Formulating SALCs with Projection Operators

Transcription

1 Formulating SALCs with Projection Operators U The mathematical form of a SALC for a particular symmetry species cannot always be deduced by inspection (e.g., e 1g and e u pi-mos of benzene). U A projection operator is a function that acts on one wave function of the basis set of functions that comprise the SALCs (e.g., one of the six p z orbitals on the carbon atoms in the ring of benzene) to project out the SALC function. U A projection operator for each symmetry species must be applied to the reference function to generate all the symmetry-allowed SALCs. U The projection operator for a given symmetry species contains terms for each and every operation of the group (not just each class of operations).

2 Full-Matrix vs. Character Form of the Projection Operator U The full form of the projection operator function for degenerate species, which often directly generates the set of all SALCs belonging to a degenerate symmetry species, requires use of the full operator matrix for each and every operation of the irreducible representation; i.e., the full-matrix form of the irreducible representation. U Because there are no generally available tabulations of the full-matrix forms of the irreducible representations for groups with degenerate species, a simpler form of the projection operator that uses only the characters for each operation is most often used. ; The character form of the projection operator for degenerate species generates only one of the degenerate SALCs, requiring other means to deduce the companion functions. L We will only use the character form of the projection operator function.

3 The Projection Operator in Characters U The projection operator in character form, P i, acting on a reference function of the basis set, φ t, generates the SALC, S i, for the ith allowed symmetry species as S i % P i φ t ' d i h j R χ R i R j φ t in which d i = dimension of the ith irreducible representation, h = order of the group, χ i R = each operation's character in the ith irreducible representation, R j = the operator for the jth operation of the group. U The term R j φ t gives one of the several basis functions of the set of functions forming the SALCs, in a positive or negative sense. U The summation is taken over all operations of the group, not all classes of operations. U The results P i φ t are not the final SALCs; they need cleaning up.

4 1. The function must be normalized. Requirements for a Wave Function N IΨΨ*dτ = 1, where N is the normalization constant. L We can routinely ignore the factor d i /h in the projection operator expression.. All wave functions must be orthogonal. IΨ i Ψ j dτ = 0 if i j

5 Mathematical Simplification When Taking Products of LCAOs U The P i φ t products have the general form U But in general (a i φ i ± a i+1 φ i+1... ± a n φ n )(b j φ j ± b j+1 φ j+1... ±b m φ m ) Iφ i φ j dτ = δ ij Thus, all φ i φ j (i j) terms vanish, and all φ i φ i or φ j φ j terms (i = j) are unity. L Ignore the cross terms when normalizing or testing for orthogonality!

6 The σ-salcs of MX 6 (O h ) Example: Generate the SALCs for sigma-bonding of six ligands to a central atom in an octahedral MX 6 complex. U From the transformation of six octahedrally arranged vectors pointing toward a central atom, the reducible representation for the six SALCs can be generated and reduced as follows: Γ σ = A 1g + E g + T 1u c d b σ 1 a σ 4 σ 6 σ 5 a σ 3 b σ d c

7 Minimizing the Work U O h has 48 operations, so each projection operator will have 48 terms. U The rotational subgroup O has half as many operations (h = 4) but still preserves the essential symmetry. L Carry out the work in O and correlate the results to O h. U In O, Γ σ = A 1 + E + T 1, which has obvious correlations to Γ σ = A 1g + E g + T 1u in O h.

8 Labeling the Symmetry Elements c d b σ 1 a σ 4 σ 6 σ 5 a σ 3 b σ d U 8 in O refers to 4 and 4 whose axes run along the cube diagonals. L Label these by the corners through which they pass: e.g., aa, bb, cc, dd. U 3C, 3C 4, and 3C 4 3 have axes that run through trans-related pairs of ligands. L Label these by the pairs of ligands through which they pass; e.g., 1, 34, 56. U 6C ' have axes that pass through the mid-points of opposite cube edges. L Label these by the two-letter designation of the cube edges through which they pass; e.g., ac, bd, ab, cd, ad, bc. U rotations are clockwise, viewed from the upper cube corner through which the axis passes. U C 4 rotations are clockwise, viewed from the lower-numbered ligand. c

