book andrei n

Upload: nandrei

Post on 14-Apr-2018

235 views

Category:

Documents


0 download

TRANSCRIPT

  • 7/29/2019 Book Andrei N

    1/155

    Table of Contents

    Chapter 1 Author: Andrei Nistreanu ....................................................................................................3Chapter 2 The tricks ............................................................................................................................3Chapter 3 List of articles:.....................................................................................................................5Chapter 4 Group theory.......................................................................................................................8

    4.1 Basis ........................................................................................................................................84.2 Rotations................................................................................................................................114.3 Reflections.............................................................................................................................134.4 Group representations............................................................................................................204.5 Unitary Matrix ......................................................................................................................224.6 Schur's Lemma......................................................................................................................234.7 Great orthogonality theorem..................................................................................................234.8 Great orthogonality theorem (application)............................................................................244.9 D3 representations and irreducible representations...............................................................254.10 Basis Functions....................................................................................................................284.11 Projection Operators (D3)....................................................................................................29

    4.12 Symmetry adapted linear combinations(C3v).....................................................................364.13 Bonding in polyatomics - constructing molecular orbitals from SALCs............................394.14 Product representations........................................................................................................404.15 Clebsch-Gordan coefficients for simply reducible groups (CGC)......................................424.16 Full matrix representations..................................................................................................494.17 Selections rules....................................................................................................................524.18 (SO3)Linear combinations of spherical harmonics of point groups....................................53

    Chapter 5 Tensors...............................................................................................................................585.1 Direct products, vectors.........................................................................................................585.2 Tensor continuation Polarizability Directions cosines..........................................................66

    Chapter 6 Fine structure.....................................................................................................................74

    6.1 Angular Momentum...............................................................................................................746.2 Atom with Many Electrons....................................................................................................816.3 The band structure of blue and yellow diamonds..................................................................886.4 optical transition ...................................................................................................................906.5 angular momentum deduction...............................................................................................91

    Chapter 7 Quantum mechanics and postulates...................................................................................997.1 The most important thing in Quantum Mechanics ...............................................................997.2 Matrix representation of operators:.......................................................................................997.3 Hermitian Operators..............................................................................................................99

    Chapter 8 Clebsch-Gordan...............................................................................................................1008.1 Addition of angular Momenta: Clebsch-Gordan coefficients I..........................................100

    8.2 Clebsch-Gordan coefficients (other source)II.....................................................................1028.3 Examples..............................................................................................................................105

    Chapter 9 Binomial relationship.......................................................................................................1089.1 Permutations........................................................................................................................1089.2 Combinations......................................................................................................................1099.3 Binomial formula ...............................................................................................................1099.4 A relationship among binomial coefficents (Wigner [55] pag 194).....................................111

    1

  • 7/29/2019 Book Andrei N

    2/155

    Chapter 10 Explicit formulas for GC or Wigner coefficients...........................................................11210.1 Equivalence between Fock states or coherent states and spinors......................................11210.2 Functions of operators ......................................................................................................11310.3 Generating Function of the States .....................................................................................11410.4 Evaluation of the Elements of the Wigner Rotation Matrix..............................................11510.5 RWF lecture.......................................................................................................................117

    10.5.1The Beta Function..........................................................................................................11710.6 Explicit expression for Wigner (CG) coefficients I...........................................................11910.7 Explicit expression for Clebsch-Gordan coefficients (CGC) (Wigner's formula)II..........12110.8 Explicit expression for Clebsch-Gordan coefficients (CGC) (Wigner's formula)III ........12510.9 Van der Waerden symmetric form of CGC .......................................................................12710.10 Wigner 3j symbols or Racah formula .............................................................................131

    Chapter 11 The formula of Lagrange and Sylvester for function F of matrix .................................13411.1 Sylvester formula for Rabi oscillations full deduction......................................................13411.2 Eigenvalues and Eigenvectors...........................................................................................13511.3 Properties of determinants.................................................................................................13611.4 The Method of Cofactors...................................................................................................13811.5 Cramer's rule proof............................................................................................................13811.6 Vandermonde determinant ................................................................................................14011.7 Vandermonde determinant and Cramer's rule result Lagrange's interpolation formula....14111.8 Cayley-Hamilton Theorem................................................................................................145

    11.8.1 The Theorem.................................................................................................................14511.8.2The Simple But Invalid Proof........................................................................................14611.8.3An Analytic Proof..........................................................................................................14611.8.4Proof for Commutative Rings........................................................................................14811.8.5Appendix A. Every Complex Matrix is Similar to an Upper Triangular Matrix...........14811.8.6APPENDIX B. A Complex Polynomial Is Identically Zero Only If All Coefficients AreZero..........................................................................................................................................14911.8.7From Wikipedia, the free encyclopedia.........................................................................149

    11.9 SYLVESTER'S MATRIX THEOREM..............................................................................15011.10 Final formula....................................................................................................................153

    Chapter 12 Time evolution two level atom Rabi frequency.............................................................15412.1 Two level atom..................................................................................................................154

    2

  • 7/29/2019 Book Andrei N

    3/155

    Chapter 1 Author: Andrei Nistreanu

    I must put the heading 4 because the numbering of formulas become long 10.10.12

    Lenef

    Chapter 2 The tricks

    At each bibliography name you must to create a button back so at the page insert the text : Insert

    Bookmark give a number

    JsMath rfs10 hhand lhand H L

    ctrl +b J

    ctrl sift+b J2

    ctrl shift +p J2

    Vezi flv-urile

    alignlpentru ecuatii

    dupa ce apesi hyperlinkul nu face ctrl-z ci Edit si vezi undo sau redoNew Paragraph

    fn apesi F3 si apare

    E=mc2 (2.1)

    Wikipedia in order to remove hyper-links: select all right click remove hyper-link

    Images , jpg from internet : click on the images after double click on the graphics from: Format styles and formatting

    ghilimelee ca text b ' and b '

    Pui caption la figuri apoi click dreapta wrap(a inveli, infasura) > optimal to page

    spelling and grammar intri in tools in options languages settings si pui ca sa faca spell check

    3 x42x1=3x122 x2=3x122x2=x14

    "" =alignl

    3

  • 7/29/2019 Book Andrei N

    4/155

    xy = 2x = 2y

    http://wiki.services.openoffice.org/wiki/Documentation/OOoAuthors_User_Manual/Writer_Guide/Creating_a_table_of_contents

    1. Mai intai configurezi headingsurile: If you want numbering by chapter, you first have to

    configure chapter numbering for the document. Your document uses no heading styles and isnot configured to number the headings. Such numbering can be configured underTools >Outline Numbering; normally you would use the "Heading 1" style for the top-level heading.

    2. Deci pentru chapter pui hedings 2 show sublevels 13. pentru paragrafe pui headings 3 - show sublevels 2 (inserezi si din coltul stanga de sus pui

    hedings 3)4. la equatii pui levelul 2 sau 3 daca pui level 2 atunci apare apare 2 cifre numarul capitolului

    punct numarul ecuatiei; daca pui level 3 apare 3 numere adica este inclus si num paragrafului5. Contents of table : Insert > Indexes and tables6. right click on the field of the tables: Edit index/tables7. if you want hyper-link on the headings 3 then chose level 3, after chose hyperlink to pe click on

    the page or on the entry

    windows Ooo 3.2.1 cu negru vede default faci dublu click pe ecuatie si schimbi la fiecare in partedac vrei din gray in negru sau scrie in windowsRight-click page este page setup margins

    Open the Navigator (pressF5).

    1. Click the + sign next toIndexes.2. Right-click on the desired index and choose Index > Edit.

    3. Delete table of contents

    In order to change the cross-reference in a Hyperlink:-make a cross reference-Insert that cross-reference > Hyperlink-Click the target in document button to the right of the Target text field, simplytype the name into the target text field box.-Click applyMai bine ctrl-B ctrl U si blue caci in pdf tot nu se vede displayed informationDar may bine hyperlink caci mai apoi cu navigatorul te intorci ina poi F5 si alegi hyperlinkuldar daca ai o mie de hyperlincuri?

    Add special character in Math first see the name in Latex e.g. \dagger or \otimes after go in insert object formula or F2 and chose catalog special edit symbolsStep 4. Add the symbol.

    (a) In the Edit symbols dialog, delete the entry under Old symbol

    (b) In the symbol field, type the command that you want to use to refer to the symbol. For example, sayyou want to use the binary order symbol . If you type succ then you will be able to access the

    4

    http://wiki.services.openoffice.org/wiki/Documentation/OOoAuthors_User_Manual/Writer_Guide/Creating_a_table_of_contentshttp://wiki.services.openoffice.org/wiki/Documentation/OOoAuthors_User_Manual/Writer_Guide/Creating_a_table_of_contentshttp://wiki.services.openoffice.org/wiki/Documentation/OOoAuthors_User_Manual/Writer_Guide/Creating_a_table_of_contentshttp://wiki.services.openoffice.org/wiki/Documentation/OOoAuthors_User_Manual/Writer_Guide/Creating_a_table_of_contents
  • 7/29/2019 Book Andrei N

    5/155

    symbol using %succ in the equation editor.

    (c) Select the font that you want to use under the font dropdown menu. The MIT math fonts are namedMath1 through Math5. You will need to browse through the catalog to find your symbol.

    (d) Select the symbol that you want to add and click add.

    ubuntu 3.2.1 cu gray dublu click si schimbi in grayp j=

    i

    a i j Ei

    pj=i

    a i j Ei pj=i

    a i j Ei pj=i

    a i j Ei

    p j=i

    a i j Ei p j=i

    a i j Ei

    latexhttp://es.wikipedia.org/wiki/LaTeX

    0=a_{11} + a_{12} 0=a11a12 correctx^{a+b}=x^ax^b x

    a b=xaxb spatiu intre a si xx'+x'' = \dot x + \ddot x x 'x ' '= x x nu se pune \E &=& mc^2 E = mc2 &=~

    You can do it with LaTeX using Eqe:\definecolor{grey}{rgb}{0.5,0.5,0.5}\textcolor{grey}{What you want in grey}or for example:\definecolor{grey}{rgb}{0.5,0.5,0.5}\textcolor{grey}{$\left(\begin {array}{c}7289\\2\end {array}\right) = 65\times10^9$}

    Chapter 3 List of articles:

    [1] A. Lenef, S. Rand , Electronic structure of the N-V center in diamond: Theory, Phys. Rev. B 5313441(1996)(Ref. Pag:34, 44, 48)

    [2] N. Manson, R. McMurtrie , Issues concerning the nitrogen-vacancy center in diamond, J.Lumin. 127, 98 103(2007)

    [3] J. Loubser and J. Van Wyk Electron spin resonance in the study of diamond, Rep. Prog. Phys.41 1201-48(1978)

    [4] M. Doherty, N. Manson, P. Delaney and L. Hollenberg , The negatively charged N-V center in

    diamond: the electronic solution, arXiv:1008.5224(2010)[5] J. Maze, A. Gali, E. Togan, Y Chu, A. Trifonov, E. Kaxiras, And M. Lukin , Propreties ofnitrogen-vacancy centers in diamond: the group theoretical approach, New J. Phys. 13025025(2011)

    [6] Jacobs P. , Group theory with applications in Chemical Physics, (Cambridge 2005)(Ref.Pag:13,13)

