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# Physical Chemistry CHEM 3510

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This 58 page Class Notes was uploaded by Mr. Clementine Gottlieb on Wednesday October 21, 2015. The Class Notes belongs to CHEM 3510 at Tennessee Tech University taught by Staff in Fall. Since its upload, it has received 28 views. For similar materials see /class/225695/chem-3510-tennessee-tech-university in Chemistry at Tennessee Tech University.

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CHEM 3510 Fall 2003 The Classical Wave Equation The classical wave equation describes various wave phenomena a vibrating string a vibrating drum head ocean waves acoustic waves Consider a ldimensional wave describing the motion of a vibrating string Q A onedimensional wave describes the motion of a vibrating string 2 2 a x90 zi 3 x5 t This is the Classical Wave Equation 6x2 L2 6t2 where uxt is the displacement of the string from the horizontal position a is the velocity or the speed which the disturbance moves along the string t is the time 7 The classical wave equation is a partial differential equation a linear partial differential equation because uxt and its derivatives appear only to the first power and there are no cross terms The x and tare independent variables and ux t is a dependent variable 7 The displacement ux t must satisfy certain physical conditions the amplitude should be zero at the end of the string u0t 0 and ult 0 7 These conditions are called boundary conditions because they satisfy the behavior at the boundaries 7 To solve the differential equation we assume that uxt factors into a function of x times a function of t this method is called separation of variables uxt XxTt 7 Substituting ux t in the equation above 2 2 3 Ttd X2xi2Xxd T2t dx u dt CHEM 3510 Fall 2003 7 Dividing by ux t X xTt 1 d2Xx LLd2T0 X06 dx2 L2 TU dtz 7 In order for this equation to be true for every x and t each side should be equal to a constant K called the separation constant 7 The problem of nding ux t transformed into two problems of nding Xx and Tt by solving the following linear differential equations with constant coefficient they are ordinary differential equations 2 d X2 KXx0 dx 2 ign Kuzrako dt 7 Solving forXx 7 trivial solution is obtained that is X x 0 if K 20 7 ifKlt 0 set K 2 is real 2 2Xx 0 7 The general solution for this equation is Xx cle x 0267in 7 Considering Euler equation em cos x i isin x 3 XxAcos xBsin x 7 Boundary conditions X00 3 A 0 Xl0 3 Bsin l 0 3 l nn39 where n 1 2 2 madam CHEM 3510 Fall 2003 7 Look more closely to the solutions Number of wavelength that ts in 2l gt gt 2l 7 By general1zat1on in eigenvalue condition n lt nn39 27239 7 The solut1ons are a set a funct1ons Xquot x Bquot s1nTx Bquot s1n7x called n eigenfunctions or characteristic functions 27wn71u 7 Also aquot 27mquot A T nwo where mo are called eigenvalues n or characteristic values CHEM 3510 Fall 2003 7 Solving for Tt but keeping in mind that quot1 72 2 d Tt 2u2Tt0 t d 2 7 Similar to above the solution is Tt D cos mnt E sin mnt where aquot u g 7 Coming back to ux t XxTt 3 uxt Fcos wntGsinwntsin n 1 2 7 There is a ux t function for each n so a better notation would be uquot xt Fn cos mnt Gn sin mntsin n l 2 7 The sum of all uquot x t solutions is also a solution of the equation This is called the principle of superposition The general solution is 00 nmc uxt 2Fn cos mnt Gquot s1n mnts1nT n 7 Make the transformation F cos at Gsin at A cosat where is the phase angle and A is the amplitude of the wave 7 Rewrite the general equation as oo oo uxt 2A cosant n sin unxt quot1 n1 7 Each uquot xt is called 7 a normal mode 7 a standing wave 7 a stationary wave 7 an eigenfunction of this problem 7 The time dependence of each mode represents a ha1monic motion of frequency aquot un L V quot 27239 21 Aquot nnu where the angular frequency is aquot u 2721quot T A CHEM 3510 Fall 2003 7 First term is Al cost 1 sin 7 First term is called fundamental mode or first harmonic 7 The frequency is V1 g 7 Second term is A2 cos27mt 2 sinZTmC N v39 39 39 t x 4 I h 39 Q 39 39 39 7 Second term is called first overtone or second harmonic 7 The frequency is V2 7 The midpoint has a zero displacement at all times and it is called a node 7 Third term is A3 cos3Tmt 3 sin 377 L I I 39 e 39 s 39 39 39 u 7 Third term is called second overtone or third harmonic 3 7 The frequency 1s V3 21 7 This term has two nodes 7 Fourth term is A4 cos4Tmt 4sin4TmC u39 w 39 39 v 1 t 1 u n u i I l n I u n u l n n z I n a I I 39 39 CHEM 3510 Fall 2003 7 Let s consider now the case of l 2 ux t cosa1ts1n Ecosa2t s1n Tm 7 This is an example of a sum of standing waves yielding a traveling wave N u 7 Thinking backwards any general wave function can be decomposed into a sum or superposition of normal modes 7 The number of allowed standing waves on a string of length 1 become more numerous as the wavelength decreases 3 the possible highfrequency oscillations outnumber the lowfrequency ones 21 7 A n 7 Consider that 1 gtgt 2 so we can approximate the set of integers n by a continuous function 712 21 21 n 3 dn 2 all A A 7 The negative sign indicates that the number of standing waves decreases as 1 increases 7 The number of standing waves in an enclosure of volume Vuse c not a for the speed dn 4 le but l andv l v A 4 2 dv id1 2 511 l dv 12 c 472V 63 47239V 12 dn 4 dv v2dv 1 c CHEM 3510 Fall 2003 Look now at a twodimensional wave equation the equation of a Vibrating membrane 2 2 2 a l6 l where uxyt 6x By u it 7 Similar to the onedimensional problem use separation of variables uxyt FOG yTt 1 d2T 1 62F 62F 2 3 2 2 2 2 5 u Tt dt F x y ax 6y 7 Use separation ofvariables for Fx y Fxy XxYy 2 2 l d Xx l d Yy 20 7 Divide by F x y X05 dx2 Yy dy2 7 Solve two equations 2 2 1 dXx zand 1 ally 2 Xx de Yy dyz where p2 q2 z 7 Solutions for X x and Y y are Xx dam n 7 12 and YyDsin m 712 2 2 n m where nm 7239 a2 b2 7 Solution for Tt Tnmt Emquot cos