13-8 Generating Representations from Basis Functions

Chapter 13 Group Theory Therefore, deciding whether or not a multidimensional representation is reducible is the same as deciding whether it can be uniformly block-diagonalized through a common unitary transformation. It turns out that there is no single unitary transformation capable of diagonalizing all the 2 × 2 matrices of 3, and so 3 is an irreducible two-dimensional representation. (We shall show later that this can be established without “trying out” an inﬁnite number of transformations.) For any group, a goal is to ﬁnd all the inequivalent, irreducible representations possible. Anything beyond this is superﬂuous information.

For our C3v (ammonia) group, 1, 2, and 3 exhaust the possibilities and constitute a complete set of inequivalent irreducible representations.

13-8 Generating Representations from Basis Functions

The reader may wonder how one goes about discovering nontrivial representations like 3.3 A convenient way to do this will now be described, and we will show at this stage the connection between quantum mechanics and the group theory of symmetry operations.

Consider the molecule shown in (III). This molecule has but one nontrivial symmetry element—a point of inversion. According to our ﬂowchart, this places it in the Ci point group. The only symmetry operations here are E and i. Now consider two functions, f1 and f2. Let f1 be located on one end of the molecule. For instance, let f1 be 1sF , a a 1s AO centered on the ﬂuorine atom on the left side of the molecule. Let f2 be a similar function on the other side of the molecule, 1sF . Now let us see what happens

b

to these functions when they are acted upon by our symmetry operations E and i: Ef1 = f1, Ef2 = f2, if1 = f2, if2 = f1

We see that f1 and f2 are interchanged by inversion. Let us try to ﬁnd numbers to represent these results. Clearly, replacing E by +1 will give the correct result. However, to obtain the effect of operation by i we need to interchange f1 and f2. We cannot achieve this by multiplying by a number, since f1 and f2 are linearly independent. If, however, we rewrite the effect of i as

f1

f2

i

=

f2

f1

it becomes clear that i can be represented by the matrix (01). Thus, use of the functions

10

1sF and 1s has generated a two-dimensional representation shown in Table 13-11.

a

Fb (The representation for E has been put into a 2 × 2 matrix form to be in dimensional agreement with the representation for i.) 1sF and 1s are called a basis for . Clearly, a

Fb 3Indeed, the reader may wonder why we even worry about all this. Patience.

Section 13-8 Generating Representations from Basis FunctionsTABLE 13-11  A Representation for the Ci Group Ci

E

i

Basis

1 0

0 1

(1sF , 1sF ) 0 1

1 0

a

b any pair of identical functions symmetrically placed with respect to the point of inversion would generate the same two-dimensional representation, and so there is nothing very special about 1sF and 1s as a basis. Is  reducible? It is easily argued that it a

Fb

must be. Any unitary 2 × 2 transformation will have no effect on the matrix for E since β−11β = β−1β = 1. We are thus at liberty to look for any unitary transformation β that diagonalizes the second matrix. Since this is a nonsingular matrix (its determinant is unequal to zero), it should be diagonalizable, and it is if we take

√

√

  √

√

1/ 2

1/ 2

√

√

0

1

1/ 2 −1/ 2 √

√

= 1

0 −1/ 2 1/ 2

1

0

1/ 2

1/ 2

0

−1

Now that our 2 × 2 representation has been diagonalized to two 1 × 1 “blocks,” we can rewrite our representations as shown in Table 13-12.

Our ﬁrst choice of basis generated a reducible representation. What bases would we need to generate the 1 × 1 representations 1 and 2? To generate 1, we need a basis function that turns into itself when it is inverted. One possibility is f1 = 1sF + 1s .

a

Fb

If we sketch this function (Fig. 13-8a), it is evident that it is regenerated unchanged by inversion. The mathematical demonstration is if1 = i(1sF + 1s ) = i1s + i1s = 1s + 1s = f a

Fb

Fa

Fb

Fb

Fa

1

TABLE 13-12  Reduced Representations for the Ci Group

Ci

E

i

Basis

1

1

1 ?

2

1

−1

?

Figure 13-8 Basis functions for irreducible representations of the Ci group.

Chapter 13 Group Theory f1 is symmetric for operations E and i. For 2, we need some function f2 that turns into minus itself upon inversion. f2 = 1sF − 1s would serve, as Fig. 13-8b shows, a

Fb

and as is demonstrated mathematically by if2 = i(1sF − 1s ) = i1s − i1s = 1s − 1s = −f a

Fb

Fa

Fb

Fb

Fa

2

f2 is symmetric for E and antisymmetric for i. (Notice that the way in which 1sFa and 1sF needed to be mixed to produce a diagonal representation is indicated by the b

coefﬁcients in the columns of the matrix we used to diagonalize the original 2 × 2 representation. The matrix β not only diagonalizes our , it also tells us how to mix the original bases in order to arrive at bases for the irreducible representations.)