    [7] B. Naydenov, F. Dolde, L. Hall, C. Shin, H Fedder, L. Hollenberg, F. Jelezco, and J. Wrachtrup,

    5

    http://es.wikipedia.org/wiki/LaTeXhttp://es.wikipedia.org/wiki/LaTeX
  • 7/29/2019 Book Andrei N

    6/155

    Dynamical decoupling of a single-electron spin at room temperature, Phys. Rev. B 83,081201(R)(2011)

    [8] A. Nizovtsev, S. Kilin, V. Pushkarchuk, A. Pushkarchuk, and S. Kuten , Quantum informatics.Quantum information processors, Optics and Spectroscopy 108,2 230(2010)

    [9] A. Abragam and B. Bleaney, Electron Paramagnetic Resonance of Transition Ions, Vol. I(Moscow 1972, Oxford 1970)

    [10] A. Lesk, Introduction to Symmetry and Group Theory for Chemists, Dordrecht (2004)[11] C. Vallance, Lecture, Molecular Symmetry, Group Theory & Applications (2006)(Ref.Pag:34,36)

    [12] R. Powell,_Symmetry, Group Theory, and the Physical Properties of Crystals Springer (2010)[13] M. Dresselhaus, Application of Group Theory to the Physics of Solids, Spring 2002[14] , Curs de matematica superioara, Chisinau 1971[15] H. Eyring and G. Kimball, Quantum Chemistry, Singapore, New-York, London (1944)[16] K. Riley, M. Hobson, Mathematical Methods for Physics and Engineering: A Comprehensive

    Guide, Second Edition, Cambridge 2002[17] D. Long, The Raman Effect A Unified Treatment of the Theory of Raman Scattering by

    Molecules , (2002)[18] L. Barron, Molecular Light Scattering and Optical Activity(2004) (Ref. Pag:134,134)[19] Eric C. Le Ru, Pablo G. Etchegoin, Principles of Surface Enhanced Raman Spectroscopy,

    (2009)(bond- polarizability model Wolkenstein)[20] D. Griffiths, Introduction to electrodynamics, New Jersey (1999)[21] M. , , Leningrad 1960[22] . . . . , . . 1 2, . .: 1949 [23] H. Heinbockel, Introduction to tensor calculus,(1996)[24] G. Arfken, H. Weber, Mathematical Methods For Physicists, International Student

    Edition(2005)[25] B. Berne and R. Pecora, Dynamic Light Scattering, New York 1990[26] Liang, ScandaloJ. Chem. Phys. 125, 194524 (2006)[27] R. Feynman, The Feynman Lectures on Physics, V I,II, III, California (1964)[28] C.J. Foot, Atomic Physics , Oxford 2005(Ref. Pag:57)[29] C. Gerry, Introductory Quantum Optics, Cambridge 2005[30] E. Kartheuser , Elements de MECANIQUE QUANTIQUE , Liege 1998?![31] M. Kuno, Quantum mechanics, Notre Dame, USA 2008(Ref. Pag:32)[32] M. Kuno, Quantum spectroscopy, Notre Dame,USA 2006[33] I. Savelyev, Fundamental of theoretical Physics,Moscow 1982[34] I. Savelyev, Physics: A general course vol. I,III,III, Moscow 1980[35] .. . [ 5, 1]

    [36] . . , M 1991[37] S. Blundell, Magnetism in condensed matter,Oxford 2001[38] P. Atkins, Physical Chemistry[39] J. Sakurai, Modern quantum mechanics, 1993[40] A. Matveev , , Moscow 1989(Ref. Pag:32)[41] . , , 2001[42] A. Messiah, Quantum mechanics[43] G. Arfken, H. Weber, Mathematical Methods For Physicists, International Student

    Edition(2005)[44] C. Clark , Angular Momentum, 2006 august (5 pag)

    6

    http://free-books.dontexist.com/book/index.php?md5=2F24D320F322DAAD028FF80B35D9E77Ahttp://free-books.dontexist.com/book/index.php?md5=2F24D320F322DAAD028FF80B35D9E77Ahttp://free-books.dontexist.com/book/index.php?md5=2F24D320F322DAAD028FF80B35D9E77Ahttp://free-books.dontexist.com/book/index.php?md5=2F24D320F322DAAD028FF80B35D9E77Ahttp://free-books.dontexist.com/book/index.php?md5=2F24D320F322DAAD028FF80B35D9E77Ahttp://free-books.dontexist.com/book/index.php?md5=2F24D320F322DAAD028FF80B35D9E77A
  • 7/29/2019 Book Andrei N

    7/155

    [45] A. Restrepo, Group theory in Quantum Mhechanics, (gt3)(2009)[46] . , , 1976[47] http://en.wikipedia.org/wiki/Lowering_operator[48] W. Nolting, Quantum thory of magnetism, Springer-Verlag Berlin Heidelberg 2009[49] I. Irodov, Problems in General Physics, English translation, Mir Publishers, 1981 Moscow[50] Abhay Kumar Singh, Solutions to Irodov's Problems in General Physics Vol. I and 2, New

    Delhi (2005)[51] G. Herzberg, Atomic Spectra and Atomic Structure, New-York (1944)[52] M. Mueller, Fundamentals of Quantum Chemistry, New-York (2002)[53] E. U. Condon, G. H. Shortley, The Theory of Atomic Spectra Cambridge (1959)[54] D. Perkins, Introduction to High Energy Physics, Cambridge (2000)[55] E. Wigner, Group Theory And Its Application To The Quantum Mechanics Of Atomic

    Spectra, New-York (1959) (Ref. Pag:117)[56] K. Schulten, Notes on Quantum Mechanics, Universities of Illinois (2000)[57] M Caola, Lat. Am. J. Phys. Educ. Vol. 4, No. 1, Jan 2010 p84 (Ref. Pag: )[58] Calais J-L., I. Jour. Of Quant. Chem. Vol.II, 715-727(1968)(Ref. Pag: )[59] L. C. Biedenharn, J.D. Louck, Angular momentum in quantum physics, London (1981) (Ref.

    Pag: 125, 130)[60] E.B. Manoukian, Quantum Theory:A Wide Spectrum , Springer (2006)(Ref. Pag: 117)[61] Smirnov V., Cours de Mathmatique Suprieure , tome III deuxime partie dition MIR

    Moscou 1972[62] Robuk V. , Nuc. Ins. and Meth. in Phys. Res. A 534, p 319-323 (2004)[63] Claude F., Atomes et lumire :Interactions Matire rayonnement, Universit Pierre et Marie

    Curie (2006)[64] Orszag M., Quantum Optics, Springer 2008[65] Scully M., Zubairy M., Quantum optics, Cambridge 2001[66] Brink Satchler Angular_momentum(1968)[67] Rieger P., Electron Spin Resonance: Analysis_and_Interpretation, Royal Society of Chemistry

    (2007)[68] Jones H. Groups, Representations, and Physics (IPP, 1998) (Ref. Pag: 23)[69] Weissbluth M., Atoms and molecules, Academic Press Inc.,U.S.(1978)(Ref. Pag:13, 25, 32, 35,

    40, 57, 55)[70] Ludwig W., Falter C., Symmetries in Physics: Group Theory Applied to Physical Problems,

    Springer Series In Solid-State Sciences 64(1988)(Ref. Pag:14, 42, 46, 48, 53, 55)[71] Inui T., Tanabe I.,Onodera I.,Group Theory and its Applications in Physics,Springer Series In

    Solid-State Sciences 78(1990)(Ref. Pag:12, 34, 52, 53)[72] Harris D., Symmetry and Spectroscopy: An Introduction to Vibrational and Electronic

    Spectroscopy, Dover (1989)(Ref. Pag:32)[73] Tinkham M., Group Theory and Quantum Mechanics, Dover(1992) (Ref. Pag:32, 132, 134)[74] Cornwell J., Group Theory in Physics, Volume 1 (Techniques of Physics) , Academic Press

    London (1984) (Ref. Pag:43)[75] Cornwell J., :Symmetry Proprieties of the Clebsch-Gordan Coefficients, phys. stat. sol.(b)

    37,225(1970) (Ref. Pag:42, 44)[76] VAN DEN Broek P.,Cornwell J., :Clebsch-Gordan Coefficients of Symmetry Groups, phys.

    stat. sol. (b) 90,211(1978) (Ref. Pag:44)[77] Sugano S., Tanabe Y., Kamimura H., Multiplets of Transition-Metal Ions in Crystals,

    Academic Press ,New-York 1970 (Ref. Pag:49, 48)

    7

    http://en.wikipedia.org/wiki/Lowering_operatorhttp://en.wikipedia.org/wiki/Lowering_operator
  • 7/29/2019 Book Andrei N

    8/155

    [78] Cornwell J., Group theory in physics ,Academic Press London (1997) (Ref. Pag:51)[79] Fitts D.D., Principles of Quantum Mechanics, as Applied to Chemistry and Chemical Physics,

    Cambridge (2002)(Ref. Pag:57)[80] Racah G., (I) Phys. Rev. 61, 186197 (1942) (Ref. Pag:128)[81] Racah G., (II)Phys. Rev. 62, 438462 (1942) (Ref. Pag:128)[82] Judd B., Operator Techniques in Atomic Spectroscopy, Princeton University Press(1998)(Ref.

    Pag:128)[83] Griffith J., The Theory of Transition-Metal Ions Cambridge (1971)(Ref. Pag:127)[84] Edmonds A., Angular momentum in quantum mechanics. [Rev. printing, 1968]Princeton

    (1957) (Ref. Pag:127)[85] Sobelman I., Atomic Spectra and Radiative Transitions(1992) (Ref. Pag:133)

    Wolkenstein, M. I94I C.R. Acad. Sci., U.R.S.S. 32, 185.Eliashevich, M. & Wolkenstein, M. I945 J. Phys., Moscow, 9, 101, 326

    Chapter 4 Group theory

    4.1 Basis

    pag 11[10]

    Definition of a groupA group consists of a set (of symmetry operations, numbers, etc.) together with a rule by which any two

    elements of the set may be combined which will be called, generically, multiplication with thefollowing four properties:1 Closure: The result of combining any two elements the product of any two elements is anotherelement in the set.2 Group multiplication satisfies the associative law: a (b c) = (a b) c for all elements a, b and c ofthe group.3 There exists a unit element, or identity, denoted E, such that E a = a for any element of the group.4 For every element a of the group, the group contains another element called the inverse, a1, suchthat a a1 = E. Note that as EE = E, the inverse of E is E itself.

    Examples of groupsElements of the group Rule of combination1. covering operations of an object Successive application2. All real numbers Addition3. All complex numbers Addition4. All real numbers except 0 Multiplication5. All complex numbers except (0, 0) Multiplication6. All integers Addition

    8

  • 7/29/2019 Book Andrei N

    9/155

    7. Even integers Addition8. The n complex numbers of the form cos 2k/n + i sin 2k/n, k = 0, . . . , n 1 Multiplication9. All permutations of an ordered set of objects Successive applicationA permutation is a specification of a way to reorder a set. For example, the group of permutations oftwo objects contains two elements: the identity, first first, second second and the exchange, firstsecond, second first. This group is isomorphic to the group formed by 1 and 1 under

    multiplication.10. All possible rotations in three-dimensional space Successive application

    Pag 15 [10]Def. A point group is the symmetry group of an object of finite extent, such as an atom or molecule.