wnmtan sin amt Gmquot cosanmt rm n2 m2 where mmquot u nm un39 a Zb 2 7 The general solution for ux yt ux y t Z Zumxyt 2 214nm cosanmt nm sin sin mb y nl ml nl ml a 7 Again the general function is a superposition of normal modes umquot x yt but in this case one obtains nodal lines instead of nodes CHEM 3510 Fall 2003 Applications onuantum Theory The Particle in a Box Translational Motion 7 The timeindependent threedimensional Schrodinger equation 2 2 2 2 h a 8 2Vxyz 7 2 2 wxyzEvxyz 239quot 8x 8y 82 7 Consider only a onedimensional motion 5 drop y andz in the equation above 7 Consider that the particle experiences no potential energy between the position 0 and a 5 Vx0 forO gxga The particle is restricted to the region 0 g x g a Potenhal energy a a x Wall Wall 2 2 l E V 2m dx2 7 The mathematical solution of the equation is Vx Acoskx Bsinkx 12 2 2 Where k ZmE 27M2mE E h k h h 87r2m 7 Apply the boundary conditions and the normalization condition to determineA and B 40 0 s A 0 Va 0 5 Bsinka 0 5 ka mt Wheren12 7The allowed energy levels are The energy is quantized hzn2 gt En 8ma2 Where n 1 2quot quot n is called quantum number CHEM 3510 Fall 2003 Classically allowed h 1 1 th b 9 energies 7T e energy eve s1ncrease as e quantum num er increases 7 The energy separation between the allowed energy levels increases as the quantum number increases 7 The energy levels and the energy separation between the energy levels increases as the size of the box decreases 7 Quantum numbers appear naturally when the boundary conditions are put in the Schrodinger equation like in the string problem and are not introduced ad hoc like in Plank model for blackbody radiation or Bohr model of hydrogen atom 7 FindB by setting the condition that the function Vx is normalized 7 jwltxwxdx 71 0 a B2 sin2 2 1 1 0 gt 321 gt B J2 B is callednormalization constant 1 7 The normalized eigenfunctions 11 x are given by 12 Lxnx sin wherenggaandnl2 1 L1 543 2 I 5 Solving the Schrodinger equation for the particle in a box problem gives a set of allowed energies eigenvalues and a set of wave functions eigenfunctions CHEM 3510 Fall 2003 Represent and investigate the energies the wave functions a and the probability densities b for the first few levels for the particle in a box N n E 1M 1 11 V MUN quotaquot n We 712 3 V n71 1 E39 917 33949 J N39s quot1 H wEA39J mun 1 1 31m 2 T A 1F 1 El nfr fulltx m C quot7 II J II x a x 1391 EM 7 The particleinabox model can be applied to electrons moving freely in a molecule free electron model 7 Example butadiene has an adsorption band at 461 X 104 cm 1 As a simple approximation consider butadiene as being a onedimensional box of length 578 A 578 pm and consider the four 11 electrons to occupy the levels 0 a 3 calculated using the particle in a box model 2 h 2 32 22 mea 7 The electronic excitation is given by AE 7 The calculated excitation energy 17 454 X 104 cm 1 compared very c 1 well with the experimental value 7 This simple freeelectron model can be quite successful CHEM 3510 Fall 2003 X2 7 The probability of nding the particle between x1 and x2 is given by Ill xzn xdx 1 71f x1 0 and x2 aZ then Prob 0x a2l2 forall n 7For nl Prob0 x a4 lt Proba4 x a2 7 As n increases for example n 20 these 2 probabilities become the same 7 More general The probability density becomes uniform as n increases This is an illustration of the Correspondence Principle that says that quantum mechanics results and classical mechanics results tend to agree in the limit of large quantum numbers The large quantumnumber limit is called the classical limit a 7 The eigenfunctions are orthogonal to each other Iw xz xdx 0 Where m i n 0 9 7 Example look at z1x and 13 x a 2a 71x 371x Ill1xl3xdx Jsin sin dx 0 0 a 0 a a 7 Look at the average values and the variances for the position and the momentum of the particle a 2 a nizx a 7 Average value of position x Illn x x yn xdx Jxsin2 0 a 0 a 2 a 2 2 2 a a 7 Average value of pos1tion square x gt Jyn x x 41quot xdx 7 0 3 2n 2722 2 2 2 7Variance in position a x2gt7ltxgt2 a j n 72 2n72 3 2 2 12 7 Standard deviation in position ox ltx2gt 7 x2 a n 7 2 2n72 3 Znna nizx nizx 7 Average value of momentum p 71h Ism cos dx 0 a2 0 a a 2 2 2 2 2 2 2 2 2 h h h 7Also 72 2mltEgt 2m 2 a ma a a 12 2 2 7 Finally oxop n 3 7 2 gt Heisenberg Uncertainty Principle V n 0 Solutions 11quot quotyquot x yz FEFsin nxm sin y y 47 2 x Z a c CHEM 3510 Fall 2003 h2 62 62 7 Cons1der a twod1mens1onal motion W W E 2m 6x23y2 Way my XxgtYygt sin x nyny Fcienual energy sin a b 2 2 2 E 11 nxny x quotmy 8m a2 b2 Panicle con ned to sudace 7 Consider a generalization to athreedimensional motion Particle in a 3dimensional box 2 2 2 2 h6 11 a 11 a 11 6y2 az 2 ELxyz 6x2 nznz sin a b c 2 2 n2 2 En n n 1 nx y quot 2 x y 2 SM 612 b2 62 7 Look at the average values of the position and the momentum of the particle a b c i A a b c ltrgtjdxjdyjdw xyzRuxyz315J k 0 0 0 where liXifjZAk a 17 c A ltpgt jdxjdyjdax xyzP wxyz 0 0 0 0 where 13 7203 ji 6 i k K 6x By 62 CHEM 3510 Fall 2003 7Lookatthecaseofabc a Dcnmmt 2 6 7Three sets of quantum numbers give same energyE2n E121 E112 Sma2 7 g 512 7 The energy level E is degenerate murmur J 8ma2 cmztzum 7 The energy level Em 2 is nondegenerate 41x SM 0 7 D generacy is due to the symmetry ofthe system Once the symmetIy is Energy level Ofpm de a be destroyed then degeneracy is lifted 7 Look at the Hamiltonian for the particle in a 3dimensional box 177 V Fy i HxHy Hz A 2 2 2 2 It can be wrinen as a sum ofterms where H 7h a h a Zrnax2 2m ayZ A 12 62 Hz 72 2 The operator is said to beseparable quot1 67 7 The eigenfunctions ynxnynl are written as a product of eigenfunctions of each operator I y and 32 and the eigenvalues Enxnynl are wrinen as a sum of the eigenvalues of each ofthe operator IQ39X y and IQ39Z 7 This is a general property in quantum mechanics If the Hamiltonian or an operator in general can be written as a sum of terms involving different coordinates ie the Hamiltonian is separable then