e sum = EJL R Sβ General objective: Reduce the complexity of the analysis by exploiting symmetry. Specific Objectives: 1. The molecular symmetry matrix S. How to populate it.. Relationships between the character tables for transitions of different symmetry and the molecular tensor. 3. The surface response for a uniaxial ensemble as a function of molecular symmetry. 4. Extension to other linear and nonlinear optical processes. Department of Chemistry Purdue University 1
Let s consider a simple example of a uniaxial surface assembly of rod-like molecules with a molecular tensor dominated by just the β zzz element. For C surface symmetry, we have 4 unique elements for vibrational SFG. R = cos θ, R = sinθcos φ; R = sinθsinφ Zz Xz Yz χ : β = N cos θ β 3 ZZZ zzz s zzz N χ : β = N cosθsin θcos φ β = cosθsin θ β Ns χxxz : βzzz = Ns cosθsin θcos φ βzzz = cosθsin θ βzzz s ZXX zzz s zzz zzz χ β θ θ φ φ β YZX : zzz = Ns cos sin sin cos zzz = 0
Bottom up: Identification of the total set of unique molecular tensor elements based on symmetry. C v E C (z) σ v (xz) σ v (yz) Lin. Quad. Worked example: C v A 1 1 1 1 1 z x,y,z -The A 1, B 1, and B A 1 1-1 -1 xy symmetries all have B 1 1-1 1-1 x xz nonzero elements for both µ and α. B 1-1 -1 1 y yz ijk ( 1) ijk ( 1) β ω ; ω, ω = S ω α µ β ω; ωω, = S ω µ α ( ) ( ) 00 3 1 n 0n n0 n ( ) ( ) 00 m 0m m0 m SFG α ij µ k SHG µ i α jk A 1 β xxz, β yyz, β zzz A 1 β zxx, β zyy, β zzz A - A - B 1 β xzx = β zxx B 1 β xxz = β xzx B β yzy = β zyy B β yyz = β yzy 3
SFG A 1 α ij µ k β xxz, β yyz, β zzz A - B 1 B β xzx = β zxx β yzy = β zyy For a B 1 transition in vibrational SFG, one unique and two nonzero tensor elements survive. xxx xxy xxz xyx xyy xyz xzx xzy xzz yxx yxy yxz yyx yyy yyz yzx yzy yzz zxx zxy zxz zyx zyy zyz zzx zzy zzz 0 0 0 0 0 0 1 0 0 0 0 0 0 0 [ βxzx ] = S β 0 0 0 0 1 0 0 0 0 0 0 0 0 Symmetry matrix S, populating the full set of 7 elements from the subset of independent tensor elements (7 1 in this case). Independent, nonzero elements within the molecular tensor. 4
Set of matrices corresponding to the C (z) and C 4 (z) symmetry operations Top down: A more general approach exemplified for D 4 ( π n) ( π n) ( π ) ( π ) cos / sin / 0 1 Cn ( z) = sin / n cos / n 0 0 0 1 D 4 E C 4 (z) Character C (z) C' table C" Lin. Quad. A 1 1 1 1 1 1 x +y,z A 1 1 1-1 -1 z - B 1 1-1 1 1-1 x -y B 1-1 1-1 1 xy E 0-0 0 (x,y) (xz,yz) Linear transformations (µ) 0 1 0 µ x µ y 1 µ ' = C4 ( z) µ = 1 0 0 µ y µ = x 0 0 1 µ z µ z 1 0 0 µ x µ x 1 µ ' = C ( z) µ = 0 1 0 µ y µ = y 0 0 1 µ z µ z -For A symmetry, the transition moment is unchanged by this operation, consistent with a transition moment with µ z as the only nonzero element. -For E symmetry, each invert in sign for the C operation, consistent with the negative sign for that operation. Both (µ x,µ y ) transform together. 5
Set of matrices corresponding to the C (z) and C 4 (z) symmetry operations Top down: A more general approach exemplified for D 4 ( π n) ( π n) ( π ) ( π ) cos / sin / 0 1 Cn ( z) = sin / n cos / n 0 0 0 1 α' = C z C z α ( ) ( ) 1 1 4 4 αyy αxy α zy = αyx αxx αzx αyz αxz α zz α' = C z C z α ( ) ( ) 1 1 αxx αyx α zx = αxy αyy αzy αxz αyz α zz D 4 E C 4 (z) Character C (z) C' table C" Lin. Quad. A 1 1 1 1 1 1 x +y,z A 1 1 1-1 -1 z - B 1 1-1 1 1-1 x -y B 1-1 1-1 1 xy E 0-0 0 (x,y) (xz,yz) Quadratic transformations (α) -For A 1 symmetry, the α matrix is unchanged by this operation. Only the diagonal elements survive both C 41 and C 43 operations with no sign change. Since the operation swaps α xx and α yy, the two must be equal. -For B 1 symmetry, swapping α xx and α yy through the C 4 operation results in a sign change, such that the two must be equal and opposite. Analogous operation of the other C rotations separate out the (xy) and (xz,yz) off-diagonal elements. 6
