Basis vectors characters

This procedure generates a representation of the group C2v, called Ttot, for which the vector of displacements forms the basis. The character of the... 

Exercise 6.14 Use the Gram-Schmidt technique of orthogonalization to find a recursive formula for an orthogonal basis ofC[—l, 1] with the property that the kth basis vector is a polynomial of degree n (for n = 0, 1, 2,. . J. Show (from general principles) that the nth basis element is precisely the character of the representation ofSU(T) on P". Use the recursive formula to calculate and /4. 

To find the symmetry of the normal modes we study the transformation of the atomic displacements xL y, z,, i 0,1,2,3, by setting up a local basis set e,i ci2 el3 on each of the four atoms. A sufficient number of these basis vectors are shown in Figure 9.1. The point group of this molecule is D3h and the character table for D3h is in Appendix A3. In Table 9.1 we give the classes of D3h a particular member R of each class the number of atoms NR that are invariant under any symmetry operator in that class the 3x3 sub-matrix r, (R) for the basis (e,i el2 cl31 (which is a 3 x 3 block of the complete matrix representative for the basis (eoi. .. e331) the characters for the representation T, and the characters for... 

It should be observed that, contrary to the orthonormal character of the AIM basis vectors, the MEC/REC vectors f M are not orthogonal, giving rise to the metric... 

The characters of these matrices for all R can be obtained in a simple manner without ever writing out the matrices. If Ta interchanges one basis function with another, then no diagonal entry appears in the rows corresponding to these basis functions. The representation so formed is reducible. We may similarly obtain a representation with the nine basis vectors Oi(2px), 0-2(2px), OsCip ) Oi(2p ), 0>(2py), 03(2p ) Oi(2px), 0 (2px), Oj(2px). These could be taken three at a time or as a nine-rowed column vector. In any particular problem it is advisable to write out some of the matrices of Tr in order to keep track of the functions by associating them with the row of a matrix. Since Oi, O2, and O3 are indistinguishable in D3 symmetry, the resulting reps correspond to states that are composites of Oi, O2, and O3 atomic symmetry. However, whatever valence description of XOl" is employed, the valence bonds, or molecular orbitals, must retain the transformation properties shown in Tables 7.3 and 7.4. 

We will need a systematic way to deduce whether a molecular property is symmetric or antisymmetric with respect to the symmetry operations for that molecule s point group, as things will quickly get more complicated. To this end, we define a number, Xp( )> called a character, which expresses the behavior of our property p when operated on by the symmetry operation, R. A collection of characters, one for each symmetry operation present in a point group, forms a representation, Fp. The property is technically referred to as the basis vector of the representation, Fp. We can define aU sorts of basis vectors, some of which have very little apparent connection to our original molecule, such as the non-symmetry operation translate along the z axis , often given the symbol z, or the non-symmetry operation rotate by an arbitrary amount about the X axis, often referred to as R. Strictly speaking, the characters are the trace of the transformation matrix for each symmetry operation, applied to the property, p. This is described in more detail in the on-line supplementary section for Chapter 2 on derivation of characters. 

Let us start with the basis vector translations. These describe movements of the whole molecule, and are therefore represented by the vectors x, y and z. We begin by working out the effect of the translational motion on the symmetry operations by hand, and then show how this information can be read directly from the character table. 

Now if we look back to the full C2Y character table (Table 2.2) it should be obvious how this information can be obtained directly, without resorting to mental gynmastics. The symmetry species for the three basis vector translations (x, y and z, sometimes denoted T, Ty and T ) and the three basis vector rotations (R, Ry and R of any molecule corresponding to C2V symmetry are in fact already given in the second-last colunm of the table. 

As we can in principle choose any basis vector we like, we might think that there is an infinite number of possible representations, Fp, each describing the behavior of one basis vector. However, in reality any basis vector we choose must generate either one of the irreducible representations present in the character table, or (as in the last example above) a reducible representation that can itself be reduced to a collection of irreducible representations. The character table therefore covers all possible modes of behavior under the symmetry operations of the point group. The character tables for a selection of point groups most likely to be of interest to chemists are included in the on-Une supplement for Chapter 2. 