the eigenfunctions of I is a product ofthe eigenfunctions of each operator constituting the sum and the eigenvalues of ii is a sum of eigenvalues of each operator constituting the sum r 7 I93lts1 392w 7 wow 7 Kmw 2 Em EK Em 31W x 7 EM 0 32 WM 7 EmuM where CHEM3510 Fall 2003 The Harm uni Oscillztnr Vibratinnal Mntinn The harrnonre oseruatorrn e1assrea1 rneehanres eForee f ekaezn kx Hooke39s Law rde where krs the fame constant i n 2 eHooke39s Law eornbrned wrth Newton39s equation mszwq 0 dt eThe solution of the equatron x0 Aeom where QJ7 eThe potenaal energy othe oseruator wr 2 2 The tota1 energy rs conserved rt rs transferred between Kand V The harrnonre motion tn a dratornre rno1eeu1e e Consrderthe movement of atorns of masses m and m 15 m m er 7 1 d1 2 I n 1er m 7km 7 Kt 7 In 122 eBy surnrnrng the two equataon above M eo Where Mmm2 and grew 122 M where er the center the W55 wardmate e Subtracting equanon 2 dwrded by m from equanon 1 dwrded m d 1q 0 where r x er etn and n W all m m where 1 rs the reduced W55 of the systern and x rs the relatxve coardmate systern wrth amass equal to the redueedrnass of the tworbody systern CHEM 3510 Fall 2003 The quantummechanical harmonic oscillator 7 The Schrodinger equation for quantummechanical harmonic oscillator if d2 7 2W mom 7 EW 21 dx But Vx kx2 Potential V dzvx 21 1 2 sh Z E7310 yx70 7 The solutions of this equation eigenvalues are Evh vijhwvljhyvij 2 2 2 Where v012 w E VL E 27 y 7 The eigenvalue representation in textbook is Wrong 7 The wave functions eigenfunctions are 7 z wvltx7NvHvltal2xe quotx Displacemem Where a k hz 14 7 The normalization constant is given by NV Z 1 2 U v 7 7 The wave functions form an orthogonal set 7 The Hva12x are polynomial functions called Hermite polynomials HV 5 in a vth degree polynomial in 5 Here are the rst few Hermite polynomials Ho 1 H1E2 11125 42 7 2 11135 7 83 712 Hug 164 7 4852 12 H3E 325 716053 120 even polynomials fx f7x odd polynomials fx 7f7x 7 Some properties of odd functions 71fthe function is continuous f0 0 A 7 Also food 0 7A CHEM3510 Fall 2003 eTlne normallzed wave funenons andthe probablllty denslty L M mull ln ml 239quot Mu l llrmr l e The exlstence of zerorpolnt energy 7 The rnrnrrnurn energy the energy ls calledthe zerapamt Eng 1 ZPE hv 2 20H 5 l lllll llllwlul groundrstate energy ls not zero even for v 0 Thls z example p 0 atthe same tlme example dlatomlc rnoleeule Ills amodel for vlbranons m dlatomlc rnoleeules 5212mm rule Av 1 frequency m the spectrum ofa dlatomlc the frequency called mdamenml yrmanmzz frequency v abs AE J vllrv 2 l Vabs ii 72 Fall 2003 CHEM 3510 7 q anuty xfrom above wru be m thrs ease the dAfference between the rnteratomre dutance dunng the mbrauon and the equrhbnum dutance x 1 71 eonstant k 272 Vom o E ltV t tt meehames regrons forbxdden by e1assrea1 meehames eThe average value of posrtron and momentum for the harmoan oseruator ltxgt Wm x WW 2ltxgt0 ltp 1w ltxgtr m jwm 2 ltp The Quantum Machanical Rigid Rntatnr Rntztinnal Mntinn eIt rs amodel for arotatrng 64amme mo1eeu1e eFor the eenter ofmass mm my I Mquot eVeloerues v n 27mm qm v2 5 27mm 54 Mm 7mm rs the angular speed where a where r 2 and a m1m2 e Srmuar to moment ofmema for a smgle rotaung pamcle 1 mt system where the mass rs replaced wrth the redueed mass 7 Angular momentum L CHEM 3510 Fall 2003 7 Solving Schrodinger equation for quantummechanical rigid rotator model 19V01I1 iv2 2 62 62 62 6x2 6y2 622 1 a 2 a 1 a a 1 62 2 I 2 I S111 6 2 r 6 quot 6 quot w r Sin 9 69 69 w r Sin 9 6 r 8 7 A special case is obtained when r is constant the rotator is rigid 19mg EYW V2 where Y 19 are the rigid rotator wave functions 2 2 h 1 sinei L 6 Y19 EY19 21 s1nt9 619 619 mm my 7 Further rearrange 2 2 sinei sin a Y 6 Y sin2 19Y 0 619 619 392 where g h 7 The rigid rotator wave functions Y 12 will be given later 7 By solving the equation above one can determine that must satisfy J J 1 7 The discrete set of allowed energy levels is given by 2 3 EJ JJ l where J 012 7 Each level has a degeneracy given by g J 2 1 7 The selection rule for the rigidrotator model A i1 only transitions between adjacent states are allowed 7 The energy and frequency for transition between levels R2 AEEJ1EJ 2 47239 I J1hv h 4721 3v Jl CHEM 3510 Fall 2003 7 The frequency of these transitions is about 1010 1011 Hz which is in the microwave range of electromagnetic radiation 3 microwave spectroscopy 7 Usually write the frequency in terms of the rotational constamB V ZBJ 1 where B 2 87239 I 7 In terms of wavenumbers 17 2 J 1 where E 2 87239 c 7 From experimental E one can determine the moment of inertia I then the bond distance r in a diatomic molecule 1 Llr2 J 4 l m 5531 3 J 39139 F 3 x 9 3565quot N 3 1 S E 1 41 I 1525 o q um 1 quot15 tammrum I n 35 39539 55 35 Fall 2003 CHEM 3510 Spherical Coordinates r 0 95 and Caltesian Coordinates x y z The position ofa point can be speci ed by using Cartesian coordinates xyz or by using spherical coordinates r t9 1 Z Z 2 E xrsint9c0s r x y 2 y rsint9sin and c0st9 Z 1 zrc0st9 x2y2zz7 tan X z oo oo oo 007r27r dVdxdydzr2sin6drdad j j j dxdydz H Irzsin drd6d 700700700 00 0 oo oo oo 00 It 27 j j j Fxyzdxdydz jrzdrjsinada jd Fr 9 wiqu 0 0 0 1 i a 2 16 262 6x2 6y2 622 6 quot 62 62 1 a 2 a 1 a a 2 r 2 s1nt9 r 6 quot 94 r s1nt96 9 6 9 r r s1n t9 CHEM 3510 Fall 2007 Unit X Molecular Symmetry and Group Theory A Introduction to Molecular Symmetry l The molecular symmetry properties can be used to a b c d Reduce the highorder secular determinants in Huckel method Determine the IR or Raman activity of Vibrational normal modes Label and designate molecular orbitals Derive selections rules for spectroscopic transitions 2 Symmetry elements and symmetry operations a b The symmetry of a molecule is described by its Each symmetry element has one or more associated with them Symmetry elements Symmetry operations Examples Description Symbol Symbol Description Identity E E No change A Rotation about the axis by C 3 C nFola axis of