

+

+

−

=∑

)1()1()1)(2(

2

)1()1)(2()1(

2:tscoefficien

,3,2,11for

exp)(,1

,1

1

1

1

0B,

BB

qllqqb

b

llqqq

bb

ln

rbnrr

rRnrr

qq

qq

n

q

qqln

…

and start with the lowest allowed state: n = 1, hence l = 0

B00,1

B00,1

with)(

exp)(

rr

ebR

rr

brR

=ρ=ρ

−=

ρ−(11.2.27)

where we introduced a dimensionless co-ordinate ρ as the radius in multiples of the Bohr-radius.Normalisation gives

12

dd)(20

0

220

0

20,1 ==ρ=ρρ ∫∫

∞

=ρ

ρ−∞

=ρ

bebR (11.2.28)

hence the normalised wavefunction for the ground state is:

ρ−=ρ eR 2)(0,1 (11.2.29)

Now the steps are repeated for n = 2, hence we have to distinguish l = 0, 1 or )(0,2 rR and

)(1,2 rR . Generally this is determined by:

( )rbbnrr

rbnr

rrR

q

qql 10

B

1

0B,2 expexp)( +

−=

−= ∑

=

(11.2.30)

where b1 can be expressed in terms of b0 by the recursion formula in à(11.2.17):

a) case: n = 2, l = 0

B

01

21

01 20122 BB

rb

bb rr −=−⋅

−=

hence, 20,2

B

00

B0,2 2

122

exp)(ρ−

ρ

−=

−

−=ρ eN

rb

brr

R (11.2.31)



b) case: n = 2, l = 1

If we would proceed as in the previous case àeq. (11.2.17) yields zero in the denominator.Therefore, we use the formula ‘backwards’

α−+β+−++

= +

)1()1()1)(2(

21

qllqqb

b qq

therefore we determine b0. and use b1 for normalisation:

022

2BB

121

10 =

−

−=

rr

bb

hence 21,2

2B11,2 )(

ρ−ρ−ρ=ρ=ρ eNerbR (11.2.32)

Note, that the final results in a) and b) contain the associated normalisation constants 0,2N and

1,2N .

Continuation of this procedure gives the following table of the radial-parts of the wavefunction for thehydrogen-like atom, which are depicted in àFig. 11.3. (Note: That Fig. 11.3 shows the radial partof the wavefunction and not a 3D-projection of the real distribution of the electron, since the angularterms have been neglected so far)

Table 11.1: The radial-parts )(, ρlnR of the wavefunction for the hydrogen-like atom as the function

of the radius in multiples of the Bohr-radius.

n l )(, ρlnR

1 0 ρ−= eR 20,1

2 0( ) 2

0,2 22

1ρ

−ρ−= eR

2 12

1,22

1ρ

−ρ= eR

3 0 ( ) 320,3 21827

272

ρ−ρ+ρ−= eR

3 1 ( ) 321,3 6

39

2ρ−

ρ−ρ= eR

3 232

2,3 272

ρ−ρ= eR



Fig. 11.3: Left: The radial part of the wavefunction Rnl(ρ) for n=1,2 and 3. Right: Thecorresponding probability density function Pnl(ρ).



The radial part of the eigen-functions of the hydrogenic atom are another ortho-normal class ofpolynomials (like the Hermite and Legendre polynomials), called associated Laguerre polynomials:They are solutions of the so-called associated Laguerre equation, in which àeq. (11.2.1) can betransformed.

( )

!)1(

d

d)1()(:spolynomial associated

)()(:solutions

0)()(1)(:equationassociated

0ix

i

kj

xxLaguerre

xxf

xfjxfxkxfxLaguerre

ikj

i

ik

kkk

j

kj

∑+

=

+−−=

=

=+′−++′′

L

L (11.2.33)

In this notation the radial wavefunction becomes:

( )!2)!1(2

1and,...3,2,1with)()(

3

121

/

lnnln

nN

nlnn

eNR

nl

lln

ln

nlnl

+−−

=

−≤=ρ

ρ

=ρ +−−

ρ− L(11.2.34)

11.3 The Complete Solution: Atomic OrbitalsThe complete solution for an electron in a Coulomb-potential (hydrogenic atom) was calculated byseparation of variables, where the radial part was solved in the previous part, after we haverecognised that the angular parts were already solved in àchapter 10 with the spherical harmonics,

),( φθlmY , as eigen-function and Ill 2/)1( +h as eigen-values (cf. àeq. (10.4.20) to (10.4.23)).

