Math Chem Course

8/4/2019 Math Chem Course

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Mathematical and Numerical Methods forChemists

Sture NordholmDepartment of Chemistry, The University of Gothenburg

August 2, 2011

Abstract

Chemists need the language and the tools of mathematics to ex-press its accumulated knowledge and to solve problems. This course isintended to help students of chemistry to acquire this language and touse mathematical methods within a chemical context. It is assumedthat the student has had exposure to university mathematics. Theconcept of the course is to briey summarize a relevant area of mathe-matics with emphasis placed on understanding the origin of its tools ofanalysis. This is then followed by selected applications of these tools

to chemical problems. In this way the twostep process of translatinga chemical problem into mathematical form and then solving it by anappropriate mathematical method is amply illustrated.

Part I

Linear Algebra andApplications

1 Chemical Algebra - A Preview

Mathematics starts with integers ...-2,-1,0,1,2,..... and we cant do withoutthem in chemistry. We need them in an interesting and nontrivial way when

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we try to balance a chemical reaction,

1C14H12 + 2O2 = 3CO2 + 4H2O. (1)

Here we are looking for the set of smallest positive integers 1; 2; 3; 4 suchthat the equation is balanced with respect to all atomic species present.As we all know this can be a frustrating task if the molecules involved aremany and large. When we then try to calculate the corresponding masses ina corresponding laboratory experiment we enter the realm of real numberswhich in mathematics we often call x, particularly before we have managed toobtain them. The concept of a "mole" is the bridge between the stochiometricequations in chemistry and the measurable masses of reactants and productsin laboratory reactions. It is by a decision of IUPAC dened as the unit of

"amount of substance" and takes stochiometeric calculations from integersto real numbers.

We know that real numbers can be added, subtracted, multiplied anddivided unless the denominator is zero. In order to be able to take squareroots of negative numbers and do other handy things we introduce complexnumbers z = x+iy. Now we have an object which is specied by two realnumbers. But the story goes on. When we want to specify the position of aparticle we need three coordinates x,y,z and we need them so often that wedecide to call this set of numbers a vector r. This particular vector is threedimensional but if we have to specify the positions of two particles we need

six real numbers x1; y1; z1;x2; y2; z2 which can be regarded as a six dimensionalvector. In the same way we can go on and nd need for vectors of very highdimension. If, for example, we would try to discuss the instantaneous stateof a gas by classical mechanics we may need vectors of Avogadros numberof dimensions. Now our ability to call these objects vectors and use vectornotation which suppresses the delineation of the components is essential be-cause we could not nd time in a lifetime to write down all these components.Nevertheless it is possible for us to work with such monstrous vectors if weknow the rules which apply to them. Thus we are getting ready to studythe rules applying to vectorspaces. A powerful but simple set of operationsinvolving vectors are described by what is called linear algebra. As it turns

out quantum mechanics is dominated by linear algebra and through quantumchemistry, which is, perhaps, the fastest growing branch of chemistry, linearalgebra has become an essential tool for chemists. For this reason we shallhave a look at how quantum mechanics relies on linear algebra and how itgets into chemistry.

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1.1 The Linear Algebra of Quantum Mechanics and

Quantum ChemistryLife may seem to get horribly complicated when we move from classical toquantum mechanics. A classical particle moving in one dimension can bedescribed by two real numbers x,v, the position and momentum. Newtonsequations of motion deal with only these two quantities. When we go toquantum mechanics we must describe the state of the same particle as awavefunction (x). A function of x is dened by its value at all x-values. Ifall such function values are compared with the components of a vector we seethat a function is a vector in an innite dimensional vectorspace so we wouldseem to have a problem worse than that of describing the classical states

of a gas of a mole of particles. It is not quite so dicult as it may sound.The time-development is described by a time-dependent wavefunction (t; x)which satises the time-dependent Schrdinger equation,

@

@t(t; x) = i

~H(t; x); (2)

where H is the Hamiltonian operator,

H = ~2

2m

@2

@x2+ V(x): (3)

Here V(x) is the potential acting on the particle. Much interest is focused on

the so-called energy eigenfunctions E(x) which satisfy the time-independentSchrdinger equation,HE(x) = EE(x): (4)

Note that the spatial probability density associated with a wavefunction (x)is

p(x) = j(x)j2 : (5)An energy eigenfunction satises the time-dependent Schrdinger equationif it is multiplied by the phasefactor eiEt=~. This means that the spatialprobability density p(x) is time-independent. Thus the energy eigenfunctionsare called stationary states. They also have well-dened energies equal to the

energy eigenvalue E while wavefunctions in general are neither stationary norof well-dened energy. In general, a wavefunction (x) can be expanded inthe energy eigenfunctions as follows,

(x) =XE

cEE(x): (6)

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In quantum chemistry one normally seeks the wavefunction of lowest energy,

the so-called groundstate wavefunction E0 which is an energy eigenfunction.It is generally found by the nite basis set method which is approximativeassuming that the groundstate can be found as a superposition of basisfunc-tions 'i; i = 1;:::N;

E0 =

NXi=1

ci'i: (7)

The coecients fcig form a vector c which is obtained by solving the time-independent Schrdinger equation in matrix form,

Hc = Ec: (8)

Here I have assumed that the basisfunctions are orthonormal, i.e., they satisfyZdx'i (x)'j(x) = ij; (9)

where ij is Kroneckers delta,

ij = 1; ifi = j; (10)

= 0; ifi 6= j:

We shall discuss the nite basis set method in greater detail in the nextchapter. Now we simply note that practical use of quantum mechanics leadsto the eigenvalue problem in matrix form.

1.1.1 The Self Consistent Field (SCF) Aproximation of QuantumChemistry

Before we leave this introduction to the linear algebra of quantum mechanicsI want to deal with one major problem of quantum chemistry, the many-electron problem, and point out the approach to it followed by chemistswith great success. We started by considering one particle moving in onedimension x. A more realistic application has one particle moving in threedimensions x,y,z. The time-independent Schrdinger equation then becomes

~2

2mr2(x;y;z) + V(x;y;z)(x;y;z) = E(x;y;z); (11)

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where

r2 =@2

@x2 +@2

@y2 +@2

@z2 : (12)

This is not too dicult to deal with. We have three coordinates to dealwith rather than one. At this point we would be able to nd the energyeigenvalues and eigenfunctions of the hydrogen atom and the H+2 -ion. Thisis not a bad start but we need to go on to H2; H2O, ..... with more than oneelectron. Now the number of coordinates grows linearly with the number ofatoms and the quantum chemistry starts to look completely intractable foranything but the smallest molecules. The way to deal with this problem isto note that dealing with many electrons is not such a problem if we couldassume that they were moving independently, i. e., without explicit coupling

to each other. If this were the case then the wavefunctions could be taken tobe products,

(r1; r2;:::) = (r1)(r2) ; (13)of one-electron wavefunctions each of which could be obtained from a one-electron Scrdinger equation with the same form of Hamiltonian operator.The corresponding many-electron energy eigenvalues would then be sums ofthe corresponding one-electron eigenvalues. This sounds brilliant but it doesnot work because the electrons interact with each other by the Coulomb re-pulsion between like charges. Moreover, the Pauli principle enters and insiststhat the total electronic wavefunction be antisymmetric with respect to in-

terchange of two electrons. These two mechanisms mean that the electronsare not moving independently. However, the scheme can be carried out ap-proximately in the following way: First we do not use product wavefunctionsdirectly but combine them into Slater determinants which satisfy the Pauliprinciple. This means that no more than two electrons of opposite spin canbe assigned the same one-electron eigenfunction (orbital) in the Slater deter-minant. In order to obtain the lowest energy we stack the electrons in thelowest available orbitals. The electrons still move without explicit coupling toeach other but the pattern of motion has been restricted by the Pauli princi-ple. But how do we choose the Hamiltonian which describes the one-particlemotion? We want this Hamiltonian, which is called the Fock operator inquantum chemistry, to represent the repulsion between the electrons in someaverage way, otherwise the resulting energy and electronic structure will becompletely unrealistic. This can be done by an iterative procedure. The Fockoperator can be found if the occupied orbitals are known. Thus we can start

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with the so-called core Hamiltonian which neglects electron-electron repul-

sion and obtain the corresponding orbitals, n(0)j o, which yield a new Fockoperator F(0). We now use this Fock operator to obtain a new set of orbitals,

which include the eect of electron-electron repulsion,n

(1)j

o: These orbitals

can, in turn, be used to nd an improved Fock operator F(1) and so on. Thevital step in this iteration is the solution of the Fock eigenvalue problem,

F(i)(i+1) = "(i+1)(i+1): (14)

Eventually, in almost all cases, this iterative procedure will converge so thatthe Fock operator eigenfunctions are the same as those orbitals used to con-

struct it to within tolerable accuracy. Then we have found the self consistenteld solution to the electronic stucture of the atom or molecule. It is not pre-cise. We have not accounted for the correlation of the electrons as they move.We have also generally used nite basis sets which cannot completely resolveeven the uncorrelated motion but the accuracy achieved in a good SCF cal-culation is often quite good and the reduction of the problem to one-electronform is so attractive that nearly all quantum chemistry is done this way. Thelanguage of chemistry is dominated by atomic and molecular orbitals whichare SCF constructs. It is often not clear to the practising chemist that theseconcepts are approximate but they are so close to the truth that almost allmethods used to unravel the subtle correlation eects start from the SCF or,

as it is more often called among quantum chemists, the Hartree-Fock theory.

