Classify – Group Theory



C'. Review: Classify Vibrational Normal Modes – Group Theory Γxyz ( Γatoms = Γ3N

Vibrations (correct for translation and rotation) Γ3N-6 = Γ3N - Γtrans - Γrot

as character: χ3N = χxyz ( χatom then reduce to get linear combination Γ3N = [pic]

To make more intuitive, separate motions, can categorize into subspaces

stretches χstr = ( equivalent bonds that do not move under operation R

bends, etc best for M-H motions, very different from bend or heavy atoms, etc.

but these may or may not span the space -- must pick carefully – to include all motion

Group Theory provides test – do you get all representation?

Alternate may use Projection Operator

pΓ(r) = (ci ri will give linear combination of equivalent “r”, could be any internal motion

Example 3: go on to CH4–Td -- did last week, just look again for method

[pic]

|Td |E |8C3 |3C2 |6S4 |6σd | |

Again χ3N = χxyz ( χatoms and χ3N-6 = χ3N -(χtrans + χrot) --> (χtrans ~χxyz & χrot ~χRxRyRz)

Reduce χ3N-6 = A1 + E + 2T2 1D + 2D + 2 • 3D ( 9 dimensional : 3N - 6 = 15 - 6 = 9

Now could choose 4 ( C-H str) + 6 ( H-C-H bend) = 10 – problem

since more internal coordinate than 3N-6, these cannot be all independent

χC-H = 4 1 0 0 2 implies 2A1 + E + 2T2 --> get one too many A1 coordinates

reduces to A1 + T2

one is not independent or in this

χHCH = 6 0 2 0 2 χHCH = α12 + α13 + α21 + α23 + α24 + α34 = 0

reduces to A1 + E + T2 can’t open all H-C-H angles at once

D. How to use the system of coordinates derived from Group Theory -- back to QM

Vibrations of polyatomics – solve 3N-dimensional Hamiltonian over Rα (nuclear coordinates)

[ TN + Ukk (Rα) ] χυ (Rα) = Eυ χυ (Rα)

now only interested in relative (or internal) motion

can remove C of M + rotation degree freedom

get 3N-6 independent coordinate but express as function of Rα’s still

Normally express as Cartesian displacement coordinate δi

( derivation from equilibrium in rotating frame

δ1 = Δx1, δ2 = Δy1 … δ3N = ΔzN

for vibration problem: mass weighted Cartesian displacement coordinate easier

q1 = m1½ Δx1, q2 = m1½ Δy2 … q3n = mN½ ΔzN

Classically: [pic]

Potential normally done in Harmonic Approximation by expansion of the nuclear potential in the displacement coordinates (same method as diatomic, but now more coordinates, 3N)

[pic]

Now same as for diatomic: Ue = constant / just shift potential E

And [pic] for minimum

1st non-zero / non-constant term is quadratic

[pic] = Vharm

but of course there are more – anharmonic terms (3rd and 4th order expansion terms)

If keep just Vharm and TN: [pic]

This is a coupled equation – multidimensional – can’t separate coordinates as written

TN is diagonal: q = [pic] TN = [pic]

TN = [pic] (direct product -- insert identity matrix - 1's on diagonal - between q's)

The potential in this form is not diagonal:

VN = [pic] VN = ½ [q1 q2 … qN ][pic][pic]

goal – find change of coordinates T = [pic]

to make both terms diagonal V = [pic]

also must span space Qi = [pic]

need transformation L [pic] ( result diagonal matrix, λi on diagonal

unitary transformation:

[pic] = matrix of eigenvectors of [pic] , L-1 = LT

λi = eigenvalues of [pic]

Method: solve secular determinant for λm:

det ([pic] - δjk λm) = 0 U(( ( 3N x 3N,

3N λm values but 6 ( zero (no potential trans and rot)

then plug λm into secular equations:

[pic](Ujk - δjk λm) ℓkm = 0 [pic] = { ℓkm }

Qi = [pic]ℓkm qk and inverse: qj = ( ℓjk Qk or matrix q = [pic] Q and Q = [pic]T q

Put it all together

2Vvib = [pic] = (LQ)T U(( (LQ) = QT LT U(( L Q

= QT ( Q or 2V = ( λm Qk2 ( diagonal

same idea:

2T = [pic] ( diagonal (LT L = 1 drops out)

can separate – then solve one coordinate at a time (i.e. H = T + V = [pic])

Classical: F = ma = [pic] = [pic] = -λ Q -- take derivative of V with one Qk

wave equation: [pic] + λQ = 0 ( Qk = Bk sin (λ½ t + bk) -- plug it in, test!

Quantum Mechanics

H = TN + VN

= ½ ( [pic] + ½ ( λk Qk2

= [pic] = [pic]hk

each hk is a 1-D harmonic oscillator Hamiltonian operator

Know solution: Hvib = [pic]hk hk χk = Ek χk

Evib = [pic](υk + ½) h νk Ek = (υk + ½) h νk

Ψvib = [pic] χυk (Qk) χk = Nke [pic] Hυk (αk½ Qk)

[Note: for frequency νk, cannot simply write k,μ ] αk = [pic]

[ but qualitatively same concepts hold ]

recall: ( summed H ( product w/f

( Total energy sum independent vibrational energies

{Note zero potential E = ½ [pic]h υk ( non trivial}

( Product functional form for vib wave/fct makes determination of [pic] easier

use Group Theory Γψvib = [pic]Γχυk

so need know representation of each vibration (you have been doing that above) and

take product ( representation of full wave/fct

For a transition only need to look at what changes, i.e. which vibrations are excited