    Pag 3. [11]A symmetry operation is an action that leaves an object looking the same after it has been carried out.For example, if we take a molecule of water and rotate it by 180 about an axis passing through thecentral O atom (between the two H atoms) it will look the same as before. It will also look the same if

    we reflect it through either of two mirror planes, as shown in the figure below.

    1. E - the identity. The identity operation consists of doing nothing, and the corresponding symmetryelement is the entire molecule. Every molecule has at least this element.2. Cn - an n-fold axis of rotation. Rotation by 360/n leaves the molecule unchanged. The H2Omolecule above has a C2 axis. Some molecules have more than one Cn axis, in which case the one withthe highest value of n is called the principal axis. Note that by convention rotations are counter-clockwise about the axis.3. - a plane of symmetry. Reflection in the plane leaves the molecule looking the same. In a moleculethat also has an axis of symmetry, a mirror plane that includes the axis is called a vertical mirror plane

    and is labelled v, while one perpendicular to the axis is called a horizontal mirror plane and is labelledh. A vertical mirror plane that bisects the angle between two C2 axes is called a dihedral mirror plane,d.4. i - a centre of symmetry. Inversion through the centre of symmetry leaves the molecule unchanged.Inversion consists of passing each point through the centre of inversion and out to the same distance onthe other side of the molecule. An example of a molecule with a centre of inversion is shown below.5. Sn - an n-fold improper rotation axis (also called a rotary-reflection axis). The rotary reflection

    9

  • 7/29/2019 Book Andrei N

    10/155

    operation consists of rotating through an angle 360/n about the axis, followed by reflecting in a planeperpendicular to the axis. Note that S1 is the same as reflection and S2 is the same as inversion. Themolecule shown above has two S2 axes.

    Cnv contains the identity, an n-fold axis of rotation, and n vertical mirror planes v.

    Cnh - contains the identity, an n-fold axis of rotation, and a horizontal reflection plane h (note that inC2h this combination of symmetry elements automatically implies a centre of inversion).

    Dn - contains the identity, an n-fold axis of rotation, and n 2-fold rotations about axes perpendicular tothe principal axis.

    Dnh - contains the same symmetry elements as Dn with the addition of a horizontal mirror plane.

    Now we will investigate what happens when we apply two symmetry operations in sequence. As anexample, consider the NH3 molecule, which belongs to the C3v point group. Consider what happens if

    we apply a C3 rotation followed by a v reflection. We write this combined operation vC3 (whenwritten, symmetry operations operate on the thing directly to their right, just as operators do in quantummechanics we therefore have to work backwards from right to left from the notation to get the correctorder in which the operators are applied).

    The combined operation vC3 is equivalent to v'', which is also a symmetry operation of the C3v pointgroup. Now lets see what happens if we apply the operators in the reverse order i.e. C 3v (v followedby C3).

    10

  • 7/29/2019 Book Andrei N

    11/155

    There are two important points that are illustrated by this example:1. The order in which two operations are applied is important. For two symmetry operations A and B,AB is not necessarily the same as BA, i.e. symmetry operations do not in general commute. In somegroups the symmetry elements do commute; such groups are said to be Abelian.2. If two operations from the same point group are applied in sequence, the result will be equivalent toanother operation from the point group. Symmetry operations that are related to each other by other

    symmetry operations of the group are said to belong to the same class. In NH3, the three mirror planesv, v' and v'' belong to the same class (related to each other through a C 3 rotation), as do the rotationsC3

    + and C3- (anticlockwise and clockwise rotations about the principal axis, related to each other by a

    vertical mirror plane).

    orPag 26 [12]Group multiplication Table

    E A B C D FE E A B C D FA A E D F B CB B F E D C AC C D F E A BD D C A B F EF F B C A E D

    Consider a group consisting of six elements represented by the letters A , B , C , D , E , and F thatobey the multiplication table shown above. The elements in the table are the product of the elementdesignating its column and the element designating its row. Following this convention, the table showsthat the identity element is a member of the group, the product of any two elements is an element of thegroup, and each group element has an element in the group that is its inverse. Each element appearsonly once in any given row or column. The associative law holds but the commutative law does nothold for all products so the group is not Abelian. The order of the group is 6.

    4.2 Rotations

    11

  • 7/29/2019 Book Andrei N

    12/155

    pag 22 [14]Rotation of Cartesian axes by an angle about thez-axis.

    = ' , = 'x=cos, y= sin

    x '= cos ' , y '=sin '

    x= cos 'cos =cos 'cos sin 'sin =x 'cos y 'sin y= sin 'cos =sin 'cos cos 'sin =y 'cos x 'sin

    finally we get

    x=x 'cosy 'sin y=x 'sin y 'cos

    z =z '

    (4.2.1)

    Another method (pag 780 [16] )that you can sum theprojections (the triangle rule x '=xy ) so we havethat:

    x '=x cos y sin see Fig 1and the same for the other

    y '=x sin ycos

    the transformation matrix are:

    R= cos sin 0sin cos 00 0 1

    (4.2.2)and from (4.2.1) we have the inverse:

    R1=cos sin 0sin cos 0

    0 0 1

    13

    (4.2.3)

    Thus we can write 321 in matrix form

    r=R1 r'rx , y , z

    28

    r'=R r see pag 51[71]

    (4.2.4)

    12

    Fig 1

    y

    x

    x cos

    y sin

    x '

  • 7/29/2019 Book Andrei N

    13/155

    pag 2 [13] also see [69] pag 61-63:

    Figure 1.1: The symmetry operations on an equilateral triangle,are the rotations by 2 / 3 about the origin 0 and the rotationsby about the axes 01, 02, and 03.

    As a simple example of a group, consider the permutation groupfor three elements, P(3). Below are listed the 3!=6 possiblepermutations that can be carried out; the top row denotes the initialarrangement of the three numbers and the bottom row denotes thefinal arrangement.

    This is for group D3:

    E= 1 2 31 2 3

    A= 1 2 32 1 3

    B = 1 2 31 3 2

    C=

    1 2 3

    3 2 1 D=

    1 2 3

    3 1 2 F=

    1 2 3

    2 3 1

    (AB)C = DC = BA(BC) = AD = Bbellow 14:

    E=1 00 1 A=1 00 1 B=1/2 3/23/ 2 1/2 C= 1/2 3 / 23/2 1 /2 D = 1 / 2 3 /23 /2 1 / 2 F=1/ 2 3/23 /2 1/2

    For C3v :we have see[6] pag 70-71:

    C3v={ E C3

    C3 d e f}

    we have two rotations using for C3

    =2

    3and for C3

    =

    23

    in eq.(4.2.3):

    C3=1 /2 3 /23 /2 1 / 2 and

    C3= 1/2 3/23/2 1 /2

    4.3 Reflectionsat page 59[6] for reflection in plane we have:from theFig. 2we put

    ](P'O )=x

    ](YOP')=y we have we need to find the angle:

    13Fig. 2 Reflection of a point P(x y) in a mirrorplane whose normal m makes an angle withy,so that the angle between and the zx plane is

    . OP makes an angle with x . P'(x'y') is the reflection of P in x= , and OP'makes an angle 2 with x .

  • 7/29/2019 Book Andrei N

    14/155

    xx= 2x (4.3.1)

    We make the system of eq.:

    {2xy=/2xy=/2 (4.3.2)The solution is

    x= (4.3.3)

    Substitution of(4.3.3)in(4.3.1) yields

    2 (4.3.4)

    x= cos and y=sin x '=cos 2 =x cos 2 y sin 2 andy '=sin 2 =x sin 2 ycos 2

    thus we got the transformation:

    [x '

    y 'z '

    ]=

    [cos 2 sin 2 0

    sin 2 cos 2 00 0 1

    ][x

    yz

    ](4.3.5)

    With the representation :

    y =[cos 2 sin 2 0sin 2 cos 2 00 0 1] (4.3.6)Thus we have three reflections:

    for d =0 ,for e =3

    and for f =3

    in eq.(4.3.6):

    d=1 00 1

    e=1/2 3 /2

    3/2 1 / 2 f= 1/2 3/23/ 2 1/2

    for representation of S3 we have

    e=1 00 1 a= 1 /2 3/23 /2 1/2 b= 1/ 2 3 / 23/2 1/ 2c=

    1 0

    0 1d=

    1 /2 3 /23/2 1/ 2

    f=1 /2 3/2

    3/2 1/2 d=D, f=F, c=A, a=B where the signs in principal diagonal is ganged, and b=C where the signs inprincipal diagonal is ganged see 13above.

    In conclusion we see that D3C3v also with S3 :see [70]pag 409 .

    14

  • 7/29/2019 Book Andrei N

    15/155

    pag 27 [10]

    Linear transformations and matrices

    in linear transformation the most important class of geometrical manipulations is the set of lineartransformations, for which the effect on a point is given by the linear equations:

    xf=a11 xia12 y ia13 ziyf=a21 xia22 yia23zi

    zf=a31x ia32 y ia33ziin which (xi, yi, zi) and (xf, yf, zf ) are vectors representing the initial and final points, and the aij are realnumbers characterizing the transformation. All the covering operations we have dealt with are of thistype.

    Matrix notation is a shorthand way of writing such systems of linear equations, that avoids tediouscopying of the symbols xi, yi and zi. The application of the identity matrix:

    xfyfzf

    =1 0 00 1 00 0 1x iy izi

    in abbreviations of system equations:

    x f=1x i0y i 0ziy f=0x i 1yi 0z iz f= 0x i 0y i1z i

    in general the array of numbers:

    a11 a12 a13a21 a22 a23

    a31 a32 a33

    is called a matrix. The numbers a ij are its elements: the first subscript specifies the row and thesecond subscript specifies the column.Each of the symmetry operations we have defined geometrically can be represented by a matrix. Theelements of the matrices depend on the choice of coordinate system.

    15

  • 7/29/2019 Book Andrei N

    16/155

    The operations v , the mirror reflection in the y-z plane, has the effect in this coordinate system ofreversing the sign of the x-coordinate of any point it operates on. This effect is specified by theequations:

    xf=x i , yf=y i , zf=zi

    or bu the matrix equation:

    xfyfzf

    =1 0 00 1 00 0 1x iyizi

    The identity operation is expressed by identity matrix:

    I=1 0 00 1 00 0 1

    Successive transformations; matrix multiplication

    Because the successive application of two linear transformations is itself a linear transformation, itmust also correspond to a matrix. The matrix corresponding to a compound transformation can becomputed directly from the matrices corresponding to the individual transformations. Let us derive theformulas; for simplicity we shall work in two dimensions. Given two linear transformations:

    xf=a11x ia12 y iyf=a21 x ia22 yi

    andx 'f=b11 x 'ib12 y 'iy '

    f

    =b21

    x 'i

    b22

    y 'i

    use the final point of the first transformation as the initial point of the second. That is, letx 'i=xf=a11x ia12 yiy 'i=yf=a21 x ia22 yi

    thenx 'f=b11 a11x ia12 y ib12 a21 x ia22 y iy 'f=b21a11 x ia12 y ib22a21x ia22 y i

    orx 'f=b11 a11 x ib12 a21x ib11a12b12 a21y iy 'f=b21 a11 x ib22 a21x ib21 a12b22 a22y i

    The last set of equations is in the standard form for a linear transformation. Its matrix form is:

    x 'fy 'f=b11 a11 x i b12 a21 b11 a12 b12 a21 b21 a11 x i b22 a21 b21 a12 b22 a22

    x iyi

    16

  • 7/29/2019 Book Andrei N

    17/155

    In more than two dimensions, the calculation is similar. The general result is that: if the matrix

    C=c11 ... c1n... ... ...

    cn1 ... cnn A= a11 ... a1n... ... ...an1 ... ann B=

    b11 ... b1n... ... ...

    bn1 ... bnn

    we say that C is the product of B and A: C=BA . The elements of C are given by the formula:

    c ik=j =1

    n

    bij ajk

    If the rows of B and the columns of A are considered as vectors, then the element c ik is the dot

    product of the i th row of B with the k th column of A:

    C=

    ... c ik ... ...