This example demonstrates an important fact:

In order to generate a one-dimensional representation for a group, we need a basis function that is either
symmetric or antisymmetric for every symmetry operation in the group. This provides the point of connection with quantum mechanics. We showed much earlier (Chapter 2) that any nondegenerate wavefunction (or MO) must be symmetric or antisymmetric for every operation that leaves the hamiltonian unchanged. Since symmetry operations leave the hamiltonian unchanged (they merely interchange identical nuclei), it follows that nondegenerate wavefunctions or orbitals are symmetric or antisymmetric for every operation in the symmetry point group of the molecule. This means that every
nondegenerate wavefunction or orbital is a basis for a one-dimensional representation
for its molecular point group. Indeed, it can be proved that every wavefunction or
orbital, even if degenerate, is a member of a basis for an irreducible representation for the point group of the molecule. The irreducible representation produced by an n-fold degenerate set of orbitals will be n-dimensional. (We indicate how this comes about in a later section.) Therefore, knowing wavefunctions enables one to generate representations. More importantly, knowing representations allows us to say something about wavefunctions. For example, the representation table (Table 13-12) for the Ci group (which is now complete—this group has only two irreducible inequivalent representations) tells us the following: (1) The molecule FClBrC–CBrClF in the conformation pictured has no degeneracies due to symmetry in its electronic states or in MO energies (if we do a calculation at the MO level). We can tell this because only one-dimensional representations exist for this group. (2) Every wavefunction for an electronic state or orbital must be either symmetric or antisymmetric for inversion. Therefore, the following AO combinations are feasible for MOs insofar as symmetry is concerned: c(1sF − 1s ), a

Fb c1(1sF + 1s ) + c + 1s ) + c + 1s )

a

Fb

2(1sCla

Clb

3(1sBra

Brb + c4(1sC + 1s ) + c + 2s ) + · · · a

Cb

5(2sFa

Fb

The following are disallowed: c1sF , a c1(2sF + 2s ) + c − 2s )

a

Fb

2(2sCla

Clb

Our halogenated ethane molecule is a good starting example because it belongs to such a simple group. But let us now return to the C3v group of the ammonia molecule and continue developing the relations between group theory and MO theory. For convenience, the molecule is again sketched (Fig. 13-9) to show its orientation with respect to Cartesian axes. (The z coordinate is the principal axis.)

Section 13-8 Generating Representations from Basis Functions

y

φ

1112 1

10

2

z

9

3

x

8

Rz 4

7

5

6

Figure 13-9

Orientation of a model of ammonia in Cartesian space.

The normal practice in group theory is to use Cartesian coordinates or linear combina tions of such coordinates as basis functions4 for generating many of the representations of a group. Therefore, we begin by examining the z coordinate to see what becomes of it under the various symmetry operations in the C3v group. The results are easily seen to be Ez = +1z, σ1z = +1z,

σ2z = +1z,

σ3z = +1z,

C+z = +1z, C−z = +1z

3

3

Thus, z is a basis for the totally symmetric representation 1 of Table 13-8. The coordinates x and y can also be used as bases. Here things get more complicated.

Clockwise rotation of x by 2π/3 radians (to give x) causes it to end up in a position where it must be expressed as a resultant of both x and y (Fig. 13-10). Simple trigonometry requires that, for a general rotation through the angle φ, the unit vector x has an x coordinate of cos φ and a y coordinate of sin φ. Similarly, a rotation of y to a new position designated y must yield a new x coordinate of −sin φ and a new y coordinate of cos φ. Thus, for a rotation through the angle φ, we have

x

x

x cos φ + y sin φ

Cφ

=

=

y

y

−x sin φ + y cos φ

Figure 13-10

Result of clockwise rotation of x by 2π/3 to produce the new vector x.

4The function corresponding to a coordinate is not exactly the same thing as the coordinate itself. The z coordinate is a ray running perpendicular to the xy plane through the coordinate origin. The function z is the altitude above (or below) the xy plane at every point, regardless of whether it is on the z axis; z2 is the square of the altitude at every point, etc. The behavior of these functions upon rotation, reﬂection, etc. is the same as that of the coordinate or product of coordinates.

Chapter 13 Group Theory The rotation operator Cφ is thus represented as

cos φ

sin φ

Cφ =

(13-1)

−sin φ cos φ

In the case at hand, clockwise rotation by 2π/3 radians is a decrease of 2π/3 in φ.

Substituting φ = −2π/3 for C +

−

3

and φ = +2π/3 for C3 gives us the following 2 × 2 representations:



√ 



√ 

−1

+ 3 −1

− 3

2

2

2

2

C+ :   C− : 



3

√

3

√

− 3 −1

+ 3 −1

2

2

2

2

For reﬂection σ1, it is easy to see that x is unmoved and y goes into minus itself.

Maintaining our dimensionality of two, this gives

1

0

σ1 : 0 −1

For σ2, x, and y again move into positions x and y, expressible as resultants of the

√

√

original x and y vectors (Fig. 13-11). Thus, x = − 1 x + 3 y, y = 3 x + 1 y, and σ

2

2

2

2

2

has the representation



√ 

−1

3

2

2

σ





2 :

√3

1

2

2

Similarly, σ3 is easily shown to have the representation



√  −1 − 3

2

2

σ





3 :

√

− 3

1

2

2

The operation E does not move either x or y and is represented by the two-dimensional unit matrix. This completes our use of x and y, and we see that together they generate the two-dimensional representation 3. What should we use next? We have not yet generated 2, and so we know that we need to look for another basis. A basis that is

Figure 13-11

Result of reﬂection through the σ2 plane of x and y to produce new vectors x and y.