E representations

Thus, our original Dl ) representation was a eombination of two Ai representations and one E representation. We say that DN) is a direet sum representation DN) = 2Ai E. a eonsequenee is that the eharaeters of the eombination representation DN) ean be obtained by adding the eharaeters of its eonstituent irredueible representations. 

This set of eharaeters is the same as Dl2) above and agrees with those of the E representation for the C3V point group. Henee, 2px and 2py belong to or transform as the E representation. This is why (x,y) is to the right of the row of eharaeters for the E representation in the C3V eharaeter table. In similar fashion, the C3V eharaeter table states... 

Using the orthogonality of eharaeters taken as veetors we ean reduee the above set of eharaeters to Ai + E. Henee, we say that our orbital set of three Ish orbitals forms a redueible representation eonsisting of the sum of and E IR s. This means that the three 1 sh orbitals ean be eombined to yield one orbital of Ai symmetry and a pair that transform aeeording to the E representation. 

In the so-calledMu/Z/ten notation, representations A and B (which does not appear in Table 7.2) are mono-dimensional, E representations are bi-dimensional, and T representations are three-dimensional. Other irreducible representations of higher order are G (three-dimensional) and H (tetra-dimensional). We will also state now that the dimension of a representation gives the degeneracy of its associated energy level. 

Typical Tafel plots for different copper materials are shown in Fig. 3.11. In all cases, an excellent linearity was obtained for n i/ip) on E representations in terms of the correlation coefficient for linear fitting. Similar results were obtained for binary or ternary mixtures of such materials where highly overlapping peaks were recorded, both using linear potential scan and square-wave voltammetries. 

The characters in the last line are those corresponding to the e representation. That representation is also that of the two 2p orbitals of the boron atom which lie in the plane. 

This leaves the possibility that it would contain an E representation, for which the equation is ... 

We see that I is the A irreducible representation. This means that the coordinate 2 forms a basis for the A representation, or, as we also say, 2 transforms as (or according to) Ax. If we examine the characters of Tuy, we find them to be those of the E representation (2 cos 2n > — -1), so that the coordinates x andy together transform as or according to the E representation. It is important to grasp that x and y are inseparable in this respect, since the representation for which they form a basis is irreducible. 

The remaining task is to assign the Cartesian coordinates, the rotations and certain algebraic functions of Cartesian coordinates to their representations. We have already seen that z belongs to the A2 representation (or, as is commonly said, has A2 symmetry ) and x and y jointly belong to the E representation. By using curved arrows about the three Cartesian axes, it is not difficult to determine the symmetries of the rotations. 

To illustrate how 6.2-6 works, let us consider the general function, xz + yz + z in the group C3t. (which is isomorphous to Gi0). We shall use the projection operators to obtain from this arbitrary function a pair of functions which form a basis for the E representation. The matrices for this representation are given in Table 6.1. Table 6.2 shows how the arbitrary function xz + yz + z2 is transformed by each of the six symmetry operators in the group. 

The second new SALC, which should be orthogonal to the first and therefore its proper partner in forming a basis for the E representation, thus has the form... 

The pairs of SALCs for the E representation are added and subtracted (dividing the result by i) to get two new orthogonal SALCs which have all real coefficients. 

The 1% and v2b modes together form the basis for the E representation of the group D3/i. Clearly the identity operation takes each component into itself, as required by the character of 2. We can express this symbolically as follows ... 

It is easy to see that the operation C2 transforms into the negative of itself and v36 into itself. Thus a matrix is obtained which has only the diagonal elements -1 and 1 and the character 0 as required by the character table. It is equally easy to see that oh carries each component of i 3 into itself, so that the matrix of the transformation has only the diagonal elements I and 1 and hence a character of 2. We could carry out similar reasoning for the remaining operation applied to v, and v36 and also with respect to the application of all of the operations in the group to v and vAh, and it would be found that they satisfy the requirements of the characters of the E representation in every respect. 

We may illustrate these rules, using the carbonate ion. We see in the D h character table that (x, y) form a basis for the E representation and z for the Ay representation. For the polarizability tensor components we see that one or more of these belong to the A, , and E" representations. Thus, for any molecule of Dv, symmetry, we have the following selection rules ... 

---