    =

    b i1 bi2 ... b i n

    a1ka2k...

    ank

    The effect on a matrix of a change in coordinate systemThe elements of the matrix that corresponds to a geometrical operation such as a rotation depend on thecoordinate system in which it is expressed. Consider a mirror reflection, in two dimensions, expressedin three different coordinate systems, as shown in Figure 52. The mirror itself is in each case vertical,independent of the orientation of the coordinate system.

    17

  • 7/29/2019 Book Andrei N

    18/155

    Figure 5.2. Expression of a reflection in a mirror plane in three different coordinate systems. Note that

    the points x i , y i and xf , yf dont change position. Only the coordinate axes change.

    The relationships between the matrices representing the reflection in different coordinate systems areexpressible in terms of the matrix S that defines the relationships between the coordinate systems

    themselves. Suppose x , y and x ' , y ' are two pairs of normalized vectors oriented along theaxes of two Cartesian coordinate systems related by a linear transformation:

    x 'y '=S xyIf the matrix A represents the mirror reflection in the x , y coordinate system, then the matrix thatrepresents the reflection in the x ' , y ' coordinate system is the triple matrix product S1A S ,where S

    1 is the inverse ofS. Such a change in representation induced by a change in coordinate

    system is called a similarity transformation.

    Traces and determinants

    Linear transformations that correspond to nonorthogonal matrices distort lengths or angles. The traceand determinant of a matrix provide partial measures of the distortions introduced. The trace of amatrix is defined as the sum of the diagonal elements.

    Tr A=i =1

    n

    aii

    deta bc d=adbc

    Analogous but more complicated formulas define the determinants of square matrices of higher

    dimensions. (A square matrix is a matrix which has the same number of rows as columns.) It is notpossible to define the determinant of a non-square matrix.

    18

  • 7/29/2019 Book Andrei N

    19/155

    What conclusions do these examples suggest? First, note that only the first six examples are orthogonaltransformations. The determinant of each of these matrices is +1 or 1. Every matrix that represents anorthogonal transformation must have determinant 1.

    19

  • 7/29/2019 Book Andrei N

    20/155

    The importance of the trace and determinant lies in their independence of the coordinate system inwhich the matrix is expressed. Recalling that a change in coordinate system leads to a change in thematrix representation of a transformation by a similarity transformation, the independence of trace anddeterminant on coordinate system is expressed by the equations:

    Tr S1A S=Tr A

    det S1

    A S=det A

    where S is an orthogonal matrix. Thus the trace and determinant provide numerical characteristics of atransformation independent of any coordinate system.

    Pag 52 [10]

    4.4 Group representations

    A representation of a symmetry group is a set of square matrices, all of the same dimension,corresponding to the elements of the group, such that multiplication of the matrices is consistent with

    the multiplication table of the group. That is, the product of matrices corresponding to two elements ofthe group corresponds to that element of the group equal to the product of the two group elements inthe group itself. Representations can be of any dimension; 11 arrays are of course just ordinarynumbers.

    If each group element corresponds to a different matrix, the representation is said to be faithful. Afaithful representation is a matrix group that is isomorphic to the group being represented. If the samematrix corresponds to more than one group element, the representation, is called unfaithful. Unfaithfulrepresentations of any group are available by assigning the number 1 to every element, or by assigningthe identity matrix of some dimension to every element. (But the number 1 is a faithful representationof the group C1.) The collection of matrices occurring in an unfaithful representation of a group, if

    taken each only once, forms a group isomorphic to a subgroup of the original group. Thus to anyunfaithful representation of a group there corresponds a faithful representation of a subgroup.

    Examples of group representations:

    20

  • 7/29/2019 Book Andrei N

    21/155

    21

  • 7/29/2019 Book Andrei N

    22/155

    4.5 Unitary Matrix

    The following complex matrix is unitary: that its set of row vectors form an orthonormal set in C3 :

    A=

    [1

    2

    1i

    2

    1

    2

    i

    3i

    31

    35 i

    2153i

    21543 i

    215 ]27 (4.5.1)

    Let r1 r2 r3 be defined as follows:

    r1= 12 , 1i2 , 12

    r2=i

    3,i

    3,1

    3 r3= 5 i215 ,

    3i215

    ,43 i215

    (4.5.2)

    The length of r1 is

    r1= r1r1 1 /2

    =[12 12 1i2 1i2 12 12 ]1/ 2

    =[1

    4

    2

    4

    1

    4 ]1/ 2

    =1

    49 (4.5.3)

    For the vectors r2 r3 the same.

    The inner product of r1 r2 is given by:

    r1r2

    = 12 i3 1i

    2 i3 1

    2 13 = 12 i3 1

    i2 i3 12 13

    =i

    23 i

    23 i

    23 i

    23 =0

    (4.5.4)

    Similarly, and r1r3=0 and r2r3=0 we can conclude that{ r1 , r2 , r3 } is an orthonormalset. (Try showing that the column vectors of A also form an orthonormal set in C3 .)see [15]pag174 :In addition the determinant of the A's is unity. The matrices representing rotations, reflections and

    22

  • 7/29/2019 Book Andrei N

    23/155

    inversions are unitary follows that the matrix representations of groups are unitary.

    4.6 Schur's Lemma

    See pag 59[68]:Lemma:

    In matrix form this states that any matrix which commutes with all the matrices of an irreduciblerepresentation must be a multiple of the unit matrix, i.e.

    BD g=D gB gG B=1 (4.6.1)

    To prove the identity , let b be an eigenvector of B with eigenvalue :

    Bb=b (4.6.2)

    Then

    B D g b=D gB b=D g b= D g b (4.6.3)You know from quantum mechanics that if the operators commutes then they have the sameeigenvalues.This means that Dgb is also an eigenvector of B, with the same eigenvalue .

    In matrix form one has to solve the equation B b= b , leading to the characteristic equationdetB 1=0 (4.6.4)

    . If we are working within the framework of complex numbers, this polynomial equation is guaranteedto have at least one root , with a corresponding eigenvector b .

    4.7 Great orthogonality theorem

    Theorem:

    Consider two unitary irreducible matrix representations i G and j G of a group G then:

    g

    i g j g =

    [G ]li

    ij 30,43 (4.7.1)

    Where [G] is order of a group G and li is the dimension of i G .

    Proof:Define the matrix M :

    MgG

    i gX j g (4.7.2)

    Where X is some unspecified matrix. Then

    23

  • 7/29/2019 Book Andrei N

    24/155

    i hM= i hg

    i gX j g

    = g

    ihgX j g

    = hg

    ihgX j h1 hg

    = g

    igX j h1g (relabing hg g)

    = g

    igX j j h

    =M j h

    (4.7.3)

    By the converse of Schur's lemma , either i=j, or M=0. If i=j then by Schu's lemma M= 1=Iwhere we find m by taking the trace .

    g

    igX j g=ij 1 (4.7.4)

    [G ]trX= li So

    g

    igX j g=[G ]li

    ij tr X1 (4.7.5)

    Or with index notations:

    g

    ig X j g =

    [G ]li

    X ij (4.7.6)

    But since X is arbitrary we chose to be also unity matrix:

    g

    ig j g =

    [ G ]li

    ij (4.7.7)

    4.8 Great orthogonality theorem (application)

    Example of the permutation group on 3 objects:The 3! permutations of three objects form a group of order 6, commonly denoted by S

    3(symmetric

    group). This group is isomorphic to the point group C3v, consisting of a threefold rotation axis and

    three vertical mirror planes. The groups have a 2-dimensional irrep (l= 2). In the case ofS3

    one usually

    labels this irrep by the Young tableau = [2,1] and in the case ofC3v one usually writes =E. In both

    cases the irrep consists of the following six real matrices, each representing a single group element:

    24

    http://en.wikipedia.org/wiki/Symmetric_grouphttp://en.wikipedia.org/wiki/Symmetric_grouphttp://en.wikipedia.org/wiki/Point_groups_in_three_dimensions#The_seven_infinite_serieshttp://en.wikipedia.org/wiki/Young_tableauhttp://en.wikipedia.org/wiki/Symmetric_grouphttp://en.wikipedia.org/wiki/Symmetric_grouphttp://en.wikipedia.org/wiki/Point_groups_in_three_dimensions#The_seven_infinite_serieshttp://en.wikipedia.org/wiki/Young_tableau
  • 7/29/2019 Book Andrei N

    25/155

    1 00 11 00 11

    2

    32

    32

    1

    2

    1

    23

    2

    32

    1

    2

    1

    2

    32

    32

    1

    2

    1

    23

    2

    32

    1

    2

    we have the flowing vectors:

    V11 V12V21 V22The normalization of the (1,1) (2,2) ,(1,2)and (2,1) elements:V11V11=

    RG

    6

    R11 R11=1

    21212 2

    12 2

    12 2

    12 2

    =3

    V12V12= RG

    6

    R12 R12=

    202 32 2

    32 2

    32 2

    32 2

    =3

    V21V21=RG

    6

    R21 R21=0

    202 32 2

    32 2

    32 2

    32 2

    =3

    V22V22= RG

    6

    R22 R22=1

    212 12 2

    12 2

    12 2

    12 2

    =3

    (4.8.1)

    The orthogonality of the (1,1) and (2,2) elements:

    V11V22= RG

    6

    R11 R22=1

    21112 12 12 12 12 2

    12 2

    =0 . (4.8.2)

    Similar relations hold for the orthogonality of the elements (1,1) and (1,2), etc. One verifies easily inthe example that all sums of corresponding matrix elements vanish because of the orthogonality of thegiven irrep to the identity irrep.