1.2 The Vibrational Modes of Molecules

Even though we often talk about the geometries of molecules as if the atomswere stationary relative to each other this is not the case. There are internalmotions in molecules in the form of rotations and vibrations. If there are Natoms in the molecule then there are 3N-3 rotations and vibrations in themolecule. For a linear molecule there are 2 rotations and 3N-5 vibrationswhile there are 3 rotations in a nonlinear molecule and 3N-6 vibrations. This

means that vibrations soon completely dominate the internal motion as thenumber of atoms increases. Again the exact treatment of the internal motionin a molecule of more than two atoms is very dicult due to the couplingbetween all the dierent motions. However, for small amplitude motion onecan enormously simplify the problem by assuming separable rotations and

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harmonic vibrations. This means that the potential is approximated by a

quadratic form in the coordinates,V(x1; x2; x3;:::::) = V11x

21 + V12x1x2 + ::::: (15)

=1

2

Xi;j

Vi;jxixj;

where Vij is the second derivative of the potential at the minimum,

Vij =@2

@xi@xjV(x1; x2; ::::::) at x1; x2;::::: = x

(min)1 ; x

(min)2 ; ::::::: . (16)

It turns out that in this harmonic approximation one can nd a coordinatetransformation such that the vibrations all separate and become indepen-dent so-called normal modes each performing harmonic oscillatory motionat a well-dened frequency. This results in an enormous simplication ofthe treatment of internal molecular dynamics which is close enough to thetruth for low energies to be of great practical value in chemistry. Thus weshall, after having learnt the necessary prerequesites of linear algebra, returnto consider how to obtain the coordinate transformation which yields thesenormal modes.

2 Vector Spaces and the Eigenvalue Problem2.1 Vector Spaces

A set of elements x;y; z:..... forms a vector space S if addition of twovectors generates another vector in S,

x + y = z 2 S; and x + y = y + x; and (x + y) + z = x + (y + z), (17)

and there exists a zero 0 such that x + 0 = x, and there exists for everyvector x a vector x such that x+ (x) = 0, and multiplication by scalar

numbers ; is possible according to the rulesx 2 S, (x) = ()x; ( + )x =x + x; (18)

(x + y) = x+y; 1 x = x, 0 x =0:

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Denition 1: The vectors x1;x2; xn are said to be linearly inde-pendent if nX

i=1

ixi = 0; ! i = 0; for all i. (19)

Denition 2: Dimension: If there is a set of N linearly independentnonzero vectors feigNi=1 but no set of N+1 such vectors then the vector spaceis N-dimensional and for any x 2 S we have

x =NXi=1

xiei: (20)

The set of vectors feigN

i=1 is called a basis in S. In a given basis the vectorx can be represented by a column matrix of its expansion coecients, e.g.0BB@x1x2x3x4

1CCA.Denition 3: Linear operator: A is a linear operator on S if for x 2 S

also Ax 2 S and A satises the linearity conditionA(x + y) = Ax+ Ay: (21)

Exercise 1 What is the dimension of our Euclidean space? Find a conve-nient basis for it.

Exercise 2 What are the vectors and the linear operators of quantum me-chanics?

Exercise 3 Is the set of vectors

0@ 111

1A ;0@ 10

1

1A ;0@ 13

1

1A a set of indepen-dent vectors?

2.1.1 Matrices

Suppose the linear operator A satises

Aei =NX

j=1

Ajiej, for i = 1, : : : N ; (22)

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then it follows that

Ax = ANXi=1

xiei =NXi=1

NXj=1

Ajixiej: (23)

Thus in a given basis a linear operator A can be represented by a matrixfAijg, where i and j are the row and column indices, respectively, and itsoperation can be described by matrix multiplication, e.g.,

Ax =

A11 A12A21 A22

x1x2

=

A11x1 + A12x2A21x1 + A22x2

: (24)

Linear operators can be multiplied by scalars,

(A)x = (Ax); (A)ij = Aij ; (25)

added,(A+ B)x = Ax + Bx; (A+ B)ij = Aij + Bij ; (26)

and multiplied,

(AB)x = A(Bx); (AB)ij =NXk=1

AikBkj : (27)

There is a null operator 0, such that 0x = 0 for all x, and an identity operatorI such that Ix = x for all x.

Denition 4: Inverse: Certain operators A have inverses A1 suchthat

AA1 = A1A = I: (28)

Denition 5: Determinant: Square matrices such as we have discussedabove posess a determinant det A dened as

detA =

A11 : : : A1N: : : : : : : : :

AN1 : : : ANN

=X

(1)aA1p1A2p2 ANpN; (29)

where the sum is over all permutations of the numbers 1,2,....N, i.e., N! per-

mutations, and the exponent a is 0 for even permutations and 1 for oddpermutations. The even and odd permutations can be distinguished by thefact that the former are constructed by an even number of pairwise inter-changes, the latter from an odd number of pairwise interchanges, startingfrom the original sequence which is even.

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Exercise 4 Find the determinants and, if possible, the inverses of the ma-

trices 1 00 1 ; 1 11 1 ; 1 23 4 :Example 1: 1,2 is an even, 2,1 an odd permutation of 1,2. 1,2,3 is an

even permutation. So are 3,1,2 and 2,3,1 but 2,1,3; 3,2,1 and 1,3,2 are oddpermutations of 1,2,3.

Matrix operations:

Transpose ofA: eA is dened by heAiij

= Aji:

Complex conjugate ofA : A is dened by [A]ij = Aij:

Hermitian conjugate ofA : Ay is dened by Ayij = Aji:Note that Hermitian conjugate is often called adjoint as in the Beta Hand-

book. There it is indicated by superscript * while complex conjugate is de-noted by a bar over the quantity. I am following the standard notation ofphysicists and chemists here.

Some terminology: The matrix is called -

real ifA = A and symmetric if eA = A, antisymmetric if

eA = A and Hermitian ifAy = A;

orthogonal ifA1 = A and unitary ifA1 = Ay; diagonal if Aij = 0 for all i6= j; idempotent ifA2 = A and normal ifAAy = AyA:Denition 6: Trace: The trace of a matrix A is denoted T r(A) and

dened as

T r (A) =NXi=1

Aii: (30)

Exercise 5 Consider the matrix a bici d , where a;b;c and d are realnumbers. Determine its i) transpose, ii) complex conjugate and iii) Her-mitian conjugate. Determine minimal conditions on a;b;c and d such thatthe matrix is iv) real, v) antisymmetric, vi) orthogonal and vii) idempotent.

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2.2 Introduction of a Metric

Denition 7: Scalar product: The bilinear operation denoted by whichtakes two vectors x and y into a scalar xy is called a scalar product if

x y = (y x) ; x (y+z) =x y + x z; x x = 0 only ifx = 0.Denition 8: Norm: jxj = px x is called the norm (or length) of the

vector x.Denition 9: Orthogonality: x and y are said to be orthogonal if

x y = 0:Denition 10: Orthonormality: The basis feigNi=1 is said to be ortho-

normal ifei ej = ij:

2.3 The Eigenvalue Problem

Denition 11: Eigenvector and eigenvalue: x is an eigenvector ofA ifAx =x and is called the eigenvalue ofA corresponding to the eigenvectorx.

How to nd the eigenvalues of an operator A? - We shall assume here

that our operator can be represented as a matrix in an orthonormal basis.Theorem 1: The eigenvalues of the matrix A can be found as the rootsof the secular equation det(A I) = 0.

- I will only sketch a proof here leaving the details to be worked outas an exercise. First we note that the determinant is unaltered if a scalartimes one of the columns of the matrix is added to another column in thematrix. This follows from the fact that matrices with two identical columnshave a vanishing determinant. In turn, this follows from the antisymmetryof the determinant with respect to the interchange of two columns. From theeigenvalue equation Ax =x follows that the columns of the matrix AIare linearly dependent. This means that by adding scalar numbers times the

other columns to one of them one can, with the proper choice of scalars, makethis column consist of only zeros in a matrix which must have unchangeddeterminant. It is then trivial to see that the determinant must be zero.

Note that the secular equation is an Nth order algebraic equation. Thusthere are N roots figNi=1 some of which may be degenerate, i.e., the same.

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Note that for each eigenvalue one can nd at least one corresponding eigen-

vector and at most a number of linearly independent eigenvectors equal tothe degeneracy (multiplicity) of the corresponding eigenvector (root of thesecular equation). Since it is trivial to see that ifx is an eigenvector so is xeigenvectors are only determined up to an arbitrary scalar prefactor. A set ofeigenvectors corresponding to dierent eigenvalues are linearly independent.

2.3.1 Properties of Hermitian matrices

Using long-hand notation for scalar product such that x y is writtenas (x;y) we have in general (x;Ay) = (Ayx;y) and for Hermitianmatrices (x;Ay) = (Ax;y).

The eigenvalues of an Hermitian operator are real and the eigenvectorscorresponding to dierent eigenvalues are orthogonal. There are Nlinearly independent eigenvectors.

The proof that the eigenvectors are real and orthogonal follows readilyfrom consideration of the scalar product (x1;Ax2) and use of the aboveHermitivity relation and the eigenvalue relation.

Exercise 6 Prove explicitly that (x;Ay) =Ayx;y

in the case of a 2x2

matrix.

Exercise 7 Prove explicitly that if Ay = A then eigenvectors of dierenteigenvalues are orthogonal and the eigenvalues are real.

How does one nd the eigenvector(s) corresponding to a given eigenvalue?- Given the eigenvalue, the eigenvalue equation Ax =x turns into a set ofcoupled linear equations which can be solved by stepwise elimination. Notethat since the eigenvector is arbitrary up to a scalar prefactor one componentof the eigenvector will be undetermined or can be set to unity unless thiscomponent turns out to be zero. If so one simply picks another componentto set to unity or some other convenient value.

Example 8 Find the eigenvalues and eigenvectors of the matrix 4 1

1 4

.

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Solution: the secular equation is -

(4 )2 1 = 0:

The eigenvalues are then 1 = 5 and 2 = 3 with corresponding eigenvectorsobtained by solving the equations -

4x1 + x2 = x1;

x1 + 4x2 = x2:

The rst equation yields -

x2 = (

4)x1;

and the second yields -x1 = ( 4)x2:

Insisting that these two equations give the same eigenvectors leads to thesecular equation and the two possible eigenvalues. A convenient way toproceed is to choose x1 = 1 and then nd the two eigenvectors -

x1 =

11

;

x2 = 11 :Note that these eigenvectors are orthogonal - as they must be for a Hermitianmatrix. In order to make them orthonormal we multiply them by 1=

p2. -

In some cases, namely when x1 = 0, the convenient way to proceed abovedoes not work. One can then instead choose x2 = 1.