symmetry Cn Cn 2 3 360n degrees C4 C6 Plane of symmetry 0v 0v 0 0h 6 0h Re ection through a plane axis of symmetry re ection through a plane mirror plane 0d 6d A Re ection through Center of symmetry i i the center of symmetry nFola Rotation about the axis by r rotation re ection A 360n degrees followed by Sn Sn 39 l improper rotation perpendicular to the axis 149 CHEM 3510 Fall 2007 c The aXis with the highest value of n is called d The planes of symmetry can be El 0V the plane of symmetry is to a unique aXis or to a principal aXis El ah the plane of symmetry is to a unique aXis or to a principal aXis El 0d the plane of symmetry bisects the angle between C2 axes that are perpendicular to a principal axis 0 ad is a special type of a UV plane e A symmetry element may have more than one symmetry operation associated with it CI The 3fold aXis of symmetry C3 has two symmetry operations associated with it A 360 0 C3 rotation With 7 120 degrees 0 C32 2 C363 rotation with 240 degrees CI The 4fold aXis of symmetry C4 has three symmetry operations A 360 0 C4 rotation With T 90 degrees 0 Ci 2 C 4639 4 rotation with 180 degrees 0 Ci 2 646464 rotation with 270 degrees 3 Point groups a A group or set of symmetry operations constitutes b Each point group consists of a number of symmetry elements c The total number of symmetry operations is called the CI The total number of symmetry operations can be greater than the total number of symmetry elements 150 CHEM 3510 Fall 2007 d Common point groups of interest to chemists Point group Symmetry elements Molecular examples C2v C3v Td E C2 20V E C3 30V E C2 139 0 E 3C2 139 30V E C3 3C2 0h S3 30V E C4 4C2 139 S4 0h 20V 20d E C6 3C2 139 S6 0h 30V 30d E S4 3C2 20 E 4C3 3C2 3S4 60d CI The number in front of a symmetry element is the number of times such a symmetry element occurs e Some point groups have some common features El El CW gt nfold axis and 11 0V mirror planes Cnh gt nfold axis and a mirror plane perpendicular to the nfold axis D gt nfold axis and n 2fold axes perpendicular to 11fold axis Dnh gt nfold axis and n 2fold axes perpendicular to 11fold axis plus a plane perpendicular to 11fold axis Dnd gt same as Dn plus ndihedral mirror planes 151 CHEM 3510 Fall 2007 f A summary of the shapes corresponding to different point groups CW A pyramid pane or bipyramid Dnd Til szn A 9 CHEM 3510 Fall 2007 g Molecule examples identify all the symmetry elements and the point group that each molecule belongs to H20 XeF4 C6H5Cl f F Fl HZCCCH2 CHzClz C6H6 transCIHCCHC1 c1 Cl H c Cg c l H m 0 H H Benzene BlransrCHCICHOl C2H4 BF3 CH3Cl CH4 C l F C1 a Cg B HH F Ethene CH7CH2 cisClHCCHCl NH3 naphthalene metaC5H4Clz 153 CHEM 3510 Fall 2007 h A ow diagram for determining the point group of a molecule D Start at the top and answer the question posed in each diamond Y yes N no Example benzene Molecule D CHEM 3510 Fall 2007 B Group Theory 1 Introduction 3 The set of symmetry operations of a molecule form a point group b A group is a set of entities A B C that satisfy certain requirements El Combining ie multiplying any 2 members of the group gives a member of the group O A group must be closed under multiplication CI The multiplication must be associative A B C A B C CI The set of entities ie the members of the group contains an identity element E such that EA AE andEBBE El For every entity in the group for example A there is an inverse Ail that is also a member of the group so that AA71A71A E andBB1B1BE 2 Group multiplication table 3 Example of a group of symmetry operations the case of water Cl There are four symmetry operations E C2 6v 6 El These four symmetry operations form the C2v point group El Consider an arbitrary vector u and investigate how each combination of symmetry operations change the vector First Operation 0 Second Operation E C2 OAK 6quot E E C2 Ov 039 v 62 62 E a on u A A A A A 0quot 0quot 0v E C2 6w 639 6W C E b The table above is called of the C2v point group 155 CHEM 3510 Fall 2007 c The four symmetry operations of C 2v satisfy the conditions of being a group and are collectively referred to as the point group C 2V d Example of a group of symmetry operations the case of ammonia El There are siX symmetry operations E C3 C32 6v 6v 0V El These siX symmetry operations form the C 3V point group First Operation A A A 2 A A A Second Operation E C3 C3 0v 0v 0v A A quot A 2 A A A H E E C3 C3 0v 0v 0v A A A 2 A C3 3 C3 E A 2 A 2 A A C3 C3 E C3 639 639 A m q lt q lt m v q m El Fill in the rest of the table 3 The character table of a point group a Symmetry operations can be represented by matrices El Example of H20 or the case of the C2v point group 0 Consider a vector u uxi uyj uzk O For the 180 degree rotation along Z aXis CZ Czux ux Czuy uy C2u2 L12 0 One can write ux ux 1 0 0 6392 My 2C2 My Where C2 0 1 0 HZ HZ 0 0 1 0 Similarly 1 0 0 1 0 0 1 0 0 E010av0 10a39v010 0 0 1 0 0 1 0 0 1 CHEM 3510 Fall 2007 A set of matrices that multiply together in the same manner as a group multiplication table is said to be a These four matrices form a representation of the C 2v point group or more specific a threedimensional representation because it consists of 3gtlt3 matrices There are an infinite number of such representations but some of them called that are called can be used to eXpress all the others El Finding the irreducible representations has been already done for all point groups The case of C 2V point group CI The irreducible representations are denoted A1 A2 BI and BZ 0 Use notation A if the representation is symmetric with respect to principal aXis C2 0 Use notation B if the representation is antisymmetric with respect to principal axis C2 El A1 is the totally symmetric onedimensional irreducible representation CI The irreducible representations of the Cgv point group E C2 0quot 6quot A1 1 1 1 1 A2 1 1 Bi 1 1 1 32 1 1 157 CHEM 3510 9 Fall 2007 The case of C 3V point group CI The irreducible representations of the C 3V point group E C 632 6w 639 6393 41 1 1 