In Table 4.4b there is not enough information in the single representation to distinguish the operations. Effectively, by only looking at the representation we are concentrating only on the orbital. If we were to consider the 2p c and 2py orbitals at the same time, then Table 4.3 shows we would get the results E (1, 1), C2 (-1,-1), Oy (1, -1) and CTy (-1, 1), where the characters are written as (2p j, 2py). Now the pairs of characters are different and the use of multiple basis vectors at the same time allows the four C2y operations to be differentiated. 

This section has shown that the matrix representation, used with a suitable set of basis vectors, can remove the difficulty of finding exactly which operation results from a given product in the same way as the introduction of basis vectors showed the difference between apparently equivalent operations. In fact, the matrix representation allows us to follow exactly what happens to a given basis under a symmetry transformation. In the next few sections we will explore the use of matrices in symmetry more fully and define the relationship between the matrix and simpler character representations. 

This is more complex than in the H2O example, since vectors are now interchanged. This makes it difficult to write down exactly what the operation does in terms of single characters, because any of the new functions could be composed from any of the original set. The matrix representation allows for this each new vector is written as a linear combination of the original basis vectors. Figure 4.6 can be used to construct the relationships required by inspection ... 

So, the character for a particular operation and basis function is just the associated diagonal element in the matrix. In the NH3 example, the C3 rotation would have a character of 0 for each of the basis vectors. 

The character we assign for a particular basis vector has been linked to the diagonal element in the operation s matrix. This can be generalized to say that the sum of the diagonal elements of a matrix representing an operation on a particular basis is the sum of the characters for the basis under that operation. In matrix algebra, the sum of the diagonal elements of a matrix A is known as the trace of the matrix, Tr(A), i.e. 

Using the trace rather than picking out particular elements of the matrix allows a single character to be assigned for the result of the operation on the entire set of basis functions being considered. This will turn out to be a useful tactic in dealing with symmetry problems, since it allows whole sets of basis vectors to be considered together. 

The examples up to now have been chosen so that each operation causes a given basis vector to be unaffected, reversed or transformed into a different basis function. In the analysis, we have shown that the matrix representation allows us to describe each of these transformations leading to the characters of +1, —1 and 0 respectively. However, if the basis vectors are not arranged to fit nicely with the symmetry elements, then the character assignment may not be so straightforward. 

We have now seen that the trace of each of these matrices is the summed characters for the basis vectors under their respective operations. In fact, it would be possible to write down the sum of characters for the basis without knowing the entire matrix representation. For any basis, we could just consider each vector in turn and ask the question How much of the original basis vector remains after the transformation Under the identity E, the X and y vectors are both unchanged, giving a total character of 2, the C2 rotation... 

For the C2v point group, we have already found all the standard labels, and the characters for the irreducible representations are given in the character table shown in Table 4.6. None of these standard representations have a 2 under the E column in fact, they all have 1. This is not a surprise, since we have already shown that x belongs to the Z i representation and y to the 82 representation by inspecting each basis vector in turn in Section 4.2. So the totalled character for the basis obtained above could be reduced to those for Bi and 82, i.e. the characters for the individual vectors. Table 4.7 shows that the sum of the characters Bi + 82 agrees with the totals laid out in Equation (4.23). The reducible representation is usually given the symbol F, and so in this case we have shown that... 

In the characters listed for the x, y, z basis in Table 4.9 there is an entry of 3 under the identity operation E. The identity operation leaves all basis vectors unchanged, so this just shows that there are three basis vectors. However, in the standard character table for the >411 point group (Table 4.8) there are no entries with 3 under the identity class coliunn, so this must be another example of a reducible representation. If we wish to assign standard labels from the character table to the x, y, z vectors, then we must simplify the situation by breaking the basis down into smaller sets in the hope of identifying which standard labels to use. 