2.4 The Generalized Eigenvalue Problem

The operator eigenvalue above could be written as a matrix equation. Suchan equation can be generated from an operator equation by use of an ortho-normal basis fejgn1 . The operator equation -

Aopx =x (31)

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can be turned into a matrix equation in the space spanned by the basis

vectors by projecting both x and the operator equation onto the space of thebasis vectors, i.e. -

x =nX

j=1

xjej ; (32)

(ei;Aopx) =

ei;Aop

nXj=1

xjej

!=

nXj=1

xj (ei;Aopej) =nX

j=1

Aijxj

= (ei;x) = nX

j=1

xj (ei; ej) = xi: (33)

In the case of a general basis which is not orthonormal the matrix equationis -

Ax = Sx; (34)nX

i=1

Aijxj = nX

i=1

Sijxj ; (35)

which is the generalized eigenvalue problem. This problem can be convertedto normal form by multiplication with the inverse of S -

S

1

Ax = S1

Sx =x; (36)Asx = x; where As = S

1A. (37)

The projection of an operator eigenvalue problem into a reduced spacespanned by a given basis and a corresponding matrix form is called theGalerkin method and used very commonly in physics and chemistry.

2.5 Linear Algebra Exercises:

1. Try to convert the chemical problem of writing a stoichiometricallybalanced equation for a reaction with given reactants and products into

a mathematical problem. a) Can molecules be treated as basis functionsin a space of all possible chemical species? b) What distinguishes thespace of molecules from a vector space? c) Can you suggest a way to usevector space methods to solve our problem of balancing a stochiometricequation?

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2. Let the matrices A and B be dened by

A =

0@ 2 1 01 0 10 1 2

1A and B =0@ 1 4 72 5 83 6 9

1A :a) Obtain the determinants of the two matrices. b) Obtain A2;B2;ABexplicitly. c) Obtain the eigenvalues ofA.

3. Verify Theorem 1 explicitly for 2 2 matrices.

4. Show that for any square matrices A and B we have a) (AB) =

eB

eA

and b) (AB)y = ByAy:

5. Show that ifU is a unitary matrix then its eigenvalues fg satisfyjj = 1:

6. Find the eigenvalues and eigenvectors of the matrix A =

1 23 7

:

7. Find the eigenvalues and eigenvectors of the matrix B =

0@ 1 0 20 5 03 0 7

1A :8. Prove that eigenvectors of a linear operator corresponding to dierent

eigenvalues must be linearly independent.

9. Under what conditions can the linear equation Ax = b be uniquelysolved? Here A is an n n matrix and x;b are n-dimensional vectors.Give proof of your conclusion.

10. Prove that (x;Ay) =Ayx;y

.

11. The concept of "linearity" is important and has many applications inchemistry. It implies a dramatic simplication by comparison with themore general "nonlinear" case. We are, e.g., lucky that the Schrdinger

equation is linear so that nearly all of quantum chemistry reduces tolinear algebra. Chemical phenomena are, however, generally nonlinear.An important illustration of this can be found in the current debateconcerning chemicals in our bodies and our environment. Imagine that

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- as a person with knowledge of chemistry - you have been asked to pro-

nounce whether a certain chemical is poisonous or not. How would youexplain the mathematical content of this question in terms of conceptsof linearity and nonlinearity? Note that nonlinearity is the complementof linearity, i.e. any relation which is not linear is nonlinear.

3 Hckel Theory of Delocalization

The Hartree-Fock theory shows us how we can, to a good approximation,reduce the problem of nding the total energy E0 of a molecule to one-electron form. We nd the Fock operator F which plays the role of a one-

electron Hamiltonian operator from which one-electron (canonical orbital)wave functions can be found by solving

F= ": (38)

The Fock operator is obtained by a self-consistent iterative procedure andit depends on the occupied electronic orbitals

j

nj=1

themselves. Once we

have chosen a basis set f'lgNl=1 in which to resolve the canonical orbitals thenthe Fock equation 38 turns into a matrix equation

Fcj = "jScj; (39)

where

Fmj =

Zdr'm(r)F 'j(r); (40)

Smj =

Zdr'm(r)'j(r): (41)

Here S is the overlap matrix which becomes equal to the unit matrix if thebasis functions are orthonormal. Note that 39 can be derived from 38 bytaking scalar products with respect to basis functions and expanding theorbital in the basis, i.e.,

j =

NXm=1

cmj'm: (42)

In the 1930s Hckel developed a very simplied form of Hartree-Focktheory for planar aromatic molecules. He noted that if the molecule liesin the x,y-plane then the carbon 2pz orbitals stick out orthogonally to the

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plane. These orbitals do not mix with the in-plane sigma orbitals. Thus

we can develop a reduced Fock matrix in the space of C 2pz-orbitals. Thecorresponding molecular orbitals obtained by diagonalizing this reduced Fockmatrix are called -orbitals and the electrons assigned to them are called -electrons. The electron assignment to orbitals follows the Aufbau principle:Fill the orbitals in order of increasing orbital energy but place no more thantwo electrons (of opposite spin direction) in each orbital not to run afoul ofthe Pauli principle.

The calculation to obtain the Fock matrix, or the -part of it, rigorouslyby the Hartree-Fock method is relatively hard numerical work. In the 30s itmust have been impossible for most molecules. Hckel had the idea that itmay be possible to construct the reduced Fock matrix for the -electrons, h

or just h below, by empirical means. He proposed to use the C 2pz orbitalson all double bonded carbons as a minimal basis set. Thus the matrix h wasa square matrix of the same order as the number of double bonded carbonatoms in the planar molecule, Nc. Next Hckel assumed that the diagonalterms in this matrix, hjj ; were equal to the carbon 2pz atomic orbital energy,

hjj = "C2p = : (43)

Then he assumed that coupling only existed between bonded carbon atoms(nearest neighbour carbon atoms). The same approximation is justied forthe overlap matrix also and we end up with a reduced Fock equation for, as

an example butadiene (C4H6 in a linear chain geometry for the carbons), ofthe type 0BB@ 0 0 00 0 0

1CCA0BB@

c1c2c3c4

1CCA = "0BB@

1 s 0 0s 1 s 00 s 1 s0 0 s 1

1CCA0BB@

c1c2c3c4

1CCA : (44)This is a generalized eigenvalue equation due to the presence of the overlapmatrix S on the left. It is not much more dicult to solve this generalizedform of the eigenvalue problem but Hckel simplied it further by settings = 0. Thus, for butadiene, he obtained0BB@

0 0 00 0 0

1CCA0BB@

c1c2c3c4

1CCA = "0BB@

c1c2c3c4

1CCA : (45)17


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Note that the coupling term ,

= hjm; with j bonded to m, (46)is taken to be independent of which bond is referred to. In order to write downthe Hckel matrices one needs to number the atoms in some reasonable wayand keep track of all coupled and uncoupled carbon atom pairs. When theindices in hjm refer to uncoupled carbon atoms the matrix element vanishesso for large molecules there will be a lot of zeroes. The coupling term isoften called the resonance integral. It is responsible for the delocalizationof -electron motion in the molecule. Thus it is also responsible for thelowering of the energy which is the cause of the resonance stabilization ofthe aromatic or conjugated molecules. Note that both and are negative

numbers representing a negative atomic C2p-energy and a coupling energy.In order to solve the Hckel eigenvalue problem one usually works in

energy units such that = 1. Then the secular equation becomes

det

0BB@" 1 0 01 " 1 00 1 " 10 0 1 "

1CCA = 0: (47)Here we have dened " as

" = ( ")= = (" )=(): (48)

The determinant can be worked out by noting that the only nonvanishingmatrix element products are the (1; 2; 3; 4), (2; 1; 3; 4), (1; 2; 4; 3), (1; 3; 2; 4),(2; 1; 4; 3) permutations leading to the secular equation -

"4 3"2 + 1 = 0: (49)We solve rst for "2 and nd -

"2 =3 p5

2: (50)

The corresponding solutions for " are obtained as -

" = s3 p52

= p5 12

! = 0:618; (51)" =

s3 +

p5

2=

p5 + 1

2

!= 1:618:

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The set of eigenvectors and eigenvalues are -

1 = 0:3717('1 + '4) + 0:6015('2 + '3), "1 = 1:618;

2 = 0:6015('1 '4) + 0:3717('2 '3), "2 = 0:618;

3 = 0:6015('1 + '4) 0:3717('2 + '3), "3 = 0:618;

4 = 0:3717('1 '4) 0:6015('2 '3), "4 = 1:618:

The energy eigenvalues are displaced down and up from the C2p energy ina symmetric fashion. Note that the eigenfunctions are symmetric or antisym-

metric to reection in the midpoint. Note also the linear growth in the num-ber of nodes with excitation. The expansion coecients change with carbonatom somewhat like the eigenfunctions of the one-dimensional particle-in-a-box problem. When we work out the contribution to the total bindingenergy we rst nd the number of electrons to be placed and then placethem in the available orbitals from the lowest and up in accord with theAufbau principle which restricts the number of electrons in an orbital to 0,1 or 2. In the case of butadiene we have four electrons which ll the twolowest orbitals. The binding energy contribution is 2(1.618+0.618)=4.472 in- units. If we instead considered the butadiene molecule to contain pair-

wise localized and uncoupled bonds, C=C-C=C, then the binding energycontribution would turn out to be -4 as will be clear below. Thus the extrabinding energy due to the delocalization of the electrons over more thantwo nuclei is -0.472 here. This we shall call the resonance stabilization ofthe butadiene molecule.

How can we obtain estimates of the two parameters and ? The rst, ,is supposed to be the C2pz orbital energy. It can be estimated as minus theionization energy of the carbon atom which is 1090 kJ/mol. This parameterdoes not play a very important role in the Hckel theory. The more impor-tant parameter is , the coupling constant, which delocalizes the electronsand stabilizes favored structures. In order to estimate we shall consider

the simplest molecule containing a -bond, i.e., ethylene C2H4. From a ta-ble of average molecular bond energies ( Table 13.5 in Zumdahl, ChemicalPrinciples) we nd that a C=C bond consisting of a and a bond theaverage bond energy is 614 kJ/mol while for the C-C bond the energy is347 kJ/mol. Thus it appears that the bond energy can be estimated for

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the ethylene molecule to be 267 kJ/mol. Let us now apply Hckel theory

to ethylene to nd the predicted bond energy in terms of . The Hckelhamiltonian ish =

: (52)

The reduced secular equation becomes -

det

" 11 "

= 0: (53)

The solution is " = 1. The lower orbital will be occupied by two electronseach contributing an equal amount to the bond energy. Thus the bondenergy is in reduced units of - equal to 2. Setting this equal to 267 kJ/molwe nd that = 134 kJ/mol in SI-units. This should be considered a veryapproximative estimate but it gives us a plausible magnitude and an exampleof how could be obtained.

