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Introductory Quantum Chemistry

Chem 570a: Lecture Notes

Prof. Victor S. Batista

Room: Sterling Chemistry Laboratories (SCL) 18Tuesdays and Thursdays 9:00 10:15 am

Yale University - Department of Chemistry

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Contents

1 Syllabus 6

2The Fundamental Postulates of Quantum Mechanics

7

3 Continuous Representations 10

4 Vector Space 104.1 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.2 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

5 Stationary States 145.1 Exercise 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145.2 Exercise 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145.3 Exercise 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

6 Particle in the Box 156.1 Exercise 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

7 Commutator 177.1 Exercise 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

8 Uncertainty Relations 178.1 Exercise 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188.2 EPR Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

9 Variational Theorem 19

10 Digital Grid-Based Representations 2110.1 Computational Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2110.2 Computational Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2210.3 Computational Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2310.4 Computational Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

11 SOFT Method 2411.1 Computational Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2411.2 Computational Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2511.3 Computational Problem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2611.4 Computational Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2611.5 Computational Problem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

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12 Tunneling Current: Landauer Formula 2712.1 WKB Transmission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

13 Time Independent Perturbation Theory 33

13.1 Exercise 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3414 Time Dependent Perturbation Theory 34

14.1 Exercise 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3714.2 Exercise 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

15 Problem Set 4015.1 Exercise 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4015.2 Exercise 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4115.3 Exercise 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4115.4 Exercise 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

16 Adiabatic Approximation 42

17 Heisenberg Representation 44

18 Two-Level Systems 45

19 Harmonic Oscillator 4919.1 Exercise 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

20 Problem Set 5320.1 Exercise 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5320.2 Exercise 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5320.3 Exercise 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5320.4 Exercise 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

21 Angular Momentum 5421.1 Exercise 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5521.2 Exercise 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5621.3 Exercise 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5821.4 Exercise 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5921.5 Exercise 24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

22 Spin Angular Momentum 6322.1 Exercise 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6522.2 Exercise 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6622.3 Exercise 27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

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22.4 Exercise 28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

23 Central Potential 6723.1 Exercise 29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

24 Two-Particle Rigid-Rotor 7024.1 Exercise 30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

25 Problem Set 7125.1 Exercise 31 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7125.2 Exercise 32 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7125.3 Exercise 33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7125.4 Exercise 34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7125.5 Exercise 35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7125.6 Exercise 36 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

25.7 Exercise 37 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

26 Hydrogen Atom 7226.1 Exercise 38 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7426.2 Exercise 39 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7526.3 Exercise 40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7526.4 Exercise 41 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7526.5 Exercise 42 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

27 Helium Atom 77

28 Spin-Atom Wavefunctions 7929 Pauli Exclusion Principle 80

30 Lithium Atom 8030.1 Exercise 44 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

31 Spin-Orbit Interaction 8231.1 Exercise 45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

32 Periodic Table 8432.1 Exercise 46 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8532.2 Exercise 47 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

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33 Problem Set 8633.1 Exercise 48 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8633.2 Exercise 49 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8633.3 Exercise 50 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

33.4 Exercise 51 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

34 LCAO Method: H +2 Molecule 8734.1 Exercise 52 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8834.2 Exercise 53 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

35 H 2 Molecule 9035.1 Heitler-London(HL) Method: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9335.2 Exercise 54 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

36 Homonuclear Diatomic Molecules 93

36.1 Exercise 55 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

37 Conjugated Systems: Organic Molecules 97

38 Self-Consistent Field Hartree-Fock Method 10038.1 Restricted Closed Shell Hartree-Fock . . . . . . . . . . . . . . . . . . . . . . . . 104

39 Empirical Parameterization of Diatomic Molecules 10939.1 Exercise 56 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11239.2 Exercise 57 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

40 Solutions to Computational Assignments 11540.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11540.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11840.3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12340.4 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12640.5 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13240.6 Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13440.7 Problem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14540.8 Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16440.9 Problem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

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1 Syllabus

The goal of this course is to introduce fundamental concepts of Quantum Mechanics with emphasison Quantum Dynamics and its applications to the description of molecular systems and their inter-

actions with electromagnetic radiation. Quantum Mechanics involves a mathematical formulationand a physical interpretation , establishing the correspondence between the mathematical elementsof the theory (e.g., functions and operators) and the elements of reality (e.g., the observable proper-ties of real systems). The presentation of the theory will be mostly based on the so-called Orthodox Interpretation , developed in Copenhagen during the rst three decades of the 20th century. How-ever, other interpretations will be discussed, including the pilot-wave theory rst suggested byPierre De Broglie in 1927 and independently rediscovered by David Bohm in the early 1950s.

Textbooks : The ofcial textbook for this class is:R1 : Levine, Ira N. Quantum Chemistry; 6th Edition; Pearson/Prentice Hall; 2009.However, the lectures will be heavily complemented with material from other textbooks including:R2 : Quantum Theory by David Bohm (Dover),R3 : Quantum Physics by Stephen Gasiorowicz (Wiley),R4 : Quantum Mechanics by Claude Cohen-Tannoudji (Wiley Interscience),R5 : Quantum Mechanics by E. Merzbacher (Wiley),R6 : Modern Quantum Mechanics by J. J. Sakurai (Addison Wesley),All these references are on-reserve at the Kline science library.References to specic pages of the textbooks listed above are indicated in the notes as follows:R1(190) indicates for more information see Reference 1, Page 190.Furthermore, a useful mathematical reference is R. Shankar, Basic Training in Mathematics. AFitness Program for Science Students, Plenum Press, New York 1995.Useful search engines for mathematical and physical concepts can be found at

http://scienceworld.wolfram.com/physics/ and http://mathworld.wolfram.com/ The lecture notes are posted online at: (http://www.chem.yale.edu/

batista/vvv/v570.pdf)

Grading : Grading and evaluation is the same for both undergraduate and graduate students.The mid-terms will be on 10/3 and 11/21. The date for the Final Exam is determined by Yalescalendar of nal exams. Homework includes exercises and computational assignments with duedates indicated in class.

Contact Information and Ofce Hours : Prof. Batista will be glad to meet with studentsat SCL 251 as requested by the students via email to [email protected], or by phone at (203)432-6672.

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2 The Fundamental Postulates of Quantum Mechanics

Quantum Mechanics can be formulated in terms of a few postulates (i.e. , theoretical principlesbased on experimental observations). The goal of this section is to introduce such principles, to-

gether with some mathematical concepts that are necessary for that purpose. To keep the notationas simple as possible, expressions are written for a 1-dimensional system. The generalization tomany dimensions is usually straightforward.

Postulate 1 : Any system in a pure state can be described by a wave-function, (t, x ) , where t isa parameter representing the time and x represents the coordinates of the system. Such a function(t, x ) must be continuous, single valued and square integrable.Note 1 : As a consequence of Postulate 4, we will see that P (t, x ) = (t, x )(t, x )dx representsthe probability of nding the system between x and x + dx at time t .

Postulate 2 : Any observable (i.e., any measurable property of the system) can be described byan operator. The operator must be linear and hermitian.

What is an operator ? What is a linear operator ? What is a hermitian operator ?

Denition 1 : An operator O is a mathematical entity that transforms a function f (x) into another function g(x) as follows, R4(96)

O f (x ) = g (x ),where f and g are functions of x.

Denition 2: An operator O that represents an observable O is obtained by rst writing the clas-sical expression of such observable in Cartesian coordinates (e.g., O = O(x, p)) and then substi-

tuting the coordinate x in such expression by the coordinate operator x as well as the momentum pby the momentum operator p = i /x .Denition 3: An operator O is linear if and only if (iff),

O(af (x) + bg(x)) = aOf (x) + bOg(x),

where a and b are constants.Denition 4: An operator O is hermitian iff,

dxn (x)Om (x) = dxm (x)On (x) ,where the asterisk represents the complex conjugate.

Denition 5: A function n (x) is an eigenfunction of O iff,

On (x) = On n (x),

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where On is a number called eigenvalue.

Property 1: The eigenvalues of a hermitian operator are real.Proof: Using Denition 4, we obtain

dxn (x)On (x) dxn (x)On (x) = 0 ,therefore,

[On On ] dxn (x)n (x) = 0 .Since n (x) are square integrable functions, then,

On = On .

Property 2: Different eigenfunctions of a hermitian operator (i.e., eigenfunctions with differenteigenvalues) are orthogonal (i.e., the scalar product of two different eigenfunctions is equal tozero). Mathematically, if On = On n , and Om = Om m , with On = Om , then dxn m = 0 .Proof: dxm On dxn Om = 0 ,and

[On Om ] dxm n = 0 .Since On = Om , then

dxm n = 0 .

Postulate 3 : The only possible experimental results of a measurement of an observable are theeigenvalues of the operator that corresponds to such observable.