Hence, the eigen-functions of the hydrogenic atom are given according to eq. (11.1.4):

Hydrogenic Atom:

( )( )

lmlnln

nh

eZE

ml

mll

lnnln

nN

e

rZe

rr

neR

rRNr

n

nlm

imlm

lm

lln

ln

nl

nlm

≤≤−−≤=ε

µ−=

+π−+

+−−

=

θ=φθ=φΦθΘ

πε

µ==ρρ

ρ

=ρ

φΦθΘ=φθψ

φ

+−−

ρ−

and1and,...3,2,1with

8

)!(4

)!(12

!2)!1(2

)(cos),()()(

4and)()(with

)()()(),,(

2220

42

3

20

2

B

121

/

PY

Lh

(11.3.1)



As expected the three degrees of freedom in motion generate three quantisation conditions,described by the three quantum numbers n, m and l. Different nomenclature is in use for them:

n principal quantum number: n = 1, 2, 3, 4,....

because n defines the ‘shell’ n = K, L, M, N, ....

l orbital quantum number (sometimes also called azimuthal quantum number): 1+≤ nl

l = 0, 1, 2, 3, 4, 5..., (n-1)

spectroscopic notation: l = s, p, d, f, g, h

m magnetic quantum number (sometimes also lm ) lml ≤≤−

m = -l, (-l+1),....-3, -2, -1, 0, 1, 2, 3,... , (l-1), l

spectroscopic notation: m = -l, (-l+1), ,...,,,,,,..., ϕδπσπδϕ , (l-1), l

The following table is illustrating the mutual dependence of the quantum numbers.

Tab. 11.2: Summary of the interdependence of the quantum numbers for the hydrogenic atom, forn = 1, 2 and 3 including nomenclature and degeneracy. (Neglecting the spin of theelectron, which is done in the next lecture)

n 1 2 3

l 0 0 1 0 1 2

m 0 0 -1 0 1 0 -1 0 1 -2 -1 0 1 2

name 1s 2s 2p 3s 3p 3d

sp.n.* 1sσ 2sσ 2p π 2pσ 2pπ 3sσ 3p π 3pσ 3pπ 3d δ 3d π 3dσ 3dπ 3dδ

D(n)** 1 4 9

*spectroscopic name including m-notation.

** D(n) degeneracy of state.

The energy is determined by n only (cf. àeq. (11.2.25)). Hence we have 2l+1 degenerate states(span of m), where the span of l is defined by n-1. Therefore the degeneracy of a state n is given by:

( ) 21

0

22

)12()32(....53112)( nnn

nnlnDn

l

==−+−++++=+= ∑−

=

(11.3.2)

Combining the equations in (11.3.1) gives the complete wavefunctions for the hydrogenic atom aslisted in Tab. 11.3:



Tab. 11.3:The complete wavefunctions, ),,( φθρψnlm , of the hydrogenic atom for

n = 1, 2 and 3.

Name n l m ),,( φθρψnlm

1sσ 1 0 0 ρ−

πe

1

2sσ 2 0 0 ( ) 2/224

1 ρ−ρ−π

e

2p π 2 1 -1 φ−ρ− θρπ

iee sin8

1 2/

2pσ 2 1 0 θρπ

ρ− cos24

1 2/e

2pπ 2 1 1 φρ− θρπ

iee sin8

1 2/

3sσ 3 0 0 ( ) 3/221827381

1 ρ−ρ+ρ−π

e

3p π 3 1 -1 ( ) φ−ρ− θρρ−π

iee sin681

1 3/

3pσ 3 1 0( ) θρρ−

πρ− cos6

81

2 3/e

3pπ 3 1 1 ( ) φρ− θρρ−π

iee sin681

1 3/

3d δ 3 2 -2 φ−ρ− θρπ

iee 223/2 sin162

1

3d π 3 2 -1 φ−ρ− θθρπ

iee cossin81

1 3/2

3dσ 3 2 0 ( )1cos3681

1 23/2 −θρπ

ρ−e

3dπ 3 2 1 φρ− θθρπ

iee cossin81

1 3/2

3dδ 3 2 2 φρ− θρπ

iee 223/2 sin162

1



For the calculation of the normalisation constant nlmN it is important to realise, that the integration is

(preferably) done in spherical co-ordinates (cf. àTab. E2.1) and the integrand can be furthersimplified according to àeq. (10.4.25), where we have learned that the φ-term behaves constant.