3.1 Hckel exercises:

1. One might expect on the basis of standard bonding pictures that themiddle carbon-carbon bond in butadiene is weaker and longer due todominant single bond character. Estimate the resonance stabilizationenergy of butadiene (i.e., that part of the binding energy due to the fur-

ther delocalization of the -electrons in the butadiene chain) in kJ/molin the case when the coupling between the two middle carbon atoms,due to a longer bond, is half the normal value. Compare with the valuefor the standard model with all couplings the same.

2. Work out the Hckel orbitals for butadiene, i.e., verify the results statedin the text showing the algebra explicitly.

3. Write down the Hckel matrix for benzene. Use symmetry to obtainthe form and the energy of the lowest lying -orbital.

4. Consider the square planar molecule cyclobutadiene C4H4 obtained bytying the ends of the butadiene molecule together with the loss of twohydrogen atoms. Write down its Hckel matrix and calculate the cor-responding -orbitals and orbital energies. What is the total resonancestabilization energy?

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5. Obtain the electron contribution to the binding energy of the squareplanar N4-molecule and estimate the corresponding resonance energydue to the further delocalization of the electrons. You may use thebond energies N-N 160 kJ/mol, N=N 418 kJ/mol, in your estimation.

4 Vibrational Modes of a Molecule

Our purpose here will be to describe the motion of the nuclei in a molecule,e.g., the water molecule H2O. The exact solution to this problem is intractabledue to coupling between the motion of electrons and nuclei and betweennuclei due to anharmonic eects. However, on the basis of a number of

simplications we shall be able to get close to reality without losing toomuch in accuracy. The most important simplications are -

the point particle model - i.e., the electrons are neglected and the atomsbecome point particles interacting by a potential V(r1; r2;....) (actuallydue to the electrons), where r1; r2;::: are the spatial coordinates of theatoms;

classical mechanics - while we really ought to use quantum mechanics; small amplitude vibrations - anharmonic eects and vibration-rotation

coupling will be neglected.

4.1 The One -Dimensional Oscillator

Let the potential acting on a particle moving in one dimension be V(x) andthe mass of the particle m. We assume that V(x) has the general characterof a bond potential in a diatomic molecule, i.e., a well dened minimum.As the temperature decreases the bondlength decreases to become equal tothe so called equilibrium value xe at T=0 K. The particle would then in oursimple picture sit motionless at the bottom of the potential well. We cannd xe by examining the stationary points satisfying

@V

@x= 0; and

@2V

@x2> 0: (54)

These conditions identify a potential minimum. If there are several minimawe choose the point where the potential is the smallest.

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We now apply the harmonic approximation by expanding the potential

in a Taylor series around the point x=xe and retaining only the rst threeterms,

V(x) = V(xe) + (x xe)V0 + 12

(x xe)2V00; (55)

= V(xe) +1

2(x xe)2V00:

Here we have used the fact that V0 vanishes at x = xe. Without loss ofgenerality, as the mathematicians like to put it, we will now choose an energyscale such that V(xe) = 0 and the origin on the x-axis such that xe = 0. Thenthe potential can be written as

V(x) = 12

kx2; with k = V00(xe). (56)

Chemists call k the force constant. A particle moving in such a potentialis called a harmonic oscillator. We shall now review the solution for theharmonic oscillator motion. We start from Newtons equation of motion -

m@2

@t2x(t) = @

@xV(x); (57)

which yieldsd2

dt2

x(t) =

k

m

x(t): (58)

In order to solve this second order ordinary dierential equation we notethat it is linear and use a method to be discussed in greater detail in a laterchapter. Shifting the RHS over to the left we rewrite the equation as

(d

dt+ i

rk

m)(

d

dt i

rk

m)x(t) = 0: (59)

It is now clear that since the factors can be written in any order any sum ofsolutions to the two rst order equations

(d

dt+ ir

k

m)x(t) = 0; or (

d

dt ir

k

m)x(t) = 0; (60)

will be a solution of the second order equation 58. The solutions to the tworst order equations are -

x(t) = aeip

k=mt; and x(t) = beip

k=mt; (61)

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respectively. Thus the harmonic oscillator solution is

x(t) = aeipk=mt + beipk=mt; (62)but since x(t) must be real the solution can be reduced to the form -

x(t) = A sin(p

k=mt + ), (63)

where A is a real amplitude, i.e., the maximal excursion of the oscillationfrom the origin, and is an arbitrary initial phase, 0 2, of theoscillator.

4.2 Multidimensional vibrations in molecules

If the molecule is made up of N atoms there are 3N spatial coordinates sothe problem is now n-dimensional where n is an integer larger than one. The3N spatial degrees of freedom can be divided into center of mass, rotationaland vibrational subsets as shown below.

# of dof linear nonlinearc of m 3 3rot 2 3vib 3N-5 3N-6

The potential now depends on the spatial coordinates, V(x1; x

2;:::;x

n): If

we are describing a molecule in eld free space then the potential must beinvariant to translation (center of mass coordinates) and rotation (rotationalcoordinates). This makes it useful to transform our analysis into a set ofinternal bond coordinates explicitly accounting for the invariance of V toexternal motions. However, the gain in lower dimensionality is more thanoset by loss of simplicity so we shall stick with our initial representationin terms of cartesian atomic coordinates. The global minimum is found bysolving for all extrema given by the set of equations -

@

@xi

V(x1; x2;:::;xn) = 0; i=1,2,.....,n. (64)

It is now a little more dicult to determine whether we have found a maxi-mum or minimum but evaluating V at the point and choosing the extremumgiving the lowest potential should work for all reasonable vibrational poten-tials. Once the global minimum is found we transform our coordinate system

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to make this point the origin. Thus we shall take x1e; x2e;::: all to vanish in

our discussion below. We also take the potential energy to be zero at theminimum. From the multidimensional Taylor series expansion (see Appen-dix) follows that in the harmonic approximation where terms of order higherthan quadratic in the coordinates vanish we get -

V(x1; x2;:::;xn) =1

2

nXi=1

nXj=1

Vijxixj ; (65)

where

Vij =

@2

@xi@xjV(x1; x2;:::;xn)

x=0

= Vji: (66)

We are now ready to consider the multidimensional motion which is de-scribed by Newtons equation suitably generalized to the multidimensionalcase,

d2

dt2xk(t) = 1

mk

@

@xkV(x1;:::xn) = 1

mk

nXl=1

Vklxl(t); for k=1,2,...,n. (67)

We have a set of n coupled linear second order dierential equations. Thissounds, perhaps, quite intractable for large n but we shall see that it ispossible to nd a coordinate transformation x ! y such that we end upwith n uncoupled equations of motion of the type 58 which we have alreadysolved, i.e.,

d2

dt2yk(t) = Ukkyk(t); for k=1,2,....,n, (68)

with solutions -

yk(t) = Ak sin(p

Ukkt + k); for k=1,2,....,n. (69)

Here the factorp

Ukk = !k is called the frequency of the kth vibrationalmode and is normally given in radians per second. We often refer to thevariables fykgnk=1 as the normal modes since they are decoupled from eachother.The coordinate transformation we shall use is linear and given by -

yk =nX

l=1

Sklxl; for k=1,2,....,n, (70)

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or in vector notation -

y = Sx: (71)The task that remains is to nd the matrix S and the frequencies f!kgnk=1 :

Note rst that Newtons equations in the x-coordinates can be rewrittenas

d2

dt2x = Rx; (72)

where

Rkl =1

mkVkl: (73)

In order to nd the corresponding equation for y we multiply both sides byS,

d2

dt2Sx = SRx = SRS1Sx; (74)

which can be rewritten asd2

dt2y = Uy: (75)

Here we used the fact that S1S = I; the identity operator, and denotedSRS1 as U:

Theorem: Let feigni=1 be the basis vectors of our vector space of posi-tions so that any vector x can be expanded in this basis,

x =nX

i=1

xiei. (76)

In our case feigni=1 will be the set of cartesian basis vectors. Suppose that Rhas n linearly independent eigenvectors fxigni=1 such that

xl =nX

i=1

xliei: (77)

Then, ifS is chosen so that

S1ei = xi; i = 1; 2;:::;n; (78)

i.e., the eigenvectors ofR become columns ofS1,S1

li

= xil = [xi]l ; (79)

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Now note that

SRS1 = SM1(MRM1)MS1 = ARA1; (87)

where

Rkl =1p

mkmlVkl = Rlk: (88)

Thus the transformation to mass weighted coordinates produces a couplingmatrix R which is Hermitian and has n orthonormal eigenvectors fxigni=1.By our theorem we then nd that Ali = xli, and from A = SM

1 followsthat

S = AM; and Skl =n

Xi=1 AkiMil =n

Xi=1 xkip

miil =p

mlxkl: (89)

The frequency corresponding to the kth mode is !k =p

Rk; where

Rxk = Rkxk: (90)

Note when the masses are dierent S is no longer orthogonal, i.e., the normalmodes are no longer orthogonal to each other. In calculations for realisticmolecular models one will nd that the normal modes corresponding to centerof mass translation and rotation correspond to zero frequency modes, ! = 0:Note that when working out normal mode coordinates we divide by someappropriate

pmass to get -

yk =nX

l=1

rmlmass

xklxl: (91)

4.2.1 Summary of steps in the normal mode analysis

1. Make sure you have masses m1;:::mn and a potential V(x1;:::xn).

2. Find minimum ofV(x1;:::xn) and double derivatives Vij = @2

@xi@xjV(x1; ::::xn)

at the minimum.

3. If the masses are not the same then we create the Hermitian matrix R

dened by -Rkl =

1pmkml

Vkl:

If the masses are all the same then we can proceed as below noting thatR =R.