9 The Effect of the Operations of O on a Reference Function σ 1 O E C C C label aa bb cc dd aa bb cc dd R j σ 1 σ 1 σ 5 σ 3 σ 6 σ 4 σ 3 σ 6 σ 4 σ 5 σ 1 σ σ C 4 C 4 C 4 C 4 3 C 4 3 C 4 3 C ' C ' C ' C ' C ' C ' ac bd ab cd ad bc σ 1 σ 5 σ 4 σ 1 σ 6 σ 3 σ σ σ 3 σ 4 σ 5 σ 6 c d b σ 1 a σ 4 σ 6 σ 5 a σ 3 b σ d c

10 Projection Operator for the A 1 Species in O (A 1g in O h ) O E C C C label aa bb cc dd aa bb cc dd R j σ 1 σ 1 σ 5 σ 3 σ 6 σ 4 σ 3 σ 6 σ 4 σ 5 σ 1 σ σ A χ ir R j σ 1 σ 1 σ 5 σ 3 σ 6 σ 4 σ 3 σ 6 σ 4 σ 5 σ 1 σ σ Summing all the χ ir R j σ 1 terms gives C 4 C 4 C 4 3 C 4 3 C 4 3 C 4 C ' C ' C ' C ' C ' C ' ac bd ab cd ad bc σ 1 σ 5 σ 4 σ 1 σ 6 σ 3 σ σ σ 3 σ 4 σ 5 σ σ 1 σ 5 σ 4 σ 1 σ 6 σ 3 σ σ σ 3 σ 4 σ 5 σ 6 P(A 1 )σ 1 % 4σ 1 + 4σ + 4σ 3 + 4σ 4 + 4σ 5 + 4σ 6 % σ 1 + σ + σ 3 + σ 4 + σ 5 + σ 6

11 SALC for A 1g Normalizing: N I(σ 1 + σ + σ 3 + σ 4 + σ 5 + σ 6 ) dτ = N I(σ 1 + σ + σ 3 + σ 4 + σ 5 + σ 6 )dτ = N ( ) = 6N / 1 Y N = 1//6 Therefore, the normalized A 1g SALC is Σ 1 (A 1g ) = 1//6(σ 1 + σ + σ 3 + σ 4 + σ 5 + σ 6 )

12 Projection Operator for the First of Two E SALCs (E u in O h ) U In O, χ R = 0 for 6C 4 and 6C '. Therefore, ignore the last 1 terms. O E C C C label aa bb cc dd aa bb cc dd R j σ 1 σ 1 σ 5 σ 3 σ 6 σ 4 σ 3 σ 6 σ 4 σ 5 σ 1 σ σ E χ ir R j σ 1 σ 1 -σ 5 -σ 3 -σ 6 -σ 4 -σ 3 -σ 6 -σ 4 -σ 5 σ 1 σ σ Summing across all χ ir R j σ 1 gives P(E)σ 1 % 4σ 1 + 4σ σ 3 σ 4 σ 5 σ 6 % σ 1 + σ σ 3 σ 4 σ 5 σ 6 Normalizing: N I(σ 1 + σ σ 3 σ 4 σ 5 σ 6 ) dτ After normalization: = N I(4σ 1 + 4σ + σ 3 + σ 4 + σ 5 + σ 6 )dτ = N ( ) = 1N / 1 Y N = 1//1 = 1/(/3) Σ (E) = 1/(/3)(σ 1 + σ σ 3 σ 4 σ 5 σ 6 )

13 Test of Orthogonality with Σ 1 (A 1 ) I[Σ 1 (A)][Σ (E)]dτ = I(σ 1 + σ + σ 3 + σ 4 + σ 5 + σ 6 )(σ 1 + σ σ 3 σ 4 σ 5 σ 6 )dτ = = 0 L But Σ (E) is only the first of two degenerate functions. ; How do we find the partner?

