Classify – Group Theory - University of Illinois at Chicago

1、Classify Group Theory Gxyz Ä Gatoms = G3N ¬ count atoms NOT momentumvibrations G3N-6 = G3N - Gtrans - Grot as character: c3N = cxyz · catom then reduce to get linear combination invedCan categorize subspacesstretchescstr = å bonds that do not ?bends, etcetc.again reduce idea thes

2、e pretty different energiesbut these may or may not span the spacemust pick carefully include all motionGroup Theory provides test do you get all representationAlternate may use Projection OperatorpG(r) = åci ri will give linear combination of equivalent “r, could be bends, ?Example 2:go on toC

3、H4TdTdE8C33C26S46sdA111111A2111-1-1E2-1200T130-11-1RxRyRzT230-1-11x y zAgain c3N = cxyz · catoms cxyz30-1-11catom52113c3N150-1-13-(ctrans + crot)60-200c3N-6901-13Reduce c3N-69A1 = 1 9 1 + 8 0 1 + 3 1 1 + 6 1 (1) + 6 3 1 = = 19A2 = 1 9 1 + 3 1 1 + 6 -1 -1 + 6 3 -1 = 09E = 2 9 1 + 3 2 1 = = 19T1

4、= 3 9 + 3 -1 1 + 6 1 -1 + 6 -1 3 = 09T2 = 3 9 + 3 -1 1 + 6 -1 -1 + 6 1 3 = = 2thus G3N-6 = A1 + E + 2T2see how 1 + 2 + 2 3 = coordinatesless obvious:NH3:cN-H = 3 0 1reduce: A1 + EcHNH = 3 0 1(Note reflection in place bisectreduce: A1 + Eangle gives +1)recallc3N-6 = 2A1 + 2E2 - 1D + 2 - 2DÞ 6 di

5、mensional3N - 6 = 12 - 6 = 6so these 3NH + 3HNH span the spacebit harderCH4 see attached Handoutc3N-6 = A1 + E + 2T21D + 2D + 2 3DÞ 9 dimensional3N - 6 = 15 - 6 = 9Now could choose C-H str 4H-C-H bend 6 10 problemsince more internal coordinate then 3N-6 these cannot be all independentcC-H = 4 1

6、 0 0 2implies 2A1 + E + 2T2reduces to A1 + T2get one too many A1 coordinatesone is not independent or in thiscHCH = 6 0 2 0 2cHCH = a12 + a13 + a21 + a23 + a24 + a34 = 0reduces to A1 + E + T2cant all open at onceNow that we have a way of getting at a system of coordinates we must look at how to use

7、themVibrations of polyatomics solve 3N-dimensional (R) TN + Ukk (R) cu (R) = Eu cu (R)now only interested in relative motioncan remove C of M + rotation degree freedomget 3N-6 independent coordinate but express as function of Rs stillNormally express as Cartesian displacementcoordinate ® deriva

8、tion from equilibrium in rotating framed1 = Dx1, d2 = Dy1 d3N = D = zNfor vibration problem mass weighted Cartesian displacement coordinate easierq1 = m1½ Dx1, q2 = m1½ Dy2 q3n = mN½ DzNClassically:Potential normally done in Harmonic Approximation (same as diatomic, more coordinates)N

9、ow same as for diatomic: Ue constant / just shift potential E for minimum1st non-zero / non-constant term is quadratic (qi qj)but of course there are more anharmonic termsIf keep just this and TN: This is coupled multidimensional cant separate as writtenTN is diagonal: q = TN = TN = (direct product

10、?)In this form:VN = VN = ½ q1 q2 qN goal change coordinatesT = V = both terms diagonal, to span spaceQi = need transformation L ® diagonal matrix, li on diagonalwriting trans form: = matrix of eigenvectors of , L-1 = LTli = eigenvalues of secular determinant:solve det ( - djk lm) = 0U¢

11、;¢ ® 3N x 3N, 3N lm values but 6 ® zeroplug lm into secular equations:(Ujk - djk lm) km = 0 = km Qi = km qkand inverse qj = å jk Qkor q = QQ = T qPut it all together2Vvib = = (LQ)T U¢¢ (LQ) = QT LT U¢¢ L Q= QT L Q or 2V = å lm Qk2 ® diagonalsame idea

12、:2T = ® diagonal (LT L = 1)can separate solve one coordinate at a time (H = )Classical: F = ma = = = -l Qwave equation: + = 0Þ Qk = Bk sin (l½ t + bk)Quantum MechanicsH = TN + VN= ½ å Qk2 + ½ å lk Qk2 = each one is a 1-D harmonic oscillator problemKnow solution:Hvi