Postulate 4 : The average value of many measurements of an observable O , when the system isdescribed by (x) as equal to the expectation value O , which is dened as follows,

O = dx(x)O(x) dx(x)(x) .Postulate 5 :The evolution of (x, t ) in time is described by the time-dependent Schr odinger equation :

i (x, t )

t= H(x, t ),

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where H = 2

2m 2

x 2 + V (x), is the operator associated with the total energy of the system, E = p22m + V (x).

Expansion P ostulate : R5(15), R4(97)The eigenfunctions of a linear and hermitian operator form a complete basis set. Therefore,

any function (x) that is continuous, single valued, and square integrable can be expanded as alinear combination of eigenfunctions n (x) of a linear and hermitian operator A as follows,

(x) = j

C j j (x),

where C j are numbers (e.g., complex numbers ) called expansion coefcients .Note that A = j C j C j a j , when (x) = j C j j (x),

A j (x) = a j j (x), and dx j (x)k(x) = jk .This is because the eigenvalues a j are the only possible experimental results of measurements of A(according to Postulate 3), and the expectation value A is the average value of many measurementsof A when the system is described by the expansion (x) = j C j j (x) (Postulate 4). Therefore,the product C j C j can be interpreted as the probability weight associated with eigenvalue a j (i.e.,the probability that the outcome of an observation of A will be a j ).

Hilbert-SpaceAccording to the Expansion Postulate (together with Postulate 1), the state of a system described

by the function (x) can be expanded as a linear combination of eigenfunctions j (x) of a linearand hermitian operator (e.g., (x) = C 11(x) + C 22(x) + . . .). Usually, the space dened by

these eigenfunctions (i.e., functions that are continuous, single valued and square integrable) hasan innite number of dimensions. Such space is called Hilbert-Space in honor to the mathematicianHilbert who did pioneer work in spaces of innite dimensionality. R4(94)

A representation of (x) in such space of functions corresponds to a vector-function,

T

E

..............................

.....

...

...

...

.(x)C 2

2(x)

C 1 1(x)

where C 1 and C 2 are the projections of (x) along 1(x) and 2(x), respectively. All other

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components are omitted from the representation because they are orthogonal to the plane denedby 1(x) and 2(x).

3 Continuous RepresentationsCertain operators have a continuous spectrum of eigenvalues. For example, the coordinate operatoris one such operator since it satises the equation x (x0 x) = x0 (x0 x), where the eigenvaluesx0 dene a continuum . Delta functions (x0 x) thus dene a continuous representation (the so-called coordinate representation) for which

(x) = dx0C x0 (x0 x),where C x0 = (x0), since

dx (x )(x) = dx dC (x ) ( x) = ( ).When combined with postulates 3 and 4, the denition of the expansion coefcients C x0 =

(x0) implies that the probability of observing the system with coordinate eigenvalues between x0and x0 + dx0 is P (x0) = C x0 C x0 dx0 = (x0)(x0)dx0 (see Note 1).

In general, eigenstates (, x ) with a continuum spectrum of eigenvalues dene continuousrepresentations,

(x) = dC (, x ),with C

=

dx(, x )(x). Delta functions and the plane waves are simply two particular

examples of basis sets with continuum spectra.Note 2 : According to the Expansion Postulate, a function (x) is uniquely and completely denedby the coefcients C j , associated with its expansion in a complete set of eigenfunctions j (x).However, the coefcients of such expansion would be different if the same basis functions jdepended on different coordinates (e.g., j (x) with x = x). In order to eliminate such ambiguityin the description it is necessary to introduce the concept of vector-ket space. R4(108)

4 Vector Space

Vector-Ket Space : The vector-ket space is introduced to represent states in a convenient spaceof vectors | j > , instead of working in the space of functions j (x). The main difference is thatthe coordinate dependence does not need to be specied when working in the vector-ket space.According to such representation, function (x) is the component of vector | > associated with

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index x (vide infra ). Therefore, for any function (x) = j C j j (x), we can dene a ket-vector

| > such that,| > =

j

C j | j >.The representation of | > in space is,

T

E

........................................

...

...

...

...

...

...

..

Ket-Space

C 2

|2 >

C 1 |1 >

| >

Note that the expansion coefcients C j depend only on the kets | j > and not on any specicvector component. Therefore, the ambiguity mentioned above is removed.In order to learn how to operate with kets we need to introduce the bra space and the concept of linear functional . After doing so, this section will be concluded with the description of Postulate5, and the Continuity Equation .

Linear functionalsA functional is a mathematical operation that transforms a function (x) into a number. This

concept is extended to the vector-ket space , as an operation that transforms a vector-ket into anumber as follows,

((x)) = n , or (| > ) = n,where n is a number. A linear functional satises the following equation,

(a(x) + bf (x)) = a ((x)) + b(f (x)) ,

where a and b are constants.Example: The scalar product, R4(110)

n = dx(x)(x),is an example of a linear functional, since such an operation transforms a function (x) into anumber n . In order to introduce the scalar product of kets, we need to introduce the bra-space .

Bra Space : For every ket | > we dene a linear functional < |, called bra-vector , as follows:11

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< |(| > ) = dx(x)(x).Note that functional < | is linear because the scalar product is a linear functional. Therefore,

<

|(a

| > + b

|f > ) = a <

|(

| > ) + b <

|(

|f > ).

Note: For convenience, we will omit parenthesis so that the notation < |(| > ) will be equivalentto < || > . Furthermore, whenever we nd two bars next to each other we can merge them intoa single one without changing the meaning of the expression. Therefore,< || > = < | > .

The space of bra-vectors is called dual space simply because given a ket | > = j C j | j > ,the corresponding bra-vector is < | = j C j < j |. In analogy to the ket-space, a bra-vector< |

is represented in space according to the following diagram:

T

E

........................................

...

...

...

...

...

...

..

Dual-Space

C 2

< 2|

C 1 < 1|

< |

where C j is the projection of < | along < j |.Projection Operator and Closure RelationGiven a ket | > in a certain basis set | j > ,

| > = j

C j | j >, (1)

where < k| j > = kj , C j = < j| > . (2)

Substituting Eq. (2) into Eq.(1), we obtain

| > = j

| j >< j | > . (3)

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From Eq.(3), it is obvious that

j| j >< j | = 1, Closure Relation

where 1 is the identity operator that transforms any ket, or function, into itself.Note that P j = | j >< j | is an operator that transforms any vector | > into a vector pointing

in the direction of | j > with magnitude < j | > . The operator P j is called the ProjectionOperator . It projects | j > according to,

P j | > = < j | > | j > .Note that P 2 j = P j , where P 2 j = P j P j . This is true simply because < j | j > = 1 .

4.1 Exercise 1

Prove thati

P jt

= [H, P j ],

where [H, P j ] = H P j P j H .Continuity Equation

4.2 Exercise 2

Prove that ((x, t )(x, t ))

t+

x j (x, t ) = 0 ,

where

j (x, t ) =

2mi(x, t )

(x, t )x (x, t )

(x, t )x

.

In general, for higher dimensional problems, the change in time of probability density, (x , t ) =(x , t )(x , t ), is equal to minus the divergence of the probability ux j ,

(x , t )t

= j.This is the so-called Continuity Equation.Note: Remember that given a vector eld j, e.g., j(x,y,z ) = j 1(x ,y,z )i+ j2(x ,y,z ) j + j3(x,y,z )k,the divergence of j is dened as the dot product of the del operator

= ( x , y , z ) and vector j

as follows:

j =j 1x

+j 2y

+j 3z

.

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5 Stationary States

Stationary states are states for which the probability density (x, t ) = (x, t )(x, t ) is constantat all times (i.e., states for which (x,t )t = 0 , and therefore j = 0 ). In this section we will showthat if (x, t ) is factorizable according to (x, t ) = (x)f (t), then (x, t ) is a stationary state.Substituting (x, t ) in the time dependent Schr odinger equation we obtain:

(x)i f (t)

t= f (t)

2

2m 2(x)

x 2+ f (t)V (x)(x),

and dividing both sides by f (t)(x) we obtain:

i f (t)

f (t)t

= 2

2m(x) 2(x)

x 2+ V (x). (4)

Since the right hand side (r.h.s) of Eq. (4) can only be a function of x and the l.h.s. can only be afunction of

tfor any

xand

t, and both functions have to be equal to each other, then such function

must be equal to a constant E. Mathematically,

i f (t)

f (t)t

= E f (t) = f (0)exp(i Et ),

2

2m(x) 2(x)

x 2+ V (x) = E H(x) = E (x) .

The boxed equation is called the time independent Schr odinger equation .Furthermore, since f (0) is a constant, function (x) = f (0)(x) also satises the time independentSchrodinger equation as follows,

H (x) = E (x) , (5)and

(x, t ) = (x)exp(i Et ).

Eq. (5) indicates that E is the eigenvalue of H associated with the eigenfunction (x).

5.1 Exercise 3

Prove that H is a Hermitian operator.

5.2 Exercise 4

Prove that -i /x is a Hermitian operator.