Top down: A more general approach exemplified for D 4 D 4 E C 4 (z) Character C (z) C' table C" Lin. Quad. This process can be extended all the way to the complete molecular tensor. A 1 1 1 1 1 1 x +y,z A 1 1 1-1 -1 z - B 1 1-1 1 1-1 x -y B 1-1 1-1 1 xy E 0-0 0 (x,y) (xz,yz) β' = C z C z C z β ( ) ( ) ( ) 1 1 1 4 4 4 γ ' = C z C z C z C z γ ( ) ( ) ( ) ( ) 1 1 1 1 4 4 4 4 etc. Qubic transformations (β) and higher All parametric processes (starting and ending in the same state with no net energy transfer), including SFG and SHG, transform with the totally symmetric representation. The nonzero tensor elements and equalities between them within the D 4 point group can be found by performing each of the individual symmetry operations and finding the set of equalities that is consistent with all. 7
Top down: A more general approach exemplified for D 4 SHG SFG D 4 (4) µ α β(-ω, ω, ω) β(-ω sum, ω 1, ω ) χ iso D A 1 - (xx+yy), zz - - - A z - - - - B 1 - (xx-yy) - - - B - xy - - - E (E 1 ) (x,y) (xz,yz) xyz=xzy=-yxz=-yzx (χ XYZ =χ YZX =χ ZXY =0) xzy=zxy=-yzx=-zyx (χ XYZ =χ YZX =χ ZXY =0) C, µ α β(-ω, ω, ω) β(-ω sum, ω 1, ω ) χ iso A z xx, yy, zz, zxx, zyy, zzz, zxy=zyx xxz, yyz, zzz, xyz=yxz xy B x,y yz, xz xxz=zxz, xyz=xzy, xzx=zxx, xzy=zxy, yxz=yzx, yyz=yzy (χ ZZZ =-χ ZXX ) yzx=zyx, yzy=zyy (χ ZZZ =-χ XXZ ) New equalities emerging between surface tensor elements 8
How are these equalities between surface tensor elements generated? The Cartesian tensor in the surface frame. χ C χ C s χ s C = N R Sβ s θψφ = N R R R θψφ θψφ θψφ β ( ) ( ) ( ) = N Q R R R R R R θ ψ θ ψ θ ψ Q= Rφ Rφ Rφ Since the distribution in φ is known in advance to be uniformly distributed in a uniaxial assembly, the averages can be separately evaluated in this expanded matrix notation. Given the constraints on β, what constraints emerge on the elements of χ C? β In crystals or assemblies of other symmetry, Q will have a different form, but can nevertheless serve a similar purpose. 9
e sum = EJL R Sβ 1. Character tables indicate the nonzero elements in µ and α, from which the nonzero elements of β can be populated.. Additional equalities between the nonzero elements can be deduced by performing the symmetry operations of the point group on µ, α, and β. 3. From these equalities, the symmetry matrix S populates the full set of 7 tensor elements from the smaller subset of unique, nonzero values. 4. For ensembles with C symmetry, equalities between the macroscopic surface tensors are indicative of molecular symmetry. References: -Begue, N. J.; Moad, A. J.; Simpson, G. J. Nonlinear optical Stokes ellipsometry (NOSE) Part I: Theoretical Framework. J. Phys. Chem. C. 009, 113, 10158-65. -Moad, A. J., Moad, C. W.; Perry, J. M.; Wampler, R. D.; Begue, N, J.; Shen, T.; Goeken, G. S.; Heiland, R.; Simpson, G. J. NLOPredict: Visualization and Data Analysis Software for Nonlinear Optics J. Computational Chem. 007, 8, 1996-00. 10