    4.9 D3 representations and irreducible representations

    See [69]pag 66:A matrix representation of a group is defined as a set of square, nonsingular matrices (matrices withnonvanishing determinants) that satisfy the multiplication table of the group when the matrices aremultiplied by the ordinary rules of matrix multiplication. There are other kinds of representations (seeSection 6.1) but unless there is a specific statement to that effect, it will be understood that arepresentation means a matrix representation.We give several examples of representations of the group D3:

    1

    E=

    1

    A=

    1

    B =

    1

    C=

    1

    D =

    1

    F=128 (4.9.1)

    This is a one-dimensional representation (lxl matrices) which obviously satisfies the groupmultiplication table. It appears to be trivial but nevertheless plays an important role in laterdevelopments. This representation is called the totally symmetric representation or the unitrepresentation. Another one-dimensional representation is

    2E= 2 A= 2B=2C=1, 2D= 2F=1 (4.9.2)

    25

  • 7/29/2019 Book Andrei N

    26/155

    A two-dimensional representation consisting of 2 x 2 matrices is:

    3 E=1 00 1 3 A=1 00 1 3 B=1/2 3/23/ 2 1/2 28,31 3 C=

    1/2 3 /2

    3/2 1/2 3 D=

    1/2 3/2

    3/2 1/2 3 F=

    1 /2 3/2

    3/2 1/2

    (4.9.3)

    We may therefore regard these matrices as a three-dimensional representation of D3:I have calculated in Mathematica (4.11.31) and here I found a lot of mistakes in Weissbluth the correctare:

    4E=1 0 00 1 00 0 1

    4A=1 0 00 1 00 0 1

    4B=1 /2 3 /2 03/2 1 /2 00 0 1

    4C= 1/ 2 3/2 03/ 2 1/2 00 0 1

    35 4D=

    1 /2 3/ 2 03 /2 1/2 0

    0 0 1 4F=1/2 3 /2 03/2 1 /2 0

    0 0 1

    (4.9.4)

    Another three-dimensional representation is the set

    5 E=1 0 00 1 00 0 1

    5 A=1 0 00 0 10 1 0

    5 C=

    0 0 1

    0 1 0

    1 0 0

    5 B=

    0 1 0

    1 0 0

    0 0 1

    5 D =0 0 11 0 00 1 0

    5 F=0 1 00 0 11 0 0

    29(4.9.5)

    The number of representations that may be constructed is without limit; as a final example we show asix-dimensional representation:

    26

  • 7/29/2019 Book Andrei N

    27/155

    6 E=

    1 0 0 0 0 0

    0 1 0 0 0 0

    0 0 1 0 0 0

    0 0 0 1 0 0

    0 0 0 0 1 0

    0 0 0 0 0 1

    6 A=

    1 0 0 0 0 0

    0 1 0 0 0 00 0 1 0 0 0

    0 0 0 1 0 00 0 0 0 1 0

    0 0 0 0 0 1

    6 B=1/ 2 3/ 2 0 0 0 03 /2 1/ 2 0 0 0 0

    0 0 1/2 3/2 0 00 0 3/2 1/2 0 00 0 0 0 1 0

    0 0 0 0 0 1

    6

    C=

    1 /2 3 /2 0 0 0 0

    3 /2 1 /2 0 0 0 00 0 1 / 2 3 /2 0 00 0 3 / 2 1 / 2 0 00 0 0 0 1 00 0 0 0 0 1

    6 D=1/ 2 3/2 0 0 0 0

    3/ 2 1/2 0 0 0 00 0 1 /2 3 /2 0 00 0 3 /2 1/2 0 00 0 0 0 1 0

    0 0 0 0 0 1

    6 F=

    1 / 2 3 / 2 0 0 0 03 /2 1 / 2 0 0 0 0

    0 0 1 /2 3 /2 0 00 0 3 / 2 1 /2 0 00 0 0 0 1 00 0 0 0 0 1

    (4.9.6)

    If the matrices belonging to a representation are subjected to a similarity transformation, theresult is a new representation ' . The two representations and ' are said to be equivalent.If and ' cannot be transformed into one another by a similarity transformation, and

    ' are then said to be inequivalent. It may be shown (eq.(4.5.1)) that for a finite group everyrepresentation is equivalent to a unitary representation (i.e., consisting entirely of unitary matrices). Weshall assume henceforth that all representations are unitary unless the contrary is explicitly stated (seealso Section 3.4).

    An important notion concerning representations is that of reducibility. Consider, for example

    27

  • 7/29/2019 Book Andrei N

    28/155

    6 B in (4.9.6):

    6 B=

    1/ 2 3/ 2 0 0 0 03 /2 1/ 2 0 0 0 0

    0 0 1/2 3/2 0 00 0 3/2 1/2 0 0

    0 0 0 0 1 00 0 0 0 0 1

    (4.9.7)

    This matrix consists of four separate blocks along the main diagonaltwo of the blocks are two-dimensional and the other two are one-dimensional. Moreover, examination of the separate blocks

    reveals that the two- dimensional blocks are identical with the representation matrix 3 B[Eq. (4.9.3)] and the one-dimensional blocks coincide with 1 B [Eq. (4.9.1)]. Therefore 6 B is said to be reducible into 2 3 B and 2 1 B . Symbolically this is expressed by

    6 B=2 3B2 1B

    6 B =2 3B 2 1 B (4.9.8)

    in which the right side of the equation signifies that 3 B appears twice and 1 B appearstwice along the main diagonal in (4.9.7). In this context the symbol + clearly has nothing to do withaddition. The symbol is also used which means direct sum.The character is denoted by j R , thus

    j R =

    j R (4.9.9)

    Table 1.1 Character table of D331

    E A B C D F

    1

    2

    3

    1

    1

    2

    1

    -1

    0

    1

    -1

    0

    1

    -1

    0

    1

    1

    -1

    1

    1

    -1

    4.10 Basis Functions

    The representations of a group are intimately connected with sets of functions called basis functions. Afew examples will help to establish the central idea. Let

    1 r=x , 2r=y (4.10.1)

    We now inquire as to how these functions are altered under the coordinate transformations E, A,. .., Fwhich are elements of the group D3. Under any coordinate transformation described by r'=R r (R isa matrix of rotations(see (4.2.4),(5.2.14))), a function f r transforms in accordance with

    PR f r= f R1 r (4.10.2)

    28

  • 7/29/2019 Book Andrei N

    29/155

    For R=E we obtain trivially,

    PE1r=1r=x , PE2r=2r=y (4.10.3)

    Or in matrix form:

    PE1, 2=1,21 00 1 (4.10.4)The general statement is that a set of linearly independent functions 1

    jr, 2j r,...,n

    j r are

    basis functions for the j representations of a group if

    PR k j r=

    =1

    n

    j r k

    j R 29,30, 40 (4.10.5)

    For all the elements of the group R.

    The basis set for a representation consists of linearly independent functions. These can always bechosen so as to be orthonormal, i.e.,

    lj rk

    j r=lk (4.10.6)

    From (4.10.5),

    ljrPRk

    jr =

    ljrk

    j r k jR= l k

    j RPR kjr=

    =1

    n

    jr k

    j(4.10.7)

    4.11 Projection Operators (D3)

    We will prove that see(4.11.31):

    5 = 3 1 31 (4.11.1)

    It is pertinent to inquire what relations, if any, exist between the basis functions of 5 and the basisfunctions of the component irreducible representations 3 and 1 . Suppose (fl,f2,f3) is the setof (normalized) basis functions of 5 [Eq. (4.9.5)]. It follows, from the definition of basisfunctions(4.10.5),that

    29

  • 7/29/2019 Book Andrei N

    30/155

    PE f1, f2, f3= f1, f2, f31 0 00 1 00 0 1

    = f1, f2, f3 PAf1, f2, f3= f1, f2, f3

    1 0 0

    0 0 1

    0 1 0

    =f1, f3, f2

    PB f1, f2, f3= f1, f2, f30 0 10 1 01 0 0

    =f3, f2, f1PC f1, f2, f3= f1, f2, f30 1 01 0 0

    0 0 1= f2, f1, f3

    PD f 1, f 2, f 3= f 1, f 2, f 3

    0 0 11 0 00 1 0

    = f 2, f 3, f 1

    PF f1, f2, f3= f1, f2, f30 1 00 0 11 0 0

    = f3, f1, f2 32

    (4.11.2)

    Where we have multiplied the row with matrix = row.

    Let kj r be a basis function belonging to the jth irreducible representation j see(4.10.5).

    Then, by definition,

    PR kjr=

    =1

    n

    j r k

    j R (4.11.3)

    Multiplying through by ' k 'j R and summing over R, we obtain

    R

    ' k ' j ' R PR k

    j r=R

    = 1

    l j

    ' k ' j ' R

    j r kj R

    =h

    lj

    j jj ' ' kk '

    =h

    lj '

    j jj ' kk '

    (4.11.4)

    in which the second and third equalities are a consequence of the orthogonality theorem (4.7.1). We

    may now define an operator

    Pk j =

    lj

    hR

    kj RPR 48 (4.11.5)

    Having the property from (4.11.4):

    30

  • 7/29/2019 Book Andrei N

    31/155

    Pk j l

    i=i ij lk (4.11.6)

    Or when i=j and l=k,

    Pk j k

    j =j

    (4.11.7)

    When = k (4.11.6) becomesPk k

    j l i=k

    i ij lk (4.11.8)

    Or

    Pk k j k

    j =k j

    Pk k j l

    j =0, i j , lk(4.11.9)

    From (4.11.9)it is seen that k j

    is an eigenvector of Pk k j

    with eigenvalue equal to one. Also

    Pk k j2 k

    j =Pk kj k

    j =k j

    (4.11.10)

    So that

    Pk k j2=Pk k

    j =lj

    hR

    k k j RPR (4.11.11)

    Pk k j

    is known as a projection operator; operators that obey a relation of the type O2 = O are said to

    be idempotent. From (3.5-14) we see that projection operators are idempotent.

    If one now sums Pk k j

    over k, then, from (4.11.11),

    k

    Pk kj Pj =

    lj

    hR

    k

    k k j RPR=

    lj

    hR

    jRPR (4.11.12)

    P j Is also a projection operator(for example kk

    3 is a matrix 2x2(4.9.3)and kk= 11,12,21,22).

    Illustration of(4.11.12):

    P 1=

    l1h

    R

    1PR=1

    6[PEPAPBPCPDPF] 32

    P 2=

    l1

    hR

    1PR=1

    6[PEPAPBPCPDPF]

    P 3= 2

    6[2PEPDPF]

    (4.11.13)

    We will prove (4.11.1) by using (4.11.12) we will generate the basis functions for 3 in thedecomposition of 5 as is given by (4.11.1) . From (4.9.3) orTable 1.1 Character table of D3 31 wehave

    3E=2, 3 A=3B=3C=0, 3D=3F=1 (4.11.14)

    31

  • 7/29/2019 Book Andrei N

    32/155

    Therefore using (4.11.2), (for example PD f2=f3 ) we get:

    P 3

    f1=2

    6[ 2PE f1PD f1PF f1 ]=

    1

    3[ 2 f1 f2 f3]h1

    P3

    f2=2

    6[2PE f2PD f2PF f2 ]=

    1

    3[2 f2f3f1]h2

    P 3 f 3= 26 [ 2PE f 3PD f3PF f 3 ]= 13 [2 f3 f1 f 2 ]h3 39

    (4.11.15)

    The three functions h1,h2, and h3 are not independent since hl + h2 + h3 = 0. Hence there are only twoindependent functions and these may be constructed in an infinite number of ways. The choicecorresponding to (3.5-3), after normalization, is g2 = h1 and g3 = h3 h2.So we have

    g2=1

    3[2 f1 f2f3] 33,39 (4.11.16)

    And

    g3= h3 h2=1

    3[2 f 3 f1 f2 ]

    1

    3[2 f2 f 3 f 1 ]= f 2 f3 33,39 (4.11.17)

    In similar way taking into account that for first equation in (4.11.13) all characters are equal to one :

    P1

    f1=l1

    hR

    1 PR f1=1

    6[PE f1PA f1PB f1PC f1PD f1PF f1 ]

    =1

    3[ f1 f2 f3 ]

    P 1

    f 2=

    1

    3 [ f1 f2 f 3 ]P

    1f 3=

    1

    3[ f 1 f 2 f3 ]

    (4.11.18)

    Thus we get for g1

    g1=1

    3[ f1 f2 f3 ] 33 (4.11.19)

    In what follows we will normalise applying the procedure described at pag 248 [72] or pag 141,248[40] or pag 63,61[31] or 214,230 [73] or pag 588[69] :

    1=

    N N d (4.11.20)

    32

  • 7/29/2019 Book Andrei N

    33/155

    1=N2

    c1 1c2 2

    c1 1c22 d

    1=N2

    c12 1

    2d2c1 c212 dc2

    2 22d

    (4.11.21)

    1=N2c1

    22c1c

    2

    1

    2dc

    2

    2

    1=N2c122c1 c2 S12c2

    2

    N=1

    c122c1 c2 S12c2

    2(4.11.22)

    Where

    S12=

    12 d (4.11.23)

    Is the overlap integral.