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5.3 Exercise 5

Prove that if two hermitian operators Q and P satisfy the equation QP = P Q, i.e., if P and Qcommute ( vide infra ), the product operator QP is also hermitian.Since H is hermitian, E is a real number

E = E (see Property 1 of Hermitian operators), then,

(x, t )(x, t ) = (x)(x).

Since (x) depends only on x, t ((x)(x)) = 0 , then, t (x, t )(x, t ) = 0 . This demonstration

proves that if (x, t ) = (x)f (t), then (x, t ) is a stationary function.

6 Particle in the Box

The particle in the box can be represented by the following diagram: R1(22)

T

E

TV (x) Box

V = V = 0 V =

0 ax

Particle

r

The goal of this section is to show that a particle with energy E and mass m in the box-potentialV(x) dened as

V (x) =0, when 0 x a,, otherwise ,

has stationary states and a discrete absorption spectrum (i.e., the particle absorbs only certaindiscrete values of energy called quanta ). To that end, we rst solve the equation H (x) = E (x),and then we obtain the stationary states (x, t ) = (x)exp(i Et ).Since (x) has to be continuous, single valued and square integrable (see Postulate 1), (0) and(a) must satisfy the appropriate boundary conditions both inside and outside the box. The bound-ary conditions inside the box lead to:

2

2m

x 2(x) = E (x), (x) = A Sin(K x). (6)

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Functions (x) determine the stationary states inside the box. The boundary conditions outside thebox are,

2

2m

x 2(x) + (x) = E (x), (x) = 0 ,

and determine the energy associated with (x) inside the box as follows. From Eq. (6), we obtain:22m AK

2 = EA, and, (a) = ASin(K a ) = 0 ,

Ka = n, with n = 1 , 2,... Note that the number of nodes of (i.e., the number of coordinates where (x) = 0) , is equal ton 1 for a given energy, and the energy levels are,

E = 2

2mn22

a2, with n = 1 , 2, ...

e.g.,

E (n = 1) = 2

2m

2

a2 ,

E (n = 2) = 2

2m42

a2, ...

Conclusion: The energy of the particle in the box is quantized! (i.e., the absorption spectrum of the particle in the box is not continuous but discrete).

6.1 Exercise 6

(i) Using the particle in the box model for an electron in a quantum dot (e.g., a nanometer sizesilicon material) explain why larger dots emit in the red end of the spectrum, and smaller dots emitblue or ultraviolet.

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(ii) Consider the molecule hexatriene CH 2 = CH CH = CH CH = CH 2 and assume thatthe 6 electrons move freely along the molecule. Approximate the energy levels using the particlein the box model. The length of the box is the sum of bond lengths with C-C = 1.54 A, C=C = 1.35A, and an extra 1.54 A, due to the ends of the molecule. Assume that only 2 electrons can occupy

each electronic state and compute:(A) The energy of the highest occupied energy level.(B) The energy of the lowest unoccupied energy level.(C) The energy difference between the highest and the lowest energy levels, and compare suchenergy difference with the energy of the peak in the absorption spectrum at MAX =268nm.(D) Predict whether the peak of the absorption spectrum for CH 2 = CH (CH = CH )n CH =CH 2 would be red- or blue-shifted relative to the absorption spectrum of hexatriene.

7 Commutator

The commutator [A, B ] is dened as follows: R4(97)

[A, B ] = AB B A.Two operators A and B are said to commute when [A, B ] = 0.

7.1 Exercise 7

Prove that [x, i x ] = i .Hint: Prove that [x,

i x

](x) = i (x), where (x) is a function of x.

8 Uncertainty Relations

The goal of this section is to show that the uncertainties A = < (A< A > )2 > and B = < (B )2 > , of any pair of hermitian operators A and B , satisfy the uncertainty rela-tion: R3(437)( A)2( B)2

14

 2 . (7)

In particular, when A = x and B = p, we obtain the Heisenberg uncertainty relation :

x p

2. (8)

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Proof:U A< A >, (, x ) (U + i V )(x),V B, I () dx(, x )(, x ) 0,

I () = dx[(A< A > )(x) + i (B )(x)][(A< A > )(x) + i (B )(x)],I () = + 2 < V |V > i (< V |U > ),I () = < |U 2| > + 2 < |V 2| > i < |UV V U | > 0, (9)

The minimum value of I (), as a function of , is reached when I/ = I/ = 0 .This condition implies that

2( B)2 = i < [A, B ] >, = > =i < [A, B ] >

2( B)2.

Substituting this expression for into Eq. (9), we obtain:

( A)2 +i2 < A, B > 2

4( B)2 i2 < A, B > 2

2( B)2 0,

( A)2( B)2 i2 < A, B > 2

4.

8.1 Exercise 8

Compute < X > , , X and P for the particle in the box in its minimum energy state andverify that X and P satisfy the uncertainty relation given by Eq. (8)?

With the exception of a few concepts (e.g., the Exclusion Principle that is introduced later in theselectures), the previous sections have already introduced most of Quantum Theory. Furthermore, wehave shown how to solve the equations introduced by Quantum Theory for the simplest possibleproblem, which is the particle in the box. There are a few other problems that can also be solvedanalytically (e.g., the harmonic-oscillator and the rigid-rotor described later in these lectures).However, most of the problems of interest in Chemistry have equations that are too complicated tobe solved analytically. This observation has been stated by Paul Dirac as follows: The underlying physical laws necessary for the mathematical theory of a large part of Physics and the wholeof Chemistry are thus completed and the difculty is only that exact application of these lawsleads to the equations much too complicated to be soluble . It is, therefore, essential, to introduceapproximate methods (e.g., perturbation methods and variational methods ).

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8.2 EPR Paradox

Gedankenexperiments (i.e. , thought experiments) have been proposed to determine hidden vari-ables. The most famous of these proposals has been the Einstein-Podolski-Rosen (EPR) gedanken-experiment [Phys. Rev. (1935) 47 :777-780] , where a system of 2 particles is initially prepared withtotal momentum pt . At a later time, when the two particles are far apart from each other, the posi-tion x1 is measured on particle 1 and the momentum p2 is measured on particle 2. The paradox isthat the momentum of particle 1 could be obtained from the difference p1 = pt p2. Therefore, thecoordinate x1 and momentum p1 of particle 1 could be determined with more precision than estab-lished as possible by the uncertainty principle, so long as the separation between the two particlescould prevent any kind of interaction or disturbance of one particule due to a measurement on theother.

The origin of the paradox is the erroneous assumption that particles that are far apart fromeach other cannot maintain instantaneous correlations. However, quantum correlations betweenthe properties of distant noninteracting systems can be maintained, as described by Bohm and

Aharonov [Phys. Rev. (1957) 108 :1070-1076] for the state of polarization of pairs of correlatedphotons. Within the Bohmian picture of quantum mechanics, these quantum correlations are estab-lished by the quantum potential V Q (q ), even when the particles are noninteracting ( i.e. , V (q ) = 0 ).

Quantum correlations between distant noninteracting photons were observed for the rst timeby Aspect and co-workers in 1982 [Phys. Rev. Lett. (1982) 49 :91-94], 47 years after the EPR para-dox was presented. These quantum correlations constitute the fundamental physics exploited byteleportation (i.e. , the transmission and reconstruction of quantum states over arbitrary large dis-tances) [ Nature (1997) 390 :575-579] and ghost imaging (i.e. , a technique where the object and theimage system are on separate optical paths) [ Am. J. Phys. (2007) 75 :343-351].

9 Variational Theorem

The expectation value of the Hamiltonian, computed with any trial wave function, is always higher or equal than the energy of the ground state . Mathematically,

< |H | >E 0,where H j = E j j .Proof: = j C j j , where { j}is a basis set of orthonormal eigenfunctions of the Hamiltonian H .

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< |H | > = j k

C k C j < k|H | j >,=

j k

C k C j E j kj ,

= j

C j C j E j E 0 j

C j C j ,

where, j C j C j = 1 .

Variational ApproachStarting with an initial trial wave function dened by the expansion coefcients {C

(0) j }, the

optimum solution of an arbitrary problem described by the Hamiltonian H can be obtained byminimizing the expectation value <

|H

| > with respect to the expansion coefcients.

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10 Digital Grid-Based Representations

The standard formulation of quantum mechanics, presented in previous sections, relies upon thetools of calculus (e.g. , derivatives, integrals, etc.) and involves equations and operations with in-

nitesimal quantities as well as states in Hilbert-space (the innite dimensional space of functionsL2). The equations, however, seldom can be solved analytically. Therefore, computational solu-tions are necessary. However, computers can not handle innite spaces since they have only limitedmemory. In fact, all they can do is to store and manipulate discrete arrays of numbers. Therefore,the question is: how can we represent continuum states and operators in the space of memory of digital computers?