14 Finding the Partner to Σ (E) - Method I Method I: Apply the E Projection operator to another reference function, here σ 3. O E C C C label aa bb cc dd aa bb cc dd R j σ 3 σ 3 σ 1 σ 6 σ σ 6 σ 5 σ 1 σ 5 σ σ 4 σ 3 σ 4 E χ ir R j σ 3 σ 3 -σ 1 -σ 6 -σ -σ 6 -σ 5 -σ 1 -σ 5 -σ σ 4 σ 3 σ 4 This gives P(E)σ 3 % σ 1 σ + 4σ 3 + 4σ 4 σ 5 σ 6 % σ 1 σ + σ 3 + σ 4 σ 5 σ 6 Orthogonal to Σ 1 (A)? I(σ 1 + σ + σ 3 + σ 4 + σ 5 + σ 6 )( σ 1 σ + σ 3 + σ 4 σ 5 σ 6 )dτ = = 0 YYes Orthogonal to Σ (E)? I(σ 1 + σ σ 3 σ 4 σ 5 σ 6 )( σ 1 σ + σ 3 + σ 4 σ 5 σ 6 )dτ = = 6 0 YNo! ; What went wrong?

15 Notes on Method I for Finding a Degenerate Partner L Changing the reference function after finding the first member of a degenerate set implicitly changes the axis orientation. The resulting function will be one of the following: A legitimate partner wave function in either its positive or negative form. [Not this time!] The negative of the first member of the degenerate set. [Not a useful result, and not what happened this time.] A linear combination of the first member with its partner(s). If this is the case, try various combinations of the two projected functions to find a function that is orthogonal. [Looks like this must be what we got!]

16 Finding the Partner to Σ (E) - Method I By trial and error with various combinations we find that this works: P(E)σ 1 = σ 1 + σ σ 3 σ 4 σ 5 σ 6 P(E)σ 3 = σ 1 σ + 4σ 3 + 4σ 4 σ 5 σ 6 3σ 3 + 3σ 4 3σ 5 3σ 6 % σ 3 + σ 4 σ 5 σ 6 This result is orthogonal to Σ (E), I(σ 1 + σ σ 3 σ 4 σ 5 σ 6 )(σ 3 + σ 4 σ 5 σ 6 )dτ = = 0 and also to Σ 1 (A 1 ), I(σ 1 + σ + σ 3 + σ 4 + σ 5 + σ 6 )(σ 3 + σ 4 σ 5 σ 6 )dτ = = 0 The normalized partner wave function, then, is Σ 3 (E) = ½(σ 3 + σ 4 σ 5 σ 6 )

17 Finding the Partner to Σ (E) - Method II Method II: Apply an operation of the group to the first function found. Principle: The effect of any group operation on a wave function of a degenerate set is to transform the function into the positive or negative of itself, a partner, or a linear combination of itself and its partner or partners. L Suppose we perform (aa) on our first degenerate function. U Effect of (aa) on the functions of the basis set: U Effect of (aa) on Σ (E): Before σ 1 σ σ 3 σ 4 σ 5 σ 6 After σ 5 σ 6 σ 1 σ σ 3 σ 4 (σ 1 + σ σ 3 σ 4 σ 5 σ 6 ) 6 (σ 5 + σ 6 σ 1 σ σ 3 σ 4 ) Is this new function orthogonal to Σ (E)? = ( σ 1 σ σ 3 σ 4 + σ 5 + σ 6 ) I(σ 1 + σ σ 3 σ 4 σ 5 σ 6 )( σ 1 σ σ 3 σ 4 + σ 5 + σ 6 )dτ = 1 1 = 10 0 YNo!