1 1 1 1 A2 1 1 1 1 1 E 01 54 r 0 1 r L 2 2 2 2 2 2 2 El Twodimensional irreducible representations are designated by E not the same as the symmetry operation E 0 It can be obtained by applying the symmetry operations to a vector in xy plane 0 Because x and y transform together the result of a given operation is written as a linear combination of x and y CI The x and y are said to form a basis for E or to belong to E Threedimensional irreducible representations are designated by For almost all applications of group theory one do not uses the complete matrices only the sum of the diagonal elements called its trace or in group theory its The characters of the irreducible representations of a point group are displayed in a table called Certain symmetry operations for example 6393 and C32 or 6V 6 and 6 in the C 3V point group are essentially equivalent have the same characters and are said to The number of classes is equal to the number of irreducible representations gt the character tables are squared For the point groups that has a center of symmetry 139 the irreducible representations are labeled as g or u to describe Whether they are symmetric or antisymmetric under the inversion 158 CHEM 3510 Fall 2007 111 Examples of character table for some useful point groups C3 v A1 A2 E xay RxRy Z R z x2 x2 y2 22 y2xy xzyz C2 v A1 A2 B1 32 x Ry y Rx C2h In 9D Cymbal xyzxy xz yz 0 x y 0 RxaRy x2 y2 22 x2 y2xy myz 159 CHEM 3510 Fall 2007 11 Description of character tables CI The second to last column of the character table lists how the three aXis x y and Z or the translation along the three aXis and how the rotation along the three aXis Rx Ry and R2 transform in that particular point group 0 It can also be said that x or y or Z for a basis of a certain irreducible representation 0 For the Cgv point group x forms a basis for the 31 representation y forms a basis for the 32 representation and Z forms a basis for the A1 representation 0 For the C 3V point group x and y form jointly a basis for the two dimensional representation E 0 Example of rotation around the Z aXis in the C 3V point group Depict the rotation around an aXis by vectors 12112 R 9022 RZ C3032 R2 ab Rz C3032 R2 M132 Rz top View The rotation along the Z aXis transforms as the A2 representation CI The last column lists how combinations of the aXis also components of the molecular polarizability x2 y2 xy etc transform in that particular point group 0 When there is no aXis that transform as a twodimensional irreducible representation the combination of the aXis is just the product between the particular aXis 160 CHEM 3510 Fall 2007 4 Mathematical properties derived from character tables a Notations that will be used in the following mathematical relations involving the characters of irreducible representations El R is an arbitrary symmetry operation El gal is the character of the matrix representation of R El zj R is the character of the jth irreducible representation of R The order of a point group can be written as h g d2 J391 J El h order of the group number of symmetry operations El dimension of the jth irreducible representation El N number of irreducible representation Considering that the dimension of the jth irreducible representation equal to the character of the jth irreducible representation for the identity operation N 2 zJE 511 h z Mm j1 The irreducible representations are orthogonal 2 RM j R 0 R z nam ltij R 0 classes 0 where n is the number of symmetry operations in the class containing 2 El Mathematically the product of two vectors is given by n 11 39V 211ka k1 161 CHEM 3510 Fall 2007 El Example A1 and 32 representations of group C 2v are orthogonal ZA1EB2E zAl 62132 62 zAl 6026132 6v zAl 6026132 6 lgtltllgtlt llgtlt llgtltl 0 If i for 5 in equations above is the totally symmetric irreducible representation A1 m1 1 22mm 2 nRzJROifJ A1 R classes For an irreducible representation zj 12 z nltRzj 122 h R classes El Looking at a ndimensional irreducible representation like a vector for which one define the length of the vector as n vv length2 Z v kl El Example A2 representation of C 3V point group 2 n1 5 j 112 IX 12 2 X 12 3 X 12 6 classes Combining the equations from d and f 2251 Rm R z Mm ltij R My R classes Reducing a reducible representation F as a sum of irreducible representations El Assume that zQQ is the character of the symmetry operator in reducible representation F El Write this as a combination of the characters of irreducible representations Z a zj J 162 CHEM 3510 Fall 2007 El Find the coef cients a multiply by zl R and sum over R EmmiR ZaJZzi RmR R j R 0 only if 139 j and considering z Zl RMJ R My R 2 aih mamR R 2 al izm m znmmRmR R classes i Example Reduce the reducible representation F 4 2 0 2 as a sum of irreducible representations belonging to Cgv group zE4 M52 2 my 0 2462 aAl any12x1ox12x12 aAz 4x12x10x 12x 11 aBI in 142 1ox12x 11 o aBZ 4x12x 10x 12x11 3F2Al 142 32 El Verification 2A1 2x1 2x1 2x1 2gtlt1 A2 1 1 1 1 B 1 1 1 1 F 4 2 0 2 163 CHEM 3510 Fall 2007 C Applications of Molecular Symmetry l Huckel theory for benzene a When using the pz orbitals on various C atoms Huckel theory leads to x10001 1x1000 01x1000 001x10 0001xl 10001x b Instead of using atomic orbitals use symmetry orbitals that are obtained as a linear combination of the pz orbitals l 1 E011 H12 lI3 lI4 lI5 lI6 l 2 E011 112 lI3 114 lI5 116 1 3 EQU1 HI2 113 2114 115 lI6 1 l 21I lI l 21 l 4 JG 1 2 3 4 5 6 l 5 E2W1 112 113 2114 115 116 l 1 E W1212 113 114 2115 116 c Using these orbitals leads to the following determinant x 2 0 0 0 0 0 0 x 2 0 0 0 0 0 0 x 1 x 0 0 0 ox 1 10 00 2 0 0 0 0x 11 x 2 0 0 0 01 xx l 2 164 CHEM 3510 Fall 2007 El This determinant is easier to solve and the solutions are x ilili2 CI The construction of those symmetry orbitals takes advantage of the symmetry of the system 2 The use of symmetry to predict the elements of the secular determinant that are zero a b Look at HU j 51de and Slj j f jdz These integrals should