Assigning Symmetries of Vibrational Modes 1 Introduction ...

1 Assigning Symmetries of Vibrational ModesC. David SherrillSchool of Chemistry and BiochemistryGeorgia Institute of TechnologyJune 2000; Revised July 20101 IntroductionGroup theory is a very powerful tool in quantum chemistry. By analyzing the symmetry propertiesof molecules, we can easily make predictions such as whether a given electronic transition should beallowed or forbidden, whether a molecule should have dipole moment, whether a given vibrationalmode should be visible in the infrared or not, etc. Here we will assume a basic familiarity withpoint groups and discuss how group theory can be used to determine the symmetry properties ofmolecular O+4 HasD2hSymmetryOur example will the O+4cation, which hasD2hpoint group symmetry, as shown in Figure 1.

2 TheCartesian coordinates of this molecule are given in Table 1, and the character table forD2hisgiven in Table 2. From the table, we can see that there are eightdistinct symmetry operationsfor this point group: the identity (E), three differentC2rotation axes, a center of inversion(i), and three mirror planes ( ). You can easily verify that O+4possesses all of these symmetryproperties. These symmetry operations form the columns of the table. There are also eight rows,orirreducible representations, labeledAg,B1g,..,B3u. The 1 s and -1 s in the table indicatewhether the irreducible representation (or irrep, for short)is symmetric or antisymmetric for thatsymmetry How Many Vibrational Modes Belong To Each Irrep?

3 From the sketch of the molecular geometry and the character table, we can fairly easily deter-mine how many Vibrational Modes there will be of each symmetrytype ( , each irreduciblerepresentation).1 OOOO+yxFigure 1: O+4 CationTable 1: O+4 Cartesian Coordinates----------------------------- -----------------------Standard Nuclear Orientation (Angstroms) 2: Character Table for Point GroupD2hD2hE C2(z)C2(y)C2(x)i (xy) (xz) (yz)Ag11111111x2, y2, z2B1g11-1-111-1-1 RzxyB2g1 -11-11-11-1 RyxzB3g1 -1-111-1-11 RxyzAu1111-1 -1-1-1B1u11-1-1 -1 -111zB2u1 -11-1 -11-11yB3u1 -1-11-111-1x2 Table 3: Character Contributions of Some Common Symmetry OperationsE3 1C2-1i-3C30 Table 4.

4 Symmetry Decomposition of Atomic MotionsE C2(z)C2(y)C2(x)i (xy) (xz) (yz)Stationary Atoms40000400 Char contrib3-1-1-1 -3111 red12 0000400 The process is as follows. Apply each of the symmetry operations of the point group (E,C2(z), etc.) to the molecule, and determinehow many atoms are not moved by the , multiply this number by the so-calledcharacter contributionof that symmetry will yeild a series ofhnumbers, wherehis the number of distinct symmetry operations inthe point group (8 forD2h).What is this mysterious character contribution? Technically speaking, it is the trace of thematrix representation inxyzCartesian coordinates of that operation.

5 However, it is usually easierjust to memorize the character contributions of the most commonly used symmetry partial table of character contributions is given in these rules, we can obtain an 8-member array of integers usually denoted red, areduciblearray. This is done in Table 3. The next step is to decompose the reducible array into a uniquelinear combination of irreducible representations (irreps). This is easily accomplished using dotproducts. For example, to get the number ofagmodes, we take the dot product of redwith therow of the character table forag, and divide by the number of operations in the group (8 forD2h).So, red ag/h= (12 + 4)/8 = 2.

6 In a similar manner, we can determine the contributions fromthe other irreps, to obtain a decomposition of redas 2ag+ 2b1g+b2g+b3g+au+b1u+ 2b2u+ , we need to subtract out the translations and rotations. Theirreps of the translations canbe found in most character tables by looking for which row containsx,y, andzon the right-handside of the table. Here, this givesb1u,b2u, andb3u. Likewise, rotations are denoted in the table3byRx, Ry, Rz, which correspond tob1g,b2g, andb3g. So, subtracting these out from red, we findthat the vibrations are described by: 2ag,b1g,au,b2u, andb3u. There are a total of six vibrations,which is correct according to the 3N 6 is the output from a Q-Chem calculation.

7 These normal Modes are sketched in Figure2, along with the irreducible representations of each.** Vibrational ANALYSIS**--------------------**VIBRATIO NAL FREQUENCIES (CM**-1) AND NORMAL Modes **INFRARED INTENSITIES (KM/MOL)** Active:YESYESYESIR Active:YESYESYESIR ** 2: Sketches of Normal Modes of O+44 Symmetry-Adapted Linear CombinationsNow, how would we have come up with these Vibrational normal Modes if wehadn thad theprogram?

8 Group theory isn t sufficient to give us normal Modes ingeneral, but in this case,it woudl almost get us there, because other thanag, no irrep has more than one vibration. Insuch cases, the normal mode issymmetry-determined. In the case of irreps with more than onevibration, group theory can at least give us a symmetry-adapted set of vectors (basis); these vectorsare mixed to form the normal Modes . It is important to point outthat we could say similar thingsabout molecular orbitals. Some MO s may be symmetry-determined (in a sufficiently small basisset), and others may be linear combinations of the symmetry-adapted AO s belonging to someirrep.

9 Thus, we need to know how to formsymmetry-adapted linear combinations(SALC s) ofbasis functions like atomic orbitals or Vibrational displacement vectors. To explore this, we willstick with our O+4vibrational example for us usedisplacement vectorsin thex,y, andzdirections on each atom as our basis functions,and then we will form SALC s from these to see what the vibrationsshouldlook like (without theneed for computations). We do this using the technique ofprojection job is somewhat easier for this example because each atom is symmetry-equivalent to allthe others. Hence, it will suffice to apply projection operators to only thex,y, andzdisplacementsononeof the atoms.

10 Let s label the displacements asxi,yi, orzi, whereiis the number of theatom displaced. See Figure 3 for the displacements of atom 1. First, we must determine what5 Figure 3: Displacement Vectors for Atom 1 in O+4 Table 5: Result of Symmetry Operations on Atomic DisplacementsE C2(z)C2(y)C2(x)i (xy) (xz) (yz)x1x1 x3 x4x2 x3x1x2 x4y1y1 y3y4 y2 y3y1 y2y4z1z1z3 z4 z2 z3 z1z2z4each symmetry operator does to each of our basis vectors. Referring to Fig. 3, we can constructTable 5, which provides the results of each symmetry operationonx1,y1, apply a projection operator, we dot each of the rows in Table 5 with the rows of thecharacter table there will be a separate projection operator for each irrep.