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Lecture 6. Character Tables. Group theory makes use of the properties of matrices Idea: When an operation, O , (proper rotation, improper rotation, reflection, inversion) is done on the function, f(x,y,z) O f(x,y,z)
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Character Tables Group theory makes use of the properties of matrices Idea: When an operation, O, (proper rotation, improper rotation, reflection, inversion) is done on the function, f(x,y,z) O f(x,y,z) the result is a value taken on by the original function at some other point (x’,y’,z’). Or O f(x,y,z) = f (x’,y’,z’) We know the function f(x,y,z) the problem is to determine the values (x’,y’,z’). The nature of the operation tells us where to look.
Character Tables Group theory makes use of the properties of matrices Each operation may be expressed as a transformation matrix: [New coordinates] = [transformation matrix][old coordinates] • Example: in Cartesian coordinate system, reflection in x = 0 plane • Changes the value of x to –x (multiplies it by -1) • Leaves y unchanged (multiplies it by 1) • Leaves z unchanged (multiplies it by 1) To see the result of the operation at (x, y, z) look at the original object at (x’, y’, z’). = Results of transformation. Transformation matrix Original coordinates
= Recall Technique of Matrix multiplication V’ M V To get an element of the product vector a row in the operation square matrix is multiplied by the original vector matrix. For example V’2 = y’ = M2,1 * V1 + M2,2 * V2 + M2,3 * V3 y’ = 0 * x + 1 * y + 0 * z = y
Character Tables - 2 The matrix representation of the symmetry operations of a point group is the set of matrices corresponding to all the symmetry operations in that group. The matrices record how the x,y,z coordinates are modified as a result of an operation. • For example, the C2v point group consists of the following operations • E: do nothing. Unchanged. • C2: rotate 180 degrees about the z axis: x becomes –x; y becomes –y and z unchanged. • sv(xz): y becomes –y • sv’ (yz): x becomes -x sv’ (yz): E C2 sv (xz):
Operations Applied to Functions - 1 Transform the coordinates. Consider f(x) = x2 sv’ (f(x)) = sv(x2) = (-x)2 = x2 = f(x) or sv’ (f(x)) = 1 * f(x) f(x) is an eigenfunction of this reflection operator with an eigenvalue of +1. This is called a symmetric eigenfunction. Similarly f(x) = x3 sv’ (f(x)) = -1 * f(x) f(x) is an eigenfunction of this reflection operator with an eigenvalue of -1. This is called a antisymmetric eigenfunction.
Plots of Functions, x2 Reflection yields. Here f(x) is x2. It can be seen to be a symmetric function for reflection at x = 0 because of mirror plane. The reflection carries out the mapping shown with the red arrows. x2 is an eigenfunction of s with eigenvalue 1
Plots of Functions, x3 Reflection yields. Here f(x) is x3. It can be seen to be a antisymmetric function for reflection at x = 0. The reflection carries out the mapping shown with the red arrows. x3 is an eigenfunction of s with eigenvalue -1
Operations Applied to Functions - 2 Now consider f(x) = (x-2)2 = x2 – 4x + 4 sv’ (f(x)) = sv(x-2)2 = (-x-2)2 = x2 + 4x + 4 f(x) = (x-2)2 is not an eigenfunction of this reflection operator because it does not return a constant times f(x). Reflection yields this function, not an eigenfunction. Neither symmetric nor antisymmetric for reflection thru x = 0.
Let’s look at Atomic Orbitals Reflection Get the same orbital back, multiplied by +1, an eigenfunction of the reflection, symmetric with respect to the reflection. The s orbital forms the basis of an irreducible representation of the operation s orbital z
Atomic Orbitals s Reflection Get the same orbital back, multiplied by -1, an eigenfunction of the reflection, antisymmetric with respect to the reflection. The p orbital behaves differently from the s orbital and forms the basis of a different irreducible representation of the operation p orbital z
Simplest ways that objects can behave for a group consisting of E and sh , the reflection plane. Irreducible Representations. Basis of the Irreducible Reps. s orbital is spherical behaves as x2 + y2 + z2. s orbital is A’. The s orbital is an eigenfunction of both E and sh. pz orbital has a multiplicative factor of z times a spherical factor. Behaves as A”. pz is an eigenfunction of both E and sh.
sp Hybrids Reflection Do not get the same hybrid back multiplied by +1 or -1 or some other constant. Not an eigenfunction. hybrid z The two hybrids form the basis of a reducible representation of the operation Recall: the hybrid can be expressed as the sum of an s orbital and a p orbital. = + Reduction: expressing a reducible representation as a combination of irreducible representations.