[ ] 1ddsin )()(2

dddsin ),,(),,(*

ddd),,(),,(*

0 0

22

)25.4.10( 2

0 0

2

0

2

)1.2(

0 0 0

2

=θθθΘπ

=φθθφθψφθψ

=ψψ

∫ ∫

∫ ∫ ∫

∫∫∫

∞

=

π

=θ

∞

=

π

=θ

π

=φ

∞ ∞ ∞

rrrRN

rrrrN

zyxzyxzyxN

r

nlm

r

nlm

E

nlm

(11.3.3)

This procedure can be easily understood, when we have a look at the s-orbitals, which according tol = 0 have spherical symmetry. Hence, we can replace the radial probability distribution function (e.g.in Fig. 11.3/right) by an integration over spherical shells of thickness dr. Hence, we have to substitute

the probing ‘volume’ (dρ) of the radial probability distribution function, ρρ=ρρ d)(d)( 200 nn RP ,

with the volume of the spherical shells (4πρ2 dρ) to get the projection of the three-dimensionaldistribution function, )(

~0 ρnP (we call it the radial distribution function), of the electron on the

radius or

ρρπρ=ρρ d)(4d)(~ 2

02

0 nn RP (11.3.4)

This principle is illustrated in àFig. 11.4 using the 1s-orbital as an example. From the plot of theradial distribution function for the 1s, 2s and 3s-orbital (see Fig. 11.5) the Bohr’s idea of shells, inwhich the electron is located can be anticipated, but we also realise, that the nth shell has only themaximum probability. The electron has a finite probability to be found in lower ‘shells’. (Closerexamination also uncovers that -except of 1s- that the maximum is not at integer multiples of rB) Wealso realise that there is a small but non-zero probability to find the electron inside the nucleus,assuming the nucleus has a finite radius. This is of great importance for nuclear reaction e.g. β+-decay via electron capture.

Finally we want to visualise the atomic orbitals, which is a difficult task, because we have a 3D-wavefunction which has a different values at the co-ordinates ρ,θ,φ . Therefore, we have to depict iteither as a cloud of varying density, or we have to pick a certain threshold of probability to which weassign a surface (isosurface). Both ways of visualisation are shown in Tab. 11.4, where the differentsigns of the lobes have been colour-coded (positive = red, negative = green).

For an on-line visualisation, you can download ‘Orbital-Viewer’ (a program from D. Manthey) fromthe àwebpage (http://wwwnmr.ukc.ac.uk/staff/pb/teach/PH502/Aos/ ao.html) of thecourse.



Fig. 11.4: Illustration for the construction of the radial probability density function for the 1s-orbital:Upper: If an infinitesimal small and constant sampling volume dρ is scanned along theradius, we will receive the radial-part of the electron density P(ρ). (If additionally thedetector would be moved on a circle, for each ρ the same value would be recorded. The1s-orbital is isotropic.) Lower: Now the electron density is measured over a sphericalshell of thickness dρ, to determine the radial probability distribution function )(~ ρP , whichhas a maximum at the Bohr radius (ρ = 1).

0

2

4

6

8

10

12

14

16

5 10 15 20 25 30 ρ

P ( )ρ30~

0

2

4

6

8

5 10 15 20 25 30 ρ

P ( )ρ20~

0

0.5

1

1.5

2

2.5

3

5 10 15 20 25 30 ρ

P ( )ρ10~

Fig. 11.5: The radial distribution function (not-normalised) for the first three s-states (l = 0).Compare with the according states in Fig. 11.3.

sampled volume 4πρ2dρ

dρ

ρ

ρ1

r=rB

dρ

P(ρ)

)(~ ρP



Tab. 11.4:Atomic orbitals for hydrogenic atoms for n = 1, 2 and 3. On the left of each cell the orbital is shown as a probability plot using 10000 points. Onthe right (for ns-orbitals in the centre) an isosurface (corresponding to a probability of 10-4) is shown. For the ns-orbitals additionally, an octant ofthis isosurface has been cut away to show the inner ‘shells’. Red colour corresponds to positive lobes, green to negative.

mn l -2 -1 0 +1 +2

1 0/s

2 0/s

1/p

3 0/s

1/p

2/d



Fig. 11.6: Energy levels for hydrogen 1H including the l ≤ 5. Energy is measured in units of

)2/( 2B

2 rµh .