In order to introduce the concept of a grid-representation , we consider the state,

0(x) =

1/ 4e2 (xx0 )2 + ip 0 (xx0 ) , (10)

which can be expanded in the innite basis set of delta functions (x x) as follows,0(x) = dxc(x) (x x), (11)

where c(x) x|0 = 0(x). All expressions are written in atomic units, so = 1 .A grid-based representation of 0(x) can be obtained, in the coordinate range x = ( xmin , xmax ),by discretizing Eq. ( 11) as follows,

0(x) = n

j =1

c j (x x j ), (12)

where the array of numbers c j x j |0 represent the state 0 on a grid of equally spaced coordi-nates x j = xmin + ( j 1) with nite resolution = ( xmax xmin )/ (n 1).Note that the grid-based representation, introduced by Eq. ( 12), can be trivially generalized toa grid-based representation in the multidimensional space of parameters ( e.g. , x j , p j , j , ... etc.)when expanding the target state 0(x) as a linear combination of basis functions x|x j , p j , j , withexpansion coefcients as c j x j , p j , j |0 .

10.1 Computational Problem 1

Write a computer program to represent the wave-packet, introduced by Eq. ( 10) on a grid of equallyspaced coordinates x j = xmin + ( j 1) with nite resolution = ( xmax xmin )/ (n 1) andvisualize the output. Choose x0 = 0 and p0 = 0 , in the range x=(-20,20), with = m, wherem = 1 and = 1 .

Next, we consider grid-based representations in momentum space:

0( p) = p|0 . (13)21


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Inserting the closure relation 1 = dx|x x| in Eq. (13), we obtain that p|0 = dx p|x x|0 = (2 )1/ 2 dxeipx x|0 . (14)

is the Fourier transform of the initial state. The second equality in Eq. ( 14) was obtained by using:x| p = (2 )1/ 2eipx , (15)

which is the eigenstate of the momentum operator p = i, with eigenvalue p, since p x| p = p x| p .The Fourier transform can be computationally implemented in O(N log(N )) steps by usingthe Fast Fourier Transform (FFT) algorithm [see, Ch. 12 of Numerical Recipes by W.H. Press,B.P. Flannery, S.A. Teukolsky and W.T. Vetterling, Cambridge University Press, Cambridge, 1986(f12-2.pdf) ] when x|0 is represented on a grid with N = 2 n points (where n is an integer).In contrast, the implementation of the Fourier transform by quadrature integration would requireO(N 2) steps.


Write a computer program to represent the initial state, introduced by Eq. ( 10), in the momentumspace by applying the FFT algorithm to the grid-based representation generated in Problem 1 andvisualize the output. Represent the wave-packet amplitudes and phases in the range p=(-4,4) andcompare your output with the corresponding values obtained from the analytic Fourier transformobtained by using:

dx exp(a2x2 + a1x + a0) =

/a 2 exp(a0 + a21/ (4a2)).

Next, we consider the grid-based representation of operators ( e.g. , x, p, V (x), and T = p2/ (2m))and learn how these operators act on states represented on grids in coordinate and momentumspaces. For simplicity, we assume that the potential is Harmonic:

V (x) =12

m2(x x)2. (16)Consider rst applying the potential energy operator to the initial state, as follows,

V (x)0(x) = V (x)0(x) 0(x). (17)Since 0(x) is just another function, Eq. ( 17) indicates that V (x) can be represented on the samegrid of coordinates as before ( i.e. , equally spaced coordinates x j = xmin + ( j 1) , with niteresolution = ( xmax xmin )/ (n 1)). Since for each x j , 0(x j ) = V (x j )(x j ), the operatorV (x) can be represented just as an array of numbers V (x j ) associated with the grid-points x j , andits operation on a state is represented on such a grid as a simple multiplication.

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Write a computer program to compute the expectation values of the position x(0) = 0|x|0 andthe potential energy V = 0|V (x)|0 , where V (x) is dened according to Eq. ( 16) for the initialwave-packet, introduced by Eq. ( 10), with various possible values of x0 and p0, with = m,where m = 1 and = 1 .

Now consider applying the momentum operator, p = i, to the initial state 0(x) as follows,G(x) = x| p|0 = i0(x). (18)

One simple way of implementing this operation, when 0(x) is represented on a grid of equallyspaced points x j = xmin + ( j 1) , is by computing nite-increment derivatives as follows:

G(x j ) = i0(x j +1 ) 0(x j1)

2. (19)

However, for a more general operator ( e.g. , T = p2/ (2m)) this nite increment derivativeprocedure becomes complicated. In order to avoid such procedures one can represent the initialstate in momentum-space (by Fourier transform of the initial state); apply the operator by simplemultiplication in momentum space and then transform the resulting product back to the coordinaterepresentation (by inverse-Fourier transform). This method can be derived by inserting the closurerelation 1 = dp| p p|, in Eq. (18),

G(x) = x| p|0 = dp x| p| p p|0 = (2 )1/ 2 dpeipx p p|0 , (20)since p|0 is dened according to Eq. ( 14) as the Fourier transform of the initial state. Note thatthe second equality of Eq. ( 20) is obtained by introducing the substitution

x| p = (2 )1/ 2eix p. (21)While Eq. ( 20) illustrates the method for the specic operator p, one immediately sees that anyoperator which is a function of p (e.g. , T = p2/ (2m)) can be computed analogously according tothe Fourier transform procedure.


Write a computer program to compute the expectation values of the initial momentum p(0) =0

| p

|0 and the kinetic energy T = 0

| p2/ (2m)

|0 by using the Fourier transform procedure,

where 0 is the initial wave-packet introduced by Eq. ( 10), with x0 = 0 , p0 = 0 , and = m,where m = 1 and = 1 . Compute the expectation value of the energy E = 0|H |0 , whereH = p2/ (2m) + V (x), with V (x) dened according to Eq. ( 16) and compare your result with thezero-point energy E 0 = / 2.

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11 SOFT Method

The Split-Operator Fourier Transform (SOFT) method is a numerical approach for solvingthe time-dependent Schr odinger equation by using grid-based representations of the time-evolving

states and operators. It relies on the Fourier transform procedure to apply operators that are func-tions of p by simple multiplication of array elements. As an example, we will illustrate the SOFTalgorithm as applied to the propagation of the harmonic oscillator, which can also be describedanalytically as follows:

t (x) = dx x|eiHt |x x|0 , (22)where the Kernel x|eiHt |x is the quantum propagator

x|eiHt |x = m2sinh (it) exp m2sinh (it ) [(x2 + x2)cosh (it ) 2xx ] . (23)The essence of the method is to discretize the propagation time on a grid tk = ( k 1) , withk = 1 ,...,n and time-resolution = t/ (n 1), and obtain the wave-packet at the intermediatetimes tk by recursively applying Eq. ( 22) as follows,

tk +1 (x) = dx x|eiH |x x|t k . (24)If is a sufciently small time-increment ( i.e. , n is large), the time-evolution operator can beapproximated according to the Trotter expansion to second order accuracy,

eiH = eiV (x) / 2ei p2 / (2m ) eiV (x) / 2 + O( 3), (25)

which separates the propagator into a product of three operators, each of them depending either onx, or p.


Expand the exponential operators in both sides of Eq. ( 25) and show that the Trotter expansion isaccurate to second order in powers of .

Substituting Eq. ( 25) into Eq. (24) and inserting the closure relation 1 = dp| p p| gives,tk +1 (x) =

dp

dxeiV (x ) / 2 x| p eip

2 / (2m ) p|x eiV (x ) / 2t k (x). (26)

By substituting p|x and x| p according to Eqs. ( 15) and (21), respectively, we obtain:tk +1 (x) = eiV (x ) / 2

1 2 dpeixp eip 2 / (2m ) 1 2 dxeipx eiV (x ) / 2tk (x). (27)

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According to Eq. ( 27), then, the computational task necessary to propagate t (x) for a time-increment involves the following steps:

1. Represent tk (x) and eiV (x ) / 2 as arrays of numbers tk (x j ) and eiV (x j ) / 2 associated

with a grid of equally spaced coordinatesx j = xmin + ( j 1)

, with nite resolution = ( xmax xmin )/ (n 1).

2. Apply the potential energy part of the Trotter expansion eiV (x ) / 2 to tk (x) by simplemultiplication of array elements:

tk (x j ) = eiV (x j ) / 2tk (x j ).

3. Fourier transform tk (x j ) to obtain tk ( p j ), and represent the kinetic energy part of theTrotter expansion eip 2 / (2m ) as an array of numbers eip 2j / (2m ) associated with a grid of equally spaced momenta p j = j/ (xmax xmin ).

4. Apply the kinetic energy part of the Trotter expansion eip 2 / (2m ) to the Fourier transformtk ( p) by simple multiplication of array elements:

tk ( p j ) = eip2j / (2m ) tk ( p j ).

5. Inverse Fourier transform tk ( p j ) to obtain t k (x j ) on the grid of equally spaced coordinatesx j .

6. Apply the potential energy part of the Trotter expansion eiV (x ) / 2 to tk (x) by simplemultiplication of array elements,

tk +1 (x j ) = eiV (x j ) / 2tk (x j ).


