Handouts - Download and Read! http://www.chem.tamu.edu/rgroup/hughbanks/ courses/67/handouts/handouts.html translation_groups.pdf translation_groups.pdf transitions.pdf Provides background for selection rules. You do not need to memorize the derivation, but results in boxes are important to know! Complex Numbers; Digression Cartesian Forms The complex plane, vector forms polar representation of complex numbers and vectors Cyclic Groups Consider C N, the cyclic group consisting of the operations C N, C N, C N,..., C N N-1, C N N = E. Because all the operations are in the different classes, there are N irreducible representations and they are all one dimensional. This means that the characters are the same as the matrices - just numbers. Character Tables for Cyclic Groups ε = exp(πi / N) C N C N C N! C N 1 N C N N = E Γ1 ε ε ε! ε N 1 ε N Γ ε ε ε 6! ε N ε N Γ ε ε6 ε 9! ε N ε N " " " " # " " Γ N 1 ε N 1 ε N εn! ε ε N(N 1) ΓN ε N ε N εn! ε 1 ε N
Example Use the group to find the characters of the reducible representation obtained using the 6 carbon pπ orbitals of benzene as a basis then find the irred. reps. spanned by this rep. Draw the complex coefficients of the orbitals for each irreducible representation. Draw real counterparts of these orbitals. Group ε = exp(πi / 6) N = E Γ 1 ε ε ε ε ε ε 6 Γ ε ε ε 6 ε 8 ε 10 ε 1 Γ ε ε 6 ε 9 ε 1 ε 1 ε 18 Γ ε ε 8 ε 1 ε 16 ε 0 ε Γ ε ε 10 ε 1 ε 0 ε ε 0 Γ 6 ε 6 ε 1 ε 18 ε ε 0 ε 6 E C N Γ 0 = Γ 6 1 1 1 1 1 1 Γ 1 1 ε ε ε ε ε Γ 1 ε ε ε 6 ε 8 ε 10 Γ 1 ε ε 6 ε 9 ε 1 ε 1 Γ 1 ε ε 8 ε 1 ε 16 ε 0 Γ 1 ε ε 10 ε 1 ε 0 ε ε = exp(πi / 6) E Γ 0 = Γ 6 1 1 1 1 1 1 Γ 1 1 ε ε ε ε ε Γ 1 ε ε ε 6 ε 8 ε 10 Γ 1 ε ε 6 ε 9 ε 1 ε 1 Γ 1 ε ε 8 ε 1 ε 16 ε 0 Γ 1 ε ε 10 ε 1 ε 0 ε Let be a basis function that belongs to Γ 1, then = ε, C 6 = ε,!, C 6 = ε Write as a combination of pπ basis functions, = φ 1 + c φ + c φ + c φ + c φ φ 6 = ( φ 1 + c φ + c φ + c φ + c φ φ 6 ) = ε φ + c φ + c φ + c φ + c φ 6 φ 1 = εφ 1 + c εφ + c εφ + c εφ + c εφ εφ 6 ε = exp(πi / 6) E Γ 0 = Γ 6 1 1 1 1 1 1 Γ 1 = E 1 1 ε ε 1 ε ε Γ = E 1 ε ε 1 ε ε Γ = B 1 1 1 1 1 1 Γ = E 1 ε * ε * 1 ε * ε * Γ = E 1 1 ε * ε * 1 ε * ε * Make the identification with chaacter tables in some books: ε = ε 9 = ε 1 = ε 1 = 1, ε 6 = ε 1 = ε 18 = ε = 1 c ε = 1 c = ε* = ε 1 c = ε 1 = c ε c = ε
Translation Group (1-dimension) The one-dimensional translation group is just a particular cyclic group of order N. The trans-polyacetylene below is an example of a system with translational symmetry. tn! t t t t H H H H C C C C C1 C1 C1 C1 H H H H! 1-D Translation Group Char. Table This has the same appearance as the C N group s character table: ε = exp(πi / N ) t t t! t N 1 t N = E Γ 1 ε ε ε! ε N 1 ε N Γ ε ε ε 6! ε N ε N Γ ε ε 6 ε 9! ε N ε N " " " " # " " Γ N 1 ε N 1 ε N ε N! ε ( N 1) ε N ( N 1) Γ N ε N ε N ε N! ε N ( N 1) ε N 1-D Translation Group Char. Table - Rearranged ε = exp(πi / N ) T N E t t! t N t N 1 " " " "! " " Γ N 1 1 1! 1 1 Γ N +1 1 ε ε! ε ε 1 " " " "! " " Γ 1 1 ε 1 ε! ε ( N ) ε Γ 0 1 1 1! 1 1 Γ 1 1 ε 1 ε! ε N ε N 1 " " " "! " " Γ N 1 1 1! 1 1 Γ N +1 1 ε ε! ε ε 1 " " " " " " " ( N 1) Character Table for T N, rewritten All the IRs of T N have the form: ε = e πi N t t t! t N 1 t N = E Γ j ε j ε j ε j! ε j 1 N + 1 < j < N We make the substitution k = 1 j a N ; where 1 a < k 1 a Making the substitution, ε j = (e πi N ) j = e πika. This is rewritten to yield E t t t! t N 1 Γ(k) 1 e πi(k a) e πi(k a) e πi(k a)! πi(k a) e