be independent on the molecule orientation the value should not change upon applying a symmetry operation 2 RSIj jR fR jdz Si 2 RHU jR fR1 IR jdr Hi The overlap integrals El Consider i is a basis for one irreducible representation Fa El Consider j is a basis for one irreducible representation H 2 M h ltwa and 1 m ltij 2 S1 a Jamar 3de h mumJ 0 So the za 1 1 product should be equal to 1 for every symmetry operation If the za R b R is equal to l gt S should be zero for the equation above to be true El Slj i 0 only if i and are bases of the same irreducible representation El Slj 0 only if i and are bases of different irreducible representations The exchange integrals CI The molecular Hamiltonian operator is symmetric under all symmetry operations 2 RHU 1wa RHR jdr a Rm11 Rm Rwy HU 165 CHEM 3510 Fall 2007 0 So za R 5 A1 1 R should be equal to 1 for all symmetry operations R in order for the Hi terms not to be zero The condition is similar as the one obtained for Si El Only the elements of the determinant that are bases of the same irreducible representation are not zero e Example H20 CI The 2px orbital of O transform as Bl CI The Zpy orbital of O transform as BZ CI The 2172 orbital of O transform as A1 El Consider a linear combination of 1s orbitals lsH A lsHB that transform as A1 El So the overlap integral of 2px orbital of oxygen with lsH A lsHB is zero because they transform differently 3 The generating operator a The symmetry orbitals are obtained using the generating operator defined as A 011 A A PJ 2751390012 R b The generating operators are used to find linear combination of atomic orbitals that are bases for irreducible representations 166 CHEM 3510 Fall 2007 D Unit Review 1 Important Terminology symmetry element symmetry operation nfold axis of symmetry plane of symmetry center of symmetry nfold rotation re ection aXis of symmetry principal aXis point group the order of the point group Cs C2v C3v C2h D2h D3h D4h D6h D2d Td group multiplication table irreducible representation reducible representation 167 CHEM 3510 Fall 2003 Bonding in Polyatomic Molecules Interpretation of Molecular Shape Using Hybrid Orbital Idea 7 Recall that carbon atom with the con guration 1s2 252 2 p16 2p may be expected to form only 2 bonds The idea of hybrid orbital was introduced in general chemistry to explain why C is tetravalent an electron is promoted from the 25 orbital into the 2 pz orbital and the four singly occupied orbitals combines to form four equivalent Sp 3 hybrid orbitals 7 Example BeH2 molecule 7 two BeH bonds that are equivalent 7 H7Be7H angle is 180 7 Beryllium electronic con guration 152252 ISO term symbol 7 The 25 and 2 pz orbitals combine to from twp Sp hybrid orbitals atomic orbitals 1 W5 EQVIZPZ 7 The molecular orbitals are formed as a Gbonding between one Sp hybrid orbital of Be and the s orbitals of one Hatom WBe7H CiWBeas t CzWBeas t CsWHas 7 Example BH3 molecule 7 three BH bonds that are equivalent 7 H7B7H angles are 120 7 The appropriate hybrid orbitals orbitals with proper orientation will be constructed by combining the 25 orbital and two 2p orbitals resulting in three sz hybrid orbitals F5111 2003 CHEM 3510 rThe orthonormal nonnahzed and onhogona1 5p2 hybnd orbxtals are 1 2 V17252Pz 3 1 1 1 121 239 V 725772 72 2 4 45 1 1 1 V 725772p1 772m rThe Omelecular orbxtal 15 formed as almear combmanon between one hybnd orbxtal on B and the 15 orbnals of one H atom eExannple CH4molecule efour CeH bonds that are equ1va1ent rHrCrH angles are 109 5 eFour 513 hybnd orbxtals are obta1neo1 eonnb1n1ng the 25 orbxtal andthree 2 p orbxtals of c The orthonormal 513 hybnd orbxtals 5n 1 w E25 2px 21 2px V2 l25 2px 21P 2p1 V3 252px e 21 Zpl w 2572px 72 e mi 57F bonds that are equlvalent 2 1 E 1 E eExannp1e 516 mo1ee111e eFeseF angles are 90 0139180 eThe geometry 15 octahedral eThe goundetate electron con gurauon of s 15 b1ta1 and the three 3p orbxtals of Sto 51 Ne13523p4 e Combme two 311 orbxtals w1th the 35 or create 6 equ1va1ent dzspzhybndorbxt 5 7An 513112 hybnduanon g1ve5 51m1151ne5u1t5 CHEM 3510 Fall 2003 7 Example H20 molecule 7 two OH bonds that are equivalent 7 H7O7H angle is 10450 7 This geometry is described by the hybrid orbitals l1045230712py 0552pz W2 045 23 07l2py 055 2pz that have 1045o between them HOH 7 These hybrid orbitals are intermediate between pure p and Sp3 hybrid orbitals 7 The other hybrid orbitals will contain the lone pairs of the oxygen 7 All of those hybrid orbitals are orthonormal 7 The s and p character of hybrid orbitals 7 Given a hybrid orbital whose wave function is a combination of s and p atomic orbitals written as 11 cmns cnpx npx cnpy npy can npz 2 the 3 character of the orbital is cm and thep character is 02 02 02 quotP npy quotP2 7 Example the hybrid orbitals in H20 have 0452 2 020 3 character and 05522 07122 080 p character 2 7 Example the hybrid orbitals in BH3 have 033 3 character and 2 LT L HE J NE 7 A hybrid orbital that is x 3 character and 100 x p character can be said to be an spaooiw x hybrid orbital 2 l 067 p character 7 Example the hybrid orbitals in H20 are Sp4 hybrid orbitals CHEM 3510 Fall 2003 Molecular Orbital Theory for Triatomics 7 MO theory explains why BeH2 molecule is linear and H20 molecule is bent 7 The number of valence electrons on the central atom Be 2 electrons O 6 electrons 7 Obtain the molecular orbitals as a combinations of atomic orbitals on Be or O and H through a LCAOMO procedure 7 Molecular orbitals of BeH2 Be H7Be7H H 7 The con guration is 2c7g2 16 2 7 Molecular orbitals of P120 7 The con guration is 2610201722 3a1 21b12 7 The labels for a bent molecule a1 a2 b1 b2 re ect the symmetry of the molecule CHEM 3510 Fall 2003 7 Create a Walsh correlation diagram a plot of the energy of a molecular orbital as a function of a change in molecular geometry for a AH2 triatomic 515 2 as m x 7339 7215 w r m Er y M 139 ii 1quot 51 E 25 HT J 3a 4 r y i quot391 Ell L quota A m 1 u 391 15 a W mm hum F B r 1 Lu 1 I iii J bi 7 