11.4 Linear Combinations: Hybrid OrbitalsWe have seen from àTab. 11.3 that the wavefunctions of the hydrogenic atom are partially real andpartially imaginary. The first case applies for m = 0 the latter for m ≠ 0. The there are n2 degeneratedeigen-functions belonging to the same n, we can create new functions by àlinear combination, whichare entirely real. This has practical applications in chemistry. Since the new functions are createdfrom the set of solutions, we call them hybrid orbitals. For instance

( ) [ ] [ ]( )

[ ] xrx

ee

ieie

p224

1cossin

282

sincossinsincossin28

1p2p2

1

B

2/(E2.22)

2/

2/2/

≡

π=φθρ

π=

φ−φθρ+φ+φθρπ

=π+ππ

ρ−ρ−

ρ−ρ−

Analogously we get:

( )

z

y

rz

ee

ry

ee

p224

1cos

281

p2

p224

1sinsin

24

1p2p2

1

B

2/)22(E2.

2/

B

2/(E2.22)

2/

≡

π=θρ

π=σ

≡

π=φθρ

π=π−π

π−

ρ−ρ−

ρ−ρ−

Hence, we receive three equivalent 2p-functions, which can be correlated to the three Cartesian co-ordinates. This is useful for geometrical considerations (e.g. chemical complexes).

Similarly we proceed with the other wavefunctions (see Tab. 11.5)



Tab. 11.5:Hybrid orbitals for hydrogenic atoms for n = 1, 2 and 3.

Symbol n l m combination hybrid wavefunction

1s 1 0 0 1sσ ρ−

πe

1

2s 2 0 0 2sσ ( ) 2/224

1 ρ−ρ−π

e

2pz 2 1 0 2pσ θρπ

ρ− cos24

1 2/e

2px 2 1 ±1 ( )π+π p2p22

1φθρ

πρ− cossin

24

1 2/e

2py 2 1 1∓ ( )π−π−

p2p22

1φθρ

πρ− sinsin

24

1 2/e

3s 3 0 0 3sσ ( ) 3/221827381

1 ρ−ρ+ρ−π

e

3pz 3 1 0 3pσ( ) θρρ−

πρ− cos6

81

2 3/e

3px 3 1 ±1 ( )π+π p3p32

1( ) φθρρ−

πρ− cossin6

81

2 3/e

3py 3 1 1∓ ( )π−π−

p3p32

1( ) φθρρ−

πρ− sinsin6

81

2 3/e

2d3 z 3 2 0 3dσ ( )1cos3681

1 23/2 −θρπ

ρ−e

3dxz 3 2 ±1 ( )π+π d3d32

1φθθρ

πρ− coscossin

81

2 3/2 e

3dyz 3 2 1∓ ( )π−π−

d3d32

iφθθρ

πρ− sincossin

81

2 3/2 e

22d3yx − 3 2 ±2 ( )δ+δ d3d3

2

1φθρ

πρ− 2cossin

281

1 23/2 e

3dxy 3 2 2∓ ( )δ−δ−

d3d32

iφθρ

πρ− 2sinsin

281

1 23/2 e

The hybrid orbitals do not look different from those in Tab. 11.4. They are just transferred back toCartesian co-ordinates (with slightly different normalisation constants).

Dr. P. Blümler: PH 502 Wave Mechanics Thanks! page: 103


Acknowledgements:I would like to thank Dr. C. Isenberg to provide access to his notes for a first layout of the course.They have turned out to be very helpful and guiding for a new lecturer. I also want to thank Mr. IanCampbell for proof-reading the first draft of the manuscript.

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