    Thus the normalisation for(4.11.19) is:

    [ f1f2 f3 ]N3 N3 [ f1 f2 f3 ]=

    N2

    9 f1f1 f2f2 f3f32 f1f22 f1f3 2 f2f3

    =N

    2

    936S =1

    thus N=3

    3 6S

    (4.11.24)

    The normalisation for(4.11.16):

    [2 f1 f2f3]N3 N3 [2 f1 f2f3]=

    N2

    9 4 f1f1 f2f2 f3f3 4 f1f2 4 f1f3 2 f2f3

    =N

    2

    9 428S2S =1

    thus N=3

    66S

    (4.11.25)

    And the normalisation for(4.11.17):

    33

  • 7/29/2019 Book Andrei N

    34/155

    [f2 f3]NN[f2 f3]=N2 f2f2 f3f32 f2f3 =N2 22S =1

    thus N=1

    22S

    (4.11.26)

    If we put all together we get (see [1])

    g1=1

    3

    3

    36S[ f 1 f 2 f3 ]

    g2 =1

    3

    3

    66S[2 f1 f 2 f3 ]

    g3=1

    22S[ f 2 f 3 ]

    (4.11.27)

    Or if we put overlap integrals to be zero we getg1=

    1

    3[ f1 f2 f3 ]

    g2 =1

    6[2 f1 f2 f 3 ]

    g3=1

    2[f2 f3]

    (4.11.28)

    Now we can construct the unitary matrix S and S1 see pag 56 [71] :

    [g1, g2, g3]=[ f1 , f2 , f3] S (4.11.29)

    1

    32

    60

    1

    3

    1

    6

    1

    21

    3

    1

    61

    2

    and see[11]pag 19:

    'G = S1 G S 42 (4.11.30)

    1) The matrix in order to obtain 5=13

    34

  • 7/29/2019 Book Andrei N

    35/155

    S={ {1

    3,

    2

    6,0 }, {

    1

    3,

    1

    6,

    1

    2},{

    1

    3,

    1

    6,

    1

    2}};

    S1=Inverse[ S] ; MatrixForm [ S1] .MatrixForm [ C ] .MatrixForm [ S]

    =MatrixForm [ S1. C. S ,TableSpacing {1,2} ]

    for 5 B

    1

    31

    31

    323

    1

    61

    6

    01

    21

    20 1 01 0 00 0 1

    1

    323

    0

    1

    31

    61

    21

    31

    61

    2=

    1 0 0

    012

    32

    032

    1

    2

    for 5 C

    1

    3

    1

    3

    1

    323

    16

    16

    01

    212

    0 0 10 1 01 0 0

    1

    3

    2

    30

    1

    316

    1

    21

    31

    61

    2 =

    1 0 0

    012

    32

    03

    2

    1

    2 for 5 A

    1

    31

    31

    32

    3

    1

    6

    1

    60

    1

    21

    2 1 0 0

    0 0 1

    0 1 0

    1

    323

    0

    1

    3

    1

    6

    1

    21

    31

    61

    2 =

    1 0 0

    0 1 0

    0 0 1

    2) We put here other matrix in order to obtain 5=31 29,26 (4.11.31)

    5 B

    here I corrected at pag 69 [69]:because Weissbluth had mistaken the correct is M 5BM1 :

    35

  • 7/29/2019 Book Andrei N

    36/155

    23

    1

    61

    6

    01

    21

    21

    3

    1

    3

    1

    3

    0 1 01 0 0

    0 0 1

    23

    01

    31

    61

    21

    31

    6

    1

    2

    1

    3

    =

    12

    32

    0

    32

    1

    20

    0 0 1 42 (4.11.32)

    for 5 D

    23

    1

    61

    6

    01

    21

    21

    31

    31

    30 1 00 0 11 0 0

    23

    01

    31

    61

    21

    31

    61

    21

    3=

    12

    32

    0

    32

    12

    0

    0 0 1

    for 5 A

    23

    1

    61

    6

    01

    21

    21

    31

    31

    31 0 00 0 10 1 0

    23

    01

    31

    61

    21

    31

    61

    21

    3=1 0 00 1 00 0 1

    4.12 Symmetry adapted linear combinations(C3v)

    See [11]:Once we know the irreps spanned by an arbitrary basis set, we can work out the appropriatelinear combinations of basis functions that transform the matrix representatives of our originalrepresentation into block diagonal form (i.e. the symmetry adapted linear combinations). Each of theSALCs transforms as one of the irreps of the reduced representation. We have already seen this in ourD3 example. The two linear combinations of A1 symmetry were sN and s1 + s2 + s3, both of which aresymmetric under all the symmetry operations of the point group. We also chose another pair offunctions, 2s1 s2 s3 and s2 s3, which together transform as the symmetry species E.

    Example: a matrix representation of the C3v point group (the ammonia molecule)

    The first thing we need to do before we can construct a matrix representation is to choose a basis. ForNH3 , we will select a basis sN, s1,s2,s3 that consists of the valence s orbitals on the nitrogen

    and the three hydrogen atoms. We need to consider what happens to this basis when it is acted on by

    each of the symmetry operations in the C3v point group, and determine the matrices that would be

    required to produce the same effect. The basis set and the symmetry operations in the C3v point

    group are summarised in the Fig. 3 below.

    36

  • 7/29/2019 Book Andrei N

    37/155

    E (sN,s1,s2,s3) (sN,s1,s2,s3)

    C3

    (sN,s1,s2,s3) (sN,s2,s3,s1)

    C3

    (sN,s1,s2,s3) (sN,s3,s1,s2)v (sN,s1,s2,s3) (sN,s1,s3,s2)v ' (sN,s1,s2,s3) (sN,s2,s1,s3)v ' ' (sN,s1,s2,s3) (sN,s3,s2,s1)

    in Matrix form:

    the matrices that carry out the same transformations are:

    (E) sN , s1,s2,s3

    1 0 0 0

    0 1 0 0

    0 0 1 0

    0 0 0 1

    =sN , s1,s2,s3

    ( C3

    ) sN , s1,s2,s3=1 0 0 0

    0 0 0 1

    0 1 0 0

    0 0 1 0=sN , s2,s3,s1

    ( C3

    ) sN , s1 , s2 , s31 0 0 0

    0 0 1 0

    0 0 0 1

    0 1 0 0=sN, s3,s1,s2

    ( v ) sN, s1,s2,s3

    1 0 0 0

    0 1 0 00 0 0 1

    0 0 1 0=sN , s1,s3,s2

    37

    Fig. 3

  • 7/29/2019 Book Andrei N

    38/155

    ( v ' ) sN, s1,s2,s31 0 0 0

    0 0 1 0

    0 1 0 0

    0 0 0 1=sN , s2,s1,s3

    ( v ' ' ) sN, s1,s2,s3

    1 0 0 0

    0 0 0 1

    0 0 1 0

    0 1 0 0=sN , s3,s2,s1

    In the table form:sN s1 s2 s3

    E sN s1 s2 s3

    C3

    sN s2 s3 s1

    C3

    sN s3 s1 s2

    v sN s1 s3 s2

    v ' sN s2 s1 s3v ' ' sN s3 s2 s1

    To determine the SALCs ofA1 symmetry, we multiply the table through by the characters of the A1irrep (all of which take the value 1). Summing the columns gives

    sN + sN + sN + sN + sN + sN = 6sNs1 + s2 + s3 + s1 + s2 + s3 = 2(s1 + s2 + s3)s2 + s3 + s1 + s3 + s1 + s2 = 2(s1 + s2 + s3)s3 + s1 + s2 + s2 + s3 + s1 = 2(s1 + s2 + s3)

    Apart from a constant factor (which doesnt affect the functional form and therefore doesnt affect thesymmetry properties), these are the same as the combinations we determined earlier.Normalising givesus two SALCs of A1 symmetry.

    1=sN

    2=1

    3s1s2s3

    (4.12.1)

    We now move on to determine the SALCs of E symmetry. Multiplying the table above by the

    appropriate characters for the E irrep gives

    sN s1 s2 s3E 2sN 2s1 2s2 2s3

    C3

    -sN -s2 -s3 -s1

    C3

    -sN -s3 -s1 -s2

    v 0 0 0 0

    38

  • 7/29/2019 Book Andrei N

    39/155

    v ' 0 0 0 0v ' ' 0 0 0 0

    Summing the columns yields2sN sN sN = 0

    2s1 s2 s32s2 s3 s12s3 s1 s2

    and this is equivalent with previous derivation (4.11.15),(4.11.16) and (4.11.17):

    3=1

    6 2s1s2s3

    4=1

    2

    s2s3

    (4.12.2)

    4.13 Bonding in polyatomics - constructing molecular orbitals fromSALCs

    In the previous section we showed how to use symmetry to determine whether two atomic orbitals canform a chemical bond. How do we carry out the same procedure for a polyatomic molecule, in whichmany atomic orbitals may combine to form a bond? Any SALCs of the same symmetry could

    potentially form a bond, so all we need to do to construct a molecular orbital is take a linearcombination of all the SALCs of the same symmetry species.The general procedure is:1. Use a basis set consisting of valence atomic orbitals on each atom in the system.2. Determine which irreps are spanned by the basis set and construct the SALCs that transform as eachirrep.3. Take linear combinations of irreps of the same symmetry species to form the molecular orbitals.

    e.g. in our NH3 example we could form a molecular orbital of A1 symmetry from the two

    SALCs that

    transform as A1 ,

    A1=c11c2 2=c1sNc2 13

    s1s2s3 (4.13.1)

    Unfortunately, this is as far as group theory can take us. It can give us the functional form of themolecular orbitals but it cannot determine the coefficients c1 and c2. To go further and obtain theexpansion coefficients and orbital energies, we must turn to quantum mechanics.

