IV- 6

C'. Review: Classify Vibrational Normal Modes – Group Theory χxyz ⊗ χatoms = χ3N Vibrations (correct for translation and rotation) χ3N-6 = χ3N - χtrans - χrot as character: χ3N = χxyz • χatom then reduce to get linear combination χ3N =

ai

i∑ χ i

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 OperatorpΓ(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

C

H1

H4 H3

H2

Td E 8C3 3C2 6S4 6σd χxyz 3 0 -1 -1 1 A1 1 1 1 1 1 χatom 5 2 1 1 3 A2 1 1 1 -1 -1 χ3N 15 0 -1 -1 3 E 2 -1 2 0 0 -(χtrans + χrot) 6 0 -2 0 0 T1 3 0 -1 1 -1 RxRyRz χ3N-6 9 0 1 -1 3 T2 3 0 -1 -1 1 x y z 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 A1 is not independent or in this case 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

IV- 7

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 = ∆x1, δ2 = ∆y1 … δ3N = ∆zN for vibration problem: mass weighted Cartesian displacement coordinate easier q1 = m1

½ ∆x1, q2 = m1½ ∆y2 … q3n = mN

½ ∆zN

Classically: ∑∑∑ === −

iiN MT 2

212

211

21 )( qqP2 &&

αα

ααα

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)

L q q q Ue U q qU

21

N3

1qU 2

+++= ∑∑∑ ∂∂∂

=∂∂

jii j

ii jii

derivatives taken at q=0 – equilibrium

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

And 0 0 qqU ==∂

∂ for minimum

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

( ) )q (q ji,

21

0 q

2 •∑=

∂∂∂

jiqq

Uji = Vharm

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

Classical form: If keep just Vharm and TN: ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛+= ∑ ∑∑ ∂∂

∂

i i jiq

Uivib i

q j0q 2

21 q q

j

2&H

IV- 8

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

TN is diagonal: q = T⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

N

21

q qq

MN =

⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

•

•

•

2

2

2

2

1

21

0

0

00

Nq

q

q

O

TN = ••

~~2

1 q q (direct product -- insert identity matrix I - 1's on diagonal - between q's)

The potential in this form is not diagonal:

VN = ½ [q1 q2 … qN ]

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

∂∂∂

∂∂∂

∂∂∂

∂∂∂

∂∂∂

NN qqU

qqU

qqU

qqU

qqU

2

22

2

21

212

2

11

2

O⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

N

21

q qq

M

~

T

~2

1 q'U' q ≈VN =

goal – find change of coordinates T = ∑•

ii

2

21 Q

to make both terms diagonal V = ∑λi

ii2

21 Q

also must span space Qi = jq k ijj

∑

need transformation L → result diagonal matrix, λ~

1 L U L Λ=′′≈≈

−

≈i on diagonal

unitary transformation: = matrix of

~L eigenvectors of

≈′′U , L-1 = LT

λi = eigenvalues of ≈′′U

Method: solve secular determinant for λm: det ( - δjkU′′ jk λm) = 0 U′′ → 3N x 3N,

3N λm values but 6 → zero (no potential trans and rot)then plug λm into secular equations: (U∑

kjk - δjk λm) ℓkm = 0

≈L = { ℓkm }

Qi = ∑ ℓk

km qk and inverse: qj = ∑ ℓjk Qk or matrix q = ≈L Q and Q =

≈L T q

IV- 9

Put it all together 2Vvib = q

~

T ′ ′ U ≈

q~ = (LQ)T U′′ (LQ) = QT LT U′′ L Q

= QT Λ Q or 2V = ∑ λm Qk2 → diagonal

same idea:

2T = ∑∑ =k

ki

i Qq 22 && → diagonal (LT L = 1 drops out)

can separate – then solve one coordinate at a time (i.e. H = T + V = ∑k

kh )

Classical: F = ma = 2

2

tQ

∂

∂ = Q

V∂∂− = -λ Q -- take derivative of V with one Qk

wave equation: 22

tQ∂∂ + λQ = 0 ⇒ Qk = Bk sin (λ½ t + bk) -- plug it in, test!

Quantum Mechanics H = TN + VN

= ½ ∑ + ½ ∑ λ2kQ& k Qk

2

= ∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛λ+

∂∂

k

2kk2

1Q2

- Q 2k

22h = ∑k

hk

each hk is a 1-D harmonic oscillator Hamiltonian operator Multi-dimensional 1-dimensional Know solution: Hvib = h∑

kk hk χk = Ek χk

Evib = ∑ (vk

k + ½) h νk Ek = (υk + ½) h νk

Ψvib = ∏ χk

υk (Qk) χk = Nkexp( 2Q 2

kkα− ) Hυk (αk½ Qk)

exp decay Hermite poly oscillation [Note: for frequency νk, cannot simply write k,µ ] αk = h

k2πν

[ but qualitatively same concepts hold ] each Qk has unique frequency, νk this is source of IR-Raman spectral analysis recall: → summed H → product w/f → Total energy sum independent vibrational energies {Note zero point (vk = 0): E = ½ ∑

kh νk → non trivial}

IV- 10

→ Product functional form for vibration wave/fct makes determination of χ′′µχ′ easier use Group Theory Γψvib = ∏ Γχ

kυk

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: <Πχk | µ | Πχj> = <Πχk’ | Πχj’><χk | µ | χj>

E. Selection rules: IR 0 dx ≠χ ′′µχ′ ρρ how determine?

expand: µ = µe + ∑ ∂µ∂

kQK

Qk + 12 ∂2µ

∂Qk ∂Qmm∑

k∑ Qk Qm + …

constant but vector nature leads to ∆J = ±1,0 rotational transitions: perm. dipole orient. rotation fct.

eg: υυυυ χ ′′χ′ψ ′′ψ′=φ′′χ ′′ψ ′′µφ′χ′ψ′ elelrel0rotel µ0 ′ Q rot cosθsinθ cosφsinθ sinφ

′ ′ Q r

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

2nd term → vibrational excitation --separate electronic and vibrational functions: rotations still involved

υυ∂µ∂

∂µ∂ χ ′′χ′ψ ′′ψ′=φ′′χ ′′ψ ′′φ′χ′ψ′∑ Q Q KeeQ

kKQ kk

<φ’|sinθcosφ, etc|φ”>

a) still has an orientation effect → rotation change with vibration: ∆J = ±1,0 b) vibration change only one quantum (harmonic oscillator restriction) → ∆υk = ±1 but only υk change, i.e. ∆υj = 0 j ≠ k

c) and 0 ≠∂∂

kQµ

i.e. the dipole moment must change along coordinate Qk

to do this Qk and µ must have same symmetry Qk ~ x,y,z

Group Theory language: 0 ≠∂∂

KQµ

Γ⎯⎯→⎯iffµ ⊂ ΓQk ⊂ Γx,y,z

so look in table for x,y,z representations if Γχ” = A1 total symmetric, allowed for Γχ’ = Γµ those are vibrational IR allowed (assume ground: χ′′ --> υ = 0)

IV- 11

Raman Spectra selection α = α + e∂α

∂Qk

k∑ Q + k

12

k, j∑ ∂2α

∂Q j∂Qk Q Qj k

same idea: Constant αe → pure rotation, transform as x2, y2, z2, xy, yz, zx ∆J = 0, ±1, ±2 Selection rule: ∆υk = ±1, ∆υj = 0 → since exact same integral for Qk : υυ χ′′χ′ Q K next: 0

rQ ≠∂

α∂ ⇒ polarizability must change to see vibrational transition

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

so allowed Raman transition from ground state: Qk ~ x2, y2, z2, xy, yz, xz