I aquot quot 77 36 quoti 339 l r L a I l 2 3 Rafa unm 45 M A h 1dquot 2 in rr in End 113 r 911quot H 5H Eli 530 10an angle mml angle 7 Whether a molecule is bent or linear depends on what energy is lower 7 Bending destabilizes 26g lau3c7g 26u orbitals 7 Bending stabilizes lnu orbital 7 BeH2 2cg 2 lau 2 con guration is more stable lower in energy than 2a12lbz2 con guration 3 The BeH2 molecule is linear 7 H20 2cg 2 16 2 Mu4 con guration is less stable higher in energy than 2a1 2 1b2 2 3a1 2 1bl 2 con guration 3 The H20 molecule is bent 7 The lnu orbital get stabilized more than the 20g and 10 gets destabilized at angles close to 180 7 The exact angle of bending for example 10450 for H20 can be determined using more complicated computational techniques 7 A Walsh correlation diagram for a XY2 triatomic X and Y are second row elements can be used to determine if C02 N02 N03 CF2 etc are linear or bent CHEM 3510 Fall 2003 Hiickel Molecular Orbital Theory Hiickel Theory 7 It is a method applied mainly to conjugated and aromatic hydrocarbons 7 It uses 727 electron approximation 7 The Sp2 hybrid orbitals of C and s orbitals of H create a 0 bonal framework The 0bond framework is in xy plane the pz orbitals are perpendicular to the plane and form 727 bonds One can say that 727 electrons are moving in some fixed effective electrostatic potential due to the electrons in the 7 framework This is the 727electron approximation 3 The problem of 727 electrons delocalized MO occupied by 727 electrons can be treated separately from the problem of 7 electrons 7 Example Ethene CH2CH2 7 The wave function for 727 orbital in ethene W7 C12PzA 62 2117213 7 Solving for 01 and 02 results in the secular determinant Ell ES11 H12 ES12 12 ES12 H22 Eszz where H y are integrals involving an effective Hamiltonian operator that includes interaction of 727 electrons with 7 electrons and S y are overlap integrals between 2 pz atomic orbitals 7 The H11 and H22 are Coulomb integrals H11 H22 in ethene while H12 is a resonance integral or exchange integral 7 Huckel assumptions 1 if i j Sy39 6y 0 If i i j H ii are assumed to be same and denoted at H B if iis bound to j same value 8 for each pair of neighbours if 0 if iisnotbound toj 7 Rewrite the secular determinant a E l3 0 l3 06 3 Eoci CHEM 3510 Fall 2003 7 To get the energy one must have 06 and Instead of calculating those values one can determine their values from comparison with experiment By doing so 3 B 75 kJmol 7 Represent the energies of the 727 orbitals considering 06 as reference 06 is the energy of an electron into a 2 pz orbital 05 7 The energy of the 727 electrons total 727 electronic energy E 2a 3207 223 7 Solve for 01 and 02 to get the orthonormal wave functions 7Conditions are 012 0 and 012 0 1 1 W1 52p2A 32p23 l 1 171 32Pm EZPZB 8 8 7 Example Butadiene CH2CH7CHCH2 7 Consider that the molecule is linear although in reality it is not 7 The wave functions for 727 orbitals 4 Vi Z Cy 2Pzj jl 7 Get to secular determinant H11 ES11 H12 ES12 H13 ES13 H14 ES14 2 3 4 12 ES12 H22 E322 H23 E323 H24 E324 0 H13 ES13 H23 E323 H33 E333 H34 ES34 14 ES14 H24 E324 H34 ES34 H44 ES44 But Hit07 HIV 3 5 CHEM 3510 Fall 2003 11451273 trams an lj3 a161213 Tne energy of the 7 e1eetrons total 7 e1eetronre energy E 2a16183 2a 0 618134a4 47213 Tne delocahzauon energy gryes tne re1anye stabxlxzauon ofthe molecule EMc En c4118 ZEnC2H4 0 47213 e35kJrno1 Solve for 5y to get the orthonormal wave funcuons 11 0 3717 2p 0 6015 2px 0 6015 2px 0 3717 2pr w 0 6015 2p 0 3717 2px 0 3717 2px 0 6015 2pr 7quot M w 0 6015 2p 0 3717 2px 0 3717 2px 0 6015 2pr w 0 3717 21 70 6015 2px 0 6015 2px 0 3717 2pr 009 I t tntt CHEM 3510 Fall 2003 eEnarnple Benzene I 3 C e eIhe cbonds between carbon atoms g m the nng and between the carbon J J J atoms and the hydrogen atoms J form the r framework ofbenzene arE ewrrte the secular detennrnant usmg the notanon re x 1 0 0 1 x eh wxzezzw eThere are 6 solunons x eThe energy of the 7 e1eetrons En 2a234a 6a83 eThe delocalxzauon energy EM En c6116 7 35 c2114 2 n 7150 kJmol e The 7 wave funeuons for benzene Mp 2p CHEM 3510 Fall 2007 Unit IV Applications of Quantum Theory A The Particle in a Box 1 Introduction a The particleina box refers to the quantum mechanical treatment of b One tries to solve the Schrodinger equation to obtain the wavefunctions and the allowed energies for the particle c The timeindependent threedimensional Schrodinger equation iiijVxyz yxyz Eyxyz 2 6x2 6y2 622 2 Particle in a onedimensional box or onedimensional motion a Consider only a onedimensional motion it along x coordinate 3 drop y and z in the equation above b Consider that the particle experiences no potential energy between the position 0 anda Vx0for0SxSa c The Schrodinger equation for this problem 0 2 2 3 Z mic ZIZEV Wall Wall Potential energy d The mathematical solution of this equation is yx AcoskxBsinkx where k 2mE1 2 27239 2mE h 2 2 3 E h k 872392m CHEM 3510 Fall 2007 e Apply the boundary conditions and the normalization condition to determine A and B qO 0 3 A 0 ya 0 3 Bsinka 0 3 ka 147239 where n 12 f The allowed energy levels are 2 2 h 14 3E 2 wherenl2 N ma 100 Classically where 14 1s called allowed 9 energies E The energy is 8 El The energy levels increase as h28m a2 D The energy separation between the allowed 3 E energy levels increases D The energy levels and the energy separation between the energy levels increases as g Quantum numbers appear naturally when the boundary conditions are put in the Schrodinger equation like in the string problem and are not introduced ad hoc like in Plank model for blackbody radiation or Bohr model of hydrogen atom h Determining the wavefunctions E Find B by setting the condition that the function 1x is normalized I 1 xwxdx 1 0 a 3B2jsin2 l3Bz l3BZ 0 a 2 a o B is called CHEM 3510 Fall 2007 D The normalized eigenfunctions x are given by l 2 2 Lynx s1n whereOSxSaandn12 a a 543 2 1 i Solving the Schrodinger equation for the particle in a box problem