Write a computer program that propagates the initial state 0(x) for a single time increment ( =0.1 a.u.). Use x0 = 2.5, p0 = 0 , and = m, where m = 1 and = 1 . Implement the SOFTmethod for the Hamiltonian H = p2/ (2m) + V (x), where V (x) is dened according to Eq. ( 16).Compare the resulting propagated state with the analytic solution obtained by substituting Eq. ( 23)into Eq. ( 22).

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Loop the computer program developed in Problem 5 with x0 = 2.5 and p0 = 0 for 100 steps with = 0 .1 a.u. For each step compute the expectation values of coordinates x(t) and momenta p(t)as done in Problems 3 and 4, respectively. Compare your calculations with the analytic solutionsobtained by substituting Eq. ( 23) into Eq. (22). Verify that these correspond to the classical trajec-tories x(t) = x + ( x0 x)cos(t) and p(t) = p0 (x0 x)m sin(t), which can be computedaccording to the Velocity-Verlet algorithm:

p j +1 = p j + ( F (x j ) + F (x j +1 )) / 2x j +1 = x j + p j /m + F (x j ) 2/ (2m).

(28)


Change the potential to that of a Morse oscillator V (x) = De(1 exp(a(x xe))) 2, with xe = 0 ,De = 8 , and a = k/ (2D

e), where k = m2

. Recompute the wave-packet propagation withx0 = 0.5 and p0 = 0 for 100 steps with = 0 .1 a.u., and compare the expectation values x(t)and p(t) with the corresponding classical trajectories obtained by recursively applying the Velocity-Verlet algorithm.


Simulate the propagation of a wave-packet with x0 = 5.5 and initial momentum p0 = 2 collidingwith a barrier potential V (x) = 3 , if abs(x) < 0.5, and V (x) = 0 , otherwise. Hint: In orderto avoid articial recurrences you might need to add an absorbing imaginary potential V a (x) =i(abs(x)

10)4, if abs(x) > 10, and V a (x) = 0 , otherwise.

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12 Tunneling Current: Landauer Formula

We consider a 1-dimensional electron tunneling

H = 2

2m 2

x 2 + eV (x) , (29)

problem described by the Hamiltonian

H = 2

2m 2

x 2+ eV (x), (30)

where e is the charge of the electron and

V (x) =V l if x < 0,V b if 0 < x < a,

V r if x > a,

(31)

where V b denes the tunneling barrier, and V = ( V l V r ) denes the voltage drop across thebarrier. Outside the tunneling interval xl < x < x r , the solutions of the Schrodinger equation aresuperpositions of plane waves since the potential is constant. For energy E > eV l and E > eV r ,there are two independent solutions l and r for incident electrons from the left and from theright, respectively.

Considering the solution for incidence from the left, we obtain:

l(x) =+l + r ll if x < 0Aeik bx + Beik bx if 0 < x < a

t r +r if x > a

(32)

where j = k1/ 2

j eik j x , are dened divided by the square root of k j so they are normalized tocarry the unit of current density /m , as shown below). The labels j = l, r indicate the left ( l) andright ( r ) side of the barrier, kl = 2m(E eV l)/ 2 and kr = 2m(E eV r )/ 2.Applying the continuity conditions for l and l /x at x = 0 and x = a, we obtain:

k1/ 2l + k1/ 2

l r l = A + B,Aeik ba + Beik ba = k1/ 2r tr eik r a ,

k1/ 2l (1

r l) = kb(A

B),

kb(Aeik ba Beik ba ) = k1/ 2r t r eik r a .(33)

The transmission amplitude t r , reection amplitude r l and coefcients A and B can be obtained bysolving for them from Eq. ( 33).

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The probability ux (or current density ) of incoming electrons from the left, described by theincident wave i(x, t ) = k1/ 2l ei(kl xwt ) with momentum kl and energy E (kl) = eV l + 2k2l / (2m),is:

j i(x, t ) =

2mii (x, t )

i (x, t )

x i (x, t )

i (x, t )

x, (34)

or

j i(x) =12

i (x, t ) i

m

xi(x, t ) + c.c.,

=12

i (x, t ) p

mi(x, t ) + c.c.,

= Re[i (x, t )vi (x, t )] =

m.

(35)

The ux of transmitted electrons described by transmitted wave t (x, t ) = t r k1/ 2r ei (kr xwt ) , withmomentum kr and energy E (kr ) = eV r + 2k2r / (2m), is:

j t (x) =12

t (x, t ) i

m

xt (x, t ) + c.c.,

= |t r |2

m.

(36)

Therefore, the transmission coefcient T l = j t /j i , dened as the transmitted ux j t over the inci-dent ux at energy E is: T l = |t r |2. The reection coefcient R l = 1 T l is the reected ux overthe incident ux.

Analogously, we consider incidence from the right of the tunneling barrier, as follows:

r (x) =r + r r +r if x > aAeik bx + Beik bx if 0 < x < at ll if x < 0

(37)

Solving for t l , we obtained the transmission coefcient T r = |t l|2, due to incidence from the right.More generally, we can consider incoming waves from both left and right (+l and r , re-spectively) with amplitudes c in = c

(l)in , c

(r )in that generate outgoing waves to the left and right ( l

and +r , respectively) with amplitudes cout = c(l)out , c

(r )out . The amplitudes of outgoing and incoming

waves are related by the linear transformation dened by the scattering matrix (or, S-matrix) S ,as follows: cout = Sc in :

cloutcrout

= r l t lt r r rclincrin

(38)

28

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Due to the conservation of probability, the S-matrix must be unitary: S1 = S. Therefore,SS = 1 :

r l t lt r r r

r l trtl rr

= 1 (39)

which gives

r lrl + t ltl = 1 . (40)In addition, SS = 1 :

r l trtl r r

r l t lt r r r

= 1 (41)

which gives

tl t l + rr r r = 1 (42)

Therefore, according to Eqs. ( 40) and (42), we obtain: 1t ltl = r lr l = rr r r . For our 1-dimensionalcase, we obtain:

|r l|2 = |r r |2 = R. (43)Under stationary state, /t = 0 , with = ||. Then, according to the continuity equation/t = j/x , we obtain: j/x = 0 . Therefore, j l for x < 0 must be equal to j l for x > a .Also, j r for x < 0 must be equal to j r for x > a :

(1 |r l|2) = |t r |2, (44)and

(1 |r r |2) = |t l|2. (45)Dividing Eq. ( 44) by Eq. (45) and using ( 43), we obtain:

t l = t r . (46)

Therefore,

T l(E ) = |t r |2,= |t l|2 = T r (E ),

(47)

so the transmission coefcient is the same for both directions of incidence and R + T = 1 .

29


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Considering that the number of electrons with energy E incident from the left and right of thebarrier are n l(E ) and n r (E ), respectively, the net ux of charge from left to right is:

I = 2e

0

dkln l(kl) kl

mT l

2e

0

dkr n r (kr ) kr

mT r ,

=2e2 0 dE T (E ) n l(E ) klm k lE n r (E ) krm krE ,

=2e2 0 dE T (E ) n l(E ) klm m 2kl n r (E ) krm m 2kr ,

=2eh 0 dE T (E ) (n l(E ) n r (E )) ,

(48)

where factor of 2 accounts for the two possible spin states, the rst term on the r.h.s. accountsfor the forward ux (i.e., from left to right) and the second term accounts for the backward ux(i.e., from right to left). Note that in the second row of Eq. ( 48) we used the following equality:1 = dE |E E | = 2 dk|k k|.At equilibrium, the population of energy levels is determined by the Fermi-Dirac distribution:

n(E ) =1

e (E E F ) + 1, (49)

where E F is the Fermi level and the factor of 2 in the numerator accounts for the 2 possible spinstates. Considering the potentials for electrons at either side of the barrier, we obtain n l(E ) =n(E eV l) and n r (E ) = n(E eV r ). Therefore, we can expand these distributions, as follows:

n l(E ) = n(E

E F ) +

n (E )

E eV l +

,

n r (E ) = n(E E F ) +n (E )

E eV r + ,

(50)

and write the Landauer formula, giving the current in the form of the Ohms law, as follows:

I =2eh 0 dE T (E ) (n l(E ) n r (E )) ,

=2e2

h dE T (E ) n (E )E V,= G(E ) V,

(51)

where G(E ) = R1 = G0 dE T (E ) n (E )E is the conductance, or inverse of the resistance R , withG0 = 2e2h = [12.906 k ]1 the quantum unit of conductance. Note that G0 denes the maximumconductance (minimum resistance) per conduction channel with perfect transmission, T (E ) = 1

30

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(i.e. , if the transport through the channel is ballistic and therefore the probability for transmittingthe electron that enters the channel is unity), as observed in experiments.At low temperature ( i.e. , ), the Fermi-Dirac distributions become step functions n l(E ) =2H (E F (E eV l)) and n r (E ) = 2 H (E F (E eV r )) , with H (x) the Heaviside function equal

to 1 for x > 0, and 0 for x < 0. Therefore, n (E )E = (E F E ), and

I =2e2

h dE T (E ) (E F E ) V,=

2e2

hT (E F ) V.