13、b = hkhk ck = Ek ckEvib = (uk + ½) h nkEk = (uk + ½) h nkYvib = cuk (Qk)ck = Nke Huk (ak½ Qk)ak = Note: cant simply write k,m nowrecall: summed H ® product w/f® Total energy sum independent vibrational energiesNote zero potential E, ½ h uk ® non ?® Product fun

14、ction makes determinant easieruse Group TheoryGyvib = Gcukso need know representation of each vibration (keep doing that) and take product ® representation of full w/flook at what changes ® unchanged no contributionsSelection rules IRhow determine?expand: m = me + Qk + + constant but vecto

15、r leads to DJ = ±1,0 rotationeg: This term only non-zero pure rotation, orientation is independent2nd term ® vibrational excitestill orientation effect ® DJ = ±1,0vibration (harmonic oscillator) ® Duk = ±1but only uk change Duj = 0 j ¹ kand dipole moment must chang

16、e along coordinate Qkto do this Qk and m must have same symmetryGroup Theory language: Gm Ì GQk Ì Gxso look in tablerepresentations for x,y,z and vibrational IR allowed (assume c¢¢ = u = ?)Raman Spectra selectiona = ae + Qk + Qjsame ideaae ® pure rotation, transform as x2, y

17、2, z2 xy, yz, zx, DJ = 0, ±1, ±2 Þ polarizability must change to see vibrational transitionDuk = ±1, Duj = 0 ® exact same ® Gvib Ì Ga Ì Gx2,y2,z2,xy, yz,xz® see Character TableHarmonic ApproximationRotation effects see Handout Banwell depends on symmetry1

18、1 vibrations ® stretch along axisC¥u ® A1 (å+)å Þ M = 0, in terms of angular momentumD¥u ® A1U (åu+), A1gDu = ±1, DJ = ±1IRJust like diatomicDJ = 0 possibleDu = ±1, DJ = 0Ramandue to K = 0if electron angular momentum vibrations ® disto

19、rt molecule from linear (bend)C¥u - E1 (x,y) ; D¥h - E1u (Pu) ® IR allowedC¥u - E1, E2 ; D¥h - E1g,2g (Pg, Dg) ® Raman allowedDu = ±1, DJ = 0, ±1 IRP,Q,R branchesDu = ±1, DJ = 0, ±1, ±2 RamanO,P,Q,R,S branchesIsotopes spin of nuclei total w/f fe

20、rmion asymmetry (-1)get intensity alteration: J even, oddexchange symmetry: bosom symmetry (+1)Note pure rotation, this would only be Raman vibration / rotation see change symmetry but population effect remainsSpherical top moleculesA1 ® not allowed ® IRTotally symmetric modes: Du = ±

21、1, DJ = 0, ±1, ±2Asymmetric modes (T2)Du = ±1, DJ = 0, ±1; DJ = 0, ±1, ±2Sort of like diatomic but degeneracy in K = (2J + 1)ADD Infra-red spectroscopyADD Banwell-Fund. MoleculeADD Infra-red spectroscopySummaryIR selection rules ¹ 0DuK = ±1, DuJ = 0 j ¹ K

22、 ¹ 0DJ = 0, ±1 DM = 0, ±1DK = 0 ¹ 0GQK Ì Gm = GxyzRaman same except:DJ = 0, ±1, ±2 since operator Y2±1¹2GQK Ì Ga = Gx2,y2,z2,xz,yz,xzIR dipole moment change / Raman polarizability changecenter of symmetry IR/Raman u + g exclusiveLinear A1 modes (E) I

23、RP,R branch, DJ ¹ 0Note: D¥h no IR for symmetry stretches / need symmetry A¢¢1uRaman can have DJ = 0, ±2q,Q,S branchE modes (P) IR P,Q,RDJ = ±1, 0Assume start ground state u = 0 Þ c0 = total symmetry / if higher temperaturecan start u = 1 on higher Þ hot bend

24、still Du ¹ 0Isotopes ® if center of symmetry, i, then spin ½ asymmetryget alternating intensity J odd, evendue to population ½ + ½ = 0Raman Polarization 2 photon ® can measure scalar or ll to excitationSpherical topà = ® polarized ® total symmetryTotal sy

25、mmetryA1 mode Raman Du = ±1DJ = 0, ±1, ±2IR not allowed (xyz T)Asymmetry MoleculesT2Du = ±1DJ = 0, ±1 IRDJ = 0, ±1, ±2 Ramanlike linear but DK = 0,K-degeneracy (2J + 1) affect intensityADDADDADDADDSymmetrical Topsparallel vibration ® Gvib = Gz Du = ±1, DJ = 0,±1, DK = 0note this is same