gives a set of allowed energies or eigenvalues and a set of wave functions or eigenfunctions j Representations including the energies and the wave functions a and the probability densities bc for the first few levels for the particle in a box N E 5 11mquot 1 WW 5m 3 l r 9 n2 rquot H 3mm L 1 b NMquot 515 I MM 3 mm 39 CHEM3510 k El F Fall 2007 I39 40 i rr T p 39 b d can b freely in a molecule also called Example Butadiene has an adsorption band at 451x10 cmquot As a simple approximation consider butadiene as being a onedimensional box oflength 578 A 578 pm and consider the four 7 electrons to occupy the levels calculated using the particle in a box model B o The electronic excitation is given by AE h 2 32 e 22 Smea o The calculated excitation energy i 3 i 454x104 cm 1 115 compares Very well With the experimental Value quot o This simple can be quite successful The probability of nding the particle between x1 andxz is given by x2 I mom mix I1 El Ifx10andxza2then Prob 03x SaZ12 for alln El Forn 1 Prob 03x 3114 ltProb 114 Sx 3112 D As It increases for example It 20 these 2 probabilities become equal Generalizing the probability density becomes uniform as This is an illustration ofthe that says that quantum mechanics results and classical mechanics results tend to agree in the limit of large quantum numbers D The largequantumnumber limit is called CHEM 3510 Fall 2007 p The eigenfunctions are ill11xlnxdx 0 Where m i n 0 El Example look at ylx and 113x a a 0 J39yfxl3xclx EJ39sin sin3 xabc 0 0 a0 a a q Average values and variances for the position and the momentum of the particle a 2 a mzx a 7 Average value ofposition x JynOc x yn xdx chsin2 dx 0 a 0 a 2 a 2 2 2 2 a a 7 Average value ofpos1t10n square x gt 11 x x z1n xdx 0 3 271272392 2 2 2 a 2 2712 7 Variance in position ax x x 2 2n7239 3 2 2 12 7 Standard deviation in position o x 1x2 x2 2a 3quot 2 717239 27m a nmc mac 7 Average value ofmomentum p ih 2Jsin cos dx 0 a 0 a 2 2 2 2 2 2 2 2 2 h h 7Also p2 2mE 2mquot 2quot 2 a 2ma a a nznz 3 12 7 Finally axap 2 gt g Heisenberg Uncertainty Principle 3 Particle in a twodimensional box or twodimensional motion a The Schrodinger equation for this problem 2 2 2 3 h Mira W Ewxy 2 6x2 6y2 Potential a b Allowed energies or eigenvalues h2 n 7 quotxny g a2 b Z Particle con ned to surlace CHIEM3510 Fall 2007 b The Wavefunctions or eigenfunctions my XXYy sin quotW San quotW a b a b 4 Particle in a threedimensional box or threedimensional motion a 57 o P The Schrodinger equation 2 2 2 2 L 6W6V6W EMWJ 2m a 62 ay2 622 Solutions r 2 2 2nx72xquoty71ynznz x z 7 7 ism s1n s1n V nxnynz y 7 le a b C 2 112 n3 quotiy r122 8m 2 b2 72 Average Values ofthe position and the momentum ofthe particle m deEfdygfdzw xyz1igxxyz ijk where 1quot 26 191 2k 3 a b ltpgt jdxjdyjdzgx xyziwxyz o where i ixhiij I I I K ax i 32 I Dogmenach mil The case ofa cubic box a b c D Three sew of quantum numbers give same energy t39 6h2 32113121E112 2 quot 8ma llllllll l r Illl l CHEM 3510 Fall 2007 6h2 CI The energy level E 1s 8ma 3h2 CI The energy level E111 2 is 8ma El Degeneracy is equal to 0 Once the symmetry is destroyed then degeneracy is lifted 5 Separation of variables a The Hamiltonian for the particle in a 3dimensional box 2 2 2 2 192 aaa Hx my 192 21 5x2 ay2 622 El It can be written as a sum of terms where 2 2 2 2 2 2 A h a A h a A h a H H x Z Qj y 2may23 2maZ2 o The operator is said to be b The eigenfunctions llnxnynz are written as a product of eigenfunctions of each operator H x H and H Z and the eigenvalues Enxnynz are ya written as a sum of the eigenvalues of each of the operator H x H y and H Z c This is a general property in quantum mechanics If the Hamiltonian or an operator in general can be written as a sum of terms involving different coordinates ie the Hamiltonian is separable then the eigenfunctions of H is a product of the eigenfunctions of each operator constituting the sum and the eigenvalues of H is a sum of eigenvalues of each operator constituting the sum d Example H H1s192w 2 wnmo w sww and Enm En Em where 1910mm Emma and 192wwmw Emumm cHEM3510 Fall 2007 B The Harmonic Oscillator 1 c Introduction I cm s u eaunent of in classical mechani 39 b The harmonic oscillator cs eForee f 7kg 71 1m Hooke39s Law Where kis thefmce cansmnt d2 7 Hooke39 s Law eornbined witn Newton39 s equation m 2Xm 0 t It rThe solution ofthis equation x0 Aeos out Where a m rThe potential energy ofthe oseiuator m 2 1 dz 2 rThe kineae energy of the oseiiiator K Em d t rThe total energy of the oseiiiator E VK kAz gtThe total energy is eonseryed itis transferred between Kand V The harmonic motion in a diatomic molecule 7 Consider the movement of atoms ofrnasses mi andmi 2 kx2nxi ln dzx 2 m2 nkOrKrln 0112 eBy summing the two equation above 2 d X 0 Where Mm m2 and X mmm1 1 0112 M where Xis the eentet tne W55 euutdinate e Subtraeting equation 2 diyided by mzfrom equation 1 dividedmi 2 n Hq 0 where xx2 7x it and n 0112 ml n2 wnere nis the tediieed mass ofthe system and x is the tezanye euutdinate D The movement ofa twobody system can be reduced to the movement of a onebody system with a mass equal to the ofthe twobody system CHEM 3510 Fall 2007 2 The quantummechanical harmonic oscillator a The Schrodinger equation for quantummechanical harmonic oscillator 2 2 h d Z Vxwx Ellx 2 dx where Vx kx2 Potential V 2 3 d wx2 dxz h2 The eigenvalues are En h nlhmnl hvnl u 2 2 2 where n012 a E Vi E 2 1 The wave functions eigenfunctions are E ka Jyx 0 Dispiacement x k 2 zjnx Aana12xeiwc 2 where a 2 E 14 1 a D The normallzation constant 1s glven by An 2 7 The wave functions form D D The H quot011 2x are polynomial functions called Hermite polynomials where H quotg in a n3911 degree polynomial in 5 0 Here are the first few Hermite polynomials Ho51 H1E2 H2lt54272 H3E8 3712 H4E16 4748 212 H5E32 57160 3120 even polynomials x ix odd polynomials x i ix D Continuous odd functions properties f 0 0 and I f xdx 0 7A

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