(52)

In this low-temperature limit, the conductance is the transmission times the quantum of conduc-tance, G(E ) = 2e

2

h T (E F ).

31
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12.1 WKB Transmission

The goal of this subsection is to show that the transmission coefcient T (E ) can be estimated,under the WKB approximation, as follows:

T (E ) = e2 R a0 dx 2m |E (x)|/ 2 , (53)where (x) = V b describes the tunneling barrier according to Eq. ( 31).

To derive Eq. ( 53), we consider the WKB approximate solution of Eq. ( 30), with the followingfunctional form:

(x) = 0ei R x

0 k(x )dx , (54)

where k(x) = 2m[E V (x)]/ 2. Note that when V (x) is constant, (x) corresponds to aparticle moving to the right with constant momentum k. Substituting (x) as dened in Eq. ( 54),into Eq. ( 30), we obtain:

2

2m

2(x)x 2

+ V (x)(x) = E (x) , (55)with = ik (x) 22m (x). Therefore, the WKB solution is a good approximation when |k(x)| a , the probability density remains constant again since (x) = (a)ei R x

a dx kr

and |(x)|2 = |(a)|2. Therefore, estimating the transmission coefcient as the ratio of the proba-bility densities to the right and to the left of the barrier, we obtain

T (E ) = |(a)|2|(0)|2

,

= e2 R

a0 dx 2m |E (x)|/ 2

.

(58)

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13 Time Independent Perturbation Theory

Consider the time independent Schr odinger equation, R2(453)

Hn (x) = E n n (x),

(59)

for a system described by the Hamiltonian H = p2/ 2m + V , and assume that all the eigenfunctionsn (x) are known. The goal of this section is to show that these eigenfunctions n (x) can be usedto solve the time independent Schr odinger equation of a slightly different problem: a problemdescribed by the Hamiltonian H = H + . This is accomplished by implementing the equationsof Perturbation Theory derived in this section.Consider the equation

(H + )n (, x ) = E n ()n (, x ), (60)

where is a small parameter, so that both n () and E n () are well approximated by rapidlyconvergent expansions in powers of (i.e., expansions where only the rst few terms are important).Expanding n () we obtain,

n (, x ) = j

C jn () j (x).

Substituting this expression in the time independent Schr odinger equation we obtain,

j

C jn ()[H j (x) + j (x)] = E n ()k

C kn ()k(x).,

therefore,C ln ()E l +

j

C jn () < l|| j > = E n ()C ln (). (61)Expanding C kj and E n in powers of we obtain,

C kj () = C (0)kj + C

(1)kj + C

(2)kj 2 + ...,

andE n () = E

(0)n + E (1)n + E (2)n 2 + ...

Substituting these expansions into Eq. ( 61) we obtain,(C (0)ln E l E

(0)n C (0)ln ) + (C

(1)ln E l + j C

(0) jn < l|| j > E

(0)n C (1)ln E

(1)n C (0)ln ) +

2(C (2)ln E l + j C

(1) jn < l|| j > E

(2)n C (0)ln E

(0)n C (2)ln E

(1)n C (1)ln ) + ... = 0 .

This equation must be valid for any . Therefore, each of the terms in between parenthesis must beequal to zero.

Zeroth order in C

(0)ln (E l E

(0)n ) = 0 ,

if l = n, then C (0)ln = 0 ,if l = n, then C (0)nn = 1 , and E l = E

(0)n .

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First order in C (1)ln (E l E

(0)n ) = E (1)n C (0)ln j C

(0) jn < l|| j >,

if l = n, then C (1)ln (E l E (0)n ) = C

(0)nn < l||n >,

if l = n, then E (1)n C (0)ln = C (0)nn < n ||n > .

Note thatC

(1)

nnis not specied by the equations listed above.

C (1)

nnis obtained by normalizing the

wave function written to rst order in .

2nd order in

C (2)ln (E l E n ) + j C (1)

jn < l|| j > = E (2)n C (0)ln + E

(1)n C (1)ln ,

if l = n, then E (2)n = j = n C (1)

jn < n || j > = j = n

(E j E (0)n )

,

if l = n, then C (2)ln (E l E (0)n ) = j C

(1) jn < l|| j > (E l E (0)n ) =

= j

(E j E (0)n )

(E lE

(0)n )

.

13.1 Exercise 9Calculate the energy shifts to rst order in for all excited states of the perturbed particle in thebox described by the following potential:

T

E

T

0 ax

Assume that the potential is described by perturbation W (x) = Sin( xa ) to the particle in thebox.

14 Time Dependent Perturbation Theory

Given an arbitary state, R2(410)

(x, t ) = j

C j j (x)ei E j t ,

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for the initially unperturbed system described by the Hamiltonian H , for which H j = E j j andi t = H , let us obtain the solution of the time dependent Schr odinger equation:

i

t= [H + (t)], (62)

assuming that such solution can be written as a rapidly convergent expansion in powers of ,

(x, t ) = j

l=0

C jl (t) l j (x)ei E j t . (63)

Substituting Eq. ( 63) into Eq. (62) we obtain,

i

l=0

C kl (t) l + C kl (t) l(i E k) e

i E k t = j

l=0

C jl (t) l (< k| j > E j + < k|| j > ) ei

Terms with 0: (Zero-order time dependent perturbation theory)

+ i [C k0 (t)ei E k t + C k0 (t)(

i E k)e

i E k t ] = j

C j 0 (t) kj E j ei E j t = C k0 (t)E ke

i E k t .

Since,C k0 (t) = 0 , C k0 (t) = C k0 (0).

Therefore, the unperturbed wave function is correct to zeroth order in .Terms with : (First-order time dependent perturbation theory)

i [C k1 (t)ei E k t + C k1 (t)(

i

E k )e

i E k t ] = j

C j1 (t) kj E j ei E j t + C j 0 (t) < k

|

| j > e

i E j t ,

C k1 (t) = i

j

C j0 (0) < k|| j > e i (E j E k )t .

Therefore,

C k1 (t) = i

j

C j 0 (0) < k|ei E k t e i E j t| j > =

i

j

C j0 (0) < k|ei Ht e i Ht | j >,

(64)Eq. (64) was obtained by making the substitution e

i Ht

|

j> = e

i E j t

|

j> , which is justied

in the note that follows this derivation. Integrating Eq. ( 64) we obtain,

C k1 (t) = i tdt j C j 0 (0) < k|e i Ht e i Ht | j > .

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36/195

which can also be written as follows:

C k1 (t) = i tdt < k|e i Ht e i Ht |0 > .

This expression gives the correction of the expansion coefcients to rst order in .Note: The substitution made in Eq. ( 64) can be justied as follows. The exponential function isdened in powers series as follows,

eA =

n =0

An

n!= 1 + A +

12!

AA + ...., R4 (169 )

In particular, when A = iHt/ ,e i Ht = 1 + (

i Ht ) +

1

2!(

i t)2H H + ....

Furthermore, sinceH | j > = E j | j >,

and,H H | j > = E j H | j > = E 2 j | j >,

we obtain,

e i Ht | j > = [1 + ( i E j t) +

12!

(i t)2E 2 j + ...]| j > = e

i E j t | j >,

which is the substitution implemented in Eq. ( 64).Terms with 2: (Second-order time dependent perturbation theory)

i [C k2 (t) + C k2 (t)(i E k)]e

i E k t = j

[C j 2 (t) kj E j + C j 1 (t) < k|| j > ]ei E j t ,

C k2 (t) = i

j

< k |ei Ht e i Ht | j > C j1 (t),

C k2 (t) = i tdt j < k|e i Ht e i Ht | j > C j 1 (t),

C k2 (t) = i

2

j tdt t

dt < k|e

i Ht ei Ht | j >< j |e

i Ht ei Ht |0 > .

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37/195

Since 1 = j | j >< j |,

C k2 (t) = i

2 tdt t

dt < k|e

i Ht ei H (t t

)ei Ht |0 > .

This expression gives the correction of the expansion coefcients to second order in .

Limiting Cases(1) Impulsive Perturbation:The perturbation is abruptly switched on: R2(412)

Tw(t)

0t

According to the equations for rst order time dependent perturbation theory,

C k1 (t) = i

j

< k || j > C j 0 (0) t0 dt e i (E j E k )t ,therefore,

C k1 (t) = ( i )

j

C j 0 (0) < k

|

| j >

i (E j E k) ei

(E j E k )t 1 .Assuming that initially: C j = lj , C j 0 = lj . Therefore,

C k1 (t) = < k||l >

(E l E k)[1e

i (E lE k )t ],

when k = l. Note that C l1 (t) must be obtained from the normalization of the wave functionexpanded to rst order in .

14.1 Exercise 10Compare this expression of the rst order correction to the expansion coefcients, due to an impulsive perturbation , with the expression obtained according to the time- independent perturbationtheory.

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38/195

(2) Adiabatic limit:The perturbation is switched-on very slowly ( dtdt e

i (E l E k )t .