Reducible Representations Use the two sp hybrids as the basis of a representation h1 h2 sh operation. E operation. h1 becomes h1; h2 becomes h2. h1 becomes h2; h2 becomes h1. = = The reflection operation interchanges the two hybrids. The hybrids are unaffected by the E operation. Proceed using the trace of the matrix representation. 0 + 0 = 0 1 + 1 = 2
Our Irreducible Representations The reducible representation derived from the two hybrids can be attached to the table. Note that G = A’ + A”
Return to polynomials: f(x) = (x-2)2 = x2 – 4x + 4 sv (f(x)) = sv(x-2)2 = (-x-2)2 = x2 + 4x + 4 =g(x) Neither f nor g is an eigenfunction of s but, taken together, they do form an reducible representation since they show what the s operator does. Approaching the problem in the same way as we did for hybrids we can carry out the reduction this way u(x) = ½ (f(x) + g(x)) = ½ (f(x) + s f(x)) = x2 + 4 symmetric, unchanged by the s operator. Behaves as A’ v(x) = ½ (f(x) - g(x)) = ½ (f(x) - s f(x)) = -4x, antisymmetric, multiplied by -1 by the s operator. Behaves as A’’
Character Table Symmetry operations, Classes Point group x, y, z Symmetry of translations (p orbitals) Rx, Ry, Rz: rotations Characters +1 symmetric behavior -1 antisymmetric Mülliken symbols Each row is an irreducible representation
Character Tables - 3 Irreducible representations are not linear combinations of other representation (Reducible representations are) # of irreducible representations = # of classes of symmetry operations Instead of the matrices, the characters are used (traces of matrices) A character Table is the complete set of irreducible representations of a point group
Effect of the 4 operations in the point group C2v on a translation in the x direction. The translation is simply multiplied by 1 or -1. It forms a basis to show what the operators do to an object.
Character Table Verify this character. It represents how a function that behaves as x, Ry, or xz behaves for C2.
Another point group, C3v. x, y, z Symmetry of translations (p orbitals) Classes of operations Rx, Ry, Rz: rotations dxy, dxz, dyz, as xy, xz, yz dx2- y2 behaves as x2 – y2 dz2 behaves as 2z2 - (x2 + y2) px, py, pz behave as x, y, z s behaves as x2 + y2 + z2
Naming of Irreducible representations • One dimensional (non degenerate) representations are designated A or B. • Two-dimensional (doubly degenerate) are designated E. • Three-dimensional (triply degenerate) are designated T. • Any 1-D representation symmetric with respect to Cn is designated A; antisymmétric ones are designated B • Subscripts 1 or 2 (applied to A or B refer) to symmetric and antisymmetric representations with respect to C2 Cn or (if no C2) to svrespectively • Superscripts ‘ and ‘’ indicate symmetric and antisymmetric operations with respect to sh, respectively • In groups having a center of inversion, subscripts g (gerade) and u (ungerade) indicate symmetric and antisymmetric representations with respect to i
Character Tables • Irreducible representations are the generalized analogues of s or p symmetry in diatomic molecules. • Characters in rows designated A, B,..., and in columns other than E indicate the behavior of an orbital or group of orbitals under the corresponding operations (+1 = orbital does not change; -1 = orbital changes sign; anything else = more complex change) • Characters in the column of operation E indicate the degeneracy of orbitals • Symmetry classes are represented by CAPITAL LETTERS (A, B, E, T,...) whereas orbitals are represented in lowercase (a, b, e, t,...) • The identity of orbitals which a row represents is found at the extreme right of the row • Pairs in brackets refer to groups of degenerate orbitals and, in those cases, the characters refer to the properties of the set
Definition of a Group • A group is a set, G, together with a binary operation : such that the product of any two members of the group is a member of the group, usually denoted by a*b, such that the following properties are satisfied : • (Associativity) (a*b)*c = a*(b*c) for all a, b, c belonging to G. • (Identity) There exists e belonging to G, such that e*g = g = g*e for all g belonging to G. • (Inverse) For each g belonging to G, there exists the inverse of g,g-1, such that g-1*g = g*g-1 = e. • If commutativity is satisfied, i.e. a*b = b*a for all a, b belonging to G, then G is called an abelian group.
Examples • The set of integers Z, is an abelian group under addition. What is the element e, identity, such that a*e = a? What is the inverse of the a element? 0 -a
As applied to our symmetry operators. For the C3v point group What is the inverse of each operator? A * A-1 = E E C3(120) C3(240) sv (1) sv (2) sv (3) E C3(240) C3(120) sv (1) sv (2) sv (3)
- Examine the matrix represetation of the C2v point group E C2 s’v(yz) sv(xz)
Most of the transformation matrices we use have the form Multiplying two matrices (a reminder)
C2 sv(xz) s’v(yz) E What is the inverse of C2? C2 = What is the inverse of sv? sv =
What of the products of operations? C2 sv(xz) s’v(yz) E C2 E * C2 = ? = sv * C2 = ? s’v =
Classes Two members, c1 and c2, of a group belong to the same class if there is a member, g, of the group such that g*c1*g-1 = c2
Properties of Characters of Irreducible Representations in Point Groups • Total number of symmetry operations in the group is called the order of the group (h). For C3v, for example, it is 6. 1 + 2 + 3 = 6 • Symmetry operations are arranged in classes. Operations in a class are grouped together as they have identical characters. Elements in a class are related. This column represents three symmetry operations having identical characters.