The complete set of matrices we have built up for the operations with the central atom X, y, z basis of Figures 4.10 and 4.11 show that it is possible to interchange x and y, but z is always left alone or reversed. For example, the and operations have the matrices shown in Equation (4.26) and these lead to the transformations (x -y,y x) and x y, y -x) respectively. This linking of the x and y basis vectors can also be seen from the diagonal elements of the matrices. Equation (4.28) also shows that the vertical mirror plane has a — 1 contribution to the trace of the matrix from the y basis function and a +1 from x, while o-y has a -1 contribution from x and -1 from y. If we tried to separate x from y the two operations will no longer fall into the same class, since the individual basis functions have different characters for a particular mirror plane. Only for the two basis functions together is it valid to have a single class for the vertical reflections, since the total character for x and y is then zero for both mirror planes. The character for z is the same in both matrices, and so it would not matter if we treated this basis vector separately. 

What the representation assignment is showing is that the x and y basis vectors on the central atom of a Z)4h complex have identical environments. This has implications for any molecular objects that have the same character set. For example, it tells us that the p t and py orbitals will form a degenerate pair in this molecular geometry, i.e. any molecular orbital containing p , will have an identical partner involving p, and the two molecular orbitals will have the same energy. 

Table 4.9 gives the total characters for x,y and z, the set of which has been labelled F this is the convention for any reducible representation. We have assigned x, y to the irreducible representation. The E characters give the x, y contribution to the character for the reducible x, y, z basis, so the z characters are simply the difference between the values for x, y, z and those for as set out in Table 4.10. We could also have obtained this result from the matrices for the x, y, z basis directly however, this type of manipulation using only the characters of operations becomes easier as the basis set size increases. Table 4.10 gives a difference with character 1 under the E operation, since it represents a single object, the z basis vector. If we compare the whole character set with the standards...

If an operation moves an atom to a symmetry-equivalent position, then all basis vectors to do with that atom will give rise to only off-diagonal elements in the transformation matrix and so contribute zero to the character for the operation. 

The ay(XZ) operation also swaps the H atoms, and so we need only consider the basis vectors on the O atom to derive the total character for this operation. This plane causes the transformation... 

So the total character for all nine basis vectors under the a., XZ) operation is 1. For the a., YZ), which is the plane of the molecule, none of the atoms are moved and we find the following characters ... 

Because y and z basis vectors are in the plane, they are unaffected by the operation and contribute 6 to the total character for the basis. However, all the x basis vectors are reversed, contributing -3. So the total character for the basis is 6 - 3 = 3. 

Finally, the E operator will leave all the vectors in the basis unchanged and so has character 9. The E operation always does nothing to the basis vectors, and so the E column in a character table gives the number of basis objects in the representation. The characters for the nine basis vectors are summarized in Table 5.1. In the Cjy group there are no... 

Table 5.1 The character set for the representation of the nine basis vectors used to represent the degrees of freedom of H2O defined by the basis of Figure 5.2.

Any collection of basis vectors that complies with the molecular symmetry can generate a character representation of the group, but in most cases it will be a reducible one and so can be simplified. In this section we will show that the simplification of a reducible representation r can be made using the data for the set of irreducible representations available in the standard character tables. 

Section 4.11 used the matrix representation to deal with a set of three basis vectors jc, y, z on the central atom of a square planar D41, complex. It was shown that this basis can be reduced to + A2U by inspection of the matrices for the operations in the Ah group. The characters for the reducible and irreducible representations are shown in Table 5.2. 

The identity operator, as always, leaves all basis vectors unchanged, and we simply count the number in the basis to arrive at a character of 5. 

Finally, the three equivalent vertical mirror planes each contain the axial ligands and one of the equatorial C=0 groups, contributing 3 to the total character for this operation. Each CTv reflection swaps the remaining two basis vectors, and so they contribute 0. 

Generate the reducible representation F of the basis by assigning characters for each basis vector according to the effect of an example operation from each class. If a basis vector is unaffected by the operation, then it contributes 1 to F if it is reversed, then it gives —1 and if it changes completely (e.g. if the atom on which the basis vector is sited moves), then a contribution of 0 results. Intermediate values occur for rotated basis vectors, as discussed in Section 4.7. F is then the sum of the characters obtained for the members of the basis set one summation for each class of operations in the point group. 

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