Integrating by parts we obtain,

C k1 (t) = ( i )

e i (E lE k )t

(i )(E l E k)< k|(t)|l >

t = t

t = tdt e

i (E l E k )t

(i )(E l E k)< k|

wt |l > ,

and, since < k |w()|l > = 0 ,C k1 (t) =

< k|(t)|l >(E l E k)

e i (E lE k )t ,

when k = l. Note that C l1 (t) must be obtained from the normalization of the wave functionexpanded to rst order in .

14.2 Exercise 11

Compare this expression for the rst order correction to the expansion coefcients, due to an adi-abatic perturbation , with the expression obtained according to the time- independent perturbationtheory.

(3) Sinusoidal Perturbation:The sinusoidal perturbation is dened as follows, (t, x ) = (x)Sin(t) when t 0 and (t, x ) =0, otherwise.

38

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(x, t )

(x)

tE

T

It is, however, more conveniently dened in terms of exponentials,

=(x)

2i[eit

eit ].

Therefore,

C k1 (t) = i t0 dt < k |e i Ht (t)e i Ht |0 >, (65)

with |0 > = j C j | j > , and H j = E j j . Substituting these expressions into Eq. (16) weobtain,C k1 (t) =

12

j

C j < k|| j > t0 dt e i [(E k E j )+ ]t e i [(E k E j ) ]t ,and therefore,

C k1 (t) =1

i2 j

C j kj1 e

i [(E k E j )+ ]tE k E j +

1 ei [(E k E j ) ]t

E k E j .

Without lost of generality, let us assume that C j = nj (i.e., initially only state n is occupied). Fork n we obtain,

|C k1 (t)|2 = |kn |2k 2

1 ei[( E k E n ) +] t

(E k E n ) + 1 ei[

( E k E n ) ]t(E k E n )

2

.

Factor ||kn determines the intensity of the transition (e.g., the selection rules ). The rst term(called anti-resonant ) is responsible for emission. The second term is called resonant and is re-sponsible for absorption.

39


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For k = n , P k1 (t) = 2|C k1 (t)|2 is the probability of nding the system in state k at time t (to rstorder in ).

P k1 (t)

= 4t

(E k E n )

t2 2 |kn |24 2

E

T..................................................

....

...

...

...

...

...

...

...

...

...

...

...

...

..

....

...

...

...

...

...

.

....

...

...

...

...

...

.E'

It is important to note that P k1


41/195

(G) When would the transition j k be forbidden?

15.2 Exercise 12

A particle in the ground state of a square box of length |a| is subject to a perturbation (t) =axet2 / .(A) What is the probability that the particle ends up in the rst excited state after a long timet >> ?(B) How does that probability depend on ?

15.3 Exercise 13

0 a x

Figure 1

V 0

E

T

(a) Compute the minimum energy stationary state for a particle in the square well (See Fig.1) bysolving the time independent Schr odinger equation.(b) What would be the minimum energy absorbed by a particle in the potential well of Fig.1?(c) What would be the minimum energy of the particle in the potential well of Fig.1?(d) What would be the minimum energy absorbed by a particle in the potential well shown in Fig.2?Assume that is a small parameter give the answer to rst order in .

0 0.5 1

Figure 2

V 0

E

T

.............................................

..............................

41

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15.4 Exercise 14

(a) Prove that P = eH is a hermitian operator.(b) Prove that P = Cos (H ) is a hermitian operator.

16 Adiabatic Approximation

The goal of this section is to solve the time dependent Schr odinger equation,

i t

= H, (66)

for a time dependent Hamiltonian,

H = 2

2m

2

+ V (x, t ), where the potential V (x, t ) undergoessignicant changes but in a very large time scale (e.g., a time scale much larger than the timeassociated with state transitions). R2(496)Since V(x,t) changes very slowly, we can solve the time independent Schr odinger equation at aspecic time t,

H (t)n (x, t ) = E n (t)n (x, t ). Assuming that nt 0, since V(x,t) changes very slowly, we nd that the function,

n (x, t ) = n (x, t )ei R t0 E n (t

)dt ,

is a good approximate solution to Eq. ( 66). In fact, it satises Eq. ( 66) exactly when nt = 0 .Expanding the general solution (x, t ) in the basis set n (x, t ) we obtain:

(x, t ) =n

C n (t)n (x, t )ei R t0 E n (t

)dt ,

and substituting this expression into Eq. ( 66) we obtain,

i n

(C n n + C n n i E n C n n )e

i R t0 E n (t )dt =

n

C n E n n ei R t0 E n (t

)dt ,

where,

C k = n C n < k|n > e i R t

0dt (E n (t )

E k (t

))

. (67)

Note that,H t

n + H n =E nt

n + E n n ,

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then,

< k|H t |n > + < k|H |n > =

E kt

kn + E n < k|n >,since < k

|H

|n > = < n

|H

|k >.

Furthermore, if k = n then,

< k|n > =< k|H t |n >

E n E k.

Substituting this expression into Eq. ( 67) we obtain,

C k = C k < k|k > n = k

C n< k|H t |n >

(E n E k)e i R

t0 dt

(E n (t )E k (t )) .

Let us suppose that the system starts with C n (0) = nj , then solving by successive approximationswe obtain that for k = j :

C k =< k|H t | j >

(E k E j )e i R

t0 dt

(E j (t )E k (t )) .

Assuming that E j (t) and E k(t) are slowly varying functions in time:

C k < k|H t | j >

i (E j E k)2[e i (E j E k )t e

i (E j E k )t0 ],

since |ei (E j E k )t e

i (E j E k )t0 | 2.Therefore,

|C k|2 4 2| < k |H t | j > |2

(E j E k)4.

The system remains in the initially populated state at all times whenever H t is sufciently small,

< k|H t | j > |


44/195

17 Heisenberg Representation

Consider the eigenvalue problem, R4(124) R3(240)

H |l = E |l ,

(69)

for an arbitrary state |l of a system ( e.g. , an atom, or molecule) expanded in a basis set { j}, asfollows:|l =

j

C ( j )l | j , (70)

where C ( j )l = j |l , and j |k = jk . Substituting Eq. ( 70) into Eq. (69) we obtain:

j

H | j C ( j )l =

j

E lC ( j )l | j .

Applying functional k| to both sides of this equation we obtain,

j

k|H | j C ( j )l =

j

E l k| j C ( j )l , (71)

where k| j = kj and k = 1, 2, ..., n .Introducing the notation H kj = k|H | j we obtain,

(k = 1) (k = 2)...(k = n)

H 11C (1)l + H 12C

(2)l + H 13C

(3)l + ... + H 1n C

(n )l = E lC

(1)l + 0 C

(2)l + ... + 0 C

(n )l ,

H 21C (1)l + H 22C

(2)l + H 23C

(3)l + ... + H 2n C

(n )l = 0C

(1)l + EC

(2)l + ... + 0 C

(n )l ,

...H n 1C

(1)l + H n 2C

(2)l + H n 3C

(3)l + ... + H nn C

(n )l = 0 C

(1)l + 0 C

(2)l + ... + EC

(n )l ,

(72)that can be conveniently written in terms of matrices and vectors as follows,

H 11 H 12 ... H 1nH 21 H 22 ... H 2n...

H n 1 H n 2 ... H nn

C (1)lC (2)l...

C (n )l

=

E l 0 ... 00 E l ... 0...0 0 ... E l

C (1)lC (2)l...

C (n )l

. (73)

This is the Heisenberg representation of the eigenvalue problem introduced by Eq. ( 69). Accordingto the Heisenberg representation , also called matrix representation , the ket |l is represented bythe vector C l , with components C

( j )l = j |l , with j = 1 ,...,n , and the operator H is represented

by the matrix H with elements H jk = j |H |k .

44


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The expectation value of the Hamiltonian,

|H | = j k

C (k)l k|H | j C ( j )l ,

can be written in the matrix representation as follows,

|H |l = C l H C l = C (1)l C (2)l ... C (n )lH 11 H 12 ... H 1nH 21 H 22 ... H 2n...

H n 1 H n 2 ... H nn

C (1)lC (2)l...

C (n )l

.

Note:(1) It is important to note that according to the matrix representation the ket-vector |l is repre-sented by a column vector with components C ( j )l = j

|l , and the bra-vector l

|is represented

by a row vector with components C ( j )l .(2) If an operator is hermitian (e.g., H ) it is represented by a hermitian matrix (i.e., a matrixwhere any two elements which are symmetric with respect to the principal diagonal are complexconjugates of each other). The diagonal elements of a hermitian matrix are real numbers, therefore,its eigenvalues are real.(3) The eigenvalue problem has a non-trivial solution only when the determinant det [H 1E ]vanishes:

det[H 1E ] = 0, where 1 is the unity matrix .This equation has n roots, which are the eigenvalues of H .(3) Finally, we note that the matrix of column eigenvectors C satisfy the equation, HC = EC ,where E is the diagonal matrix of eigenvalues:

H 11 H 12 ... H 1nH 21 H 22 ... H 2n...