18 Finding the Partner to Σ (E) - Method II The new function must be a combination of Σ (E) and the partner function. By trial-and-error: Σ : σ 1 σ + σ 3 + σ 4 + σ 5 + σ 6 {Σ R[ (aa)]} +σ 1 + σ + σ 3 + σ 4 4σ 5 4σ 6 +3σ 3 + 3σ 4 3σ 5 3σ 6 This is the same result as found by Method I: % σ 3 + σ 4 σ 5 σ 6 Σ 3 (E) = ½(σ 3 + σ 4 σ 5 σ 6 )

19 Projection Operator for the First of Three T 1u SALCs Note: In T 1 of the group O, for 8 χ = 0; therefore, skip the first eight terms in the operator expression. O E C C C label aa bb cc dd aa bb cc dd R j σ 1 σ 1 σ 5 σ 3 σ 6 σ 4 σ 3 σ 6 σ 4 σ 5 σ 1 σ σ T χ ir R j σ 1 3σ 1 -σ 1 -σ -σ C 4 C 4 C 4 3 C 4 3 C 4 3 C 4 C ' C ' C ' C ' C ' C ' ac bd ab cd ad bc σ 1 σ 5 σ 4 σ 1 σ 6 σ 3 σ σ σ 3 σ 4 σ 5 σ σ 1 σ 5 σ 4 σ 1 σ 6 σ 3 -σ -σ -σ 3 -σ 4 -σ 5 -σ 6 Summing across the χ ir R j σ 1 : P(T 1 )σ 1 % 4σ 1 4σ % σ 1 σ We can readily show that this is orthogonal to the previous three functions for A and E, and on normalization we obtain the SALC G 4 (T 1 ) = 1// (σ 1 σ )

20 Finding the Partner T 1u SALCs The partner functions become apparent from considering the form of G 4 (T 1 ). c d b σ 1 a a b σ d c L The other two functions must have the same form along the other two orthogonal axes: G 5 (T 1 ) = 1// (σ 3 σ 4 ) G 6 (T 1 ) = 1// (σ 5 σ 6 ) U Alternately, note the effects of (aa) and (aa) on (σ 1 σ ): R[ (aa)] (σ 1 σ ) = (σ 5 σ 6 ) Y G 6 (T 1 ) R[ (aa)] (σ 1 σ ) = (σ 3 σ 4 ) Y G 5 (T 1 )

21 Summary: The Six σ-salcs of MX 6 (O h ) Σ 1 (A 1g ) = 1//6(σ 1 + σ + σ 3 + σ 4 + σ 5 + σ 6 ) Σ (E u ) = 1/(/3)(σ 1 + σ σ 3 σ 4 σ 5 σ 6 ) Σ 3 (E u ) = ½(σ 3 + σ 4 σ 5 σ 6 ) G 4 (T 1g ) = 1// (σ 1 σ ) G 5 (T 1g ) = 1// (σ 3 σ 4 ) G 6 (T 1g ) = 1// (σ 5 σ 6 )

22 The σ-salcs of CH 4 (T d ) B z A y x D C Γ = A 1 + T U The A 1 SALC is obvious, so only use the projection operator for the T SALCs. U The group T d has h = 4, but its rotational subgroup T has h = 1. Therefore, do the work in T, where the T species corresponds to T of T d. U For the irreducible representation T, the only non-zero characters are E, C (x), C (y), C (z), so the T operator has only four non-vanishing terms: T E... C (x) C (y) C (z) R j s A s A... s C s D s B T χ ir R j s A 3s A... s C s D s B

23 The σ-salcs of CH 4 (T d ) P(T)s A % 3s A s C s D s B U This function is orthogonal to the A SALC, Φ 1 = ½(s A + s B + s C + s D ), and could be normalized to give the function Φ(T) = 1/(/3)(3s A - s C - s D - s B ) ; This SALC is supposed to match with one of the p AOs on carbon. Φ(T) makes no sense geometrically for such overlap! ( P(T)s A must be a combination of all three functions: P(T z )s A % s A + s B s C s D P(T y )s A % s A s B s C + s D P(T x )s A % s A s B + s C s D P(T)s A % 3s A s B s C s d U After normalization, the three T SALCs of MX 4 (T d ) are: Φ = ½{s A + s B s C s D } Φ 3 = ½{s A s B s C + s D } Φ 4 = ½{s A s B + s C s D }