( parts unchanged – no contributions to integral—just bring it out

if J and K are only changes: =

E. Selection rules: IR [pic] how determine?

expand: μ = μe + [pic] Qk + [pic] + …

constant but vector leads to ΔJ = (1,0 rotation

eg: [pic][pic]

This term only non-zero pure rotation, orientation is important part of μ

2nd term ( vibrational excitation --separate electronic and vibrational functions:

[pic]

a) still has an orientation effect ( rotation change with vibration: ( ΔJ = (1,0

b) vibration change only one quantum (harmonic oscillator restriction) ( Δυk = (1

but only υk change, i.e. ( Δυj = 0 j ( k

c) and [pic] i.e. the dipole moment must change along coordinate Qk

to do this Qk and μ must have same symmetry

Group Theory language: [pic] [pic] Γμ ( ΓQk ( Γx,y,z

so look in table for x,y,z representations

those are vibrational IR allowed (assume ground: χ(( --> υ = 0)

Raman Spectra selection α = αe + [pic] Qk + [pic] Qj

same idea: Constant αe ( pure rotation, transform as x2, y2, z2, xy, yz, zx ( ΔJ = 0, (1, (2

Selection rule: Δυk = (1, Δυj = 0 ( exact same integral for Qk : [pic]

next: [pic] ( polarizability must change to see vibrational transition

( Γvib ( Γα ( Γx2,y2,z2,xy, yz,xz ( see Character Table

F. Harmonic Approximation

Rotation effects – see Handout – Banwell depends on symmetry

Linear Molecules – (review) – McHale, Ch 9 & 10.6

11 vibrations ( stretch along axis

C(( ( A1 ((+) ( ( M = 0, in terms of angular momentum

D(( ( A1u ((u+), A1g

Δυ = (1, ΔJ = (1 IR Just like diatomic DJ = 0 possible

Δυ = (1, ΔJ = 0 Raman due to K = 0 if electron angular momentum

( vibrations ( distort molecule from linear (bend)

C(( - E1 (x,y) ; D(h - E1u ((u) ( IR allowed

C(( - E1, E2 ; D(h - E1g,2g ((g, Δg) ( Raman allowed

Δυ = (1, ΔJ = 0, (1 – IR P,Q,R branches

Δυ = (1, ΔJ = 0, (1, (2 – Raman O,P,Q,R,S branches

Isotopes – spin of nuclei – total w/f fermion – asymmetry (-1)

get intensity alteration: J – {even, odd exchange symmetry: bosom – symmetry (+1)

Note – pure rotation, this would only be Raman

– vibration / rotation – see change symmetry but population effect remains

Spherical top molecules

A1 ( not allowed ( IR

Totally symmetric modes: Δυ = (1, ΔJ = 0, (1, (2

Asymmetric modes (T2) Δυ = (1, ΔJ = 0, (1; ΔJ = 0, (1, (2

Sort of like diatomic but degeneracy in K = (2J + 1)

[ADD Infra-red spectroscopy]

[ADD Banwell-Fund. Molecule]

[ADD Infra-red spectroscopy]

Summary

IR selection rules [pic] ( 0 ΔυK = (1, ΔυJ = 0 j ( K

[pic] ( 0 ΔJ = 0, (1 ΔM = 0, (1 ΔK = 0

[pic] ( 0 ΓQK ( Γμ = Γxyz

Raman – same except: ΔJ = 0, (1, (2 since operator Y2(1(2

ΓQK ( Γα = Γx2,y2,z2,xz,yz,xz

IR – dipole moment change / Raman – polarizability change

center of symmetry – IR/Raman u + g – exclusive

Linear – A1 modes (E) – IR P,R branch, ΔJ ( 0

Note: D(h – no IR for symmetry stretches / need symmetry A((1u

E modes (() – IR P,Q,R ΔJ = (1, 0

Raman – can have ΔJ = 0, (2 O,Q,S branch

Assume start ground state υ = 0 ( χ0 = total symmetry / if higher temperature

can start υ = 1 on higher ( hot bend still Δυ ( 0

Isotopes ( if center of symmetry, i, then spin ½ – asymmetry

get alternating intensity J {odd, even} due to population ½ + ½ = 0

Raman Polarization 2 photon ( can measure scalar ( or ll to excitation

Spherical top ρ = [pic] ( polarized ( total symmetry

Total symmetry A1 mode Raman Δυ = (1 ΔJ = 0, (1, (2

IR – not allowed (xyz ~ T)

Asymmetry Modes T2 Δυ = (1 ΔJ = 0, (1 – IR

ΔJ = 0, (1, (2 – Raman

like linear but ΔK = 0, K-degeneracy (2J + 1) – affect intensity

Symmetrical Tops

parallel vibration ( Γvib = Γz [pic] Δυ = (1, ΔJ = 0,(1, ΔK = 0

note this is same situation as t-linear

perpendicular vibration ( Γvib = Γxy (E) Δυ = (1, ΔJ = 0,(1, ΔK = (1

( this will give multiple sets of P,Q,R lines, differences depend on (A–B)

ex: CH3I A >> B ( Q branch separation much greater than R,P

Handout CH3I

Raman ( Δυ = (1, ΔJ = 0,(1, (2, ΔK = 0 symmetry ΔK = (1 asymmetry

Asymmetry tops ( like prolate, like oblate

band contours distinguish types of modes

A B C – A – strong center, weak wing

B – minimum center, strong wing

C – strong center and wing

Anharmonicity ( Δυ = 0,(1,(2 … Δυi = (1, Δυj = (1

Intensity – Group Theory just tells us on/off, but allowed transition may be weak or strong

IR – [pic] ( bigger dipole change

for motion of change C = 0 strong, C = C weak

Raman – [pic] ( large polarizability – delocalized parts – aromatics, heavy atoms

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