1-D Energy Level Diagram 1-D s-band Dispersion curve Examples Find the characters of the reducible representation obtained using the N hydrogen 1s orbitals of a hypothetical H-atom chain (with N atoms) as a basis then find the irreducible reps. spanned by this rep. Follow the same procedure (i) using the longitudinal stretching vectors as a basis, (ii) using the transverse stretching vectors as a basis, (iii) using p σ orbitals as a basis. [() ] +x α p α d = 8 β ; β = 1 β dd = β β pp = β β dp = 1.β a a a Chains in K [() ] and K [() ]Br 0. H O [() ] - : d(-) = a =.8 Å [() ] -1.7 : d(-) = a =.88 Å The tetracyanoplatinates crystallize such that square planar () x species stack upon each other as indicated in the illustration below. (Steric factors cause the square planar ions to stack in a staggered fashion, but we ll proceed as if the stacking is eclipsed, i.e., as if there is just one () x ion per unit cell.) - distances are markedly shortened (from.8 Å to.88 Å) when the platinum is oxidized by reaction with Br that results in the intercalation of some additional bromide ions (Br ) into voids between the chains in the solid state structure.
[()] +x [()] +x Consider only the largest - σ overlaps involving the dz orbital (occupied for this d8 complex) and the 6pz orbital (a highlying unoccupied orbital that is stabilized to some extent by overlap with the π* orbitals). Set up the k-dependent Hückel-like secular equation and solve it to obtain analytical k-dependent expressions for each of the two band curves. Draw a one-dimensional band dispersion diagram that includes bands that derive from the dz and the 6pz orbitals. Use these parameters: α p αd = 8 β β dd = β β pp = β ; β = 1 β dp = 1.β Mark the Fermi levels for both systems. Explain why the - distances shrink upon oxidation. Show the lowest energy allowed optical transitions for both systems. α d + (eπ ik a + eπ ik a )β E eπ ik a ( 1.β ) + eπ ik a (1.β ) eπ ik a ( 1.β ) + eπ ik a (1.β ) α p + (eπ ik a + eπ ik a )( β ) E β cos π ka E iβ sin π ka =0 iβ sin π ka 8β β cos π ka E for convenience let β = 1 cos π ka E isin π ka =0 isin π ka 8 + cos π ka E (E + cos π ka)[e (8 + cos π ka)] 9sin π ka = 0 E = + cos π ka ± + cos π ka [()] +x =0 -dimensional Layers Bloch s Theorem in or D ϕ k (r + R) = e π ik Rϕ k (r) R = ua + vb + wc k = ka a + ka b + ka c k R = kau + kb v + kc w Orbitals and bands for a square net of Hydrogen atoms Orbitals and bands for graphite
Selection Rules for Crystals: Vertical Transitions -D Square H array - Band ky 1/a Intensity, I ψ i * H (t)ψ f d τ where H (t) is the perturbation of the molecule (solid) caused by the electric field of the radiation, and the electromagnetic wave propagating in the z-direction E0 π ik z π ik photon zi iω t H (t) = x qi xi e photon i e iω t + e e i (See Eqs. 16 & 17 in Handout on "Transitions Between Stationary States") { ψ i uk (r)eπ ik i r ψ f uk (r)e π ik f r E } Important terms to consider ~ I =0 iω t e e unless π i(k f k i +k photon ) r ky dτ -1/a 1/a kx kx k f k i + k photon = 0 but k photon k f, k i (λ photon λ f, λi ), transition forbidden unless k f k i! 0-1/a Allowed transitions are vertical, k f! k i -D Graphite (Graphene) π Bands E 1/a ka 1/b kb
PES Measurements of Graphite Densities of States for 1D, -D, -D 1/ 1/ k k