H n 1 H n 2 ... H nn

C (1)1 C (1)2 C

(1)n

C (2)1 C (2)2 C

(2)n

... ... ... ...C (n )1 C

(n )2 C

(n )n

=

E 1 0 ... 00 E 2 ... 0...0 0 ... E n

C (1)1 C (1)2 C

(1)n

C (2)1 C (2)2 C

(2)n

... ... ... ...C (n )1 C

(n )2 C

(n )n

.

(74)

18 Two-Level Systems

There are many problems in Quantum Chemistry that can be modeled in terms of the two-levelHamiltonian (i.e., a state-space with only two dimensions). Examples include electron transfer,proton transfer, and isomerization reactions.

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Consider two states |1 > and |2 > , of a system. Assume that these states have similarenergies, E 1 and E 2, both of them well separated from all of the other energy levels of the system,H 0|1 > = E 1|1 >,H 0|2 > = E 2|2 > .

In the presence of a perturbation,

W = 0 0 ,

the total Hamiltonian becomes H = H 0 + W . Therefore, states |1 > and |2 > are no longereigenstates of the system.The goal of this section is to compute the eigenstates of the system in the presence of the

perturbation W. The eigenvalue problem,

H 11 H 12H 21 H 22 C (1)l

C (2)l= E l 00 E l C

(1)l

C (2)l,

is solved by nding the roots of the characteristic equation, (H 11 E l)(H 22 E l) H 12H 21 = 0 .The values of E l that satisfy such equation are,

E l =(E 1 + E 2)

2 E 1 E 22 2 + 2.These eigenvalues E l can be represented as a function of the energy difference (E 1 E 2), accord-ing to the following diagram:

E

T

c

r r r r r r r r r r r r r r r r r r r r r r r r r r r r r

.......................................................

.......................................................

E m +

E m

E m = 12 (E 1 + E 2)

E +

E 1

E 1 E 2

E 2E

046

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Note that E 1 and E 2 cross each other, but E and E + repel each other. Having found the eigen-values E , we can obtain the eigenstates | > = C

(1)

|1 + C (2)

|2 by solving for C (1)

and C (2)

from the following equations:C (1)

(H 11

E ) + C

(2)

H 12 = 0 ,

2 j =1 C ( j ) C ( j ) = 1 .We see that in the presence of the perturbation the minimum energy state | > is always morestable than the minimum energy state of the unperturbed system.

Example 1 . Resonance Structure

d d

d d d d

1 2

E 1 = E 2 = E m

d d d d

d d

The coupling between the two states makes the linear combination of the two more stable thanthe minimum energy state of the unperturbed system.

Example 2 . Chemical Bond

uH + re

uH + |2 >

uH +

re

uH +

|1 >

The state of the system that involves a linear combination of these two states is more stable thanE m because < 1|H |2 > = 0 .

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Time EvolutionConsider a two level system described by the Hamiltonian H = H 0 + W , with H 0 | 1 > = E 1 |1 > . Assume that the system is initially prepared in state | (0) > = | 1 > . Due to the presenceof the perturbation W , state | 1 > is not a stationary state. Therefore, the initial state evolves intime according to the time-dependent Schr odinger Equation,

i | >

t= ( H 0 + W ) | >,

and becomes a linear superposition of states |1 > and |2 > ,

|(t) > = C 1(t)|1 > + C 2(t)|2 > .State | (t) > can be expanded in terms of the eigenstates | > as follows,

|(t) > = C + (t)

|+ > + C (t)

| >,

where the expansion coefcients C (t) evolve in time according to the following equations,

i C + (t)

t= E + C + (t),

i C (t)

t= E C (t).

Therefore, state |(t) > can be written in terms of | > as follows,

|(t) > = C + (0)ei E + t |+ > + C (0)e

i E t| > .The probability amplitude of nding the system in state |2 > at time t is,

P 12(t) = | < 2|(t) > |2 = C 2(t)C 2(t),which can also be written as follows,

P 12(t) = |C 2+ C + (0)|2 + |C 2C (0)|2 + 2 Re[C 2+ C + (0)C 2C (0)ei (E E + )t ],

where C 2 = < 2 | > . The following diagram represents P 12(t) as a function of time:

Rabi Oscillations T

E

P 12(t)

0 tE + E

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The frequency = ( E + E )/ ( ) is called Rabi Frequency . It is observed, e.g., in theabsorption spectrum of H +2 (see Example 2). It corresponds to the frequency of the oscillatingdipole moment which uctuates according to the electronic congurations of |1 > and |2 > ,respectively. The oscillating dipole moment exchanges energy with an external electromagneticeld of its own characteristic frequency and, therefore, it is observed in the absorption spectrum of the system.

19 Harmonic Oscillator

Many physical systems, including molecules with congurations near their equilibrium positions,can be described (at least approximately) by the Hamiltonian of the harmonic oscillator: R4(483)R1(62) click

H =P 2

2m+

1

2m2x2.

In order to nd the eigenfunctions of H we introduce two operators called creation a+ and annihi-lation a , which are dened as follows:

a+ 1 2 (x i p), and a 1 2 (x + i p), where x = x m , and p = p m .Using these denitions of a+ and a, we can write H as follows,H = ( a+ a + 12 ) .

Introducing the number operator N , dened in terms of a+ and a as follows,N a+ a ,we obtain that the Hamiltonian of the Harmonic Oscillator can be written as follows,H = ( N + 1 / 2) .

19.1 Exercise 15

Show that if is an eigenfunction of H with eigenvalue E , then is an eigenfunction of N with eigenvalue = E 12 . Mathematically, if H | > = E | > , then N | > = | > , with = E 12 .

Theorem IThe eigenvalues of N are greater or equal to zero, i.e.,

0.

Proof: dx| < x |a| > |2 0,< |a+ a| > 0, < | > 0.49
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As a consequence: a|0 > = 0 ,1 2 [x m + i p m ]|0 > = 0 , p = i x ,x0(x) + m

0 (x)x = 0 ,

ln0(x) = m xx,0(x) = A exp(

m 2

x2),

where A = 4 m . The wave function 0(x) is the eigenfunction of N with = 0 (i.e., the ground state wave function because 0).Theorem II If > 0 , state a| > is an eigenstate of N with eigenvalue equal to ( -1) .Proof:In order to prove this theorem we need to show that,

N a| > = ( 1)a| > . (75)We rst observe that,

[N , a] = a.Therefore,

[N , a] = a+ aa aa+ a,[N , a] = [a+ , a]a,[N , a] = 1a, because [a+ , a] = 1,[a

+

, a] =1

2 (xx + ix pi px + p p(xx ix p + i px + p p)),[a+ , a] = i2 2[x, p] = 1, since [x, p] = i .Applying the operator a to state | > we obtain,(N a aN )| > = a| >,and, therefore,N a| > a | > = a| > , which proves the theorem.A natural consequence of theorems I and II is that is an integer number greater or equal to zero.

The spectrum of N is therefore discrete and consists of integer numbers that are 0. In order todemonstrate such consequence we rst prove that,N a p

| > = (

p)a p

| > . (76)

In order to prove Eq. ( 76) we apply a to both sides of Eq. ( 75):

aN a| > = ( 1)a2| >,

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and since [N , a] = a we obtain,(a + N a)a| > = ( 1)a2| >,

andN a2| > = ( 2)a2| > . (77)

Applying a to Eq. (77) we obtain,

aN a2| > = ( 2)a3| >,and substituting aN by a + N a we obtain,

N a3| > = ( 3)a3| > .Repeating this procedure p times we obtain Eq. ( 76).

Having proved Eq. ( 76) we now realize that if = n , with n an integer number,

a p|n > = 0 ,when p > n . This is because state an |n > is the eigenstate of N with eigenvalue equal to zero,i.e., an |n > = |0 > . Therefore a|0 > = a p|n > = 0 , when p > n . Note that Eq. ( 77) wouldcontradict Theorem I if was not an integer, because starting with a nonzero function | > itwould be possible to obtain a function a p| > different from zero with a negative eigenvalue.

Eigenfunctions of N In order to obtain eigenfunctions of N consider that,

N | > = | >,and

N a| +1 > = a| +1 > .Therefore, a| +1 > is proportional to | > ,

a

| +1 > = C +1

| > (78)

Applying a+ to Eq. (78) we obtain,

N | +1 > = C +1 a+ | >,

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| +1 > =C +1

( + 1)a+ | >,

< +1 | +1 > = 1 =C 2 +1

( + 1) 2< |N + 1 | >,

C +1 = + 1 .Therefore,

| +1 > =1

+ 1 a+ | > =

(a+ ) +1

( + 1)! |0 >The eigenfunctions of N can be generated from |0 > as follows,

| > =1

! x m i p m |0 >, (x) =

1 ! x m m x

0(x).

For example,

1(x) = x m + m m x Aem2 x2 ,1(x) = 2 x m 4 m

em2 x2 .

The pre-exponential factor is the Hermite polynomial for = 1 .

Time Evolution of Expectation

Quimica Cuantica en El Chesee

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