Properties of Characters of Irreducible Representations in Point Groups - 2 The number of irreducible reps equals the number of classes. The character table issquare. 1 + 2 + 3 = 6 3 by 3 1 1 22 6 The sum of the squares of the dimensions of the each irreducible rep equals the order of the group, h.
Properties of Characters of Irreducible Representations in Point Groups - 3 For any irreducible rep the squares of the characters summed over the symmetry operations equals the order of the group, h. A1: 12 + (12 + 12 ) + = 6 A2: 12 + (12 + 12 ) + ((-1)2 + (-1)2 + (-1)2 ) = 6 E: 22 + (-1)2 + (-1)2 = 6
Properties of Characters of Irreducible Representations in Point Groups - 4 Irreducible reps are orthogonal. The sum of the products of the characters for each symmetry operation is zero. For A1 and E: 1 * 2 + (1 *(-1) + 1 *(-1)) + (1 * 0 + 1 * 0 + 1 * 0) = 0
Properties of Characters of Irreducible Representations in Point Groups - 5 Each group has a totally symmetric irreducible rep having all characters equal to 1
Reduction of a Reducible Representation Irreducible reps may be regarded as orthogonal vectors. The magnitude of the vector is h-1/2 Any representation may be regarded as a vector which is a linear combination of the irreducible representations. Reducible Rep = S (ai * IrreducibleRepi) The Irreducible reps are orthogonal. Hence S(character of Reducible Rep)(character of Irreducible Repi) = ai * h Or ai =S(character of Reducible Rep)(character of Irreducible Repi) / h Sym ops Sym ops
Irreducible representations These are block-diagonalized matrices (x, y, z coordinates are independent of each other) Reducible Rep
C2v Character Table to be used for water Symmetry operations Point group Characters +1 symmetric behavior -1 antisymmetric Mülliken symbols Each row is an irreducible representation
Let’s use character tables! Symmetry and molecular vibrations
Symmetry and molecular vibrations A molecular vibration is IR active only if it results in a change in the dipole moment of the molecule A molecular vibration is Raman active only if it results in a change in the polarizability of the molecule In group theory terms: A vibrational mode is IR active if it corresponds to an irreducible representation with the same symmetry of a x, y, z coordinate (or function) and it is Raman active if the symmetry is the same as A quadratic function x2, y2, z2, xy, xz, yz, x2-y2 If the molecule has a center of inversion, no vibration can be both IR & Raman active
How many vibrational modes belong to each irreducible representation? You need the molecular geometry (point group) and the character table Use the translation vectors of the atoms as the basis of a reducible representation. Since you only need the trace recognize that only the vectors that are either unchanged or have become the negatives of themselves by a symmetry operation contribute to the character.
A shorter method can be devised. Recognize that a vector is unchanged or becomes the negative of itself if the atom does not move. A reflection will leave two vectors unchanged and multiply the other by -1 contributing +1. For a rotation leaving the position of an atom unchanged will invert the direction of two vectors, leaving the third unchanged. Etc. Apply each symmetry operation in that point group to the molecule and determine how many atomsare not moved by the symmetry operation. Multiply that number by the character contribution of that operation: E = 3 s = 1 C2 = -1 i = -3 C3 = 0 That will give you the reducible representation
Finding the reducible representation E = 3 s = 1 C2 = -1 i = -3 C3 = 0 3x3 9 1x-1 -1 3x1 3 1x1 1 (# atoms not moving x char. contrib.) G
G 9 -1 3 1 Now separate the reducible representation into irreducible ones to see how many there are of each type S A1 = 1/4 (1x9x1 + 1x(-1)x1 + 1x3x1 + 1x1x1) = 3 A2 = 1/4 (1x9x1 + 1x(-1)x1 + 1x3x(-1) + 1x1x(-1)) = 1
Symmetry of molecular movements of water Vibrational modes
Raman active IR active Which of these vibrations having A1 and B1 symmetry are IR or Raman active?
Often you analyze selected vibrational modes Example: C-O stretch in C2v complex. n(CO) 2 x 1 2 0 x 1 0 2 x 1 2 0 x 1 0 G Find: # vectors remaining unchanged after operation.
G 2 0 2 0 Both A1 and B1 are IR andRamanactive = A1 + B1 A1 = 1/4 (1x2x1 + 1x0x1 + 1x2x1 + 1x0x1) = 1 A2 = 1/4 (1x2x1 + 1x0x1 + 1x2x-1 + 1x0x-1) = 0 B1 = 1/4 (1x2x1 + 1x0x1 + 1x2x1 + 1x0x1) = 1 B2 = 1/4 (1x2x1 + 1x0x1 + 1x2x-1 + 1x0x1) = 0