    39

  • 7/29/2019 Book Andrei N

    40/155

    4.14 Product representations

    Consider the two matrices(see pag 83 [69] )

    The direct product of two matrices (given the symbol ) is a special type of matrix product that generatesa matrix of higher dimensionality if both matrices have dimension greater than one. The easiest way todemonstrate how to construct a direct product of two matrices A and B is by an example:

    AB=a11 a 12a21 a 22b11 b12b21 b 22 =

    a11B a12 B

    a21 B a22 B

    =a11b11 b12b21 b22 a12

    b11 b12b21 b22

    a21b11 b12b21 b22 a22b11 b12b21 b22

    =

    a11 b11 a11 b12 a12 b11 a12b12

    a11b21 a11b22 a12 b21 a12 b22a21 b11 a21 b12 a22 b11 a22b12a21 b21 a21 b22 a22 b21 a22b22The characters of A , B and AxB are evidently related by

    AB =A B (4.14.1)

    Now let 1, 2 be a set of basis function,s for a two dimensional irreducible representation

    of a group G and let 1, 2 similarly belong to of G. This mean , according to (4.10.5)

    that:

    PR1=1 11R2 21

    R

    PR 2=1 12R2 22 R (4.14.2)

    With analogous relations for 1, 2 .Therefore,

    PR 11= [1 11 R 2 21 R ] [ 1 11 R 2 21 R ]= 11 11

    R 11 R 1 2 11

    R 21 R

    21 21 R 11

    R 2 2 21 R 21

    R

    (4.14.3)

    And continuing in the same fashion it is found that

    PR1 1 ,12, 2 1, 22

    = 11,12, 21 ,2211

    11

    11

    12

    12

    11

    12

    12

    11 21

    11 22

    12 21

    12 22

    21 11

    21 12

    22 11

    22 12

    21 21

    21 22

    22 21

    22 22

    41 (4.14.4)

    We can see that (4.14.3) is the first column of(4.14.4).

    40

  • 7/29/2019 Book Andrei N

    41/155

    Hence it may be concluded that the four product functions 1 1, 1 2 ,2 1, 2 2 are basisfunctions for a representationcalled a product representationof the group G. The product

    representation is generally reducible even if and are irreducible. Asin (4.14.1),

    R=R R (4.14.5)

    This result can be formalised by writingPR j

    l =

    ik

    i k

    ij R kl

    R

    =ik

    i k

    [ R R ]ik , jl

    =ik

    i k

    ik , jlR

    43 (4.14.6)

    And

    R =ij

    ii R jj

    R (4.14.7)

    It is sometimes useful to regard

    1,

    2

    and

    1,

    2

    the basis sets for

    and

    ,respectivelyas the components of two-dimensional vectors.

    An important case arises when = ; then the two representations forming the productrepresentation are the same. As before, let both 1, 2 and 1, 2 belong to

    (=

    ). It should be kept in mind that 1 and 2 are partners, as are 1 and 2 , butthere is no partnership relation between any of the 's with any of the 's. From (4.14.4) ,(4.14.3) we find (with the superscript suppressed)

    PR 11=11[11R]21221 11R 21 R

    22[21R]2 (4.14.8)

    PR 1 1 2 1=2 1 1 11 R 12 R 1 2 2 1 [ 11 R 22 R 12 R 21 R ]2 22 21 R 22 R

    (4.14.9)

    PR 2 2=11[ 12R]21 22 1 12R 22R

    22[22R]2 (4.14.10)

    PR1 12 1=1221[ 11 R 22 R12 R 21 R] (4.14.11)

    The products 1 1 , 1 22 1 , 2 2 are symmetric products in the sense that they remainunchanged under an interchange of indices. The product 1 22 1 is antisymmetric. From(4.14.8)-(4.14.11)it is seen that the symmetric products transform among themselves and do not mixwith the antisymmetric product and vice versa. This feature may be expressed by writing

    = + - (4.14.12)

    where + stands for the product representation whose basis functions are symmetric

    41

  • 7/29/2019 Book Andrei N

    42/155

    products; + is called the symmetric product representation; - , the productrepresentation whose basis functions are antisymmetric, is called the antisymmetric productrepresentation.

    4.15 Clebsch-Gordan coefficients for simply reducible groups (CGC)

    See pag 61 [70] :Equivalent representationsWe have seen in (4.11.30) and example (4.11.32) that on changing the basis by a nonsingular matrix S ,

    an operator P transforms according to P '=S1P S . This leads us to the following definition:

    Two REPs ={PR|R2G} and ' ={P'R| RG } of a group G

    are equivalent or similar , if every PR and P'R there exists a nonsingular operator S such that

    P '=S1P S for each a2G

    Let { fi '} be the new basis . Then in the matrix REP we have,

    f i '=j

    Sij f j (4.15.1)

    PR

    fi

    '=

    JS

    ij

    PR

    fi

    =

    jkS

    ji

    kj

    R fk

    =jkl

    Sji kj RS1 fl'=

    l

    S1 R Sli fl'(4.15.2)

    Therefore

    'R=S1 R S = (4.11.30) (4.15.3)

    The reduction of is achieved according to (4.15.3)by a non-singular matrix C (we will

    use C-Clebsch instead of S ), wich transform the basis {ei ek } into a new basis

    {e

    l

    s

    s

    =1,2,...,

    }. The matrix elements of

    C

    are denoted by

    i k s

    l

    and called

    Clebsch-Gordan or Wigner coefficients (CGC). Then

    e l s=

    ik i ksl e i e k (4.15.4)

    Or in notations of[75] :

    42

  • 7/29/2019 Book Andrei N

    43/155

    l s =

    ik i ksl i k 48 (4.15.5)

    Diagonal form

    C 1 C ={ ,1

    , s

    ...} (4.15.6)If the basis systems {ei ek } and {el , s } are orthonormal then C is unitary and the CGCobey

    ik i k ' s 'l ' i ksl = 'ss' l l '

    s l

    i k

    s

    l

    i ' k '

    s

    l

    = ii'kk '

    (4.15.7)

    And therefore also

    e i

    ek=

    ik i ksl

    els

    (4.15.8)

    Or in notations of[75] :

    i k

    =ik i ksl ls (4.15.9)

    Multiplying both sides of(4.15.6) with C

    and C

    1

    we get =C , sC 1 (4.15.10)

    But from(4.14.6) we have also

    ik , jlR= ij

    R kl R (4.15.11)

    Equating (4.15.10) and(4.15.11) we obtain,

    ij R kl

    R= s ,mn

    i ksm mnsR j lsn

    (4.15.12)

    Multiplying this by m' n' ' R and summing over all RG, we obtain with(4.7.1):

    s i k

    sm j l

    sn =

    l

    h

    RG

    ij R kl

    R mn R (4.15.13)

    As particular case for(4.15.13) see pag 104 [74]with j=i , l=k and n=m have:

    43

  • 7/29/2019 Book Andrei N

    44/155

    j l nn ={l

    h

    RG

    jj R l l

    R nn R}

    1 / 2

    (4.15.14)

    Substituting in (4.15.14) we get:

    s

    i k

    s

    m

    {

    l

    h

    RG

    jj R l l

    R nn R

    }

    1/ 2

    =l

    h

    RG

    ij R kl

    R mn R

    i ksm =l

    h

    RG

    ij R kl

    R mn R

    {lh RG jj R l l R nn R}1 /2

    i ksm =lh RGij R kl R mn R

    RG

    jj R l l

    R nn R

    51

    (4.15.15)

    For other forms of(4.15.15) see[75] and [76] .

    Illustration (4.15.15):we have a table from Lenef[1]:

    Table 1.2 Irreducible representations of the group C3v

    C3v E C3 C3 ' va vb vc

    A1 1 1 1 1 1 1

    A2 1 1 1 -1 -1 -1

    E 1 00 1 1

    2

    32

    32

    1

    2

    12

    32

    32

    1

    2

    1 0

    0 1 1

    23

    2

    32

    1

    2

    1

    2

    32

    32

    1

    2

    we will multiply approximately(or slightly different ) like in great orthogonality theorem

    E E11 A1s1 =l

    hR=1

    6

    11ER 11

    ER 11A1R

    R=1

    6

    11ER 11

    ER 11A1R

    l=1 this is the dimension of A1h=6 number of the groups

    44

  • 7/29/2019 Book Andrei N

    45/155

    i k=11 1221 22 for E

    E E

    11

    A1s

    1

    =

    1

    6

    [111 12 12 112 12 1111 12 12 1 12 12 1]

    111

    12

    12

    112

    12

    111112

    12

    112

    12

    1

    =16[1 14 14 1 14 14 ]1 14 14 1 14 14

    =16 33 =123 33 = 12

    E E22 A1s1 =16[111 12 12 112 12 1111 12 12 1 12 12 1]

    111

    1

    2 1

    2 1

    1

    2 1

    2 1111

    1

    2 1

    2 1

    1

    2 1

    2 1

    =16[1 14 14 1 14 14 ]1 14 14 1 14 14

    =16 33 =123 33 = 12

    E E1 2 A1s1 =l

    hR=1

    6

    12ER 21

    ER 11A1R

    R=1

    6

    22ER 11

    ER 11A1R

    E E1 2 A1s1 =16[001 32 32 132 32 100132 32 1 32 32 1]

    11112 12 112 12 1111 12 12 1 12 12 1=16

    [ 34 34 34 34 ]1 14 14 1 14 14

    =0

    0

    we have 00

    and in this case we use the formula(4.15.13)

    45

  • 7/29/2019 Book Andrei N

    46/155

    E E1 2 A1 s1 E E2 1 A1 s1

    =

    16 [00132 32 132 32 1001 32 32 1 32 32 1 ]=

    1

    6 [ 3

    4

    3

    4

    3

    4

    3

    4 ]= 0

    Thus we get:

    E Ei kA11 = 12 1 00 1 (4.15.16)

    for A2See pag 85[70] we take i= j , k= l , m=n in (4.15.13)and obtain at least one CGC which isdifferent from zero, then we keep j , l ,n constant and vary i ,k ,m :

    E E11 A2 s1 =111 12 12 1 12 12 1111 12 12 112 12 1=1 14 141 14 14=0

    thus tis square root is not good.

    E E1 2 A2 s1 =l

    h R=1

    6

    12E

    R 21E

    R 11A1

    R

    R=1

    6

    22ER 11

    ER 11A1R

    (4.15.17)

    46

  • 7/29/2019 Book Andrei N

    47/155

    E E12 A2s1 =E E21 A2s1 =

    16[001 32 32 132 32 100132 32 1 32 32 1]

    111

    1

    2

    1

    2

    1

    1

    2

    1

    2

    1111

    1

    2

    1

    2

    1

    1

    2

    1

    2

    1

    =16[34 34 34 34 ]

    1 14 14 1 14 14=16 33 =123 33 = 12

    Thus we find a sure root different from zero and we fiw j=2,l=1,n=1 thus

    E Ex y A2 s1 =l

    hR=1

    6

    x2ER y1

    ER 11A1 R

    R=1

    6

    22ER 11

    ER 11A1R

    (4.15.18)

    xy=11 From (4.15.18) :

    E E11 A2 s1 =

    16[01132 12 132 12 101132 12 1 32 12 1]

    111 12 12 1 12 12 1111 12 12 1 12 12 1

    = 0

    xy=2 2 From (4.15.18) we get:

    E E22 A2 s1 =0xy=2 1 From (4.15.18) we get:

    47

  • 7/29/2019 Book Andrei N

    48/155

    E E21 A2s1 =

    16[111 12 12 112 12 1111 12 12 1 12 12 1]

    111

    1

    2

    1

    2

    1

    1

    2

    1

    2

    1111

    1

    2

    1

    2

    1

    1

    2

    1

    2

    1

    =16[1 14 14 1 14 14 ]1 14 14 1 14 14

    =16 33 =123 33 = 12thus we get again:

    E Ei k A11 = 12 0 11 0 (4.15.19)

    The same procedure for E

    l=2 this is the dimension of E 2x2 matrix:

    E Ei kE2 = 12 0 11 0

    E Ei kE1 = 12 1 00 1(4.15.20)

    If f1 , f2 and g1 , g2 are basis functions belonging to E , then basis functions in EE ( EE=A1A2E )are from (4.15.5):

    A1 = 1

    2 f 1g1 f 2g2

    A2 = 1

    2 f 1g2 f2 g1

    1E= 1

    2 f1 g2 f 2g1 2

    E=1

    2 f1g1 f2g2 53

    (4.15.21)

    Or if you want

    {f1, f2}={1E

    1,2E

    1} and {g1, g2}={1E

    2, 2E

    2} (4.15.22)You can see eq (11) and (12a,b,c,d) from [1], actually he solve the problem 4.11 at pag 86 using table8.1 pag 196 from [70]. Where problem 4.11 from [70] is another way for finding CGC by applyingprojector operators (4.11.5)to (4.15.22).where CGC are from (4.15.5) after ortho-normalization. Thisprocedure you can also see in[77]pag 47,48.