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4. Solve the eigenvalue problem Rx = Rjxj for n orthonormal eigenvec-

tors with corresponding real eigenvalues.5. Obtain the vibrational frequencies as !k =

pRk and the corresponding

normal mode coordinate as yk =Pn

l=1

pml

massxklxl where mass can betaken to be any convenient mass, e.g., one of the atomic masses orthe average atomic mass. Note that the masstretched components ofthe eigenvectors of the coupling matrix R become the normal modecoordinates.

4.3 Example 1 - A Two-Dimensional Vibration

Suppose that two particles of mass m1 = m and m2 = 3m, respectively, areperforming vibrations in a potential V(r1; r2) = 10 + r1 2r2 + r21=2 + 2r22 +r1r2: Let us nd the normal modes of this motion. We begin by nding theequilibrium geometry of the vibrational motion, i.e., the potential minimum.

@

@r1V = 1 + r1 + r2 = 0; (92)

@

@r2V = 2 + 4r2 + r1 = 0;

r2;eq = 1; r1;eq = 2:

Next we rewrite the potential in new coordinates x1; x2 dened as x1 =r1 r1;eq = r1 + 2 and x2 = r2 r2;eq = r2 1. We get -V(x1; x2) = 8 + x

21=2 + 2x

22 + x1x2: (93)

The V-matrix is then -

V =

1 11 4

; (94)

and the symmetrized R-matrix is -

R =1

m 1 1=

p3

1=p

3 4=3 : (95)The matrix mR has the eigenvalues and eigenvectors -

1 1=p

3

1=p

3 4=3

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, eigenvectors: 1p

3 1 + p13 =6 $ 7 +p

13 =6; 1p

3 1 p13 =6 $7 p13 =6. Thus the vibrational frequencies are - !1 = q76 + 16p13=pm

and !2 =q

76 1

6

p13=

pm. The corresponding normal mode coordinates

(unnormalized) are -

y1 = x1 +p

3

p3

7

6+

p13

6

!

p3

!x2 = x1 +

1 +

p13

2

x2;

y2 = x1 +p

3

p3

7

6

p13

6

!

p3

!x2 = x1 +

(1 p13)2

x2: (96)

4.4 Example 2 - Normal Modes of a Linear Chain

Molecule

Let us consider a molecule consisting of three atoms attached to each otherby chemical bonds represented by Morse potentials, (r) = D(e2(rre) 2e(rre)); and conned to move in one dimension. We shall take the massesto be the same. The coordinate r is the bondlength. The Morse potentialreaches its minimum potential energy of D, the bond dissociation energy, atthe equilibrium bondlength re: The full potential can then be written as

V(r1; r2; r3) = D(e

2(r2r1re)

2e(r2r1re)

+e

2(r3r2re)

2e(r3r2re)

):(97)The global potential minimum occurs when r2 r1 = r3 r2 = re; wherethe total potential energy is 2D. The minimum we shall choose to expandaround is r1 = 0; r2 = re; r3 = 2re. Our x-coordinates measuring deviationfrom the equilibrium positions are then x1 = r1; x2 = r2 re; x3 = r3 2re:In these coordinates the potential becomes

V(x1; x2; x3) = D(e2(x2x1) 2e(x2x1) + e2(x3x2) 2e(x3x2)): (98)

The rst derivatives are@V

@x1= 2D(e2(x2x1)

e(x2x1)); (99)

@V

@x2= 2D(e2(x2x1) + e(x2x1) + e2(x3x2) e(x3x2));

@V

@x3= 2D(e2(x3x2) e(x3x2)):

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Note that the derivatives vanish at x1 = x2 = x3 = 0: Using the denition

Vij =@2

@xi@xjV; (100)

where the double derivative is evaluated at the origin, we get

V11 = 2D2 = V33; (101)

V22 = 4D2;

V12 = 2D2 = V21 = V23 = V32;V13 = V31 = 0:

The corresponding R matrix is

R =1

m2D2

0@ 1 1 01 2 10 1 1

1A = 2D2m

bR: (102)Now we are ready to nd the eigenvectors of bR: We have -

det

0@ 1 1 01 2 10 1 1

1A = (1 )2(2 ) 2(1 ) (103)=

3 + 42

3:

The roots are 1 = 0; 2 = 3; 3 = 1: The corresponding orthonormal eigen-vectors are

x1 =1p

3

0@ 111

1A ;x2 = 1p6

0@ 121

1A ;x3 = 1p2

0@ 101

1A : (104)It follows that the normal modes are

y1 =1p

3x1 +

1p3

x2 +1p

3x3; !1 = 0; translation, (105)

y2 =1p

6x1 2p

6x2 +

1p6

x3; !2 = r6D

m; antisymmetric vibration,

y3 =1p

2x1 1p

2x3; !3 =

r2D

m; symmetric vibration.

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4.5 Exercises on Normal Mode Analysis

1. A particle of mass m is hanging vertically from the ceiling in a har-monic spring described by the wall-particle potential V(x1) = kx21=2.The gravitational potential can be taken to be VG(x1) = gmx1: a)What is the equilibrium position and vibrational frequency of the ver-tical vibration? b) Obtain the equilibrium geometry, vibrational modesand corresponding frequencies for the case when a second particle ofthe same type is attached to the rst particle by a harmonic springcorresponding to the potential V(x1; x2) = k(x1 x2)2=2:

2. Find the vibrational frequencies of the triatomic one-dimensional chainin the example above in the case when the mass of the central particle

is four times larger (4m) but all other parameters of the system remainthe same.

3. Obtain the normal modes and corresponding frequencies of a systemconsisting of two identical particles of mass m moving in one dimensionx and interacting by the Lennard-Jones potential -

V(x1; x2) = 4"

"

x12

12

x12

6#;

where x12 = jx1 x2j.4. Reconsider the system in Exercise 3 above generalized to the case when

the particle masses are not equal m1 6= m2.5. Find the harmonic vibrational frequency of the diatomic molecule A2

with the bond potential V(r) = D (exp (2ar2) 2 exp(ar2)) :

4.6 Appendix - Multidimensional Taylor Series Expan-sion

Recall the form of the one-dimensional Taylor series expansion,

V(r0 + x) = V(r0) + x@V

@rr=r0

+12

x2@2V

@r2r=r0

+ :::::: (106)

=1Xn=0

1

n!xn

@nV

@rn

r=r0

;

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where r = r0+x: The corresponding multidimensional Taylor series expansion

is

V(r0 + x) = V(r0) +

NXi=1

xi

@V

@ri

r=r0

+1

2

NXi=1

NXj=1

xixj

@2V

@ri@rj

r=r0

:::::::::(107)

=1Xn=0

1

n!

x @

@r

nV(r)

r=r0

:

Probably the easiest way to prove this expansion is to convert it to a one-dimensional expansion by setting x = s

bx, where

bx is a directional vector and

s tells us how far in this direction we have gone. Then we can, of course,

expand with respect to s to get

V(r0 + sbx) = 1Xn=0

1

n!sn

@nV

@sn

s=0

: (108)

But now we note that

@

@sV(r0 + sbx) = NX

i=1

bxi @V@ri

r=r0

; (109)

@2

@s2V(r0 + sbx) =NXi=1

NXj=1bxibxj @2V@ri@rj r=r0 ; (110)

and so on. Thus we can write

sn@n

@snV(r0 + sbx) = x @

@r

nV(r0+x): (111)

Insertion of this result in 108 above yields the multidimensional form of theTaylor series expansion.

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are all orthogonal in our Hilbert space, i.e.,

Z

dx sin mx cos nx = 0; for all m,n, (113)Z

dx sin mx sin nx =

Z

dx cos m cos nx = m;n, for m,n>0,

cos0x = 1 !Z

dx cos0x cos mx = 20;m;Z

dx cos0x sin nx = 0; for n > 0.

Using these functions as basis functions we can expand f(x) as

f(x) = a0 +1Xn=1

(an cos nx + bn sin nx) ; - < x < ; (114)

where the expansion cecients can be found by taking scalar products ofboth sides of the equality 114 with respect to each basis function in turn.We get

a0 =1

2

Z

dxf(x); (115)

an =1

Z dx cos nxf(x); n = 1,.....1;bn =

1

Z

dx sin nxf(x); n = 1,......1:

This denes the Fourier expansion of f(x) on [; ] : The identity of theoriginal and the expanded function is not absolute but of a weak sensemeaning that

(g; f) =

g; a0 +

1

Xn=1(an cos nx + bn sin nx)

!(116)

for any function g(x) in our vector space. Although the mathematicians callthis sense of equality weak we nd in physics and chemistry that it is quitestrong. We rarely need to distinguish between weak and absolute point bypoint equality.

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Note that all the basis functions are periodic with the period 2; i.e., for

an integer m we getcos(nx + 2m) = cos nx; n = 0; 1; .....1; (117)sin(nx + 2m) = sin nx; n = 1; .....1:

It follows that the expanded function also is periodic with the period 2;

f(x + 2m) = f(x): (118)

Thus, irrespective of what the original function was up to, the Fourier ex-pansion outside the chosen interval [; ] produces a periodic function bythe relation 118.

5.2 Arbitrary Interval

The Fourier expansion above can easily be generalized to an arbitrary niteinterval [x0; x0 + L] : This is done by recognizing that the variable transforma-tion y=(2(x-x0)=L) transforms the interval [x0; x0 + L] back to the inter-val [; ] : Thus our new basis functions are fcos(2n(x x0)=L) )g1n=0[fsin(2n(x x0)=L) )g1n=1 : However, using the trigonometric relations

cos( ) = cos(); (119)sin( ) = sin();

and the fact that the sign of a basis function can be changed without dif-culty we can simplify the basis to the form fcos(2n(x x0)=L))g1n=0 [fsin(2n(x x0)=L))g1n=1 : The Fourier expansion then becomes

f(x) = a0+1Xn=1

(an cos(2n(xx0)=L)+bn sin(2n(xx0)=L)); x0 < x < x0+L;(120)

where the expansion coecients are obtained as

a0 =1

L Zx0+L

x0

dxf(x); (121)

an =2

L

Zx0+Lx0

dx cos(2n(x x0)=L)f(x); n = 0; 1; .....1;

bn =2

L

Zx0+Lx0

dx sin(2n(x x0)=L)f(x); n = 1; .....1:

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When can a function be expanded in a Fourier series on a nite interval?

There are two answers to this question:Theorem - Weak formulation: The Fourier basis functions form a com-plete set on the nite interval for which they were chosen.

Theorem - Strong formulation - Dirichlets theorem: Suppose f(x) iswell-dened on [x0; x0 + L], is bounded, has only a nite number of maximaand minima and discontinuities. Let f(x) for other values of x be dened byperiodicity, f(x+nL)=f(x) for n=integer, then the Fourier series 120 for f(x)converges to (f(x+)+f(x-))/2 for all x.

The rst theorem simply states that the Fourier expansion gives a uniquemapping off(x) into a vector in our vector space of functions such that allscalar products with functions in our vector space are reproduced by the

Fourier expansion. As noted above, this is nearly always sucient for us.The second theorem is deeper and describes what happens at each point xgiven some conditions on the function.

5.3 Complex form

At the minor cost of working with complex basis functions we can give theFourier series expansion its most general form. Suppose that we want toconsider a function f(x) on the interval [L=2; L=2] : We can then use thebasis functions fexp(i2nx=L)g1n=1 which satisfy the orthogonality relation

(ei2mx=L; ei2nx=L) =ZL=2L=2

dxei2mx=Lei2nx=L = Lm;n for m,n=integers.

(122)The Fourier expansion then takes the form

f(x) =1X

n=1

dnei2nx=L; (123)

with expansion coecients dened by

dn = 1LZL=2L=2

dxei2nx=Lf(x); n=-1::: 1; 0; 1; :::::1: (124)

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2. Obtain the Fourier expansion off(x) = x2 on the interval [r; r] usingrst the real form then the complex form of the expansion.

3. Obtain the Fourier expansion of the function f(x) = x on the interval[; ]. Draw f(x) and its approximation fn(x) = a0 + :::an cos nx +bn sin nx for n = 0; 1 and 2. Discuss the observed approach of fn to fwith increasing n. Why might this particular function be a harsh testof the convergence of a Fourier expansion?

6 Fourier Transforms

The Fourier series expansion is applicable on a nite interval and to periodic

functions on an innite interval. Naturally one would like to be able tohandle functions on the entire line -1 < x < 1 whether or not they areperiodic. The required generalization to accomplish this will lead us to theFourier transform. We shall start from the complex form of the Fourier seriesexpansion and write it in a suggestive form as follows,

f(x) =1X

n=1

k ef(kn)eiknx; (127)where kn = 2n=L and k = kn kn1 = 2=L. The coecient dn has beenrenamed k ef(kn); i.e.,ef(kn) = 1

2

ZL=2L=2

dxeiknxf(x): (128)

Insertion into 127 yields

f(x) =1X

n=1

k1

2

ZL=2L=2

dx0eiknx0

f(x0)eiknx: (129)

Now we are ready to take the limit as L ! 1: Instead of the discretecoecients kn we now get a continuous parameter k and the sum over nbecomes an integral over k. If all integrals converge properly we have

f(x) =Z11

dk ef(k)eikx; (130)ef(k) = 1

2

Z11

dxf(x)eikx:

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Note that we have found a pair of functions which are images in either x- or

k-space of one function. If we know f(x) we can generate ef(k) and vice versa.We say that ef(k) is the Fourier transform of f(x). The expressions for f(x)and ef(k) look very similar and can be made to look even more similar if wesymmetrize the transform. We dene a new transform f(k) by

f(k) =p

2 ef(k): (131)The new transform can then be seen to satisfy the relations

f(x) =1p2

Z11

dkf(k)eikx; (132)

f(k) = 1p2

Z11

dxf(x)eikx:

Example 2: Let us obtain the Fourier transform of the function -

f(x) = exp(ax); x > 0;= 0; x 0:

Ifa is real and greater than zero we have -

f(k) =1

p2 Z1

1

dxeikxf(x) =1

p2 Z1

0

dxeikxax

=1p2

1a + ik

e(a+ik)x10

=1p2

1

a + ik:

Finally, to more clearly expose the real and imaginary parts, we might writethe result above in the form -

f(k) =1p2

a ika2 + k2

:

6.1 The -Function

By inserting the expression for the Fourier transform in the expression forthe function f(x) we get

f(x) =

Z11

dk1

2

Z11

dx0f(x0)eikx0

eikx: (133)

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By inverting order of integration we nd then that

f(x) = Z11

dx0f(x0)(x x0); (134)

where the so called function is dened by

(x x0) = 12

Z11

dkeik(xx0): (135)

The denition of a -function requires that it satisfy the integral relation

g(x) =

Zb

a

dx0(x x0)g(x0), ifa < x < b, (136)

but there are many forms for it other than that in 135. The -function is ageneralized sort of function, - a limit of a sequence of functions, e.g.,

(x x0) = lim!1

1

2ejxx

0j: (137)

Next we shall consider the lengths of f(x) and its Fourier transform f(k)in their respective Hilbert spaces.

Parsevals Theorem: The L2-norms of the function f(x) and its Fouriertransform f(k) are the same, i.e.,

Z11

dx jf(x)j2 = Z11

dk f(k)2 : (138)Proof:Z1

1

dx jf(x)j2 =Z11

dxf(x)f(x) (139)

=

Z11

dx1p2

Z11

dkf(k)eikx1p2

Z11

dk0f(k0)eik0x

=

Z11

dk

Z11

dk0f(k)f(k0)

1

2

Z11

dxei(kk0)x

= Z11

dk f(k)2 ;since

1

2

Z11

dxei(kk0)x = (k k0): (140)

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Note that according to the L2-metric f(x) and f(k) have the same length.

Existence Theorem: The Fourier transform off(x), i.e., either ef(k) orf(k), exists iff(x) satises the Dirichlet theorem on any nite interval andmoreover

R11 dx jf(x)j < M < 1:

6.2 Properties of the Fourier Transform

1. Linearity: Let F T(g) be either eg or g and ; two scalar numbers,then

F T(f + g) = F T(f) + F T(g): (141)

2. Fourier transform of derivatives:

F T(dn

dxnf(x) = (ik)nF T(f(x)): (142)

3. Convolution theorem: If the function g(x) is a convolution off and h,

g(x) =

Z11

dyf(x y)h(y); (143)

then the Fourier transform of g satises

eg(k) = 2 ef(k)eh(k); (144)g(k) = p2f(k)h(k):6.3 Fourier Series and Transforms in Higher Dimen-

sion

Both the Fourier series expansion and the Fourier transform are readily gen-eralized to higher dimension. This is accomplished by forming products ofone-dimensional Fourier basis functions and using these products as basisfunctions in higher dimensional spaces. Suppose x is an n-dimensional vectorand the domain D of the function g(x) is rectangular, Li=2 < xi < Li=2;for i=1,....n , then if g(x) is a properly behaved function on D it can beexpanded as

g(x) =1X

j1=1

::::1X

jn=1

dj1;:::jnei2(j1x1=L1+::::+jnxn=Ln); (145)

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where the expansion coecients are given by

dj1;:::jn =1

L1 LnZD

dxg(x)ei2(j1x1=L1+::::+jnxn=Ln): (146)

This is the higher dimensional form of the Fourier series expansion. Removingthe restriction to a nite domain we can dene the Fourier transform as

eg(k) = 1(2)n

Zdxg(x)eikx; (147)

where both k and x are n-dimensional vectors. The function g(x) can thenbe expressed as

g(x) = Zdkeg(k)eikx: (148)The symmetrized Fourier transform can be obtained by noting that

g(k) = (2)n=2eg(k) (149)in which case we have -

g(x) =1

(2)n=2

Zdkg(k)eikx: (150)

6.4 Exercises on Fourier Transforms:1. What is the Fourier transform of a -function? Note how in 137 the

-function is dened as a limit

(x x0) = lim!1

f(x x0)

with a particular choice of function f(xx0). Can you think of anotherchoice of function f which will still generate the -function in the samelimit? Show that your chosen form of f is valid.

2. Prove the convolution theorem, i.e., starting from 143 show 144.

3. Calculate the Fourier transform of the function f(x) = sinpx, L x L; f(x) = 0, jxj > L: Draw simple gures to show how thetransform ef(k) varies with L:

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4. Prove Plancherels formula -

Z11

dxf(x)g(x) = Z11

dkf(k)g(k);

i.e., show that the scalar product (L2) is preserved as we go from func-tion to Fourier transform (symmetric) space.

(Chapter head:)Applications of Fourier Series and Transforms

7 Diraction

Natural areas of application of Fourier series and transforms are quantumdynamics and electrodynamics since the plane wave,

F(t;x) = F0ei(kx!t); (151)

is a solution of both the Schrdinger and the Maxwell equations, respectively,in the absence of a variable external eld. The phenomenon of diractionarises when particles are placed in the path of propagating plane waves rep-resenting either particles or electromagnetic radiation. We shall assume thatwe are dealing with radiation remembering that our results could apply aswell to plane wave particle motion. It is important to note that both the

wave function in the case of particle motion and the electromagnetic eldwe shall discuss below are amplitudes. Thus the probability density or thelight intensity are obtained by taking the absolute magnitude squared,

I(t;x) = jF(t;x)j2 : (152)

For the plane wave above the intensity turns out to be a constant. We shallsee below, however, that scattered light is not characterized by a constantintensity. The particles in a sample scatter the radiation in all directionsand a detector placed at a large distance measures an intensity of radiationwhich is sensitive to the orientation if the particles in the irradiated sample

are ordered as in a crystal. Even if the particles are disordered as in a glassor a liquid the intensity tells us something about the structure of the sample.

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7.1 Single Particle Scattering

If a small spherically symmetric particle is placed in the path of a propa-gating plane wave it will scatter some of the radiation, i.e., it will absorband reemit the radiation becoming itself a radiative source. The emittedradiation will have spherical symmetry but it retains the wavelength andfrequency of the original plane wave. This follows from our assumption thatthe scattering process is elastic, i.e., energy conserving. Thus the amplitudeof the electromagnetic eld will have the form

Fj(t;x) = Fjei(kjxxjj!t)= jx xj j ; (153)

where xj is the location of the particle, k is the wave vector of the planewave, its length k is related to the wave length by the relation k = 2=;and ! is the frequency of the radiation. Note that ! = 2 = kc. Thecorresponding intensity of the radiation from a single particle is

Ij(t;x) = jFj j2 = jx xj j2 : (154)

Note that the denominator is the square of the distance from the particle.It accounts for the fact that the same ux of radiation intensity is passingthrough a spherical surface which grows like 4 jx xj j2 with distance. Fluxconservation then dictates the form of the intensity. The amplitude factor

Fjis given by

Fj = jeikxj ; (155)

where the exponential phase factor accounts for the phase of the plane waveat the particle and j is a constant independent of the particle location xj :It will reect the size of the particle (its cross section) and the process ofabsorbtion and reemission.

7.2 Many-Particle Scattering - Crystal Diraction

We now consider the case of scattering from a crystal, i.e., from a three-dimensional array of particles located at the positions

fxj

g. The scattered

eld Fs(t;x) is now a sum over the elds of all the particles,

Fs(t;x) =Xj

Fj(t;x) =Xj

jjx xjje

ikxjei(kjxxjj!t): (156)

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If the particles are identical as they might be in a crystal then j is indepen-

dent of j, j = .We now apply an approximation valid when the irradiated sample is ofmuch smaller length scale than the distance from sample to detector. Notethat x is the vector taking us from the origin which we shall place in thesample and the detector. The vector xj takes us from the sample originto the specic particle j. These two vectors lie in a plane and the anglebetween them we shall call j. It is easy to see that if x>>xj then to a goodapproximation we have

jx xj j

e=x xj cos j : (157)

We then nd the much simplied form for the scattered eld -

Fs(t;x) = x

ei(kx!t)Xj

eikxj ; (158)

where k = (kx=x) k and we have assumed identical scatterers as in amonatomic crystal. Here we have neglected all terms of order xj=x or smaller.

7.2.1 Constructive and destructive interference:

The observed intensity at x is given by the absolute magnitude squared ofthe eld,

I(x) =

jFs(t;x)

j2 =

j

j2

jx

j2

jA

j2 ; (159)

where all dependence on particle positions is collected in the amplitude factor,

A =Xj

eikxj : (160)

In a crystal with a monatomic unit cell, for simplicity, the particle positionscan be given by

xj = mja+ njb +pjc; (161)

where a;b; c are primitive translational vectors and mj; nj; pj are integers.For simplicity we assume that the crystal sample is described by the integersm = 0;::::;M; n = 0; ::::::N;p = 0; :::::P; We then get

A =MX

m=0

eimkaNX

n=0

einkbPX

p=0

eipkc (162)

=1 ei(M+1)ka

1 eika1 ei(N+1)kb

1 eikb1 ei(P+1)kc

1 eikc :

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But note now that

MXm=0

eimka = Order(M1=2); generally, (163)

= M + 1; for k a =2q; q=integer.

Thus if M;N ;P are large integers, as they would be for a typical crystalsample in a plane wave beam of macroscopic width, then we would ndstrong intensity peaks when x and k are such that

ak = 2q; for q=integer, (164)b

k = 2r; for r=integer,

ck = 2s; for s=integer.These are called Laues equations. When they are all satised we get astrong peak in the intensity. For which k-vectors will the Laue conditionsall be satised? - They will be satised for k of the form

k = qbA + rbB+ sbC; (165)where

bA = 2(b c)=a(b c); (166)bB = 2(c a)=a(b c);bC = 2(a b)=a(b c):Note that bA; bB; bC serve as primitive translational vectors in k-space andthey determine the primitive translational vectors a;b; c and thereby thecrystal structure. The vectors bA; bB; bC are dened with the help of the vectorproduct concept, e.g., bc, which is a vector orthogonal to b and c and of alength bcsin where is the angle between the vectors. The denominator isthe volume of the parallelipiped formed by the vectors a;b; c, possibly witha negative sign. Thus we can enter the vectors in any order without aecting

the Laue conditions. See the Beta Handbook Section 3.4.Example 1 - Simple cubic crystal: - The simple cubic crystal has

orthogonal primitive translational vectors a;b; c of equal length. Let us takethis length, i.e., the separation between neighboring atoms to be 4 . We canalso let the directions ofa;b; c be the three axial directions in our Cartesian

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8 Fourier Spectroscopy

Our purpose here is to determine the frequency spectrum of a light source,i.e., the intensity I(!) as a function of the frequency !: The method is touse the interference pattern arising when a plane wave light beam is splitinto two parts traveling paths of length L and L before being recombinedat the detector. The interference pattern as a function ofL is in Fourierspectroscopy used to reveal the frequency spectrum. Thus we shall considerhow to determine I(!) from the intensity as a function ofL, I(L).

8.1 Monochromatic Light Source

Let us rst note that a thin beam of light can be passed by a path determinedby mirrors from light source to detector. If the path length is L then theamplitude of the electromagnetic eld at the detector is given by

FD(t;x) = F0ei(kL!t): (172)

Here we have assumed that we have a point source and an innitely sharpbeam. If the beam is not sharp there will be a distribution of path lengthsso that the amplitude at the detector is instead given by

FD(t;x) = F0

ZdL(L)ei(kL!t); (173)

where (L) is a probabilty density satisfyingZdL(L) = 1: (174)

We shall always assume that in the absence of a beam splitter our beam issharp.

Suppose now that we introduce a beam splitter which splits the beam intotwo parts of equal amplitude but travelling dierent path lengths L1 and L2,respectively. The dierence in path length is L, i.e., L1 = L, L2 = L + L.The amplitude at the detector will then be the sum of the contributions from

the two paths,

FD(t) =1

2F0e

i(kL1!t) +1

2F0e

i(kL2!t) (175)

=1

2F0e

i(kL!t)(1 + eikL):

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The corresponding intensity is

ID(t; L) = jFD(t; L)j2 = 14

jF0j2 1 + eikL2 (176)=

1

2jF0j2 (1 + cos kL):

Noting that for radiation propagating in vacuum we have ! = kc, where c isthe velocity of light, we can write

ID(t; L) =1

2jF0j2 (1 + cos !t); (177)

where t is the dierence between the times of propagation along the twopaths, t = L=c: Thus the intensity shows a sinusoidal variation with Lor t as shown below in a plot of y = (1 + cos x):

-10 -8 -6 -4 -2 0 2 4 6 8 10

0.5

1.0

1.5

2.0

x

y

If we identify x in this plot with t then this would be the shape of theintensity variation for ! = 1: In general, the separation between neighbour-ing peaks would be 2=!: This allows the frequency of the radiation to beidentied from the intensity as a function of t:

8.1.1 Several frequencies:

If the radiation is made up of intensity at a number of well-dened frequenciesthen the amplitude without the beam splitter becomes

FD(t) =Xj

Fjei(kjL!jt); (178)

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and the corresponding intensity is

ID(t) = Xm

Xn

FmFn ei((knkm)L(!n!m)t) (179)=

Xm

jFmj2 +Xm

Xn6=m

FmFn exp(i((kn km)L (!n !m)t)):

But note that the latter term above gives rise to a uctuation in intensity intime. The long time average of this uctuation vanishes, i.e.,

ID = limT!1

1

T

ZT0

dtI(t) =Xm

jFmj2 : (180)

After introduction of the beam splitter we get

FD(t) =1

2

Xj

Fjei(kjL!jt)(1 + eikjL); (181)

I(t) =1

4

Xm

Xn

FmFnei((knkm)L(!n!m)t)(1 + eiknL)(1 + eikmL); (182)

I(t) =1

2

Xm

jFmj2 (1 + cos !mt): (183)

We shall now use a method which has the character of a mathematical form

of ltering. Suppose now that I(t) has been measured over the interval0


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Note now that for ! = !m we have

sin(! !m)R(! !m) = R; (187)

which, if R is suciently large, gives rise to a blip in B(!) at ! = !m;

B(R; !m)e=14

ImR:

This maximum in B(R; !) can be used to identify the frequencies !m presentand the corresponding intensities Im = jFmj2 :

Example 2: Suppose we have a light source of two spectral lines with

freqencies ! = 1 and ! = 3 and unit intensities. Draw the cosine transformB(R; !) for R = 5 and 10. - Note that B(R; !) for R = 5 takes the form -

B(R; !) =1

2(2

sin5!

!+

sin5(! + 1)

2(! + 1)+

sin5(! 1)2(! 1) +

sin5(! + 3)

2(! + 3)+

sin5(! 3)2(! 3) )

-5 -4 -3 -2 -1 1 2 3 4 5

1

2

3

4

x

y

and the shape shown in the gure above. If R = 10 we get for B(R; !) -

B(R; !) =

1

2(2

sin10!

! +

sin 10(! + 1)

2(! + 1) +

sin 10(!

1)

2(! 1) +sin 10(! + 3)

2(! + 3) +

sin 10(!

3)

2(! 3) )

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-5 -4 -3 -2 -1 1 2 3 4 5

2

4

6

8

x

y

and the shape as shown above. Note the sharper positive peaks at ! = 1and 3.

8.2 Exercises on Fourier Spectroscopy:

1. Consider the interpretation of the function B(R; !) as derived above.One problem in determining the frequencies and intensities from thepeaks ofB(R; !) is to make sure that the peak is not just a uctuationin the background, i.e., not due to ! = !m as suggested. How couldone guard against this possibility? Propose a method that as far aspossible eliminates background peaks from a set of chosen high peaks.

2. In the case of a continuous light source we have

I(s) =1

2

Z10

d!(!)(1 + cos !s); (188)

where (!) is the light intensity as a function of the frequency. Obtainthe cosine transform B(R; !) for this type of light. Supposing thatwe can obtain I(s) over the interval 0 < s < R < 1 suggest a wayby which (!) could be obtained at least approximately from B(R; !).

You may use the fact that -

limR!1

1

sin((x x0)R)x x0 = (x x0):

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9 Laplace Transforms and Applications

9.1 Derivation and Properties

With the help of Fourier transforms we can work with functions on thewhole real axis but there are still many functions that we cannot apply theFourier transform to due to the requirement of absolute integrability, i.e.,R11 dxf(x) < M < 1: Thus we cannot dene the Fourier transform for the

functions , xn; n 1. For this and other reasons we shall continue todevelop the Fourier transform into the Laplace transform.

Suppose we are interested in f(x) on the interval [0; 1] : In order to beable to apply the Fourier transform we shall rst apply two operations to the

function f(x): Multiply it by the Heaviside step function H(x) dened by

H(x) = 0; for x 0 such that

limx!1

exf(x) = 0; (191)

and c is picked suciently large. The Fourier transform of g(c; x) is

eg(c; k) =

1

2

Z11

dxH(x)e(c+ik)xf(x) =1

2

Z10

dxe(c+ik)xf(x); (192)

and the corresponding expression for the function g(c; x) is

g(c; x) =

Z11

dkeg(c; k)eikx: (193)54


55/116

By multiplication of this last equation by ecx we can recover f(x),

f(x) =Z11

dkeg(c; k)e(c+ik)x; for x > 0. (194)Since we are dealing with complex numbers anyway we shall take the libertyto dene the complex variable s = c + ik. Moreover, we dene the Laplacetransform bf(s) as 2eg(s), i.e.,

bf(s) = Z10

dxesxf(x); for Re(s) = c = large enough. (195)

The corresponding expression for f(x) can be obtained as

f(x) =1

2i

Zc+i1ci1

dsesxbf(s); for x > 0. (196)Here we have changed variable of integration from k to s and noted that if weintegrate in the complex plane along a line parallel to the imaginary axis fromc i1 to c + i1 then dk = ds=i. The form of the inverse Laplace transformin 196 invites the use of the residue theorem (see the Beta Handbook, section14.2). However, it is more common to do the inversion directly from a tableof Laplace transforms, perhaps with the aid of some of the many simplifyingproperties of the Laplace transform described below.

9.1.1 Properties of the Laplace transform: LP= Laplace trans-form

1. Linearity: LP(f(x) + h(x)) = LP(f(x)) + LP(h(x)):

2. Derivative theorem: LP( ddxf(x)) = sLP(f) f(0):3. Integral theorem: LP(

Rx0

dtf(t)) = 1s

LP(f):

4. Convolution theorem: Ifg(x) =

Rx

0dtf(t)h(xt) then LP(g) = LP(f)LP(h).

5. Exponential shift theorem: LP(epxf(x)) = bf(s +p):The linearity follows directly from the linearity of the Fourier transform.

The derivative and the integral theorems can be proven by use of partial

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where k is the unimolecular rate coecient. Find the time development ofc

from the initial value c(0). - This is a linear rst order dierential equation.We want c(t) for t > 0 so we apply the Laplace transform to both sides ofthe equation,

sbc(s) c(0) = kbc(s): (199)Here we have used the derivative law. This equation for the Laplace trans-form can be solved to yield

bc(s) = c(0)s + k

: (200)

From our small table of Laplace transforms it follows that

c(t) = c(0)ekt

: (201)Example 2 - Coupled chemical reactions: Consider a set of coupled

chemical reactions of the type A ! B, B ! C, C ! product. The corre-sponding time dependent concentrations are cA(t), cB(t); cC(t) and the rateequations are

d

dtcA(t) = k1cA(t); (202)

d

dtcB(t) = k1cA(t) k2cB(t);

d

dtcC(t) = k2cB(t) k3cC(t):These are coupled linear rst order equations. We apply the Laplace trans-form to both sides of all three equations to obtain -

sbcA(s) cA(0) = k1bcA(s); (203)sbcB(s) cB(0) = k1bcA(s) k2bcB(s);sbcC(s) cC(0) = k2bcB(s) k3bcC(s):

The rst equation can be solved as in the example above. We get -

bcA(s) = cA(0)s + k1 ; and cA(t) = cA(0)ek1t: (204)Insertion in the second equation yields -

bcB(s) = cB(0)s + k2

+k1cA(0)

(s + k1)(s + k2): (205)

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In order to nd this transform by linear combinations of transforms in our

small table we note that -1

(s + k1)(s + k2)=

1

k1 k2 (1

s + k2 1

s + k1): (206)

Now it is straightforward to nd the inverse Laplace transform. We get -

cB(t) = cB(0)ek2t +

k1cA(0)

k1 k2 (ek2t ek1t): (207)

Finally, we solve for bcC(s) and nd -bcC(s) =

cC(0)

s + k3+

k2bcB(s)

s + k3(208)

=cC(0)

s + k3+

k2s + k3

(cB(0)

s + k2+

k1cA(0)

(s + k1)(s + k2))

=cC(0)

s + k3+

k2s + k3

(cB(0)

s + k2+

k1cA(0)

k1 k2 (1

s + k2 1

s + k1))

=cC(0)

s + k3+

k2cB(0)

k2 k3 (1

s + k3 1

s + k2) +

k1k2cA(0)

k1 k2

1

k2 k3 (1

s + k3 1

s + k2) 1

k1 k3 (1

s + k3 1

s + k1)

:

At this point the transform is of a form such that we can immediately identifythe terms in our table. We nd that the concentration of species C decaysby a triple exponential time dependence, i.e.,

cC(t) = a1ek1t + a2e

k2t + a3ek3t; (209)

a1 =k1k2cA(0)

(k1 k2)(k1 k3) ;

a2 =k2cB(0)

k3 k2 +k1k2cA(0)

(k1 k2)(k3 k2) ;

a3 = cC(0) +k2cB(0)

k2

k3+

k1k2cA(0)

k1

k2(

1

k2

k3 1

k1

k3)

= cC(0) +k2cB(0)

k2 k3 +k1k2cA(0)

(k2 k3)(k1 k3) : (210)

Example 3 - Harmonic oscillator: We have already encountered theharmonic oscillator in our discussion of normal modes of molecules in Chapter

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2. The equation of motion is -

md2

dt2x(t) + kx(t) = 0; (211)

where m is the mass and k the force constant. We shall now see that we canreadily solve this equation by Laplace transformation but rst we must notethat the derivative theorem can be iterated to apply to higher derivatives:

LP(d2

dt2x(t)) = sLP(

d

dtx(t)) v(0) (212)

= s2

bx(s) sx(0) v(0);

where v is the velocity, i.e., the time derivative ofx. Using this extension ofthe derivative theorem we can take the Laplace transform of the equation ofmotion for the harmonic oscillator and obtain -

ms2bx(s) msx(0) mv(0) + kbx(s) = 0: (213)This equation yields -

bx(s) = msx(0) + mv(0)ms2 + k

=sx(0) + v(0)

s2 + km

(214)

=sx(0)

s2 + km+

v(0)

s2 + km:

Now we can identify the terms in our small table of Laplace transforms andnd -

x(t) = x(0) cos(

rk

mt) + v(0)

rm

ksin(

rk

mt): (215)

Example 4 - Debye Hckel theory: Now we shall consider the screen-ing of an ion in an electrolyte solution. Let (r) be the average electrostaticpotential at the distance r from the ion. It must be spherically symmetricso it depends on the distance but not on the direction. In the absence ofother ions the eld would have been of Coulombic form, (r) / q=r, whereq is the charge of the central ion. In the presence of the mobile ions in thesolution the eld is screened by the attraction of counterions and repulsionof coions. According to the Debye-Hckel analysis the concentration of anion of species i is altered by the eld to the form -

ci(r) = ciBeEi=kBT = ciBe

qi(r)=kBT; (216)

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where qi is the charge of the ionic species in the screening atmosphere, ciB is

its bulk concentration, kB is Boltzmanns constant and T is the temperaturein Kelvin. From electrostatic theory we know that there is a direct rela-tionship between the eld and the charge density expressed by Poissonsequation,

(@2

@x2+

@2

@y2+

@2

@z2) = r2 = =""0; (217)

which in the case of spherical symmetry becomes

1

r2d

dr(r2

d

dr(r)) = (r)=""0: (218)

The charge density can be expressed in terms of the concentrations of thecharged species, i.e.,

(r) =Xi

qici(r) =Xi

qiciBeqi(r)=kBT: (219)

If we now insert this expression for in Poissons equation we get the socalled Poisson-Boltzmann equation, which in spherical symmetry takes theform -

1

r2d

dr(r2

d

dr(r)) =

Xi

qiciB""0

eqi(r)=kBT: (220)

At this point we note that the Poisson-Boltzmann equation is nonlinearin the eld and therefore dicult to solve. Debye and Hckel investigatedthe weak coupling limit when the interaction energy is small compared tokBT; i.e.,

jqi(r)=kBTj 1: (221)Then we can linearize the Boltzmann factor and get -

(r) =Xi

qiciBeqi(r)=kBT =

Xi

(qiciB q2i ciB(r)=kBT) (222)

= Xi

q2i ciB(r)=kBT;

since by electroneutrality in the bulk we have -Xi

qiciB = 0: (223)

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The corresponding linearized Poisson-Boltzmann equation takes the form -

1r2

ddr

(r2 ddr

(r)) = 2(r); (224)

where2 =

Xi

q2i ciB=(""0kBT): (225)

Now we have a second order linear dierential equation to solve. Beforewe apply the Laplace transform method we change dependent variable tou(r) = r(r): The linearized Poisson-Boltzmann equation then becomes -

d2

dr2

u(r) = 2u(r): (226)

By Laplace transformation of both sides we get -

s2bu(s) su(0) u0(0) = 2bu(s); (227)bu(s) = su(0) + u0(0)

s2 2 :

Recalling that -

1

s2 2 =1

(s + )(s ) =1

2(

1

s 1

s + ); (228)

s

s2 2 =s

+

(s + )(s ) =1

s + +1

2(1

s 1

s + )

=1

2(

1

s +1

s + );

we nd that u(r) has the form -

u(r) =1

2(u(0) u

0(0)

)er +

1

2(u(0) +

u0(0)

)er: (229)

However, for physical reasons we cannot tolerate the exponentially growingterm so we must have u0(0) = u(0). Thus we get the physical solution -

u(r) = u(0)er; (230)

and, nally, the eld is given by -

(r) =u(0)

rer: (231)

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Part III

Dierential Equations

10 Ordinary Dierential Equations

Consider the equation -

d3

dx3y +

d

dxy + x2yn = g(x) (234)

for the the function y(x). It is called an ordinary dierential equation(ODE) because there appear only derivatives with respect to one unknown

variable called x in this case. It is said to be of 3rd order because the highestorder derivative to appear is of this order. If

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