    48

  • 7/29/2019 Book Andrei N

    49/155

    At pag 85[70]is not correct see pag 287 Sugano [77].

    A1 A2 E E

    u v

    u1

    20

    1

    20

    v 01

    20

    1

    2

    u 0 12

    01

    2

    v1

    20

    1

    20

    CEE=

    1

    20

    1

    20

    01

    20

    1

    2

    0 1

    20

    1

    21

    20

    1

    20

    (4.15.23)Therefore (4.15.23) Is a unitary matrix(you can verify by applying eq. (4.5.3) ).Finally you can verify that:

    CEE1 EE CEE=A1A2E (4.15.24)

    4.16 Full matrix representations

    We list here full matrix representations for several groups. Abelian groups are omitted, as their irrepsare one-dimensional and hence all the necessary information is contained in the character table. Wegive C3v (isomorphic with D3) and C4v (isomorphic with D4 and D2d). By employing higher l valuespherical harmonics as basis functions it is straightforward to extend these to Cnv for any n, even or

    49

  • 7/29/2019 Book Andrei N

    50/155

    odd. We note that the even n Cnv case has four nondegenerate irreps while the odd n Cnv case has onlytwo.

    Table 1.3 C3v and D3 Matrix irreducible representations:

    D3 E C3 C32

    C2a

    C2b

    C2c

    C3v E C3 C32

    va

    vb

    vc

    A1 A1 1 1 1 1 1 1

    A2 A2 1 1 1 -1 -1 -1

    E11 E11 1 1

    2

    1

    21

    1

    2

    1

    2

    E21 E21 0 32

    32

    03

    2

    32

    E12 E12 0 32

    32

    03

    2

    32

    E22 E22 1 1

    2

    1

    2-1 1

    2

    1

    2

    50

  • 7/29/2019 Book Andrei N

    51/155

    Table 1.4 D4 , D2d and D4v Matrix irreducible representations:

    D4 E C2 C4 C43

    C2x

    C2y

    C2135

    C245

    D2d E C2 S4 S43

    C2x

    C2y

    C2135

    245

    D4v E C2

    C4

    C4

    3

    xz

    yz 2

    45

    2

    135

    A1

    A2

    B1

    B2

    E11

    E21

    E12

    E22

    A1

    A2

    B1

    B2

    E11

    E21

    E12

    E22

    A1

    A2

    B1

    B2

    E11

    E21

    E12

    E22

    1 1 1 1 1 1 1 1 1

    1 1 1 1 -1 -1 -1 -1 2

    1 1 -1 -1 1 1 -1 -1 3

    1 1 -1 -1 -1 -1 1 1 4

    1 -1 0 0 1 -1 0 0 5

    0 0 1 -1 0 0 1 -1 5

    0 0 -1 1 0 0 -1 1

    5

    1 -1 0 0 -1 1 0 0 5

    The numbers 45 and 135 refer to the angles (in degrees) the particular axis makes with the positive xaxis, or the plane makes with the xz plane.

    Clebsch-Gordan coefficients for the crystallographic point group D4See pag 76-77[78] :

    Using the matrices of the two-dimensional irreducible representation 5 specified in Table 1.4 , theClebsch-Gordan coefficients corresponding to the series 5 4 5 are given by Equation(4.15.15) (with j=2,l=1,n=2 ) as

    5411512 =5 42 1511 =15 41 15 11 =5 42 15 12 =0

    Similarly for 535 (with j = 2, l=1, n=2 )

    5311512 =5321511 =15311511 =5 32 1512 =0

    Likewise for 5512 34 .Equation (4.15.15) implies that all the Clebsch-Gordan

    51

  • 7/29/2019 Book Andrei N

    52/155

    coefficients are zero except the following:

    5511111 =5 52 21 11 = 125511211 =55222 11 = 12

    5 5

    2 13 1

    1 =5 5

    1 23 1

    1 =1

    2

    5521411 =5 51 2411 = 12

    4.17 Selections rules

    See pag 109[71] :

    when calculating various matrix elements in quantum mechanics some of them happen to vanish forreason of symmetry. Representation theory greatly helps in judjing whether or not a given matrixelement vanishes by symmetry.Let us examine a matrix element like

    m

    , Hl (4.17.1)

    Where H is the hamiltonian or any other operator that is invariant under the symmetry group G .

    Invariance of H means that Hl

    transform in exactly the same way as l

    . then the

    orthogonality relationm

    , l

    = ml constant independent of m and l (4.17.2)

    Indicates that the above matrix element is non-vanishing only when = and m= l and that it isindependent of m. That is, the Hamiltonian has non-vanishing matrix elements only between thefunctions that have the same transformations properties.To consider matrix elements of general operators it is convenient to define irreducible tensor operators.

    In parallel with the basis functions m

    transforming like

    R m =

    m'

    m' m'

    R (4.17.3)

    The operators Tm

    which transform likeR Tm

    R

    1=m'

    Tm' m'

    R (4.17.4)

    Are called irreducible tensors operators. For instance in the group C3v components of the vector r

    , momentum p and angular momentum l vectors constitute irreducible tensor operatorsbelonging to the following irreducible representations:

    52

  • 7/29/2019 Book Andrei N

    53/155

    A1 : z , pzA2 : lzE : {x , y} , {px, py } , {lx , ly} .

    Whether or not a matrix element of the irreducible tensor operator Tj

    m

    , Tj l

    (4.17.5)

    vanishes by symmetry is examined in the following way. The functions Tj l transform as basisfunctions of the product representation . So the function Tj

    l

    contains in itself

    irreducible components that are obtained by reduction of the product representation . Inorder that the matrix element (4.17.5) should not vanish by symmetry it has to contain the m

    component because of the orthogonality relation (4.17.2). This requires that the representation

    should appear at least once in the reduction of the product representation

    Product representations are reducible in general . If has the irreducible decomposition

    =

    q

    (4.17.6)

    Then the values ofq

    may be calculated by means ofq=

    1

    g

    G

    G G=1

    g

    G

    G G G (4.17.7)

    Here q gives the number of the times the representation apears in the reduction of .

    Thus the matrix element vanishes by symmetry if q=0 .see pag 67 [71]is equivalent with (4.15.21):As an example let us examine the selection rules for photoabsorbtion by an atom placed in a field of

    C3v symmetry. Interaction of the electron with light has the form p , where stands for thepolarisation of the electric field . Suppose that the initial state belongs to the irreducible representation i . For light polarised in the z direction the optical transition takes place trough pz .Since theoperator pz belongs to A1 representation of C3v the final state must be A1 i=i . So

    pz causes transitions between states with the same symmetry.

    When the light is polarized in the xy-plane the transitions takes place trough px and py wich

    belong to the representation E . The symmetry of final states is obtained by reducing E i . Theresult of this reduction turns out to be E , E , A1A2E for i=A1,A2,E .

    4.18 (SO3)Linear combinations of spherical harmonics of point

    groups

    See pag 88[70] :

    For the point groups that are subgroups of SO3 we obtain the standard REPs via the spherical

    harmonics Yml

    . These Yml

    are the basis functions of the Irs l of the rotation group SO3 andconstitute a generally reducible basis for REPs of point groups.The spherical harmonics ( , : angles of spherical polar coordinates)

    53

  • 7/29/2019 Book Andrei N

    54/155

    Yml , =

    1

    2 m

    l e i m, m0 56 (4.18.1)

    ml = 2 l12 lm !lm !

    12 1

    m

    2ll !

    sinm cos

    lm

    cos2 1l

    Yml =1mYml, lmlconstitute a 2 l1 dimensional linear space for the REPs of point groups.

    Often it is useful to chose real functions, thus m0

    Ym, cl =

    1

    2[1m Yml Yml ]=1m

    1

    m

    lcosm

    Ym , sl =

    i

    2[1 m Yml Yml ]=1m

    1

    m

    lsin m

    (4.18.2)

    For those point groups with one main axis thee real functions are bases for the standard REPs . In thecase of cubic point groups we have to chose linear combinations of spherical harmonics adptet to thecubic symmetry: the cubic harmonics are different from real spherical harmonics for l3 . Thespherical harmonics are denoted as s-, p-, d-, f-, , or more detailed px , py ,..., according to their

    use in quantum mechanics see the Table 1.5:

    54

  • 7/29/2019 Book Andrei N

    55/155

    Table 1.5 Linear combinations of spherical harmonics of point groups according to (4.18.1),(4.18.2)

    for l=0 to l=3 pag 88 [70]:

    Yml Normalization Spherical coordinates Cartesian coordinates

    Y00 s 1/4 1 1

    Y01

    Y1 c1

    Y1 s1

    pz

    px

    py

    3/4

    3/4

    3/4

    cos sin cos

    sin sin

    z

    x

    y

    Y02

    Y1 c2

    Y1 s2

    Y2 c2

    Y2s2

    d0

    d1c

    d1s

    d2c

    d2s

    5/16

    15/4

    15/4

    15/16

    15/16

    3cos2 1

    sin cos cos

    sin cos sin

    sin2 cos2

    sin2sin 2

    3z 21= 2z 2x2y 2

    zx

    zy

    x2y2

    2 xy

    Y03

    Y1 c3

    Y1 s3

    f0

    f1c

    f1s

    7/16

    21/32

    21/32

    cos 5cos2 3

    sin 5cos21cos

    sin 5cos2 1sin

    z5z23=z2z23x23y2

    x 5z 21=x 4z 2x2y2

    y 5z21=y 4z2x2y 2

    Y2c3

    Y2s3

    Y3 c3

    Y3 s3

    f2c

    f2s

    f3c

    f3s

    105/16

    105/16

    35/32

    35/32

    sin2 cos cos2

    sin2 cos sin2

    sin3 cos3

    sin3 sin3

    zx2y2

    2xyz

    x x23y 2

    y 3x2y 2

    This is equivalent with

    Table 1.6 Character table of C3v with assigned functions

    E 2C3 3 v s p d L

    1 A1

    2 A2

    3 E

    1

    1

    2

    1

    1

    -1

    1

    -1

    0

    1 z

    x , y

    z2 : singlets

    doublets:

    2xy , x2y2 ; xz , yz

    Lz

    Lx, Ly

    The spherical harmonics see pag 3 [69]: